LIFE OF PHYSICS BY OFFREDI TCHIENGANG NGAMGOUM

OFFREDI ZOE’S COLLECTION o

LIFE OF PHYSICS FIRST EDITION AUTHOR:O.B TCHIENGANG NGAMGOUM MATHEMATICS LECTURER , RESEARCHER AT UNISA,MEMBER OF SHAW ACADEMY ONLINE PROGRAMS(WEBSITE:WWW.SHAWACADEMY.COM EMAIL:Studentsupport@shawskills.com);GUTSY AND BRIGHTSPARK.

2019

“BELIEVE IN YOUR DREAM AND STRENGTH,DO NOT LET ANYONE DISCOURAGE YOU JUST BECAUSE THEY ARE UNINSPIRED” O.B TCHIENGANG NGAMGOUM

PREFACE

This book is intended for students of science and engineering; it aims to develop both an understanding of the important concepts of physics and some analytical skill in the solutions of problems. The mathematical level of the book is such that it may be used by students who are taking a course in calculus concurrently. The notations and methods of the calculus and trigonometry are used to proof some formulas because there is a proverb from the Author that said:”If I know where you are coming from,I will never forget your identity forever….” O.B TCHIENGANG NGAMGOUM.The Author of this Book ,considered the fact that the introduction to pure mathematics and Calculus are the prerequisites for the entrance in any University such as: Vaal University Of Technology(VUT),University Of South Africa(UNISA),SHAW ACADEMIC Etc… PUBLISHED BY:REACH PUBLISHERS P.O BOX 1684 WANDSBECK,SOUTH AFRICA 3631 Edited by Dr ALEXI NGOUNOU and Mr BASSIE MOTSELEBANE Printed by Minuteman Press Cresta tel:0116788250 ISBN 978-0-620-82782-9 WARNING!!

All rights reserved according to the South African copyright act.No part of this book may be reproduced by photocopying or any other method without written permission of the

publisher ,author or Zoe Ngamgoum.Any person who exercises any unauthorized act in relation to this publication may be subject to criminal prosecution and civil claims against damage.

Acknowledgements: The author and publisher wish to thank the following for granting permission to reproduce few photographs: http://hyperphysics.phy-astr.qsu.edu/hbase/phyopt/totintt.html The author uses the opportunity to thank Mr and Mrs Nsaman,Dr BERNARD,Mme Therese,Mr and Mrs YENWONG,Prof Okosun,Dr SIKAKANA,Mss MUDAU,Dr Ngamgoum Claude,Dr NGOUNOU,Dr KAHAPI,Dr(specialist) Meyet,Bar Frans,Mme Ovono,Mr Gerald,Mr WORKU,Mss Zoe,Mme Sandra,and all my beloved family….for your encouragement and support!!!

CONTENTS UNIT 1:UNITS CONVERSION

pge6

UNIT 2:WAVES AND SOUND

pge11

UNIT 3: The Principle of linear Superposition and Interference Phenomena pge28 UNIT 4:Electromagnetic Waves

pge 47

UNIT 5: Interference and the wave nature of light pge62 UNIT 6: The Reflection of Light: Mirrors

pge81

UNIT 7: The Refraction of Light: Lenses and optical instruments pge97

UNIT 8: Vectors and scalars: Introduction and Mathematical Concept components

pge110

UNIT 9: Kinematics in One Dimension

pge129

UNIT 10: Forces and Newton’s Law of Motion

pge142

UNIT 11: Work and Energy

pge169

UNIT 12: Impulse and Momentum

pge189

UNIT 13: Electric Forces and Electric Fields

pge209

UNIT 14: Electric Potential Energy and the Electric Potential pge245 UNIT 15: Electric circuits UNIT 16:

Fluids

pge262

pge345

UNIT 17: Temperature and heat

pge370

UNIT 18: The transfer of heat

pge389

UNIT 19: Nuclear Physics and Radioactivity Pge394

UNIT 1: CONVERSION OF THE UNITS 1.INTRODUCTION : Historically, many different systems of units have been used, where a system of units is defined as a collection of units of measurement with rules that relate them to each other. A unit of measurement is a defined magnitude of a quantity that it used as a standard for measurement for the same kind of quantity, such as measurements of length, weight, and volume.

In the past, many systems of measurement were defined on a local level, and could be based on factors as arbitrary as the length of a king's thumb. While this may work on a local level, when considering trade, as well as science, having systems of units based on units that others may not be able to relate to or understand makes interaction difficult. As such, the development of more universal and consistent systems developed over time. Today, some of the systems of units in use include the metric system, the imperial system, and the United States customary units.

The International System of Units (SI) is the standard metric system that is currently used, and consists of seven SI base units of length, mass, time, temperature, electric current, luminous intensity, and amount of substance. Although SI is used almost universally in science (including in the US), some countries such as the United States still use their own system of units. This is partly due to the substantial financial and cultural costs involved in changing a measurement system compared to the potential benefit of using a standardized system. Since US customary units (USC) are so entrenched in the United States, and SI is already used in most applications where standardization is important, everyday use of USC is still prevalent in the United States, and is unlikely to change. To understand fully the conversion,the student should be able to manipulate the conversion tables and examples below.

1.1)METER,,LITER AND GRAM… In the United States, both the U.S. customary measurement system and the metric system are used, especially in medical, scientific, and technical fields. In most other countries, the metric system is the primary system of measurement. If you travel to other countries, you will see that road signs list distances in kilometers and milk is sold in liters.

People in many countries use words like “kilometer,” “liter,” and “milligram” to measure the length, volume, and weight of different objects. These measurement units are part of the metric system.Note:A kilogram is 1,000 times larger than one gram (so 1 kilogram = 1,000 grams). A centimeter is 100 times smaller than one meter (so 1 meter = 100 centimeters). A dekaliter is 10 times larger than one liter (so 1 dekaliter = 10 liters).The table below regularly used in the French system can help to understand easily the manipulation of the conversion .The author was inspired to introduce this unit ,during the marking of students scripts .The notion of the units conversion has been carelessly studied by some students in the primary and secondary school .Thus,he urge each student to carefully revise this unit with lot of attentions and focuseness!!!.

1.2 Table Of Conversion symbols

k

h

da

u

d

c

m

name

kilo

hecto

deca

unit

deci

centi

milli

Meter(m)

km

hm

dam

m

dm

cm

mm

Liter(l)

kl

hl

dal

l

dl

cl

ml

hg

dag

g

dg

cg

mg

Gramme(g) kg

1.3 Examples: Example1 :convert 1km to m

so 1km=1000m.

1kPa=1000Pa 1kΩ=1000Ω……

Example 2:convert 5g to kg

So 5g=0,005kg 5g=0,05hg Example3:convert 1L to Ml

So 1L=1000Ml 1.4 Meter square (m²) Example4:convert4hm² to m²

So 4hm²=40000m² Example 5:Cubic meter m3

NB:THE TABLES ABOVE WORK FOR TIME TOO little tricks for the speed unit To covert from Km/h to m/s ,the student or lecturer or researcher should just multiply the given number by the constant “0,278 “ .so 1km/h=1x0,278m/s=0,278m/s To reverse back from m/s to km/h…just divide by 0,278 TUTORIAL

EXERCISE 1 I;

EXERCISE 2

CONVERT THE FOLLOWING UNITS :

0.9 g/cm3=……………………………………………………………………. kg/m3 3 m/s=…………………………………………………………………………..km/h 100 km/h =…………………………………………………………… ………. m/s 3 days =……………………………………………………………………………… s 350 g =……………………………………………………………………………….Kg 1251,524L=……………………………………………………………………………m3 1,2558x104 hPa=……………………………………………………………………….Pa

Exercise 3 The former King of BATCHINGOU village (MR TCHIENGANG NICOLAS) was coming back from KEKEM for an important meeting with some elders. On theirs way back to BATCHINGOU, the king saw a bushmeat crossing the road with a speed of 5m/s. He suddenly order the guards to pursue the animal .One of the guard started running at the speed of 36km/h. Could him catch the animal?

If yes, show that the coefficient of proportionality between theirs speeds is equal to 2.

UNIT 2:

WAVES AND SOUND

INTRODUCTION:

A wave is a disturbance on the surface of a liquid or body,as the sea or a lake, in the form of a moving ridge or swell.A wave carries energy from place to place.

Sound it's almost impossible to imagine a world without it. It's probably the first thing you experience when you wake up in the morning—when you hear birds chirping or your alarm clock bleeping away. Sound fills our days with excitement and meaning, when people talk to us, when we listen to music, or when we hear interesting programs on the radio and TV. Sound may be the last thing you hear at night as well when you listen to your heartbeat and drift gradually into the soundless world of sleep. Sound is fascinating—let's take a closer look at how it works.

2.1 TYPES OF THE WAVES: a)Transversal wave :a wave in which the disturbance occurs perpendicular to the direction of the motion of the wave .Example: wave on the non slinky string ,electromagnetic waves etc… b) Longitudinal wave :a wave in which the disturbance occurs parallel to the direction of the motion of the wave. Example: sound wave, wave in a slinky spring and seismic P-waves (created by earthquakes and explosions). In the longitudinal wave,waves can be either straight or round. A wave along the length of a stretched Slinky toy, where the distance between coils increases and decreases, is one of the good visualization.

These waves could be visualized on the figure 1 below. By acronym, "longitudinal waves" and "transverse waves" were occasionally abbreviated by some authors as "L-waves" and "Twaves" respectively for their own convenience. While these two acronyms have specific meanings in seismology (L-wave for Love wave or long wave[3]) and electrocardiography (google T wave), some authors chose to use "l-waves" (lowercase 'L') and "t-waves" instead, although they are not commonly found in physics writings except for some popular science books.

Figure 1:Longitudinal and Transversal Waves c) periodic waves :periodic waves consist of cycles or patterns that are produced over with repetition consecutively.

The figure below illustrate it clearly.

Figure 2:Graphic representation of the Sinusoidal Waveform

‘’A’’ stand for the amplititude or maximum excursion of a particle of the medium from the particle’s undisturbed position.

Example:transversal wave.

Pressure amplitude is the magnitude of the maximum change in pressure,measured relatively to the undisturbed or atmospheric pressure.Example:longitudinal wave. a)Wavelength(⋋):the horizontal length of one cycle/oscillation.The unit is in meter(m). b)Period(T):The time of one complete cycle/oscillation.The unit is in second(s). c)Frequency(f):The number of cycle/oscillation per second.The unit is hertz or turn/s. f

or v

⋋

d)The characteristics of the sound waves are : 1. Frequency is directly related to musical pitch or tone 2. Pressure amplitude is directly related to the loudness of a sound wave.

Figure 3:Graphic representation of the frequency In this unit we are going to consider two periodicities a)Spatial(⋋) b)Temporal(T)

2.2 SPEED OF A WAVE ON A STRING The speed of a traveling wave in a stretched stringis determined by the tension and the mass per unit length of the string. for a string of length cm and mass/length = Kg/m. For such a string, the fundamental frequency would be Hz. Any of the highlighted quantities can be calculated by clicking on them.

The speed of a wave pulse traveling along a string or wire is determined by knowing its mass per unit length and its tension.

In this formula, the ratio mass/length is read "mass per unit length" and represents the linear mass density of the string. This quantity is measured in kilograms/meter.

Tension is the force conducted along the string and is measured in newtons, N. The maximum tension that a string can withstand is called its tensile strength.

The formula given above tells us that the "tighter" the string (that is, the greater the tension placed on the string) the faster the waves will travel down its length. It also tells us that the "lighter" the string, that is, the smaller its mass/length ratio, the faster the waves will travel down its length.

2.3 THE MATHEMATICS DESCRIPTION OF A WAVE When a wave travels through a medium, it displaces the particles of the medium from their undisturbed positions. Suppose a particle is located at a distance x from a coordinate origin. We would like to know the displacement y of this particle from its undisturbed position at any time t as the wave passes. For periodic waves that result from simple harmonic motion of the source displacement involves a sine or cosine, a fact that is not surprising. y(t)=Asin( ω t+φ ),where φ= ±

; Angular frequency ω=2πf

Here A is the amplitude of the wave,i.e. the maximum height of the wave; f the frequency, i.e. is the number of oscillations (cycles) that occur each second of time; φ, the phase, specifies (in radians) where in its cycle the oscillation is at t=0. . (+) if the wave is moving in the negative x-direction . (–) if the wave is moving in the positive x-direction Note that ω t+φ is measured in radian.

2.4 THE NATURE OF THE SOUND : Sound is one kind of longitudinal wave, in which the particles oscillate to and fro in the same direction of wave propagation. Sound waves cannot be transmitted through vacuum. The transmission of sound requires at least a medium, which can be solid, liquid, or gas.

Figure 4: Propagation of the sound wave NB: The wavelength, Îť is the distance between two successive rarefactions or condensations.

2.5 SOUND INTENSITY (I): a)Discussion about intensity vs amplitude The amplitude of a sound wave can be quantified in several ways, all of which are a measure of the maximum change in a quantity that occurs when the wave is propagating through some region of a medium. •

Amplitudes associated with changes in kinematic quantities of the particles that make up the medium o

The displacement amplitude is the maximum change in position.

o

The velocity amplitude is the maximum change in velocity.

o

The acceleration amplitude is the maximum change in acceleration.

•

Amplitudes associated with changes in bulk properties of arbitrarily small regions of the medium o

The pressure amplitude is the maximum change in pressure (the maximum gauge pressure).

o

The density amplitude is the maximum change in density.

Measuring displacement might as well be impossible. For typical sound waves, the maximum displacement of the molecules in the air is only a hundred or a thousand times larger than the molecules themselves — and what technologies are there for tracking individual molecules anyway? The velocity and acceleration changes caused by a sound wave are equally hard to measure in the particles that make up the medium.

Density fluctuations are minuscule and short lived. The period of a sound wave is typically measured in milliseconds. There are some optical techniques that make it possible to image the intense compressions are rarefactions associated with shock waves in air, but these are not the kinds of sounds we deal with in our everyday lives.

Pressure fluctuations caused by sound waves are much easier to measure.

Animals (including humans) have been doing it for several hundred million years with devices called ears. Humans have also been doing it electromechanically for about a hundred years with devices called microphones. All types of amplitudes are equally valid for describing sound waves mathematically, but pressure amplitudes are the one we humans have the closest connection to.

In any case, the results of such measurements are rarely ever reported.

Instead, amplitude measurements are almost always used as the raw data in some computation. When done by an electronic circuit (like the circuits in a telephone that connect to a microphone) the resulting value is called intensity.

When done by a neuronal circuit (like the circuits in your brain that connect to your ears) the resulting sensation is called loudness.

The intensity of a sound wave is a combination of its rate and density of energy transfer. It is an objective quantity associated with a wave. Loudness is a perceptual response to the physical property of intensity. It is a subjective quality associated with a wave and is a bit more complex. As a general rule the larger the amplitude, the greater the intensity, the louder the sound. Sound waves with large amplitudes are said to be "loud". Sound waves with small amplitudes are said to be "quiet" or "soft". The word "low" is sometimes also used to mean quiet, but this should be avoided. Use "low" to describe sounds that are low in frequency.

2.6 DEFINITION :

the intensity (I) of any wave is the time-averaged power (⟨P⟩) it transfers per area (A) through some region of space. The traditional way to indicate the time-averaged value of a varying quantity is to enclose it in angle brackets (⟨⟩).

These look similar to the greater than and less than symbols but they are taller and less pointy. That gives us an equation that looks like this…

The SI unit for I is W/m2 2.7 Sound intensity inverse square law: The Inverse Square Law teaches us that for every doubling of the distance from the sound source in a free field situation, the sound intensity will diminish by 6 decibels. ... The intensity of the sound is inversely proportional to the square of the distance of the wavefront from the signal source.

Where: = I1 sound intensity 1 at closer distance r1 from the sound source I2

=

r = 1

sound intensity 2 at more far distance r2 from the sound source closer distance r1 from the sound source

more far distance r2 from the sound source Note! Since the sound intensity level is difficult to measure, it is common r = to use sound pressure level measured in decibels instead.

Doubling the sound pressure raises the sound pressure level (SPL) by 6 dB. 2 Doubling the sound intensity raises the sound intensity level by 3 dB.

Figure 5:GRAPHIC REPRESENTATION OF THE INVERSE SQUARE LAW The other examples of inverse square law behavior: Gravity, Electric, Radiation, Light.

2.8 SOUND INTENSITY LEVEL(DECIBELS) Sound intensity levels are quoted in decibels (dB) much more often than sound intensities in watts per meter squared. Decibels are the unit of choice in the scientific literature as well as in the popular media. The reasons for this choice of units are related to how we perceive sounds. How our ears perceive sound can be more accurately described by the logarithm of the intensity rather than directly to the intensity. The sound intensity level in decibels of a sound having an intensity in watts per meter squared is defined to be

where Io = 10^12 w/m2 is a reference intensity. In particular, I0 is the lowest or threshold intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz. Sound intensity level is not the same as intensity.

Because is defined in terms of a ratio, it is a unitless quantity telling you the level of the sound relative to a fixed standard ( Io = 10-12 w/m2, in this case).

The units of decibels (dB) are used to indicate this ratio is multiplied by 10 in its

definition. The bel, upon which the decibel is based, is named for Alexander Graham Bell, the inventor of the telephone.

2.9 CHANGE IN SOUND INTENSITY level β2 − β1 =Δβ or S2 –S1 =(ΔS)2n If two sonorous sensations are directly compared using the expression seen in the preceding item ("The decibel - Sound intensity level"), without referring to any intensity threshold,

2.10 Proof: S1 =10.log(I1/I0) and S2 =10.log(I2/I0) S2-S1=10{ log(I2/I0)- log(I1/I0)} =10.log(I2/I1) ,Note that in mathematics log(a)-log(b)=log(a/b) i.e ΔS=10.log(I2/I1)

Figure 6:Diagram of the description of sound

2.11 TAKE-HOME INVESTIGATION: FEELING SOUND Find a CD player and a CD that has rock music. Place the player on a light table, insert the CD into the player, and start playing the CD. Place your hand gently on the table next to the speakers. Increase the volume and note the level when the table just begins to vibrate as the rock music plays. Increase the reading on the volume control until it doubles.

What has happened to the vibrations?

2.12 CHECK YOUR UNDERSTANDING Part 1 Describe how amplitude is related to the loudness of a sound.

Solution Amplitude is directly proportional to the experience of loudness. As amplitude increases, loudness increases.

Part 2 Identify common sounds at the levels of 10 dB, 50 dB, and 100 dB. Solution 10 dB: Running fingers through your hair.

50 dB: Inside a quiet home with no television or radio.

100 dB: Take-off of a jet plane.

Examples : 1. Point A and B located at 4 meters and 9 meters from a source of the sound.

If IA and IB are intensity at point A and point B, then IA : IB =‌ Known : The distance of point A from a source of sound (rA) = 4 meters A distance of point B from the source of sound (rB) = 9 meters The intensity of sound at point A = IA The intensity of sound at point B = IB Wanted: IA : IB Solution : IA rA2 = IB rB2 IA 4 2 = IB 9 2 IA 16 = IB 81 IA / IB = 81/16

2. The intensity of a source of sound is 10−9 Wm−2. Io = 10−12 Wm−2. What is the sound level of 10 sources of sounds?

Solution : Known : I = 10-9 W/m2 Io = 10-12 W/m2 x = 10 Known: Sound level (β) Solution :

Thus knowing that:

3. The sound level of a source of sound is 10 dB. What is the intensity of 1000 sources of sound? The minimum intensity Io = 10−12 Wm−2.

Known : β = 10 dB Io = 10-12 W/m2 x = 1000 Wanted : Intensity Solution : The intensity of a source of sound :

The intensity of 1000 sources of sound : I = (1000)(10-11) = (103)(10-11) I = 10-8 W/m2 4. The sound level of A is 40 dB, and the sound level of B is 60 dB. Io = 10-12 W m-2.

Determine 100βA : 10βB. Known : The sound level of A = 40 dB The sound level of B = 60 dB Io = 10-12 W m-2.

Wanted : 100βA : 10βB Solution : The sound level of 100 A : β = 40 + 10 log 100 β = 40 + 10 log 102 β = 40 + (2)(10)(log 10) β = 40 + (2)(10)(1) β = 40 + 20 β = 60 dB The sound level of 10 B : β = 60 + 10 log 10 β = 60 + 10 log 101 β = 60 + (1)(10)(log 10) β = 60 + (1)(10)(1) β = 60 + 10 β = 70 dB βA : βB 60 : 70

5) With one violin playing, the sound level at a certain place is measured as 50 dB. If four violins play equally loudly, what will the sound level most likely be at this place?

Let's call the sound intensity produced by the one violin playing, I1. Then the sound intensity level from this violin that we hear is LI(1) = 10 log( I1 / I0 ).

When four violins play, assuming they each produce sound intensity I 1 at our ear, the total sound intensity is just the sum of the intensities produced separately from

each violin, or Itotal = 4 I1 We can then find the sound intensity level when four violins play LI (total) = 10 log( Itotal / I0 ) = 10 log( 4I1 / I0 ) We now make use of a property of logarithms, log(A x B) = log A + log B, to write the total sound intensity level as LI (total) = 10 log( 4 ) + 10 log( I1 / I0 ) The second term is just the sound intensity level from a single violin, L I(1) = 10 log( I1 / I0 ). This means that the sound intensity level from four violins playing equally loudly is LI (total) = 6 dB + 50 dB = 56 dB The first term is 10 log( 4 ) = 6 dB. The second term is the sound intensity level from one violin.

S22 The sound level from a single violin playing is 60 dB. We can find its intensity from the definition of level in terms of intensity L = 10 log ( I / I0) or, 60 dB = 10 log ( I / I0) or, 6 = log ( I / I0) meaning, I / I0 = 106 or, I = 106 x I0 = 106 x 10-12 W/m2 finally, this gives I = 10-6 W/m2 Now, when two violins play together, the total intensity is 2 x 10-6 W/m2. NOTE: WE DO NOT ADD THE SOUND LEVELS FROM THE TWO VIOLINS PLAYING SEPARATELY!!! This would make no sense if we did add the sound levels, since we would conclude that two violins playing would produce a level of 120 dB, equal to the loudness of a nearby jet engine!

The proper thing to do is to add the INTENSITIES for the two sounds: I2 violins = 2 x I = 2 x 10-6 W/m2 This means that the sound level is, L2 violins = 10 log (2I / I0) or, L2 violins = 10 log(2) + 10 log(I / I0) or, L2 violins = 3 dB + 60 dB = 63 dB S23 Each of four people talking, when speaking individually produce an unknown sound level L1, corresponding to an unknown intensity I 1. When all four talk together, the sound level is 70 dB and the total intensity is 4 x I1. So, from the definition of

sound level L4 people = 70 dB = 10 log(4 I1 / I0) or, using a rule for logarithms: L4 people = 70 dB = 10 log(4) + 10 log(I1 / I0) or, L4 people = 70 dB = 10 log(4) + L1 or, 70 dB - 6 dB = L1 finally, L1 = 64 dB TUTORIAL QUESTIONS: 1. A source is emitting sound uniformly in all directions. There are no reflections anywhere. A flat surface faces the source. Is the sound intensity the same at all points on the surface? Give our reasoning.

2. If two people talk simultaneously and each creates an intensity level of 65 dB at a certain point, does the total intensity level at this point equal 130 dB? Account for your answer.

3. A typical adult ear has a surface area of 2.1 Ă— 10−3 m2. The sound intensity during a normal conversation is about 3.2 Ă— 10−6 W/m2 at the listener’s ear.

Assume that the sound strikes the surface of the ear perpendicularly. How much power is intercepted by the ear? đ?&#x;”. đ?&#x;• Ă— đ?&#x;?đ?&#x;Žâˆ’đ?&#x;— W 4.What is the intensity in watts per meter squared of 85.0-dB sound? đ?&#x;‘. đ?&#x;?đ?&#x;” Ă—10-4 N/m 4. The warning tag on a lawn mower states that it produces noise at a level of 91.0 dB. What is this in watts per meter squared? đ?&#x;?. đ?&#x;?đ?&#x;” Ă— đ?&#x;?đ?&#x;Žâˆ’đ?&#x;‘ W/m2 5. . A sound wave traveling in 20 °C air has a pressure amplitude of 0.5 Pa. What is the intensity of the wave? đ?&#x;‘. đ?&#x;Žđ?&#x;’ Ă— đ?&#x;?đ?&#x;Žâˆ’đ?&#x;’ W/m2 6. What intensity level does the sound in the preceding problem correspond to?

85 Db 7. What sound intensity level in dB is produced by earphones that create an intensity of 4.00 Ă— 10−2 W/m2? 106 dB 8. (a) What is the intensity of a sound that has a level 7.00 dB lower than a 4.00 Ă— 10−9 W/m2 sound? (b) What is the intensity of a sound that is 3.00 dB higher than a 4.00 Ă— 10−9 W/m2 sound? đ?&#x;–. đ?&#x;Žđ?&#x;Ž Ă— đ?&#x;?đ?&#x;Žâˆ’đ?&#x;?đ?&#x;Ž W/m2, đ?&#x;–. đ?&#x;Žđ?&#x;Ž Ă— đ?&#x;?đ?&#x;Žâˆ’đ?&#x;— W/m2 9. People with good hearing can perceive sounds as low in level as –8.00 dB at a frequency of 3000 Hz. What is the intensity of this sound in watts per meter squared? đ?&#x;?. đ?&#x;“đ?&#x;– Ă— đ?&#x;?đ?&#x;Žâˆ’đ?&#x;?đ?&#x;‘ W/m2 10. If a large housefly 3.0 m away from you makes a noise of 40.0 dB, what is the

noise level of 1000 flies at that distance, assuming interference has a negligible effect? 70.0 dB 11. An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-diameter eardrum so exposed? đ?&#x;?. đ?&#x;’đ?&#x;“ Ă— đ?&#x;?đ?&#x;Žâˆ’đ?&#x;‘ J 12. The bellow of a territorial bull hippopotamus has been measured at 115 dB above the threshold of hearing. What is the sound intensity? 0.316 W/m2 13. Humans can detect a difference in sound intensity levels as small as 1.0 dB. What is the ratio of the sound intensities? 1.3 14. Ultrasound equipment used in the medical profession uses sound waves of a frequency above the range of human hearing. If the frequency of the sound produced by the ultrasound machine is f = 30 kHz, what is the wavelength of the ultrasound in bone, if the speed of sound in bone is v = 3000 m/s?

15. The speed of light in air is approximately v = 3.00 x 10 8 m/s and the speed of light in glass is v = 2.00 x 108 m/s. A red laser with a wavelength of ΝΝ = 633.00 nm shines light incident of the glass, and some of the red light is transmitted to the glass. The frequency of the light is the same for the air and the glass. (a) What is the frequency of the light? (b) What is the wavelength of the light in the glass?

16. A radio station broadcasts radio waves at a frequency of 101.7 MHz. The radio waves move through the air at approximately the speed of light in a vacuum. What is the wavelength of the radio waves?

17. A sunbather stands waist deep in the ocean and observes that six crests of periodic surface waves pass each minute. The crests are 16.00 meters apart.

What is the wavelength, frequency, period, and speed of the waves?

18. A tuning fork vibrates producing sound at a frequency of 512 Hz. The speed of sound of sound in air is v = 343.00 m/s if the air is at a temperature of 20.00 °C. What is the wavelength of the sound?

19. A motorboat is traveling across a lake at a speed of v b = 15.00 m/s. The boat bounces up and down every 0.50 s as it travels in the same direction as a wave. It bounces up and down every 0.30 s as it travels in a direction opposite the direction of the waves. What is the speed and wavelength of the wave?

20. Use the linear wave equation to show that the wave speed of a wave modeled with the wave function y(x, t) = 0.20 m sin(3.00 m−1 x + 6.00 s−1 t) is v = 2.00 m/s. What are the wavelength and the speed of the wave?

21. Given the wave functions y1(x, t) = A sin(kx − ωt) and y2(x, t) = A sin(kx − ωt + Ď•) with ϕ≠π/2, show that y1(x, t) + y2(x, t) is a solution to the linear wave

equation with a wave velocity of v = √ω/k.

22. A transverse wave on a string is modeled with the wave function y(x, t) = 0.10 m sin(0.15 m−1 x + 1.50 s−1 t + 0.20). (a) Find the wave velocity. (b) Find the position in the y-direction, the velocity perpendicular to the motion of the wave, and the acceleration perpendicular to the motion of the wave, of a small segment of the string centered at x = 0.40 m at time t = 5.00 s.

23. A sinusoidal wave travels down a taut, horizontal string with a linear mass density of μ = 0.060 kg/m. The magnitude of maximum vertical acceleration of the wave is ay max = 0.90 cm/s2 and the amplitude of the wave is 0.40 m. The string is under a tension of FT = 600.00 N . The wave moves in the negative xdirection. Write an equation to model the wave.

24. A transverse wave on a string (μ = 0.0030 kg/m) is described with the equation y(x, t) = 0.30 m sin(2π/4.00m(x−16.00m/st)). What is the tension under which the string is held taut 25. Shown below is the plot of a wave function that models a wave at time t = 0.00 s and t = 2.00 s. The dotted line is the wave function at time t = 0.00 s and the solid line is the function at time t = 2.00 s. Estimate the amplitude, wavelength, velocity, and period of the wave.

UNIT 3:The principle of linear Superposition and Interference Phenomena INTRODUCTION: Superposition of oscillations in a non-linear medium or system introduces complications .In this unit our attention shall be focused on the linear superposition only.

3.The Principle of Linear Superposition Principle Of Superposition. The Principle Of Superposition states that when two waves of the same kind meet at a point in space, the resultant displacement at that point is the vector sum of the displacements that the two waves would separately produce at that point.

Interference refers to the superposing of two or more coherent waves to produce regions of maxima and minima in space, according to the principle of superposition.

Figure 7 :Combination of Constructive and Destructive Interference of Sound Waves Constructive interference occurs when two or more waves arrive at the screen in phase with each other, such that the amplitude of the resultant wave is the sum of the amplitude of the individual waves.

FIGURE 8:GRAPHIC REPRESENTATION OF THE CONSTRUCTIVE INTERFERENCE Destructive interference occurs when the two or more waves arrive π out of phase with each other. a positive displacement of one wave is cancelled exactly by a negative displacement of the other wave. The amplitude of the resulting wave is zero.

FIGURE 9: GRAPHIC REPRESENTATION OF THE DESTRUCTIVE INTERFERENCE 3.1 PATH DIFFERENCE: The path difference is the difference in distance traveled by the two waves.

Proof: Let y1(t)=Acos

be the equation of the source S1 and

y2(t)=Acos(

the equation of the source S2.The resultant is

y(t)=y1(t)+y2(t) ,where t

. with .

Using the trigonometry formula sinp+sinq=2sin(

,and Fresnel

construction, y(t)=2Acos ( AR=2Acos (

where the amplitude is .When the resultant source vibrates in phase,the amplitude

AR is maximal,therefore

so d2-d1=k take m=k When the resultant source vibrates out of phase ,AR=0 ie 2Acos

0

i.e

so d2-d1=

take m=2k+1{odd numbers}

If the sources have same phase,Then d2-d1=2k take m=2k{even numbers}.

In general ,to make the calculation easy,consider this notes: Difference in path lengths is IN PHASE ➢ n=0,2,4,6,8…{even numbers only} constructive interference takes place. ➢ n=1,3,5,7,….{odds numbers only} destructive interference takes place.

OUT OF PHASE • •

n=0,2,4,6,8…{even numbers only} destructive interference takes place. n=1,3,5,7,….{odds numbers only} constructive interference takes place.

3.2 DIFFRACTION : Diffraction happens when a wave hits an obstacle or gap, diffraction is greatest when the gap is about the same size as the wavelength of the wave.

The waves bend round the object or spread out when they pass through the gap, this is called diffraction.

3.3 Single Slit Diffraction When monochromatic laser light is shone through a narrow single slit a diffraction pattern is produced consisting of light and dark fringes. It produces a wide central bright fringe. The other bright fringes get dimmer as you move away from the centre.

Dsinθ = λ for single slit – first minimum

D sinθ =1, 22 λ for circular opening – first minimum

3.4 STANDING WAVE A standing wave is a pattern of interference that occurs when identical waves travel in opposite directions through the same medium. (This is most often a result of waves approaching a boundary and at the same time reflecting off of that boundary in the opposite direction.)

In any standing wave there will be nodes and antinodes: A node is a point in the medium at which there is zero disturbance. A node is a site of continuous and total destructive interference and the medium does not oscillate at that point.

An antinode is a point in the medium at which the amount of disturbance is maximized. An antinode is a site where constructive interference occurs and the oscillations of the medium are greatest.

The wavelength of any standing wave is equal to twice the distance between successive nodes or twice the distance between successive antinodes.

The speed of a standing wave is the speed at which that particular type of wave travels through the medium. (Even though the standing wave pattern is stationary it is being produced by waves that are traveling with a particular speed through the medium.)

Any standing wave system (such as a string or column of air) can only sustain standing waves at particular frequencies. These patterns of vibration that are unique to the system are called its harmonics (also sometimes called overtones). The harmonic with the lowest frequency is called the fundamental. It is a property of standing waves that any harmonic frequency is an integer multiple of the fundamental frequency.

FIGURE 10:GRAPHIC REPRESENTATION OF THE STANDING WAVE

String or cable fixed rigidly at both ends: Examples: guitar string, suspension cable, electrical cable between poles, etc.

Each end is a node.

Number of antinodes = n Number of nodes = n + 1

n = 1, 2, 3, 4 . . .

Pipe open at one end, closed at other end: Examples: reed instrument, certain organ pipes, straw in a drink, etc.

Open end is an antinode.

Closed end is a node.

Number of antinodes = Number of nodes =

n = 1, 3, 5, 7 . . . (only odd harmonics occur!)

.

Standing waves can form under a variety of conditions, but they are easily demonstrated in a medium which is finite or bounded. A phone cord begins at the base and ends at the handset. (Or is it the other way around?) Other simple examples of finite media are a guitar string (it runs from fret to bridge), a drum head (it's bounded by the rim), the air in a room (it's bounded by the walls), the water in Lake Michigan (it's bounded by the shores), or the surface of the Earth (although not bounded, the surface of the Earth is finite). In general, standing waves can be produced by any two identical waves traveling in opposite directions that have the right wavelength. In a bounded medium, standing waves occur when a wave with the correct wavelength meets its reflection. The interference of these two waves produces a resultant wave that does not appear to move.

Standing waves don't form under just any circumstances. They require that energy be fed into a system at an appropriate frequency. That is, when the driving frequency applied to a system equals its natural frequency. This condition is known as resonance. Standing waves are always associated with resonance.

Resonance can be identified by a dramatic increase in amplitude of the resultant vibrations. Compared to traveling waves with the same amplitude, producing standing waves is relatively effortless. In the case of the telephone cord, small motions in the hand result will result in much larger motions of the telephone cord.

Any system in which standing waves can form has numerous natural frequencies.

The set of all possible standing waves are known as the harmonics of a system. The simplest of the harmonics is called the fundamental or first harmonic. Subsequent standing waves are called the second harmonic, third harmonic, etc. The harmonics above the fundamental, especially in music theory, are sometimes also called overtones. What wavelengths will form standing waves in a simple, onedimensional system? There are three simple cases. one dimension: two fixed ends If a medium is bounded such that its opposite ends can be considered fixed, nodes will then be found at the ends. The simplest standing wave that can form under these circumstances has one antinode in the middle. This is half a wavelength. To make the next possible standing wave, place a node in the center. We now have one whole wavelength. To make the third possible standing wave, divide the length into thirds by adding another node. This gives us one and a half wavelengths. It should become obvious that to continue all that is needed is to keep adding nodes, dividing the medium into fourths, then fifths, sixths, etc. There are an infinite number of harmonics for this system, but no matter how many times we divide the medium up, we always get a whole number of half wavelengths ((1/2)λ, (2/2)λ,(3/2)λ,…,(n/2)λ).

There are important relations among the harmonics themselves in this sequence.

The wavelengths of the harmonics are simple fractions of the fundamental wavelength. If the fundamental wavelength were 1 m the wavelength of the second harmonic would be (½)m, the third harmonic would be (1/3)m, the fourth (1/4)m, and so on. Since frequency is inversely proportional to wavelength, the frequencies are also related. The frequencies of the harmonics are whole-number multiples of the fundamental frequency. If the fundamental frequency were 1 Hz the frequency of the second harmonic would be 2 Hz, the third harmonic would be 3 Hz, the fourth 4 Hz, and so on. one dimension: two free ends If a medium is bounded such that its opposite ends can be considered free, antinodes will then be found at the ends. The simplest standing wave that can form under these circumstances has one node in the middle. This is half a wavelength. To make the next possible standing wave, place another anti-node in the center. We now have one whole wavelength. To make the third possible standing wave, divide the length into thirds by adding another anti-node. This gives us one and a half wavelengths. It should become obvious that we will get the same relationships for the standing waves formed between two free ends that we have for two fixed ends.

The only difference is that the nodes have been replaced with antinodes and vice versa. Thus when standing waves form in a linear medium that has two free ends a whole number of half wavelengths fit inside the medium and the overtones are whole number multiples of the fundamental frequency.

one dimension: one fixed end — one free end When the medium has one fixed end and one free end the situation changes in an interesting way. A node will always form at the fixed end while an anti-node will always form at the free end. The simplest standing wave that can form under these circumstances is one-quarter wavelength long. To make the next possible standing wave add both a node and an anti-node, dividing the drawing up into thirds. We now have three-quarters of a wavelength. Repeating this procedure we get five-quarters of a wavelength, then seven-quarters, etc. In this arrangement, there are always an odd number of quarter wavelengths present. Thus the wavelengths of the harmonics are always fractional multiples of the fundamental wavelength with an odd number in the denominator. Likewise, the frequencies of the harmonics are always odd multiples of the fundamental frequency.

The three cases above show that, although not all frequencies will result in standing waves, a simple, one-dimensional system possesses an infinite number of natural frequencies that will. It also shows that these frequencies are simple multiples of some fundamental frequency. For any real-world system, however, the higher frequency standing waves are difficult if not impossible to produce. Tuning forks, for example, vibrate strongly at the fundamental frequency, very little at the second harmonic, and effectively not at all at the higher harmonics.

3.5 filtering The best part of a standing wave is not that it appears to stand still, but that the amplitude of a standing wave is much larger that the amplitude of the disturbance driving it. It seems like getting something for nothing. Put a little bit of energy in at the right rate and watch it accumulate into something with a lot of energy. This ability to amplify a wave of one particular frequency over those of any other frequency has numerous applications.

.

Basically, all non-digital musical instruments work directly on this principle. What

gets put into a musical instrument is vibrations or waves covering a spread of frequencies (for brass, it's the buzzing of the lips; for reeds, it's the raucous squawk of the reed; for percussion, it's the relatively indiscriminate pounding; for strings, it's plucking or scraping; for flutes and organ pipes, it's blowing induced turbulence).

What gets amplified is the fundamental frequency plus its multiples. These frequencies are louder than the rest and are heard. All the other frequencies keep their original amplitudes while some are even de-amplified. These other frequencies are quieter in comparison and are not heard. • You don't need a musical instrument to illustrate this principle. Cup your hands together loosely and hold them next to your ear forming a little chamber.

You will notice that one frequency gets amplified out of the background noise in the space around you. Vary the size and shape of this chamber. The amplified

•

•

pitch changes in response. This is what people hear when the hold a seashell up to their ears. It's not "the ocean" but a few select frequencies amplified out of the noise that always surrounds us.

During speech, human vocal cords tend to vibrate within a much smaller range that they would while singing. How is it then possible to distinguish the sound of one vowel from another? English is not a tonal language (unlike Chinese and many African languages). There is little difference in the fundamental frequency of the vocal cords for English speakers during a declarative sentence. (Interrogative sentences rise in pitch near the end. Don't they?) Vocal cords don't vibrate with just one frequency, but with all the harmonic frequencies. Different arrangements of the parts of the mouth (teeth, lips, front and back of tongue, etc.) favor different harmonics in a complicated manner. This amplifies some of the frequencies and de-amplifies others. This makes "EE" sound like "EE" and "OO" sound like "OO".

The filtering effect of resonance is not always useful or beneficial. People that work around machinery are exposed to a variety of frequencies. (This is what noise is.) Due to resonance in the ear canal, sounds near 4000 Hz are amplified and are thus louder than the other sounds entering the ear. Everyone should know that loud sounds can damage one's hearing. What everyone may not know is that exposure to loud sounds of just one frequency will damage one's hearing at that frequency. People exposed to noise are often experience 4000 Hz hearing loss. Those afflicted with this condition do not hear sounds near this frequency with the same acuity that unafflicted people do. It is often a precursor to more serious forms of hearing loss.

3.6 NATURAL FREQUENCIES:

The frequency or frequencies at which an object tends to vibrate with when hit, struck, plucked, strummed or somehow disturbed is known as the natural frequency of the object. ... All objects have a natural frequency or set of frequencies at which they vibrates.

fn=n(

with n=1,2,3… string fixed at both ends or tube open at both ends.

fn=n(

with n=1,3,5… tube open at one end.

The above diagrams show the relationship between L and . Observe that an open tube has all possible harmonics while the tube with a closed end can only have the odd harmonics.

Remember the velocity of sound is temperature dependent. v = (331 + 0.60T) .

Also If the strings are uniform, a special equation exist.

L1 f2 ----- = ----L2 f1

NB:For n=1 (fundamental and 1st harmonic);n=2 (2nd harmonic and 1st overtone);n=3(3rd harmonic and 2nd overtone) ect‌.

3.7 Beats frequency The beat frequency refers to the rate at which the volume is heard to be oscillating from high to low volume. For example, if two complete cycles of high and low volumes are heard every second, the beat frequency is 2 Hz. The beat frequency is always equal to the difference in frequency of the two notes that interfere to produce the beats. So if two sound waves with frequencies of 256 Hz and 254 Hz are played simultaneously, a beat frequency of 2 Hz will be detected. A common physics demonstration involves producing beats using two tuning forks with very similar frequencies. If a tine on one of two identical tuning forks is wrapped with a rubber band, then that tuning forks frequency will be lowered. If both tuning forks are vibrated together, then they produce sounds with slightly different frequencies.

These sounds will interfere to produce detectable beats. The human ear is capable of detecting beats with frequencies of 7 Hz and below.

A piano tuner frequently utilizes the phenomenon of beats to tune a piano string. She will pluck the string and tap a tuning fork at the same time. If the two sound sources the piano string and the tuning fork - produce detectable beats then their frequencies are not identical. She will then adjust the tension of the piano string and repeat the process until the beats can no longer be heard. As the piano string becomes more in tune with the tuning fork, the beat frequency will be reduced and approach 0 Hz. When beats are no longer heard, the piano string is tuned to the tuning fork; that is, they play the same frequency. The process allows a piano tuner to match the strings' frequency to the frequency of a standardized set of tuning forks. fbeat= Where, f1 and f2 are frequency of two waves.

Example 1: Calculate the beat frequency if the two frequencies of waves are 750Hz and 380Hz respectively?

Solution:

Given parameters are, f1 = 750Hz and f2 = 380Hz The beat frequency is given by,

= =

=

370Hz

Example 2: Determine the beat frequency if the wave frequencies are 550Hz, 1000Hz respectively?

Solution: Given parameters are, f1 = 550Hz and f2 = 1000Hz The beat frequency is given by,

=

= 450Hz

= TUTORIAL MULTIPLE CHOICES:

1) Destructive interference occurs when the crest of one wave lines up with the _____ of another wave. trough amplitude

crest

wavelength

frequency

2) Several positions along the medium are labeled with a letter. Categorize each labeled position along the medium as being a position where either constructive or destructive interference occurs.

•

3) Twin water bugs Jimminy and Johnny are both creating a series of circular waves by jiggling their legs in the water. The waves undergo interference and create the pattern represented in the diagram at the right. The thick lines in the diagram represent wave crests and the thin lines represent wave troughs.

Several of positions in the water are labeled with a letter. Categorize each labeled position as being a position where either constructive or destructive interference occurs.

4) What is a standing wave?

A vibration of a system where some points remain fixed, while others vibrate with the maximum amplitude .

A wave that doesn't move.

A wave where the whole thing is fixed in place

.

Where a peak and a trough combine together to form a flat line.

5)What is the superposition principle?

A principle that describes what will happen when you put two waves on top of each other .

A principle that describes how superheroes move.

A principle that tells you what happens when you put any two objects on top of

each other .

A principle about human responses to stimuli.

6.Speakers A and B are vibrating in phase. They are directly facing each other, are 7.80 m apart, and are each playing a 73.0-Hz tone. The speed of sound is 343 m/s. On the line between the speakers there are three points where constructive inter ference occurs. What are the distances of these three points from speaker A? (1,55m) 7.Two loudspeakers are vibrating in phase. They are set up as in the figure below, and point C is located as shown there. The speed of sound is 343 m/s. The speakers play the same tone. What is the smallest frequency that will produce destructive

interference at point C? (107Hz)

8) (a) A cello string has a fundamental frequency of 65.40 Hz. What beat frequency is heard when the cello string is bowed at the same time as a violin string with frequency of 1 96.0IIz? (Hint: The beats occur between the third harmonic of the cello string and the fundamental of the violin.)(0,2Hz) (b) A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.330 m long and has a mass of 9.60 g. It is fixed at both ends and oscillates in its fundamental mode. By resonance, itselfs the air column in the tube into oscillation at the tubes fundamental flequency. Find that frequency and the tension in the wire.( 71,5 Hz; 64,8 N) 9) A wave with a frequency of 14 Hz has a wavelength of 3 meters. At what speed will this wave travel?

10) The speed of a wave is 65 m/sec. If the wavelength of the wave is 0.8 meters, what is the frequency of the wave?

11) A wave has a frequency of 46 Hz and a wavelength of 1.7 meters. What is the speed of this wave?

12) A wave traveling at 230 m/sec has a wavelength of 2.1 meters. What is the frequency of this wave?

13) A wave with a frequency of 500 Hz is traveling at a speed of 200 m/s. What is the wavelength?

14) A wave has a frequency of 540 Hz and is traveling at 340 m/s. What is its wavelength?

15) A wave has a wavelength of 125 meters is moving at a speed of 20 m/s. What is it’s frequency?

16) A wave has a frequency of 900 Hz and a wavelength of 200 m. At what speed is this wave traveling?

17) A wave has a wavelength of 0.5 meters and a frequency of 120 Hz. What is the wave’s speed?

18) Radio waves travel at a speed of 300,000,000 m/s. WFNX broadcasts radio waves at a frequency of 101,700,000 Hertz. What is the wavelength of WFNX’s radio waves?

19) A sound wave travels 2.462 km in 8.73 s.

What is the velocity of the sound? ( 282 m/s) 20) An electromagentic wave travels for 21.38 ms.

How far does it travel in kilometers? ( 6.41 x 103 km) 21) What is the frequency of your dribbling if you bounce a ball 27 times in 20 seconds? (1.4 Hz) 22) What is the frequency of a mechanical wave that has a velocity of 1.77 m/s and a wavelength of 12.05 m ? ( 0.147 Hz)

23) You count 17 complete waves pass a certain point in 120 s and you measure 5.50 m from a crest to a trough on the passing waves.

What is the velocity of these waves? ( 1.56 m/s) 24) What is the period of 60Hz electrical power?(16,7ms) 25)If your heart rate is 150 beats per minute during strenuous exercise, what is the time per beat in units of seconds? (0,400s/beats)

26)Find the frequency of a tuning fork that takes 0,0025s to complete one oscillation. (400 Hz) 27) A stroboscope is set to flash every 0,00008s. What is the frequency of the flashes?(12,500 Hz) 28)Determine the frequency of: a)a ball bouncing 50 times in 38s b)an atom vibratin 1,3.1010 times in 2,5s c)a sound wave from a guitar string with a period of 3,50.10 -3 s d)a tuning fork which completes 2048cycles in 8s e)an electric jigsaw making 3200slices per minute

(Hint:

)

29) Two pure tones are sounded together. The drawing shows the pressure variations of the two sound waves, measured with respect to atmospheric pressure.

What is the beat frequency? (8Hz)

30) On a cello, the string with the largest linear density (1.56 × 10–2 kg/m) is the C string. This string produces a fundamental frequency of 65.4 Hz and has a length of 0.800 m between the two fixed ends. Find the tension in the string. (171N) 31)The two speakers in the drawing are vibrating in phase, and a listener is standing at point P. Does constructive or destructive interference occur at P when the speakers produce sound waves whose frequency is (a) 1466 Hz and (b) 977 Hz? Justify your answers with appropriate calculations. Take the speed of sound to be 343 m/s.

32) A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be in a speaker where the desired diffraction angle is 75° and a 9100 Hz sound is generated? The speed of sound is 343 m/s.

33)A man wishes to find out how far down the water level is in the iron pipe leading into an old well. Being blessed with perfect pitch, he merely hums musical sounds at

the mouth of the pipe and notice that the lowest-frequency resonance is 94 Hz. About how far from the top of the pipe is the water level? (0.91 m)

ASSIGNMENT:

UNIT 4: ELECTROMAGNETIC WAVES INTRODUCTION Definition: electromagnetic waves are one of the waves that are propagated by simultaneous periodic variations of electric and magnetic field intensity and that include radio waves, infrared, visible light, ultraviolet, X-rays, thermal radiation and gamma rays.The electromagnetic waves are transversal.

These waves were discovered by famous physicist James Maxwell. They can also be defined as "When electric and magnetic fields fluctuate together they lead to formation of the propagating waves called Electromagnetic waves.".

4.1 Properties of Electromagnetic Waves Electromagnetic radiations are basically streams of photons. It is a transverse wave.

In a transverse wave the medium has particles that vibrate in a direction perpendicular to the direction of the propagation of wave.

The various properties of Electromagnetic waves are: 1. The Velocity of electromagnetic wave in vacuum is 3 ×× 108 m / s.

2. The existence of medium is not essential for propagation.

3. In vacuum, E.M waves travel with light velocity.

4. E.M waves can be polarized.

5. E.M waves are transverse in nature.

6. E.M waves have momentum.

7. There is no deflection on account of magnetic or electric field.

8. They can exhibit diffraction and interference.

4.2 Examples of Electromagnetic Waves Radio waves, Light waves, thermal radiation, X ray, visible light, microwave, infrared, gamma rays etc. are the example of electromagnetic waves. These waves together form the electromagnetic spectrum.

4.3 Speed of Electromagnetic Waves Electromagnetic waves can have a wave or particle nature. When wave it is characterized by frequency and wavelength. In vacuum the waves travel with a

speed of light = 3 ×× 108 m/s. If a medium is considered then it would depend upon the refractive index of the medium.

4.4 Electromagnetic Spectrum The Electromagnetic spectrum consists of all the possible frequencies of electromagnetic radiations.

The electromagnetic spectrum starts from radio waves and goes beyond the gamma rays. It is very vast range of frequencies.

Electromagnetic Spectrum Waves EM spectrum consists of different types of waves with different wavelength and frequencies. The wavelength and the frequency of the wave are inversely proportional to each other. The following are the types of Electromagnetic Waves in the spectrum: 1. Radio Waves 2. Micro Waves 3. Infrared Rays 4. Visible Light 5. Ultraviolet 6. X rays 7. Gamma Rays

4.5 Types of Radiation There are seven types of electromagnetic waves which can be classified as follows:

4.6 Radio Waves : 1. These are the waves that are emitted by radio stations, during transmission or

by TV stations etc. they are also emitted by stars.

2. They Can range from few millimeters to large wavelength.

3. They are used by antennas.

4. They are used for data transmission via modulation.

4.7 Micro Waves : 1. The waves emitted by the microwave oven, yes they are the microwaves!

2. They are used by astronomers to study the structures of galaxies.

3. They are of two types Super high frequency and extremely high frequency.

4. They are used by Wi-fi.

4.8 Infrared Rays : They are emitted by everything that is warm and hot.

They are of three types viz.

1. Far infrared rays 2. Mid infrared rays 3. Near infrared rays.

4.9 Visible Light : 1. These are emitted by everything from a bulb to particle striking other particles.

These waves are detected by human eye.

2. The wavelength lies between 380 nm to 760 nm 3. They are used in optical fiber communication medium.

4.10 Ultraviolet : 1. They are emitted by sun, is harmful for human body. They cause skin burn.

2. They can ionize atoms.

3. They can cause skin cancer.

4. They cause mutagen 5. They are absorbed by ozone layer mostly.

4.11 X rays :

1. They are used by doctors.

2. They can interact with matter.

3. They cause Compton Effect.

They are of two types viz. •

Hard ray

•

Soft X-ray.

4.12 Gamma Rays : 1. They are emitted by radioactive substances.

2. They are used in treatment of many diseases.

4.13 Do Electromagnetic Waves Need a Medium?

Electromagnetic waves can travel through vacuum.

There are many theories behind this.

1. One says that as the electromagnetic waves are made up of electric and magnetic fields, oscillating perpendicular to each other and so at the point where they meet they push each other and hence propagate without any mediums need.

2. Electromagnetic waves possess both wave as well as particle nature and hence can travel without medium.

Thus the Traveling Electromagnetic Wave do not need medium for propagation.

4.14 Transfer of Energy by Electromagnetic Waves Electromagnetic radiation is the transfer of energy by electromagnetic waves.

Basically it is the movement of waves from one place to another. For example: X rays and Gamma rays. In order to transfer energy E. R act on the particular stuff to cause of initiate the chemical reactions. It can break bonds by exceeding the activation energy. They can break or make bonds.

For example: the microwave oven transferring energy for making pop corns. Also the light rays causing the reaction on the retina. The colors and everything else is an example. The energy carried is called Radiant energy.

4.15 Electromagnetic Spectrum Radio Waves

•

They are given out by transmitters

•

They are formed as a result of thunder, lightning etc.

•

They have the lowest frequency in the electromagnetic spectrum.

•

They are used in communication mostly.

•

They are of 4 types viz.

1. Long wave 2. Medium wave 3. UHF and 4. VHF. •

Much exposure can cause cancer.

•

They are used by antennas.

•

They are used for data transmission via modulation.

•

These are the waves that are emitted by radio stations, during transmission or

by TV stations etc. they are also emitted by stars.

•

They can range from few millimeters to large wavelength.

•

They are used in Phones, TV, radio etc.

Is Sound an Electromagnetic Wave?

The radio waves are used to carry the sound waves to the larger distance. It is composed of long radio waves, AM and FM waves. So, from this a question “Is Sound an Electromagnetic Wave” could be positively answered. Yes the sound is electromagnetic waves. In fact, we can say that any radiation of whatsoever the wavelength it is the part of the EM spectrum.

4.16 Energy in Electromagnetic Waves The Energy in electromagnetic waves are half due to electric field and half due to magnetic field.

Electromagnetic radiation is the transfer of energy by electromagnetic waves.

Basically it is the movement of waves from one place to another.

For example: X rays and Gamma Rays.

Electromagnetic Field Electromagnetic fields are produced by electromagnetic waves. Basically they are produced by every charge particle that is moving. Now as the waves have a particle as well as the wave nature so this field is also produced by electromagnetic waves. It is produced in the area surrounding the wave. It is the fundamental force of nature.

Basically the field is produced by electric and magnetic field, by static as well as moving charges. It is described by Maxwell equations.

Radio Waves The Radio waves are the very starting point of the EM spectrum. They have wavelength in the range of 1 millimeter to several kilometers. The Radio wave frequency lies between as low as 3 Hz to as high as 300 GHz. The radio waves are generally used in communication systems.

Infrared Rays The infrared waves are of wavelength well below the visible region of the EM spectrum. The infrared are not visible to our eye since they are outside the narrow window of the visible frequencies. In other words, we can say that we are blind for all the spectrum except the visible spectrum.

The heat emitted by the body after the flame extinguishes is an examples of the infrared waves.

Ultraviolet Rays In the Electromagnetic spectrum the rays having frequencies higher than the violet color of the visible light and frequency lower than the x rays is termed as the ultraviolet rays.

In terms of wavelength we can define the ultraviolet rays as the rays having wavelength less than the violet visible light and greater than the x rays. Ultraviolet rays wavelength varies from 10 nanometer to the 400 nanometer.

Ultraviolet spectrum of the electromagnetic spectrum is basically divided into three parts.

1. Near ultraviolet region

2. Far ultraviolet region 3. Extreme ultraviolet region.

It show three regions are made on the basis of the increasing energy Gamma Rays

The gamma rays are discovered by a French chemist and physicist, Paul Ulrich Villard, in 1900.

1. The gamma rays are the high frequency, high energy and shorter wavelength rays.

2. They lie above the X – rays in the electromagnetic spectrum and are distinguishable from the X – rays by the source of their origins.

They have too many application in medical sciences and nuclear physics.

4.17 X Rays After LAB BY Dr NSAMAN Alain Wilhelm Roentgen named the rays he discovered as X – rays. The X – rays belongs to the electromagnetic spectrum. The X – rays extends from the ultraviolet band to gamma rays band. Although their is very thin line between the X-ray and gamma rays but they are distinguished by the source of their generation. They have their own applications in medical science in diagnosis.

Just as the use of X-Rays has revolutionized surgery and dentistry, so now these wonderful rays that can penetrate the toughest metal are beginning to work their miracles in industry.

Dr Nsaman Alain After an X-RAY

Electromagnetic Spectrum

Types of Radiation

Radio Waves

Infrared Waves

Ultraviolet Rays

Gamma Rays

X Rays

Examples of electromagnetic wave in a Sentence Recent Examples on the Web ➢ In the wall’s appliance-detection mode, the power is turned off, and the electrodes act as an antenna to passively pick up electromagnetic waves emitted by nearby devices.(— Prachi Patel, Scientific American, "How to Convert Your Wall into a Giant Touch Screen," 1 June 2018). ➢ Around the same time, Giuseppe Cocconi and Philip Morrison from Cornell University theorized that aliens would contact us via electromagnetic waves and over the frequency of 1420 MHz(.— Matt Blitz, Popular Mechanics, "We Can Explore Instead of Believe: Inside the Epic Hunt for E.T.," 30 Apr. 2018).

FIGURE 11 :An Electromagnetic spectrum is the spectrum containing all possible range of frequencies coming in the electromagnetic wave. These waves do not need the medium for their Propagation. These waves can be polarized and are transverse in nature. Medium is not required for their propagation.

4.18 SPEED OF LIGHT IN VACCUM All waves of electromagnetic radiation travel at the speed of 3 x 108 m/s (in a vacuum). When traveling in other media, such as air or water, the speed of electromagnetic waves is only slightly slower. λf = c In free space is the speed of light c = 3*108 m/s An electromagnetic wave such as light consists of a coupled oscillating electric field and magnetic field which are always perpendicular; by convention, the "polarization" of electromagnetic waves refers to the direction of the electric field.

In linear polarization, the fields oscillate in a single direction. In circular or elliptical polarization, the fields rotate at a constant rate in a plane as the wave travels. The

rotation can have two possible directions; if the fields rotate in a right hand sense with respect to the direction of wave travel, it is called right circular polarization, or, if the fields rotate in a left hand sense, it is called left circular polarization.

Light or other electromagnetic radiation from many sources, such as the sun, flames, and incandescent lamps, consists of short wave trains with an equal mixture of polarizations; this is called unpolarized light. Polarized light can be produced by passing unpolarized light through a polarizing filter, which allows waves of only one polarization to pass through. The most common optical materials (such as glass) are isotropic and do not affect the polarization of light passing through them; however, some materials—those that exhibit birefringence, dichroism, or optical activity—can change the polarization of light. Some of these are used to make polarizing filters. Light is also partially polarized when it reflects from a surface.

According to quantum mechanics, electromagnetic waves can also be viewed as streams of particles called photons. When viewed in this way, the polarization of an electromagnetic wave is determined by a quantum mechanical property of photons called their spin. A photon has one of two possible spins: it can either spin in a right hand sense or a left hand sense about its direction of travel. Circularly polarized electromagnetic waves are composed of photons with only one type of spin, either right- or left-hand. Linearly polarized waves consist of equal numbers of right and left hand spinning photons, with their phase synchronized so they superpose to give oscillation in a plan .Polarization is an important parameter in areas of science dealing with transverse waves, such as optics, seismology, radio,and microwaves. Especially impacted are technologies such as lasers, wireless and optical fiber telecommunications, and radar.

4.19 Etienne-Louis MALUS’ LAW: Malus's Law According to malus, when completely plane polarized light is incident on the analyzer, the intensity S of the light transmitted by the analyzer is directly proportional to the square of the cosine of angle between the transmission axes of the analyzer and the polarizer. i.e S ∞ cos2θ

Suppose the angle between the transmission axes of the analyzer and the polarizer is θ. The completely plane polarized light form the polarizer is incident on the analyzer. If E0 is the amplitude of the electric vector transmitted by the polarizer, then intensity S0 of the light incident on the analyzer is S ∞ E02 The electric field vector E0 can be resolved into two rectangular components i.e E0 cosθ and E0sinθ. The analyzer will transmit only the component ( i.e E 0 cosθ ) which is parallel to its transmission axis. However, the component E 0sinθ will be absorbed by the analyser. Therefore, the intensity I of light transmitted by the analyzer is, S ∞ ( E0 x cosθ )2 S / S0 = ( E0 x cosθ )2 / E02 = cos2θ S = S0 x cos2θ Note :

Polarized light is S0=½Si

Si(incoming unpolarized light)

S-Average intensity of light leaving the analyzer S0-Average intensity of light entering the analyzer Θ-angle between transmission axis of polarizer and analyzer

Therefore, S ∞ cos2θ. This proves law of malus.

When θ = 0° ( or 180° ), S = S0 cos20° = S0 That is the intensity of light transmitted by the analyzer is maximum when the transmission axes of the analyzer and the polarizer are parallel.

When θ = 90°, S = S0 cos290° = 0 That is the intensity of light transmitted by the analyzer is minimum when the transmission axes of the analyzer and polarizer are perpendicular to each other.

Example: Unpolarized light of intensity I0 is incident on a series of three polarizing filters. The axis of the second filter is oriented at 45 o to that of the first filter, while the axis of the third filter is oriented at 90o to that of the first filter. What is the intensity of the light transmitted through the third filter?

A. 0 B. I0/8 C. I0/4 D. I0/2 E. I0/√2 Solution: Polarizer: The first filter always reduces the intensity of the light to half, I1=½ I0 The next filter reduces the intensity by In = I(n-1) cos2 θ θ is the angle with respect to the nth filter.

Thus,

I1 = ½ I0 I2 = I1 (cos 45)2 = I1 (½ √2)2 = ½ I1 = ¼ I0 I3=I2 (cos 45)2 = ½ I2 = I0/8

Tutorial: 1)Which of the following describes the meaning of polarization?

The direction that a light wave vibrates (or oscillates).

The direction that light travels.

What happens when light bounces off a lake.

The direction opposite to the way light vibrates.

A type of filter.

2) Consider the three pairs of sunglasses below. Identify the pair of glasses is capable of eliminating the glare resulting from sunlight reflecting off the calm waters of a lake? _________ Explain. (The polarization axes are shown by the straight lines.)

3) Suppose that light passes through two Polaroid filters whose polarization axes are parallel to each other. What would be the result?

4) Are the electromagnetic waves solutions of Maxwell’s equations in a vacuum? a)yes b)no 5)Unpolarized light with an intensity of so = 16 W/m2 is incident on a pair of polarizers.

The first polarizer has its transmission axis aligned at 50 o from the vertical. The second polarizer has its transmission axis aligned at 20 o from the vertical.

What is the intensity of the light when it emerges from the first polarizer?

1.

2.

3.

4.

4 W/m2 16 cos250o 8 W/m2 12 W/m2

6) A polaroid sheet and an analyzer are placed such that their transmission axes are co-linear. The analyzer is then rotated by 22.5o. What is the irradiance of the transmitted light as a fraction of its previous value?

7) You have three ideal linear polarizers. Light of irradiance 1000 W/m2 is shone through two of the polarizers, with their transmission axes placed at a relative angle of 40o. What is the intensity of the transmitted light? Now place the third polarizer at an angle of 20o between the other two. What is the irradiance?

8) What angle is needed between the direction of polarized light and the axis of a polarizing filter to reduce its intensity by ?(71,6deg) 9)You have three polarizers. Polarizer A has its transmission axis at 0° relative to the vertical; polarizer B has its transmission axis at 30° to the vertical; and polarizer C has its transmission axis at 90° to the vertical. [4 points] (a) You can arrange the three polarizers in any order you wish. If the incident light is unpolarized, in what order should you place the three polarizers so as to maximize the intensity of the light emerging from the last polarizer in the sequence? [ ] Polarizer A, then B, then C [ ] Polarizer C, then B, then A [ X ] Either of the above, they’d give the same final intensity [ ] Polarizer B, then A, then C [ ] Polarizer B, then C, then A [ ] It doesn’t matter what order they’re in, none of the wave gets through Briefly justify your answer.

10) A particular plane polarized electromagnetic wave, with a frequency of 100 MHz, is traveling through a vacuum in a direction we can call the x-axis. At t = 0, the electric field due to this wave at x = 0 has a magnitude of 300 V/m. (a) What is the wavelength of this wave? (b) If this wave entered your eye, would you see anything? Explain why or why not.

11) A beam of polarized light passes through a polarizing filter. When the angle between the polarizing axis of the filter and the direction of polarization of the light is θ, the intensity of the emerging beam is S If you now want the intensity to be S/2, what should be the angle (in terms of θ) between the polarizing angle of the filter and the original direction of polarization of the light?

12)Unpolarized light of intensity 21.2W/cm2 is incident on two polarizing filters. The axis of the first filter is at an angle of 24.6 degrees counterclockwise from the vertical (viewed in the direction the light is traveling) and the axis of the second filter is at 62.2 degrees counterclockwise from the vertical. What is the intensity of the light after it has

passed through the second polarizer?(6,65W/cm2) 13)Three polarizing filters are stacked, with the polarizing axis of the second and third filters at angles of 23.6 degrees and 62.2 degrees, respectively, to that of the first. If unpolarized light is incident on the stack, the light has an intensity of 75.5W/cm2 after it passes through the stack.

If the incident intensity is kept constant, what is the intensity of the light after it has passed through the stack if the second polarizer is removed?(32,02 W/cm2) 14) Two polarizing sheets have their transmission axes parallel so that the intensity of unpolarized light transmitted through both of them is a maximum.

Through what angle must either sheet be rotated if the transmitted intensity is 25 % of the incident intensity? (45 degree)

15) An FM radio station generates radio waves that have a frequency of 95.50 MHz. The frequency of the waves from a competing station have a frequency of 102.70 MHz..

What is the difference in wavelength between the waves emitted from the two stations? (0.22m)

16) At what angle will light reflected from diamond be completely polarized? ( đ?&#x;”đ?&#x;•. đ?&#x;”°)

17) Verify that the intensity of polarized light is reduced to 90.0% of its original value by passing through a polarizing filter with its axis at an angle of 18.4° to the direction of polarization. (90.0%)

UNIT 5:INTERFERENCE AND THE WAVE NATURE OF LIGHT INTRODUCTION Light is a transverse,electromagnetic wave that can be seen by the typical human.The wave nature of light was first illustrated through experiments on diffraction and interference .Like all electromagnetic waves,light can travel through a vacuum.

5.1 YOUNG’S DOUBLE SLIT EXPERIMENT In 1801 Thomas Young was able to offer some very strong evidence to support the wave model of light.

• • •

He placed a screen that had two slits cut into it in front of a monochromatic (single color) light.

The results of Young's Double Slit Experiment should be very different if light is a wave or a particle.

Let’s look at what the results would be in both situations, and then see how this experiment supports the wave model.

If light is a particle…

We set up our screen and shine a bunch of monochromatic light onto it.

•

•

If light is a particle, then only the couple of rays of light that hit exactly where the slits are will be able to pass through. o Imagine it as being almost as though we are spraying paint from a spray can through the openings.

Since they are little particles they will make a pattern of two exact lines on the viewing screen (Figure 12).

Figure 12

If light is a wave…

If light is a wave, everything starts the same way, but results we get are very different.

•

There are still only two light rays that actually go through the slits, but as soon as they pass through they start to diffract. o Remember, diffraction is when light passes through a small opening and starts to spread out. This will happen from both openings (Figure 2).

Figure 13 Crossing of the monochromatic light through screen with two slits •

Notice that at some points the two sets of waves will meet crest to crest, at other spots crest meets trough. o Where crest meets crest, there will be constructive interference and the waves will make it to the viewing screen as a bright spot. o

•

Where crest meets trough there will be destructive interference that cancel each other out… a black spot will appear on the screen.

When this experiment is performed we actually see this, as shown in Figure 3.

Figure 14 viewing of the screen We must conclude that light is made up of waves, since particles can not diffract.

Calculations When you set up this sort of an apparatus, there is actually a way for you to calculate where the bright lines (called fringes) will appear.

•

There is always a middle line, which is the brightest. We call it the central fringe.

Figure 15: Note I’ve only coloured the lines differently here so that they will stand out better. For monochromatic light they would all have the same colour… that of the original light.

•

In the formula we will use, there is a variable, “n”, that is a count of how many bright fringes you are away from the central fringe. o The central fringe is n = 0. o The fringe to either side of the central fringe has an order of n = 1 (the first order fringe). o The order of the next fringe out on either side is n = 2(the second order fringe). o

And so on, as shown in Figure 4.

The formula that we will use to figure out problems involving double slit experiments is easy to mix up, so make sure you study it carefully.

λ = wavelength of light used (m) x = distance from central fringe (m) d = distance between the slits (m) n = the order of the fringe L = length from the screen with slits to the viewing screen (m)

It is very easy to mix up the measurements of x, d, and L. • • •

Make sure to look at Figure 5 and see the different things each is measuring.

If you mix up x and d it's not so bad, since they are both on top in the formula.

If you were to mix them up with L, you would get the wrong answer.

Almost all questions that you will see for this formula just involve sorting out what each variable is... you might find it helpful to write out a list of givens.

Example 1: A pair of screens are placed 13.7m apart. A third order fringe is seen on the screen 2.50cm from the central fringe. If the slits were cut 0.0960 cm apart, determinethe wavelength of this light. Roughly what colour is it?

Just to make sure you’ve got all the numbers from the question matched with the correct variables…

L = 13.7 m n=3 x = 2.50cm = 0.0250 m d = 0.0960cm = 9.60e-4 m

It’s probably a yellow light being used given the wavelength we've measured.

If a white light is used in the double slit experiment, the different colours will be split up on the viewing screen according to their wavelengths. •

The violet end of the spectrum (with the shortest wavelengths) is closer to the central fringe, with the other colours being further away in order.

There is also a version of the formula where you measure the angle between the central fringe and whatever fringe you are measuring.

• •

The formula works the same way, with the only difference being that we measure the angle instead of x and L. Make sure that your calculator is in degree mode before using this version of the formula.

Example 2: If a yellow light with a wavelength of 540 nm shines on a double slit with the slits cut 0.0100 mm apart, determine what angle you should look away from the central fringe to see the second order fringe?

Do not forget to:

1. Change the wavelength into metres.

2. Change the slit separation into metres.

3. "Second order" is a perfect number and has an infinite number of sig digs.

The Single Slit

A surprising experiment is that you can get the same effect from using a single slit instead of a double slit.

•

Light from different ends of the slit will be traveling to the same spot on the

•

screen and reach there either in or out of sync. o In Figure 7, the blue path has to travel further than the red path... if this difference is equal to half a wavelength, they will meet each other out of sync.

If they meet crest to crest or trough to trough they will be in sync, but if

• •

they meet crest to trough they will be out of sync.

Being in sync will result in constructive interference, while meeting out of sync will result in destructive interference.

The result is the same interference pattern, although the effect isn’t nearly as dramatic or clear as the double slit experiment.

After the first couple of fringes (n = 1 and 2), the edges start getting really fuzzy, so you have a hard time measuring anything.

The only real difference in calculations is that "d" is now the width of the single opening. o

•

•

If two slits work better than one, would more than two slits work better? This is a question that we will answer in the next section.

Example 3: For a single slit experiment apparatus like the one described above, determine how far from the central fringe the first order violet (λ = 350nm) and red (λ = 700nm) colours will appear if the screen is 10 m away and the slit is 0.050 cm wide.

We need to solve the formula for “x”, the distance from the central fringe.

For the violet light…

For the red light…

NOTE:

ou can actually do the single slit experiment wherever you are right now! Hold two of your fingers very close together; there should be only the tiniest little gap between them that you can barely see through. Look towards a light source, light a light bulb, through the gap in your fingers. In the gap between your fingers you shold see very faint gray lines that run parallel to your fingers... these are the destructive interference "dark" fringes!

Diffraction Diffraction happens when a wave hits an obstacle or gap, diffraction is greatest when the gap is about the same size as the wavelength of the wave. The

waves bend round the object or spread out when they pass through the gap, this is called diffraction.

Single Slit Diffraction When monochromatic laser light is shone through a narrow single slit a diffraction pattern is produced consisting of light and dark fringes. It produces a wide central bright fringe. The other bright fringes get dimmer as you move away from the centre.

Note that for the single slit The variation in wave intensity can be mathematically modeled. From the center of the slit, the diffracting waves propagate radially. The angle of the minimum intensity (θ min) can be related to wavelength (λ) and the slit’s width (w) such that: wsinθmin=λ Diffraction grating.

A diffraction grating is a piece of glass with lots of closely spaced parallel lines on it each of which allows light to pass through it, this is a transmission diffraction grating.

Diffraction gratings are used in spectrometers. The diffraction grating splits up the light into a spectra.

d = grating spacing in metres (m) θ= angle of diffraction n = order number Ν = wavelength on the light in metres (m) This equation can be rearranged to

If there are N lines per metre on a diffraction grating then d can be calculated using

with N= # slits per meter Diffraction gratings are used in the spectral analysis of light from stars.

Double slit formula Interference Fringes Image is in the public domain m = 0, ±1, ±2, ... dsin(θ)=(m + 1/2)λ, destructive interference(dark) m = 0, ±1, ±2, ... dsin(θ) = mλ, constructive interference(bright) tan( θ)= , With θ in degrees or radians.Take x=L

Figure 16 Graphic combinasion of the single and double slit NB:To calculate the distance between two fringes ,the below graph should be considered.

xtan(θ1 )=y1 and xtan(θ2 )=y2 y2- y1= xtan(θ2 )- xtan(θ1 )=x{tan(θ2 )-tan(θ1 )}

Circular opening: sin θ = 1.22 (λ/D),Where D is the diameter of the hole.Where θ= , θ in radians.

Rayleigh criterium for resolution: The Rayleigh criterion is the generally accepted criterion for the minimum resolvable detail - the imaging process is said to be diffraction-limited when the first diffraction minimum of the image of one source point coincides with the maximum of another.

Resolving power: The ability to distinguish between two closely spaced objects.The first minimum is at an angle of θ=1.22Ν/D, so that two point objects are just resolvable if they are separated by the angle

θ=1.22λ/D where λ is the wavelength of light (or other electromagnetic radiation) and D is the diameter of the aperture, lens, mirror, etc., with which the two objects are observed.

In this expression, θ has units of radians. This angle is also commonly known as the diffraction limit.

Example The primary mirror of the orbiting Hubble Space Telescope has a diameter of 2.40 m. Being in orbit, this telescope avoids the degrading effects of atmospheric distortion on its resolution. (a) What is the angle between two just-resolvable point light sources (perhaps two stars)? Assume an average light wavelength of 550 nm. (b) If these two stars are at a distance of 2 million light-years, which is the distance of the Andromeda Galaxy, how close together can they be and still be resolved? (A light-year, or ly, is the distance light travels in 1 year.)

Solution: Strategy The Rayleigh criterion stated in Equation θ=1.22λ/D, gives the smallest possible angle θ between point sources, or the best obtainable resolution. Once this angle is known, we can calculate the distance between the stars, since we are given how far away they are.

Solution 1. The Rayleigh criterion for the minimum resolvable angle is θ=1.22λD. Entering known values gives θ=1.22(550×10−9m)/2.40m=2.80×10−7rad.

2. The distance s between two objects a distance r away and separated by an angle θ is s=rθ. Substituting known values gives s=(2.0×106ly)(2.80×10−7rad)=0.56ly.

Significance The angle found above is extraordinarily small (less than 1/50,000 of a degree), because the primary mirror is so large compared with the wavelength of light. As noticed, diffraction effects are most noticeable when light interacts with objects having sizes on the order of the wavelength of light. However, the effect is still there, and there is a diffraction limit to what is observable. The actual resolution of the Hubble Telescope is not quite as good as that found here. As with all instruments, there are other effects, such as non uniformities in mirrors or aberrations in lenses that further limit resolution. However, this gives an indication of the extent of the detail observable with the Hubble because of its size and quality, and especially because it is above Earth’s atmosphere.

5.2 HUYGENS PRINCIPLE: Christiaan Huygens was a Dutch scientist who developed a useful technique for determining how and where waves propagate. In 1678, he proposed that every point that a luminous disturbance touches becomes itself a source of a spherical wave. The sum of the secondary waves (waves that are a result of the disturbance) determines the form of the new wave. shows secondary waves traveling forward from their point of origin. He was able to come up with an explanation of the linear and spherical wave propagation, and derive the laws of reflection and refraction (covered in previous atoms ) using this principle. He could not, however, explain what is commonly known as diffraction effects. Diffraction effects are the deviations from rectilinear propagation that occurs when light encounters edges, screens and apertures.These effects were explained in 1816 by French physicist Augustin-Jean Fresnel.

where S is the distance, v is the propagation speed, and t is time.

TUTORIAL : A)SINGLE-SLIT DIFFRACTION: 1. (a) At what angle is the first minimum for 550-nm light falling on a single slit

of width 1.00μm?(33,4 degrees) (b) Will there be a second minimum?(no) 2. (a) Calculate the angle at which a 2.00−μm2.00−μm-wide slit produces its first minimum for 410-nm violet light.( (b) Where is the first minimum for 700-nm red light?

3. (a) How wide is a single slit that produces its first minimum for 633-nm light at an angle of 28.0°28.0°?( 1.35×10−6m) (b) At what angle will the second minimum be?( 69.9°) 4. (a) What is the width of a single slit that produces its first minimum at 60.0°60.0° for 600-nm light? (b) Find the wavelength of light that has its first minimum at 62.0°62.0°.

5. Find the wavelength of light that has its third minimum at an angle of 48.6°48.6° when it falls on a single slit of width 3.00μm3.00μm.( 750 nm) 6. (a) Sodium vapor light averaging 589 nm in wavelength falls on a single slit of width 7.50μm7.50μm. At what angle does it produces its second minimum? (b) What is the highest-order minimum produced?

7. Consider a single-slit diffraction pattern for λ=589nm λ=589nm, projected on a screen that is 1.00 m from a slit of width 0.25 mm. How far from the center of the pattern are the centers of the first and second dark fringes?( . 2.4 mm, 4.7 mm) 8. (a) Find the angle between the first minima for the two sodium vapor lines, which have wavelengths of 589.1 and 589.6 nm, when they fall upon a single slit of width 2.00μm2.00μm. (b) What is the distance between these minima if the diffraction pattern falls on a screen 1.00 m from the slit? (c) Discuss the ease or difficulty of measuring such a distance.

9. (a) What is the minimum width of a single slit (in multiples of λ) that will produce a first minimum for a wavelength λ?( 1.00λ) (b) What 50.0λ)

is

its

minimum

width

if

it

produces

50

minima?(

(c) 1000 minima?( 1000λ) 10. (a) If a single slit produces a first minimum at 14.5°14.5°, at what angle is the second-order minimum?

(b) What is the angle of the third-order minimum? (c) Is there a fourth-order minimum? (d) Use your answers to illustrate how the angular width of the central maximum is about twice the angular width of the next maximum (which is the angle between the first and second minima).

11. If the separation between the first and the second minima of a single-slit diffraction pattern is 6.0 mm, what is the distance between the screen and the slit? The light wavelength is 500 nm and the slit width is 0.16 mm.( 1.92 m) 12. A water break at the entrance to a harbor consists of a rock barrier with a 50.0-m-wide opening. Ocean waves of 20.0-m wavelength approach the opening straight on. At what angles to the incident direction are the boats inside the harbor most protected against wave action?

13. An aircraft maintenance technician walks past a tall hangar door that acts like a single slit for sound entering the hangar. Outside the door, on a line perpendicular to the opening in the door, a jet engine makes a 600-Hz sound.

At what angle with the door will the technician observe the first minimum in sound intensity if the vertical opening is 0.800 m wide and the speed of sound is 340 m/s?( 45.1°) 14) Two slits separated by 2.00 x 10-5 m are illuminated by light of wavelength 500 nm. If the screen is 8.00 m from the slits, what is the distance between the m = 0 and m = 1 bright fringes? (0.2 m)

B)DOUBLE SLIT DIFFRACTION: 14. Two slits of width 2μm2μm, each in an opaque material, are separated by a center-to-center distance of 6μm6μm. A monochromatic light of wavelength 450 nm is incident on the double-slit. One finds a combined interference and diffraction pattern on the screen. (a) How many peaks of the interference will be observed in the central maximum of the diffraction pattern? (b) How many peaks of the interference will be observed if the slit width is doubled while keeping the distance between the slits same? (c) How many peaks of interference will be observed if the slits are separated by twice the distance, that is, 12μm, while keeping the widths of the slits same? (d) What will happen in (a) if instead of 450-nm light another light of wavelength 680 nm is used?

(e) What is the value of the ratio of the intensity of the central peak to the intensity of the next bright peak in (a)? (f) Does this ratio depend on the wavelength of the light? (g) Does this ratio depend on the width or separation of the slits?

15. A double slit produces a diffraction pattern that is a combination of singleand double-slit interference. Find the ratio of the width of the slits to the separation between them, if the first minimum of the single-slit pattern falls on the fifth maximum of the double-slit pattern. (This will greatly reduce the intensity of the fifth maximum.)( 0.200) 16. For a double-slit configuration where the slit separation is four times the slit width, how many interference fringes lie in the central peak of the diffraction pattern?

17. Light of wavelength 500 nm falls normally on 50 slits that are 2.5×10−3mm2.5×10−3mm wide and spaced 5.0×10−3mm5.0×10−3mm apart. How many interference fringes lie in the central peak of the diffraction pattern?(3) 18. A monochromatic light of wavelength 589 nm incident on a double slit with slit width 2.5μm2.5μm and unknown separation results in a diffraction pattern containing nine interference peaks inside the central maximum. Find the separation of the slits.

19. When a monochromatic light of wavelength 430 nm incident on a double slit of slit separation 5μm5μm, there are 11 interference fringes in its central maximum. How many interference fringes will be in the central maximum of a light of wavelength 632.8 nm for the same double slit?(9) C)DIFFRACTION GRATING: 20. A diffraction grating has 2000 lines per centimeter. At what angle will the first-order maximum be for 520-nm-wavelength green light?( 5.97°) 21. Find the angle for the third-order maximum for 580-nm-wavelength yellow light falling on a difraction grating having 1500 lines per centimeter.

22. How many lines per centimeter are there on a diffraction grating that gives a first-order maximum for 470-nm blue light at an angle of 25.0°?( 8.99×103) 23. What is the distance between lines on a diffraction grating that produces a second-order maximum for 760-nm red light at an angle of 60.0°60.0°?

24. Calculate the wavelength of light that has its second-order maximum at 45.0°45.0° when falling on a diffraction grating that has 5000 lines per centimeter.( 707 nm) 25. An electric current through hydrogen gas produces several distinct

wavelengths of visible light. What are the wavelengths of the hydrogen spectrum,if they form first-order maxima at angles 24.2°,25.7°,29.1°,24.2°,25.7°,29.1°, and 41.0°when projected on a diffraction grating having 10,000 lines per centimeter?

26. (a) What do the four angles in the preceding problem become if a 5000-line per centimeter diffraction grating is used?( 11.8°,12.5°,14.1°,19.2°) (b) Using this grating, what would the angles be for the second-order maxima?( 24.2°,25.7°,29.1°,41.0°) (c) Discuss the relationship between integral reductions in lines per centimeter and the new angles of various order maxima.( Decreasing the number of lines per centimeter by a factor of x means that the angle for the x-order maximum is the same as the original angle for the first-order maximum.)

27. What is the spacing between structures in a feather that acts as a reflection grating, giving that they produce a first-order maximum for 525-nm light at a 30.0° angle?

28. (a) Find the maximum number of lines per centimeter a diffraction grating can have and produce a maximum for the smallest wavelength of visible light.( 26,300 lines/cm;) (b) Would such a grating be useful for ultraviolet spectra?( yes;) (c) For infrared spectra?(no)

D)CIRCULAR OPENING AND RESOLUTION: 28. The 305-m-diameter Arecibo radio telescope pictured in Figure 4.20 detects radio waves with a 4.00-cm average wavelength. (a) What is the angle between two just-resolvable point sources for this telescope? (b) How close together could these point sources be at the 2 million lightyear distance of the Andromeda Galaxy?

29. Assuming the angular resolution found for the Hubble Telescope in the example from the notes , what is the smallest detail that could be observed on the moon?( 107 m) 30. Diffraction spreading for a flashlight is insignificant compared with other limitations in its optics, such as spherical aberrations in its mirror. To show this, calculate the minimum angular spreading of a flashlight beam that is originally

5.00 cm in diameter with an average wavelength of 600 nm.

31. (a) What is the minimum angular spread of a 633-nm wavelength He-Ne laser beam that is originally 1.00 mm in diameter?( 7.72×10−4rad) (b) If this laser is aimed at a mountain cliff 15.0 km away, how big will the illuminated spot be?( 23.2 m;) (c) How big a spot would be illuminated on the moon, neglecting atmospheric effects? (This might be done to hit a corner reflector to measure the round-trip time and, hence, distance.)( 590 km) 32. A telescope can be used to enlarge the diameter of a laser beam and limit diffraction spreading. The laser beam is sent through the telescope in opposite the normal direction and can then be projected onto a satellite or the moon. (a) If this is done with the Mount Wilson telescope, producing a 2.54-m-diameter beam of 633-nm light, what is the minimum angular spread of the beam? (b) Neglecting atmospheric effects, what is the size of the spot this beam would make on the moon, assuming a lunar distance of 3.84×108m3.84×108m. ?

33. What is the minimum angular separation of two stars that are just-resolvable by the 8.1-m Gemini South telescope, if atmospheric effects do not limit resolution? Use 550 nm for the wavelength of the light from the stars.

34. The headlights of a car are 1.3 m apart. What is the maximum distance at which the eye can resolve these two headlights? Take the pupil diameter to be 0.40 cm.

35)The limit to the eye’s acuity is actually related to diffraction by the pupil. (a) What is the angle between two just-resolvable points of light for a 3.00mm-diameter pupil, assuming an average wavelength of 550 nm?( 2.24×10−4rad) (b) Take your result to be the practical limit for the eye. What is the greatest possible distance a car can be from you if you can resolve its two headlights, given they are 1.30 m apart?( 5.81 km;) (c) What is the distance between two just-resolvable points held at an arm’s length (0.800 m) from your eye?( 0.179 mm;) (d) How does your answer to (c) compare to details you normally observe in everyday circumstances?( can resolve details 0.2 mm apart at arm’s length) 36. White light falls on two narrow slits separated by 0.40 mm. The interference pattern is observed on a screen 3.0 m away. (a) What is the separation between the first maxima for red light (λ=700nm) and violet light (λ=400nm)?( 2.2 mm) (b) At what point nearest the central maximum will a maximum for yellow light (λ=600nm) coincide with a maximum for violet light? Identify the order for each maximum.( 0.172°, second-order yellow and third-order violet coincide).

UNIT 6:THE REFLECTION OF THE LIGHT:MIRRORS INTRODUCTION 6.1 Wave Fronts and Rays:

Imagine you throw a rock into a pond. Seen from the side, at the level of the water, the ripples look like this:

The distance between one ripple and the next is called the wavelength, Îť. The high points are called crests, and the low points are the troughs.

If instead you look down on the pond, as if from a hovering helicopter, the ripples are round, and spreading outwards (diverging). The technical term for ripples is wavefronts. The arrows are pointing in the direction the waves are moving, and they are called rays.

Notice that the rays are always perpendicular to the wavefronts. In other words, the wavefront always moves in a direction at right angles to itself.

As the waves move farther and farther from the center, where the rock hit the water, the wavefronts are larger and larger circles. But if you look at a small piece of the wavefront, it nearly looks flat.

This is why rays are often (especially in basic textbooks) drawn parallel to each other when entering a lens — if the light source is very far away, the wavefronts hitting the lens from one point of the object are essentially flat (plane waves), and the rays are

parallel to each other.

Images Notice that in the discussion above, it was assumed that the wavefronts were spreading outwards, diverging. This jives with our common sense notion of what waves do — ripples naturally leave the point where the rock hit the water, making larger and larger circles. There are some very special devices, however, that can change the direction of those waves and make them come back together again (converge). These devices are called mirrors and lenses.

When these are used to bring light waves back together, they can create a replica or real image of the original source of the light. We'll be discussing this process in considerable detail in the coming days, and learning how lenses and mirrors make such useful things as eyes, telescopes and microscopes.

6.2 The Reflection of The Light: 6.3 The law of reflection Objects can be seen by the light they emit, or, more often, by the light they reflect. Reflected light obeys the law of reflection, that the angle of reflection equals the angle of incidence.

For objects such as mirrors, with surfaces so smooth that any hills or valleys on the surface are smaller than the wavelength of light, the law of reflection applies on a large scale. All the light travelling in one direction and reflecting from the mirror is reflected in one direction; reflection from such objects is known as specular reflection.

Most objects exhibit diffuse reflection, with light being reflected in all directions.

All objects obey the law of reflection on a microscopic level, but if the irregularities on the surface of an object are larger than the wavelength of light, which is usually the case, the light reflects off in all directions.

6.4 Plane mirrors A plane mirror is simply a mirror with a flat surface; all of us use plane mirrors every day, so we've got plenty of experience with them. Images produced by plane mirrors have a number of properties, including:

1. the image produced is upright 2. the image is the same size as the object (i.e., the magnification is m = 1) 3. the image is the same distance from the mirror as the object appears to be (i.e., the image distance = the object distance) 4. the image is a virtual image, as opposed to a real image, because the light rays do not actually pass through the image. This also implies that an image could not be focused on a screen placed at the location where the image is.

6.5 A little geometry Dealing with light in terms of rays is known as geometrical optics, for good reason: there is a lot of geometry involved. It's relatively straight-forward geometry, all based on similar triangles, but we should review that for a plane mirror.

Consider an object placed a certain distance in front of a mirror, as shown in the diagram. To figure out where the image of this object is located, a ray diagram can be used. In a ray diagram, rays of light are drawn from the object to the mirror, along with the rays that reflect off the mirror. The image will be found where the reflected rays intersect. Note that the reflected rays obey the law of reflection. What you notice is that the reflected rays diverge from the mirror; they must be extended back to find the place where they intersect, and that's where the image is.

Analyzing this a little further, it's easy to see that the height of the image is the same as the height of the object. Using the similar triangles ABC and EDC, it can also be seen that the distance from the object to the mirror is the same as the distance from the image to the mirror.

6.6 Spherical mirrors Light reflecting off a flat mirror is one thing, but what happens when light reflects off a curved surface? We'll take a look at what happens when light reflects from a spherical mirror, because it turns out that, using reasonable approximations, this analysis is fairly straight-forward. The image you see is located either where the reflected light converges, or where the reflected light appears to diverge from.

A spherical mirror is simply a piece cut out of a reflective sphere. It has a center of curvature, C, which corresponds to the center of the sphere it was cut from; a radius of curvature, R, which corresponds to the radius of the sphere; and a focal point (the point where parallel light rays are focused to) which is located half the distance from the mirror to the center of curvature.

The focal length, f, is therefore: focal length of a spherical mirror : f =

R/2

This is actually an approximation. Parabolic mirrors are really the only mirrors that focus parallel rays to a single focal point, but as long as the rays don't get too far from the principal axis then the equation above applies for spherical mirrors. The diagram shows the principal axis, focal point (F), and center of curvature for both a concave and convex spherical mirror.

Spherical mirrors are either concave (converging) mirrors or convex (diverging) mirrors, depending on which side of the spherical surface is reflective. If the inside surface is reflective, the mirror is concave; if the outside is reflective, it's a convex mirror. Concave mirrors can form either real or virtual images, depending on where the object is relative to the focal point. A convex mirror can only form virtual images. A real image is an image that the light rays from the object actually pass through; a virtual image is formed because the light rays can be extended back to meet at the image position, but they don't actually go through the image position.

6.7 Ray diagrams To determine where the image is, it is very helpful to draw a ray diagram. The image will be located at the place where the rays intersect. You could just draw random rays from the object to the mirror and follow the reflected rays, but there are three rays in particular that are very easy to draw.

Only two rays are necessary to locate the image on a ray diagram, but it's useful to add the third as a check. The first is the parallel ray; it is drawn from the tip of the object parallel to the principal axis. It then reflects off the mirror and either passes through the focal point, or can be extended back to pass through the focal point.

The second ray is the chief ray. This is drawn from the tip of the object to the mirror through the center of curvature. This ray will hit the mirror at a 90° angle, reflecting back the way it came. The chief and parallel rays meet at the tip of the image.

The third ray, the focal ray, is a mirror image of the parallel ray. The focal ray is drawn from the tip of the object through (or towards) the focal point, reflecting off the mirror parallel to the principal axis. All three rays should meet at the same point.

A ray diagram for a concave mirror is shown above. This shows a few different things. For this object, located beyond the center of curvature from the mirror, the image lies between the focal point (F) and the center of curvature. The image is inverted compared to the object, and it is also a real image, because the light rays actually pass through the point where the image is located.

With a concave mirror, any object beyond C will always have an image that is real, inverted compared to the object, and between F and C. You can always trade the object and image places (that just reverses all the arrows on the ray diagram), so any object placed between F and C will have an image that is real, inverted, and beyond C. What happens when the object is between F and the mirror? You should draw the ray diagram to convince yourself that the image will be behind the mirror, making it a virtual image, and it will be upright compared to the object.

6.8 The Mirror Equation and the Magnification equation: Concave Mirror: Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a concave mirror. The use of these diagrams was demonstrated earlier . Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and object size. To obtain this type of numerical information, it is necessary to use the Mirror Equation and the Magnification Equation. The mirror equation expresses the quantitative relationship between the object distance (d o), the image distance (di), and the focal length (f). The equation is stated as follows:

The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (h i) and object height (ho). The

magnification equation is stated as follows:

These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known.

As a demonstration of the effectiveness of the mirror equation and magnification equation, consider the following example problem and its solution.

Example 1 A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size.

Like all problems in physics, begin by the identification of the known information(collection of datas). ho = 4.0 cm

do = 45.7 cm

f = 15.2 cm

Next identify the unknown quantities that you wish to solve for. di = ???

hi = ???

To determine the image distance, the mirror equation must be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

1/f = 1/do + 1/di 1/(15.2 cm) = 1/(45.7 cm) + 1/di 0.0658 cm-1 = 0.0219 cm-1 + 1/di 0.0439 cm-1 = 1/di di = 22.8 cm The numerical values in the solution above were rounded when written down, yet un-rounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.

hi/ho = - di/do hi /(4.0 cm) = - (22.8 cm)/(45.7 cm) hi = - (4.0 cm) • (22.8 cm)/(45.7 cm) hi = -1.99 cm The negative values for image height indicate that the image is an inverted image. As is often the case in physics, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image height, a negative value always indicates an inverted image.

From the calculations in this problem it can be concluded that if a 4.00-cm tall object is placed 45.7 cm from a concave mirror having a focal length of 15.2 cm, then the image will be inverted, 1.99-cm tall and located 22.8 cm from the mirror. The results of this calculation agree with the principles discussed earlier in this lesson. In this case, the object is located beyond the center of curvature (which would be two focal lengths from the mirror), and the image is located between the center of curvature and the focal point.

Example 2 A 4.0-cm tall light bulb is placed a distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. (NOTE: this is the same object and the same mirror, only this time the object is placed closer to the mirror.) Determine the image distance and the image size.

Again, begin by the identification of the known information. ho = 4.0 cm

do = 8.3 cm

f = 15.2 cm

Next identify the unknown quantities that you wish to solve for. di = ???

hi = ???

To determine the image distance, the mirror equation will have to be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

1/f = 1/do + 1/di 1/(15.2 cm) = 1/(8.3 cm) + 1/di

0.0658 cm-1 = 0.120 cm-1 + 1/di -0.0547 cm-1 = 1/di di = -18.3 cm The numerical values in the solution above were rounded when written down, yet un-rounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below. hi/ho = - di/do hi /(4.0 cm) = - (-18.2 cm)/(8.3 cm) hi = - (4.0 cm) • (-18.2 cm)/(8.3 cm) hi = 8.8 cm The negative value for image distance indicates that the image is a virtual image located behind the mirror. Again, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image distance, a negative value always means behind the mirror. Note also that the image height is a positive value, meaning an upright image. Any image that is upright and located behind the mirror is considered to be a virtual image.

From the calculations in the second example problem it can be concluded that if a 4.0-cm tall object is placed 8.3 cm from a concave mirror having a focal length of 15.2 cm, then the image will be magnified, upright, 8.8-cm tall and located 18.3 cm behind the mirror. The results of this calculation agree with the principles discussed earlier in this lesson. In this case, the object is located in front of the focal point (i.e., the object distance is less than the focal length), and the image is located behind the mirror.

The +/- Sign Conventions The sign conventions for the given quantities in the mirror equation and magnification equations are as follows: • • • • • •

f is + if the mirror is a concave mirror f is - if the mirror is a convex mirror di is + if the image is a real image and located on the object's side of the mirror. di is - if the image is a virtual image and located behind the mirror. hi is + if the image is an upright image (and therefore, also virtual) hi is - if the image an inverted image (and therefore, also real)

Convex Mirror:

Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a mirror. The use of these diagrams was demonstrated earlier in this lesson. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and image size. To obtain this type of numerical information, it is necessary to use the Mirror Equation and the Magnification Equation. The mirror equation expresses the quantitative relationship between the object distance (d o), the image distance (di), and the focal length (f). The equation is stated as follows:

The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (h i) and object height (ho). The magnification equation is stated as follows:

These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known. Their use was demonstrated in this lesson for concave mirrors and will be demonstrated here for convex mirrors. As a demonstration of the effectiveness of the Mirror equation and Magnification equation, consider the following example problem and its solution.

Example 1 A 4.0-cm tall light bulb is placed a distance of 35.5 cm from a convex mirror having a focal length of -12.2 cm. Determine the image distance and the image size.

Like all problems in physics, begin by the identification of the known information.

ho = 4.0 cm

do = 35.5 cm

f = -12.2 cm

Next identify the unknown quantities that you wish to solve for. di = ???

hi = ???

To determine the image distance (di), the mirror equation will have to be used.

The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

1/f = 1/do + 1/di 1/(-12.2 cm) = 1/(35.5 cm) + 1/di -0.0820 cm-1 = 0.0282 cm-1 + 1/di -0.110 cm-1 = 1/di di = -9.08 cm

The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

To determine the image height (hi), the magnification equation is needed.

Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below. hi/ho = - di/do hi /(4.0 cm) = - (-9.08 cm)/(35.5 cm) hi = - (4.0 cm) • (-9.08 cm)/(35.5 cm) hi = 1.02 cm The negative values for image distance indicate that the image is located behind the mirror. As is often the case in physics, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image distance, a negative value always indicates the existence of a virtual image located behind the mirror. In the case of the image height, a positive value indicates an upright image. Further information about the sign conventions for the variables in the Mirror Equation and the Magnification Equation can be found in this lesson.

From the calculations in this problem it can be concluded that if a 4.0-cm tall object is placed 35.5 cm from a convex mirror having a focal length of -12.2 cm, then the image will be upright, 1.02-cm tall and located 9.08 cm behind the mirror. The results of this calculation agree with the principles discussed earlier in this lesson. Convex mirrors always produce images that are upright, virtual, reduced in size, and located behind the mirror.

 1  do

Proof of the equation : 

 1 +   di

 1 = .  f

From the ray that reflects from the center of the mirror,we have: tan  =

m=

tan  =

ho hi = ; d o di

ho di = . hi do

ho hi = . ( do + r ) ( r − di )

When we divide the two equations, we get

( do

+ r)

do

=

( r − di ) ; di

 r  r  1 +   =   − 1, or  d o   di   r  r  −  d o   di

  = − 2; 

 1   1  − 2 −1 r = , with f = .   − = 2 f  d o   di  r

From the ray diagram, we see that di  0.

If we consider f to be negative, we have

 1  1 1   + = .  d o   di  f

The little note below should be memorized by the student M positive

Upright image (Virtual image)

M negative Up side down image (Real image) di positive

Image is Real

di negative Image is Virtual IMI<1

Image smaller than object

IMI>1

Image bigger than object

IMI=1

Image same size as object

TUTORIAL: 1. A convex mirror has a focal length of -10.8 cm. An object is placed 32.7 cm from the mirror's surface. Determine the image distance.( -8.1 cm) 2. Determine the focal length of a convex mirror that produces an image that is 16.0 cm behind the mirror when the object is 28.5 cm from the mirror.( -36.6 cm) 3. A 2.80-cm diameter coin is placed a distance of 25.0 cm from a convex mirror that has a focal length of -12.0 cm. Determine the image distance and the diameter of the image.( di = -8.1 cm and hi = 0.909 cm)

4. A focal point is located 20.0 cm from a convex mirror. An object is placed 12 cm from the mirror. Determine the image distance.( di = -7.5 cm) 5.Determine the image distance and image height for a 5.00-cm tall object placed 45.0 cm from a concave mirror having a focal length of 15.0 cm.( di = 22.5 cm and hi = -2.5 cm)

6. Determine the image distance and image height for a 5.00-cm tall object placed 30.0 cm from a concave mirror having a focal length of 15.0 cm.( di = 30.0 cm and hi = -5.0 cm)

7. Determine the image distance and image height for a 5.00-cm tall object placed 20.0 cm from a concave mirror having a focal length of 15.0 cm.( di = 60.0 cm and hi = -15.0 cm)

8. Determine the image distance and image height for a 5.00-cm tall object placed 10.0 cm from a concave mirror having a focal length of 15.0 cm.( di = -30.0 cm and hi = +15.0 cm) 9. Is it possible to use a convex mirror to produce an image that is larger than the object? Provide a reason for your answer.

10. A clown is using a concave mirror to get ready for a show and is 27 cm from the mirror. The image of his face is 65 cm behind the mirror. Calculate (a) the focal length, and (b) the magnification of the mirror. (46 cm, 2.4)

11. A convex mirror in an amusement park has a radius of curvature of 3.00 m. A man stands in front of the mirror so that his image is half as tall as his actual height.

At what distance must the man focus his eyes in order to see his image? (2.25 m)

12. A spherical concave mirror has a radius of curvature of 6.0 cm. At what distance from the mirror should a 6.0-cm object be placed to obtain an image that is 48 cm tall? (3.375 cm or 2.625 cm from the mirror)

13. An object is 1.0 m in front of a mirror. A virtual image is formed 10.0 m behind the mirror. What is the radius of curvature of the mirror? (2.2 m)

14. A 0.127m pencil is oriented perpendicular to the principal axis of a concave spherical mirror that has a radius of curvature of 0.300 m. What are the image distance and the image height if the pencil is 0.250 m from the mirror? (0.375 m, 0.191 m) 15. A Star Wars action figure, 8.0 cm tall, is placed 30.0 cm in front of a concave mirror with a focal length of 10.0 cm. Where is the image? How tall is the image?

What are the characteristics of the image?

(15 cm; -4.0 cm; real, inverted, 4.0 cm high, and 15.0 cm in front of the mirror) 16.The same Star Wars action figure, 8.0 cm tall, is placed 6.0 cm in front of a convex mirror with a focal length of -12.0 cm. Where is the image in this case, and

what are the image characteristics?( -4.0 cm; 5.3 cm; virtual, upright compared to the object, and 4.0 cm behind the mirror) 17.An object is 15 cm in front of a converging lens with a focal length of 10 cm. Use ray tracing to determine the location of the image. What are the image characteristics: (i) real or virtual?, (ii) upright or inverted?, (iii) smaller, larger or same?, and (iv) the image location?

18.A 2.0-cm tall object is 15 cm in front of a converging lens that has a 20 cm focal length. Where is the image located and what is the height of the image? What are the characteristics of the image (real or virtual, upright or inverted, enlarged or reduced)?

19.A light bulb is 60 cm from a concave mirror with a focal length of 40 cm. A 5-cmlong mascara brush is held upright 20 cm from the mirror. Use ray tracing to determine the location of the image. What are the image characteristics: (i) real or virtual?, (ii) upright or inverted?, (iii) smaller, larger or same?, and (iv) the image location?

20. If the focal length of a concave mirror is 60 cm, what is the radius of curvature?

(1,2m) 21. If an object is placed 50 cm in front of a concave mirror of 60 cm radius, where does the image form?

22. Given a spherical mirror whose radius of curvature is +20 cm. What is the focal length of this mirror? Does it form a real or virtual image?( 0.44 m) 23. A concave mirror of radius 60 cm is placed so that a luminous object is 20 cm in front of the mirror. Where does the image form?

24. A man 2.2 m tall stands 10 m in front of a convex mirror which has a radius of curvature of 5 m. How tall is the image?( 0.44 m) 25. What is the radius of curvature of a convex mirror which forms an image one fourth of the size of an object, when the object is placed 6 m in front of the mirror?

26. Where must an object be placed in front of a concave mirror of radius R in order for the image to be superimposed upon the object? Is this image real or virtual?( real) 27. An ornamental silvered ball 6 cm in diameter forms an image of an object 2 m in front of the ball. Locate the image.

28. A 2.2 m tall man is standing 5 m in front of a plane mirror. Locate the image.(5m) 29. A man looks into a convex spherical mirror of radius 30 cm. If his face is 10 cm from the vertex of the mirror, where is his image located? What is the magnification?

30. The radius of curvature of a convex mirror is 30 cm, what is its focal length?( . 15 cm) 31. A 3 cm object is placed in front of a concave mirror and the image formed is 9 cm in height. What is the magnification of this object? In general terms where was the object placed (beyond C, at C, between C and F, or between F and the mirror)?

32. A convex mirror has a radius of curvature of 60 cm. If a 2.2 m tall object is placed 4 m in front of the mirror. Where is the image located and how tall is it? 14. An image is formed 20 cm from a convex mirror of radius 30 cm. Where was the object placed? ( -0.279 m, 0.152 m) 33. An object 0.1 m tall is placed 0.4 m from a convex mirror with a focal length of 0.3 m. What is the height of the image?( 0.0428 m)

34. An object 0.5 m tall is placed 0.9 m from a convex mirror. An image is formed 0.7 m from the mirror. What is the focal length of the mirror?

35. An object is placed 37.5 cm in front of a concave mirror with a radius of curvature of 75 cm, where is the image formed?( infinity (no image is formed)) 36. An object is located 45 cm in front of a convex mirror which has a radius of curvature of 20 cm. What is the magnification?

37. An object is located 30 cm in front of a concave mirror which has a focal length of 40 cm. What is the magnification? ( 4) 38. An object is placed 1.2 m from a concave mirror with a radius of curvature of 60 cm. What is the image distance?

39. A concave mirror is designed to have a magnification of 4 when an object is placed 60 cm in front of it. What is the radius of curvature of the mirror?( 160 cm) 40. A woman looks at herself in a magnifying converging mirror whose focal length is 20 cm. If her face is 10 cm from the mirror, where is her image located? What is the magnification?

41. A child looks at his reflection in a spherical Christmas tree ornament 8 cm in diameter, and sees that the image of his face is reduced by one-half. How far is his face from the ornament?( 2 cm) 42. A converging mirror has a focal length of 15 cm. Where would you place an object in order to produce an upright, virtual image twice as tall as the object?

43. A 2 cm high candle is placed 15 cm in front of a converging mirror with a focal length of 30 cm. How far â&#x20AC;&#x153;behindâ&#x20AC;? the mirror does the candle appear, and how large is it? (. -30 cm, 4 cm)

UNIT 7:The Reflection Refraction of Light: Lens and optical instruments 7.1 Law of Reflection: The law of reflection governs the reflection of light-rays off smooth conducting surfaces, such as polished metal or metal-coated glass mirrors.

Consider a light-ray incident on a plane mirror.Snell’s first law or law of reflection states that the incident ray, the reflected ray, and the normal to the surface of the mirror all lie in the same plane. Furthermore, the angle of reflection is equal to the angle of incidence . Both angles are measured with respect to the normal to the mirror.

Figure 17: The law of reflection The law of reflection also holds for non-plane mirrors, provided that the normal at any point on the mirror is understood to be the outward pointing normal to the local tangent plane of the mirror at that point. For rough surfaces, the law of reflection remains valid. It predicts that rays incident at slightly different points on the surface are reflected in completely different directions, because the normal to a rough surface varies in direction very strongly from point to point on the surface. This type of reflection is called diffuse reflection, and is what enables us to see non-shiny objects.

Important Terminology ● The normal is an imaginary line perpendicular (at 90 o ) to the surface. All angles are measured from the normal. ● The incident ray is the original light ray ● The reflected ray is the ray that is reflected off the surface. ● i is the angle of incidence. It is the angle between the normal the incident

light ray ● r is the angle of reflection. It is the angle between the normal the reflected light ray

Refraction Refraction is the bending of light as it passes from one medium into another.

7.2 Law of Refraction The law of refraction, which is generally known as Snell's law, governs the behaviour of light-rays as they propagate across a sharp interface between two transparent dielectric media.

Figure 18: The law of refraction. .Snell’s second law or law of refraction : n1 sin θ 1 = n2 sin θ2

Here θ1 is the angle of incidence between the light ray and the normal in the first medium. Similarly, the light ray makes an angle of refraction θ2 with respect to the normal in the second medium. The first medium has an index of refraction of n1, and the index of refraction in the second medium is n2.

The picture we have drawn above is not completely accurate. Not all of the light

striking a surface travels across the surface; some is reflected.

You have probably seen this happen when you looked out a window. Often you can see reflections of objects inside the room at the same time as someone outside the window could see the objects. The light you see is reflected; the light the person outside sees has been refracted through the window. In this module, we are primarily concerned with the light that is transmitted and refracted, but we must remember that some light is reflected as well.

7.3 Total internal reflection: When light travels from one medium to another it changes speed and is refracted. If the light rays are travelling for a less dense material to a dense medium they are refracted towards the normal and if they are travelling from a dense to less dense medium they are refracted away from the normal.

For total internal reflection to occur the light must travel from a dense medium to a less dense medium (e.g. glass to air or water to air).

As the angle of incidence increases so does the angle of refraction. When the angle of incidence reaches a value known as the critical angle the refracted rays travel along the surface of the medium or in other words are refracted to an angle of 90°. The critical angle for the angle of incidence in glass is 42°.

7.4 The Dispersion of the light Prisms and rainbows:

Everyone enjoys the spectacle of a rainbow glimmering against a dark stormy sky.

How does sunlight falling on clear drops of rain get broken into the rainbow of colors we see? The same process causes white light to be broken into colors by a clear

glass prism or a diamond.

We see about six colors in a rainbowâ&#x20AC;&#x201D;red, orange, yellow, green, blue, and violet; sometimes indigo is listed, too. Those colors are associated with different wavelengths of light, as shown in the Figure 20 below. When our eye receives purewavelength light, we tend to see only one of the six colors, depending on wavelength.

The thousands of other hues we can sense in other situations are our eyeâ&#x20AC;&#x2122;s response to various mixtures of wavelengths. White light, in particular, is a fairly uniform mixture of all visible wavelengths. Sunlight, considered to be white, actually appears to be a bit yellow because of its mixture of wavelengths, but it does contain all visible wavelengths. The sequence of colors in rainbows is the same sequence as the colors plotted versus wavelength . What this implies is that white light is spread out according to wavelength in a rainbow. Dispersion is defined as the spreading of white light into its full spectrum of wavelengths. More technically, dispersion occurs whenever there is a process that changes the direction of light in a manner that depends on wavelength. Dispersion, as a general phenomenon, can occur for any type of wave and always involves wavelength-dependent processes.

Dispersion is defined to be the spreading of white light into its full spectrum of wavelengths.

Figure 19. Even though rainbows are associated with seven colors, the rainbow is a continuous distribution of colors according to wavelengths.

Refraction is responsible for dispersion in rainbows and many other situations.

The angle of refraction depends on the index of refraction, as we saw in The Law of Refraction. We know that the index of refraction n depends on the medium. But for a given medium, n also depends on wavelength. (See Table 1. Note that, for a given medium, n increases as wavelength decreases and is greatest for violet light. Thus violet light is bent more than red light, as shown for a prism in Figure 20b, and the light is dispersed into the same sequence of wavelengths as seen in Figure 17 and Figure 18.

Table 1. Index of Refraction n in Selected Media at Various Wavelengths

Medium

Red (660 nm)

Orange (610 nm)

Yellow (580 nm)

Green (550 nm)

Blue (470 nm)

Violet (410 nm)

Water

1.331

1.332

1.333

1.335

1.338

1.342

Diamond

2.410

2.415

2.417

2.426

2.444

2.458

Glass, crown

1.512

1.514

1.518

1.519

1.524

1.530

Glass, flint

1.662

1.665

1.667

1.674

1.684

1.698

Polystyrene

1.488

1.490

1.492

1.493

1.499

1.506

Quartz, fused

1.455

1.456

1.458

1.459

1.462

1.468

Figure 20. (a) A pure wavelength of light falls onto a prism and is refracted at both surfaces. (b) White light is dispersed by the prism (shown exaggerated). Since the index of refraction varies with wavelength, the angles of refraction vary with wavelength. A sequence of red to violet is produced, because the index of refraction increases steadily with decreasing wavelength.

Rainbows: Figure21

Rainbows are produced by a combination of refraction and reflection. You may have noticed that you see a rainbow only when you look away from the sun. Light enters a drop of water and is reflected from the back of the drop, as shown in Figure 21. The light is refracted both as it enters and as it leaves the drop. Since the index of refraction of water varies with wavelength, the light is dispersed, and a rainbow is

observed, as shown in Figure 20a. (There is no dispersion caused by reflection at the back surface, since the law of reflection does not depend on wavelength.) The actual rainbow of colors seen by an observer depends on the myriad of rays being refracted and reflected toward the observer’s eyes from numerous drops of water. The effect is most spectacular when the background is dark, as in stormy weather, but can also be observed in waterfalls and lawn sprinklers. The arc of a rainbow comes from the need to be looking at a specific angle relative to the direction of the sun, as illustrated in Figure 21b. (If there are two reflections of light within the water drop, another “secondary” rainbow is produced. This rare event produces an arc that lies above the primary rainbow arc) Real and Apparent Depth: d'=d(n2/n1) ; d’ = Apparent depth (m) d = Real depth (m) n1 = Refractive index of the material in which the object is. n2 = Refractive index of the material in which the observer is Proof:

Consider a ray of light incident normally along OA. It passes straight along OAA'. Consider another ray from O (the object) incident at an angle i along OB. This ray gets refracted and passes along BC. On producing this ray BC backwards, it appears to come from the point I, and hence, AI represents the apparent depth, which is less than the real depth AO. Since AO and BNI are parallel and OB is the transversal, ∠ AOB = ∠OBNI (Alternate angles) ∠BIAI = ∠CBN (Corresponding angles) In ΔBAO, Sin i = BA/OB In ΔIAB, Sin r = BA/IB We know that refractive index of air w.r.t the medium

Since the point B is very close to point A, i.e the object is viewed from a point vertically above the object. ∴ IB = IA and OB = OA. Hence, aμm = OA/IA = Real depth/Apparent depth Example1:

a Îź g = đ?&#x2018;&#x2026;đ?&#x2018;&#x2019;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122; đ??ˇđ?&#x2018;&#x2019;đ?&#x2018;?đ?&#x2018;Ą â&#x201E;&#x17D; /đ??´đ?&#x2018;?đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ą đ?&#x2018;&#x2018;đ?&#x2018;&#x2019;đ?&#x2018;?đ?&#x2018;Ą â&#x201E;&#x17D;â&#x20AC;&#x2122; or 1.54 = 3 đ??´đ?&#x2018;?đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ą đ?&#x2018;&#x2018;đ?&#x2018;&#x2019;đ?&#x2018;?đ?&#x2018;Ą â&#x201E;&#x17D; Apparent depth = 3/ 1.54 = 1.94 cm Height through which image is raised = 3 â&#x20AC;&#x201C; 1.94 = 1.06 cm Example 2:Light travels from air into an optical fiber with an index of refraction of 1.44. (a) In which direction does the light bend? (b) If the angle of incidence on the end of the fiber is 22o, what is the angle of refraction inside the fiber? (c) Sketch the path of light as it changes media.

Solution:

(a) Since the light is traveling from a rarer region (lower n) to a denser region (higher n), it will bend toward the normal. (b) We will identify air as medium 1 and the fiber as medium 2. Thus, n1 = 1.00, n2 = 1.44, and θ/font>1 = 22o. Snell's Law then becomes (1.00) sin 22o = 1.44 sin θ2. sin θ2 = (1.00/1.44) sin 22o = 0.260 θ2 = sin-1 (0.260) = 15o. (c) The path of the light is shown in the figure below.

TUTORIAL : 1. Light traveling through an optical fiber (n=1.44) reaches the end of the fiber and exits into air. (a) If the angle of incidence on the end of the fiber is 30 o, what is the angle of refraction outside the fiber? (b) How would your answer be different if the angle of incidence were 50 o?( 46o, 1.103 This equality cannot be met, so light cannot exit the fiber under these conditions.

The situation in part (b) is an example of total internal reflection) 2.You are approached by a man on the street who offers to sell you a rather impressive gen-u-ine diamond. He produces a certificate of authenticity while telling you of the hardships forcing him to part with his wifes engagement stone (a family heirloom). Being skeptical, you decide to verify the composition of the stone with the laser pointer attached to your key ring. (a) What is the critical angle for a diamond/air interface? (24.6o)For a glass/air interface?( 41.8o )(b) How could you use this information to determine the composition of the stone?( We can observe the path of laser light as it travels through the stone. By rotating the stone until TIR is just reached for the light exiting the stone, we can observe the critical angle. If the critical angle for light incident on the back face of the stone is fairly close to 45 o (halfway between face and normal to face so "easy" to reference), the stone is glass. If, on the other hand, the critical angle for the light incident on the back face is significantly closer to the normal than halfway, the stone might actually be a diamond. Of course observing TIR on the back face of a cut stone might be difficult unless the stone is very large and/or has few facets.)

3. A step-index fiber has a core index of refraction of n1 = 1.425. The cut-off angle for light entering the fiber from air is found to be 8.50 o. (a) What is the numerical aperture of the fiber? (b) What is the index of refraction of the cladding of this fiber? (c) If the fiber were submersed in water, what would be the new numerical aperture and cut-off angle?( 0.148; 1.417; 6.38o ) Hint: Light entering a fiber at angles less than the cut-off angle θ0max will be trapped in the fiber. This cut-off angle is related to the numerical aperture, NA, which is a property of the fiber: n0 sin θ0max = (n12 - n22)1/2 = NA. 4) When you look through an aquarium window at a fish, is the fish as close as it appears? Explain, and use a diagram to assist your explanation.

5) A point source of light is submerged 2.2 m below the surface of a lake and emits rays in all directions. On the surface of the lake, directly above the source, the area illuminated is a circle. Calculate the maximum radius that this circle could have. (2.5 m) 6) A 6.0 cm object is placed 30.0 cm from a lens. The resulting image height has a magnitude of 2.0 cm, and the image is inverted. What is the focal length of the lens?

(7.5 cm) 7) A ray of light impinges from air onto a block of ice (n = 1.309) at a 60.0° angle of incidence. Assuming that this angle remains the same, find the difference θ2,ice – θ2,water in the angles of refraction when the ice turns to water (n = 1.333). (0.9 degree)

8) A light ray travels from air (n = 1.0) into water (n = 1.33). The angle of incidence is 34°. What is the angle of refraction?

9). A light ray travels from water into air. If the angle of refraction is 56°, what is the angle of incidence?

10) What is the index of refraction of a medium if the angle of incidence in air is 40o and the angle of refraction is 29 deg?

11). What is the angle of incidence for a light ray traveling from water into flint glass (n=1.70), if the angle of refraction is 30 deg ?

12). A ray of light in air strikes a block of quartz at an angle of incidence of 30°. The angle of refraction is 20°. What is the index of refraction of the quartz?

13) . A light wave traveling in air passes into the water in a swimming pool at an angle of incidence of 35˚. Calculate the angle of refraction of the light in water.

14). Light entering a block of glass at an angle of incidence of 18.5˚ leaves the boundary between the air and the glass at an angle of 12.0˚. What is the index of refraction of this type of glass?

15). Light is incident on diamond at an angle of 10.0˚. At what angle will it refract? 4.

A transparent material has a refractive index of 1.27. What is the angle of incidence in air when the angle of refraction in the substance is 43˚?

16). What is the index of refraction of a material if the angle of incidence in air is 50˚ and the angle of refraction in the material is 40˚?

17). A ray of light passes from water into carbon disulphide ( 1.63 r n = ) with an angle of incidence of 30˚. What is the angle of refraction in the carbon disulphide?

18. Green light traveling in air has an angle of incidence of 50˚ as it passes into a certain glass. The refracted angle in the glass is 33˚. What is the index of refraction for this type of glass?

19.Light travels from crown glass (n=1.52) into air (n=1.00). The angle of refraction in air is 60o. What is the angle of incidence in glass?

20.Light travels from crown glass (n=1.52) into water (n=1.33). The angle incidence in crown glass is 40o. What is the angle of refraction in water?

21.If the index of refraction for diamond is 2.42, what will be the angle of refraction in diamond for an angle of incidence, in water, of 60 o?

22.A ray of light in air (n=1.00) strikes a block of quartz at an angle of incidence of 30o. The angle of refraction of refraction is 20o. What is the index of refraction of the

quartz?

23.Light travels from air (n=1.00) into water (n=1.33). If the angle of refraction is 30 o, what is the angle of incidence?

24.A ray of light passes from air (n=1.00) into water (n=1.33) at an angle of incidence of 50o. What is the angle of refraction?

25.One ray of light in air (n=1.00) strikes a diamond (n=2.42) and another strikes a piece of fused quartz (n=1.46), in each case at an angle of incidence of 40 o. What is the difference between the angles of refraction?

26. . What is the angle of incidence on when light travels from air to glass if the angle of refraction in the glass (n = 1.52) is 25 degrees?

â&#x20AC;˘

16 degrees

â&#x20AC;˘

30 degrees

•

40 degrees

•

45 degrees

27. Which of the following statements is INCORRECT about the following illustration?

•

Medium 1 is denser than medium 2

•

Light travels faster through medium 1 than through medium 2

•

Light travels slower through medium 2 than through medium 1

•

Light bends towards the normal when it goes through medium 2

UNIT 8: VECTORS AND SCALARS •

Scalar Quantities: The physical quantities which are specified with the magnitude or size alone are scalar quantities. For example, length, speed, work, mass, density, etc.

•

Vector Quantities: Vector quantities refer to the physical quantities characterized by the presence of both magnitude as well as direction. For example, displacement, force, torque, momentum, acceleration, velocity, etc.

Characteristics of Vectors The characteristics of vectors are as followed •

They possess both magnitudes as well as direction.

•

They do not obey the ordinary laws of Algebra.

•

These change if either the magnitude or direction change or both change.

Equal Vectors Vectors A and B are equal if | A | = | B | as well as their directions, are same.

Zero Vectors Zero vector is a vector with zero magnitudes and an arbitrary direction is a zero vector. It can be represented by O and is a Null Vector.

VECTORS ADDITION AND SUBTRACTION: A variety of mathematical operations can be performed with and upon vectors.

One such operation is the addition of vectors. Two vectors can be added together to determine the result (or resultant). This process of adding two or more vectors has already been discussed in an earlier unit. Recall in our discussion of Newton's laws of motion, that the net force experienced by an object was determined by computing the vector sum of all the individual forces acting upon that object. That is the net force was the result (or resultant) of adding up all the force vectors. During that unit, the rules for summing vectors (such as force vectors) were kept relatively simple. Observe the following summations of two force vectors:

These rules for summing vectors were applied to free-body diagrams in order to determine the net force (i.e., the vector sum of all the individual forces).

Sample applications are shown in the diagram below.

In this unit, the task of summing vectors will be extended to more complicated cases in which the vectors are directed in directions other than purely vertical and horizontal directions. For example, a vector directed up and to the right will be added to a vector directed up and to the left. The vector sum will be determined for the more complicated cases shown in the diagrams below.

There are a variety of methods for determining the magnitude and direction of the result of adding two or more vectors. The two methods that will be discussed in this lesson and used throughout the entire unit are: â&#x20AC;˘ â&#x20AC;˘ â&#x20AC;˘

the Pythagorean theorem and trigonometric methods the head-to-tail method using a scaled vector diagram analytical methods of vector addition and subtraction employ geometry

The Pythagorean Theorem The Pythagorean theorem is a useful method for determining the result of adding two (and only two) vectors that make a right angle to each other. The method is not applicable for adding more than two vectors or for adding vectors that are not at 90-degrees to each other. The Pythagorean theorem is a mathematical equation that relates the length of the sides of a right triangle to the length of the hypotenuse of a right triangle.

To see how the method works, consider the following problem: ZOE NGAMGOUM leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine ZOE's resulting displacement.

This problem asks to determine the result of adding two displacement vectors that are at right angles to each other. The result (or resultant) of walking 11 km north and 11 km east is a vector directed northeast as shown in the diagram to the right. Since the northward displacement and the eastward displacement are at right angles to each other, the Pythagorean theorem can be used to determine the resultant (i.e., the hypotenuse of the right triangle).

The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15.6 km. Later, the method of determining the direction of the vector will be discussed.

Using Trigonometry to Determine a Vector's Direction The direction of a resultant vector can often be determined by use of trigonometric functions. Most students recall the meaning of the useful mnemonic SOH CAH TOA from their course in trigonometry. SOH CAH TOA is a mnemonic that helps one remember the meaning of the three common trigonometric functions - sine, cosine, and tangent functions. These three functions relate an acute angle in a right triangle to the ratio of the lengths of two of the sides of the right triangle. The sine function relates the measure of an acute angle to the ratio of the length of the side opposite the angle to the length of the hypotenuse. The cosine function relates the measure of an acute angle to the ratio of the length of the side adjacent the angle to

the length of the hypotenuse. The tangent function relates the measure of an angle to the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. The three equations below summarize these three functions in equation form.

These three trigonometric functions can be applied to the hiker problem in order to determine the direction of the hiker's overall displacement. The process begins by the selection of one of the two angles (other than the right angle) of the triangle. Once the angle is selected, any of the three functions can be used to find the measure of the angle. Write the function and proceed with the proper algebraic steps to solve for the measure of the angle. The work is shown below.

Once the measure of the angle is determined, the direction of the vector can be found.

The Calculated Angle is Not Always the Direction The measure of an angle as determined through use of SOH CAH TOA is not always the direction of the vector. The following vector addition diagram is an example of

such a situation. Observe that the angle within the triangle is determined to be 26.6 degrees using SOH CAH TOA. This angle is the southward angle of rotation that the vector R makes with respect to West. Yet the direction of the vector as expressed with the CCW (counterclockwise from East) convention is 206.6 degrees.

Use of Scaled Vector Diagrams to Determine a Resultant The magnitude and direction of the sum of two or more vectors can also be determined by use of an accurately drawn scaled vector diagram. Using a scaled diagram, the head-to-tail method is employed to determine the vector sum or resultant. A common Physics lab involves a vector walk. Either using centimetersized displacements upon a map or meter-sized displacements in a large open area, a student makes several consecutive displacements beginning from a designated starting position. Suppose that you were given a map of your local area and a set of 18 directions to follow. Starting at home base, these 18 displacement vectors could be added together in consecutive fashion to determine the result of adding the set of 18 directions. Perhaps the first vector is measured 5 cm, East. Where this measurement ended, the next measurement would begin. The process would be repeated for all 18 directions. Each time one measurement ended, the next measurement would begin. In essence, you would be using the head-to-tail method of vector addition.

The head-to-tail method involves drawing a vector to scale on a sheet of paper beginning at a designated starting position. Where the head of this first vector ends, the tail of the second vector begins (thus, head-to-tail method). The process is repeated for all vectors that are being added. Once all the vectors have been added head-to-tail, the resultant is then drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish. Once the resultant is drawn, its length can be measured and converted to real units using the given scale. The direction of the resultant can be determined by using a protractor and measuring its counterclockwise angle of rotation from due East.

A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below.

1.

2.

3.

4.

5.

6.

7.

Choose a scale and indicate it on a sheet of paper. The best choice of scale is one that will result in a diagram that is as large as possible, yet fits on the sheet of paper.

Pick a starting location and draw the first vector to scale in the indicated direction. Label the magnitude and direction of the scale on the diagram (e.g., SCALE: 1 cm = 20 m).

Starting from where the head of the first vector ends, draw the second vector to scale in the indicated direction. Label the magnitude and direction of this vector on the diagram.

Repeat steps 2 and 3 for all vectors that are to be added Draw the resultant from the tail of the first vector to the head of the last vector.

Label this vector as Resultant or simply R. Using a ruler, measure the length of the resultant and determine its magnitude by converting to real units using the scale (4.4 cm x 20 m/1 cm = 88 m).

Measure the direction of the resultant using the counterclockwise convention.

An example of the use of the head-to-tail method is illustrated below. The problem involves the addition of three vectors:

20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg.

SCALE: 1 cm = 5 m

The head-to-tail method is employed as described above and the resultant is determined (drawn in red). Its magnitude and direction is labeled on the diagram.

SCALE: 1 cm = 5 m

Interestingly enough, the order in which three vectors are added has no effect upon either the magnitude or the direction of the resultant. The resultant will still have the same magnitude and direction. For example, consider the addition of the same three vectors in a different order.

15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg.

SCALE: 1 cm = 5 m

When added together in this different order, these same three vectors still produce a resultant with the same magnitude and direction as before (20. m, 312 degrees). The order in which vectors are added using the head-to-tail method is insignificant.

SCALE: 1 cm = 5 m

Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods.

Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made. Analytical methods are limited only by the accuracy and precision with which physical quantities are known.

Resolving a Vector into Perpendicular Components:

Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are independent.

We very often need to separate a vector into perpendicular components.

For example, given a vector like A in the Figure below, we may wish to find which two perpendicular vectors, Ax and Ay, add to produce it.

Figure :The vector A, with its tail at the origin of an x, y-coordinate system, is shown together with its x- and y-components, Ax and Ay. These vectors form a right triangle. The analytical relationships among these vectors are summarized below.

Ax and Ay are defined to be the components of A along the x- and y-axes. The three vectors A, Ax and Ay form a right triangle: Ax+Ay=A Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include both magnitude and direction).

The relationship does not apply for the magnitudes alone. For example, if Ax=3m east, Ay=4m north, and A=5m north-east, then it is true that the vectors Ax+Ay=A. However, it is not true that the sum of the magnitudes of the vectors is also equal. That is, 3m+4mâ&#x2030; 5m Thus,

Ax+Ay≠A If the vector AA is known, then its magnitude AA (its length) and its angle θθ (its direction) are known. To find Ax and Ay, its x- and y-components, we use the following relationships for a right triangle.

Ax=Acosθ and Ay=Asinθ.

Figure : The magnitudes of the vector components Ax and Ay can be related to the resultant vector A and the angle θ with trigonometric identities. Here we see that Ax=Acosθ and Ay=Asinθ.

Suppose, for example, that A is the vector representing the total displacement of the person walking in a city considered in Kinematics in Two Dimensions: An Introduction and Vector Addition and Subtraction: Graphical Methods.

Figure : We can use the relationships Ax=Acosθ and Ay=Asinθ to determine the magnitude of the horizontal and vertical component vectors in this example.

Then A=10.3 blocks and θ=29.1º , so that Ax=Acosθ=(10.3blocks)(cos29.1º)=9.0 blocks Ay=Asinθ=(10.3blocks)(sin29.1º)=5.0 blocks.

Calculating a Resultant Vector If the perpendicular components Ax and Ay of a vector A are known, then A can also be found analytically. To find the magnitude A and direction θ of a vector from its perpendicular components Ax and Ay, we use the following relationships:

Figure : The magnitude and direction of the resultant vector can be determined once the horizontal and vertical components Ax and Ay have been determined.

To see how to add vectors using perpendicular components, consider the figure below, in which the vectors A and B are added to produce the resultant R.

Vectors A and B are two legs of a walk, and R is the resultant or total displacement.

You can use analytical methods to determine the magnitude and direction of R. If A and B represent two legs of a walk (two displacements), then R is the total displacement. The person taking the walk ends up at the tip of R. There are many ways to arrive at the same point. In particular, the person could have walked first in the x-direction and then in the y-direction. Those paths are the xâ&#x20AC;&#x201C; and y-components of the resultant, Rx and Ry If we know Rx and Ry, we can find R and θ using the equations A=

and θ=tan–1(Ay/Ax). When you use the analytical method of vector addition, you can determine the components or the magnitude and direction of a vector.

Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes.

Use the equations Ax=Acosθ and Ay=Asinθ to find the components. In the figure below, these components are Ax, Ay, Bx, and By. The angles that vectors A and B make with the x-axis are θA and θB, respectively.

To add vectors A and B, first determine the horizontal and vertical components of each vector. These are the dotted vectors Ax, Ay, Bx and By shown in the image.

Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis.

That is, as shown in the figure below, Rx=Ax+Bx and Ry=Ay+By.

The magnitude of the vectors Ax and Bx add to give the magnitude Rx of the resultant vector in the horizontal direction. Similarly, the magnitudes of the vectors Ay and By add to give the magnitude Ry of the resultant vector in the vertical direction.

Components along the same axis, say the x-axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the y-axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.)

So resolving vectors into components along common axes makes it easier to add them. Now that the components of R are known, its magnitude and direction can be found.

Step 3. To get the magnitude R of the resultant, use the Pythagorean theorem: R= Step 4. To get the direction of the resultant: θ=tan−1(Ry/Rx).

The following example illustrates this technique for adding vectors using perpendicular components.

Example: Adding Vectors Using Analytical Methods Add the vector A to the vector B shown in the figure below, using perpendicular components along the x– and y-axes. The x– and y-axes are along the east–west and north–south directions, respectively.

Vector A represents the first leg of a walk in which a person walks 53.0 m in a direction 20.0º north of east. Vector B represents the second leg, a displacement of 34.0 m in a direction 63.0º north of east.

Vector A has magnitude 53.0 m and direction 20.0º north of the x-axis. Vector B has magnitude 34.0 m and direction 63.0º north of the x-axis. You can use analytical methods to determine the magnitude and direction of R.

Strategy The components of A and B along the x– and y-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.

Solution Following the method outlined above, we first find the components of A and B along the x– and y-axes. Note that A=53.0m, θA=20.0º, B=34.0m, and θB=63.0º. We find the x-components by using Ax=Acosθ, which gives Ax=A cos θA=(53.0 m)(cos 20.0∘) Ax=(53.0 m)(0.940)=49.8 m and Bx=B cos θB=(34.0 m)(cos 63.0∘) Bx=(34.0 m)(0.454)=15.4 m Similarly, the y-components are found using Ay=AsinθA: Ay=A sin θA=(53.0 m)(sin 20.0∘) Ay=(53.0 m)(0.342)=18.1 m and By=B sin θA=(34.0 m)(sin 63.0∘) By=(34.0 m)(0.891)=30.3m The x– and y-components of the resultant are thus Rx=Ax+Bx=49.8m+15.4m=65.2m and Ry=Ay+By=18.1m+30.3m=48.4m Now we can find the magnitude of the resultant by using the Pythagorean theorem: R=

so that

R=81.2 m.

Finally, we find the direction of the resultant: θ=tan−1(Ry/Rx)=+tan−1(48.4/65.2) Thus, θ=tan−1(0.742)=36.6º.

TUTORIAL :

1) A 2.5-m ladder leans against a wall and makes an angle with the wall of 32° as shown in the figure. What is the height h above the floor where the ladder makes contact with the wall? (2.2 m) 2) A bug crawls 4.25 m along the base of a wall. Upon reaching a corner, the bug's direction of travel changes from south to west. The bug then crawls 3.15 m before stopping. What is the magnitude of the bug's displacement? (5.29 m) 3) A commuter airplane flies from an airport and takes the following route: First it flies to city A located 175 km in a direction 30o north of east. Next it flies 150 km 20o west of north to city B. Finally, it flies 190 km due west to city C. Find the magnitude and direction of the resultant vector R from the airport to city C (W of N 55 .68) 4)Are the following quantities vectors or scalars? Explain. (a) The cost of a theater ticket. (b) The current in a river. (c) The initial flight path from Houston to Dallas. (d) The population of the world.

5). Of the following quantities, which are vectors? If it is, explain why. If it isn’t, explain why. (a) The speed of a car. (b) The velocity of a car. (c) How long it takes to do a physics assignment. (d) The magnitude of the force required to push a pumpkin off an abandoned overpass. (e) The acceleration of said pumpkin.

6) . A plane flies with a velocity of 52 m/s east through a 12 m/s cross wind blowing the plane south. Find the magnitude and direction (relative to due north) of the

resultant velocity at which it travels.

7) Give the x-component and y component of the following vectors: a. A = 7.0 cm, E b. B = 5.7 cm, S c. C = 5.5 cm, 30 degrees E of N d. D = 5.5 cm, 60 degrees S of E Get the resultant of the four vectors, R = A + B + C + D 8) A laser beam is aimed 15.95° above the horizontal at a mirror 11,648 m away. It glances off the mirror and continues for an additional 8,570. m at 11.44° above the horizon until it hits its target. What is the resultant displacement of the beam to the target?

9) Three forces act on a point: 3 N at 0°, 4 N at 90°, and 5 N at 217°. What is the net force?

10) A cyclist heads due west on a straight road. The wind is blowing from 248° at 10 m/s. a. Is this wind more like a headwind or a tailwind? b. What is the headwind/tailwind speed? c. What is the crosswind speed?

11) A boy pulls a sled along level ground by means of a rope under 78 N of tension.

The rope makes an angle of 37° with the horizontal. What is the magnitude of the component of the tension that… a.actually moves the sled? b.is essentially wasted?

12) Four forces act on an object: 70 N at 0°, 90 N at 90°, 30 N at 180°, and 60 N at 270°. Determine the following quantities (rounded to the nearest whole number)… a.the net force in the x direction (include the mathematical sign) b.the net force in the y direction (include the mathematical sign) c.the magnitude of the net force d.the direction of the net force as a standard position angle 13) Zoe walks 3.50 m south, then 8.20 m at an angle 24.6 degrees north of east, and finally 15.0 m west. What is the magnitude of the Zoe's total displacement (m)? What is the direction of the man's total displacement where directly east is taken as zero degrees and counter-clockwise is positive (degrees)?

14) Three forces are applied to an object, as indicated in the drawing. Force F1 has a magnitude of 21.0 N and is directed 30.0° to the left of the + y axis. Force F2 has a magnitude of 15.0 N and points along the + x axis. What must be the magnitude and direction (specified by the angle q in the drawing) of the third force F3 such that the vector sum of the three forces is 0 N? (18.7 N, 76 o S of W)

15) Two forces F1 and F2 with magnitudes 20 and 30â&#x20AC;&#x2030;lb , respectively, act on an object at a point P as shown in the figure below. Find the resultant forces acting at P .

16) An ambitious hiker walks 25 km west and then 35 km south in a day. Find the magnitude and direction of the hiker's resultant displacement (relative to due west) 17)Alfred went for a walk the other day. He went four avenues east (0.80 miles), then twenty-four streets south (1.20 miles), then one avenue west (0.20 miles), and finally eight streets north (0.40 miles). a.What distance did him travel? b.What's his resultant displacement (magnitude and direction relative to due east)?

18)

UNIT 9:KINEMATIC IN ONE DIMENSION Kinematics Kinematics is the study of the motion of objects without concern for the forces causing the motion. These familiar equations allow students to analyze and predict the motion of objects, and students will continue to use these equations throughout their study of physics. A solid understanding of these equations and how to employ them to solve problems is essential for success in physics. This unit is a purely mathematical exercise designed to provide a quick review of how the kinematics equations are derived using algebra.

Displacement: A displacement is a vector whose length is the shortest distance from the initial to the final position of a point P. It quantifies both the distance and direction of an imaginary motion along a straight line from the initial position to the final position of the point.

│s│= │v│Δt Where Δt is the time interval t2 – t1. s = the displacement vector, the magnitude of the displacement is the distance, s = │s│ = d (vectors are indicated in bold; the same symbol not in bold represents the magnitude of the vector) Δ indicates change, t = time t1 = the initial time t2 = the final time Speed and Velocity: SPEED Speed can be defined as “how fast something moves” or it can be explained more scientifically as “the distance covered in a unit of time”. In daily life we use the first definition and say the faster object has higher speed. Speed does not show us the direction of the motion it just gives the magnitude of what distance taken in a given time. In other words it is a scalar quantity. We use a symbol v to show speed. v = where v = speed (m.s-1) s = distance (m) t = time (s) VELOCITY

Velocity can be defined as “speed having direction”. As you can understand from the definition velocity is a vector quantity having both magnitude and direction. In daily life we use speed and velocity interchangeably but in physics they have different meanings. We can define velocity as the “rate of change of displacement” whereas “the speed is rate of change of distance”. While we calculate speed we look at the total distance, however, in calculating velocity we must consider the direction and in short we can just look at the change in position not the whole distance traveled. If a man walks 5m to east and then 5m to west speed of that man calculated by dividing total distance traveled which is 10m to the time elapsed, however, velocity calculated by dividing the displacement to the elapsed time, which is 0m divided elapsed time gives us zero. In other words, if the displacement is zero we can not talk about the velocity. v = s/t where v = velocity (m.s-1) s = distance (m) t = time (s) Direction: Velocity is in the same direction as displacement.

Accelaration: Acceleration is the rate of change of velocity of an object with respect to time. An object's acceleration is the net result of all forces acting on the object, as described

by Newton's Second Law. The SI unit for acceleration is meter per second squared.

Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.

Since accelerating objects are constantly changing their velocity, one can say that the distance traveled/time is not a constant value. A falling object for instance usually accelerates as it falls. If we were to observe the motion of a free-falling object (free fall motion will be discussed in detail later), we would observe that the object averages a velocity of approximately 5 m/s in the first second, approximately 15 m/s in the second second, approximately 25 m/s in the third second, approximately 35 m/s in the fourth second, etc. Our free-falling object would be constantly accelerating.

Given these average velocity values during each consecutive 1-second time interval, we could say that the object would fall 5 meters in the first second, 15 meters in the second second (for a total distance of 20 meters), 25 meters in the third second (for a total distance of 45 meters), 35 meters in the fourth second (for a total distance of 80 meters after four seconds). These numbers are summarized in the table below.

Time Interval

Velocity Change During Interval

Ave. Velocity During Interval

Distance Traveled During Interval

Total Distance Traveled from 0 s to End of Interval

0 – 1.0 s

0 to ~10 m/s

~5 m/s

~5 m

~5 m

1.0 – 2.0 s

~10 to 20 m/s

~15 m/s

~15 m

~20 m

2.0 – 3.0 s

~20 to 30 m/s

~25 m/s

~25 m

~45 m

3.0 – 4.0 s

~30 to 40 m/s

~35 m/s

~35 m

~80 m

Note: The ~ symbol as used here means approximately.

This discussion illustrates that a free-falling object that is accelerating at a constant rate will cover different distances in each consecutive second. Further analysis of the first and last columns of the data above reveal that there is a square relationship between the total distance traveled and the time of travel for an object starting from rest and moving with a constant acceleration. The total distance traveled is directly proportional to the square of the time. As such, if an object travels for twice the time, it will cover four times (2^2) the distance; the total distance traveled after two seconds is four times the total distance traveled after one second. If an object travels for three times the time, then it will cover nine times (3^2) the distance; the distance traveled after three seconds is nine times the distance traveled after one second. Finally, if an object travels for four times the time, then it will cover 16 times (4^2) the distance; the distance traveled after four seconds is 16 times the distance traveled after one second. For objects with a constant acceleration, the distance of travel is directly proportional to the square of the time of travel.

Calculating the Average Acceleration The average acceleration (a) of any object over a given interval of time (t) can be calculated using the equation

This equation can be used to calculate the acceleration of the object whose motion is depicted by the velocity-time data table above. The velocity-time data in the table shows that the object has an acceleration of 10 m/s/s. The calculation is shown below.

Acceleration values are expressed in units of velocity/time. Typical acceleration units include the following: m/s/s mi/hr/s km/hr/s m/s2 These units may seem a little awkward to a beginning physics student. Yet they are very reasonable units when you begin to consider the definition and equation for acceleration. The reason for the units becomes obvious upon examination of the acceleration equation.

Since acceleration is a velocity change over a time, the units on acceleration are velocity units divided by time units - thus (m/s)/s or (mi/hr)/s. The (m/s)/s unit can be mathematically simplified to m/s2.

The Direction of the Acceleration Vector Since acceleration is a vector quantity, it has a direction associated with it. The direction of the acceleration vector depends on two things: â&#x20AC;˘ â&#x20AC;˘

whether the object is speeding up or slowing down whether the object is moving in the + or - direction The general principle for determining the acceleration is:

If an object is slowing down, then its acceleration is in the opposite direction of its motion.

This general principle can be applied to determine whether the sign of the acceleration of an object is positive or negative, right or left, up or down, etc.

Consider the two data tables below. In each case, the acceleration of the object is in the positive direction. In Example A, the object is moving in the positive direction (i.e., has a positive velocity) and is speeding up. When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object has a positive acceleration. In Example B, the object is moving in the negative direction (i.e., has a negative velocity) and is slowing down. According to our general principle, when an object is slowing down, the acceleration is in the opposite direction as the velocity.

Thus, this object also has a positive acceleration.

This same general principle can be applied to the motion of the objects represented in the two data tables below. In each case, the acceleration of the object is in the negative direction. In Example C, the object is moving in the positive direction (i.e., has a positive velocity) and is slowing down. According to our principle, when an object is slowing down, the acceleration is in the opposite direction as the velocity.

Thus, this object has a negative acceleration. In Example D, the object is moving in the negative direction (i.e., has a negative velocity) and is speeding up. When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object also has a negative acceleration.

Observe the use of positive and negative as used in the discussion above (Examples A - D). In physics, the use of positive and negative always has a physical meaning. It is more than a mere mathematical symbol. As used here to describe the velocity and the acceleration of a moving object, positive and negative describe a direction. Both velocity and acceleration are vector quantities and a full description of the quantity demands the use of a directional adjective. North, south, east, west, right, left, up and down are all directional adjectives. Physics often borrows from mathematics and uses the + and - symbols as directional adjectives. Consistent with the mathematical convention used on number lines and graphs, positive often means to the right or up and negative often means to the left or down. So to say that an object has a negative acceleration as in Examples C and D is to simply say that its acceleration is to the left or down (or in whatever direction has been defined as negative). Negative accelerations do not refer acceleration values that are less than 0. An acceleration of -2 m/s/s is an acceleration with a magnitude of 2 m/s/s that is directed in the negative direction.

Example:Consider the following problems and the corresponding solutions. Use the equation for acceleration to determine the acceleration for the following two motions.

Practice B: Use a = (vf-vi) / t and pick any two points. a = (0 m/s - 8 m/s) / (4 s) a = (-8 m/s) / (4 s) a = -2 m/s/s=-2m/s2 Practice A: Use a = (vf - vi) / t and pick any two points. a = (8 m/s - 0 m/s) / (4 s) a = (8 m/s) / (4 s) a = 2 m/s/s=2m/s2 Averages velocity: Average Velocity is displacement over total time. The average speed of an object is defined as the distance traveled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time.

NB: It is given by Vav. Average Velocity Formula varies based on the given problem.

Examples Example 1: Calculate the average velocity at a particular time interval of a particle if it is moves 5 m at 2 s and 15 m at 4s along x-axis?

Solution:

Given: Initial distance traveled, xi = 5 m, Final distance traveled, xf = 15 m, Initial time interval ti = 2s, Final time interval tf = 4s.

Average Velocity Vav = (xf−xi)/(tf−ti ) = (15−5)/(4−2) = 10/2 = 5m/s.

Example 2: A car is moving with initial velocity of 20 m/s and it reaches its destiny at 50 m/s. Calculate its average velocity.

Solution:

Given: Initial Velocity U = 20 m/s,

Final velocity V = 50 m/s.

Average Velocity Vav =( U+V)/2 = (20m/s+50m/s)/2 = 35 m/s.

Average speed:

Total distance is the distance traveled by the object at the constant speed. Elapsed time is the time the object took to cover the total distance. In most instances, an object will travel with varying speeds over a certain distance. For example, a car traveling from one city to another will rarely move at a constant speed. It is more likely that the car's speed will fluctuate during the trip. The car might travel at 65 mph for some time and then slow down to 25 mph. It's possible that at certain times, the car is even at full stop (such as, at a red light). To calculate the car's average speed, we don't really care about the fluctuations in its speed. We only care about the total distance traveled by the car and the elapsed time to cover that distance.

Example1: Suppose a freight train travels a distance of 120 miles in 3 hours. What is the average speed of the train?

Answer: Its average speed is

Example 2: Suppose a truck travels in segments that are described in the following table: Segment Distance (miles) Time (hours) 1

30

1

2

45

2

3

50

1

4 65 2 What is the average speed of the truck?

Deriving The Equations of motion: Considering the fact that the integrations are applied in mathematics 1,lets have fun by applying it here. constant acceleration Calculus is an advanced math topic, but it makes deriving two of the three equations of motion much simpler. By definition, acceleration is the first derivative of velocity with respect to time. Take the operation in that definition and reverse it. Instead of differentiating velocity to find acceleration, integrate acceleration to find velocity. This gives us the velocity-time equation. If we assume acceleration is constant, we get the so-called first equation of motion [1]. a

=

dv

=

a dt

= v â&#x2C6;&#x2019; v0

=

at

v

=

v0 + at [1]

Again by definition, velocity is the first derivative of position with respect to time.

Reverse the operation in the definition. Instead of differentiating position to find velocity, integrate velocity to find position. This gives us the position-time equation for constant acceleration, also known as the second equation of motion [2]. v= ds = v dt ds = (v0 + at) dt s ⌠ ds ⌡ s0

=

t ⌠ (v + at) dt ⌡ 0 0

s − s0 = v0t + ½at2 s = s0 + v0t + ½at2 [2] Unlike the first and second equations of motion, there is no obvious way to derive the third equation of motion (the one that relates velocity to position) using calculus.

We can't just reverse engineer from the definitions. We need to play a rather sophisticated trick.

The first equation of motion relates velocity to time. We essentially derived it from this derivative… dv =a dt The second equation of motion relates position to time. It came from this derivative…

=v

The third equation of motion relates velocity to position. By logical extension, it should come from a derivative that looks like this…

=?

But what does this equal? Well nothing by definition, but like all quantities it does equal itself. It also equals itself multiplied by 1. We'll use a special version of 1 (dt/dt) and a special version of algebra (algebra with infinitesimals). Look what happens when we do this. We get a derivative equal to acceleration (dv/dt) and another equal to the inverse of velocity (dt/ds).

=

1

= = = a

Next step, separation of variables. Get things that are similar together and integrate them. Here's what we get when acceleration is constant… = v dv

=

a ds

v ⌠ v dv ⌡ v0

s ⌠ a ds ⌡ s0

=

½(v2 − v02)

=

a(s − s0)

v2

=

v02 + 2a(s − s0) [3]

Certainly a clever solution, and it wasn't all that more difficult than the first two derivations. However, it really only worked because acceleration was constant — constant in time and constant in space. If acceleration varied in any way, this method would be uncomfortably difficult. We'd be back to using algebra just to save our sanity. Not that there's anything wrong with that. Algebra works and sanity is worth saving. v = v0 + at

[1] + 2 s = s0 + v0t + ½at [2] = 2 2 v = v0 + 2a(s − s0) [3] Example 1: An airplane accelerates down a runway at 3.20 m/s 2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

Solution: a = +3.2 m/s2 t = 32.8 s vi = 0 m/s Find: d = ?? d = vi*t + 0.5*a*t2 d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2 d = 1720m

Example 2: A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

Solution: d = 110 m

t = 5.21 s

vi = 0 m/s

Find: a = ??

2

d = vi*t + 0.5*a*t 110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2 110 m = (13.57 s2)*a a = (110 m)/(13.57 s2) a = 8.10 m/ s2 TUTORIAL: 1.Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?(25,5m/s;33,1m).

2.A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds.

Determine the acceleration of the car and the distance traveled.(11,2m/s2 )

3.A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.(1,3s)

4.Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.83 seconds, then what is the acceleration and what is the distance that the sled travels?(243m/s2 ;406m)

5.A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.(0,712m/s2 ) 6.An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s 2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?(704m) 7.A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding

distance of the car (assume uniform acceleration).(28,6m)

8.A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.(7,17m/s)

9.If Zoe Ngamgoum has a vertical leap of 1.29 m, then what is her takeoff speed and her hang time (total time to move upwards to the peak and then return to the ground)?(5,03m/s;1,03s) 10.A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).(1,62.105 m/s2 ) 11.A baseball is popped straight up into the air and has a hang-time of 6.25 s.

Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one-half the total hang-time.)(48,0m) 12.The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.(8,69s)

13.A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)(-1,08.106 m/s2 ) 14.A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.( -57.0 (57.0 meters deep) )

15.It was once recorded that a Jaguar left skid marks that were 290 m in length.

Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.( 47.6 m/s)

16.A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed. .(2,86m/s2 ;30,8s)

17.A dragster accelerates to a speed of 112 m/s over a distance of 398 m.

Determine the acceleration (assume uniform) of the dragster.( 15.8 m/s/s) 18.With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.( 94.4 mi/hr) 19.A motor scooter travels east at a speed of 13 m/s. The driver then reverses direction and heads west at 17 m/s. What was the change in velocity of the scooter?( 30 m/s) 20.A cement block accidentally falls from rest from the ledge of a 53.0-m-high building. When the block is 14.0 m above the ground, a man, 2.00 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way ?

21.A landing airplane makes contact with the runway with a speed of 78.0 m/s and moves toward the south. After 18.5 seconds, the airplane comes to rest. What is the magnitude and direction of acceleration of the airplane during the landing? (4.22 m/s2 towards north) 22. A car is moving at a constant velocity when it is involved in a collision. The car comes to rest after 0.450 s with an average acceleration of 65.0 m/s2 in the direction opposite that of the car's velocity. What was the speed, in km/h, of the car before the collision? (105.3 km/h) 23. A gun is fired straight downward from the edge of a cliff that is 15 m above the ground. The bullet strikes the ground with a speed of 27 m/s. How far above the cliff edge would the bullet have gone had the gun been fired straight up. (22 m)

UNIT 10:FORCES AND NEWTON’S LAW OF MOTION INTRODUCTION 10.1 A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force. Forces only exist as a result of an interaction.

10.2 Contact versus Action-at-a-Distance Forces For simplicity sake, all forces (interactions) between objects can be placed into two broad categories: • •

contact forces, and forces resulting from action-at-a-distance Contact forces are those types of forces that result when the two interacting objects are perceived to be physically contacting each other. Examples of contact forces include frictional forces, tensional forces, normal forces, air resistance forces, and applied forces. These specific forces will be discussed in more detail later in Lesson 2 as well as in other lessons.

Action-at-a-distance forces are those types of forces that result even when the two interacting objects are not in physical contact with each other, yet are able to exert a push or pull despite their physical separation. Examples of action-at-a-distance forces include gravitational forces. For example, the sun and planets exert a gravitational pull on each other despite their large spatial separation. Even when your feet leave the earth and you are no longer in physical contact with the earth, there is a gravitational pull between you and the Earth. Electric forces are action-at-a-distance forces. For example, the protons in the nucleus of an atom and the electrons outside the nucleus experience an electrical pull towards each other despite their small spatial separation. And magnetic forces are action-at-a-distance forces. For example, two magnets can exert a magnetic pull on each other even when separated by a distance of a few centimeters.

Examples of contact and action-at-distance forces are listed below.

Contact Forces

Action-at-a-Distance Forces

Frictional Force

Gravitational Force

Tension Force

Electrical Force

Normal Force

Magnetic Force

Air Resistance Force Applied Force Spring Force The Newton

Force is a quantity that is measured using the standard metric unit known as the Newton. A Newton is abbreviated by an "N." To say "10.0 N" means 10.0 Newton of force. One Newton is the amount of force required to give a 1-kg mass an acceleration of 1 m/s/s. Thus, the following unit equivalency can be stated: 1 Newton = 1 kg â&#x20AC;˘ m/s2 Force is a Vector Quantity A force is a vector quantity. As learned in an earlier unit, a vector quantity is a quantity that has both magnitude and direction. To fully describe the force acting upon an object, you must describe both the magnitude (size or numerical value) and the direction. Thus, 10 Newton is not a full description of the force acting upon an object. In contrast, 10 Newton, downward is a complete description of the force acting upon an object; both the magnitude (10 Newton) and the direction (downward) are given.

Because a force is a vector that has a direction, it is common to represent forces using diagrams in which a force is represented by an arrow. Such vector diagrams were introduced in an earlier unit and are used throughout the study of physics. The size of the arrow is reflective of the magnitude of the force and the direction of the arrow reveals the direction that the force is acting. (Such diagrams are known as free-body diagrams and are discussed later in this lesson.) Furthermore, because forces are vectors, the effect of an individual force upon an object is often canceled by the effect of another force. For example, the effect of a 20-Newton upward force acting upon a book is canceled by the effect of a 20-Newton downward force acting upon the book. In such instances, it is said that the two individual forces balance each other; there would be no unbalanced force acting upon the book.

Other situations could be imagined in which two of the individual vector forces cancel each other ("balance"), yet a third individual force exists that is not balanced by another force. For example, imagine a book sliding across the rough surface of a table from left to right. The downward force of gravity and the upward force of the table supporting the book act in opposite directions and thus balance each other. However, the force of friction acts leftwards, and there is no rightward force to balance it. In this case, an unbalanced force acts upon the book to change its state of motion.

The exact details of drawing free-body diagrams are discussed later. For now, the emphasis is upon the fact that a force is a vector quantity that has a direction. The importance of this fact will become clear as we analyze the individual forces acting upon an object later in this lesson.

Newtonâ&#x20AC;&#x2122;s Law of Motion :

Newton's First Law of Motion: Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.

• • • •

•

1.

2.

3.

a.

4.

5.

•

•

Examples: One's body movement to the side when a car makes a sharp turn.

Tightening of seat belts in a car when it stops quickly.

A ball rolling down a hill will continue to roll unless friction or another force stops it.

If pulled quickly, a tablecloth can be removed from underneath of dishes.

The dishes have the tendency to remain still as long as the friction from the movement of the tablecloth is not too great.

Shaking a bottle of ketchup. When bringing the bottom down, the suddenly stopping it, inertia is what causes the ketchup to come out of .

The plates and cutlery in your kitchen cabinets will remain in place, until you lift them out.

You will remain in your chair until you lift yourself out of it, or until someone picks you up, or until you fall off a stool because of gravity.

Satellites remain in orbit without burning fuel because they maintain a constant speed, and gravity pulls on them constantly, bending their path around the Earth.

Without Newton's first law, they might slow down or speed up without warning, unpredictably. The Earth itself might be alone, no longer circling the sun, and the Moon may have gone off in some other random direction.

Voyager 1 will continue in a near-perfectly straight line for the next 40,000 years or so, until it gets within 1.5 light-years of the next star it encounters.

Maybe it will appear to be going in a straight line as it passes by that star, since gravity is still extremely weak at that "close approach" and won't apply much force.

This tragic news story from Quebec on July 6, 2013 is still fresh in my mind: A freight train loaded with petroleum products was parked on a siding and left unattended on a relatively flat grade with the engines running and airbrakes applied. It lost power when the engines mysteriously stopped running, and all of it's air pressure leaked out, releasing the brakes. Then it rolled several kilometers down the tracks in the middle of the night, gathering speed along the way, and derailed in a huge fireball as rail car after rail car plunged into the wreck. A good portion of a small town was completely destroyed, and several lives were lost.

Had Newton's first law not applied to this incident, that freight train would have remained in place where it had been parked, even with no brakes applied. Gravity acting on the train is what started the train moving, since it was not parked on a perfectly flat section of tracks. The entire train would have stopped once the engines derailed, but Newton's first law ensured that each rail car following would plunge into the disaster zone at full speed.

Reference:http://www.cbc.ca/montreal/featu...

Newton's Second Law of Motion: The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector.

Examples: â&#x20AC;˘

If you use the same force to push a truck and push a car, the car will have more acceleration than the truck, because the car has less mass.

â&#x20AC;˘

It is easier to push an empty shopping cart than a full one, because the full shopping cart has more mass than the empty one.

This is the most powerful of Newton's three Laws, because it allows quantitative calculations of dynamics: how do velocities change when forces are applied. Notice the fundamental difference between Newton's 2nd Law and the dynamics of Aristotle: according to Newton, a force causes only a change in velocity (an acceleration); it does not maintain the velocity as Aristotle held.

This is sometimes summarized by saying that under Newton, F = ma, but under Aristotle F = mv, where v is the velocity. Thus, according to Aristotle there is only a velocity if there is a force, but according to Newton an object with a certain velocity maintains that velocity unless a force acts on it to cause an acceleration (that is, a change in the velocity). As we have noted earlier in conjunction with the discussion of Galileo, Aristotle's view seems to be more in accord with common sense, but that is because of a failure to appreciate the role played by frictional forces. Once account is taken of all forces acting in a given situation it is the dynamics of Galileo and Newton, not of Aristotle, that are found to be in accord with the observations.

Static Friction Formula Static friction is a force that keeps an object at rest. It must be overcome to start moving the object. Once an object is in motion, it experiences kinetic friction. If a small amount of force is applied to an object, the static friction has an equal magnitude in the opposite direction. If the force is increased, at some point the value of the maximum static friction will be reached, and the object will move. The coefficient of static friction is assigned the Greek letter "mu" (Îź), with a subscript "s".

The maximum force of static friction is μs times the normal force on an object. force of static friction ≤ (coefficient of static friction)(normal force) maximum force of static friction = (coefficient of static friction)(normal force) Fs ≤ μs η , and Fsmax = μs η Fs = force of static friction μs = coefficient of static friction η = normal force (Greek letter "eta") ≤ means "less than or equal to" Fsmax = maximum force of static friction Static Friction Formula Questions: 1) A 5500 N force is applied to a sled full of firewood in a snow-covered forest. The skis of the sled have a coefficient of static friction μs = 0.75 with the snow. If the fullyloaded sled has a mass of 700 kg, what is the maximum force of static friction, and is the force applied enough to overcome it?

Answer: On a flat surface, the normal force on an object is η = mg. Using this, the maximum force of static friction can be found: Fsmax = μs η Fsmax = μs mg Fsmax = (0.75)(700 kg)(9.8 m/(s2)) Fsmax = 5145 kg∙m/s2 Fsmax = 5145 N The maximum force of static friction is 5145 N, and therefore the applied force of 5500 N is enough to overcome it, and start moving the sled.

2) A person building a brick-making machine wants to measure the coefficient of static friction between brick and wood. To do this, she places a 2.00 kg brick on a flat piece of wood, and gradually applies more and more force until the brick moves. She finds that the brick moves when exactly 11.8 N of force is applied. What is the coefficient of static friction?

Answer: The force that was applied was exactly the right amount to overcome static friction, so it is equal to Fsmax. On a flat surface, the normal force on an object is η = mg. The coefficient of static friction can be found by rearranging the formula for the maximum force of static friction:

μs = 0.6020...

Give the answer with three significant digits, to match the other numbers in the equation: μs ≈ 0.602 We find the coefficient of static friction between the brick and wood to be 0.602.

Examples solved:

1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to accelerate the object.

Known : Mass (m) = 1 kg Acceleration (a) = 5 m/s2 Wanted : net force (∑F) Solution : We use Newton’s second law to get the net force. ∑F = m a ∑F = (1 kg)(5 m/s2) = 5 kg m/s2 = 5 Newton 2. Mass of an object = 1 kg, net force ∑F = 2 Newton. Determine the magnitude and direction of the object’s acceleration….

Known : Mass (m) = 1 kg Net force (∑F) = 2 Newton Wanted : The magnitude and direction of the acceleration (a) Solution : a = ∑F / m a=2/1 a = 2 m/s2 The direction of the acceleration = the direction of the net force (∑F) 3. Object’s mass = 2 kg, F1 = 5 Newton, F2 = 3 Newton. The magnitude and direction of the acceleration is…

Known : Mass (m) = 2 kg F1 = 5 Newton F2 = 3 Newton Wanted : The magnitude and direction of the acceleration (a) Solution : net force : ∑F = F1 – F2 = 5 – 3 = 2 Newton The magnitude of the acceleration : a = ∑F / m a=2/2 a = 1 m/s2 Direction of the acceleration = direction of the net force = direction of F1

4. Object’s mass = 2 kg, F1 = 10 Newton, F2 = 1 Newton. The magnitude and direction of the acceleration is…

Known :

Mass (m) = 2 kg F2 = 1 Newton F1 = 10 Newton F1x = F1 cos 60o = (10)(0.5) = 5 Newton Wanted : The magnitude and direction of the acceleration (a) Solution : Net force : ∑F = F1x – F2 = 5 – 1 = 4 Newton The magnitude of the acceleration : a = ∑F / m a=4/2 a = 2 m/s2 Direction of the acceleration = direction of the net force = direction of F1x 5. F1 = 10 Newton, F2 = 1 Newton, m1 = 1 kg, m2 = 2 kg. The magnitude and direction of the acceleration is…

Known : Mass 1 (m1) = 1 kg Mass 2 (m2) = 2 kg F1 = 10 Newton F2 = 1 Newton Wanted : The magnitude and direction of the acceleration (a) Solution : The net force : ∑F = F1 – F2 = 10 – 1 = 9 Newton The magnitude of the acceleration : a = ∑F / (m1 + m2) a = 9 / (1 + 2) a=9/3 a = 3 m/s2

The direction of the acceleration = the direction of the net force = direction of F1 6.

A 40-kg block accelerated by a force of 200 N. Acceleration of the block is 3 m/s2.

Determine the magnitude of friction force experienced by the block.

A. 15 N B. 40 N C. 43 N D. 80 N Known : Mass (m) = 40 kg Force (F) = 200 N Acceleration (a) = 3 m/s2 Wanted: Friction force (Fg) Solution : The equation of Newton’s second law of motion ∑F = m a ∑F = net force, m = mass, a = acceleration The direction of force F rightward, the direction of friction force leftward (the direction of friction force is opposite with the direction of object’s motion).

Choose rightward as positive and leftward as negative. ∑F = m a F – Fg = m a 200 – Fg = (40)(3) 200 – Fg = 120 Fg = 200 – 120 Fg = 80 Newton The correct answer is D. 7. Block A with a mass of 100-gram place above block B with a mass of 300 gram, and then block b pushed with a force of 5 N vertically upward. Determine the normal forceexerted by block B on block A.

A. 1 N B. 1.25 N C. 2 N D. 3 N Known : Force (F) = 5 Newton Mass of block A (mA) = 100 gram = 0.1 kg Mass of block B (mB) = 300 gram = 0.3 kg

Acceleration of gravity (g) = 10 m/s2 Weight of block A (wA) = (0.1 kg)(10 m/s2) = 1 kg m/s2 = 1 Newton Weight of block B (wB) = (0.3 kg)(10 m/s2) = 3 kg m/s2 = 3 Newton Wanted : Normal force exerted by block B to block A Solution :

There are several forces that act on both block, as shown in figure.

F = push force (act on block B) wA = weight of block A (act on block A) wB = weight of block B (act on block B) NA = normal force exerted by block B on block A (Act on block A) NA’ = normal force exerted by block A on block B (Act on block B) Apply Newton’s second law of motion on both blocks : ∑F = m a F – wA – wB + NA – NA’ = (mA + mB) a NA and NA’ are action-reaction forces that have the same magnitude but opposite in direction so eliminated from the equation.

F – wA – wB = (mA + mB) a 5 – 1 – 3 = (0.1 + 0.3) a 5 – 4 = (0.4) a 1 = (0.4) a a = 1 / 0.4 a = 2.5 m/s2 Apply Newton’s second law of motion on block A : ∑F = m a NA – wA = mA a NA – 1 = (0.1)(2.5) NA – 1 = 0.25 NA = 1 + 0.25 NA = 1.25 Newton The correct answer is B. 8. An object with weight of 4 N supported by a cord and pulley. A force of 2 N acts on the block and one end of the cord pulled by a force of 9 N. Determine the net force acts on object X.

A. 3 N upward B. 4 N downward

C. 9 N upward D. 9 N downward Known : Weight of X (wX) = 4 Newton Pull force (Fx) = 2 Newton Tension force (FT) = 9 Newton Wanted: Net force acts on object X Solution : Vertically upward forces that act on object X : The tension force has the same magnitude in all part of the cord. So the tension force is 9 N. Vertically downward forces that act on object X : There are two forces that act on object X and both forces are vertically downward, the horizontal component of weight wx and the horizontal component of force Fx. Net force act on the object X : FT – wX – Fx = 9 – 4 – 2 = 9 – 6 = 3 The net force act on the object X is 3 Newton, vertically upward.

The correct answer is A. 9. An object initially at rest on a smooth horizontal surface. A force of 16 N acts on the object so the object accelerated at 2 m/s2. If the same object at rest on a rough horizontal surface so the friction force acts on the object is 2 N, then determine the acceleration of the object if the same force of 16 N acts on the object.

A. 1.75 m/s2 B. 1.50 m/s2 C. 1.00 m/s2 D. 0.88 m/s2 Known : Force (F) = 16 Newton = 16 kg m/s2 Acceleration (a) = 2 m/s2 Friction force (Ffric) = 2 Newton = 2 kg m/s2 Wanted : Object’s acceleration ?

Solution : Smooth horizontal surface (no friction force) :

∑F = m a F=ma 16 = (m) 2 m = 16 / 2 m = 8 kg Mass of object is 8 kilogram.

Rough horizontal surface (there is a friction force) :

∑F = m a F – Ffric = m a 16 – 2 = 8 a 14 = 8 a a = 14 / 8

a = 1.75 m/s2 Object’s acceleration is 1.75 m/s2.

The correct answer is A. 10. FREDDY and ALFRED push an object on the smooth floor. FREDDY push the object with a force of 5.70 N. If the mass of the object is 2.00 kg and acceleration experienced by the object is 2.00 ms-2, then determine the magnitude and direction of force act by Freddy.

A. 1.70 N and its direction is opposite with force acted by Alfred B. 1.70 N and its direction same as force acted by Alfred C. 2.30 N and its direction is opposite with force acted by Alfred.

D. 2.30 N and its direction same as force acted by Alfred.

Known : Push force acted by Alfred (F1) = 5.70 Newton Mass of object (m) = 2.00 kg Acceleration (a) = 2.00 m/s2 Wanted : Magnitude and direction of force acted by freddy (F2) ?

Solution : Apply Newton’s second law of motion : ∑F = m a F1 + F2 = m a 5.70 + F2 = (2)(2) 5.70 + F2 = 4 F2 = 4 – 5.70 F2 = – 1.7 Newton Minus sign indicated that (F2) is opposite with push force act by Alfred (F1).

The correct answer is A. 11. If the mass of the block is the same, which figure shows the smallest acceleration?

Solution Net force A : ΣF = 4 N + 2 N – 3 N = 6 N – 3 N = 3 Newton, leftward Net force B : ΣF = 2 N + 3 N – 4 N = 5 N – 4 N = 1 Newton, rightward Net force C : ΣF = 4 N + 3 N – 2 N = 7 N – 2 N = 5 Newton, rightward Net force D : ΣF = 3 N + 4 N + 2 N = 9 Newton, rightward The equation of Newton’s second law : ΣF = m a

a = ΣF / m a = acceleration, ΣF = net force, m = mass Based on the above formula, the acceleration (a) is directly proportional to the net force (ΣF) and inversely proportional to mass (m). If the mass of an object is the same, the greater the resultant force, the greater the acceleration or the smaller the resultant force, the smaller the acceleration.

Based on the above calculation, the smallest net force is 1 Newton so the acceleration is also smallest.

The correct answer is B.

12. Some forces act on an object with a mass of 20 kg, as shown in the figure below.

Determine the object’s acceleration.

Known : Mass of object (m) = 20 kg Net force (ΣF) = 25 N + 30 N – 15 N = 40 N Wanted: Acceleration of an object Solution : Object’s acceleration calculated using the equation of Newton’s second law : ΣF = m a a = ΣF / m = 40 N / 20 kg = 2 N/kg = 2 m/s2

Newton's Third Law of Motion: For every action there is an equal and opposite reaction.

Action and reaction are equal and opposite.

This law is exemplified by what happens if we step off a boat onto the bank of a lake: as we move in the direction of the shore, the boat tends to move in the opposite direction (leaving us facedown in the water, if we aren't careful!).

Newton’s law of universal gravitation: Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

where the constant G is what Cavendish measured, and found to be 6.67 x 10-11 N.m2 .kg-2

The proportionalities expressed by Newton's universal law of gravitation are represented graphically by the following illustration. Observe how the force of gravity is directly proportional to the product of the two masses and inversely proportional to the square of the distance of separation.

Example 1

Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is standing at sea level, a distance of 6.38 x 106m from earth's center.

The solution of the problem involves substituting known values of G (6.673 x 10-11 N m2/kg2), m1(5.98 x 1024 kg), m2 (70 kg) and d (6.38 x 106 m) into the universal gravitation equation and solving for F grav. The solution is as follows:

Example 2 Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is in an airplane at 40000 feet above earth's surface. This would place the student a distance of 6.39 x 106 m from earth's center.

The solution of the problem involves substituting known values of G (6.673 x 10 -11 N m2/kg2), m1(5.98 x 1024 kg), m2 (70 kg) and d (6.39 x 106 m) into the universal gravitation equation and solving for F grav. The solution is as follows:

Two general conceptual comments can be made about the results of the two sample calculations above. First, observe that the force of gravity acting upon the student (a.k.a. the student's weight) is less on an airplane at 40 000 feet than at sea level.

This illustrates the inverse relationship between separation distance and the force of gravity (or in this case, the weight of the student). The student weighs less at the higher altitude. However, a mere change of 40 000 feet further from the center of the Earth is virtually negligible. This altitude change altered the student's weight changed by 2 N that is much less than 1% of the original weight. A distance of 40 000 feet (from the earth's surface to a high altitude airplane) is not very far when compared to a distance of 6.38 x 106 m (equivalent to nearly 20 000 000 feet from the center of the earth to the surface of the earth). This alteration of distance is like a drop in a bucket when compared to the large radius of the Earth. As shown in the diagram below, distance of separation becomes much more influential when a significant variation is made.

The second conceptual comment to be made about the above sample calculations is that the use of Newton's universal gravitation equation to calculate the force of gravity (or weight) yields the same result as when calculating it using the equation presented in the previous unit: Fgrav = mâ&#x20AC;˘g = (70 kg)â&#x20AC;˘(9.8 m/s2) = 686 N Both equations accomplish the same result because the value of g is equivalent to the ratio of (Gâ&#x20AC;˘Mearth)/(Rearth)2.

Some of definitions and explanations: State of equilibrium: An object is in equilibrium when it has zero acceleration (in rest or move with constant velocity). OR A rigid body (by definition distinguished from a particle in having the property of extension) is considered to be in equilibrium if, in addition to the states listed for the particle above, the vector sum of all torques acting on the body equals zero so that its state of rotational motion remains constant.

Inertia: Inertia is a quality of all objects made of matter that possess mass. They keep doing what they are doing until a force changes their speed or direction. A ball sitting still on a table won't start rolling around unless something pushes on it, be it your hand, a gust of air, or vibrations from the surface of the table. If you tossed a ball in the frictionless vacuum of space, it would travel on at the same speed and direction forever unless acted on by gravity or another force such as a collision.

Normal force: It is a force which acts perpendicular to an area. It is developed when external loads tend to push or pull two segments of a body. It is denoted by the symbol N. When two surfaces are in contact then normal force acts on them. When two segments are not in contact, then normal force does not act on either of them.

Figure (24) shows two surfaces (a table and a box) that are not in contact.

Figure (24) shows two surfaces (a table and a box) that are in contact.

The formula of normal force on a horizontal plane is: N = mg Here, normal force is N, mass of any object is m, and acceleration due to gravity is g.

The formula to calculate the normal force on an incline is: N = mg cos θ Here, the angle of the inclined object measured from the horizontal is θ.

There are two normal forces acting on the object; a downward and an upward force.

These two forces are equal and opposite of each other. According to Newton's second law, one force will be positive and the other will be negative.

Weight: The weight of an object is the force of gravity on the object and may be defined as the mass times the acceleration of gravity, w = mg. Since the weight is a force, its SI unit is the newton.

Mass: Mass is a measure of inertia. Objects of higher mass resist changes in motion more than objects of lower mass. A more massive ball, such as one made of lead, will take more of a push to start it rolling. A styrofoam ball of the same size but low mass may be set in motion by a puff of air.

FREE-BODY DIAGRAM: Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation. A free-body diagram is a special example of the vector diagrams that were discussed in an earlier unit. These diagrams will be used throughout our study of physics. The size of the arrow in a free-body diagram reflects the magnitude of the force. The direction of the arrow

shows the direction that the force is acting. Each force arrow in the diagram is labeled to indicate the exact type of force. It is generally customary in a free-body diagram to represent the object by a box and to draw the force arrow from the center of the box outward in the direction that the force is acting. An example of a free-body diagram is shown below.

The free-body diagram above depicts four forces acting upon the object. Objects do not necessarily always have four forces acting upon them. There will be cases in which the number of forces depicted by a free-body diagram will be one, two, or three. There is no hard and fast rule about the number of forces that must be drawn in a free-body diagram. The only rule for drawing free-body diagrams is to depict all the forces that exist for that object in the given situation. Thus, to construct free-body diagrams, it is extremely important to know the various types of forces. If given a description of a physical situation, begin by using your understanding of the force types to identify which forces are present. Then determine the direction in which each force is acting. Finally, draw a box and add arrows for each existing force in the appropriate direction; label each force arrow according to its type. If necessary, refer to the list of forces and their description in order to understand the various force types and their appropriate symbols.

Apparent Weight (Normal Force) : .If object is accelerating in vertical direction weight appears different .Accelerating up, increases apparent weight .Accelerating down decreases apparent weight 10.4 The equation for measuring apparent weight is F = mg + ma. F represents apparent weight in newtons, m is the mass of the object, g is the acceleration due to gravity (9.8 meters per second squared on Earth's surface) and a is the acceleration of the object.

The apparent weight of an object when it is at rest, free falling or moving at constant speed is equal to its normal weight because the object's acceleration is zero. When the object is moving upward, for example in an elevator, the object's apparent weight is heavier. This is due to the additional force necessary to push the object up, represented by mass time acceleration with positive acceleration value. If the object is moving downward, its apparent weight is lighter, since the object is actually decelerating and thus has a negative acceleration value.

Examples: A book is at rest on a tabletop. A free-body diagram for this situation looks like this:

A gymnast holding onto a bar, is suspended motionless in mid-air. The bar is supported by two ropes that attach to the ceiling. Diagram the forces acting on the combination of gymnast and bar. A free-body diagram for this situation looks like this:

An egg is free-falling from a nest in a tree. Neglect air resistance. A free-body diagram for this situation looks like this:

Additional notes: Mathematical modelling in mechanics: Mathematical modelling is used to describe the process of solving real world problems through the use of mathematics. The process begins when a problem statement passes through 3 main phases, formulation, solution and review, and culminates in a report. Mathematical models are often revised after the review stage and the formulation, solution, review process repeated. Each of these three phases is now considered in more detail.

10.5 Formulation Before embarking on the formulation stage it is important to make sure that you understand the problem that has been posed. The first part of the formulation stage is to make a list of all the features of the problem that could influence the solution you will eventually give.

Some of the features identified may be very important while others are much more trivial. The second stage is to simplify the problem and to state a set of assumptions on which to base the problem.

The use of assumptions simplifies the problem considerably and allows a mathematical problem to be stated. The assumptions are important because they describe the conditions under which the mathematical solution will be obtained and they allow the problem to be simplified to an extent where it is possible to define a mathematical problem that can be solved.

10.6 Mathematical Solution At this stage the mathematical problem defined in the formulation stage is solved to give a solution to the problem. This stage will involve many of the mathematical skills and techniques that you will have encountered in your mathematics and applied mathematics modules.

Review This stage contains a number of elements which include interpretation, comparison with reality, criticism of results and reformulation. In some cases it will be important for interpretation of the solution to be given.

Comparison of results with real situations can be very useful for determining the validity of a solution.

Criticism of results is important to show any weaknesses of the solution, which may well be due to the initial assumptions.

In the light of the review of the solution it will very often be desirable to refine the solution to the problem and the first stage in this will be to reformulate the problem to take account of new factors or to revise the assumptions. The three stages of mathematical modelling are then often repeated until a satisfactory solution is obtained. The diagram shows how mathematical modelling has a cycle that can be repeated several times.

An Example of Mathematical Modelling in Mechanics Problem Statement A car at an accident skids down a gentle slope leaving a skid mark of 20m before it collides with a stationary vehicle. How fast was the car travelling when it began to skid and was it breaking the 30mph speed limit?

Formulation The first step is to draw up a feature list to include all those factors that might affect the solution of the problem. The list below includes a number of important factors.

• The road and tyre conditions.

• The gradient of the slope.

• The speed of impact between the two cars. The conditions of the road.

• Were all the wheels locked?

Are there any further factors that you think should be included?

The assumptions below allow a simple model to be formulated so that a mathematical problem can be defined.

• The road is horizontal.

• The speed of impact is zero.

• There is constant friction force and no air resistance.

• The coefficient of friction between the tyres and the road is 0.8. The car is to be modelled as a particle.

Mathematical Problem A car skids 20m to rest on a horizontal road. If the coefficient of friction is 0.8 find the initial speed of the car.

Mathematical Solution

As the car is skidding the friction force will take its limiting value of R .

So, F=− mg=−0.8 10m where m is the mass of the car and g is taken as 10 ms−2.

The acceleration of the car is given by a=F/m=−8 ms-2.

Now the speed can be found using V2=u2+2as 02=u2+2 (−8) 20

u2 =320 u=17.9 ms-1

Interpretation The initial speed predicted can be converted to give 40mph. This figure clearly suggests that the car was breaking the speed limit.

Compare with Realitv

The Highway Code provides a useful source of data that can be used to compare the results with reality. The Highway Code quotes a distance of 80 feet or 24m as the braking distance for 40mph. This compares favourably with the prediction made above.

Criticism of the Model The two major criticisms that can be made of this model are that the road is not horizontal and that the speed of impact is not taken into account.

Reformulation

The diagram shows the forces acting when the road is assumed to be at an angle of 50 to the horizontal. The normal reaction now has the magnitude, R=mgcos50 and the resultant force up the slope on the car is, mgcos50−mgsin50=mg( cos50−sin50 ) So the acceleration is a=g( cos50−sin50) =9.8(0.8cos50−sin50) =−6.96 ms-2 Using this value for awith an impact speed of zero gives.

02=u2+2 (−6.96) 20 u2=278.4 u=16.7 ms-1 or 37 mph Notice that taking account of the hill reduces the initial speed As a final refinement to the model an impact speed 9 ms−1 or approximately 20 mph could be introduced. There was only minor damage to the vehicles which suggests that such a value is reasonable. Using this gives a revised initial speed 92=u2+2 (−6.96) 20

u2=359.4 u=19.0 ms-1 or 42 mph

It is interesting to note how close the two revised estimates are to the original prediction and that the conclusion that the car was breaking the speed limit was

confirmed by both revisions.

TUTORIAL: 1. Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is doubled, what is the new force of attraction between the two objects? 4 units 2. Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is reduced in half, then what is the new force of attraction between the two objects? 64 units 3. Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of both objects was doubled, and if the distance between the objects remained the same, then what would be the new force of attraction between the two objects? 64 units 4. Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of both objects was doubled, and if the distance between the objects was doubled, then what would be the new force of attraction between the two objects? 16 units 5. Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of both objects was tripled, and if the distance between the objects was doubled, then what would be the new force of attraction between the two objects?( (16 units) * 9 / 4 = 36 units) 6. Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of object 1 was doubled, and if the distance between the objects was tripled, then what would be the new force of attraction between the two objects?( (16 units) â&#x20AC;˘ 2 / 9 = 3.56 units) 7. As a star ages, it is believed to undergo a variety of changes. One of the last phases of a star's life is to gravitationally collapse into a black hole. What will happen to the orbit of the planets of the solar system if our star (the Sun shrinks into a black hole)? (And of course, this assumes that the planets are unaffected by prior stages of the Sun's evolving stages.).( No affect The shrinking of the sun into a black hole would not influence the amount of force with which the sun attracted the Earth since neither the mass of the sun nor the distance between the Earth's and sun's centers would change.)

8. Having recently completed her first Physics course, Dawn Well has devised a new business plan based on her teacher's Physics for Better Living theme. Dawn learned that objects weigh different amounts at different distances from Earth's center. Her plan involves buying gold by the weight at one altitude and then selling it at another altitude at the same price per weight. Should Dawn buy at a high altitude and sell at a low altitude or vice versa?( Buy high and sell low The mass of the purchased gold would be the same at both altitudes. Yet it would weight less at higher altitudes. So to make a profit, Dawn should buy at high altitudes and sell at low altitudes. She would have more gold (by weight) to sell at the lower

altitudes.)

9. Anita Diet is very concerned about her weight but seldom does anything about it.

After learning about Newton's law of universal gravitation in Physics class, she becomes all concerned about the possible effect of a change in Earth's mass upon her weight. During a (rare) free moment at the lunch table, she speaks up "How would my weight change if the mass of the Earth increased by 10%?" How would you answer Anita?( "Anita - that's a great question! Since your weight is directly dependent upon the mass of the Earth, you would weigh 10% more. But don't worry honey. You wouldn't look any different than you do now since your mass would remain as is.")

10. When comparing mass and size data for the planets Earth and Jupiter, it is observed that Jupiter is about 300 times more massive than Earth. One might quickly conclude that an object on the surface of Jupiter would weigh 300 times more than on the surface of the Earth. For instance, one might expect a person who weighs 500 N on Earth would weigh 150000 N on the surface of Jupiter. Yet this is not the case. In fact, a 500-N person on Earth weighs about 1500 N on the surface of Jupiter. Explain how this can be.( he affect of the greater mass of Jupiter is partly offset by the fact that the radius of Jupiter is larger. An object on Jupiter's surface is 10 times farther from Jupiter's center than it would be if on Earth's surface. So the 300-fold increase in force (due to the greater mass) must be divided by 100 since the separation distance is 10 times greater.)

11.What is the magnitude of the resultant force due to two forces of 30 Newtons and 40 Newtons acting on an object at a point such that the two forces are at right angle to each other?

12.Two forces F1 and F2 act on a mass m located at the origin of a system of axis x y and R is the resultant of the two forces. F1 makes an angle of 33deg with the x axis and has a magnitude of 10 Newtons. F2 makes an angle of 123deg with the x axis and has a magnitude of 20 Newtons. What angle does the resultant R makes with force F1?

13.An object of mass m rests on a horizontal plane. If we start inclining the plane at an angle θ with the horizontal, at which angle will the object start moving if the coefficient of friction is equal to 1?

14.A 2-kilogram box is accelerated by a horizontal force F. If the magnitude of the acceleration is 2m/s2 and the force of friction is 44 N, what is the magnitude of F ?

15. A lift and its passengers have a total mass of 300 kg. Find the tension in the lift cable if;

a) it accelerates upwards at 0.2 ms−2,( 3000 N ) b) it accelerates downwards at 0.05 ms−2.( 2925 N ) 16. A car accelerates at 1.8 ms−2 along a straight horizontal road. The mass of the car is 1200 kg. A forward force of magnitude 3000 N acts on the car. Find the magnitude of the resistance force that also acts on the car.( 840 N) 17. A van has mass 2500 kg. A forward force of 5000 N acts on the van and a resistance force of 2000 N also acts. Find the acceleration of the van on a horizontal surface(1.2 ms-2 ) 18. A van, of mass 1200 kg, rolls down a slope, inclined at 3 to the horizontal and experiences a resistance force of magnitude 400N. Find the acceleration of the car.( 0.180 ms-2 (to 3 sf) ) 19) A package with mass 300 kg is lifted vertically upwards. Find the tension in the cable which lifts the package, when the package, a) accelerates upwards at 0.1 ms−2, b) accelerates downwards at 0.2 ms−2, c) travels upwards with a retardation of 0.1 ms−2.

20) A car of mass 1 tonne travels along a horizontal road and brakes from 50 ms−1 to rest in a distance of 300 m. Find the magnitude of the braking force on the car.( 4167 N) 21) A particle of mass 5 kg slides down a smooth plane, inclined at an angle of 30deg to the horizontal. Find the acceleration of the particle down the plane.

(4.9 ms−2 ){ Note this result is true independent of the mass of the particle. The fact that all bodies slide down a smooth plane with the same acceleration was an important discovery in the development of modern mechanics.} 22) A particle of mass 6 kg starts from rest and accelerates uniformly. The resultant force on the particle has magnitude 15 N. Find the time taken to reach a speed of 10 ms−1.

(4 s ) 23) A block of mass 20 kg is pulled across a rough horizontal plane by a string, inclined at 30 to the horizontal. If the tension in the string is 50 N and the acceleration produced is 0.5 ms−2 find the friction force on the block and the coefficient of friction.( 0.195 ) 24) A particle, of mass 12 kg slides down a rough slope inclined at 40 to the horizontal. The coefficient of friction between the particle and the slope is 0.2. Find

the acceleration of the particle.( 4.80 msâ&#x2C6;&#x2019;2 ) 25) A particle of mass 3 kg is pulled across a rough horizontal plane, by a string inclined at 30 to the horizontal by a force of 40 N. The coefficient of friction between the particle and the plane is 0.5. Find the acceleration of the particle.( 9.98 msâ&#x2C6;&#x2019;2 ). \26)

A crate of mass 200 kg is on a horizontal surface. The coefficient of friction between the crate and the surface is 0.4. The crate is pulled by a rope, as shown in the diagram, so that the crate moves at a constant speed in a straight line. Find the tension in the rope.( 728 N ) 27) A skier, of mass 60 kg, skis down a slope inclined at to the horizontal at a constant speed. If the coefficient of friction between her skis and the slope is 0.1, find the magnitude of the air resistance force acting on her.( 243 N ) 28)A cyclist, of mass 75 kg, freewheels down a slope inclined at to the horizontal at a constant speed. Find the magnitude of the resistance force acting on the cyclist.

29) A helicopter of mass 880 kg is rising vertically at a constant rate. Find the magnitude of the lift force acting on the helicopter. How would your answer change if the helicopter was descending at a constant rate?( 8624 N) 30) Draw a diagram to show the forces acting on a ladder leaning against a wall.(hint: Note it has been assumed that the wall is vertical and the ground horizontal) 31) The coefficient of friction between a crate, of mass 50 kg, and the floor is 0.7.

Find the magnitude of the friction force acting on the crate if it is on a horizontal surface and;(for the maximum friction inequality F

R)

a) no horizontal force acts on the crate, b) a horizontal force of 80 N acts on the crate, c) A horizontal force of 400 N acts on the crate.

32) The diagram shows a block that is at rest on a horizontal plane. Two light ropes, attached to the block pass over pulleys and are also attached to masses as shown in the diagram. Find the tension in each rope and the friction between the block and the plane and an inequality that the coefficient of friction must satisfy.(hint:divide the system into 3 sub-systems)( T1=78.4 N; T2=117.6 N; F=39.2 N ; friction inequlity F R gives, 39.2 196 19639.2=0.2 )

33) Two boxes are placed on top of each other on a horizontal surface as shown in the diagram. The top box has mass 5 kg and the lower box has mass 10 kg. Draw diagrams to show the forces acting on each box and calculate the magnitude of each of these forces.

34) A 20 kg box is pushed up a rough sloping ramp (30o relative to the incline) with a force of 300 N, parallel to the horizontal with constant speed. Calculate (a) the normal force on the block, (319.7 N) (b) the friction force on the block(161.8 N) (c) the friction coefficient on the block. (0.51) 35) A 10-kg block is set moving with an initial speed of 6 m/s on a rough horizontal surface. If the force of friction is 20 N, approximately how far does the block travel before it stops? (9 m)

UNIT 11: WORK AND ENERGY INTRODUCTION Work: In physics, work is defined as a force causing the movement—or displacement—of an object. In the case of a constant force, work is the scalar product of the force acting on an object and the displacement caused by that force.

Though both force and displacement are vector quantities, work has no direction due to the nature of a scalar product (or dot product) in vector mathematics. This definition is consistent with the proper definition because a constant force integrates to merely the product of the force and distance.

Read on to learn some real-life examples of work as well as how to calculate the amount of work being performed.

Examples of Work There are many examples of work in everyday life. Few notes : a horse pulling a plow through the field; a father pushing a grocery cart down the aisle of a grocery store; a student lifting a backpack full of books upon her shoulder; a weightlifter lifting a barbell above his head; and an Olympian launching the shot-put.

In general, for work to occur, a force has to be exerted on an object causing it to move. So, a frustrated person pushing against a wall, only to exhaust himself, is not doing any work because the wall does not move. But, a book falling off a table and hitting the ground would be considered work, at least in terms of physics, because a force (gravity) acts on the book causing it to be displaced in a downward direction.

What's Not Work Interestingly, a waiter carrying a tray high above his head, supported by one arm, as he walks at a steady pace across a room, might think he's working hard. (He might even be perspiring.) But, by definition, he is not doing any work. True, the waiter is using force to push the tray above his head, and also true, the tray is moving across the room as the waiter walks. But, the force—the waiter's lifting of the tray—does not cause the tray to move. "To cause a displacement, there must be a component of force in the direction of the displacement," notes The Physics Classroom.

Calculating Work The basic calculation of work is actually quite simple: W = Fd Here, "W" stands for work, "F" is the force, and "d" represents displacement (or the distance the object travels). Example problem: A baseball player throws a ball with a force of 10 Newtons. The ball travels 20 meters. What is the total work?

To solve it, you first need to know that a Newton is defined as the force necessary to provide a mass of 1 kilogram (2.2 pounds) with an acceleration of 1 meter (1.1 yards) per second. A Newton is generally abbreviated as "N." So, use the formula: W = Fd Thus: W = 10 N .20 meters (where the symbol "." represents times) So: Work = 200 joules A joule, a term used in physics, is equal to the kinetic energy of 1 kilogram moving at 1 meter per second.

NB:Alpha is the angle between the constant force and the displacement/direction of the displacement.The angle is measured anti-clockwise from the displacement.

WORK DONE BY THE FRICTION FORCE: Friction: It is the force which opposes the motion.

It is always applied against the motion.

And Work done is given by the formula W=F.s (vector) Since the direction of the force and the displacement suffered by the body is in opposite direction, W=F.s or W=Fs cos becomes negative.If we consider f as frictional force (ÂľN), then, it is and since they both are opposite (180 degree in this representation, cos = -1 and work becomes negative.)

NB:No work: If the net force is perpendicular to the motion then no work is

done. If you push on an object and it doesn’t move, then no work is done. If an object’s kinetic energy doesn’t change, then no work is done.

• Work done by gravity(W g ): • If the object is moving down-ward Wg =+mgh • If the object is moving up-ward Wg=-mgh m= Mass ;g=Gravity;h=Height 11.1 Kinetic Energy and the Work-Energy Theorem: Kinetic Energy: Kinetic energy is the energy of motion. An object that has motion - whether it is vertical or horizontal motion - has kinetic energy. There are many forms of kinetic energy - vibrational (the energy due to vibrational motion), rotational (the energy due to rotational motion), and translational (the energy due to motion from one location to another). To keep matters simple, we will focus upon translational kinetic energy. The amount of translational kinetic energy (from here on, the phrase kinetic energy will refer to translational kinetic energy) that an object has depends upon two variables: the mass (m) of the object and the speed (v) of the object. The following equation is used to represent the kinetic energy (KE) of an object.

KE = 0.5 • m • v2 where m = mass of object v = speed of object This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. That means that for a twofold increase in speed, the kinetic energy will increase by a factor of four. For a threefold increase in speed, the kinetic energy will increase by a factor of nine. And for a fourfold increase in speed, the kinetic energy will increase by a factor of sixteen. The kinetic energy is dependent upon the square of the speed. As it is often said, an equation is not merely a recipe for algebraic problem solving, but also a guide to thinking about the relationship between quantities.

Kinetic energy is a scalar quantity; it does not have a direction.

Unlike velocity, acceleration, force, and momentum, the kinetic energy of an object is completely described by magnitude alone. Like work and potential energy, the standard metric unit of measurement for kinetic energy is the Joule. As might be implied by the above equation, 1 Joule is equivalent to 1 kg*(m/s)^2.

1 Joule = 1 kg • m2/s2 Examples: Eample 1: 1.Use your understanding of kinetic energy to answer the following questions. Then click the button to Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s.

Answers: KE = 0.5*m*v2 KE = (0.5) * (625 kg) * (18.3 m/s)2 KE = 1.05 x105 Joules 2. If the roller coaster car in the above problem were moving with twice the speed, then what would be its new kinetic energy?

Answer: If the speed is doubled, then the KE is quadrupled. Thus, KE = 4 * (1.04653 x 105 J) = 4.19 x 105 Joules. or KE = 0.5*m*v2 KE = 0.5*625 kg*(36.6 m/s)2 KE = 4.19 x 105 Joules 3. Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12 000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then what is her speed?

Answer: KE = 0.5*m*v2 12 000 J = (0.5) * (40 kg) * v2 300 J = (0.5) * v2 600 J = v2 v = 24.5 m/s 4. A 900-kg compact car moving at 60 mi/hr has approximately 320 000 Joules of kinetic energy. Estimate its new kinetic energy if it is moving at 30 mi/hr. (HINT: use the kinetic energy equation as a "guide to thinking.")

Answer: KE = 80 000 J The KE is directly related to the square of the speed. If the speed is reduced by a factor of 2 (as in from 60 mi/hr to 30 mi/hr) then the KE will be reduced by a factor of 4. Thus, the new KE is (320 000 J)/4 or 80 000 J.

Work-Energy Theorem: Work kinetic energy theorem derivation We consider not the work done on a particle by a single force,but the net work W net done by all the forces that act on the particle.There are two ways to

find the net work.The first is to find the net force, that is, the vector sum of all the forces that act on the particle: Fnet=F1+F2+F3+……..(1) And then treat this net force as a single force in calculating the work according to the equation: We know that a net unbalanced force applied to a particle will change its state of motion by accelerating it,let us say from initial velocity v i to final velocity vf. What is the effect of the work done on the particle by this net unbalanced force?

We first look at the answer to this question in the case of the constant force in one dimension.Under the influence of this force , the particles move from xi to xf , and it accelerates uniformly from vi to vf.The work done is: Wnet=Fnet(xf-xi)=ma(xf –xi) Because the acceleration a is constant,we can use equation: to obtain: That is,the result of the net work on the particle has to bring about a change in the value of the quantity from point i to point f.This quantity is called the kinetic energy k of the particle,with a definition In terms of the kinetic energy k,we can rewrite equation (2) as: Wnet=Kf –Ki=∆k ……….(4) Equation (4) is the mathematical representation of an important result called the work-energy theorem, which in words can be stated as follows: The net work done by the forces acting on a particle is equal to the change in the kinetic energy of the particle.The work-energy theorem is useful, however, for solving problems in which the net work is done on a particle by external forces is easily computed and in which we are interested in finding the particles speed at certain positions.Of even more significance is the work-energy theorem as a starting point for a broad generalization of the concept of energy and how energy can be stored or shared among the parts of a complex system.

General proof of work energy theorem If a non-constant force acts on the object in one dimension, then the work done by the force on the object can find out by using the expression: From Newton’s Second Law of motion, we know that F = ma, and because of the definition of acceleration we can say that

If we multiply both sides by the same thing, we haven’t changed anything, so we multiply by v:

But remember that v = dx/dt:

We rearrange and integrate: F dx = mv dv

Fx = m(½v2) = ½mv2 = Ek But Fx = Work; therefore Work = ΔEk.

This is the expression of kinetic energy for the object under the action of nonconstant force.

What are Limitations of work energy theorem?

We derived the work-energy theorem directly from Newton’s second law, which,in the form in which we have stated it, applies only to particles.Hence the work-energy theorem,as we have presented so far, likewise applies only to particles.We can apply this important theorem to real objects only if those objects behave like particles.Previously,we considered an object to behave like a particle if all parts of the object move in exactly the same way.In the use of work-energy theorem,we can treat an extended object as a particle if the only kind of energy it has is directed kinetic energy.

Consider for example,a test car that is crashed head-on into a heavy, rigid concrete barrier.The directed kinetic energy of the car certainly decreases as the car hits the barrier, crumples up,and comes to rest.However, there are forms of energy other than directed kinetic energy that enter in this situation.There is internal energy associated with the bending and crumpling of the body of the car; some of this internal energy may appear, for instance as an increase in the temperature of the car, and some may be transferred to the surroundings as heat.Note that, even though the barrier may exert a large force on the car during the crash, the force does no work because the point of application of the force on the car does not move.

11.2 Potential Energy: An object can store energy as the result of its position. For example, the heavy ball of a demolition machine is storing energy when it is held at an elevated position. This stored energy of position is referred to as potential energy. Similarly, a drawn bow is able to store energy as the result of its position. When assuming its usual position (i.e., when not drawn), there is no energy stored in the bow. Yet when its position is altered from its usual equilibrium position, the bow is able to store energy by virtue of its position. This stored energy of position is referred to as potential energy. Potential energy is the stored energy of position possessed by an object.

11.3 Gravitational Potential Energy The two examples above illustrate the two forms of potential energy to be discussed in this course - gravitational potential energy and elastic potential energy. Gravitational potential energy is the energy stored in an object as the result of its vertical position or height. The energy is stored as the result of the gravitational attraction of the Earth for the object. The gravitational potential energy of the massive ball of a demolition machine is dependent on two variables - the mass of the ball and the height to which it is raised. There is a direct relation between gravitational potential energy and the mass of an object. More massive objects have greater gravitational potential energy. There is also a direct relation between gravitational potential energy and the height of an object. The higher that an object is elevated, the greater the gravitational potential energy. These relationships are expressed by the following equation: PEgrav = mass • g • height PEgrav = m • g • h In the above equation, m represents the mass of the object, h represents the height of the object and g represents the gravitational field strength (9.8 N/kg on Earth) sometimes referred to as the acceleration of gravity.

To determine the gravitational potential energy of an object, a zero height position must first be arbitrarily assigned.

Typically, the ground is considered to be a position of zero height. But this is merely an arbitrarily assigned position that most people agree upon. Since many of our labs are done on tabletops, it is often customary to assign the tabletop to be the zero height position. Again this is merely arbitrary. If the tabletop is the zero position, then the potential energy of an object is

based upon its height relative to the tabletop. For example, a pendulum bob swinging to and from above the tabletop has a potential energy that can be measured based on its height above the tabletop. By measuring the mass of the bob and the height of the bob above the tabletop, the potential energy of the bob can be determined.

Since the gravitational potential energy of an object is directly proportional to its height above the zero position, a doubling of the height will result in a doubling of the gravitational potential energy. A tripling of the height will result in a tripling of the gravitational potential energy.

Example 1: Use this principle to determine the blanks in the following diagram.

Knowing that the potential energy at the top of the tall platform is 50 J, what is the potential energy at the other positions shown on the stair steps and the incline?

Answers: A: PE = 40 J (since the same mass is elevated to 4/5-ths height of the top stair) B: PE = 30 J (since the same mass is elevated to 3/5-ths height of the top stair) C: PE = 20 J (since the same mass is elevated to 2/5-ths height of the top stair) D: PE = 10 J (since the same mass is elevated to 1/5-ths height of the top stair) E and F: PE = 0 J (since the same mass is at the same zero height position as shown for the bottom stair).

Example 2: A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of the loaded cart is 3.0 kg and the height of the seat top is 0.45 meters, then what is the potential energy of the loaded cart at the height of the seat-top?

Answer: PE = m.g.h PE = (3 kg ) . (9.8 m/s/s) .(0.45 m) PE = 13.2 J

CONSERVATIVE VERSUS NON-CONSERVATIVE FORCES: In physics, it’s important to know the difference between conservative and nonconservative forces. The work a conservative force does on an object is pathindependent; the actual path taken by the object makes no difference. Fifty meters up in the air has the same gravitational potential energy whether you get there by taking the steps or by hopping on a Ferris wheel. That’s different from the force of friction, which dissipates kinetic energy as heat. When friction is involved, the path you take matters — a longer path will dissipate more kinetic energy than a short one. For that reason, friction is a non-conservative force.

For example, suppose you and some buddies arrive at Mt. Newton, a majestic peak that rises h meters into the air. You can take two ways up — the quick way or the scenic route. Your friends drive up the quick route, and you drive up the scenic way, taking time out to have a picnic and to solve a few physics problems. They greet you at the top by saying, “Guess what is our potential energy compared to before is mgh greater.”

“Mine, too,” you say, looking out over the view. You pull out this equation:

This equation basically states that the actual path you take when going vertically from hi to hf doesn’t matter. All that matters is your beginning height compared to your ending height. Because the path taken by the object against gravity doesn’t matter, gravity is a conservative force.

Here’s another way of looking at conservative and non-86ikconservative forces. Say you’re vacationing in the Alps and your hotel is at the top of Mt. Newton. You spend the whole day driving around down to a lake one minute, to the top of a higher peak the next. At the end of the day, you end up back at the same location: your hotel on top of Mt. Newton.

What’s the change in your gravitational potential energy? In other words, how much net work did gravity perform on you during the day? Gravity is a conservative force, so the change in your gravitational potential energy is 0. Because you’ve experienced no net change in your gravitational potential energy, gravity did no net work on you during the day.

The road exerted a normal force on your car as you drove around, but that force was always perpendicular to the road (meaning no force parallel to your motion), so it didn’t do any work, either.

Conservative forces are easier to work with in physics because they don’t “leak” energy as you move around a path ,if you end up in the same place, you have the same amount of energy. If you have to deal with nonconservative forces such as friction, including air friction, the situation is different. If you’re dragging something over a field carpeted with sandpaper, for example, the force of friction does different amounts of work on you depending on your path. A path that’s twice as long will involve twice as much work to overcome friction.

What’s really not being conserved around a track with friction is the total potential and kinetic energy, which taken together is mechanical energy. When friction is involved, the loss in mechanical energy goes into heat energy. You can say that the total amount of energy doesn’t change if you include that heat energy. However, the heat energy dissipates into the environment quickly, so it isn’t recoverable or convertible.

For that and other reasons, physicists often work in terms of mechanical energy.

11.4 MECHANICAL ENERGY(E): What is Mechanical Energy

In physics, mechanical energy (Emech) is the energy associated with the motion and position of an object usually in some force field (e.g. gravitational field). Mechanical energy (and also the thermal energy) can be separated into two categories, transient and stored. Transient energy is energy in motion, that is, energy being transferred from one place to another. Stored energy is the energy contained within a substance or object. Transient mechanical energy is commonly referred to as work. Stored mechanical energy exists in one of two forms: kinetic or potential: • Potential energy. Potential energy, U, is defined as the energy stored in an object subjected to a conservative force. Common types include the gravitational potential energy of an object that depends on its mass and its

•

distance from the center of mass of another object.

Kinetic energy. The kinetic energy KE, is defined as the energy stored in an object because of its motion. It depends on the speed of an object and is the ability of a moving object to do work on other objects when it collides with them.

Fnet=Fconservative+Fnon-conservative(the work done by these forces yields)=Wconservative+Wnon conservative(using the work-energy theorem)=Wconservative+Wnonconservative(using Wconservative=−ΔPE)=−ΔEp+Wnonconservative=ΔEk+ΔEp

An isolated system is one in which no external force causes energy changes. If only conservative forces act on an object and EP is the potential energy function for the total conservative force,then Emech = Ep + KE The total mechanical energy (defined as the sum of its potential and kinetic energies) of a particle being acted on by only conservative forces is constant.

According to the principle of conservation of mechanical energy, the total mechanical energy of a system is conserved i.e., the energy can neither be created nor be destroyed; it can only be internally converted from one form to another, if the forces doing work on the system are conservative in nature.

In order to understand this statement more clearly, let us consider an example of a one-dimensional motion of a system. Here a body, under the action of a conservative force F, gets displaced by Δx, then from the work-energy theorem we can say that the net work done by all the forces acting on a system is equal to the change in the kinetic energy of the system.

Mathematically, ΔKE = F(x) Δx Where, ΔK is the change in kinetic energy of the system.Considering only conservative forces are acting on the system W net = W c.

Thus Wnetc = ΔKE where W net =Wc+Wnc Also, If conservative forces do the work in a system, the system loses potential energy equal to the work done. Hence, W c = -PE. Which implies that, the total kinetic energy and potential energy of a system remains constant if the process involves only conservative forces.If Wnc=0 then, KE + PE = constant KEi+ PE i= KEf + PEf Where, denotes the initial values and f denotes the final values of KE and PE. This law applies only to the extent that the forces are conservative in nature.

Mechanical energy of the system is defined as the total kinetic energy plus the total potential energy. In a system that comprises of only conservative forces, each force is associated with a form of potential energy and the energy only changes between the kinetic energy and different types of potential energy, such that, the total energy remains constant.

Example 1

Let us understand this principle more clearly with the following example. Let us say, a ball of mass m is dropped from a cliff of height H, as shown above.

At height H: Potential energy (PE) = m×g×H Kinetic energy (K.E.) = 0 Total mechanical energy = mgH At height h: Potential energy(PE) = m×g×h Kinetic energy (K.E.) =1/2(mv^2) Using the equations of motion, the velocity v 1 at a height h for an object of mass m falling from a height H can be written as Hence the kinetic energy can be given as,

Total mechanical energy = (mgH – mgh) – mgh = mgH At height zero: Potential energy: 0 Kinetic energy: 1/2(mv^2) Using the equations of motion we can see that velocity v at the bottom of the cliff, just before touching the ground is Hence, the kinetic energy can be given as,

Total mechanical energy: mgH

We saw the total mechanical energy of the system is constant throughout.

POWER: The quantity work has to do with a force causing a displacement. Work has nothing to do with the amount of time that this force acts to cause the displacement. Sometimes, the work is done very quickly and other times the work is done rather slowly. For example, a rock climber takes an abnormally long time to elevate her body up a few meters along the side of a cliff. On the other hand, a trail hiker (who selects the easier path up the mountain) might elevate her body a few meters in a short amount of time.

The two people might do the same amount of work, yet the hiker does the work in considerably less time than the rock climber. The quantity that has to do with the rate at which a certain amount of work is done is known as the power. The hiker has a greater power rating than the rock climber.

Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using the following equation.

Power = Work / time or P=W/t

The standard metric unit of power is the Watt. As is implied by the equation for power, a unit of power is equivalent to a unit of work divided by a unit of time. Thus, a Watt is equivalent to a Joule/second. For historical reasons, the horsepower is occasionally used to describe the power delivered by a machine. One horsepower is equivalent to approximately 750 Watts.

Most machines are designed and built to do work on objects. All machines are typically described by a power rating. The power rating indicates the rate at which that machine can do work upon other objects. Thus, the power of a machine is the work/time ratio for that particular machine. A car engine is an example of a machine that is given a power rating. The power rating relates to how rapidly the car can accelerate the car. Suppose that a 40-horsepower engine could accelerate the car from 0 mi/hr to 60 mi/hr in 16 seconds. If this were the case, then a car with four times the horsepower could do the same amount of work in one-fourth the time. That is, a 160-horsepower engine could accelerate the same car from 0 mi/hr to 60 mi/hr in 4 seconds. The point is that for the same amount of work, power and time are inversely proportional. The power equation suggests that a more powerful engine can do the same amount of work in less time.

A person is also a machine that has a power rating. Some people are more power-full than others. That is, some people are capable of doing the same amount of work in less time or more work in the same amount of time. A common physics lab involves quickly climbing a flight of stairs and using mass, height and time information to determine a student's personal power. Despite the diagonal motion along the staircase, it is often assumed that the horizontal motion is constant and all the force

from the steps is used to elevate the student upward at a constant speed. Thus, the weight of the student is equal to the force that does the work on the student and the height of the staircase is the upward displacement. Suppose that Ben Pumpiniron elevates his 80-kg body up the 2.0-meter stairwell in 1.8 seconds. If this were the case, then we could calculate Ben's power rating. It can be assumed that Ben must apply an 800-Newton downward force upon the stairs to elevate his body. By so doing, the stairs would push upward on Ben's body with just enough force to lift his body up the stairs. It can also be assumed that the angle between the force of the stairs on Ben and Ben's displacement is 0 degrees. With these two approximations, Ben's power rating could be determined as shown below.

BEN's power rating is 871 Watts. He is quite a horse.

Another Formula for Power The expression for power is work/time. And since the expression for work is force*displacement, the expression for power can be rewritten as (force*displacement)/time. Since the expression for velocity is displacement/time, the expression for power can be rewritten once more as force*velocity. This is shown below.

Example1: A tired squirrel (mass of approximately 1 kg) does push-ups by applying a force to elevate its center-of-mass by 5 cm in order to do a mere 0.50 Joule of work.

If the tired squirrel does all this work in 2 seconds, then determine its power.

Answer: The tired squirrel does 0.50 Joule of work in 2.0 seconds. The power rating of this squirrel is found by P = W / t = (0.50 J) / (2.0 s) = 0.25 Watts Example 2:A small object of mass m = 234 g slides along a track with elevated ends and a central flat part, as shown in below figure. The flat part has a length L = 2.16 m.

The curved portions the tracks are frictionless; but in traversing the flat part, the object loses 688 mJ of mechanical energy, due to friction. The object is released at point A, which is a height h = 1.05 m above the flat part of the track. Where does the object finally come to rest?

Concept: The potential energy Ep is defined as, Ep= mgh Here, m is the mass of the body, g is the free fall acceleration and h is the fall of height.

Solution:To find the potential energy of the object at the point A, substitute 234 g for mass of the object m, 9.81 m/s2 for free fall acceleration g and 1.05 m for height h in the equation Ep = mgh, Ep = mgh = (234 g) (9.81 m/s2) (1.05 m) = (234 gĂ&#x2014;10-3 kg/1 g) (9.81 m/s2) (1.05 m) = 2.41 kg. m2/s2 = (2.41 kg. m2/s2) (1 J/1 kg. m2/s2) = 2.41 J The curved portions of the track are frictionless; but in traversing the flat part, the object losses 688 mJ of mechanical energy, due to friction.

The number of times (n) that the particle will move back and forth across the flat portion is the ratio of the potential energy (2.41 J) of the object at the point A to the mechanical energy (688 mJ)that the object losses due to friction.

So, n = 2.41 J/688 mJ = (2.41 J/688 mJ) (1 mJ/10-3 J) = 3.50 The number of times n signifies that, the particle will come to a rest at the center of the flat part while attempting one last right to left journey

TUTORIAL: 1.Consider the situation shown where a football player slides to a stop on level ground. Using energy considerations, calculate the distance the 65,0kg football player slides, given that his initial speed is 6,00m·s^−1 and the force of friction against him is a constant 450N.(2,60m)

2. As a 100 g hockey puck moves across the ice, its speed drops from 20 m/s to 18 m/s. (a) What is the change in kinetic energy?(-3,8J) (b) What work is done on the puck? (-3,8J) 3. A person pushes a 10 kg crate 3 m up a 30° incline with a 80-N force parallel to the incline. The frictional force is 22 N. Find the work done (a) by the person;(240J) (b) by gravity;(-150J) (c) by friction;(-66J) (d) by the normal force.(0J) (e) What is the change of the kinetic energy of the box?(24J) 4. A 1.8 kg block is moved at constant speed over a surface for which UK= 0.25.

The displacement is 2 m. It is pulled by a force directed at 45° to the horizontal as shown. Find the work done on the block by

(a) the force

;(7,20J)

(b) by friction;(-7,20J) (c) by gravity.

5. What is the work needed to lift 15 kg of water from a well 12 m deep? Assume the water has a constant upward acceleration of 0.7 m/s2 .(1926J) 6. A man pushes a lawnmower 10 m. There is a resistive force of 10 N acting on the lawnmower as shown in the diagram below. (a) Find the work done by each force acting on the lawnmower, then the total work done.(652J) (b) Find the net force acting on the lawnmower and then the work done by the net force.(652J)

7. Given the following force-position graph for a 200 g particle, (a) find the work done in going from 2 m to 10 m.( work done is zero) (b) If the particle has a velocity of 2 m/s at 2 m, what is its velocity at 10 m?( no change in velocity)

8. Given the following force-position graph for a 2 kg particle, (a) find the work done in going from 0 m to 10 m.(30J) (b) If the particle has a velocity of 2 m/s at 0 m, what is its velocity at 10 m?(5,83m/s) .

9. Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12 000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then what is her speed?

10. A 900-kg compact car moving at 60 mi/hr has approximately 320 000 Joules of kinetic energy. Estimate its new kinetic energy if it is moving at 30 mi/hr. (HINT: use the kinetic energy equation as a "guide to thinking.")

11. Two physics students, FREDDY and ALFRED, are in the weightlifting room. Will lifts the 100-pound barbell over his head 10 times in one minute; Ben lifts the 100pound barbell over his head 10 times in 10 seconds. Which student does the most work? ______________ Which student delivers the most power? ______________ Explain your answers.

12. During a physics lab, Jack and Jill ran up a hill. Jack is twice as massive as Jill; yet Jill ascends the same distance in half the time. Who did the most work? ______________ Who delivered the most power? ______________ Explain your answers.

13. A tired squirrel (mass of approximately 1 kg) does push-ups by applying a force to elevate its center-of-mass by 5 cm in order to do a mere 0.50 Joule of work. If the tired squirrel does all this work in 2 seconds, then determine its power.( 0.25 Watts)

14. When doing a chin-up, a physics student lifts her 42.0-kg body a distance of 0.25 meters in 2 seconds. What is the power delivered by the student's biceps?( 51.5 Watts)

.

15. An escalator is used to move 20 passengers every minute from the first floor of a department store to the second. The second floor is located 5.20 meters above the first floor. The average passenger's mass is 54.9 kg. Determine the power requirement of the escalator in order to move this number of passengers in this amount of time.( 933 W) 16. A 10.0-g bullet traveling horizontally at 755 m/s strikes a stationary target and stops after penetrating 14.5 cm into the target. What is the average force of the target on the bullet? (1.97 x 104 N) 17. Larry's gravitational potential energy is 1870 J as he sits 2.20 m above the ground in a sky diving airplane. What is Larry's gravitational potential energy when be begins to jump from the airplane at an altitude of 923 m? (784 550 J) 18. The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determine the magnitude and the direction of the non-conservative force acting on the box as it slides. (59.375 N, east) 19. A 9.0-kg box of oranges slides from rest down a frictionless incline from a height of 5.0 m. A constant frictional force, introduced at point A, brings the block to rest at point B, 19 m to the right of point A. Calculate: (a) The speed of the block just before it reaches point A? (b) The coefficient of kinetic friction, μk, of the surface from A to B? (9.9 m/s, 0.263)

20. Maman ELISE has made a long gun. The bullet has a mass of 10 grams. A force of 3 Newtons pushes the bullet down the barrel, but the barrel has a frictional force of 1 Newton in the opposite direction. The barrel is 5 meters long and the bullet starts off at rest. What is the final speed of the bullet as it leaves the barrel?( 44.7 m/s) 21. Djomo Ariane Nina is sitting on top of a big slide. The slide has a vertical drop of 10 meters. If there is no friction between her and the slide, what is her speed at the bottom of the slide?

22. An absent-minded professor does the ”pendulum-of-death” demonstration. He takes a bowling ball of mass 3 Kg and ties it to a long rope. For the demonstration he is supposed to let the ball go at his nose with no speed. The ball swings across the room and returns stopping at his nose. However, he forgets and gives the ball an initial speed of 1 m/s. He is lucky however. There is a lot of air friction and the ball still returns just in front of his nose. How much work was done by air friction?( −1.5 Joules) 23. Dora wants to determine the coefficient of friction between a block and a particular surface. She places a block of mass m on an inclined plane. the plane makes an angle of 37◦ with respect to the horizontal. The block starts from rest a

distance of 10 meters up the incline. she lets the block go, and it slides down the plane and along a table top. The block comes to rest 10 meters from the start of the incline. The coefficient of friction is the same on the incline and the tabletop. What is the coefficient of friction?( 1/3) 24. Alan pushes on a 10 Kg block on a frictionless surface in the ”x” direction. The block starts from rest at x = 0, and Alan pushes with a force given by Fx = 6x Newtons, where x is in meters. From this formula, one can see that Alan’s force increases linearly with the distance that the block is from the origin. He keeps applying this force for 2 meters, i.e. till x = 2. What is the speed of the block at x = 2 meters?(hint:use the integral to find the net work done)( 1.55 m/s)

UNIT 12: IMPULSE AND MOMENTUM INTRODUCTION 12.1 Momentum An object which is moving has momentum. The amount of momentum (p) possessed by the moving object is the product of mass (m) and velocity (v). In equation form: p=m•v An equation such as the one above can be treated as a sort of recipe for problemsolving. Knowing the numerical values of all but one of the quantities in the equations allows one to calculate the final quantity in the equation. An equation can also be treated as a statement which describes qualitatively how one variable depends upon another. Two quantities in an equation could be thought of as being either directly proportional or inversely proportional. Momentum is directly proportional to both mass and velocity. A two-fold or three-fold increase in the mass (with the velocity held constant) will result in a two-fold or a three-fold increase in the amount of momentum possessed by the object. Similarly, a two-fold or three-fold increase in the velocity (with the mass held constant) will result in a two-fold or a three-fold increase in the amount of momentum possessed by the object. Thinking and reasoning proportionally about quantities allows you to predict how an alteration in one variable would effect another variable.

Impulse-Momentum Change Equation In a collision, a force acts upon an object for a given amount of time to change the object's velocity. The product of force and time is known as impulse. The product of mass and velocity change is known as momentum change. In a collision the impulse encountered by an object is equal to the momentum change it experiences.

Impulse = Momentum Change F • t = mass •

v

Several problems in this set of problems test your understanding of the above relationship. In many of these problems, a piece of extraneous information is provided. Without an understanding of the above relationships, you will be tempted to force such information into your calculations. Physics is about conceptual ideas and relationships; and problems test your mathematical understanding of these relationships. If you treat this problem set as a mere exercise in the algebraic manipulation of physics equations, then you are likely to become frustrated quickly.

As you proceed through this problem set, be concepts-minded. Do not strip physics of its conceptual meaning.

Rebounding Several of the problems in this set of problems demand that you be able to calculate the velocity change of an object. This calculation becomes particularly challenging when the collision involves a rebounding effect - that is, the object is moving in one direction before the collision and in the opposite direction after the collision. Velocity

is a vector and is distinguished from speed in that it has a direction associated with it.

This direction is often expressed in mathematics as a + or - sign. In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before. Mathematically, the before-collision velocity would be + and the after-collision velocity would be - in sign. Ignoring this principle will result in great difficulty when analyzing any collision involving the rebounding of an object.

The Momentum Conservation Principle In a collision between two objects, each object is interacting with the other object.

The interaction involves a force acting between the objects for some amount of time.

This force and time constitutes an impulse and the impulse changes the momentum of each object. Such a collision is governed by Newton's laws of motion; and as such, the laws of motion can be applied to the analysis of the collision (or explosion) situation. So with confidence it can be stated that ...

In a collision between object 1 and object 2, the force exerted on object 1 (F 1) is equal in magnitude and opposite in direction to the force exerted on object 2 (F 2). In equation form: F1 = - F2 The above statement is simply an application of Newton's third law of motion to the collision between objects 1 and 2. Now in any given interaction, the forces which are exerted upon an object act for the same amount of time. You can't contact another object and not be contacted yourself (by that object). And the duration of time during which you contact the object is the same as the duration of time during which that object contacts you. Touch a wall for 2.0 seconds, and the wall touches you for 2.0 seconds. Such a contact interaction is mutual; you touch the wall and the wall touches you. It's a two-way interaction - a mutual interaction; not a one-way interaction. Thus, it is simply logical to state that in a collision between object 1 and object 2, the time during which the force acts upon object 1 (t 1) is equal to the time during which the force acts upon object 2 (t 2). In equation form: t1 = t2 The basis for the above statement is simply logic. Now we have two equations which relate the forces exerted upon individual objects involved in a collision and the times over which these forces occur. It is accepted mathematical logic to state the following: If A = - B and C = D then A • C = - B • D The above logic is fundamental to mathematics and can be used here to analyze our collision.

If F1 = - F2 and t1 = t2 then F1 • t1 = - F2 • t2

The above equation states that in a collision between object 1 and object 2, the impulse experienced by object 1 (F1 • t1) is equal in magnitude and opposite in direction to the impulse experienced by object 2 (F 2 • t2). Objects encountering impulses in collisions will experience a momentum change. The momentum change is equal to the impulse. Thus, if the impulse encountered by object 1 is equal in magnitude and opposite in direction to the impulse experienced by object 2, then the same can be said of the two objects' momentum changes. The momentum change experienced by object 1 (m1 • Delta v1) is equal in magnitude and opposite in direction to the momentum change experienced by object 2 (m2 • Delta v2). This statement could be written in equation form as m1 •

v1 = - m2 •

v2

This equation claims that in a collision, one object gains momentum and the other object loses momentum. The amount of momentum gained by one object is equal to the amount of momentum lost by the other object. The total amount of momentum possessed by the two objects does not change. Momentum is simply transferred from one object to the other object.

Put another way, it could be said that when a collision occurs between two objects in an isolated system, the sum of the momentum of the two objects before the collision is equal to the sum of the momentum of the two objects after the collision. If the system is indeed isolated from external forces, then the only forces contributing to the momentum change of the objects are the interaction forces between the objects. As such, the momentum lost by one object is gained by the other object and the total system momentum is conserved. And so the sum of the momentum of object 1 and the momentum of object 2 before the collision is equal to the sum of the momentum of object 1 and the momentum of object 2 after the collision. The following mathematical equation is often used to express the above principle. m1 • v1 + m2 • v2 = m1 • v1' + m2 • v2' The symbols m1 and m2 in the above equation represent the mass of objects 1 and 2.

The symbols v1and v2 in the above equation represent the velocities of objects 1 and 2 before the collision. And the symbols v1' and v2' in the above equation represent the velocities of objects 1 and 2 after the collision. (Note that a ' symbol is used to indicate after the collision.)

12.2 Direction Matters Momentum is a vector quantity; it is fully described by both a magnitude (numerical value) and a direction. The direction of the momentum vector is always in the same direction as the velocity vector. Because momentum is a vector, the addition of two momentum vectors is conducted in the same manner by which any two vectors are added. For situations in which the two vectors are in opposite directions, one vector is considered negative and the other positive. Successful solutions to many of the problems in this set of problems demands that attention be given to the vector nature of momentum.

Example 1: Two-Dimensional Collision Problems A two-dimensional collision is a collision in which the two objects are not originally

moving along the same line of motion. They could be initially moving at right angles to one another or at least at some angle (other than 0 degrees and 180 degrees) relative to one another. In such cases, vector principles must be combined with momentum conservation principles in order to analyze the collision. The underlying principle of such collisions is that both the "x" and the "y" momentum are conserved in the collision. The analysis involves determining pre-collision momentum for both the x- and the y- directions. If inelastic, then the total amount of system momentum before the collision (and after) can be determined by using the Pythagorean theorem.

Since the two colliding objects travel together in the same direction after the collision, the total momentum is simply the total mass of the objects multiplied by their velocity.

12.3 Momentum and Vector Components Momentum is "mass in motion", or a measure of how much motion an object has.

Algebraically, momentum is defined as the product of an object's mass and velocity, . Momentum is a vector - the direction of the momentum matters.

Any vector can be resolved into components. Generally, we resolve vectors into horizontal (x) and vertical (y) components. The diagram to the right shows a vector, v, resolved into its x and y components.

Using the diagram to the right and some basic trigonometric identities, answer the following questions.

1. What is the expression that gives vx as a function of v and θ?

2. What is the expression that gives vy as a function of v and θ?

3. Write a general expression for v, if both vx and vy are known.

4. If both vx and vy are known, what is the general expression for θ?

12.4 Velocity of the Centre of Mass 5. What is the basic definition for location of the cm of mass (express suing vector notion)?

6. What is the mathematical relation between velocity and position? Use this to show that the velocity of the centre of mass of a two particle system is given

by the equation

12.5 2D Collisions and Conservation of Momentum Analyzing Collisions In the previous section you discovered that the total momentum of a system is conserved, as long as there are no external forces acting on a system. Let's use the conservation of momentum to analyze the following collisions. When solving multistep questions, it is useful to follow a four-step method: • •

• •

Plan-it - make sure you understand the question. It helps to draw a diagram and predict what you think will happen.

Set-it-Up - list all the known and unknown information and determine which laws or equations to use. By rearranging equations, develop expressions for the unknown(s).

Solve-it - substitute in known values and solve for the unknown(s) Think-About-it - think about your answer. Does it make sense? If possible, check you answer with the applet.

Let's do an example question together: Example 1: An 8.0 kg mass collides elastically with a 5.0 kg mass that is at rest. Initially, the 8.0 kg mass was travelling to the right at 4.5 m/s. After the collision, it is moving with a speed of 3.65 m/s and at an angle of 27° to its original direction. What is the final speed and direction of motion for the 5.0 kg mass?

Solution: 1. Plan-it: Initially, the 8.0 kg mass (mass 1) moves to the right and the 5.0 kg mass (mass 2) is at rest. Thus, the total momentum is in the x-direction (to the right). After the collision, mass 1 travels at an angle of 27° to its original direction of motion - it now has momentum in the y-direction and the xdirection. For momentum to be conserved, mass 2 must have momentum in the x and y directions.

2. Set-it-up: Motion to the right is positive. Following convention, angles are measured counterclockwise, from the horizontal. Our unknown information is the final velocity of mass 2 and its direction of motion, . Furthermore, since this is a 2D problem, we should include the x and y components of the velocities:

We will use the Law of Conservation of Momentum to solve this question.

Since momentum is conserved in both the x and y directions, we can set up two sets of equations, one to solve for v2fxand v2fy: Conservation of momentum in the xdirection:

Conservation of momentum in the y-direction:

3.

4. Solve-it: Now we substitute known values into equations 1 and 2: Mass 2's final velocity in the x-direction:

Mass 2's final velocity in the y-direction:

5.

Once we know the final velocity in the x and y directions, we can easily determine the final velocity and direction of motion for mass 2. Remember, do not round off during your calculations - only at the end.

Mass 2's final velocity:

Direction of motion:

6.

7. Think-About-It: Our answer seems to make sense - we know that mass 2 has to move to the right and down in order to conserve momentum.

Furthermore, this answer can be verified with the applet (check it yourself!).

Now it's your turn! Solve each question using the four-step method. Use the applet to help visualize the collisions and also to check your answers. Remember that to set the scatter angle, you must actually adjust the impact parameter (while viewing the data box, adjust the impact parameter until the scatter angle is correct).

Summary In this lesson you looked at collision that occur in two-dimensions. You saw that the total momentum of a system is conserved during a collision. Moreover, the total momentum in the x-direction and in the y-direction is also conserved during a collision. The key points you looked at are: â&#x20AC;˘ â&#x20AC;˘

Momentum is a vector and can be resolved into components During a collision, the total momentum for the components is conserved:

â&#x20AC;˘

Some energy can be lost in the collision but, in an isolated frame, the CM energy MUST be conserved!

Example 2:

Question: Two objects slide over a frictionless horizontal surface. The first object, mass m1 =15kg , is propelled with speed toward the second object, mass m2 =2,5kg , which is initially at rest. After the collision, both objects have velocities which are directed on either side of the original line of motion of the first object. What are the final speeds of the two objects? Is the collision elastic or inelastic?

Answer: Let us adopt the coordinate system shown in the diagram. Conservation of momentum along the -axis yields

Likewise, conservation of momentum along the

-axis yields

The above pair of equations can be combined to give

and

The initial kinetic energy of the system is

The final kinetic energy of the system is

Since Ki =Kf , the collision is elastic.Remember K is the kinetic energy.

Example 3: Question: An object of mass

=2kg, moving with velocity

on with a stationary object whose mass is

=12m/s, collides head-

=6kg . Given that the collision is elastic,

what are the final velocities of the two objects. Neglect friction.

Answer: Momentum conservation yields

Where

and

are the final velocities of the first and second objects,

respectively. Since the collision is elastic, the total kinetic energy must be the same before and after the collision. Hence,

Let

and

. Noting that

, the above two equations reduce to

and

Example 4: Question: A bullet of mass m=12g strikes a stationary wooden block of mass M=5,2kg standing on a frictionless surface. The block, with the bullet embedded in it, acquires a velocity of . What was the velocity of the bullet before it struck the block? What fraction of the bullet's initial kinetic energy is lost (i.e., dissipated) due to the collision with the block?

Answer: Let be the initial velocity of the bullet. Momentum conservation requires the total horizontal momentum of the system to be the same before and after the bullet strikes the block. The initial momentum of the system is simply , since the block is initially at rest. The final momentum is (M+m) , since both the block and the bullet end up moving with velocity

. Hence,

giving

The initial kinetic energy of the bullet is

The final kinetic energy of the system is

Hence, the fraction of the initial kinetic energy which is dissipated is

Eliminating

between the previous two expressions, we obtain

or

which has the non-trivial solution The corresponding solution for

. is

It follows that the final velocity of the first object is

.

The minus sign indicates that this object reverses direction as a result of the collision.

Likewise, the final velocity of the second object is

Example 5: Question: A skater of mass M=120kg is skating across a pond with uniform velocity . One of the skater's friends, who is standing at the edge of the pond, throws a medicine ball of mass m=20kg with velocity u=3m/s to the skater, who catches it. The direction of motion of the ball is perpendicular to the initial direction of motion of the skater. What is the final speed of the skater? What is the final direction of motion of the skater relative to his/her initial direction of motion?

Assume that the skater moves without friction.

Answer: Suppose that the skater is initially moving along the -axis, whereas the initial direction of motion of the medicine ball is along the -axis. The skater's initial momentum is

Likewise, the initial momentum of the medicine ball is

After the skater catches the ball, the combined momentum of the skater and the ball is

This follows from momentum conservation. The final speed of the skater (and the ball) is

The final direction of motion of the skater is parameterized by the angle above diagram), where

(see the

Example 6: Question: A softball of mass m=0,35kg is pitched at a speed of batter hits the ball directly back to the pitcher at a speed of

. The . The bat acts

on the ball for . What impulse is imparted by the bat to the ball? What average force is exerted by the bat on the ball?

Answer: The initial momentum of the softball is

, whereas its final momentum

is . Here, the final direction of motion of the softball is taken to be positive. Thus, the net change in momentum of the softball due to its collision with the bat is

By definition, the net momentum change is equal to the impulse imparted by the bat,

so

The average force exerted by the bat on the ball is simply the net impulse divided by the time interval over which the ball is in contact with the bat. Hence,

Example7: Question: A cannon is bolted to the floor of a railway carriage, which is free to move without friction along a straight track. The combined mass of the cannon and the carriage is M=1200kg . The cannon fires a cannonball, of mass , horizontally with velocity

. The cannonball travels the length of the carriage, a

distance , and then becomes embedded in the carriage's end wall. What is the recoil speed of the carriage right after the cannon is fired? What is the velocity of the carriage after the cannonball strikes the far wall? What net distance, and in what direction, does the carriage move as a result of the firing of the cannon?

Answer: Conservation of momentum implies that the net horizontal momentum of the system is the same before and after the cannon is fired. The momentum before the cannon is fired is zero, since nothing is initially moving. Hence, we can also set the momentum after the cannon is fired to zero, giving

where

is the recoil velocity of the carriage. It follows that

The minus sign indicates that the recoil velocity of the carriage is in the opposite direction to the direction of motion of the cannonball. Hence, the recoil speed of the carriage is .

Suppose that, after the cannonball strikes the far wall of the carriage, both the cannonball and the carriage move with common velocity . Conservation of momentum implies that the net horizontal momentum of the system is the same before and after the collision. Hence, we can write

However, we have already seen that M

. It follows that

: in other

words, the carriage is brought to a complete halt when the cannonball strikes its far wall.In the frame of reference of the carriage, the cannonball moves with velocity

after the cannon is fired. Hence, the time of flight of the cannonball is

The distance moved by the carriage in this time interval is

TUTORIAL: 1. An object travels with a velocity 4m/s to the east. Then, its direction of motion and magnitude of velocity are changed. Picture given below shows the directions and magnitudes of velocities. Find the impulse given to this object.( -21kg.m/s) 2. Ball having mass 4kg and velocity 8m/s travels to the east. Impulse given at point O, makes it change direction to north with velocity 6m/s.

Find the given impulse and change in the momentum.( 40kg.m/s)

3. A ball having mass 500g hits wall with a10m/s velocity. Wall applies 4000 N force to the ball and it turns back with 8m/s velocity. Find the time of ball-wall contact.( 0,00025s) 4.A 50 kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s. a) What is the initial momentum of the mass?( 0 kg⋅m/s) b) What is the final momentum of the mass?( 150 kg⋅m/s) c) What was the force acting on the mass?( 75 kg⋅m/s2 ) d) What was the impulse acting on the mass?( 150 Ns or 150 kg⋅m/s)

4. A 325 kg motorcycle is moving at 140 km/h, south. a) Find its momentum.(12,6.103 kg.m/s;south) b) At what velocity is the momentum of a 1754 kg car equal to that of the motorcycle?(25,9km/h;south)

5. The net force on a body varies with time according to Fx=3.0t+0.5t^2 where Fx is in Newtons and t is in seconds. What is the impulse imparted to the body during the time interval 0<t<3.0s?

6. Calculate the momentum of the Titanic (m = 4.2 x 10 7 kg) moving at 14

knots (1 kt = 1.852 km/h).

7.If the momentum of the NASA space shuttle as it leaves the atmosphere is 3.75 x 108 kgâ&#x20AC;˘m/s and its mass is 75000 kg, what is its speed?

8.A 59 kg physics student jumps off the back of her Laser sailboat (42 kg). After she jumps, the Laser is found to be travelling at 1.5 m/s. What is the speed of the physics student?

9.The same physics student jumps off the back of her Laser again, but this time the Laser is already travelling at 3.1 m/s before she jumps. If the physics student jumps off with a speed of 2.1 m/s, how fast is the Laser going after she jumps?

10.The Titanic hit an iceberg estimated to be half of her mass. Before hitting the iceberg, the Titanic was estimated to be going 22 kts (11.3 m/s). After hitting the iceberg, the Titanic was estimated to be going about 6.0 knots (3.1 m/s). How fast was the iceberg going after the collision? (Assume a head-on collision) (Hint: first find the momentum of the Titanic before the collision. Then write the expression for the total momentum after the collision, which is the sum of the momentum of the Titanic and the momentum of the iceberg. Make the expression equal to the momentum before and solve.)

11.A 59 kg physics student is riding her 220 kg Harley at 12 m/s when she has a head-on collision with a 2.1 kg pigeon flying the opposite direction at 44 m/s. The bird is still on the motorcycle after the collision. How fast is the motorcycle going after the collision?

12.The Millersburg Ferry (m = 13000.0 kg loaded) is travelling at 11 m/s when the engines are put in reverse. The engine produces a force of 1.0 x 10 6 N for a period of 20.0 seconds. a)

What is the magnitude of the impulse from the engine?

b)

What is the new momentum of the ferry?

c)

What is the new velocity of the ferry?

13.

The Millersburg Ferry (m = 13000.0 kg loaded) puts its engines in full reverse and stops in 65 seconds. If the speed before braking was 2.0 m/s, what was the braking force supplied by the propellers? Solve using impulse and momentum.

14. A 1.2 x 103 kg car traveling 33 m/s collides head-on with a 1.8 x 103 car traveling 22 m/s in the opposite direction. If the cars stick together, what is the velocity of the wreckage immediately after impact?

15. A 6.0 x 103 kg railway car is coasting along the track at 7.0 m/s. Suddenly a 2.0 x 103 kg load of coal is dumped into the car. What is its new velocity?

16. A bullet of mass m is fired into a steel wall with a velocity v, and then rebounds with -1/2v velocity. A second bullet of the same mass m is fired into a wall covered with Plasticine, and it does not rebound. Which bullet exerts a greater force on the wall: the one that rebounds or the one that does not? Show your reasoning!

17. a) What impulse must be imparted to a 100.0 g baseball to change its velocity from 40.0 m/s south to 50.0 m/s north?

18. To measure the speed of a bullet, a physicist fired the 45 g bullet into a large block of Plasticine that rested on a metal disk floating on a large air table. The combined mass of the Plasticine and disk was 18.0 kg. By using strobe photography, the speed of the disk was measured to be 0.900 m/s immediately after the impact of the bullet. What was the speed of the bullet?

19. . A 2.0 x 103 kg car traveling 15 m/s â&#x20AC;&#x153;rear endsâ&#x20AC;? another car of mass 1.0 x 10 3 kg.

The second car was initially moving 6.0 m/s in the same direction as the first car.

What is the common velocity of the two cars after the collision if they lock together during impact?

20. Determine the impulse imparted during the interaction represented in each graph below.

21. A car of mass 3000 kg moving due South at 15 m/s collides completely inelastic (i.e. stick together) with a second car of mass 2000 kg moving West at 12 m/s. a) Draw a vector diagram representing the total momentum before and after the collision. b) What is the total momentum after the collision (both magnitude and direction)? c) What is the velocity of the combined vehicles after the collision?

22. A car moving North at 12 m/s strikes a stationary car of equal mass. The first car moves off after the collision at an angle of 30 East of North with a speed of 8.0 m/s. a) What is the velocity of the struck car just after the collision? b) Show that the collision is inelastic. c) Explain how dents, skid marks etc. show that kinetic energy has been lost.

23.A steel ball of mass 10 kg moves due East at 5.0 m/s. It collides with a rubber ball of mass 5.0 kg moving at 10 m/s due North. After the collision the steel ball moves at an angle of 60 24.A child’s ball, of mass 250 g, rolls due East along the ground at 4.2 m/s towards a stationary bowling ball, of mass 3.2 kg. After the collision, the bowling ball travels in a direction 32 South of East at 38.7 cm/s. a) What is the speed of the child’s ball after the collision? b) What is the direction in which the child’s ball travels? c) What fraction of the original kinetic energy is transformed into heat?

25.A bowling ball, of mass 4.00 kg, strikes a bowling pin of mass 110.0 g obliquely.

Before the collision, the ball was traveling at 3.10 m/s down the alley; after the collision it travels at 2.99 m/s and its direction has changed by 1.3. a) Calculate the speed of the pin immediately after the collision. b) In what direction does the pin travel immediately after the collision? c) What fraction of the ball’s kinetic energy is transferred to the pin?

26. A 10.0-g bullet traveling horizontally at 755 m/s strikes a stationary target and stops after penetrating 14.5 cm into the target. What is the average force of the target on the bullet? (1.97 x 104 N) 27. A 0.5 kg ball is dropped from rest at a point 1.2 m above the floor. The ball rebounds straight upward to a height of 0.7 m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?

(0.578 N.s upward) 28. A person stands in a stationary canoe and throws a 5-kg stone with a velocity of 8 m.s-1 at an angle of 30 deg above the horizontal. The person and canoe have a combined mass of 105 kg. Ignoring all resistive effects, find the recoil velocity of the canoe. (0.33 m/s) 29. A hockey puck B rests on a smooth ice surface and is truck by a second puck, A, which was originally traveling at 30 m.s-1 and which is deflected 30o from its original

direction. Puck B acquires a velocity at 45o relative to the original velocity of A. The two pucks have equal mass. Write down the equations to determine the unknown velocities after the collision. (0 + 30 =v1cos30+ v2cos 45, 0 =v1sin30+ v2sin 45) 30. A boy weighs 490N runs at a speed of 10 m/s and jumps onto a cart as shown in the figure. The cart is initially at rest. If the speed of the cart and the boy on it is 2.5 m/s, what is the mass of the cart? (m = 150 kg)

t u b e o p e n a t o n e e n d .

UNIT 13:ELECTRIC FORCES AND ELECTRIC FIELDS INTRODUCTION 13.1 ORIGIN OF THE ELECTRICITY Electricity is a kind of energy that can only be valued by the effects it gives. It is a fundamental part of nature and it is one of the commonly used forms of energy. This word comes from the Greek word elektron which means amber.

Earlier, electricity generation began over 100 years ago, houses were lit with kerosene lamps, food was chilled in iceboxes, and rooms were warmed by wood-burning or coal-burning stoves. It was started with Benjamin Franklin's experiment with a kite one stormy night in Philadelphia, the idea of electricity slowly became implicit.

In the mid-1800s, there is a life changed with the invention of the electric light bulb. In the year 1879, electricity had been used in arc lights for outdoor lighting. The light bulbâ&#x20AC;&#x2122;s invention used electricity to bring indoor lighting to our homes.

Early History of Electricity The history of electricity started with William Gilbert, a physician who served Queen Elizabeth the first of England. Before William Gilbert, all knowledge about electricity and magnetism was that the lodestone has magnetic properties and that rubbing amber and jet would draw bits of stuff to create sticking. Motivated and learned by William Gilbert several Europeans inventors, Otto von Guericke of Germany, Charles Francois Du Fay of France, and Stephen Gray of England, extended the knowledge.

Otto von Guericke though his research proved that a vacuum could exist.

Making a vacuum was necessary for all kinds of research into electronics. In 1660, Otto von Guericke created a machine that produced static electricity. In 1729, Stephen Gray revealed the principle of the conduction of electricity. In 1733, Charles Francois du Fay discovered that electricity comes in two forms which he called resinous (-) and vitreous (+), now called negative and positive terminals.

Otto von Guericke 13.2 How the Electricity is Produced Electricity is a type of energy which involves the flow of electrons. All matter is made up of atoms and the center of it is called a nucleus. The nucleus has positively charged particles known as protons and uncharged particles called neutrons.

The nucleus of an atom is bounded by negatively charged particles known as electrons. The negative charge of an electron is the same to the positive charge of a proton, and the number of electrons in an atom is equal to the number of protons.

When the equilibrium force between protons and electrons is disturbed by an outside force, an atom may gain or lose an electron. When electrons are lost from an atom, the free movement of these electrons constitutes an electric current.

How Electricity is Measured Electricity is measured in units of power called watts represented by (W). It was named to honor James Watt, the discovered of the steam engine. A kilowatt represents 1,000 watts. Kilowatt-hours are computed by multiplying the number of kW needed by the number of hours of use. For example, if a 40-watt light bulb used 5 hours a day, it used 200 watts of power, or 0.2 kilowatt-hours of electrical energy.

Conductors and Insulators: This power adapter uses metal wires and connectors to conduct electricity from the wall socket to a laptop computer. The conducting wires allow electrons to move freely through the cables, which are shielded by rubber and plastic. These materials act as insulators that donâ&#x20AC;&#x2122;t allow electric charge to escape outward. (credit: Evan-Amos, Wikimedia Commons)

Some substances, such as metals and salty water, allow charges to move through them with relative ease. Some of the electrons in metals and similar conductors are not bound to individual atoms or sites in the material. These free electrons can move through the material much as air moves through loose sand. Any substance that has free electrons and allows charge to move relatively freely through it is called a conductor. The moving electrons may collide with fixed atoms and molecules, losing some energy, but they can move in a conductor. Superconductors allow the movement of charge without any loss of energy. Salty water and other similar conducting materials contain free ions that can move through them. An ion is an atom or molecule having a positive or negative (nonzero) total charge. In other words, the total number of electrons is not equal to the total number of protons.

Other substances, such as glass, do not allow charges to move through them. These are called insulators. Electrons and ions in insulators are bound in the structure and cannot move easilyâ&#x20AC;&#x201D;as much as times more slowly than in conductors. Pure water and dry table salt are insulators, for example, whereas molten salt and salty water are conductors.

An electroscope is a favorite instrument in physics demonstrations and student laboratories. It is typically made with gold foil leaves hung from a (conducting) metal stem and is insulated from the room air in a glass-walled container. (a) A positively charged glass rod is brought near the tip of the electroscope, attracting electrons to the top and leaving a net positive charge on the leaves. Like charges in the light flexible gold leaves repel, separating them. (b) When the rod is touched against the ball, electrons are attracted and transferred, reducing the net charge on the glass rod but leaving the electroscope positively charged. (c) The excess charges are evenly distributed in the stem and leaves of the electroscope once the glass rod is removed.

13.3 Charging by Contact Shows an electroscope being charged by touching it with a positively charged glass rod. Because the glass rod is an insulator, it must actually touch the electroscope to transfer charge to or from it. (Note that the extra positive charges reside on the surface of the glass rod as a result of rubbing it with silk before starting the experiment.) Since only electrons move in metals, we see that they are attracted to the top of the electroscope. There, some are transferred to the positive rod by touch, leaving the electroscope with a net positive charge.

Electrostatic repulsion in the leaves of the charged electroscope separates them. The electrostatic force has a horizontal component that results in the leaves moving apart as well as a vertical component that is balanced by the gravitational force. Similarly, the electroscope can be negatively charged by contact with a negatively charged object.

Charging by Induction It is not necessary to transfer excess charge directly to an object in order to charge it.Shows a method of induction wherein a charge is created in a nearby object, without direct contact. Here we see two neutral metal spheres in contact with one another but insulated from the rest of the world. A positively charged rod is brought near one of them, attracting negative charge to that side, leaving the other sphere positively charged.

This is an example of induced polarization of neutral objects. Polarization is the separation of charges in an object that remains neutral. If the spheres are now separated (before the rod is pulled away), each sphere will have a net charge. Note that the object closest to the charged rod receives an opposite charge when charged by induction. Note also that no charge is removed from the charged rod, so that this process can be repeated without depleting the supply of excess charge.

Another method of charging by induction is shown in . The neutral metal sphere is polarized when a charged rod is brought near it. The sphere is then grounded, meaning that a conducting wire is run from the sphere to the ground. Since the earth is large and most ground is a good conductor, it can supply or accept excess charge easily. In this case, electrons are attracted to the sphere through a wire called the ground wire, because it supplies a conducting path to the ground. The ground connection is broken before the charged rod is removed, leaving the sphere with an excess charge opposite to that of the rod. Again, an opposite charge is achieved when charging by induction and the charged rod loses none of its excess charge.

Charging by induction. (a) Two uncharged or neutral metal spheres are in contact with each other but insulated from the rest of the world. (b) A positively charged glass rod is brought near the sphere on the left, attracting negative charge and leaving the other sphere positively charged. (c) The spheres are separated before the rod is removed,

thus separating negative and positive charge. (d) The spheres retain net charges after the inducing rod is removedâ&#x20AC;&#x201D;without ever having been touched by a charged object.

Charging by induction, using a ground connection. (a) A positively charged rod is brought near a neutral metal sphere, polarizing it. (b) The sphere is grounded, allowing electrons to be attracted from the earthâ&#x20AC;&#x2122;s ample supply. (c) The ground connection is broken. (d) The positive rod is removed, leaving the sphere with an induced negative charge.

Both positive and negative objects attract a neutral object by polarizing its molecules. (a) A positive object brought near a neutral insulator polarizes its molecules. There is a slight shift in the distribution of the electrons orbiting the molecule, with unlike charges being brought nearer and like charges moved away. Since the electrostatic force decreases with distance, there is a net attraction. (b) A negative object produces

the opposite polarization, but again attracts the neutral object. (c) The same effect occurs for a conductor; since the unlike charges are closer, there is a net attraction.

Neutral objects can be attracted to any charged object. The pieces of straw attracted to polished amber are neutral, for example. If you run a plastic comb through your hair, the charged comb can pick up neutral pieces of paper.shows how the polarization of atoms and molecules in neutral objects results in their attraction to a charged object.

When a charged rod is brought near a neutral substance, an insulator in this case, the distribution of charge in atoms and molecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction. Thus a positively charged glass rod attracts neutral pieces of paper, as will a negatively charged rubber rod. Some molecules, like water, are polar molecules. Polar molecules have a natural or inherent separation of charge, although they are neutral overall. Polar molecules are particularly affected by other charged objects and show greater polarization effects than molecules with naturally uniform charge distributions.

Check Your Understanding Can you explain the attraction of water to the charged rod in the figure below?

Answer Water molecules are polarized, giving them slightly positive and slightly negative sides. This makes water even more susceptible to a charged rodâ&#x20AC;&#x2122;s attraction. As the water flows downward, due to the force of gravity, the charged conductor exerts a net attraction to the opposite charges in the stream of water, pulling it closer.

FEW DEFINITIONS free electron an electron that is free to move away from its atomic orbit conductor a material that allows electrons to move separately from their atomic orbits insulator a material that holds electrons securely within their atomic orbits grounded when a conductor is connected to the Earth, allowing charge to freely flow to and from Earthâ&#x20AC;&#x2122;s unlimited reservoir induction the process by which an electrically charged object brought near a neutral object creates a charge in that object polarization slight shifting of positive and negative charges to opposite sides of an atom or molecule

electrostatic repulsion the phenomenon of two objects with like charges repelling each other

13.4 CHARGED OBJECTS AND THE ELECTRIC FORCE

Charge: there are two kinds of charge, positive and negative like charges repel, unlike charges attract positive charge comes from having more protons than electrons; negative charge comes from having more electrons than protons charge is quantized, meaning that charge comes in integer multiples of the elementary charge e charge is conserved Probably everyone is familiar with the first three concepts, but what does it mean for charge to be quantized? Charge comes in multiples of an indivisible unit of charge, represented by the letter e. In other words, charge comes in multiples of the charge on the electron or the proton. These things have the same size charge, but the sign is different. A proton has a charge of +e, while an electron has a charge of -e.

Electrons and protons are not the only things that carry charge. Other particles (positrons, for example) also carry charge in multiples of the electronic charge. Those are not going to be discussed, for the most part, in this course, however.

Putting "charge is quantized" in terms of an equation, we say: q=ne q is the symbol used to represent charge, while n is a positive or negative integer, and e is the electronic charge, 1.60 x 10-19 Coulombs.

The Law of Conservation of Charge The Law of conservation of charge states that the net charge of an isolated system remains constant.

If a system starts out with an equal number of positive and negative charges, thereÄąs nothing we can do to create an excess of one kind of charge in that system unless we bring in charge from outside the system (or remove some charge from the system).

Likewise, if something starts out with a certain net charge, say +100 e, it will always

have +100 e unless it is allowed to interact with something external to it.

Charge can be created and destroyed, but only in positive-negative pairs.

Table of elementary particle masses and charges:

13.5 Electrostatic charging Forces between two electrically-charged objects can be extremely large. Most things are electrically neutral; they have equal amounts of positive and negative charge. If this wasnÄąt the case, the world we live in would be a much stranger place. We also have a lot of control over how things get charged. This is because we can choose the appropriate material to use in a given situation.

Metals are good conductors of electric charge, while plastics, wood, and rubber are not. They are called insulators. Charge does not flow nearly as easily through insulators as it does through conductors, which is why wires you plug into a wall socket are covered with a protective rubber coating. Charge flows along the wire, but not through the coating to you.

Materials are divided into three categories, depending on how easily they will allow charge (i.e., electrons) to flow along them. These are: conductors - metals, for example semi-conductors - silicon is a good example insulators - rubber, wood, plastic for example Most materials are either conductors or insulators. The difference between them is that in conductors, the outermost electrons in the atoms are so loosely bound to their atoms that they are free to travel around. In insulators, on the other hand, the electrons are much more tightly bound to the atoms, and are not free to flow. Semiconductors are a very useful intermediate class, not as conductive as metals but considerably more conductive than insulators. By adding certain impurities to semiconductors in the appropriate concentrations the conductivity can be well-controlled.

There are three ways that objects can be given a net charge. These are: Charging by friction - this is useful for charging insulators. If you rub one material with another (say, a plastic ruler with a piece of paper towel), electrons have a tendency to be transferred from one material to the other. For example, rubbing glass with silk or saran wrap generally leaves the glass with a positive charge; rubbing PVC rod with fur generally gives the rod a negative charge.

Charging by conduction - useful for charging metals and other conductors. If a charged object touches a conductor, some charge will be transferred between the object and the conductor, charging the conductor with the same sign as the charge on the object.

Charging by induction - also useful for charging metals and other conductors. Again, a charged object is used, but this time it is only brought close to the conductor, and does not touch it. If the conductor is connected to ground (ground is basically anything neutral that can give up electrons to, or take electrons from, an object), electrons will either flow on to it or away from it. When the ground connection is removed , the conductor will have a charge opposite in sign to that of the charged object.

An example of induction using a negatively charged object and an initially-uncharged conductor (for example, a metal ball on a plastic handle).

(1) bring the negatively-charged object close to, but not touching, the conductor.

Electrons on the conductor will be repelled from the area nearest the charged object.

(2) connect the conductor to ground. The electrons on the conductor want to get as far away from the negatively-charged object as possible, so some of them flow to ground.

(3) remove the ground connection. This leaves the conductor with a deficit of electrons.

(4) remove the charged object. The conductor is now positively charged.

A practical application involving the transfer of charge is in how laser printers and photocopiers work. This is a good web page that gives a nice description of how a photocopier works: Why is static electricity more apparent in winter?

You notice static electricity much more in winter (with clothes in a dryer, or taking a sweater off, or getting a shock when you touch something after walking on carpet) than in summer because the air is much drier in winter than summer. Dry air is a relatively good electrical insulator, so if something is charged the charge tends to stay. In more humid conditions, such as you find on a typical summer day, water molecules, which are polarized, can quickly remove charge from a charged object.

Try this at home See if you can charge something at home using friction. I got good results by rubbing a Bic pen with a piece of paper towel. To test the charge, you can use a narrow stream of water from a faucet; if the object attracts the stream when it's brought close, you know it's charged. All you need to do is to find something to rub - try anything made out of hard plastic or rubber. You also need to find something to rub the object with -

potential candidates are things like paper towel, wool, silk, and saran wrap or other plastic.

Coulomb's law The force exerted by one charge q on another charge Q is given by Coulomb's law:

r is the distance between the charges.

Remember that force is a vector, so when more than one charge exerts a force on another charge, the net force on that charge is the vector sum of the individual forces.

Remember, too, that charges of the same sign exert repulsive forces on one another, while charges of opposite sign attract.

An example Four charges are arranged in a square with sides of length 2.5 cm. The two charges in the top right and bottom left corners are +3.0 x 10-6 C. The charges in the other two corners are -3.0 x 10-6C. What is the net force exerted on the charge in the top right corner by the other three charges?

To solve any problem like this, the simplest thing to do is to draw a good diagram showing the forces acting on the charge. You should also let your diagram handle your signs for you. Force is a vector, and any time you have a minus sign associated with a vector all it does is tell you about the direction of the vector. If you have the arrows giving you the direction on your diagram, you can just drop any signs that come out of the equation for Coulomb's law.

Consider the forces exerted on the charge in the top right by the other three:

You have to be very careful to add these forces as vectors to get the net force. In this problem we can take advantage of the symmetry, and combine the forces from charges 2 and 4 into a force along the diagonal (opposite to the force from charge 3) of magnitude 183.1 N. When this is combined with the 64.7 N force in the opposite direction, the result is a net force of 118 N pointing along the diagonal of the square.

The symmetry here makes things a little easier. If it wasn't so symmetric, all you'd have to do is split the vectors up in to x and y components, add them to find the x and y components of the net force, and then calculate the magnitude and direction of the net force from the components.

The parallel between gravity and electrostatics An electric field describes how an electric charge affects the region around it. It's a powerful concept, because it allows you to determine ahead of time how a charge will be affected if it is brought into the region. Many people have trouble with the concept of a field, though, because it's something that's hard to get a real feel for. The fact is, though, that you're already familiar with a field. We've talked about gravity, and we've even used a gravitational field; we just didn't call it a field.

When talking about gravity, we got into the (probably bad) habit of calling g "the acceleration due to gravity". It's more accurate to call g the gravitational field produced by the Earth at the surface of the Earth. If you understand gravity you can understand electric forces and fields because the equations that govern both have the same form.

The gravitational force between two masses (m and M) separated by a distance r is given by Newton's law of universal gravitation:

A similar equation applies to the force between two charges (q and Q) separated by a distance r:

The force equations are similar, so the behavior of interacting masses is similar to that of interacting charges, and similar analysis methods can be used. The main difference is that gravitational forces are always attractive, while electrostatic forces can be attractive or repulsive. The charge (q or Q) plays the same role in the electrostatic case that the mass (m or M) plays in the case of the gravity.

A good example of a question involving two interacting masses is a projectile motion problem, where there is one mass m, the projectile, interacting with a much larger mass M, the Earth. If we throw the projectile (at some random launch angle) off a 40meter-high cliff, the force on the projectile is given by: F = mg This is the same equation as the more complicated equation above, with G, M, and the radius of the Earth, squared, incorporated into g, the gravitational field.

So, you've seen a field before, in the form of g. Electric fields operate in a similar way.

An equivalent electrostatics problem is to launch a charge q (again, at some random angle) into a uniform electric field E, as we did for m in the Earth's gravitational field g.

The force on the charge is given by F = qE, the same way the force on the mass m is given by F = mg.

We can extend the parallel between gravity and electrostatics to energy, but we'll deal with that later. The bottom line is that if you can do projectile motion questions using gravity, you should be able to do them using electrostatics. In some cases, you will need to apply both; in other cases one force will be so much larger than the other that you can ignore one (generally if you can ignore one, it'll be the gravitational force).

13.6 ELECTRIC FIELDS(E):

Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge.

The Force per Charge Ratio

Electric field strength is a vector quantity; it has both magnitude and direction. The magnitude of the electric field strength is defined in terms of how it is measured. Let's suppose that an electric charge can be denoted by the symbol Q. This electric charge creates an electric field; since Q is the source of the electric field, we will refer to it as the source charge. The strength of the source charge's electric field could be measured by any other charge placed somewhere in its surroundings. The charge that is used to measure the electric field strength is referred to as a test charge since it is used to test the field strength. The test charge has a quantity of charge denoted by the symbol q. When placed within the electric field, the test charge will experience an electric force - either attractive or repulsive. As is usually the case, this force will be denoted by the symbol F. The magnitude of the electric field is simply defined as the force per charge on the test charge.

If the electric field strength is denoted by the symbol E, then the equation can be rewritten in symbolic form as

.

The standard metric units on electric field strength arise from its definition. Since electric field is defined as a force per charge, its units would be force units divided by charge units. In this case, the standard metric units are Newton/Coulomb or N/C. In the above discussion, you will note that two charges are mentioned - the source charge and the test charge. Two charges would always be necessary to encounter a force. In the electric world, it takes two to attract or repel. The equation for electric field strength (E) has one of the two charge quantities listed in it. Since there are two charges involved, a student will have to be ultimately careful to use the correct charge quantity when computing the electric field strength. The symbol q in the equation is the quantity of charge on the test charge (not the source charge). Recall that the electric field strength is defined in terms of how it is measured or tested; thus, the test charge finds its way into the equation. Electric field is the force per quantity of charge on the test charge.

The electric field strength is not dependent upon the quantity of charge on the test charge. If you think about that statement for a little while, you might be bothered by it.

(Of course if you don't think at all - ever - nothing really bothers you. Ignorance is bliss.) After all, the quantity of charge on the test charge (q) is in the equation for electric field. So how could electric field strength not be dependent upon q if q is in the equation? Good question. But if you think about it a little while longer, you will be able to answer your own question. (Ignorance might be bliss. But with a little extra

thinking you might achieve insight, a state much better than bliss.) Increasing the quantity of charge on the test charge - say, by a factor of 2 - would increase the denominator of the equation by a factor of 2. But according to Coulomb's law, more charge also means more electric force (F). In fact, a twofold increase in q would be accompanied by a twofold increase in F. So as the denominator in the equation increases by a factor of two (or three or four), the numerator increases by the same factor. These two changes offset each other such that one can safely say that the electric field strength is not dependent upon the quantity of charge on the test charge.

So regardless of what test charge is used, the electric field strength at any given location around the source charge Q will be measured to be the same.

Another Electric Field Strength Formula The above discussion pertained to defining electric field strength in terms of how it is measured. Now we will investigate a new equation that defines electric field strength in terms of the variables that affect the electric field strength. To do so, we will have to revisit the Coulomb's law equation. Coulomb's law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between their centers. When applied to our two charges - the source charge (Q) and the test charge (q) - the formula for electric force can be written as

If the expression for electric force as given by Coulomb's law is substituted for force in the above E =F/q equation, a new equation can be derived as shown below.

Note that the derivation above shows that the test charge q was canceled from both numerator and denominator of the equation. The new formula for electric field strength (shown inside the box) expresses the field strength in terms of the two variables that affect it. The electric field strength is dependent upon the quantity of charge on the source charge (Q) and the distance of separation (d) from the source charge.

An Inverse Square Law Like all formulas in physics, the formulas for electric field strength can be used to algebraically solve physics word problems. And like all formulas, these electric field strength formulas can also be used to guide our thinking about how an alteration of one variable might (or might not) affect another variable. One feature of this electric field strength formula is that it illustrates an inverse square relationship between electric field strength and distance. The strength of an electric field as created by source charge Q is inversely related to square of the distance from the source. This is known as an inverse square law.

Electric field strength is location dependent, and its magnitude decreases as the distance from a location to the source increases. And by whatever factor the distance is changed, the electric field strength will change inversely by the square of that factor. So if separation distance increases by a factor of 2, the electric field strength decreases by a factor of 4 (2^2). If the separation distance increases by a factor of 3, the electric field strength decreases by a factor of 9 (3^2). If the separation distance increases by a factor of 4, the electric field strength decreases by a factor of 16 (4^2).

And finally, if separation distance decreases by a factor of 2, the electric field strength increases by a factor of 4 (2^2).

The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge. Since electric field is a vector quantity, it can be represented by a vector arrow. For any given location, the arrows point in the direction of the electric field and their length is proportional to the strength of the electric field at that location. Such vector arrows are shown in the diagram below. Note that the lengths of the arrows are longer when closer to the source charge and shorter when further from the source charge.

A more useful means of visually representing the vector nature of an electric field is through the use of electric field lines of force. Rather than draw countless vector arrows in the space surrounding a source charge, it is perhaps more useful to draw a pattern of several lines that extend between infinity and the source charge. These pattern of lines, sometimes referred to as electric field lines, point in the direction that a positive test charge would accelerate if placed upon the line. As such, the lines are

directed away from positively charged source charges and toward negatively charged source charges. To communicate information about the direction of the field, each line must include an arrowhead that points in the appropriate direction. An electric field line pattern could include an infinite number of lines. Because drawing such large quantities of lines tends to decrease the readability of the patterns, the number of lines is usually limited. The presence of a few lines around a charge is typically sufficient to convey the nature of the electric field in the space surrounding the lines.

Rules for Drawing Electric Field Patterns There are a variety of conventions and rules to drawing such patterns of electric field lines. The conventions are simply established in order that electric field line patterns communicate the greatest amount of information about the nature of the electric field surrounding a charged object. One common convention is to surround more charged objects by more lines. Objects with greater charge create stronger electric fields. By surrounding a highly charged object with more lines, one can communicate the strength of an electric field in the space surrounding a charged object by the line density. This convention is depicted in the diagram below.

Not only does the density of lines surrounding any given object reveal information about the quantity of charge on the source charge, the density of lines at a specific location in space reveals information about the strength of the field at that location. Consider the

object shown at the right. Two different circular cross-sections are drawn at different distances from the source charge. These cross-sections represent regions of space closer to and further from the source charge. The field lines are closer together in the regions of space closest to the charge; and they are spread further apart in the regions of space furthest from the charge. Based on the convention concerning line density, one would reason that the electric field is greatest at locations closest to the surface of the charge and least at locations further from the surface of the charge.

Line density in an electric field line pattern reveals information about the strength or magnitude of an electric field.

A second rule for drawing electric field lines involves drawing the lines of force perpendicular to the surfaces of objects at the locations where the lines connect to object's surfaces. At the surface of both symmetrically shaped and irregularly shaped objects, there is never a component of electric force that is directed parallel to the surface. The electric force, and thus the electric field, is always directed perpendicular to the surface of an object. If there were ever any component of force parallel to the surface, then any excess charge residing upon the surface of a source charge would begin to accelerate. This would lead to the occurrence of an electric current within the object; this is never observed in static electricity. Once a line of force leaves the surface of an object, it will often alter its direction. This occurs when drawing electric field lines for configurations of two or more charges as discussed in the section below.

A final rule for drawing electric field lines involves the intersection of lines. Electric field lines should never cross. This is particularly important (and tempting to break) when drawing electric field lines for situations involving a configuration of charges (as in the section below). If electric field lines were ever allowed to cross each other at a given location, then you might be able to imagine the results. Electric field lines reveal information about the direction (and the strength) of an electric field within a region of space. If the lines cross each other at a given location, then there must be two distinctly different values of electric field with their own individual direction at that given location. This could never be the case. Every single location in space has its own electric field strength and direction associated with it. Consequently, the lines representing the field cannot cross each other at any given location in space.

Electric Field Lines for Configurations of Two or More Charges In the examples above, we've seen electric field lines for the space surrounding single point charges. But what if a region of space contains more than one point charge?

How can the electric field in the space surrounding a configuration of two or more charges be described by electric field lines? To answer this question, we will first return to our original method of drawing electric field vectors.

Suppose that there are two positive charges - charge A (QA) and charge B (QB) - in a

given region of space. Each charge creates its own electric field. At any given location surrounding the charges, the strength of the electric field can be calculated using the expression kQ/d2. Since there are two charges, the kQ/d2 calculation would have to be performed twice at each location - once with kQA/dA2 and once with kQB/dB2 (dA is the distance from that location to the center of charge A and dB is the distance from that location to the center of charge B). The results of these calculations are illustrated in the diagram below with electric field vectors (EA and EB) drawn at a variety of locations. The strength of the field is represented by the length of the arrow and the direction of the field is represented by the direction of the arrow.

Since electric field is a vector, the usual operations that apply to vectors can be applied to electric field. That is, they can be added in head-to-tail fashion to determine the resultant or net electric field vector at each location. This is shown in the diagram below.

The diagram above shows that the magnitude and direction of the electric field at each location is simply the vector sum of the electric field vectors for each individual charge. If more locations are selected and the process of drawing EA, EB and Enet is repeated, then the electric field strength and direction at a multitude of locations will be known. (This is not done since it is a highly time intensive task.) Ultimately, the

electric field lines surrounding the configuration of our two charges would begin to emerge. For the limited number of points selected in this location, the beginnings of the electric field line pattern can be seen. This is depicted in the diagram below. Note that for each location, the electric field vectors point tangent to the direction of the electric field lines at any given point.

The construction of electric field lines in this manner is a tedious and cumbersome task. The use of a field plotting computer software program or a lab procedure produces similar results in less time (and with more phun). Whatever the method used to determine the electric field line patterns for a configuration of charges, the general idea is that the pattern is the resultant of the patterns for the individual charges within the configuration. The electric field line patterns for other charge configurations are shown in the diagrams below.

In each of the above diagrams, the individual source charges in the configuration possess the same amount of charge. Having an identical quantity of charge, each source charge has an equal ability to alter the space surrounding it. Subsequently, the pattern is symmetrical in nature and the number of lines emanating from a source charge or extending towards a source charge is the same. This reinforces a principle discussed earlier that stated that the density of lines surrounding any given source charge is proportional to the quantity of charge on that source charge. If the quantity

of charge on a source charge is not identical, the pattern will take on an asymmetric nature, as one of the source charges will have a greater ability to alter the electrical nature of the surrounding space. This is depicted in the electric field line patterns below.

After plotting the electric field line patterns for a variety of charge configurations, the general patterns for other configurations can be predicted. There are a number of principles that will assist in such predictions. These principles are described (or redescribed) in the list below.

Electric field lines always extend from a positively charged object to a negatively charged object, from a positively charged object to infinity, or from infinity to a negatively charged object.

Electric field lines never cross each other.

Electric field lines are most dense around objects with the greatest amount of charge.

At locations where electric field lines meet the surface of an object, the lines are perpendicular to the surface.

Electric Field Lines as an Invisible Reality The concept of the electric field was first introduced by 19th century physicist Michael Faraday. It was Faraday's perception that the pattern of lines characterizing the electric field represents an invisible reality. Rather than thinking in terms of one charge affecting another charge, Faraday used the concept of a field to propose that a charged object (or a massive object in the case of a gravitational field) affects the space that surrounds it. As another object enters that space, it becomes affected by the field established in that space. Viewed in this manner, a charge is seen to interact

with an electric field as opposed to with another charge. To Faraday, the secret to understanding action-at-a-distance is to understand the power of charge-field-charge.

A charged object sends its electric field into space, reaching from the "puller to the pullee." Each charge or configuration of charges creates an intricate web of influence in the space surrounding it. While the lines are invisible, the effect is ever so real. So as you practice the exercise of constructing electric field lines around charges or configuration of charges, you are doing more than simply drawing curvy lines. Rather, you are describing the electrified web of space that will draw and repel other charges that enter it.

Electric Field Inside a Conductor Shielding: In conducting materials such as copper, electric charges move readily in response to the forces that electric fields exert. This property of conducting materials has a major effect on the electric field that can exist within and around them. Suppose that a piece of copper carries a number of excess electrons somewhere within it, as in Figure 27a.

Each electron would experience a force of repulsion because of the electric field of its neighbors. And, since copper is a conductor, the excess electrons move readily in response to that force. In fact, as a consequence of the 1/r2 dependence on distance in Coulombâ&#x20AC;&#x2122;s law, they rush to the surface of the copper. Once static equilibrium is established with all of the excess charge on the surface, no further movement of charge occurs, as part b of the drawing indicates. Similarly, excess positive charge also moves to the surface of a conductor. In general, at equilibrium under electrostatic conditions, any excess charge resides on the surface of a conductor.

Figure 27 (a) Excess charge within a conductor (copper) moves quickly (b) to the surface.

Now consider the interior of the copper in Figure 27 b. The interior is electrically

neutral, although there are still free electrons that can move under the influence of an electric field. The absence of a net movement of these free electrons indicates that there is no net electric field present within the conductor. In fact, the excess charges arrange themselves on the conductor surface precisely in the manner needed to make the electric field zero within the material. Thus, at equilibrium under electrostatic conditions, the electric field is zero at any point within a conducting material. This fact has some fascinating implications.

Figure 28 (a) shows an uncharged, solid, cylindrical conductor at equilibrium in the central region of a parallel plate capacitor. Induced charges on the surface of the cylinder alter the electric field lines of the capacitor. Since an electric field cannot exist within the conductor under these conditions, the electric field lines do not penetrate the cylinder. Instead, they end or begin on the induced charges.

Consequently, a test charge placed inside the conductor would feel no force due to the presence of the charges on the capacitor. In other words, the conductor shields any charge within it from electric fields created outside the conductor. The shielding results from the induced charges on the conductor surface.

Figure 29 (a) A cylindrical conductor (shown as an end view) is placed between the oppositely charged plates of a capacitor. The electric field lines do not penetrate the conductor. The blowup shows that, just outside the conductor, the electric field lines are perpendicular to its surface. (b) The electric field is zero in a cavity within the conductor.

Since the electric field is zero inside the conductor, nothing is disturbed if a cavity is cut from the interior of the material, as in part b of the drawing. Thus, the interior of the cavity is also shielded from external electric fields, a fact that has important applications, particularly for shielding electronic circuits. â&#x20AC;&#x153;Strayâ&#x20AC;? electric fields are produced by various electrical appliances (e.g., hair dryers, blenders, and vacuum cleaners), and these fields can interfere with the operation of sensitive electronic circuits, such as those in stereo amplifiers, televisions, and computers. To eliminate such interference, circuits are often enclosed within metal boxes that provide shielding from external fields.

The blowup in Figure 29 (a) shows another aspect of how conductors alter the electric field lines created by external charges. The lines are altered because the electric field just outside the surface of a conductor is perpendicular to the surface at equilibrium

under electrostatic conditions. If the field were not perpendicular, there would be a component of the field parallel to the surface. Since the free electrons on the surface of the conductor can move, they would do so under the force exerted by that parallel component. In reality, however, no electron flow occurs at equilibrium.

Therefore, there can be no parallel component, and the electric field is perpendicular to the surface.

The preceding discussion deals with features of the electric field within and around a conductor at equilibrium under electrostatic conditions. These features are related to the fact that conductors contain mobile free electrons and do not apply to insulators, which contain very few free electrons.

Conceptual Example A Conductor in an Electric Field

A charge +q is suspended at the center of a hollow, electrically neutral, spherical conductor, as Figure 18.31 illustrates. Show that this charge induces (a) a charge of â&#x20AC;&#x201C; q on the interior surface and (b) a charge of +q on the exterior surface of the conductor.

Figure 30 A positive charge + q is suspended at the center of a hollow spherical conductor that is electrically neutral. Induced charges appear on the inner and outer surfaces of the conductor. The electric field within the conductor itself is zero.

Reasoning and Solution For (a)

Electric field lines emanate from the positive charge +q. Since the electric field

inside the metal conductor must be zero at equilibrium under electrostatic conditions, each field line ends when it reaches the conductor, as the picture shows. Since field lines terminate only on negative charges, there must be an induced negative charge on the interior surface of the conductor. Furthermore, the lines begin and end on equal amounts of charge, so the magnitude of the total induced charge is the same as the magnitude of the charge at the center.

Thus, the total induced charge on the interior surface is â&#x20AC;&#x201C;q.

For (b)

Before the charge +q is introduced, the conductor is electrically neutral.Therefore, it carries no net charge.

We have also seen that there can be no excess charge within the metal.

Thus, since an induced charge of â&#x20AC;&#x201C;q appears on the interior surface, a charge of +q must be induced on the outer surface. The positive charge on the outer surface generates field lines that radiate outward (see the drawing) as if they originated from the central charge and the conductor were absent.

The conductor does not shield the outside from the field produced by the charge on the inside.

DISCUSSION ABOUT THE NET ELECTRIC FIELD AND FORCE: NB:The net electric field between two charges is zero at the side of the smallest charge in absolute value,and out of the segment formed by the two charges.

Example 1: Where is the Field Equal to Zero?

Two charges, +3Q and -Q, are separated by 4 cm. Is there a point along the line passing through them (and a finite distance from the charges) where the net electric field is zero? If so, where?

More specifically, is the field equal to zero at some point in one of these three regions: to the left of both charges (Region I), in between both charges (Region II), and/or to the right of both charges (Region III)?

The field is zero at a point in: Region I Region II Region III two of the above

all of the above In Region I, to the left of both charges, the fields from the two charges are in opposite directions, which is what we need for them to cancel. However, in region I we are always closer to the larger charge. Both the smaller r and the larger q act to make the field from the positive charge significantly larger than that from the negative charge, so they can't cancel one another.

In Region II, between the charges, both vectors point in the same direction so there is no possibility of cancelling out.

In Region III, the fields again point in opposite directions and there is a point where their magnitudes are the same. It is at this point where the net electric field is zero.

What happens at this point? Because F = qE, if there is no electric field at a point then a test charge placed at that point would feel no force.

How can we calculate where the point is? If the point is a distance x from the +3Q charge, then it is x-4 away from the -Q charge. If we define right as positive, we can write this as: k (3Q / x2) - k (Q / (x - 4)2) = 0 where the minus sign in front of the second term is not the one associated with the charge but the one associated with the direction of the field from the charge.

The k's and Q's cancel. Re-arranging gives: 3 / x^2 = 1 / (x - 4)^2 Cross multiplying and expanding the brackets: 3(x^2 - 8x + 16) = x2 X^2 - 12x + 24 = 0 Solving this using the quadratic equation gives two answers: x = 2.54 cm and x = 9.46 cm. Which answer should we keep?

2.54 cm 9.46 cm both answers are correct If you decide to throw out an answer, state why it showed up as a solution.

The answer we want is x = 9.46 cm because this represents a point in

Region III. The other answer represents the point between the charges where the magnitudes are the same. The net field is not zero there, though, because the fields point in the same direction.

This is also applied to the electric force and electric potential.

TUTORIAL: 1.Figure shows an electric field extending over three regions,labelled ǁ,ǁǁ, and ǁǁǁ (a) Are there any isolated charges? If so, in what region and what are their signs? (b) Where is the field strongest? (c) Where is it weakest? (d) Where is the field the most uniform?

2. Is the object in Figure 18.8.9 a conductor or an insulator? Justify your answer.

3.If the electric field lines in the figure above were perpendicular to the object, would it necessarily be a conductor? Explain.

4. The discussion of the electric field between two parallel conducting plates, in this module states that edge effects are less important if the plates are close together. What does close mean? That is, is the actual plate separation crucial, or is the ratio of plate separation to plate area crucial?

5. Would the self-created electric field at the end of a pointed conductor, such as a lightning rod, remove positive or negative charge from the conductor? Would the same sign charge be removed from a neutral pointed conductor by the application of a similar externally created electric field?

(The answers to both questions have implications for charge transfer utilizing points.)

6. Are you relatively safe from lightning inside an automobile? Give two reasons.

7. Using the symmetry of the arrangement, show that the net Coulomb force on the charge qq at the center of the square below (Figure) is zero if the charges on the four corners are exactly equal.

Four point charges qa, qb, qc, and qd lie on the corners of a square and q is located at its center. (a) Using the symmetry of the arrangement, show that the electric field at the center of the square in Figure zero if the charges on the four corners are exactly equal. (b) Show that this is also true for any combination of charges in which qa=qd and qb=qc 8. Considering Figure in question 7, suppose that qa=qd and qb=qc. First show that qq is in static equilibrium. (You may neglect the gravitational force.) Then discuss whether the equilibrium is stable or unstable, noting that this may depend on the signs of the charges and the direction of displacement of qq from the center of the square.

9. Suppose that Zoe carries an excess charge. To maintain her charged status can she be standing on ground wearing just any pair of shoes? How would you discharge her? What are the consequences if she simply walks away?

10. Common static electricity involves charges ranging from nanocoulombs to microcoulombs. (a) How many electrons are needed to form a charge of –2.00nC(1.25×10^10) (b) How many electrons must be removed from a neutral object to leave a net charge of 0.500μC ?( 3.13×10^12) 11. An amoeba has 1.00×10^16 protons and a net charge of 0.300 pC.

(a) How many fewer electrons are there than protons? (b) If you paired them up, what fraction of the protons would have no electrons?

12. A 50.0 g ball of copper has a net charge of 2.00μC. What fraction of the copper’s electrons has been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5.)( 9.09×10−13) 13. What net charge would you place on a 100 g piece of sulfur if you put an extra electron on 1 in 10121012 of its atoms? (Sulfur has an atomic mass of 32.1.)

14. How many coulombs of positive charge are there in 4.00 kg of plutonium, given its atomic mass is 244 and that each plutonium atom has 94 protons?( 1.48×10^8C) 15. What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC? 16. (a) How strong is the attractive force between a glass rod with a 0.700μC charge and a silk cloth with a −0.600μC charge, which are 12.0 cm apart, using the approximation that they act like point charges?( 0.263N) (b) Discuss how the answer to this problem might be affected if the charges are distributed over some area and do not act like point charges.( If the charges are distributed over some area, there will be a concentration of charge along the side closest to the oppositely charged object. This effect will increase the net force.)

17. Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a factor of three?

18. Two point charges are brought closer together, increasing the force between them by a factor of 25. By what factor was their separation decreased?( The separation decreased by a factor of 5.).

19. How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them?

20. A test charge of +2μC is placed halfway between a charge of +6μC and another of +4μC separated by 10 cm. (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the +6μC charge)?

21. Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated protons separated by 2.00 nm (a typical distance between gas atoms). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.( 3.45×10^16m/s^2) 22. (a) By what factor must you change the distance between two point charges to change the force between them by a factor of 10?( 3.2) (b) Explain how the distance can either increase or decrease by this factor and still cause a factor of 10 change in the force.(If the distance increases by 3.2, then the force will decrease by a factor of 10 ; if the distance decreases by 3.2, then the force will increase by a factor of 10. Either way, the force changes by a factor of 10.)

23. At what distance is the electrostatic force between two protons equal to the weight of one proton?

24. A certain five cent coin contains 5.00 g of nickel. What fraction of the nickel atoms’ electrons, removed and placed 1.00 m above it, would support the weight of this coin? The atomic mass of nickel is 58.7, and each nickel atom contains 28 electrons and 28 protons.( 1.02×10−1 ) 25. Point charges of 5.00μc and −3.00μc are placed 0.250 m apart. (a) Where can a third charge be placed so that the net force on it is zero?( 0.859 m beyond negative charge on line connecting two charges) (b) What if both charges are positive?( 0.109 m from lesser charge on line connecting two charges) 26. What is the magnitude and direction of an electric field that exerts a 2.00×10^−5N upward force on a −1.75μC charge?

27. What is the magnitude and direction of the force exerted on a 3.50μC charge by a 250 N/C electric field that points due east?( 8.75×10^−4) 28. Find the magnitude and direction of the total electric field due to the two point charges, q1 and q2, at the origin of the coordinate system as shown in Figure 18.5(1.26×105N/C; 63.4deg)

Figure 18.5: The electric fields E1 and E2 at the origin O add to E tot 29.(a) Sketch the electric field lines near a point charge +q. (b) Do the same for a point charge â&#x2C6;&#x2019;3.00q.

30. Sketch the electric field lines a long distance from the charge distributions shown in Figure (a) and (b) 31. Figure shows the electric field lines near two charges q1 and q2. What is the ratio of their magnitudes? (b) Sketch the electric field lines a long distance from the charges shown in the figure.

32. .Sketch the electric field lines in the vicinity of two opposite charges, where the negative charge is three times greater in magnitude than the positive. (See Figure 18.6.8 for a similar situation).

33. What is the force on the charge located at x=8.00cm in Figure (a) given that q=1.00μC?

(a) Point charges located at 3.00, 8.00, and 11.0 cm along the x-axis. (b) Point charges located at 1.00, 5.00, 8.00, and 14.0 cm along the x-axis.

34. (a) Find the total electric field at x=1.00cm in Figure(−∞) (b) given that q=5.00nC. (b) Find the total electric field at x=1.00cm in Figure 18.8.15(b).( 12×10^5N/C) (c) If the charges are allowed to move and eventually be brought to rest by friction, what will the final charge configuration be? (That is, will there be a

single charge, double charge, etc., and what will its value(s) be?( one charge of +q) 35. (a) Find the electric field at x=5.00com in Figure 18.8.15(a), given that q=1.00μC. (b) At what position between 3.00 and 8.00 cm is the total electric field the same as that for −2q alone? (c) Can the electric field be zero anywhere between 0.00 and 8.00 cm? (d) At very large positive or negative values of x, the electric field approaches zero in both (a) and (b). In which does it most rapidly approach zero and why? (e) At what position to the right of 11.0 cm is the total electric field zero, other than at infinity? (Hint: A graphing calculator can yield considerable insight in this problem.)

36. Two charges, +10 (left) and -3 µC (right) are fixed in place and separated by 5 m. a. Where along the line joining and going through the two charges will the net electric field be zero? Locate this point relative to the positive charge.

Explain your reasoning. b. Calculate the electric field strength, half-way between the two charges. c. Calculate the electric field strength 0.5 m left of the +10 µC charge. d. Calculate the resultant force on a -2 µC charge, placed half-way between the two charges. e. Calculate the resultant force on a -5µC charge placed 0.5 m left of the +10 µC charge.

37. Two small ”point” objects are separated by a distance d as shown in the figure. The object on the left has a net chare of 9q, and the object on the right has a net charge of −4q, where q is some positive amount of charge. At what location in the figure is the electric field equal to zero?( 2d) 38. Suppose there are two small ”point” objects located on the ”y-axis”.

One object has a charge of +q and is located at the coordinates (0,+a). The other ”point” object has a charge of −q and is located at the coordinates (0,-a). Find:

i: The electric field vector for locations on the y-axis (0,y) such that |y| > a. ii: The electrid field vector for locations on the x-axis (x,0).

UNIT 14:ELECTRIC POTENTIAL AND ELECTRICAL POTENTIAL ENERGY

INTRODUCTION Electric potential energy (EPE) is the energy stored in a charge due to its location in an electric field. For example, moving two like charges close together stores potential energy, since the charges repel each other. Separating opposite charges also stores potential energy, since the charges attract each other. It is similar idea to gravitational potential energy.

14.1 Electric Potential Difference: Electric potential is a location-dependent quantity that expresses the amount of potential energy per unit of charge at a specified location. When a Coulomb of charge (or any given amount of charge) possesses a relatively large quantity of potential energy at a given location, then that location is said to be a location of high electric potential. And similarly, if a Coulomb of charge (or any given amount of charge) possesses a relatively small quantity of potential energy at a given location, then that location is said to be a location of low electric potential. As we begin to apply our concepts of potential energy and electric potential to circuits, we will begin to refer to the difference in electric potential between two points. This part of Lesson 1 will be devoted to an understanding of electric potential difference and its application to the movement of charge in electric circuits.

Consider the task of moving a positive test charge within a uniform electric field from location A to location B as shown in the diagram at the right. In moving the charge against the electric field from location A to location B, work will have to be done on the charge by an external force. The work done on the charge changes its potential energy to a higher value; and the amount of work that is done is equal to the change in the potential energy. As a result of this change in potential energy, there is also a difference in electric potential between locations A and B. This difference in electric potential is represented by the symbol Î&#x201D;Vand is formally referred to as the electric potential difference. By definition, the electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy. In equation form, the electric potential difference is

The standard metric unit on electric potential difference is the volt, abbreviated V and named in honor of Alessandro Volta. One Volt is equivalent to one Joule per Coulomb. If the electric potential difference between two locations is 1 volt, then one Coulomb of charge will gain 1 joule of potential energy when moved between those two locations. If the electric potential difference between two locations is 3 volts, then

one coulomb of charge will gain 3 joules of potential energy when moved between those two locations. And finally, if the electric potential difference between two locations is 12 volts, then one coulomb of charge will gain 12 joules of potential energy when moved between those two locations. Because electric potential difference is expressed in units of volts, it is sometimes referred to as the voltage.

Consider a parallel plate capacitor that produces a uniform electric field between its large plates. This is accomplished by connecting each plate to one of the terminals of a power supply (such as a battery).

Figure 1: An electric field is set up by the charged plates separated by a distance l. The charges on the plates are +Q and â&#x20AC;&#x201C;Q.

Figure 2: An electric charge q is moved from point A towards point B with an external force T against the electric force qE.

Figure 3, 4: When it is moved through a distance d, its potential energy at the point B is qE, d distance relative to the point A.

Figure 5: When released from B (T = 0), it will accelerate toward the lower plate. As it is moving toward the lower plate, its potential energy decreases and its Kinetic energy increases. When it reaches the lower plate (where we can choose the Potential energy to be zero), its potential energy at A is completely converted to Kinetic Energy at point B:

Note that qEd is the work done by the field as the charge moves under the force qE from B to A. Here m is the mass of the charge q, and v is its velocity as it reaches point A. Here we assumed that electric field is uniform! Work done by E field:

Letâ&#x20AC;&#x2122;s remember Kinetic Energy-Work theorem (Work Energy principle):

where we introduced the concept of potential energy and conservative force ( a force under which one can define a potential energy so that the work done only depends the differences of the potential energy function evaluated at the end points).

A rule of thumb for deciding whether or not EPE is increasing: If a charge is moving in the direction that it would normally move, its electric potential energy is decreasing. If a charge is moved in a direction opposite to that of it would normally move, its electric potential energy is increasing. This situation is similar to that of constant gravitational field (g = 9,8 m/s2). When you lift up an object, you are increasing its gravitational potential energy. Likewise, as you are lowering an object, its gravitational energy is decreasing.

A General Formula for Potential Difference: The work done by an E field as it act on a charge q to move it from point A to point B is defined as Electric Potential Difference between points A and B:

Clearly, the potential function V can be assigned to each point in the space surrounding a charge distribution (such as parallel plates). The above formula provides a simple recipe to calculate work done in moving a charge between two points where we know the value of the potential difference. The above statements and the formula are valid regardless of the path through which the charge is moved. A particular interest is the potential of a point-like charge Q. It can be found by simply performing the integration through a simple path (such as a straight line) from a point A whose distance from Q is r to infinity. Path is chosen along a radial line so that

becomes simply Edr. Since the electric field of Q is kQ/r2,

This process defines the electric potential of a point-like charge. Note that potential function is a scalar quantity as oppose to electric field being a vector quantity. Now, we can define the electric potential energy of a system of charges or charge distributions. Suppose we compute the work done against electric forces in moving a charge q from infinity to a point a distance r from the charge Q. The work is given by:

Note that if q is negative, its sigh should be used in the equation! Therefore, a system consisting of a negative and a positive point-like charge has a negative potential energy.

A negative potential energy means that work must be done against the electric field in moving the charges apart!

Now consider a more general case, which deals with the potential in the neighborhood of a number of charges as depicted in the picture below:

Let r1,r2,r3 be the distances of the charges to a field point A, and r12, r13, r23 represent the distance between the charges. The electric potential at point A is:

Example: If we bring a charge Q from infinity and place it at point A the work done would be:

The total Electric Potential Energy of this system of charges namely, the work needed to bring them to their current positions can be calculated as follows: first bring q1 (zero work since there is no charge around yet), then in the field of q1 bring q2, then in the fields of q1 and q2 bring q3. Add all of the work needed to compute the total work. The result would be:

Finding Electric Field from Electric Potential:

The component of E in any direction is the negative of the rate of change of the potential with distance in that direction: Equi potential Surfaces: These are imaginary surfaces surrounding a charge distribution. In particular, if the charge distribution is spherical (point charge, or uniformly charged sphere), the surfaces are spherical, concentric with the center of the charge distribution. Electric field lines are always perpendicular to the equi potential surfaces. The equation implies that due to the negative sign, the direction of E is opposite to the direction in which V increases; E is directed from higher to lower levels of V (from higher potential to lower potential). Another words, the gradient of a scalar (in this case E field) is normal to a surface of constant value (equi potential surface) of the scalar and in the direction of maximum rate of change of constant scalar.

Example 1: 1. An electron is accelerated from rest through a potential difference 12 V. What is the change in electric potential energy of the electron?

Known : The charge on an electron (e) = -1.60 x 10-19 Coulomb Electric potential = voltage (V) = 12 Volt Wanted: The change in electric potential energy of the electron (Î&#x201D;PE) Solution : Î&#x201D;PE = q V = (-1.60 x 10-19 C)(12 V) = -19.2 x 10-19 Joule The minus sign indicates that the potential energy decreases.

Example 2. Two parallel plates are charged. The separation between the plates is 0.050 m and the magnitude of the electric field between the plates is 500 Volt/meter.

What is the change in potential energy of the proton when accelerated from the positively charged plate to the negatively charged plate.

Known : The magnitude of the electric field between the plates (E) = 500 Volt/meter The distance between the plates (s) = 2 cm = 0,02 m The charge on an proton = +1.60 x 10-19 Coulomb Wanted : The change in electric potential energy (ΔPE) Solution : Electric potential : V=Es V = (500 Volt/m)(0.02 m) V = 10 Volt The change in electric potential energy : ΔPE = q V ΔPE = (1,60 x 10-19 C)(10 V) ΔPE = 16 x 10-19 Joule ΔPE = 1.6 x 10-18 Joule Example 3. Two point charges are separated by a distance of 10 cm. Charge on point A is +9 μC and charge on point B is -4 μC . k = 9 x 109 Nm^2C^−2, 1 μC = 10^−6 C. What is the change in electric potential energy of charge on point B if accelerated to point A ?

Known : Charge A (q1) = +9 μC = +9 x 10−6 C Charge B (q1) = -4 μC = -4 x 10−6 C k = 9 x 109 Nm2C−2 The distance between charge A and B (r) = 10 cm = 0.1 m = 10-1 m Wanted : The change in electric potential energy (ΔEP) Solution :

Example 4: The charges, 10q, 8q and -5q are placed in the corners of a right triangle. Find total electric potential energy of the system.

Important Advise:To start solving any question from the tutorial below ,please do make sure that you read and re-read the notes carefully.

If you encounter any

challenges do not forget to consult your lecturer.

TUTORIAL: 1.

Top of Form

If 600 J of work is required to move 50 coulombs of charge between point A and point B in an electric field with an intensity of 120 N/C , what is the distance between these points?( 10 m) 5m 10 m 15 m 18 m 25 m 2. If 60. joules of work is required to move 5.0 coulombs of charge between two points in an electric field, what is the potential difference between these points?( 12 V) 5.0 V 12 V 60. V 300 V 3. A proton moves through a potential difference of 1,000 volts. The change in the

proton’s potential energy will be(1000 eV.) 1,000 eV 2,000 eV 3,000 eV 4,000 eV 4. An electron moves through a potential difference of 100 volts in an electric field.

The change in the electron’s potential energy is:(100 eV) 10eV 20 eV 50 eV 100 eV 150 eV 5.Top of Form

If the potential difference from point A to B is 45 V, what’s the electric field intensity of the uniform electric field?( 150 N/C) 13.5 N/C 150 N/C 135 N/C 45 N/C

1.35 N/C 6. How much work is required to move a single electron through a potential difference of 100. volts?( 1.6 Ă&#x2014; 10-17 J) 1.6 x 10-2 J 1.6 x 10-19 J 1.6 x 10-17 J 1.0 x102 J 7. Moving a point charge of 3.2 x 10-19 C between points A and B in an electric field requires 4.8 x 10-19 J of energy. What is the potential difference between these two points?( 1.5 V) 0.67 V 2.0 V 3.0 V 1.5 V 8. . Moving an electron within an electric field would change the ____ the electron.( C ) a. mass of

b. amount of charge on

c. potential energy of

9. If an electrical circuit were analogous to a water circuit at a water park, then the battery voltage would be comparable to _____.(D ) a. the rate at which water flows through the circuit b. the speed at which water flows through the circuit c. the distance that water flows through the circuit d. the water pressure between the top and bottom of the circuit e. the hindrance caused by obstacles in the path of the moving water 10. If the electrical circuit in your Walkman were analogous to a water circuit at a water park, then the battery would be comparable to _____.(C) a. the people that slide from the elevated positions to the ground

b. the obstacles that stand in the path of the moving water c. the pump that moves water from the ground to the elevated positions d. the pipes through which water flows e. the distance that water flows through the circuit 11. Which of the following is true about the electrical circuit in your flashlight?( E) a. Charge moves around the circuit very fast - nearly as fast as the speed of light. b. The battery supplies the charge (electrons) that moves through the wires. c. The battery supplies the charge (protons) that moves through the wires. d. The charge becomes used up as it passes through the light bulb. e. The battery supplies energy that raises charge from low to high voltage. f. ... nonsense! None of these are true.

12. If a battery provides a high voltage, it can ____.( B ) a. do a lot of work over the course of its lifetime b. do a lot of work on each charge it encounters c. push a lot of charge through a circuit d. last a long time 13. The diagram below at the right shows a light bulb connected by wires to the + and - terminals of a car battery. Use the diagram in answering the next four questions.( A ) 6. Compared to point D, point A is _____ electric potential. a. 12 V higher in b. 12 V lower in c. exactly the same d. ... impossible to tell

14. The energy required to move +2 C of charge between points D and A is ____ J.( E ) justification: A 12 volt battery would supply 12 Joules of electric potential energy per every 1 Coulomb of charge which moves between its negative and positive terminals. The ratio of the change in potential energy to charge is 12:1. Thus, 24 Joules would be the difference in potential energy for 2 Coulombs of charge.)

a. 0.167

b. 2.0

c. 6.0

d. 12

e. 24

Bottom of Form 15.The drawing shows six point charges arranged in a rectangle. The value of q is 8.80 ÎźC, and the distance d is 0.168 m. Find the total electric potential at location P, which is at the center of the rectangle.

16. The following circuit consists of a D-cell and a light bulb. Use >, <, and = symbols to compare the electric potential at A to B and at C to D. Indicate whether the devices add energy to or remove energy from the charge.

Answer:

17.Four identical charges (2C each) are brought from infinity and fixed to a straight line. The charges are located 0.40 m apart. Determine the electric potential energy of this group. (0,090J)

18. Use your understanding of the mathematical relationship between work, potential energy, charge and electric potential difference to complete the following statements: a. A 9-volt battery will increase the potential energy of 1 coulomb of charge by ____ joules. b. A 9-volt battery will increase the potential energy of 2 coulombs of charge by ____ joules. c. A 9-volt battery will increase the potential energy of 0.5 coulombs of charge by ____ joules. d. A ___-volt battery will increase the potential energy of 3 coulombs of charge by 18 joules. e. A ___-volt battery will increase the potential energy of 2 coulombs of charge by 3 joules. f. A 1.5-volt battery will increase the potential energy of ____ coulombs of charge by 0.75 joules. g. A 12-volt battery will increase the potential energy of ____ coulombs of charge by 6 joules.

19.Two point charges q1 = -88.0 nC & q2 = 360 nC are located at the fixed positions at x1 = 22.0 cm and x2 = 96.0 cm on the x-axis.

(A)What is the electric potential due to both charges at x = 50.0 cm?

(B)Where on the x-axis is the electric potential due to both charges equal to zero?

(C)How much work is required to move a third charge q3 =13.0 mC from the origin to x = 50.0 cm along any path (that does not go through one of the charges)?

(C)If a fourth charge q4 with a charge of -11.0 mC and a mass of 3.90 gm is released from rest at the origin, what is the charge's final speed and direction when it is very far away from the other two charges?(Hint:Apply the conservation of energy) 20. A proton is accelerated from rest through a potential of 500 volts. What is its final kinetic energy?

21. What is the final velocity of the proton in the previous problem? (The mass of the proton is 1.67 x 10â&#x20AC;&#x201C;27 kg.)

22. If Q1 and Q2 are equal and oppositely charged, what is the potential at point B?

23. If Q1 is twice Q2 and both are positive, where can a point of zero potential be found?

24. An electric field is created by a Van de Graaf generator. A test charge of (q) is placed in the field at a distance r. a) If the test charge has a magnitude of +1q write an expression for i) the potential energy between the charges ii) the potential of the system b) if the test charge has a magnitude of +2q write an expression for i) the potential energy between the charges ii) the potential of the system

25. The potential difference between a storm cloud and the ground is 100-million volts. If a charge of 2.0 C transfers (via a lightning bolt), how much potential energy was transferred?

26. A system involves a +6μC charge and a +4μC charge initially separated by an infinite distance. a) How much work is required to bring them 150 cm from each other? b) What is the total voltage at a point in the middle of the system?

27. For the uniform field shown on the right, an electron is released from rest from plate B. Determine the: a) work done by the field b) force on the charge c) acceleration of the charge d) the speed of the charge when it reaches plate A e) change in potential energy

+

E = 500 N/C

+

-

+

5 mm

+ A

B

+

28. A charge of -4.0μC is initially 400 mm from a fixed charge or -6.0μC and is then moved to a position 90 cm from the fixed charge. a) What is the change in the mutual potential energy of the charges? b) Does this change depend on the path through which the one charge is moved?

29. proton moving directly toward another fixed proton has a speed of 5.0 m/s when the two are 100 cm apart. How close to the stationary proton will the moving proton be when it stops and reverses course?

30.Two charges, -16 and +4 C are fixed in place and separated by 3 m. Where along the line joining and going through the two charges will the absolute potential be zero?

Locate this point relative to the positive charge.

31. How much work is done to move three 7 µC charges from infinity to the corners of an equilateral triangle with sides 15 cm?

32.Four point charges qA = 2 μC, qB = −5 μC, qC= 2 μC, and qD = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

UNIT 15: ELECTRIC CIRCUITS INTRODUCTION

The information in this unit is essential, if you are to progress to the point where you can operate as a competent electrician.

Objectives By the end of this unit you will be able to: Understand the basic theory of electricity List common conducting materials List common insulating materials Recognise and use important circuit symbols State the units of Current, Voltage and Resistance Calculate circuit values using Ohm’s Law Understand basic circuit protection Understand basic circuit control Explain the term “short circuit” Explain the term “open circuit” Understand the basic effects of an electric current Measure circuit currents using an ammeter Measure circuit voltages using a voltmeter Measure resistor values using an ohmmeter Check continuity of circuits using an ohmmeter Understand and use the basic resistor colour code Perform calculations involving indices Electric power Series and Parallel circuits Reasons

Structure of Matter Our world is made up of various materials. It contains soil, water, rock, sand etc. It is surrounded by an invisible layer of gas, which we call air. The scientific name given to each of these materials is matter. â&#x20AC;&#x153;Matter is anything which occupies space and which has massâ&#x20AC;?

Matter can exist in one of three forms i.e. as a solid, a liquid or a gas. Wood is a solid, water is a liquid and air is a gas. These three forms of matter are known as the States of Matter.

15.1 States of Matter Many of these materials are found in the earth. Coal is mined, oil is found underground and many metals can be found in the rocks of the earth. Some materials come from plants. Sugar comes from beet and rubber comes from a tree. The main properties of each state can be summarised as follows: Solids have a definite shape. Each solid has its own shape and will retain this shape unless it is subjected to heating or considerable force.

Solids have a definite volume. It is very difficult to squeeze them into smaller bulks.

Solids are said to be almost incompressible.

Solids do not flow. They do not spread over a surface.

Liquids have no definite shape. They always adopt the shape of the container into which they are placed.

Liquids have a definite volume. Like solids, they are almost incompressible.

Liquids flow and evaporate When spilled they usually spread over the surface. Most liquids evaporate from open containers i.e. they change to a gaseous form, ( vapour ) at the surface.

Gases 1.Gases have no definite shape. They always take up the shape of the container into which they are placed.

2.Gases have no definite volume. They always spread out in all directions to fill the container into which they are placed. This spreading out of gas to fill all the available space is called diffusion.

3.Gases can be compressed easily. A given volume of gas can be squeezed ( pressurised ) into a smaller volume.

Refer to Figure 1a:

In a solid the particles are arranged in a regular pattern. They cannot be moved out of position. Therefore the solid has a definite shape.

Refer to Figure 1b: In a liquid the particles can slide over one another. Since there are no regular arrangements of particles, the liquid has no shape of its own. A liquid always takes up the shape of its container.

Figure 1a

Figure 1b

Refer to Figure 1c: In a gas the particles are much further apart than in a liquid or a solid. Particles in a gas move into all the space available.

Figure 1c

All matter is composed of atoms. An atom is the basic building block of matter. There are different types of atoms, but all atoms are extremely small.

Atoms are made up of smaller particles called Protons, Neutrons and Electrons.

Definitions Atom:The smallest portion of a material that still exhibits all the characteristics of that material.

Proton:The Proton has a Positive ( + ) charge of electricity. It is situated in the nucleus ( or core ) of the atom.

Neutron:The Neutron is electrically Neutral. It is also situated in the nucleus of the atom.

Electron:The Electron has a Negative ( - ) charge of electricity. Electrons orbit the nucleus of the atom at great speed.

Simplified Representation of Atoms The models of three different atoms are shown in Figures 2a, 2b and 2c. They illustrate how the electron(s) are arranged around the nucleus.

The simplest atom of all - the Hydrogen atom, consists of a single electron orbiting a nucleus, which, is composed of a single proton.

P

E

Hydrogen Atom 1 Electron, 1 Proton

Figure 2a

The carbon atom consists of, 6 electrons orbiting a nucleus of 6 protons and 6 neutrons.

Figure 2b The copper atom consists of, 29 electrons orbiting a nucleus of 29 protons and 35 neutrons.

E E

E

E

E

E

E E

E

E E

E

E E

E

E

E

E E

E

E

E E

E

E

E

E E

E

Copper Atom 29 Electrons, 29 Protons, 35 Neutrons Figure 2c Electrons orbit the nucleus of the atom in shells. The inner shell cannot have any more than two electrons. The copper atom has four shells.

The outer shell is known as the valence shell. The electrons in the outer shell are

more easily dislodged from the atom than the electrons in the inner shells. An atom cannot have more than eight electrons in its outer or valence shell.

15.2 Laws of Electric Charge There are basic laws of nature, which describes the action of electric charges. These laws state:

1.Like charges repel each other Unlike charges attract each other.

Two Negative charges

Figure 3a. Electrons Repel

Two Positive charges

Figure 3b. Protons Repel

A Negative charge and a Positive charge

Figure 3c. Electrons and Protons Attract The Balanced Atom In the previously mentioned examples ( hydrogen, carbon and copper ) you may have noticed that the number of electrons was always equal to the number of protons.

This is normally true of any atom. When this is the case, the atom is said to be neutral, balanced or normal. However, external forces can upset this state.

The Unbalanced Atom An atom that has gained or lost one or more electrons is no longer balanced. An unbalanced atom is called an ion.

The atom that has lost an electron has an overall Positive charge.

The atom that has gained an electron has an overall Negative charge.

Conductors In some materials the electrons in the outer shells are easily dislodged. They can move from atom to atom inside the material. This movement of electrons is electric current flow. Materials, through which electric current can flow freely, are called conductors. Some typical conductors are copper, aluminium, brass, steel, silver and gold.

15.3 Insulators In other materials the electrons are tightly bound to their own particular atoms.

Electric current cannot flow freely through them. These materials are known as insulators. Some typical insulators are ( Poly-vinyl chloride ), PVC, rubber, plastic, glass, porcelain and magnesium oxide.

Graphical Symbols

Cell

Battery

Resistor

Incandescent Lamp

Fuse

One Way Switch

Ammeter

Voltmeter

Ohmmeter

Electric Current Flow Electric current is the movement of free electrons. These electrons have a negative charge and are attracted to a positive charge. When the terminals of a cell are connected via a conductor as shown in Figure 4, free electrons drift purposefully in one direction only. This flow of current, is known as Direct Current ( DC ).

Electron

Flow

Conductor Electron

+

-

Cell

Figure 4 The electrons close to the positive plate of the cell are attracted to it. Each electron that enters the positive plate causes an electron to leave the negative plate and move through the conductor. The number of electrons in the conductor remains constant.

The movement of electrons through a conductor is from negative to positive. Long before this theory was discovered, it was thought that current flowed from positive to negative. This direction of current flow is called conventional current flow.

Electron

+

-

Electron Flow from - to +

Flow

See Figure 5.

Conventional Current Flow

+

-

Conventional Current Flow from + to -

Figure 5 This movement of electrons through a conductor is known as an electric current and is measured in Amperes.

The Ampere

The symbol for current flow is I. The Ampere ( Amp ), is the unit of measurement for current flow.

When 6.28 x 1018 electrons pass a given point in one Second, a current of one Ampere is said to flow.

See Figure 6.

Point

6,280,000,000,000,000,000 Electron Flow

+

Cell

Figure 6

The Coulomb

This number of electrons is exceptionally large. A unit called the Coulomb is used to represent this figure.

A coulomb is the quantity ( charge ) of electricity, which passes a point when a steady current of 1 Ampere flows for one Second.

The symbol for Quantity of electricity is Q.

Formula

Where:

Q

Q

=

=

I x t

Quantity of electricity ( Coulombs )

I

=

Current flow in Amperes ( Amps )

t

=

the time for which current flows, measured in Seconds.

Examples

A current of 8.5 Amps passes through a point in a circuit for 3 Minutes. What quantity of electricity is transferred?

Solution:

Q

=

I x t

I

=

8.5 Amps

t

=

3 Minutes

Q

=

8.5 x 3 x 60

Q

=

1530 Coulombs

=

3 x 60 Seconds

2.A current of 1.5 Amps transfers a charge of 1800 Coulombs. For what period of time did the current flow?

Solution:

Q

=

I x t

I

=

1.5

Q

=

1800 Coulombs

Amps

To find t, the formula must be transformed. Divide both sides by I.

Q

=

I x t

Q t

=

— I

1800 t

=

—— 1.5

t

=

1200 Seconds

1200 t

=

——

Minutes

60

t

=

20

Minutes

3.What current will flow in a circuit if 540 Coulombs is transferred in 2 Minutes?

Solution:

Q

=

I x t

Q

=

540 Coulombs

t

=

2 Minutes

=

2 x 60 Seconds

To find I, the formula must be transformed. Divide both sides by t.

Q

=

I x t

Q —

=

I

t

Q I

=

— t

540 I =

—— 2 x 60

I

=

4.5 Amps

Sample Questions A current of 15 Amps flows for 3 Minutes. What charge is transferred?

For how long must a current of 3 Amps flow so as to transfer a charge of 240 Coulombs?

What current must flow if 100 Coulombs is to be transferred in 5 Seconds?

The Electrical Circuit For continuous current flow, we must be a complete circuit. If the circuit is broken, by opening a switch for example, electron movement and therefore the current will stop immediately. To cause a current to flow through a circuit, a driving force is required,

just as a circulating pump is required to drive water around a central heating system.

See Figure 7.

Radiator

Pump

WaterPressure or

Flow

Driving Force

Figure 7

This driving force is the electromotive force ( abbreviated to EMF ). It is the energy, which causes current to flow through a circuit. Each time an electron passes through the source of EMF, more energy is provided to keep it moving.

See Figure 8.

A circuit must have:

1. A source of supply ( EMF ).

2. A load ( Lamp ).

3. Connecting cables ( Conductors ).

Figure 8.

1.

The source of supply is always associated with energy conversion.

(a)

Generator ( converts mechanical energy to electrical energy )

(b)

Cell or Battery ( converts chemical energy to electrical energy )

The source of supply will provide pressure called Electromotive Force or Voltage.

The symbol for voltage is U. 2.The load is any device that is placed in the electrical circuit that produces an effect when an electric current flows through it. When an electric current flows through an incandescent lamp, the lamp gives off light from heat.

3.The connecting leads or cables complete the circuit. The cable consists of the conductor to carry the current and insulation to prevent leakage. A water pipe must have a bore to carry the water and the pipe material ( e.g. copper ) to prevent leakage.

15.4 Circuit Analogy The simplest analogy of an electric circuit is to consider a hosepipe connected to a tap. The rate of flow of water from the end of the hosepipe will depend upon: The water pressure at the tap.

The diameter of the hosepipe The restriction / resistance of the inner walls of the hosepipe.

The degree of any bends or kinks in the hosepipe.

If there are many restrictions, the water will flow out of the hosepipe at a reduced rate.

See Figure 9.

Off

On

Pressure

Rate of Flow

Kink or Reduced Pipe Size = Opposition or Restriction

Figure 9.

In much the same way, current flows through conductors by means of electric pressure provided by a battery or generating source. This source of electric pressure, electromotive force ( EMF ), provides the energy required to push current through the circuit. It can be referred to as the supply voltage.

Every circuit offers some opposition or restriction to current flow, which is called circuit resistance. The unit of resistance is the Ohm, symbol ď &#x2014;, pronounced Omega.

At this stage, conductor resistance is ignored and the load resistance is treated as the total opposition to current flow.

For a stable supply voltage, the current ( I ) which flows, is determined by resistance ( R ) of the circuit. There will be a voltage drop across different parts of the circuit and this is called Potential Difference ( PD ).

See Figure 10.

Figure 10 Unlike the hosepipe analogy, the electric circuit requires a “go” and “return” conductor to form a closed loop or complete circuit. These conductors must offer a low resistance path to the flow of current. Most metallic conductors satisfy this requirement.

Ohm’s Law George Ohm discovered the relationship between, current flowing in a circuit and the pressure applied across that circuit. This became known as Ohm’s Law. Ohm’s Law states that the current ( I ) flowing through a circuit is directly proportional to the potential difference ( U ), across that circuit, and inversely proportional to the resistance ( R ), of that circuit, provided the temperature remains constant.

U I

=

— R

To find U, transpose the formula by multiplying both sides of the equation by R.

U I

=

— R

U x R I x R=

——— R

I x R=

U

or: U

=

I x R

To find R, divide both sides by I as follows:

U

U

=

I x R

=

R

=I x R

U — I or

R=U/I

15.5 The Magic Triangle

U I

R

Now consider any circuit in which you know the values of any two of the three factors - voltage, current and resistance - and you want to find the third.

The rule for working the â&#x20AC;&#x153;Magic Triangleâ&#x20AC;? to give the correct formula is as follows: Place your thumb over the letter in the triangle whose value you want to know - and the formula for calculating that value is given by the two remaining letters.

U= U=

I=

I

R

U R

R=

U I

To Confirm Ohmâ&#x20AC;&#x2122;s Law Experiment Refer to Figure 11. In this arrangement the resistor value is kept constant whilst the voltage is increased in steps of two volts and current readings are taken.

In this Circuit the Resistance is Constant

Figure 11.

The following results were obtained from the experiment and plotted in graph form as shown below.

Voltage

=

I

x

R

2

=

0.02

x

100

4

=

0.04

x

100

6

=

0.06

x

100

8

=

0.08

x

100

10

=

0.10

x

100

Voltage

10 9 8 7 6 5 4 3 2 1 .01

.02 .03 .04 .05 .06 .07 .08 .09

.10

Amps Figure 12

The above graph and experiment illustrates, that current flow increases proportionally as the applied voltage is increased.

Examples

1.An electrical lamp used on a 230 Volt supply takes a current of 0.42 Amps. What is the resistance of the lamp?

Solution: U=RI U

=

230 Volts

I

=

0.42 Amps

230 R

=

â&#x20AC;&#x201D;â&#x20AC;&#x201D; 0.42

R

=

547 Ohms ( Hot resistance )

An immersion heater connected to a 230 Volts supply takes a current of 13.5 Amps.

Calculate the resistance of the element.

Solution: U R

=

— I

U

=

230 Volts

I

=

13.5 Amps

230 R

=

—— 13.5

R

=

17 Ohms

An electric heater has a resistance of 23  and is connected to a 230 Volts supply.

Calculate the current the heater will take.

Solution: U=RI U

=

230 Volts

R

=

23 Ohms

230 I

=

—— 23

I

=

10 Amps

An electrical circuit has a resistance of 23  and takes a current of 5 Amps. Calculate the voltage applied to the circuit.

Solution:

U

=

I x R

I

=

5 Amps

R

=

23 Ohms

U

=

5 x 23

U

=

115 Volts

From the previous exercises it will be noticed that the amount of current that flows in a circuit is directly proportional to the voltage and inversely proportional to the resistance.

With a fairly constant supply of 230 Volts, the load or resistance of the circuit will determine the amount of current that will flow.

Electrical Circuit Requirements Circuit Protection One of the basic requirements that a circuit must have is overcurrent protection. This is essential for protection of the cables and accessories in the circuit. A fuse or circuit breaker is used to provide this protection. It is fitted as close to the origin of the circuit as possible to cut off the supply if too much current flows in the circuit. This is called circuit protection.

See Figure 13.

Internal resistance

Internal resistance model of a source of voltage A practical electrical power source which is a linear electric circuit may, according to ThĂŠvenin's theorem, be represented as an ideal voltage source in series with an impedance. This impedance is termed the internal resistance of the source. When the power source delivers current, the measured voltage output is lower than the noload voltage; the difference is the voltage drop (the product of current and resistance) caused by the internal resistance. The concept of internal resistance applies to all kinds of electrical sources and is useful for analyzing many types of electrical circuits.

Internal resistance usually means the electrical resistance inside batteries and power supplies that can limit the potential difference that can be supplied to an external load.

Circuit Control

Another basic requirement is that the circuit can be controlled. A switch must be fitted to turn the supply on or off. This is called circuit control.

The principles of circuit protection and circuit control are illustrated in Figure .

Circuit Load Control Circuit Protection Source

Switch Fuse

of Cell supply

Figure .

Basic Circuits Concepts

Resistor

Open Circuit An open circuit exists, when there is a break in a circuit. This break results in an extremely high resistance in the circuit. This will stop current flow. This value of extremely high resistance is referred to as infinity. It is denoted by the symbol .

Examples: See Figure 14.

A switch is open.

A fuse is blown or a circuit breaker is tripped.

A physical break in the resistor or element.

A connecting cable is broken.

1

3

2 4

Figure 14.

Assume a supply of 230 Volts and the circuit resistance of 1,000 Ohms.

U From Ohm’s Law I

=

— R

230 I

=

—— 1,000

I

=

0.23 Amps

Assume a supply of 230 Volts and the circuit resistance of 1,000.000 Ohms.

U From Ohm’s Law I

=

— R

230 I

=

———— 1,000,000

I

=

0.00023 Amps

Short Circuit

Current will flow through the path of least resistance or opposition in a circuit.

A short circuit occurs when the resistance or opposition in a circuit is very low.

Examples: See Figures 15, 15A and 15B.

The “load” is shorted out and the current takes the path of least resistance.

Short Circuit 230V Supply

Figure 15.

The connecting cables are damaged prior to or after the wiring process.

230V Supply

Short Circuit

Figure 15A.

The connecting cables in the circuit are connected together during the wiring process.

230V Supply

Short Circuit

Figure 15B. A short circuit usually results in a dangerously high current flowing. A fuse or circuit breaker is the deliberate “weak link” in a circuit. Either will open the circuit when too much current flows.

Assume a supply of 230 Volts and the circuit resistance of 1 Ohm.

U From Ohm’s Law I

=

— R

230 I

=

—— 1

I

=

230 Amps

Assume a supply of 230 Volts and the circuit resistance of 0.1 Ohm.

U From Ohm’s Law I

=

— R

230 I

=

—— 0.1

I

=

2300 Amps

Assume a supply of 230 Volts and the circuit resistance of 0.01 Ohm.

U From Ohm’s Law I

=

— R

230 I

=

—— 0.01

I

=

23,000 Amps

The Effects of an Electric Current

When an electric current flows in a circuit it can have one or more of the following effects:

Heating Effect

Magnetic Effect

Chemical Effect

The Heating Effect

The movement of electrons in a circuit, which is the flow of an electric current, causes an increase in the temperature of the load resistance. The huge number of electrons being pushed through the load resistance, results in high friction and collision of these electrons. This generates heat. The amount of heat generated depends upon the type and dimensions of the load resistance wire and the value of current flowing. By changing these variables, a length of resistance wire may be operated at different temperatures to give different effects, e.g. an ordinary light bulb or an electric heater.

See Figure 16.

Switch Fuse

Load Resistance

MCB Cell

Figure 16.

The Magnetic Effect

Whenever a current flows in a conductor a magnetic field is set up around that conductor.

See Figure 17 below.

Switch Direction of Current Flow

Magnetic field lines

Figure 17.

This magnetic field increases in strength if the current is increased and collapses if the current is switched off. A “current carrying conductor”, wound in the form of a solenoid ( coil ), produces a magnetic field very similar to that of a permanent magnet, but has the advantage in that it can be switched on or off by any switch controlling the circuit current.

The magnetic effect of an electric current is the principle upon which electric bells, relays, moving coil instruments, motors and generators work.

The strength of the magnetic field is directly proportional to the current in the circuit.

The “clamp on” type ammeter measures the strength of the magnetic field and produces a reading in Amps.

The Chemical Effect

When an electric current flows through an electrolyte ( conducting liquid / paste ), this electrolyte is separated into chemical parts. The two conductors, which make contact with the electrolyte, are called the anode ( positive plate ) and the cathode ( negative plate ).

An anode or cathode of dissimilar metals placed in an electrolyte can react chemically and produce an EMF. When a load is connected across the anode and cathode, a current will flow in the circuit.

The chemical effect of an electric current is the principle upon which electric cells operate.

See Figure 18.

R +

Cathode

Anode

Liquid Figure 18.

SI Units

A unit is what we use to indicate the measurement of a quantity. For example, the unit of current is the Ampere. The unit of length could be the Inch or the Metre. However, the Metre is the SI unit of length.

In order that we all work to a common standard, an international system is used. It is known as the SI system ( System International ). This system is used throughout the course. A number of prefixes will be used e.g. microamps and milliamps, millivolts and kilovolts, kilohms and megohms.

Metric Prefixes

Symbol

Multiplying factor

Power index

mega*

M

1000000

106

kilo*

k

1000

103

hecto

h

100

102

deca

da

unity

10

101

1

102

deci

d

0,1

10-1

centi

c

0,01

10-2

milli*

m

0,001

10-3

micro*

ď ­

0.000 001

10-6

nano

n

0,000000001

10--9

pico

p

0,00000000000 1

10-12

denotes frequently used prefixes in the Electrical Trade.

Examples To convert amps to milliamps multiply by 1,000 (103) To convert amps to microamps multiply by 1,000,000 (106)

To convert milliamps to amps multiply by 0.001 (10-3) To convert microamps to milliamps multiply by 0.001 (10-3) Examples

Convert the following:

1.

13,000,000 Ohms to

megohms

2.

500,000 Ohms

to

megohms

3.

0.5 Volts

to

millivolts

4.

100 Volts

to

millivolts

5.

15,000 Volts to

kilovolts

6.

0.4 kilovolts to

Volts

7.

600 milliamps

to

Amps

8.

0.03 Amps

to

milliamps

=

1 megohm

Solutions

1.

1,000,000 Ohms 13,000,000 Ohms =

=

13 megohms

=

1M 13M

2.

500,000 Ohms

=

0.5 megohms

=

0.5M

3.

1 Volt

=

1000 millivolts

=

1000mV

0.5 Volts

=

500 millivolts

=

500mV

4.

100 Volts

=

100,000 millivolts= 100,000mV

5.

1,000 Volts

=

1 kilovolt

15,000 Volts =

15 kilovolts

1 kilovolt

=

6.

1,000 Volts

= =

1kV

15kV

=

1000V

7.

8.

0.4 kilovolts =

400 Volts

=

400V

1,000 milliamps

=

1 Amp

600 milliamps

=

0.6 Amps

1 Amp

=

1,000 milliamps

=

1000mA

0.1 Amps

=

100 milliamps

=

100mA

0.01 Amps

=

10 milliamps

=

10mA

0.03 Amps

=

30 milliamps

=

30mA

=

1A

= 0.6A

Electrical Measuring Instruments

The Ammeter An ammeter is a device used to measure current flowing through a circuit or part of a circuit. It indicates, in terms of amperes, the number of coulombs passing a given point in a circuit, in one second. The ammeter shown in Figure 1 has a Full Scale Deflection ( FSD ) of 40A.

Figure 19.

The ammeter must be physically connected into the circuit. It can then measure all the electrons passing through it.

The ammeter in Figure 20 is said to be in series with the resistor or load.

+

A

-

Ammeter

+

Resistor or Load

-

Figure 20.

Note: The ammeter may be damaged if incorrectly connected.

The positive and negative terminals of a direct current ( DC ) meter must be connected correctly as shown in Figure 2. This is referred to as correct polarity. It allows the meter to read up-scale. Reversed polarity causes the meter to read down scale. This forces the pointer against the stop at the left, which may damage the meter.

Meter Range and Selection

The current required to be measured should be estimated. Select a meter having a FSD a good deal higher than the estimated current. Any current value in excess of the FSD of the meter will not only fail to register properly on the scale, but will probably cause serious damage to the meter.

On the other hand, any current value very low in relation to the FSD will not cause the pointer to move. An accurate reading cannot be obtained in this situation. The useful range of any meter never, in fact, extends right down to zero on its scale. It only goes down to the point at which readings can be distinguished from zero with reasonable accuracy.

Activity

Apprentices to install an ammeter into a circuit as shown in Figure 21 and measure the current.

Figure 21.

The Voltmeter

A voltmeter measures electromotive force ( EMF ) or potential difference ( PD ). It must be connected across the supply or load resistance in order to record the voltage. That is, it must be connected in parallel with the component as in Figure 22.

Figure 22.

Note: The voltmeter may be damaged if incorrectly connected.

A DC voltmeter ( analogue type ) must be connected with the correct polarity for the

meter to read up scale. When selecting a meter to measure voltage, choose one having a maximum range a good deal higher than the value of any voltage you expect to be measuring. The reason for this is that a voltage in excess of the maximum rated value of the meter will not only fail to register properly on the scale, but will probably cause serious damage to the meter.

Activity

Measure the voltage at the supply terminals and the potential difference across the load of the circuit shown in Figure b.

Figure b.

The Ohmmeter

An ohmmeter is used to check the electrical continuity of components and to measure their resistance. It is powered by its own internal battery. A typical ohmmeter is depicted in Figure 24. Before connecting an ohmmeter in circuit it is important to ensure that:

There is no voltage across the component ( supply disconnected ).

The component to be measured is not connected in parallel with any other component.

The instrument has been set to infinity ď&#x201A;Ľ with the leads separated.

The instrument has been set to Zero Ohms with the leads connected together.

Scale

Infinity Adjustment Scale Multiplier Switch Zero Ohms Adjustment

Figure 24.

Measurements are taken by connecting the meter across the unknown resistor as shown in Figure 725. It is important to select the most suitable scale for the resistance under test. The current flowing through the unknown resistance will cause the pointer to deflect. This deflection, when multiplied by the scale factor ( range selection switch setting ), will give the value of the resistance being measured. The ohmmeter scale is normally non-linear with zero on the opposite side of the scale to that used for voltage and current measurement.

ď &#x2014; Figure 25.

The Multimeter

A multimeter is an instrument that can be set or programmed to measure voltage, current or resistance. Most modern multimeters have a number of other functions such as diode checking and capacitance measurement. These instruments can measure Direct Current ( DC ) or Alternating Current ( AC ) over several ranges.

The main types are: Volt-Ohm-Milli-ammeter ( VOM ) and Digital Multimeter ( DMM )

Comparison of Analogue and Digital Multimeters

Analogue Type

Digital Type

Features:

Features:

Analogue Display

Digital Display

Manual Range Setting

Range Setting Automatic / Manual

It is important that the instruction manual supplied with a meter is studied prior to operation of the meter. These manuals normally contain warnings and information, which must be followed to ensure safe operation and retain the meter in a safe condition.

Meter Operating Suggestions Ensure that the instrument is set to measure the desired unit e.g. Volts to measure voltage.

Set the range switch to the proper position before making any measurement.

Ensure the instrument test leads are connected to the appropriate jack sockets. When the voltage, current or resistance to be measured is not known, always start with the highest range first and work your way down to a lower range that gives an accurate reading.

Always observe correct test lead polarity when making DC voltage and current measurements.

For most accurate readings, look at the scale from a position where the pointer and its reflection on the mirror come together to avoid parallax error. Wherever possible, use a range setting, which results in a reading in the centre 1/3rd of the meter scale.

Set the range selector switch to the "Off" position when the tester is not in use or during transit.

Remove the battery before storing the meter for a long period of time.

Great care must be taken to ensure that the instrument range setting is not exceeded when measuring a voltage or current.

Resistors

All materials have some resistance to the flow of an electric current. In general, the term resistor describes a conductor specially chosen for its resistive properties.

Resistors are the most commonly used electronic components. They are made in a variety of ways to suit particular types of application. They are usually manufactured as either carbon composition or carbon film. In both cases the base resistive material is carbon. The general appearance is of a small cylinder with leads protruding from each end.

Resistor Colour Code

The value of the resistor and its tolerance may be marked on the body of the component. This may be done either by direct numerical indication or by using a standard colour code. The coloured bands are located on the component towards one end. If the resistor is turned so that this end is towards the left, then the bands are read from left to right, See Figure 26.

First Band

Second Third Band Band

Fourth Band

Figure 26.

Explanation of Colour Coding Bands ( 4 Band Resistors )

1st Band Colour

-

1st Digit

2nd Band Colour -

2nd Digit

3rd Band Colour

-

Multiplier ( in effect the number of zeros )

4th Band Colour

-

Tolerance

Resistor Colour Code Values

Black

0

Brown

1

Red

2

Orange

3

Yellow

4

Green

5

Blue

6

Violet

7

Grey

8

White

9

Resistor Tolerance

The fourth resistor colour band indicates the resistor tolerance. This is commonly gold or silver, indicating a tolerance of 5% or 10% respectively. Sometimes the coloured bands are not clearly oriented towards one end. In this case, first identify the tolerance band and turn the resistor so that this is to the right. Then read the colour code as described below.

The tolerance band indicates the maximum tolerance variation in the declared value of resistance. Thus a 100  resistor with a 5% tolerance will have a value of somewhere between 95  and 105 , since 5% of 100  is 5 .

If no fourth band exists this indicates a tolerance of + / - 20%.

Tolerance indicators

Gold

Silver

No band

5%

10%

20%

Reading a resistor colour code.

Blue

Gold Grey

Red

Figure 27.

If a resistor has colour code bands of blue, grey, red, gold, what is its ohmic value?

Solution: Blue

=

6

1st Digit

Grey

=

8

2nd Digit

Red

=

2

Number of zeros

Gold

=

5%

Tolerance

Resistor value is 6,800 ď &#x2014; with a 5% tolerance

This resistor will have an ohmic value between 6,460 and 7,140 ohms.

Activity

Practice reading the colour code of different values of resistors and calculate the tolerance range.

Resistor Colour Code Mnemonic

Mnemonic to Remember the Resistor Colour Code: Better Be Right Or Your Great Big Venture Goes Wrong

Resistor Preferred Values It is difficult to manufacture small electronic resistors to exact values by mass production methods. This is not a disadvantage as in most electronic circuits the values of the resistors is not critical. Manufacturers produce a limited range of PREFERRED resistance values rather than an overwhelming number of individual resistance values. Therefore, in electronics we use the preferred value closest to the actual value required.

A resistor with a preferred value of 100  and a 10% tolerance could actually have any value between 90  and 110 . The next largest preferred value, which would give the maximum possible range of resistance values without too much overlap, would be 120 . This could have any value between 108  and 132 .

Table 1 indicates the preferred values between 10  and 82 , but the larger values, which are manufactured, can be obtained by multiplying these preferred values by a factor of 10.

Example:

47 ,

( 47 x 10 )

=

470 

=

470R

470 ,( 470 x 10 )

=

4,700 

=

4.7k

=

47k

4,700 ,

( 4,700 x 10 ) =

47,000 

47,000 

( 47,000 x 10 )

=

470,000 

470,000 

( 470,000 x 10 )

=

4,700,000 

=

470k =

4.7M

For a number of reasons the decimal point is not used. Take the 4·7k resistor, this may be written 4k7. The k is positioned so that it indicates where the decimal point should be. The R and M can be used in the same manner as in the examples below.

Examples:

18 

=

18R

2·7 k

=

2k7

(2,700R)

39 k

=

39k

(39,000R)

5·6 M

=

5M6

(5,600,000R)

82 M

=

82M

(82,000,000R)

E12 Series of Preferred Values

( 10% Tolerance )

Rated Value

Possible Range of Values

in Ohms

in Ohms

10

9.0 to 11.0

12

10.8

to 13.2

15

13.5

to 16.5

18

16.2

to 19.8

22

19.8

to 24.2

27

24.3

to 29.7

33

29.7

to 36.3

and so on in multiples of 10

Table 1 Indices

It is very important to understand what Indices are and how they are used. Without such knowledge, calculations and manipulation of formulae are difficult and frustrating.

So, what are Indices? Well, they are perhaps most easily explained by example. If we multiply two identical numbers, say 2 and 2, the answer is, clearly, 4, and this process is usually expressed thus:

22 = 4

However, another way of expressing the same condition is:

22 = 4

The upper 2 simply means that the lower 2 is multiplied by itself. The upper 2 is known as the index. Sometimes this situation is referred to as ‘two raised to the power of two’. So 23 means ‘two multiplied by itself three times’.

i.e.

222 = 8

Do not be misled by thinking that 23 is 2 x 3 24 = 2 x 2 x 2 x 2 = 16 ( not 2 x 4 = 8 ) 242 = 24 x 24 = 576 ( not 24 x 2 = 48 )

Examples:

33 = 3 x 3 x 3 = 27 92 = 9 x 9 = 81 43 = 4 x 4 x 4 = 64

105 = 10 x 10 x 10 x 10 x 10 = 100,000

A number by itself, say 3, has an invisible index, 1, but it is not shown. Now consider this:

22 x 22 x 2 may be written as 2 x 2 x 2 x 2 x 2, or as 25, which means, that the indices 2 and 2 or the invisible index 1 have been added together. So the rule is, when multiplying like figures, add the indices.

Examples:

4 x 42 = 41 x 42 = 43 = 4 x 4 x 4 = 64 32 x 33 = 35 = 3 x 3 x 3 x 3 x 3 = 243 10 x 103 = 104 = 10 x 10 x 10 x 10 = 10,000 Multiplication of Indices

RULE 1

32

x

33

=

35

YM

x

YN

=

YM+N

When multiplying powers of the same number add the indices

Let us advance to the following situation:

1 104 x

104

—

is the same as

102

— 102

10 x 10 x 10 x 10 =

——————— 10 x 10

10 x 10 x 10 x 10 ————————

Cancelling out the 10s

10 x 10

We get 10 x 10 = 10^2

which means that the indices have been subtracted i.e. 4 - 2.

YM —

=

YM-N

YN

RULE 2

When dividing powers of the same number, subtract the indices

How about this: 4 - 2 is either 4 subtract 2, or 4 add -2, and remember, the addition of indices is used with multiplication. So from this we should see that 104 divided by 102 is the same as 104 multiplied by 10-2. So

1 —

is the same as 10-2

10^2 Examples:

1 —

1 = 3-4

34

—

= 2-6

26

and conversely:-

1 —

= 102

10-2

Hence we can see that indices may be moved above or below the line providing the sign is changed.

Examples:

1.10^6 x 10^7 x 10^-3

10^13 x 10^-3

——————

=

————

10^4 x 10^2

=

10^6

10^10 ——

= 10^10 x 10^-6

=

10^4

=

10,000

10^6

2.

10^4 x 10^-6 ————

=

10^4 x 10^-6 x 10^-1

=

10

=10^4 x 10^-7

= 10-3

=

1/10^3

=

1 = ——

=

0.001

1000 Examples

1.

What is the resistance of a circuit if the voltage across it is 50 kV and a current of 500 mA is flowing?

Solution:

I

=

R=

U

=

500 mA

=

50 kV

=

50 x 103 Volts

500 x 10-3 Amps

U I

50  103 R= 500  10 − 3 50  106 R= 500 R=

50,000,000 500

R=

500,000 5

R = 100,000 Ohms

2.What current will flow through a circuit if the supply voltage is 50 kV and the circuit resistance is 5 megohms?

Solution: U

=

R

=

50 kV

5 M=

=

50 x 10^3 Volts

5 x 10^6 Ohms

I=

U R

3 50  10 I= 6 5  10

I=

3 −6 50  10  10 5

I=

50  10

−3

5

I=

10  10

−3

1 I = 0.01 Amps I = 10 milliamps

3.What voltage will cause a current of 50 mA to flow through a resistance of 1 k ?

Solution: I

R

=

50 mA

=

50 x 10^-3 Amps

=

1 k

=

1 x 10^3 Ohms

U = IR U = 50  10 − 3  1  103 U = 50 Volts Electric power(P):

Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit. The SI unit of power is the watt, one joule per second.

Electric power is usually produced by electric generators, but can also be supplied by sources such as electric batteries. It is usually supplied to businesses and homes (as domestic mains electricity) by the electric power industry through an electric power grid. Electric power is usually sold by the kilowatt hour (3.6 MJ) which is the product of the power in kilowatts multiplied by running time in hours. Electric utilities measure power using an electricity meter, which keeps a running total of the electric energy delivered to a customer.

Electrical power provides a low entropy form of energy and can be carried long distances and converted into other forms of energy such as motion, light or heat with high energy efficiency.

Electric power is transmitted on overhead lines like these, and also on underground high-voltage cables.

Power Formula 1 – Electrical power equation: Power P = I × V = R × I^2 = V^2 ⁄ R where power P is in watts, voltage V is in volts and current I is in amperes (DC).

If there is AC, look also at the power factor PF = cos φ and φ = power factor angle (phase angle) between voltage and amperage.

Electric Energy is E = P × t − measured in watt-hours, or also in kWh. 1J = 1N×m = 1W×s Power Formula 2 – Mechanical power equation: Power P = E ⁄ t where power P is in watts, Power P = work / time (W ⁄ t). Energy E is in joules, and time t is in seconds. 1 W = 1 J/s.

Power = force multiplied by displacement divided by time P = F × s / t or Power = force multiplied by speed (velocity) P = F × v.

Undistorted powerful sound is not found in these formulas. Please, mind your ears!

The eardrum and microphone diaphragms are really only moved by the waves of the sound pressure. That does not do either the intensity, nor the power or the energy.

If you are in the audio recording business, it is wise not to care much about the energy, power and intensity as the cause, care more about the effect of sound pressure p and sound pressure level at the ears and at the microphones and look at the corresponding audio voltage V ~ p; see: Sound pressure and Sound power − Effect and Cause Very loud sounding speakers will have a lot of power, but better look closer at the very important efficiency of loudspeakers. This includes the typical question: How many decibels (dB) are actually twice or three times as loud?

There is really no RMS power. The words "RMS power" are not correct. There is a calculation of power which is the multiplication of a RMS voltage and a RMS current.

Watts RMS is meaningless. In fact, we use that term as an extreme shorthand for power in watts calculated from measuring the RMS voltage. Please, read here: Why there is no such thing as 'RMS watts' or 'watts RMS' and never has been.

"RMS" power is a rather silly term which has gathered currency among audio people.

Power is the amount of energy that is converted to a unit of time. Expect to pay more when demanding higher power.

Parallel Circuits A parallel circuit is one that has two or more paths for the electricity to flow, the loads are parallel to each other. If the loads in this circuit were light bulbs and one blew out, there is still current flowing to the others because they are still in a direct path from the negative to positive terminals of the battery.

UNDERSTANDING & CALCULATING PARALLEL CIRCUITS EXPLANATION A Parallel circuit is one with several different paths for the electricity to travel. It's like a river that has been divided up into smaller streams, however, all the streams come back to the same point to form the river once again.

The parallel circuit has very different characteristics than a series circuit. For one, the total resistance of a Parallel Circuit is NOT equal to the sum of the resistors (like in a series circuit). The total resistance in a parallel circuit is always less than any of the branch resistances. Adding more parallel resistances to the paths causes the total resistance in the circuit to decrease. As you add more and more branches to the circuit the total current will increase because Ohm's Law states that the lower the resistance, the higher the current.

BASIC RULES A Parallel circuit has certain characteristics and basic rules: A parallel circuit has two or more paths for current to flow through.

Voltage is the same across each component of the parallel circuit.

The sum of the currents through each path is equal to the total current that flows from the source.

You can find total resistance in a Parallel circuit with the following formula: 1/Rt = 1/R1 + 1/R2 + 1/R3 +...

Rt = R (t)otal If one of the parallel paths is broken, current will continue to flow in all the other paths.

1. "A parallel circuit has two or more paths for current to flow through."

Simply remember that PARALLEL means two paths up to thousands of paths. The flow of electricity is divided between each according to the resistance along each route.

2. "Voltage is the same across each component of the parallel circuit."

You may remember from the last section that the voltage drops across a resistor in series. Not so with a parallel circuit. The voltage will be the same anywhere in the circuit.

3. "The sum of the currents through each path is equal to the total current that flows from the source."

If one path is drawing 1 amp and the other is drawing 1 amp then the total is 2 amps at the source. If there are 4 branches in this same 2 amp circuit, then one path may draw 1/4A (.25A), the next 1/4A (.25), the next 1/2A (.5A) and the last 1A. Don't worry, the next rule will show you how to figure this out.

Simply remember for now that the branch currents must be equal to the source current.

4."You can find TOTAL RESISTANCE in a Parallel circuit with the following formula: 1/Rt = 1/R1 + 1/R2 + 1/R3 + ... " Before we get into the calculations, remember what we said at the start of this section: "The total resistance of a parallel circuit is NOT equal to the sum of the resistors (like in a series circuit).

We will use a parallel circuit with 3 paths as an example (it could be 2, 4 or a 1000 resistors in parallel). The power source is providing 12 volts and the value of the resistors are 5 Ohms, 5 Ohms and 2 Ohms.

Let's summize this EXAMPLE: Voltage = 12V R1 = 4 Ohm R2 = 4 Ohm R3 = 2 Ohm Remember that "Rt" means Total resistance of the circuit.

R1, R2, etc. are Resistor one, Resistor two, etc.

Now we will apply the formula above to this example: 1 1 1 1 —= —+ —+ — Rt R1 R2 R3 Therefore: 1 1 1 1 —= —+ —+ — Rt 4 4 2 It is easiest to change the fractions into decimal numbers (example 1 divide by 4 equals .25): 1/Rt = .25 + .25 + .5 1/Rt = 1 Now you have to get rid of the 1 on the left side so...

Rt = 1/1 Rt = 1 Ohms NOW, Let's try a more complex one: Voltage = 12 Volts R1 = 10 Ohms R2 = 20 Ohms R3 = 10 Ohms R4 = 1 Ohms 1/Rt = 1/10 + 1/20 + 1/10 + 1/1

1/Rt = .1 + .05 + .1 + 1 1/Rt = 1.25 Rt = 1/1.25 = .8 Ohms Try this examples

.

The equation for calculating total resistance in a parallel circuit (for any number of

parallel resistances) is sometimes written like this:

Rtotal = (R1−1 + R2−1 + …Rn−1)−1 Re-write this equation in such a way that it no longer contains any exponents

.

Determine the amount of voltage impressed across each resistor in this circuit:

Hint: locate all the points in this circuit that are electrically common to one another!

. According to Ohm’s Law, how much current goes through each of the two resistors in this circuit?

Draw the paths of all currents in this circuit.

Example 2: Consider three resistors of 2 , 6 , and 10  connected in series with a 9 volt battery.

Draw a schematic diagram of the circuit which includes an ammeter to measure the current through the 2  resistor, and a voltmeter to measure the voltage across the 10  resistor, and indicate the reading on the ammeter and voltmeter.

I1 A 2Ω

ε=9V

6Ω

I2

10Ω

I3 V

The current through each resistor is equal to the total current in the circuit, so the ammeter will read the total current regardless of where it is placed, as long as it is placed in series with the resistances.

V 9V I total = total = = 0.5 A Rtotal 2 + 6 + 10 The voltmeter will read the voltage across the 10  resistor, which is NOT 9 volts. The 9 volts provided by the battery is divided proportionally among the resistances. The voltage across the 10  resistor is V10 = I10R10 = (0.5 A)(10 ) = 5 V.

Before we move on to the last rule here's how easy it is to calculate the amperage through each path using OHM'S LAW. In the example we see a 12 and 24 ohm resistor in parallel with a 12 volt source. First we figure out the total resistance of the circuit: 1/Rt = 1/12 + 1/24 Rt = 8 Ohms

Now that you know this you can figure out the total amperage (It) using Ohm's Law: I total (It) = 12V / 8 Ohms = 1.5 Amps Therefore the total amperage between the two resistive paths must equal 1.5 Amps (Rule 3). Now we can figure out exactly what each path is pulling using Ohm's Law once more. Remember that the voltage is the same everywhere in a parallel circuit.

So we know the voltage and the resistance: I1 = 12V / 12 Ohm = 1 A I2 = 12V / 24 Ohm = .5 A We figured the total amperage (It) previously, so now we can double check if the figures are correct: I1 + I2 = It 1A + .5A = 1.5A â&#x20AC;&#x201C; check

5. "If one of the parallel paths is broken, current will continue to flow in all the other paths."

The best way to illustrate this is also with a string of light bulbs in paralallel. If one is burnt out, the others stay lit.

EXAMPLE: RESISTORS IN PARALLEL Question In the following schematic diagram, find the total current, I.

Figure 1 Example Problem: Resistors in Parallel Hints You will need Ohm's Law. How are resistors related when connected in parallel?

What is the potential drop across each resistor?

How does current behave in parallel branches?

Solution We know the total potential of this circuit, = 12.0 V So, between points A and B, the potential must drop 12.0V. Also, the potential drop across branches of a circuit are equal. That is,

We can use Ohm's Law V = IR or I = V/R to find the current across each resistor.

Recall that the currents through branches of a parallel circuit add to give the total

current. That is, the total current 'splits up' so that part of the total current travels down each branch. Because of conservation of charge, the sum of the currents in each branch must equal the amount going into the branch. (This is Kirchhoff's Current Law.) So, adding up the three currents, we get:

So, the total current is I = 12.0A.

USEFUL FORMULA TO BE MEMORIZED:

Series Circuit A series circuit is one with all the loads in a row. There is only ONE path for the electricity to flow. If this circuit was a string of light bulbs, and one blew out, the remaining bulbs would turn off.

UNDERSTANDING & CALCULATING SERIES CIRCUITS BASIC RULES Select the rule below to see an expanded explanation The same current flows through each part of a series circuit.

The total resistance of a series circuit is equal to the sum of individual resistances.

Voltage applied to a series circuit is equal to the sum of the individual voltage drops.

The voltage drop across a resistor in a series circuit is directly proportional to the size of the resistor.

If the circuit is broken at any point, no current will flow.

"1. The same current flows through each part of a series circuit."

In a series circuit, the amperage at any point in the circuit is the same. This will help in calculating circuit values using Ohm's Law. You will notice from the diagram that 1 amp continually flows through the circuit. We will get to the calculations in a moment.

"2. The total resistance of a series circuit is equal to the sum of individual resistances."

In a series circuit you will need to calculate the total resistance of the circuit in order to figure out the amperage. This is done by adding up the individual values of each component in series.

In this example we have three resistors. To calculate the total resistance we use the formula: RT = R1 + R2 + R3 2 + 2 + 3 = 7 Ohms R total is 7 Ohms

Now with these two rules we can learn how to calculate the amperage of a circuit.

Remember from Ohms Law that I = V / R. Now we will modify this slightly and say I = V / R total.

Lets follow our example figure: RT = R1 + R2 + R3 RT = 7 Ohms I = V / RT I = 12V / 7 Ohms I = 1.7 Amp If we had the amperage already and wanted to know the voltage, we can use Ohm's Law as well.

V = I x R total V = 1.7 A x 7 Ohms V = 12 V

"Voltage Drops" Before we go any further let's define what a "voltage drop" is. A voltage drop is the amount the voltage lowers when crossing a component from the negative side to the positive side in a series circuit. If you placed a multimeter across a resistor, the voltage drop would be the amount of voltage you are reading. This is pictured with the red arrow in the diagram.

Say a battery is supplying 12 volts to a circuit of two resistors; each having a value of 5 Ohms. According to the previous rules we figure out the total resistance.: RT = R1 + R2 = 5 = 5 = 10 Ohms Next we calculate the amperage in the circuit: I = V / RT = 12V / 10 Ohms = 1.2 Amp Now that we know the amperage for the circuit (remember the amperage does not change in a series circuit) we can calculate what the voltage drops across each

resistor is using Ohm's Law (V = I x R).

V1 = 1.2A x 5 Ohms = 6 V V2 = 1.2A x 5 Ohms = 6V

"3. Voltage applied to a series circuit is equal to the sum of the individual voltage drops."

This simply means that the voltage drops have to add up to the voltage coming from the battey or batteries.

V total = V1 + V2 + V3 ...

In our example above, this means that 6V + 6V = 12V. "4. The voltage drop across a resistor in a series circuit is directly proportional to the size of the resistor."

This is what we described in the Voltage Drop section above.

Voltage drop = Current times Resistor size.

"5. If the circuit is broken at any point, no current will flow."

The best way to illustrate this is with a string of light bulbs. If one is burnt out, the whole thing stops working.

EXAMPLES: RESISTORS IN SERIES Question The current flowing in a circuit containing four resistors connected in series is I = 1.0 A. The potential drops across the first, second and third resistors are, respectively: V = 5 V, V = 8 Vand V = 7 V. The equivalent resistance of the circuit is R = 30 .

Find the total voltage supplied by the battery, and also current, voltage drop, and resistance of each resistor in the circuit.

Figure 1 Example Problem: Resistors in series Hints How are resistors related when connected in series?

What is true about potential drops of resistors when connected in series?

You will need to use Ohm's Law. Solution

First, let's label the diagram with the information given in the question.

There are several ways of solving this problem (see alternate solutions), but this tutorial will only go through one of these ways.

Figure 2 Example Problem, with given data Because the resistors are connected in series, then the same current flows through each one. Using the Ohm's Law, we can find the resistances of the first, second and third resistors.

Now, using the equivalent resistance, we can find the resistance in the fourth resistor.

This is a series circuit, so the equivalent resistance is the sum of the individual resistances.

The current flowing through the fourth resistor is also I=1.0A. Using Ohm's Law again, we find the voltage across this resistor.

The total voltage supplied by the battery must equal to the total voltage drop across the circuit (this is known as Kirchhoff's Voltage Law). So, we must sum up the voltage drops across the resistors.

USEFUL FORMULA TO BE MEMORIZED:

Since V = I R, then

RESISTORS IN COMBINATION CIRCUITS: Resistors in Series and Parallel Example 1 Find the equivalent resistance, REQ for the following resistor combination circuit.

Again, at first glance this resistor ladder network may seem a complicated task, but as before it is just a combination of series and parallel resistors connected together.

Starting from the right hand side and using the simplified equation for two parallel resistors, we can find the equivalent resistance of the R8 to R10 combination and call it RA.

RA is in series with R7 therefore the total resistance will be RA + R7 = 4 + 8 = 12Ί as shown.

This resistive value of 12立 is now in parallel with R6 and can be calculated as RB.

RB is in series with R5 therefore the total resistance will be RB + R5 = 4 + 4 = 8立 as shown.

This resistive value of 8立 is now in parallel with R4 and can be calculated as RC as shown.

RC is in series with R3 therefore the total resistance will be RC + R3 = 8立 as shown.

This resistive value of 8Ί is now in parallel with R2 from which we can calculated RD as:

RD is in series with R1 therefore the total resistance will be RD + R1 = 4 + 6 = 10Ί as shown.

Then the complex combinational resistive network above comprising of ten individual resistors connected together in series and parallel combinations can be replaced with just one single equivalent resistance ( REQ ) of value 10Ί.

When solving any combinational resistor circuit that is made up of resistors in series and parallel branches, the first step we need to take is to identify the simple series and parallel resistor branches and replace them with equivalent resistors.

This step will allow us to reduce the complexity of the circuit and help us transform a complex combinational resistive circuit into a single equivalent resistance remembering that series circuits are voltage dividers and parallel circuits are current dividers.

However, calculations of more complex T-pad Attenuator and resistive bridge networks which cannot be reduced to a simple parallel or series circuit using

equivalent resistances require a different approach. These more complex circuits need to be solved using Kirchhoffâ&#x20AC;&#x2122;s Current Law, and Kirchhoffâ&#x20AC;&#x2122;s Voltage Law which will be developed in physic 2.

EXAMPLE 2 Question

Hints Which resistors are in parallel and which are in series?

Is this circuit composed of small groups of parallel resistors, all connected in series?

Or is it composed of groups of series resistors, connected in parallel?

Figure 1 Combination Circuit 1 Solution This circuit is composed of 3 'elements' connected in series: the group of parallel resistors between A and B, the single resistor R3, and the group of parallel resistors between C and D. First, we will find the equivalent resistance between A and B. Here, we have two resistors, R1 and R2, connected in parallel. Using the formula for resistors connected in parallel:

we can find the equivalent resistance between points A and B. Let's call this equivalent resistance RAB.

Now, we'll find the equivalent resistance between C and D, and will call it RCD. Using the equation from above for resistors connected in parallel,

Replacing the two parallel sections with their equivalent resistances, and redrawing the circuit, we get the circuit in Figure 2. We see that there are three resistances connected in series: RAB, R3, and RCD. Using the formula for resistors in series,

we can find the equivalent resistance of the circuit.

Figure 2 Simplified Version of Circuit 1

So the equivalent resistance of this circuit is R = 6.7

.

TUTORIAL: 1. Use your understanding of equivalent resistance to complete the following statements: a. Two 3- resistors placed in series would provide a resistance which is equivalent to one _____- resistor. b. Three 3- resistors placed in series would provide a resistance which is equivalent to one _____- resistor. c. Three 5- resistors placed in series would provide a resistance which is equivalent to one _____- resistor.

d. Three resistors with resistance values of 2- , 4- , and 6- are placed in series.

These would provide a resistance which is equivalent to one _____- resistor. e. Three resistors with resistance values of 5- , 6- , and 7- are placed in series.

These would provide a resistance which is equivalent to one _____- resistor. f. Three resistors with resistance values of 12- , 3- , and 21- are placed in series.

These would provide a resistance which is equivalent to one _____- resistor.

2. As the number of resistors in a series circuit increases, the overall resistance __________ (increases, decreases, remains the same) and the current in the circuit __________ (increases, decreases, remains the same).

4. Three identical light bulbs are connected to a battery as shown shown at the right.

Which one of the following statements is true? a. All three bulbs will have the same brightness. b. The bulb between X and Y will be the brightness.

c. The bulb between Y and Z will be the brightest. d. The bulb between Z and the battery will be the brightest.

5. Three identical light bulbs are connected to a battery as shown at the right. Which adjustments could be made to the circuit that would increase the current being measured at X? List all that apply. a. Increase the resistance of one of the bulbs. b. Increase the resistance of two of the bulbs. c. Decrease the resistance of two of the bulbs. d. Increase the voltage of the battery. e. Decrease the voltage of the battery. f. Remove one of the bulbs.

6. Three identical light bulbs are connected to a battery as shown at the right. W, X,Y and Z represent locations along circuit. Which one of the following statements is true?

the

a. The potential difference between X and Y is greater than that between Y and Z. b. The potential difference between X and Y is greater than that between Y and W. c. The potential difference between Y and Z is greater than that between Y and W. d. The potential difference between X and Z is greater than that between Z and W. e. The potential difference between X and W is greater than that across the battery. f. The potential difference between X and Y is greater than that between Z and W. 7. Compare circuit X and Y. Each are powered 12-volt battery. The voltage drop across the 12resistor in circuit Y is ____ the voltage drop the single resistor in X.

by a ohm across

a. smaller than b. larger than c. the same as 8. A 12-V battery, a 12-ohm resistor and a light bulb are connected as shown in circuit X below. A 6-ohm resistor is added to the 12-ohm resistor and bulb to create circuit Y as shown. The bulb will appear ____. a. dimmer in circuit X b. dimmer in circuit Y c. the same brightness in both circuits 9. Three resistors are connected in series. If placed in a circuit with a 12-volt power supply. Determine the equivalent resistance, the total circuit current, and the voltage drop across and current through each resistor.

10.The Figure below shows part of a circuit. It consists of resistors combined in both parallel and series configurations. Find the equivalent resistance.

11. The Figure below shows part of a circuit. It consists of resistors combined in both parallel and series configurations. Find the equivalent resistance and the current I2 in each figure below.

Figure a

Figure b

Figure c 12. Which graph indicates the relationship between the potential difference with the electric current?Explain.

13. An emf source of 6.0V is connected to a purely resistive lamp and a current of 2.0 amperes flows. All the wires are resistance-free. What is the resistance of the lamp 14. What is the power of the 20Ί resistor?

15. What is the total power of R2?

16.

If the resistor shown in the diagram above has 150 coulombs of charge passing through it in 1 minute, how much current has passed through the resistor?

17.(a) If a wire with current 2.0A has 12 coulombs of charge pass through it, how much time has passed? (b) What is the voltage of a circuit with power 12 W and current 2 A? 18. What is the power of the 20 Ω resistor in the circuit?

19. In a hearing aid a battery supplies a current of 25.0 mA through a resistance of 400 ΩΩ. When the volume is increased, the resistance is changed to 100 ΩΩ and the current rises to 60 mA. What is the emf and internal resistance of the cell?(10v; 1,1428×10−21,1428×10−2 Ω)

20. When a cell is connected directly across a high resistance voltmeter the reading is 1,5 V. When the cell is shorted through a low resistance ammeter the current is 2,5A. What is the emf and internal resistance of the cell?(1,5v; 0,6 Ί) 21. For the following circuit, calculate: a) the potential differences V1, V2 and V3 across the resistors R1, R2, and R3 b) the resistance of R3. c) the resistance of R3. d) If the internal resistance is 0,1 Ί, what is the emf of the battery and what power is dissipated by the internal resistance of the battery?

Answers: V1=2,0 V V2=6,0 V V3=10,0 V R3=7,5Ί E=23,2 V

Pr=0,4 W

22.Given the following circuit:

The current leaving the battery is 1,07A, the total power dissipated in the external circuit is 6,42 W, the ratio of the total resistances of the two parallel networks RP1:RP2 is 1/2, the ratio R1:R2 is 3:5 and R3=7,00 Ί Determine the: potential difference of the battery, the power dissipated in RP1 and RP2, and If the battery is labelled as having an emf of 6,50V what is the value of the resistance of each resistor and the power dissipated in each of them.

23. The headlamp and two IDENTICAL tail lamps of a scooter are connected in parallel to a battery with unknown internal resistance as shown in the simplified circuit diagram below. The headlamp has a resistance of 2,4Ί and is controlled by switch S1.

The tail lamps are controlled by switch S2. The resistance of the connecting wires may be ignored.

The graph below shows the potential difference across the terminals of the battery before and after switch S1 is closed (whilst switch S2 is open). Switch S1 is closed at time t1.

1.Use the graph below to determine the emf of the battery.

WITH ONLY SWITCH S1 CLOSED, calculate the following: 1.Current through the headlamp 2.Internal resistance, r, of the battery 3. BOTH SWITCHES S1 AND S2 ARE NOW CLOSED. The battery delivers a current of 6 A during this period.Calculate the resistance of each tail lamp.

4. How will the reading on the voltmeter be affected if the headlamp burns out? (Both switches S1 and S2 are still closed.)Write down only INCREASES, DECREASES or REMAINS THE SAME.Give an explanation.

24. A battery charger is connected to a dead battery and delivers a current of 6.0 A for 5.0 hours, keeping the voltage across the battery terminals at 12 V in the process.

How much energy is delivered to the battery? (1.3.10^6 J )

25. There is current of 0.40 A through a light bulb. (a) How much charge is transported through the bulb in 3 h? (b) How many electrons flow through the bulb in this time?

(4320 C, 2.697 x 10^22)

26. A certain car battery maintains a current of 2.0 A for 7.0 h. How much charge flows from the battery during this time? (50400 C)

27.A 40-W fluorescent lamp is designed to operate on 120 V. In operation, (a) how much current does it draw and (b) what is its resistance? (0.333 A, 360 Ί)

UNIT 16: Fluids

INTRODUCTION: Fluids really know how to move! As flowing materials, fluids have some unique properties that help us understand and describe them. In this lesson you'll explore what fluids are and how they are defined in the world of physics.

What Is a Fluid?

Let's take a trip to the beach. It's quite relaxing here, isn't it? The white sand, the blue water, the fresh air - just thinking about it makes me want to do some physics! Okay, that was a mean trick, but while we're here, we might as well use this great location and learn a little bit about the different phases of matter.

You are probably quite familiar with these already because they are solids (the sand), liquids (the water) and gases (the fresh air). Solids are pretty different from liquids and gases because they hold their shape. You can't put a square block into a round hole because the block is solid and won't conform to the shape of the hole.

Liquids and gases are more similar to each other because, unlike solids, they both conform to the shape of their container. They also both flow when moved around, and any material that flows is called a fluid. When you put air in your car tires, they stay inflated because the gas particles fill the entire space inside. Likewise, when you fill your bathtub, you can't keep the water on just one side because it flows to cover all the space of the tub.

16.1 Properties of Fluids There are certain properties that fluids share, though the specifics of these may be slightly different for each type of fluid. The main difference between the two fluids mentioned here is that gas particles are much farther apart than the particles of a liquid. Both will spread out to fill their container, but a liquid only does so beneath its surface.

This important difference helps us understand that a gas is compressible, which means its volume can easily be increased or decreased, while a liquid is incompressible, meaning its volume cannot easily be changed. In other words, you can more easily press gas particles together than you can the particles of a liquid.

This is because there is more space between those gas particles, while the liquid particles are already about as close together as they can get.

One property that all fluids do share is that they have density. This is simply the amount of matter in a given space for that substance. Another way of saying this is that the density is the amount of matter per unit volume, or in equation form: density = mass/volume.

Definition: a fluid is a substance that continually deforms (flows) under an applied shear stress, or external force. Fluids are a phase of matter and include liquids, gases and plasmas.

Mass density: The ratio of the mass of a substance to its volume is known as the mass densityor, simply, the density of the substance. Density is expressed in units of mass per volume, such as g/mL or kg/m3. Because the density of a substance does not depend on the amount of substance present, density is an â&#x20AC;&#x153;intensive propertyâ&#x20AC;?.

.

Density is an intensive property of a substance that doesnâ&#x20AC;&#x2122;t depend on the amount of substance present. Thus, density can be used to identify an unknown pure substance if a list of reference densities is available, and the experimenter can choose a convenient amount of substance to work with when measuring density.

To measure the density of a sample of a substance, it is necessary to measure its mass and volume. Mass is typically measured using an analytical balance, a precise instrument that relies on the force exerted by the sample due to gravity. The container to hold the sample (also used to measure volume) is weighed and tared, so only the sample mass appears on the balance display when the sample is added to the container.

For liquids, this container is typically a volumetric flask, which has one marking that corresponds to a specific volume. The container is filled to the line with the liquid sample and weighed again after the empty flask has been tared. The measured density is the ratio of the measured mass to the volume indicated on the flask.

Most solid substances are irregularly shaped, which complicates volume determination. It is inaccurate, for example, to determine the volume of a powder by measuring its dimensions. Instead of directly measuring dimensions or using glassware like a volumetric flask, it is necessary to make use of a liquid displacement method to measure the volume of an irregularly shaped solid. A graduated cylinder containing a known volume of liquid (in which the solid is insoluble) is tared. The solid is added to the cylinder, and the total mass is weighed again to determine the mass of the solid. Addition of the solid causes an upward displacement of the liquid, resulting in a new volume reading. The volume of the solid is equal to the change in volume due to liquid displacement (i.e., the difference in liquid volume before and after adding solid).

As for liquids, the measured density of a solid sample is the ratio of the measured mass to the measured volume.

PROCEDURE 1. Determination of the Density of Liquid Ethanol 1.1.Place a clean and dry 50-mL volumetric flask on an analytical balance.

1.2.Press the “Tare” or “Zero” button on the balance. The balance should read 0.000 g.

1.3.Use a buret funnel to add 45 mL of liquid ethanol to the volumetric flask.

1.4.Use a Pasteur pipette to add the final 5 mL of liquid, just until the bottom of the liquid’s meniscus touches the marking on the flask.

1.5..Weigh the volumetric flask again and record the mass of the ethanol.

1.6.For best results, repeat steps 1.1 – 1.5 twice more to obtain two additional density measurements.

2. Determination of the Density of Solid Zinc Metal 2.1.Add approximately 40 mL of water to a clean and dry 100-mL graduated cylinder.

Record the exact volume of the water.

2.2.Place the cylinder and water on an analytical balance. Press the “Tare” or “Zero” button on the balance. The balance should read 0.000 g.

2.3.Add approximately 10 zinc pellets to the graduated cylinder. Record the new volume of the water plus zinc pellets using the liquid level after addition of the zinc (Figure a ).

Figure a :LAB EXPERIMENT CONDUCTED BY OFFREDI BICLAIR AND BASSIE( UNISA MASTER STUDENT )Zinc added to the cylinder on the right causes the water level to be displaced upward.

Weigh the cylinder, water, and zinc pellets on the balance. Record the mass of the zinc pellets.

For best results, repeat steps 2.1 – 2.4 twice more to obtain two additional density measurements.

NB:In the SI always take

 water = 1.000  103 kg

m3

Pressure: Pressure is defined as force per unit area. It is usually more convenient to use pressure rather than force to describe the influences upon fluid behavior. The

standard unit for pressure is the Pascal, which is a Newton per square meter.

For an object sitting on a surface, the force pressing on the surface is the weight of the object, but in different orientations it might have a different area in contact with the surface and therefore exert a different pressure.

NB:In the SI always take

Patm=1.013 ď&#x201A;´ 105 Pa = 1 atm

There are many physical situations where pressure is the most important variable. If you are peeling an apple, then pressure is the key variable: if the knife is sharp, then the area of contact is small and you can peel with less force exerted on the blade. If you must get an injection, then pressure is the most important variable in getting the needle through your skin: it is better to have a sharp needle than a dull one since the smaller area of contact implies that less force is required to push the needle through the skin.

When you deal with the pressure of a liquid at rest, the medium is treated as a continuous distribution of matter. But when you deal with a gas pressure, it must be approached as an average pressure from molecular collisions with the walls.

Pressure in a fluid can be seen to be a measure of energy per unit volume by means of the definition of work. This energy is related to other forms of fluid energy by the Bernoulli equation.

Pressure in a Static Fluid and depth in a static fluid:

The formula that gives the P pressure on an object submerged in a fluid is:

P=ρ*g*h where ρ (rho) is the density of the fluid, g is the acceleration of gravity h is the height of the fluid above the object If the container is open to the atmosphere above, the added pressure must be included if one is to find the total pressure on an object. The total pressure is the same as absolute pressure on pressure gauges readings, while the gauge pressure is the same as the fluid pressure alone, not including atmospheric pressure.

Ptotal = Patmosphere + Pfluid

Ptotal = Patmosphere + (ρ * g * h ) A Pascal is the unit of pressure in the metric system. It represents 1 newton/m2 Example: Find the pressure on a scuba diver when she is 12 meters below the surface of the ocean. Assume standard atmospheric conditions.

Solution: The density of sea water is 1.03 X 10 3 kg/m3 and the atmospheric pressure is 1.01 x 105 N/m2.

Pfluid = ρ g h = (1.03 x10 3 kg/m3) (9.8 m/s2) (12 m) = 1.21 x 105 Newtons/m2 Ptotal = Patmosphere + Pfluid = (1.01 x 105) + (1.21 x 105 ) Pa = 2.22 x 10 2 kPa (kilo Pascals) In general, the Pressure in a Static Fluid is given by the formula: P2 = P1 +  gh Pressure Gauge: Gauge pressure is a relative pressure measurement which measures pressure relative to atmospheric pressure and is defined as the absolute pressure minus the atmospheric pressure. Most pressure measuring equipment give the pressure of a system in terms of gauge pressure as opposed to absolute pressure.

If you limp into a gas station with a nearly flat tire, you will notice the tire gauge on the airline reads nearly zero when you begin to fill it. In fact, if there were a gaping hole in

your tire, the gauge would read zero, even though atmospheric pressure exists in the tire. Why does the gauge read zero? There is no mystery here. Tire gauges are simply designed to read zero at atmospheric pressure and positive when pressure is greater than atmospheric.

Similarly, atmospheric pressure adds to blood pressure in every part of the circulatory system. (As noted in Pascal’s Principle, the total pressure in a fluid is the sum of the pressures from different sources—here, the heart and the atmosphere.)

But atmospheric pressure has no net effect on blood flow since it adds to the pressure coming out of the heart and going back into it, too. What is important is how much greater blood pressure is than atmospheric pressure. Blood pressure measurements, like tire pressures, are thus made relative to atmospheric pressure.

In brief, it is very common for pressure gauges to ignore atmospheric pressure—that is, to read zero at atmospheric pressure. We therefore define gauge pressure to be the pressure relative to atmospheric pressure. Gauge pressure is positive for pressures above atmospheric pressure, and negative for pressures below it.

Gauge pressure is positive for pressures above atmospheric pressure, and negative for pressures below it.In fact, atmospheric pressure does add to the pressure in any fluid not enclosed in a rigid container. This happens because of Pascal’s principle.

The total pressure, or absolute pressure, is thus the sum of gauge pressure and atmospheric pressure: Pabs =Pg +Patm, where Pabs is absolute pressure, Pg is gauge pressure, and Patm is atmospheric pressure. For example, if your tire gauge reads 34 psi (pounds per square inch), then the absolute pressure is 34 psi plus 14.7 psi (Patm in psi), or 48.7 psi (equivalent to 336 kPa).

Absolute Pressure Absolute pressure is the sum of gauge pressure and atmospheric pressure.

For reasons we will explore later, in most cases the absolute pressure in fluids cannot be negative. Fluids push rather than pull, so the smallest absolute pressure is zero.

(A negative absolute pressure is a pull.) Thus the smallest possible gauge pressure is Pg =-Patm (this makes Pabs zero). There is no theoretical limit to how large a gauge pressure can be.

There are a host of devices for measuring pressure, ranging from tire gauges to blood pressure cuffs. Pascal’s principle is of major importance in these devices. The undiminished transmission of pressure through a fluid allows precise remote sensing of pressures. Remote sensing is often more convenient than putting a measuring device into a system, such as a person’s artery.

This aneroid gauge utilizes flexible bellows connected to a mechanical indicator to measure pressure.

An entire class of gauges uses the property that pressure due to the weight of a fluid is given by P= ρgh Consider the U-shaped tube shown in fig below, for example. This simple tube is called a manometer. In fig (a), both sides of the tube are open to the atmosphere. Atmospheric pressure therefore pushes down on each side equally so its effect cancels. If the fluid is deeper on one side, there is a greater pressure on the deeper side, and the fluid flows away from that side until the depths are equal.

16.2 Relative density Relative density is also known as specific gravity. The relative density of a substance is a pure number without any unit. It tells how many times a substance is heavier than water. Relative density(R.D or ρr) of a substance can be calculated by dividingdensity of a substance with the density of water. Ρr= ρs/ ρw ρr = Relative density ρs = Density of substance ρw = Density of water

Example :

Density: If 5 m3 of certain oil weighs 45 kN calculate the specific weight, specific gravity and mass density of the oil.

Solution :

Given data: Volume = 5 m3 Weight = 45 kN

Answer:

; 0.917;

Let us examine how a manometer is used to measure pressure. Suppose one side of the U-tube is connected to some source of pressure such as the toy balloon in fig (b) or the vacuum-packed peanut jar shown in fig (c). Pressure is transmitted undiminished to the manometer, and the fluid levels are no longer equal. In fig (b), Pabs is greater than atmospheric pressure, whereas in fig (c), Pabs is less than atmospheric pressure. In both cases, Pabs differs from atmospheric pressure by an amount hgρ, where h is the density of the fluid in the manometer. In fig (b), Pabs can support a column of fluid of height h, and so it must exert a pressure hg ρ greater than atmospheric pressure (the gauge pressure is positive). In fig (c), atmospheric pressure can support a column of fluid of height h, and so Pabs is less than atmospheric pressure by an amount hg ρ (the gauge pressure is negative). A manometer with one side open to the atmosphere is an ideal device for measuring gauge pressures. The gauge pressure is and is found by measuring .

An open-tube manometer has one side open to the atmosphere. (a) Fluid depth must be the same on both sides, or the pressure each side exerts at the bottom will be unequal and there will be flow from the deeper side. (b) A positive gauge pressure transmitted to one side of the manometer can support a column of fluid of height . (c) Similarly, atmospheric pressure is greater than a negative gauge pressure by an amount hgρ. The jar’s rigidity prevents atmospheric pressure from being transmitted to the peanuts.

Mercury manometers are often used to measure arterial blood pressure. An inflatable cuff is placed on the upper arm as shown below. By squeezing the bulb, the person making the measurement exerts pressure, which is transmitted undiminished to both the main artery in the arm and the manometer. When this applied pressure exceeds blood pressure, blood flow below the cuff is cut off. The person making the measurement then slowly lowers the applied pressure and listens for blood flow to resume. Blood pressure pulsates because of the pumping action of the heart, reaching a maximum, called systolic pressure, and a minimum, called diastolic pressure, with each heartbeat. Systolic pressure is measured by noting the value of when blood flow first begins as cuff pressure is lowered. Diastolic pressure is measured by noting when blood flows without interruption. The typical blood pressure of a young adult raises the mercury to a height of 120 mm at systolic and 80 mm at diastolic. This is commonly quoted as 120 over 80, or 120/80. The first pressure is representative of the maximum output of the heart; the second is due to the elasticity of the arteries in maintaining the pressure between beats. The density of the mercury fluid in the manometer is 13.6 times greater than water, so the height of the fluid will be 1/13.6 of that in a water manometer. This reduced height can make measurements difficult, so mercury manometers are used to measure larger pressures, such as blood pressure. The density of mercury is such that .

Systolic Pressure Systolic pressure is the maximum blood pressure.

Diastolic Pressure Diastolic pressure is the minimum blood pressure.In routine blood pressure measurements, an inflatable cuff is placed on the upper arm at the same level as the heart. Blood flow is detected just below the cuff, and corresponding pressures are transmitted to a mercury-filled manometer. (credit: U.S. Army photo by Spc. Micah E. Clare\4TH BCT).

Pascal’s Principle: in any fluid, the static pressure is exerted on the walls of the container and in the fluid. These forces act perpendicular to the walls of the container.

When an external pressure is applied to the fluid, the pressure is distributed uniformly in all parts of the fluid and this is known as Pascal’s principle named after the Physicist Blaise Pascal.The Pascal’s principle refers to only the external pressure and within the fluid the pressure at the bottom is greater than the top.

According to Pascal’s principle, the force per unit area describes an external pressure which is transmitted through fluid and the formula is written as,

Example 1: In a hydraulic system, a piston has a cross-sectional area of 20 square centimeters pushes on an incompressible liquid with a force of 30 newtons. The other end of the hydraulic pipe connects to the second piston with a cross-sectional surface area of 100 square centimeters. Calculate the force on the second piston.

Solution: Pascal’s principle is given by,

We know that, F1 = 30 newtons, A1 = 20 square centimeters A2 = 100 square centimeters Insert the given values to calculate the force on the second piston: F2 = A2 x F1/A1 = 100 x 30 / 20 = 150 N

Example 2: A piston has a cross-sectional area of 40 square centimeters in a hydraulic system and pushes an incompressible liquid with a force of 50 newtons. The other end of the hydraulic pipe is connected to a second piston with a cross-sectional surface area of 50 square centimeters. Determine the force on the second piston.

Solution: Pascalâ&#x20AC;&#x2122;s principle is given by, We know that, F1 = 50 newtons, A1 = 40 square centimeters A2 = 50 square centimeters Insert the given values to calculate the force on the second piston: F2 = A2 x F1/A1 = 50 x 50 / 40 = 62.5 N Steady flow: The Steady Flow Energy Equation (SFEE) is used for open systems to determine the total energy flows. It is assumed that the mass flow through the system is constant. It is also assumed that the total energy input to the system is equal to the total energy output and the velocity of the fluid particles at any point is constant.

16.3 Incompressible fluid:

The density of the fluid remains constant as the pressure changes. When the value of Mach number crosses above 0.3, density begins to vary and the amplitude of variation spikes when Mach number reaches and exceeds unity.

The behavior of control volume (CV) for incompressible and compressible flow is depicted in the image below.

It can be seen that the CV remains constant for a flow that is incompressible and CV is squeezed for compressible flow.

Bernoulli's equation is applicable only when flow is assumed to be incompressible. In case of compressible flow, Bernoulli's equation becomes invalid since the very basic assumption for Bernoulli's equation is density r is constant .For compressible flow,

.

Discharge(Q):The amount of fluid passing a section of a stream in unit time is called the discharge. If v is the mean velocity and A is the cross sectional area, the discharge Q is given by.

Q = Av =velocity .Area =VOLUME/TIME The unit of the discharge is m3 /s The equation of continuity: A continuity equation in physics is an equation that describes the transport of some quantity. It is particularly simple and powerful when applied to a conserved quantity, but it can be generalized to apply to any extensive quantity. ... Continuity equations are a stronger, local form of conservation laws.

When a fluid is in motion, it must move in such a way that mass is conserved. To see how mass conservation places restrictions on the velocity field, consider the steady flow of fluid through a duct (that is, the inlet and outlet flows do not vary with time). The inflow and outflow are one-dimensional, so that the velocity V and density \rho are constant over the area A (figure 14).

Derivation of Streamline Flow Consider three parts (P, R, and Q) in planes that are present in the perpendicular direction to the fluid. Refer to the diagram below

We will be able to determine the boundaries the selected points in the plane by the same set of the streamlines. Hence, the particles of fluid passing through the surfaces at the three-point P, R and Q will be the same. Now consider the area at the three points as AP, AR, and AQ. Consider the speed of the fluid particles as vP, vR, and vq.

Now, we will calculate the mass of fluids. The mass of the fluid mP crossing at the area AP at a small time interval t will be “ρP AP vP t”. Similarly, the mass of fluid mR and mQ will be “ρRAR vR t” and “ρQ AQ vQ t” passing at AP and AQ respectively at a small interval of time Ät.

In all the three cases, the mass of liquid flowing in and flowing will be equal. Therefore, we can write the equation as ρP AP vP t = ρRAR vR t= ρQ AQ vQ t If we consider the fluids as incompressible in nature then “ρP = ρR = ρQ“ will be equal. So the above equation can be rewritten as AP vP = AR vR = AQ vQ (after elimination of ρ) The above equation is the equation of continuity. The equation also represents conservation of mass in case of the flow of the incompressible liquids.

General Equation of Continuity Av = Constant where Av is the flow rate of the liquid or volume flux of the liquid. The flow rate in case of streamline flow remains constant throughout the flow of liquid through the pipe.

Hence, the streamlines in narrower regions are present closely thereby resulting in the increase of velocity and vice versa.

Figure 14. One-dimensional duct showing control volume.

Now we apply the principle of mass conservation. Since there is no flow through the side walls of the duct, what mass comes in over A_1 goes out of A_2, (the flow is steady so that there is no mass accumulation). Over a short time interval \Delta t,

Thus ,A1 V 1=A 2 V2 , A = Cross sectional area (m2) v = fluid speed (m.s-1) This is a statement of the principle of mass conservation for a steady, onedimensional flow, with one inlet and one outlet. This equation is called the continuity equation for steady one-dimensional flow. For a steady flow through a control volume with many inlets and outlets, the net mass flow must be zero, where inflows are negative and outflows are positive.

16.4 Bernoulli’s equation: Bernoulli’s theorem which is known as Bernoulli’s principle, states that an increase in the speed of moving air or a flowing fluid is accompanied by a decrease in the air or fluid’s pressure. Swiss scientist, Daniel Bernoulli (1700-1782), demonstrated that, in most cases the pressure in a liquid or gas decreases as the liquid or gas move faster. This is an important principle involving the movement of a fluid through the pressure difference. Suppose a fluid is moving in a horizontal direction and encounters a pressure difference. This pressure difference will result in a net force, which is by Newton’s Second Law will cause an acceleration of the fluid.

Bernoulli’s theorem states that the total energy (pressure energy, potential energy and kinetic energy) of an incompressible and non-viscous fluid in steady flow through a pipe remains constant throughout the flow, provided there is no source or sink of the

fluid along the length of the pipe. This statement is due to the assumption that there is no loss energy due to friction. This theorem deals with the facts that when there is slow flow in a fluid, there will be increase in pressure and when there is increased flow in a fluid, there will be decrease in pressure. If the elevation remains constant, velocity and pressure, energy to or from the system can be calculated by this equation Static pressure + dynamic pressure = total pressure = constant Static pressure + 1/2 x density x velocity2 = total pressure = constant . i.e P1+ρv 1 2 /2

+ρgh1=P2+ρv 2 2 /2 +ρgh2

The variables P1, v1, h1 refer to the pressure, speed, and height of the fluid at point 1, whereas the variables P2, v2, and h2 refer to the pressure, speed, and height of the fluid at point 2 as seen in the diagram below. The diagram below shows one particular choice of two points (1 and 2) in the fluid, but Bernoulli's equation will hold for any two points in the fluid.

When using Bernoulli's equation, how do you know where to choose your points?

Choosing one of the points at the location where you want to find an unknown variable is a must. Otherwise how will you ever solve for that variable? You will typically choose the second point at a location where you have been given some information, or where the fluid is open to the atmosphere, since the absolute pressure there is known to be atmospheric pressure Patm=1.01×10 5 Pa. Note that the h refers to the height of the fluid above an arbitrary level that you can choose in any way that is convenient. Typically it is often easiest to just choose the lower of the two points (1 or 2) as the height where h=0.

The P refers to the pressure at that point. You can choose to use gauge pressure or absolute pressure, but whichever kind of pressure you choose (gauge or absolute) must also be used on the other side of the equation. You can't insert the gauge pressure at point 1, and the absolute pressure at point 2.

Similarly, if you insert the gauge pressure at point 1 and solve for the pressure at point 2, the value you obtain will be the gauge pressure at point 2 (not the absolute pressure).

Fig 1:The figure below shows the Bernoulli effect

PROOF :Deriving Bernoulli’s Equation Mechanism of fluid flow is a complex process. However, it is possible to get some important properties with respect to streamline flows by using the concept of conservation of energy. Let us take an example of any fluid moving inside a pipe. The pipe has different cross-sectional areas in different parts and is present in different heights. Refer to the diagram below.

Now we will consider that an incompressible fluid will flow through this pipe in a steady motion. As per the concept of the equation of continuity, the velocity of the fluid should change. However, to produce acceleration, it is important to produce a force. This is possible by the fluid around it but the pressure must vary in different parts.

Bernoulli’s equation is the general equation that describes the pressure difference in two different points of pipe with respect to velocity changes or change in kinetic energy and height changes or change in potential energy. The relationship was given by Swiss Physicist and Mathematician “Bernoulli” in the year 1738.

General Expression of Bernoulli’s Equation Let us consider two different regions in the above diagram. Let us name the first region as BC and the second region as DE. Now consider the fluid was previously present in between B and D. However, this fluid will move in a minute (infinitesimal) interval of time (∆t).

If the speed of fluid at point B is v1 and at point D is v2. Therefore, if the fluid initially at B moves to C then the distance is v1∆t. However, v1∆t is very small and we can consider it constant across the cross-section in the region BC. Similarly, during the same interval of time ∆t the fluid which was previously present in the point D is now at E. Thus, the distance covered is v2∆t. Pressures, P1 and P2, will act in the two regions, A1 and A2, thereby binding the two parts. The entire diagram will look something like the figure given below.

Finding the Work Done First, we will calculate the work done (W1) on the fluid in the region BC. Work done is W1 = P1A1 (v1∆t) = P1∆V Moreover, if we consider the equation of continuity, the same volume of fluid will pass through BC and DE. Therefore, work done by the fluid on the right-hand side of the pipe or DE region is W2 = P2A2 (v2∆t) = P2∆V Thus, we can consider the work done on the fluid as – P2∆V. Therefore, the total work done on the fluid is W1 – W2 = (P1 − P2) ∆V The total work done helps to convert the gravitational potential energy and kinetic energy of the fluid. Now, consider the fluid density as ρ and the mass passing through

the pipe as ∆m in the ∆t interval of time.

Hence, ∆m = ρA1 v1∆t = ρ∆V Change in Gravitational Potential and Kinetic Energy Now, we have to calculate the change in gravitational potential energy ∆U.

Similarly, the change in ∆K or kinetic energy can be written as

Calculation of Bernoulli’s Equation Applying work-energy theorem in the volume of the fluid, the equation will be

Dividing each term by ∆V, we will obtain the equation

Rearranging the equation will yield

The above equation is the Bernoulli’s equation. However, the 1 and 2 of both the sides of the equation denotes two different points along the pipe. Thus, the general equation can be written as

Thus, we can state that Bernoulli’s equation state that the Pressure (P), potential energy (ρgh) per unit volume and the kinetic energy (ρv2/2) per unit volume will remain constant.

Important Points to Remember It is important to note that while deriving this equation we assume there is no loss of energy because of friction if we apply the principle of energy conservation. However, there is actually a loss of energy because of internal friction caused during fluid flow.

This, in fact, will result in the loss of some energy.

Limitations of the Applications of Bernoulli’s Equation

One of the restrictions is that some amount of energy will be lost due to internal friction during fluid flow. This is because fluid has separate layers and each layer of fluid will flow with different velocities. Thus, each layer will exert some amount of frictional force on the other layer thereby losing energy in the process.

The proper term for this property of the fluid is viscosity. Now, what happens to the kinetic energy lost in the process? The kinetic energy of the fluid lost in the process will change into heat energy. Therefore, we can easily conclude that Bernoulli’s principle is applicable to non-viscous fluids (fluids with no viscosity).

Another major limitation of this principle is the requirement of the incompressible fluid.

Thus, the equation does not consider the elastic energy of the fluid. However, elastic energy plays a very important role in various applications. It also helps us to understand the concepts related to low viscosity incompressible fluids.

Furthermore, Bernoulli’s principle is not possible in turbulent flows. This is because the pressure and velocity are constantly fluctuating in case of turbulent flow.

What will happen to Bernoulli’s equation if a fluid is at rest or the velocity is zero?

When the velocity is zero, the equation will become

This equation is the same as the equation of pressure with depth, that is, P2 − P1 = ρgh.

EXAMPLE 1: Water at a gauge pressure of 3.8 atm at street level flows in to an office building at a speed of 0.06 m/s through a pipe 5.0 cm in diameter. The pipes taper down to 2.6cm in diameter by the top floor, 20 m above. Calculate the flow velocity and the gauge pressure in such a pipe on the top floor. Assume no branch pipe and ignore viscosity.

Solution: By continuity equation: v2 = (A1v1) / A2 = (π (5.0 / 2)2 (0.60) ) / ( π (2.6 / 2)2) v2 = 2.2 m/s

By Bernoulli’s Equation: P1 + ρgh1 + ½ρ(v1)2 = P2 + ρgh2 + ½ρ(v2)2

(Po = atmospheric pressure)

P2 = (3.8 x Po) + Po + ½(1000)(0.6)2 – (1000)(9.8)(20) – (1000)½(2.2)2 P2 = 2.8 x 105 Pa EXAMPLE 2: What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10–2 N m–1?

If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 105 Pa).

Solution: Excess pressure inside the soap bubble is 20 Pa; Soap bubble is of radius, r = 5.00 mm = 5 × 10–3 m The surface tension of the soap solution, S = 2.50 × 10–2 Nm–1 A relative density of the soap solution = 1.20 Density of the soap solution, ρ = 1.2 × 103 kg/m3 Air bubble formed at a depth, h = 40 cm = 0.4 m Radius of the air bubble, r = 5 mm = 5 × 10–3 m 1 atmospheric pressure = 1.01 × 105 Pa Acceleration due to gravity, g = 9.8 m/s2 Hence, the excess pressure inside the soap bubble is: P=4S/r =(4×2.5×10-2)/5×10-3 =20Pa Therefore, the excess pressure inside the soap bubble is 20 Pa. The excess pressure inside the air bubble is : P’=2S/r =(2×2.5×10-2)/5×10-3 =10Pa Therefore, the excess pressure inside the air bubble is 10 Pa. At a depth of 0.4 m, the total pressure inside the air bubble, Atmospheric pressure + hρg + P’ =1.01×105+0.4×1.2×103x9.8+10 =1.057×105Pa =1.06×105Pa Therefore, the pressure inside the air bubble is 1.06×105 Pa.

TUTORIAL 1. Water flows through a pipe of varying diameter, A to B and then to C. The ratio of A to C is 8 / 3. If the speed of water in pipe A is v, what is the speed of water in pipe C.( 8/3 v)

3. If the speed of water in pipe with a diameter of 12 cm is 10 cm/s, what is the speed of water in a pipe with a diameter of 8 cm?( 22.5 cm/s) 4.Water flows through a pipe of varying diameter, as shown in figure below. If area 1 (A1) = 8 cm2, A2 = 2 cm2 and the speed of water in pipe 2 = v2 = 2 m/s then what is the speed of water in pipe 1 = v1.( 0.5 m/s)

5. If the diameter of the larger pipe is 2 times the diameter of smaller pipe, what is the speed of fluid at the smaller pipe?.

6.A storm approaches, and the air outside suddenly drops to 0.96 x 105 Pa from 1.013 x 105 Pa. Before the pressure inside had time to change, what is the magnitude of the net force on a window that measures 2.0 m x 3.1 m? (32 860 N) 7. For the system shown in figure, the cylinder on the left, at L, has a mass of 600 kg and a cross-sectional area of 800 cm2. The piston on the right, at S, has crosssectional area 25 cm2 and negligible weight. If the apparatus is filled with oil (r = 0.78 g/cm3), find the force F required to hold the system in equilibrium as shown in figure( 31 N)

8. A siphon tube is discharging a liquid of specific gravity 0.9 from a reservoir as shown in the figure (a) Find the velocity of the liquid through the siphon(10m/s) (b) Find the pressure at the highest point B.( 41.5 kN/m2 )

(c) Find the pressure at the points A(outside the tube) and C.( 64 kN/m2.)

State and explain the following : (d) Would the rate of flow be more, less or the same if the liquid were water (e) Is there a limit on the maximum height of B above the liquid level in the reservoir?( pB > pvapour ) (f) Is there a limit on the vertical depth of the right limb of the siphon?

9. A garden hose has an inside cross-sectional area of 3.60 cm2, and the opening in the nozzle is 0.250 cm2. If the water velocity is 50 m/s in a segment of the hose that lies on the ground (a) With what velocity does the water come from the nozzle when it is held 1.50 m above the ground and(7.2msâ&#x2C6;&#x2019;1 ) (b) What is the water pressure in the hose on the ground?( 1.41 ´ 105 Pa)

10.A hydraulic lift has two connected pistons with cross-sectional areas 25 cm2 and 420 cm2. It is filled with oil of density 730 kg/m3. a) What mass must be placed on the small piston to support a car of mass 1000 kg at equal fluid levels? (59.523kg) b) With the lift in balance with equal fluid levels, a person of mass 70 kg gets into the car.

What is the equilibrium height difference in the fluid levels in the pistons? c) How much did the height of the car drop when the person got in the car?

11. A hydraulic lift has two connected pistons with cross-sectional areas 20 cm220 cm2 and 640 cm2640 cm2. It is filled with oil of density 650 kg/m3650 kg/m3. (a) What mass must be placed on the small piston to support a car of mass 1200 kg at equal fluid levels?( 37.5 kg) (b) With the lift in balance with equal fluid levels, a person of mass 70 kg gets into the car.

What is the equilibrium height difference in the fluid levels in the pistons?( 1.68 m.) (c) How much did the height of the car drop when the person got in the car?( 5.1 cm) 12. What is the difference between the hydrostatic pressure of blood between the brain and the soles of the feet of a person whose height 165 cm (suppose the density of blood = 1.0 Ă&#x2014; 103 kg/m3, acceleration due to gravity = 10 m/s2).( 1.65 x 104 N/m2) 13. A U pipe is initially filled with water than on one pipe filled with oil, as shown in the figure below. The density of water is 1000 kg/m3. If the height of oil is 8 cm and the height of the water is 5 cm, what is the density of oil?( 625 kg.m-3 )

14. A U pipe was first filled with kerosene then added water. If the mass of kerosene is 0.8 grams/cm3 and the density of water is 1 gram/cm3 and the cross sectional area is 1.25 cm2. Determine how much water should be added so that the height difference of the kerosene surface is 15 cm ( 15 ml)

15. A pipe U filled with water with density of 1000 kg/m3. One column of pipe U filled with glyserin with density of 1200 kg/m3. If the height of glyserin is 4 cm, determine the height difference of both columns of the pipe.( 0.8 cm) 16. A pipe U has two ends are open filled with water with a mass of 1 g/cm3. The sectional area along the pipe is the same, that is 1 cm2. Someone blows on one end of the foot of the pipe so that the surface of the water at the other foot rises 10 cm from its original position.

If the acceleration due to gravity is 10 m/s2 then determine the force in kilodyne acted by that person.( 10 kilodyne) {Hint: 1 Newton = 105 dyne} 17. A Y-shaped tube is inserted upside down so that the left foot and right foot are immersed in two kinds of liquid. After both feet are immersed in the liquid, then the top of the Y pipe is closed with the finger and pulled upwards, so that the two legs of the Y pipe are filled with a column of different high-density liquids. If the density of the first liquid is 0.80 gram.cm-3 and the second density is 0.75 gram.cm-3, and the lower liquid column is 8 cm, then determine the height difference between the two liquid columns on U pipe.( 0.5333 cm)

18. An object float in a liquid where 2/3 of the object in the liquid. If the density of the object is 0.6 gr cm3, then what is the density of water.( 900 kg/ m3) 19. The hydraulic oil in a car lift has a density of 830,00 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are0.0070 m and 0.12500 m, respectively. What input force F is needed to support the 24 500 N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.3000 m above that of the input piston? (93.0 N and 94.9 N ) 20.The blood (Ď = 1060,00 kg/m3) speed in a normal segment of a horizontal artery is 0.110 m/s. An abnormal segment of the artery is narrowed down by an arteriosclerotic plaque to one-fourth the normal cross-sectional area. What is the difference in blood pressures between the normal and constricted segments of the artery? (96 Pa) 21.A liquid is flowing through a horizontal pipe whose radius is 0.020000 m. The pipe bends straight upward through a height of 10.00 m and joins another horizontal pipe whose radius is 0.04000 m. What volume flow rate will keep the pressures in the two horizontal pipes the same? (1.81*10^-2 m 3 /s ) 22. Measured along the surface of the water, a rectangular pool has a length of 15,00 m.

Along this length, the flat bottom of the pool slopes downward at an angle of 11,00degree below the horizontal, from one end to the other. By how much does the pressure at the bottom of the deep end exceed the pressure at the bottom of the shallow end?

UNIT 17: Temperature and Heat INTRODUCTION: 17.1 Temperature scales In the USA, the Fahrenheit temperature scale is used. Most of the rest of the world uses Celsius, and in science it is often most convenient to use the Kelvin scale.

The Celsius scale is based on the temperatures at which water freezes and boils. 0°C is the freezing point of water, and 100° C is the boiling point. Room temperature is about 20° C, a hot summer day might be 40° C, and a cold winter day would be around -20° C. To convert between Fahrenheit and Celsius, use these equations:

The two scales agree when the temperature is -40°. A change by 1.0° C is a change by 1.8° F. The Kelvin scale has the same increments as the Celsius scale (100 degrees between the freezing and boiling points of water), but the zero is in a different place. The two scales are simply offset by 273.15 degrees. The zero of the Kelvin scale is absolute zero, which is the lowest possible temperature that a substance can be cooled to. Several physics formulas involving temperature only make sense when an absolute temperature (a temperature measured in Kelvin) is used, so the fact that the Kelvin scale is an absolute scale makes it very convenient to apply to scientific work.

17.2 Measuring temperature A device used to measure temperature is called a thermometer, and all thermometers exploit the fact that properties of a material depend on temperature. The pressure in a sealed bulb depends on temperature; the volume occupied by a liquid depends on temperature; the voltage generated across a junction of two different metals depends on temperature, and all these effects can be used in thermometers.

17.3 Linear thermal expansion The length of an object is one of the more obvious things that depends on temperature.

When something is heated or cooled, its length changes by an amount proportional to the original length and the change in temperature:

The coefficient of linear expansion depends only on the material an object is made from.

If an object is heated or cooled and it is not free to expand or contract (it's tied down at both ends, in other words), the thermal stresses can be large enough to damage the object, or to damage whatever the object is constrained by. This is why bridges have expansion joints in them (check this out where the BU bridge meets Comm. Ave.). Even sidewalks are built accounting for thermal expansion.

Holes expand and contract the same way as the material around them.

Example This is similar to problem 12.20 in the text. Consider a 2 m long brass rod and a 1 m long aluminum rod. When the temperature is 22 °C, there is a gap of 1.0 x 10 -3 m separating their ends. No expansion is possible at the other end of either rod. At what temperature will the two bars touch?

The change in temperature is the same for both, the original length is known for both, and the coefficients of linear expansion can be found from Table 12.2 in the textbook.

Both rods will expand when heated. They will touch when the sum of the two length changes equals the initial width of the gap. Therefore:

So, the temperature change is:

If the original temperature was 22 °C, the final temperature is 38.4 °C. 17.4 Thermal expansion : expanding holes Consider a donut, a flat, two-dimensional donut, just to make things a little easier. The donut has a hole, with radius r, and an outer radius R. It has a width w which is simply w = R - r.

What happens when the donut is heated? It expands, but what happens to the hole? Does it get larger or smaller? If you apply the thermal expansion equation to all three lengths in this problem, do you get consistent results? The three lengths would change as follows:

The final width should also be equal to the difference between the outer and inner radii. This gives:

This is exactly what we got by applying the linear thermal expansion equation to the width of the donut above. So, with something like a donut, an increase in temperature causes the width to increase, the outer radius to increase, and the inner radius to increase, with all dimensions obeying linear thermal expansion. The hole expands just as if it's made as the same material as the hole.

17.5 Volume thermal expansion When something changes temperature, it shrinks or expands in all three dimensions. In some cases (bridges and sidewalks, for example), it is just a change in one dimension that really matters. In other cases, such as for a mercury or alcohol-filled thermometer, it is the change in volume that is important. With fluid-filled containers, in general, it's how the volume of the fluid changes that's important. Often you can neglect any expansion or contraction of the container itself, because liquids generally have a substantially larger coefficient of thermal expansion than do solids. It's always a good idea to check in a given situation, however, comparing the two coefficients of thermal expansion for the liquid and solid involved.

The equation relating the volume change to a change in temperature has the same form as the linear expansion equation, and is given by:

Where β = 3 α The volume expansion coefficient is three times larger than the linear expansion coefficient.

Proof:let

Example: Suppose that the steel gas tank in your car is completely filled when the temperature is 14° C....?

How many gallons will spill out of the twenty-gallon tank when the temperature rises to 32° C? Solution: for steel β = 12 x 10^-6 for gasoline β = 950 x 10^-6 ΔT = 32 - 14 = 18 β = (ΔV / V) / ΔT ΔV = VβΔT for gasoline ΔV = 20(950 x 10^-6)(18) ΔV = 0.342 gallons for steel ΔV = 20(12 x 10^-6)(18) ΔV = 0.00432 gallons so total spillage will be the expansion of the gasoline minus the expansion of the steel

envelope.

0.342 - 0.004 = 0.338 gallons spillage Overflow Overflow Overflow=V0 (β

Liquid –V0 (β

Tank=V0 {βliquid –βcontainer }

; with βcontainer = 3 α

Suppose your 60.0-L (15.9-gal) steel gasoline tank is full of gas, so both the tank and the gasoline have a temperature of 15.0ºC. How much gasoline has spilled by the time they warm to 35.0ºC? Strategy The tank and gasoline increase in volume, but the gasoline increases more, so the amount spilled is the difference in their volume changes. (The gasoline tank can be treated as solid steel.) We can use the equation for volume expansion to calculate the change in volume of the gasoline and of the tank.

Solution: 1.

Use the equation for volume expansion to calculate the increase in volume of the steel tank: ΔVs = βsVsΔT. 2. The increase in volume of the gasoline is given by this equation: ΔVgas = βgas Vgas ΔT. 3. Find the difference in volume to determine the amount spilled as Vspill=ΔVgas − ΔVs. Alternatively, we can combine these three equations into a single equation. (Note that the original volumes are equal.)

Vspill=(βgas−βs)VΔT=[(950−35)×10−6/∘C](60.0L)(20.0∘C)=1.10LVspill=(βgas−βs)VΔT=[(95 0−35)×10−6/∘C](60.0L)(20.0∘C)=1.10L conclusion This amount is significant, particularly for a 60.0-L tank. The effect is so striking because the gasoline and steel expand quickly. The rate of change in thermal properties is discussed in the chapter Heat and Heat Transfer Methods.

If you try to cap the tank tightly to prevent overflow, you will find that it leaks anyway, either around the cap or by bursting the tank. Tightly constricting the expanding gas is equivalent to compressing it, and both liquids and solids resist being compressed with extremely large forces. To avoid rupturing rigid containers, these containers have air gaps, which allow them to expand and contract without stressing them.

Heat and Internal Energy 17.6 Difference between internal energy and heat

Internal Energy of a System

Any system (solid, liquid or gaseous) is made up of atoms and molecules. These molecules may have mutual interaction and consequently may have their potential and kinetic energies exchanged.

When the system is in equilibrium the total mechanical energy of the system remains constant. Consequently the temperature of the system remains constant. This means that the thermal energy of the system remains constant. This thermal energy is known as the Internal energy of the system.

Let us take an example for that:

Suppose that there is a gas in the cylinder and it has its own Kinetic energy. Now, if the gas is pushed using the piston to the cylinder, gas acquires additional Kinetic energy and also the potential energy. This leads to the increase in Internal energy of the system.

Also if heat is provided to the Cylinder then the internal energy of the system increases.

When heat is given to the system, the atoms and molecules of the cylinder gain energy, collisions between the molecules and atoms take place and they start moving faster.

Molecules and atoms have internal Kinetic energies and internal potential energies. Hence we can conclude that: Internal energy is stated as the sum of all of the energies of the molecules in the system considered.

We can discuss internal energy by taking another example of one gram water at zero degrees Celsius and compare it with one gram of copper at the same temperature. Then both systems have the same kinetic energy but not internal energy. The internal energy is greater for greater potential energy, so water has greater internal energy then copper. This is because the internal energy of system is the total energy of a system which consist both the kinetic and potential energy. It is identified with disorder of a system. Let’s discuss more about internal energy. Since the system has constant volume (ΔV=0) the term -PΔV=0 and work is equal to zero. Thus, in the equation ΔU=q+w w=0 and ΔU=q.

The internal energy is equal to the heat of the system. ... The value of Internal Energy will

be the negative value of the heat absorbed by the surroundings.

Heat Transfer: heat transfer is the process of transfer of heat from high temperature reservoir to low temperature reservoir. There are three basic ways in which heat is transferred. In fluids, heat is often transferred by convection, in which the motion of the fluid itself carries heat from one place to another. Another way to transfer heat is by conduction, which does not involve any motion of a substance, but rather is a transfer of energy within a substance (or between substances in contact). The third way to transfer energy is by radiation, which involves absorbing or giving off electromagnetic waves.

Temperature, Specific Heat and Latent Heat: • •

Specific heat: = energy required to change a unit mass of a material by 1°C. Units: energy per unit mass per degree.

Latent heat = energy required to change the state (gas, liquid, solid) of a unit mass of material. Units: energy per unit mass.

Both specific heat and latent heat are properties of a given material. In other words, every time the material is heated/cooled, no matter how quickly or by what heating process, the same amount of heat is transferred to reach the same state. : – Temperature: • The temperature of an object is a measure of the energy per molecule of an object.

• To raise the temperature, energy must be added • to lower the temperature, energy has to be removed.

• This thermal energy is internal, in the sense that it is associated with the motion of the atoms and molecules making up the object.

• When objects of different temperatures are brought together, the temperatures will tend to equalize. Energy is transferred from hotter objects to cooler objects; this transferred energy is known as heat.

– Specific heat capacity : When objects of different temperature are brought together, and heat is transferred from the higher-temperature objects to the lower-temperature objects, the total internal energy is conserved. Applying conservation of energy means that the total heat transferred from the hotter objects must equal the total heat transferred to the cooler objects.

If the temperature of an object changes, the heat (Q) added or removed can be found using the equation: – where m is the mass, and C is the specific heat capacity, a measure of the heat required to change the temperature of a particular mass by a particular temperature.

The SI unit for specific heat is J / (kg °C). – This applies to liquids and solids.

Generally, the specific heat capacities for solids are a few hundred J / (kg °C), and for liquids they're a few thousand J / (kg °C). For gases, the same equation applies, but there are two different specific heat values. The specific heat capacity of a gas depends on whether the pressure or the volume of the gas is kept constant; there is a specific heat capacity for constant pressure, and a specific heat capacity for constant volume. Q=m×C×ΔT where m is the mass of the substance (in grams), C is the specific heat capacity, and ΔT is the change in temperature during the heat transfer. Note that both mass and specific heat capacity can only have positive values, so the sign of the heat Q will depend on the sign of ΔT. We can calculate ΔT using the following equation: ΔT=Tfinal−Tinitial where Tfinal and Tinitial can have units of either C or K. Based on this equation, if Q is positive (energy of the system increases), then our system increases in temperature and Tfinal>Tinitial. If Q is negative (energy of the system decreases), then our system's temperature decreases and Tfinal<Tinitial.

17.7 Law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.[1] This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.In other word the heat loss equal to the heat gained. i.e Q gained=Q lost Valid for NO phase change!

Phases of matter The states in which matter can exist: as :a solid, liquid,or gas. When temperature changes, matter can undergo a phasechange, shifti ng from one form to another.

Examples of phase changes are melting (changing from a solid to a liquid), freezing(changing from a liquid to a solid) , evaporation (changing from a liquid to a gas), and condensation (changing from a gas to a liquid).

Latent Heat

When a substance changes phase, that is it goes from either a solid to a liquid or liquid to gas, the energy, it requires energy to do so. The potential energy stored in the interatomics forces between molecules needs to be overcome by the kinetic energy the motion of the particles before the substance can change phase.

If we measure the temperature of the substance which is initially solid as we heat it we produce a graph like Figure 1.

Figure : Temperature change with time. Phase changes are indicated by flat regions where heat energy used to overcome attractive forces between molecules Starting a point A, the substance is in its solid phase, heating it brings the temperature up to its melting point but the material is still a solid at point B. As it is heated further, the energy from the heat source goes into breaking the bonds holding the atoms in place. This takes place from B to C. At point C all of the solid phase has been transformed into the liquid

phase. Once again, as energy is added the energy goes into the kinetic energy of the particles raising the temperature, (C to D). At point D the temperature has reached its boiling point but it is still in the liquid phase. From points D to E thermal energy is overcoming the bonds and the particles have enough kinetic energy to escape from the liquid. The substance is entering the gas phase. Beyond E, further heating under pressure can raise the temperature still further is how a pressure cooker works.

17.8 Latent Heat of Fusion and Vaporisation The energy required to change the phase of a substance is known as a latent heat. The word latent means hidden. When the phase change is from solid to liquid we must use the latent heat of fusion, and when the phase change is from liquid to a gas, we must use the latent heat of vaporisation.

The energy require is Q= m L, where m is the mass of the substance and L is the specific latent heat of fusion or vaporisation which measures the heat energy to change 1 kg of a solid into a liquid.

The specific latent heat of water is: •

for fusion (solid→liquid) or freezing (liquid→solid)

•

for vaporisation (liquid→gas) or condensation (gas→liquid)

The table below shows the specific latent heat of fusion of some substances Substance

Specific latent heat of fusion kJ.kg-1

°C

Specific latent heat of vaporisation kJ.kg-1

°C

Water Ethanol Ethanoic acid Chloroform Mercury Sulphur Hydrogen Oxygen Nitrogen

334 109 192

0 -114 17

2258 838 395

100 78 118

74 11 54 60 14 25

-64 -39 115 -259 -219 -210

254 294 1406 449 213 199

62 357 445 -253 -183 -196

17.9 Steam Steam is water in the gas phase, which is formed when water boils. Steam is invisible; however, "steam" often refers to wet steam, the visible mist or aerosol of water droplets

formed as this water vapour condenses. At lower pressures, such as in the upper atmosphere or at the top of high mountains, water boils at a lower temperature than the nominal 100 °C (212 °F) at standard pressure. If heated further it becomes superheated steam.

How Steam works In liquid water, H2O molecules are constantly being joined together and separated. As the water molecules are heated, however, the bonds connecting the molecules start breaking more rapidly than they can form. Eventually, when enough heat is supplied, some molecules will break free. These 'free' molecules form the transparent gas we know as steam, or more specifically dry steam.

Dry and Wet Steam

• •

In steam-using industries, two commonly referred to types of steam are dry steam (also called ”saturated steam”) and wet steam.

Dry steam applies to steam when all its water molecules remain in the gaseous state.

It's a transparent gas.

Wet steam applies to steam when a portion of its water molecules have given up their energy (latent heat) and condense to form tiny water droplets.

Take the example of a kettle boiling water. Water is first heated using an element. As water absorbs more and more heat from the element, its molecules become more agitated and it starts to boil. Once enough energy is absorbed, part of the water vaporizes, which can represent an increase as much as 1600X in molecular volume.

Sometimes a mist can be seen coming out of the spout. This mist is an example of how dry steam, when released into the colder atmosphere, loses some of its energy by transferring it to the ambient air. If enough energy is lost that intermolecular bonds start forming again, tiny airborne droplets can be seen. This mixture of water in the liquid state (tiny droplets) and gaseous state (steam) is called wet steam.

•

For more information about the nature and various types of steam, read the following article: TYPES OF STEAM Steam as a source of power Steam played a vital role in the industrial revolution. The modernization of the steam engine in the early 18th century led to major breakthroughs such as the invention of the steam locomotive and the steamboat, not to mention the steam furnace and the steam hammer.

The latter is not a reference to water hammer found in steam piping, but rather to a steampowered hammer used to shape forgings.

Nowadays, however, internal combustion engines and electricity have often replaced steam as a power source. Even so, steam is still being widely used in electrical power plants and for some large scale industrial applications.

Steam as a Source of Heat Steam is now mostly known for its heating applications, as both a source of direct and

indirect heat.

Direct Steam Heating The direct steam heating method refers to processes where steam is in direct contact with the product being heated.

Example: of heat energy is removed from a mass of steam at to produce water at .

Calculate the mass of water produced. steamâ&#x2020;&#x2019;water is condensation so Q=mlv

So

of water is produced

17.10 Phase equilibrium Phase equilibrium is the most important physical equilibrium : the equilibrium exist between the phase of a system. Phase equilibrium is the study of the equilibrium which exists between or within different states of matter namely solid, liquid and gas. Equilibrium is defined as a stage when chemical potential of any component present in the system stays steady with time. Phase is a region where the intermolecular interaction is spatially uniform or in other words physical and chemical properties of the system are same throughout the region. Within the same state, a component can exist in two different phases such as allotropes of an element. Also, two immiscible compounds in same liquid state can coexist in two phases.

Example : the equilibrium exists between water and steam.

Phase equilibrium has wide range of applications in industries including production of different allotropes of carbon, lowering of freezing point of water by dissolving salt (brine), purification of components by distillation, usage of emulsions in food production, pharmaceutical industry etc. Solid-solid phase equilibrium has a special place in metallurgy and is used to make alloys of different physical and chemical properties. For instance, melting point of alloys of copper and silver is lower than melting point of either copper or silver.

Phase diagrams Phase diagrams are used to understand the relationship between different phases and are usually represented as the change in the phase of a system as a function of temperature, pressure or composition of the components in a system. The system exists in a phase where

Gibbs free energy of the system is least. At equilibrium, temperature, pressure and chemical potential of constituent component molecules in the system have to be same throughout all the phases. Figure 1 gives a general schematic of phase diagram of a single component system (Lue, 2009).

The curves shown in the figure represent the coexistence of two phases. Melting curve is the curve in the phase diagram along which solid and liquid phase of a system stays in equilibrium. Liquid and gas phase of a system stay in equilibrium along the vaporization curve while sublimation curve represents the equilibrium stage between solid and gas phase.

Triple point is point on the graph where all the three states coexist and is unique for every component.

GUYS let try this examples Example 1: A metal rod heated from 30oC to 80oC. The final length of the rod is 115 cm. The coefficient of linear expansion is 3.10-3 oC-1. What is the initial length of the metal rod?

Known : The initial temperature (T1) = 30oC The final temperature (T2) = 80oC The change in temperature (ΔT) = 80oC – 30oC = 50oC The coefficient of linear expansion (α) = 3.10-3 oC-1 The final length of the metal (L) = 115 cm Wanted : The initial length of the metal rod (Lo) Solution : The equation of the linear expansion : L = Lo + ΔL L = Lo + α Lo ΔT L = Lo (1 + α ΔT) 115 = Lo (1 + 3.10-3.50) 115 = Lo (1 + 150.10-3) 115 = Lo (1 + 0.15) 115 = Lo (1.15) Lo = 115 / 1.15 Lo = 100 cm Example 2

The initial length of a brass rod is 40 cm. After heated, the final length of the brass is 40.04 cm and the final temperature is 80oC. If the coefficient of linear expansion of the brass is 2.0 x 10-5 oC–1, what is the initial temperature of the brass rod.

Known : The final temperature (T2) = 80oC The initial length (Lo) = 40 cm The final length (L) = 40.04 cm The increase in length (ΔL) = 40.04 cm – 40 cm = 0.04 cm The coefficient of linear expansion (α) = 2.0 x 10-5 oC-1 Solution : Initial temperature (T1) Solution ; The equation of the linear expansion : L = Lo + α Lo ΔT L – Lo = α Lo ΔT ΔL = α Lo ΔT ΔL = α Lo (T2 – T1) 0.04 = (2.0 x 10-5)(40)(80 – T1) 0..04 = (80 x 10-5)(80 – T1) 0.04 = 0.0008 (80 – T1) 0.04 = 0.064 – 0.0008 T1 0..0008 T1 = 0.064 – 0.040 0.0008 T1 = 0.024 T1 = 30oC Example 3 A 200-gram water at 20°C placed in 50-gram ice at -2°C. If the change of heat just between water and ice, what is the final temperature of the mixture? The specific heat of water is 1 cal/gr°C, the specific heat of ice is 0.5 cal/gr°C, the heat of fusion for ice is 80 cal/gr.

Known : Mass of water (mwater) = 200 gram Temperature of water (Twater) = 20oC The specific heat of water (cwater) = 1 cal/gr°C Mass of ice (mice) = 50 gram Temperature of ice (Tice) = -2oC The specific heat of ice (cice) = 0.5 cal/gr°C The heat of fusion for ice (L) = 80 cal/gr Solution : Heat to increases ice from -2oC to 0oC : Q = m c ΔT Q = (50 gram)(0.5 cal/gr°C)(0oC – (-2oC)) Q = (50)(0.5 cal)(2) Q = 50 calorie Heat for melting all ice : Q = m L = (50 gram)(80 cal/gram) = 4000 calorie Heat for decrease temperature of all water from 20oC to 0oC : Q = m c ΔT Q = (200 gram)(1 cal/gr°C)(0oC – (20oC)) Q = (200)(1 cal)(-20) Q = -4000 calorie Plus sign indicates that the heat is added, the minus sign indicates that heat released.

50-calorie of heat needed to increase the temperature of ice to 0oC and 4000-calorie needed

to melting all ice. Total heat = 4050 calorie. The heat released by water is 4000 calorie.

Example 4 A 200-gram aluminum at 20oC placed in 100-gram water at 80oC in a container. The specific heat of aluminum is 0.22 cal/g oC and the specific heat of water is 1 cal/g oC. What is the final temperature of aluminum?

Known : Mass of aluminum = 200 gram Temperature of aluminum = 20oC Mass of water = 100 gram Temperature of water = 80oC The specific heat of aluminum = 0.22 cal/g oC The specific heat of water = 1 cal/g oC Wanted: The final temperature of aluminum Solution : Aluminum and water in thermal equilibrium so that the final temperature of aluminum = the final temperature of water.

Heat released by hot water (Q release) = heat absorbed by aluminum (Q absorb) mwater c (ΔT) = maluminum c (ΔT) (100)(1)(80 – T) = (200)(0.22)(T – 20) (100)(80 – T) = (44)(T – 20) 8000 – 100T = 44T – 880 8000 + 880 = 44T + 100T 8880 = 144T T = 62oC 11. A 50-gram metal at 85 °C placed in 50 gram water at 29.8 °C. The specific heat of water = 1 cal.g —1 .°C—1. The final temperature is 37 °C. What is the specific heat of metal.

Known : Mass of metal (mmetal) = 50 gram Temperature of metal = 85oC Mass of water (mwater) = 50 gram Temperature of water = 29,8oC The specific heat of water (cwater) = 1 cal.g -1 .°C-1 The final temperature of water = 37oC Wanted : The specific heat of metal (c metal) Solution : Heat released by hot metal (Q release) = heat absorbed by water (Q absorb) mmetal c (ΔT) = mwater c (ΔT) (50)(c)(85 – 37) = (50)(1)(37 – 29.8) (c)(85 – 37) = (1)(37 – 29.8) 48 c = 7.2 c = 0.15 cal.g -1 .°C-1 Example 5 A block of ice with mass of 50-gram at 0°C and 200-gram water at 30°C, placed in a container. . If the specific heat of water is 1 cal.g– 1 °C –1 and the heat of fusion for ice is 80 cal.g –1. What is the final temperature of the mixture.

Known : Mass of ice (mice) = 50 gram The temperature of ice = 0°C Mass of water (mwater) = 200 gram Temperature of water = 30oC

The specific heat of water (cwater) = 1 cal.g– 1 °C –1 The heat of fusion for ice (Lice) = 80 cal.g –1 Wanted: The final temperature Solution : Estimate the final condition : Heat released by water to decrease its temperature from 30oC to 0oC : Qrelease = mwater cwater (ΔT) = (200)(1)(30-0) = (200)(30) = 6000 Heat needed to melting all ice : Q = mice Lice= (50)(80) = 4000 Heat used to melting all ice is 4000, while heat released by water is 6000. Can be concluded that the final temperature of the mixture above 0oC. Black principle : Heat released by water = heat for melting all ice + heat to increase the temperature of ice. (mwater)(cwater)(ΔT) = (mice)(Lice) + (mice)(cwater)(ΔT) (200)(1)(30-T) = (50)(80) + (50)(1)(T-0) (200)(30-T) = (50)(80) + (50)(T-0) 6000 – 200T = 4000 + 50T – 0 6000 – 4000 = 50T + 200T 2000 = 250T T = 2000/250 T = 8oC NB: while solving the steam problem this diagram should be memorized to overcome some challenges.

TUTORIAL: 1.Two matters have specific heat capacities c and 2c. If we give Q and 4Q heat to these matters, changes in the temperatures of them become equal. If the matter A has mass m, find the mass of matter B in terms of m.(mB=2m).

2.A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20 c to 80 c. (a) How much heat is required? What percentage of the heat is used to raise the temperature of: (b) the pan and (c) the water?

3. If 25 kJ is necessary to raise the temperature of a block from 25 deg celsus to 30 deg celsus, how much heat is necessary to heat the block from 45 deg celsus to 50 deg celsus?

The heat transfer depends only on the temperature difference. Since the temperature

differences are the same in both cases, the same 25 kJ is necessary in the second case.

4.Ice at –10.0 °C and steam at 130 °C are brought together at atmospheric pressure in a perfectly insulated container. After thermal equilibrium is reached, the liquid phase at 50.0 °C is present. Ignoring the container and the equilibrium vapor pressure of the liquid at 50.0 °C, find the ratio of the mass of steam to the mass of ice. The specific heat capacity of steam is 2020 J/(kg·C°).( 0.223) 5. An aluminum baseball bat has a length of 0.86 m at a temperature of 17 °C. When the temperature of the bat is raised, the bat lengthens by 0.000 16 m. Determine the final temperature of the bat. (250 C) 6. Suppose that the steel gas tank in your car is completely filled when the temperature is 17 °C. How many litres will spill out of the twenty-litres tank when the temperature rises to 35 °C? (0,33litter) 7. A 12.9 gram sample of an unknown metal at 26.5°C is placed in a Styrofoam cup containing 50.0 grams of water at 88.6°C. The water cools down and the metal warms up until thermal equilibrium is achieved at 87.1°C. Assuming all the heat lost by the water is gained by the metal and that the cup is perfectly insulated, determine the specific heat capacity of the unknown metal. The specific heat capacity of water is 4.18 J/g/°C.(0.40 J/g/°C) 8. Njandjouo Elise Berthe places 48.2 grams of ice in her beverage. What quantity of energy would be absorbed by the ice (and released by the beverage) during the melting process?

The heat of fusion of water is 333 J/g.(16.1kJ) 9. What is the minimum amount of liquid water at 26.5 degrees that would be required to completely melt 50.0 grams of ice? The specific heat capacity of liquid water is 4.18 J/g/°C and the specific heat of fusion of ice is 333 J/g. Rounde the answer to three significant digits ( 1.50x102 g ) 10. Explain why large bodies of water such as Lake Michigan can be quite chilly in early July despite the outdoor air temperatures being near or above 90°F (32°C).

11. . Water has an unusually high specific heat capacity. Which one of the following statements logically follows from this fact? a. Compared to other substances, hot water causes severe burns because it is a good conductor of heat. b. Compared to other substances, water will quickly warm up to high temperatures when heated. c. Compared to other substances, it takes a considerable amount of heat for a sample of water to change its temperature by a small amount.

12. An 11.98-gram sample of zinc metal is placed in a hot water bath and warmed to 78.4°C. It is then removed and placed into a Styrofoam cup containing 50.0 mL of room temperature water (T=27.0°C; density = 1.00 g/mL). The water warms to a temperature of 28.1°C. Determine the specific heat capacity of the zinc.( 0.382 J/g/°C) 13. MUDAU grabs a can of soda from the closet and pours it over ice in a cup. Determine the amount of heat lost by the room temperature soda as it melts 61.9 g of ice (ΔH fusion = 333 J/g).( 20.6 Kj) 14. The heat of sublimation (ΔHsublimation) of dry ice (solid carbon dioxide) is 570 J/g.

Determine the amount of heat required to turn a 5.0-pound bag of dry ice into gaseous carbon dioxide. (Given: 1.00 kg = 2.20 lb)( 1300 kJ )

15. Determine the amount of heat required to increase the temperature of a 3.82-gram sample of solid para-dichlorobenzene from 24°C to its liquid state at 75°C. Paradichlorobenzene has a melting point of 54°C, a heat of fusion of 124 J/g and specific heat capacities of 1.01 J/g/°C (solid state) and 1.19 J/g/°C (liquid state).( 680 J) 16.How much heat is required to change 1.0 kg of ice, originally at –20.0°C, into steam at 110.0°C? Assume 1.0 atm of pressure. (3.1 * 10^6 J) Useful constants: cwater = 4186 J/(kg Co) cice = 2.00 × 103 J/(kg Co) csteam = 2.00 × 103 J/(kg Co) Lf = 33.5 × 104 J/kg Lv = 22.6 × 105 J/kg 17. What mass of water at 95 0C must be mixed with 150.0 g of ice at -5 0 C, in a thermally insulated container, to produce liquid water at 50 0C?(441g) Useful constants: cwater = 4186 J/(kg Co) cice = 2.00 × 103 J/(kg Co) Lf = 33.5 × 104 J/kg Lv = 22.6 × 105 J/kg 18. 5.0 g of copper was heated from 20°C to 80°C. How much energy was used to heat Cu? (Specific heat capacity of Cu is 0.092 cal/g °C) .( 27.6 cal) 20.How much heat is absorbed by 20g granite boulder as energy from the sun causes its temperature to change from 10°C to 29°C? (Specific heat capacity of granite is 0.1 cal/gºC) .( 38 cal) 21.How much heat is released when 30 g of water at 96°C cools to 25°C? The specific heat of water is 1 cal/g°C. (2130 cal (endothermic) 22.If a 3.1g ring is heated using 10.0 calories, its temperature rises 17.9°C. Calculate the specific heat capacity of the ring.( 0.18 cal/g °C ) 23.The temperature of a sample of water increases from 20°C to 46.6°C as it absorbs 5650 calories of heat. What is the mass of the sample? (Specific heat of water is 1.0 cal/g °C)( 212g) 24.The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 47 calories of heat. What is the specific heat of iron?( 0.185 cal/g° C (endothermic))

25.A 4.50 g coin of copper absorbed 54 calories of heat. What was the final temperature of the copper if the initial temperature was 25°C? The specific heat of copper is 0.092 cal/g°C.( 155° C) 26.A 155 g sample of an unknown substance was heated from 25°C to 40°C. In the process, the substance absorbed 569 calories of energy. What is the specific heat of the substance?( 0.24 cal/g °C) 27.What is the specific heat of an unknown substance if a 2.50 g sample releases 12 calories as its temperature changes from 25°C to 20°C?( 0.96 cal/g °C) 28. A steel petrol tank of volume 0.070 m3 is filled to the top with petrol at 20 °C. The tank is placed inside a chamber with an interior temperature of 50 °C. The coefficient of volume expansion for petrol is 9.50 x 10-4/C°; and the coefficient of linear expansion of steel is 12 x 10-6/C°. After the tank and its contents reach thermal equilibrium with the interior of the chamber, how much petrol has spilled? (1.92 x 10 -6 m 3) 29. Steam at 100 C is bubbled through 250 g of water in a 400 g glass beaker (cglass = 840 J.kg-1.oC-1). If the temperature of the water and beaker changes from 4 deg C to 44 deg C, calculate the mass of steam that condenses. (ms = 0.0222 kg)

UNIT 18: Transfer of heat Introduction: If you have been following along since the beginning of UNIT 17, then you have been developing a progressively sophisticated understanding of temperature and heat. You should be developing a model of matter as consisting of particles which vibrate (wiggle about a fixed position), translate (move from one location to another) and even rotate (revolve about an imaginary axis). These motions give the particles kinetic energy. Temperature is a measure of the average amount of kinetic energy possessed by the particles in a sample of matter. The more the particles vibrate, translate and rotate, the greater the temperature of the object. You have hopefully adopted an understanding of heat as a flow of energy from a higher temperature object to a lower temperature object. It is the temperature difference between the two neighboring objects that causes this heat transfer. The heat transfer continues until the two objects have reached thermal equilibrium and are at the same temperature. The discussion of heat transfer has been structured around some everyday examples such as the cooling of a hot mug of coffee and the warming of a cold can of pop.

Finally, we have explored a thought experiment in which a metal can containing hot water is placed within a Styrofoam cup containing cold water. Heat is transferred from the hot water to the cold water until both samples have the same temperature.

There are three modes of heat transfer : conduction, convection, and radiation.

Any energy exchange between bodies occurs through one of these modes or a combination of them. Conduction is the transfer of heat through solids or stationery fluids. Convection uses the movement of fluids to transfer heat.

18.1 CONDUCTION Conduction occurs when two object at different temperatures are in contact with each other.

Heat flows from the warmer to the cooler object until they are both at the same temperature.

Conduction is the movement of heat through a substance by the collision of molecules. At the place where the two object touch, the faster-moving molecules of the warmer object collide with the slower moving molecules of the cooler object. As they collide, the faster molecules give up some of their energy to the slower molecules. The slower molecules gain more thermal energy and collide with other molecules in the cooler object. This process continues until heat energy from the warmer object spreads throughout the cooler object.

Some substances conduct heat more easily than others. Solids are better conductor than liquids and liquids are better conductor than gases. Metals are very good conductors of heat, while air is very poor conductor of heat. You experience heat transfer by conduction whenever you touch something that is hotter or colder than your skin e.g. when you wash your hands in warm or cold water.

A thermal infrared image of a coffee cup filled with a hot liquid. Notice the rings of color showing heat traveling from the hot liquid through the metal cup. You can see this in the metal spoon as well. This is a good example of conduction 18.2 CONVECTION: In liquids and gases, convection is usually the most efficient way to transfer heat. Convection occurs when warmer areas of a liquid or gas rise to cooler areas in the liquid or gas. As this happens, cooler liquid or gas takes the place of the warmer areas which have risen higher.

This cycle results in a continous circulation pattern and heat is transfered to cooler areas.

You see convection when you boil water in a pan. The bubbles of water that rise are the hotter parts of the water rising to the cooler area of water at the top of the pan. You have probably heard the expression "Hot air rises and cool air falls to take its place" - this is a description of convection in our atmosphere. Heat energy is transfered by the circulation of the air.

This thermal infrared image shows hot oil boiling in a pan. The oil is transfering heat out of the pan by convection. Notice the hot (yellow) centers of rising hot oil and the cooler outlines of the sinking oil. Image courtesy of K.-P. Mรถllmann and M. Vollmer, University of Applied Sciences Brandenburg/Germany.

18.3 RADIATION:Both conduction and convection require matter to transfer heat. Radiation is a method of heat transfer that does not rely upon any contact between the heat source and the heated object. For example, we feel heat from the sun even though we are not touching it. Heat can be transmitted though empty space by thermal radiation. Thermal radiation (often called infrared radiation) is a type electromagnetic radiation (or light).

Radiation is a form of energy transport consisting of electromagnetic waves traveling at the speed of light. No mass is exchanged and no medium is required.

Objects emit radiation when high energy electrons in a higher atomic level fall down to lower energy levels. The energy lost is emitted as light or electromagnetic radiation. Energy that is absorbed by an atom causes its electrons to "jump" up to higher energy levels. All objects absorb and emit radiation. When the absorption of energy balances the emission of energy, the temperature of an object stays constant. If the absorption of energy is greater than the emission of energy, the temperature of an object rises. If the absorption of energy is less than the emission of energy, the temperature of an object falls.

A thermal infrared image of the center of our galaxy. This heat from numerous stars and interstellar clouds traveled about 24,000 light years (about 150,000,000,000,000,000 miles!) through space by radiation to reach our infrared telescopes.

TUTORIAL 1.Which one of the following is not an example of convection? a. Smoke rises above a fire. b. An eagle soars on an updraft of wind. c. A person gets a suntan on a beach. d. Water cooks spaghetti. e. An electric heater warms a room.

2. Suppose you are sitting next to a fireplace in which there is a fire burning. One end of a metal poker has been left in the fire. Which one of the following statements concerning this situation is true? a. You can feel the heat of the fire primarily because of convection. b. The other end of the poker is warmed through conduction. c. Heat escapes through the chimney primarily through conduction. d. You can feel the heat of the fire primarily because of conduction. e. The other end of the poker is warmed through convection.

3. Complete the following statement: Most of the heat that is lost to space from the earth occurs by a. conduction. b. convection.

c. radiation. d. both conduction and radiation. e. both conduction and convection 4. Which statement is the best example of heat energy transfer by conduction?

eat energy is transferred from the Earth's surface to the upper atmosphere.

Heat energy is transferred from the surface soil to the rocks below.

Heat energy is transferred from the bottom to the top of a lake.

Heat energy is transferred from the Sun to the Earth.

5. Conduction is the transfer of heat energy by

density differences electromagnetic waves movement through a vacuum molecular contact

6. The collision of molecules within a substance results in the transfer of energy by

convection Conductio -n insolation radiation

7. During which process of energy exchange does cold air displace warmer air?

radiation absorption conduction convection

8.The process by which air flows from one location in the atmosphere to another is called

convection conduction radiation absorption

9.During which process does heat transfer occur because of density differences?

convection radiation Conduction reflection

10.What method of energy transfer requires no medium for transfer?

radiation conduction advection convection 11.Which action would help an air-conditioner use less energy on a hot, sunny summer day?

turning on lights and heat-producing appliances adding insulation in the walls and ceiling replacing light-colored roofing materials with darkcolored ones opening draperies and blinds

UNIT 19:Nuclear Physics and Radioactivity INTRODUCTION Nuclear physics is the field of physics that studies atomic nuclei and their constituents and interactions. Other forms of nuclear matterare also studied.[1] Nuclear physics should not be confused with atomic physics, which studies the atom as a whole, including its electrons.

Discoveries in nuclear physics have led to applications in many fields. This includes nuclear power, nuclear weapons, nuclear medicine and magnetic resonance imaging, industrial and agricultural isotopes, ion implantation in materials engineering, and radiocarbon dating in geology and archaeology. Such applications are studied in the field of nuclear engineering.

Particle physics evolved out of nuclear physics and the two fields are typically taught in close association. Nuclear astrophysics, the application of nuclear physics to astrophysics, is crucial in explaining the inner workings of stars and the origin of the chemical elements.

19.1 Nuclear Notation Standard nuclear notation shows the chemical symbol, the mass number and the atomic number of the isotope. Example: the isotopes of carbon. The element is determined by the atomic number 6. Carbon-12 is the common isotope, with carbon13 as another stable isotope which makes up about 1%.

The mass of an element that is numerically equal to the atomic mass A in grams is called a mole and will contain Avogadro's number NA of nuclei. If the density Ď of the material is known, then the number of nuclei per unit volume n can be calculated from n =Ď NA/A. This is useful in calculating the cross section for nuclear scattering.

Example: the isotopes of carbon. The element is determined by the atomic number 6. Carbon12 is the common isotope, with carbon-13 as another stable isotope which makes up about 1%. Carbon 14 is radioactive and the basis for carbon dating.

According to the present views, a nucleus consists of nucleons: protons and neutrons. As the mass of a nucleon is about 2000 times the mass of an electron the nucleus carries practically all the mass of an atom A nuclid is a specific combination of a number of protons and neutrons. The complete symbol for a nuclide is written as: A Z

X

where X is the chemical symbol of the element, Z is the atomic number, giving the number of protons in the nucleus. A is the totla number of nucleons in the nuclues. It is also known as the mass number. N = A − Z is the number of neutrons.

In nucleus physics it is said that the proton and the neutron are two charge states of the same particle, the nucleon. The proton is the protonic state of the nucleon with a charge +e, and the neutron is its neutronic state with zero charge. According to the latest data, the rest mass of a proton and of a neutron respectively is

mp =1.0075975±0.000001 amu=(1836.09±0.01)me mn =1.008982±0.000003 amu=(1838.63±0.01)me The proton and the neutron have the same mass number equal to unity. In the nucleus, the nucleons are in states substantially differing from their free states. This is because in all nuclei, except that of ordinary hydrogen, there are at least two nucleons between which a special nuclear interaction or coupling exists.

The proton-neutron model of the nucleus accounts for both the observed values of isotopic masses and, the magnetic moments of the nuclei. For, since the magnetic moments of the proton and the neutron are of the same order of magnitude as the nuclear magneton, it follows that a nucleus built up of nucleons should have a magnetic moment of the same order as the nuclear magneton. Therefore, with protons and neutrons as the building blocks of nuclei, the magnetic moment should be of the same order of magnitude. Observations have confirmed this.

1 fm (femto meter = fermi) = 10−10 m is the typical length scale of nuclear physics.

Also with protons and neutrons as the constituents of nuclei, the uncertainty principle leads to reasonable value of energy for these particles in a nucleus, in full agreement with the observed energies per particle Finally, with the assumption that nuclei are composed of neutrons and protons, the difficulty arising from nuclear spin has likewise been resolved. For if a nucleus contains an even number of nucleons, it has integral spin (in units of ). With an odd number of nucleons, its spin will be half-integral (in units of ).

Nuclear Binding Energy Atomic nuclei containing positively charged protons and uncharged neutrons make up stable systems despite the fact that the protons experience Coulomb repulsion. The stability of nuclei is an indication that there must be some kind of binding force between the nucleons. The binding force can be investigated on the energy basis alone, without invoking any considerations concerning the nature and properties of nuclear forces.

An idea about the strength of a system can be gleaned from the effort required to break it up i.e. to do work against the binding. This approach leads to several important facts about the forces that hold the nucleons in a nucleus.

The energy required to remove any nucleon from the nucleus is called the binding (or separation) energy of that nucleon in the nucleus. It is equal to the work that must be done in order to remove the nucleon from the nucleus without imparting it any kinetic energy. The total binding energy of a nucleus is defined as the amount of work that must be done in order to break up the nucleus into its constituent nucleons. From the law of conservation of energy it follows that in forming a nucleus, the same amount of energy must be released as is put in to break it up.

The magnitude of the binding energy of nuclei may be estimated from the following considerations. The rest mass of any permanently stable nucleus has been found to be less than the sum of the rest masses of the nucleons that it contains. It appears as if in “packing up’’ to form a nucleus the protons and neutrons lose some of their masses.

An explanation of this phenomenon is given by the special theory of relativity. This fact is accounted for by the conversion of part of the mass energy of the particles into binding energy. The rest energy of a body, E0 , is related to its rest mass m0 by:

E 0 =m0c 2 where c is the velocity of light in a vacuum. Designating the energy given upon the formation of a nucleus as Eb , then the mass equivalent of the total binding energy

Δm 0 =ΔE b /c 2 is the decrease in the rest mass as the nucleons combine to make up the nucleus. The quantity mo is also known as mass defect or mass decreament. If a nucleus of mass M is composed of a number Z of protons with a mass m p and of a number A-Z of neutrons with a mass mn , the quantity m0 is given by

Δ m o =Zm p +(A-Z)m n -M The quantity m0 gives a measure of the binding energy:,

ΔEb =Δm0c2 =[Zmp +(A-Z)mn -M]c2 In nuclear physics, energies are expressed in atomic energy units (aeu) corresponding to atomic mass units:

1aeu = c2 1amu = ( 9 1016 m 2 /s 2 ) 1.660kg = 1.49110-10 J = 931.1MeV Thus, in order to find the binding energy in MeV, one should use the equation

ΔE b =[Zm p +(A-Z)m n -M]  931.1MeV or Δm =

of the protons +

the neutrons – mass of the nucleus.BE (J) = Δmc2 BE = Binding energy (J) Δm = mass defect (kg) c = speed of light (3 x 108 m/s)

mass of

Δm = of the protons + of the neutrons – mass of the nucleus BE(MeV) = 931.5 Δm Δm = Mass defect in atomic mass units NB: In most of the cases, the atomic mass (and not the nucleus mass) is given, it includes the number of electrons that atom has. In such a case, the mass of the Hydrogen atom ( ), and not the mass of a proton should be used in the calculation of the mass defect. That would cancel the electrons of the atom. H11 The above equations can also be used to calculate the energy released in a nuclear reaction. Δm = Mass of the particles before the reaction – Mass of the particles after nuclear reaction Where the masses of the nucleons and the mass of the nucleus are expressed in atomic mass units. On the average, the binding energy per nucleon is about 8MeV, which is a fairly large amount.

10

Cu-63

Binding Energy per Nucleon (MeV)

9

Sn-120

Al-27

Th-232

8

Pt-195

U-238

He-4 7

B-10 6

Li-6 5

4

3

He-3

2

H-1 1

0 0

40 20

80 60

120 100

160 140

200 180

240 220

Mass Number (A) Figure 1: A plot of the binding energy per nucleon as a function of mass number A As is seen from the plot, the strength of binding varies with the mass number of the nuclei.

The binding is at its strongest in the middle of the periodic Table, in the range 28<A<138, that is, from

28 14

Si to

138 56

Ba. In these nuclei, the binding energy is very close to 8.7 MeV. With

further increases in the number of nucleons in the nucleus, the binding energy per nucleon decreases. For the nuclei at the end of the periodic Table (for example, uranium), Δε b is about 7.6 MeV.

In the region of small mass numbers, the binding energy per nucleon shows characteristic maxima and minima. Minima in the binding energy per nucleon are shown by nuclei containing an odd number of protons and neutrons, such as 63 Li,

10 5

B and

14 7

N

Maxima in the binding energy per nucleon are associated with nuclei having an even number of protons and neutrons, such as 42 He,

12 6

C and 168 O.

The general course of the curve gives a clue to the mechanisms by which nuclear energy is released. We find that nuclear energy can be released either by the fission of heavy nuclei and the fusion of light nuclei from still lighter ones. It is clear from general considerations that energy will be released in nuclear reactions for which the binding energy per nucleon in the end products exceeds the binding energy per nucleon in the original nuclei.

19.2 Transmutation A process called transmutation, took place if the parent and the daughter nuclei are different.

The ancient dreams of the alchemists...

Examples

The figure above shows two examples of nuclear transmutations sparked off by neutrons in nuclear reactors. Above, a neutron capture causes a nucleus of plutonium 239 to split and generate two smaller nuclei, known as fission products. The plutonium and the fission products are different atoms. Below, a second neutron is captured by an iodine 129 nucleus that transforms into iodine 130. Iodine 130 then emits a beta particle, causing

it to transmute into a stable nucleus of xenon 130, a noble gas atom quite different also from Iodine.

19.3 Radiation Radiation is energy traveling in the form of particles or waves in bundles of energy called photons. Some everyday examples are microwaves used to cook food, radio waves for radio and television, light, and x-rays used in medicine. Radioactive materials emit (gave out) radiation as they decay. There are several different types of radiation. Two of the most common ways in which radioactive elements decay are by emitting alpha or beta particles (also known as alpha and beta decay). Sometimes, a third form of radiation, known as gamma rays, may also be produced. Radioactivity is a natural and spontaneous process by which the unstable atoms of an element emit or radiate excess energy in the form of particles or waves. These emissions are collectively called ionizing radiation. Depending on how the nucleus loses this excess energy either a lower energy atom of the same form will result, or a completely different nucleus and atom can be formed.

19.4 Nuclear decays, Radioactivity Rutherford makes classification/nomenclature 4

-rays penetrates the least produces most ionization (of air), is nuclei of 4He , i.e. 2 He arise from tunneling, i.e. have quantum mechanical explanation, Gamow, Gurney and Condon, 1928 β-rays between the extremes, is either electrons (or positrons, but Rutherford does not

know positrons exist, only discovered 1932), so there is really β- rays, i.e. electron rays and β+ rays, i.e. positronsγ penetrates the most produces the least ionization (of air), is electromagnetic radiation of extremely short wavelength (10-11 to 10-13 m, 0.1-10 MeV, arise from relaxation of exited states within the nucleolus (there is some overlap in the

energy with X-rays, if it comes due to processes within the nucleolus, we call it Îł-ray, if it comes out of X-ray formation mechanism, we call it X-rays.figure12.7 X- RAY FORMATION).

19.5 Radioactivity, Discovery and Laws: Pierre and Marie Curie found that the radiation from pitchblende was four times as strong as from uranium. This led to an intensive search for the source of this stronger radiation. Finally, in 1898, the curies succeeded in discovering two new substances which they named polonium, 210 84

Po , and radium,

226 88

Ra

The substances emitting the newly discovered radiation were called radioactive, and the newly discovered property was named radioactivity by Mme M. Curie.

It was soon found that the rays from these radioactive substances were of three kinds, called alpha-rays, beta rays and gamma rays.

Alpha rays are positive; , beta-rays are negative, and gamma rays are uncharged.

Further investigations showed that alpha-rays were helium nuclei. A glass vial holding a sample of radon, a radioactive gas

(

222 86

Rn ) was placed in a glass vessel from which

practically all air had been evacuated. The alpha-particles emitted by the radon sample were absorbed by the walls of the vessel, each captured two electrons, and turned to helium

atoms. These were driven from the walls of the vessel by heating.

The spectrum of the gas in the vessel was found to be identical with the emission spectrum of helium, and this confirmed that the alpha-particles emitted by the radon sample turned to helium. Applying the methods of magnetic and electrostatic deflection. Rutherford q determined the specific charge, , of alpha particles (where m is the mass of an alpham particle) and found that their charge was 2e and the mass the same as that of the nucleus of the helium isotope, 42 He .

Beta-rays are streams of very fast electrons whose velocity exceeds that of ordinary cathode (electron) rays and approaches that of light in a vacuum. Their energy is 10 MeV. The character of betarays has been confirmed by measuring their specific charge, q / m , where

m  is the mass of a beta-particle.

Gamma-rays are a hard electromagnetic radiation much more penetrating of all radioactive rays. The properties of gamma-rays mostly from their absorption and scattering by substances. It has been found that they cause a weak ionization in the material they traverse.

Since they have higher frequencies (that is, shorter wavelengths) than X-rays, their quantummechanical properties stand out with special clarity.

Experiments have shown that all radioactive radiations casue: • chemical effects, • blacken photographic plates, • ionize gases and, some solids and liquids to fluoresce.

These properties are at the basis of experimental techniques for the detection and investigation of radioactive rays .

Alpha decay

A Z

X → AZ−−42Y + 24He

4 U → 234 Th + 90 2 He

238 92

A is sum of protons and neutrons, Z is number of protons also called atomic number, the principle number that defines the place of an element in the periodic table, X is called parent nucleus for example and Y is called daughter nucleus

remember, -decaying atoms had variations in their half-lives of more than 10 orders of magnitude, it was due to a tunneling effect Beta decays and electron capturing

A Z

X→ Y +e A Z +1

−

Cu→ Ni + e

14 6

14 7

−

or simply n0 → p+ + β-+ 

neutron (n) without a charge, proton (p) with charge +, beta particle with negative charge is an electron



A Z

is the electron neutrino

X →Z −A1Y + e+

p+ → n0 + β++

64 Cu→ 28 Ni + e +

64 29



beta particle with positive charge is positron,



is the positron neutrino

process that competes with β decay is electron capturing from the innermost shell, this electron is simply absorbed

A Z

X + e− → Z −A1Y

64 Cu + e − → 28 Ni

64 29

into the nucleolus, where it does not exist as such and

p+ + β- → n0 Ni is more stable because it has an even number of protons the neutron that is formed out of it and the proton that captured it is not simply composed of both particles .

Radioactive decay laws quantum mechanically it is just oscillating expectation values (in the dipole model) as we had them earlier for radio-active transitions when we derived selection rules rate at which radioactive nuclei decay is called activity (R): number of decays per second (some texts A for activity some other texts)

SI unit, 1 Bq = 1 decay per second, pretty slow

1 Ci = 3.7 1010 Bq, per definition, pretty fast, so usual mCi or μCi (1 Ci originally the rate of decay of 1 gram of pure Ra) decay constant say we have an activity (R) of 1 Ci, i.e. 3.7 1010 nuclei decay every second, and a sample of one mole, i.e. 6.022 10 23 atoms so we can also say: each nucleolus in this sample has a probability of decaying during each second of 3.7 1010 nuclei s-1 divided by 6.022 1023 nuclei total = 6.14 10-14 this quantity is called decay constant λ (and really is a probability per second) decay constant (probability) does not depend on the history of the sample, we can not predict when any one atom (we may watch) will decay, we just watch the whole process for a large collection of atoms and get a pretty constant decay constant it is a fully statistical process: only for a large sample we will measure the prescribed decay constant, for a small sample it may be higher only to change in the next instant to be smaller that the prescribed decay rate activity, on the other hand, does depend on number of radioactive nuclei (N) (mole number) and the probability for each nucleolus to decay

R=λN

the more radioactivity nuclei you have to more radio-activity comes out of the sample

now both R and N are functions of time (t) as sample decays, N decreases, there are fewer radioactive nuclei left,

as λ is taken to be a constant, activity R, i.e. decays per second, must decrease as well, so a radioactive sample becomes less and less active (and dangerous) over time, because there are fewer and fewer radioactive nuclei we can regard R as the change in the number of radioactive nuclei per unit time

R=

−

dN dt

N is decreasing with time, that’s where the negative sign comes from

dN − = N Now dt

as R = λN above, which we can rearrange as

dN = − dt N t dN N0 N = − 0 dt N

where N0 is number of radioactive nuclei at t = 0

N ln( ) = − t N0

N = e − t N0

so finally we have

N = N 0e

so we have an exponential (pretty fact) decay over time

− t

now N and No are notoriously difficult to measure so we multiply both sides with the “constant λ” , i.e. the decay constant and get the law for the time dependency of the activity

R = N 0e

− t

= R0e

− t

where R0 is the original activity

now we define half-live: it is the time it takes for half of a given number of radioactive nuclei to decay

we set N =

N0 2

and t = T½ and get

− T 1 N0 = N 0e 2 2

divided by N0 we get

− T 1 1 =e 2 2

taking the ln from both sides gives

or

2=e

T1

2

T1 = 2

ln 2





0.693



which is measured in seconds or any multiple of it

this is convenient for relating half-life to decay constant

e.g. half life of a radio isotope is 5 hour, what is it decay constant

=

0.693 0.693 = = 3.85  10 −5 s −1 T 1 2 5  3600s

that is the probability for any one

nucleolus to decay per second, some will decay within the first millisecond after t 0 some will not decay for the next 100,000 years, (if we do have a large initial amount N0 say mole or more the statistics will be pretty accurate)

this means if the sample is large enough, that is N 0 is large, the actual fraction of it that decays in a certain time span will be very close to the probability for any individual nucleolus to decay to say that a certain radio-isotope has a half life of 5 h means that every radioactive nucleolus has a 50 % chance of decaying every five hours period so on average, 50 % of all the nuclei will actually do so – but this does by no means imply that there is a 100 % probability of decaying of all the nuclei in 10 hours!! the nucleolus has no internal clock that tells him how long he is already around and “kills” him off finally - it’s a quantum mechanical object While quantum mechanics was constructed to describe the world of the very small, it is also needed to explain some macroscopic phenomena such as super-conductors ,and super-fluids.

The word quantum derives from the Latin, meaning "how great" or "how much".[19] In quantum mechanics, it refers to a discrete unit assigned to certain physical quantities such as the energy of an atom at rest . The discovery that particles are discrete packets of energy with wave-like properties led to the branch of physics dealing with atomic and subatomic systems which is today called quantum mechanics. It underlies the mathematical framework of many fields

of physics and chemistry, including condensed matter physics,

so after one half time T½ ,

N0 2

radioactive nuclei remain, per definition

after two half times 2T½ ,

N0 4

radioactive nuclei remain, per definition

N0 after three half times 3T½ , 8

radioactive nuclei remain, per definition

Quantum weirdness: the decay probability of each nucleolus (λ) is constant over time until it actually does decay half life 5 h implies 75 % decay in 10 h, 87.5 % decay in 15 h, 93.75 % in 20 h, 95.3125 % in 25 hours 96.09375 % in 30 hours it’s approaching 100 % even more slowly the longer after t0 Many decays in nature follow exponential laws, but quantum objects are different to macroscopic and living nature,

for example “decaying of people” decaying can be “manipulated”, e.g. better medicine, healthier conditions, avoiding of “bad” cholesterol, wearing seat belts in cars, delays the “decay” of people – no such manipulations for quantum objects as λ is a constant rephrased for the purpose of “cheating” you on your live insurance policy: in principle a sufficiently detailed analysis of all of the relevant circumstances of a person’s health, environment, .. and medical history allows one to predict when this person is going to die “of old age and all other prevailing causes such as strike by lightening, … ”, there are physiological mechanism within individuals that make a kind of internal clock no such thing as an internal clock in quantum objects, decaying is purely statistical and time of decay event impossible to predict for any one atom, all we ever know from a quantum mechanical calculation is the probability of decay – which is stationary – never changes in time if the average life expectancy of people were, say 75 years and we select a group of 100 years old people, we can be pretty sure that they are not going to live for another 75 years, with quantum objects it is again different, a single atom may “live” 100000000000000000000000000000000000000 or any number you chose of life times without decaying, provided there is enough radioactive material

1 = there is also mean half live τ also called life time, per definition:  this means N(τ) = N0 e-1

life time is the time required for the number of particles in the excited state to decrease for any initial value by a factor of e  2.7182818…

same reasoning for activity, if t = τ, R=R0 e-1  0.3678794 R0

also means life time (mean half life) it the time when the activity has dropped to about 37 %. 

=

T 1

2 ln 2

T 1 = ln 2   2

So is radioactivity really only a “bad thing” a decaying of our “nice” elements beyond 83 Bi? Discussion The earth is about 4.5 109 years old, it is believed to have formed as a cold aggregate of smaller bodies, containing a lot of Fe, Ni, and silicates, then the whole thing melted due to the energy released by radioactive processes!!!.The metals went down (as they are heavier) and make up the core, they are believed to be still molten - because we think the earth spinning about an axis produced the earth magnetic field we observe), the silicates went up as they are lighter and cooled down, forming the crust that supports the biosphere and us.

There are lots of radioactive isotopes trapped in the silicates, the main culprit is 238U (99.28 % relative abundance of all U), there is still lots of it left as 238U has a half life of 4.47 109 years 235

U isotope has only (0.72 % relative abundance) and a half life of 7.04 10 8 years

The energy released by radioactivity drives all geological processes, moving of tectonic plates, folding up of the Tibetan mountains, the great San Francisco Earth quake, break out of Mount St. Helens, …

Radioactivity is also bothering our health there is lots of charlatans and scaremonger out there, so I try to be fair unit of radiation dosage is sieverts (Sv), 1 Sv radiation has the same effect as 1 kg body tissue absorbing 1 J of energy from γ- or X-rays chances of dying from cancer are 1 in 20 for a dose of 1Sv per year, 1 in 10,000 for a dose of the 2 mSv per year as given above in the pie chart, so 9999 people die of other causes, 1 as a direct result of the 2 mSv “background”, it’s a fact of live you can’t really do much about “natural background”, decide on where you want to live, e.g. better not on rocky granite ground, e.g. Aberdeen in Scotland where people do have as life expectancy that is about 5 years shorter than in the rest of the U.K., where there are high levels or the inert gas radon in the houses coming out from the ground do better not live at really high elevation, (you catch more than your fair share of cosmic radiation), frequent flyer is in comparison really a minor issue, but pilots have here a major an occupational hazard problem is, there is a long time between exposure to radioactivity and outbreak of disease, so people do not tend to worry about it much

taking ionizating radiation collectively: the one thing you can do is refuse to have too many Xrays taken, it’s really bad for your health, e.g. the screening of healthy woman for breast cancer killed more women directly that it saved by early cancer treatment – don’t trust the physicians or dentist, don’t let them X-ray you with old equipment, 1 computer tomography image is 8 mSv, one top modern X-ray may be only 0.02 Sv, so do not have more than 5 X-rays or so taken per year, if you can avoid it Medical X-ray personal and researcher into X-ray diffraction and spectroscopy are allowed to get 20 mSv exposure, so 10 times the “natural background” dose and they statistically do die earlier than the rest of the population in comparison, nuclear power plants and everything that goes with them like storage, waste disposal, … is comparatively save only major accidents may cause trouble, but mainly locally, the cousin of my father (lives in Canada and was involved all his professional live with nuclear power research and safety at Canada’s prime research site in Chalk River), was actually at one time heading an international committee on safety of nuclear power plants, according to the information he has, the effect of Chernobyl is negligible outside the Ukraine, maybe 10 extra deaths from cancer in all of North America of all the people that were living at that time the accident happened.

Nuclear bomb testing above ground in the 50 and 60 is a much much greater health risk that everybody my age has, worldwide, that’s why it got banned internationally! we are all in the same boot and there better be no nuclear war of whatever kind over ground Radiometric dating 14 for living things 6

C is radioactive, constantly produces on earth

14 7

N + 01n→146C +11H

gets incorporated in all organic (living) beings, when these beings die there is no exchange 14 with sounding any more, so the level of 6

C in the “corpse” (literally or in the paper, cloths,

…) does not get replenished, it decays with half live 5730 years

14 limit of applicability of 6

C dating is about 50,000 years

for geological dating nuclei with larger half lives are required, e.g.

235 40

U → 207Pb

K → 40Ar,

238

U → 206Pb

0.7 10 9 years 1.3 10 9 years 4.5 10 9 years, about the age of the earth

there is a series of decay series ending is stable daughter nuclei, e.g.

Important to know :ď Ą- and β- - decays (i.e. 4He and electron emissions) are most frequently observed decaying processes How old is the moon? we think about 4.5 109 years, just the same as the earth, but the

“youngest rocks” were only about 3 109 years, so igneous activity such as volcanic eruptions must have stopped at that time, it’s a pretty “dead” object, except for some recent foot prints

who invented radiometric dating of biological material, sure no archeologist or biologist, a bunch of physicists of course, they figured out cosmic rays and how the resultant radiocarbon may be used, along the same lines, who discovered X-rays? no physician, a physicist of course!

nuclear decays/reactions are all to be described by quantum mechanical tunneling

4 remember -decay is 2

He

tunneling out of the finite ( 25 MeV) potential energy well of

the nucleolus beta decay and weak interaction while, γ and -decay could be explained from tunneling β-decay seemed to fly in the face of all conservation principles: n → p + e- + (electron neutrino, with is an antiparticle)? conservation of mass/energy, kinetic energy of emerging photon should be as calculated, but hardy ever was observed to be so

conservation of linear momentum, angular momentum, recoil directions of electron and neutron were not opposite conservation of angular momentum, neutron, proton, electron have all spin, so “new spin” seemed to have been created

to resolve: Pauli in 1930 “desperate measure” in β-decay, there is also an additional particle with zero or tiny mass, no charge, spin 1/2" as this particle - that became to be known as neutrino “the little one” in Italian after Fermi – has little mass and no charge, it is extremely difficult to detect, nevertheless, it has been observed indirectly by the “inverse β-decay” from the neutrons coming out of a nuclear reactor in 1953 p +  → n + e+ the positron meets an electron and gets annilated in 2 characteristic γ-rays both about 0.51 MeV, the masses of both particles) the neutron got captured by a Cd nucleus, that emits a 8 MeV γ-rays time delayed to sum up: a proton converted into a positron and a neutron, a antineutrino (also called electron neutrino) must have triggered this reaction !! in order to describe β-decay, the weak force had also to be introduced It is believed that just four forces explain all interactions in the universe, characteristic time is time it takes for particles that are within the force action range to “react”, propositional to the strength of the force.

What is the weak force?

The weak force is one of the four fundamental forces that govern all matter in the universe (the other three are gravity, electromagnetism and the strong force). While the other forces hold things together, the weak force plays a greater role in things falling apart, or decaying.

The weak force, or weak interaction, is stronger than gravity, but it is only effective at very short distances. It acts on the subatomic level and plays a crucial role in powering stars and creating elements. It is also responsible for much of the natural radiation present in the universe, according to the Thomas Jefferson National Accelerator Facility (Jefferson Lab).

Italian physicist Enrico Fermi devised a theory in 1933 to explain beta decay, which is the process by which a neutron in a nucleus changes into a proton and expels an electron, often called a beta particle in this context. "“He defined a new type of force, the so-called weak interaction, that was responsible for decay, and whose fundamental process was transforming a neutron into a proton, an electron and a neutrino," which was later determined to be an antineutrino, wrote Giulio Maltese, an Italian physics historian, in "Particles of Man," an article published in 2013 in the journal Lettera Matematica.

The table below the “List of equations in nuclear and particle physics”gives some ranges,characteristic times and strength of some particles.

List of equations in nuclear and particle physics N0 = N+ ND N= Number of atoms remaining at time t N0 = Initial number of atoms at time t= 0 ND = Number of atoms decayed at time t ,r=r0 A 1/3

r0 ≈ 1.2 fm

hence (approximately) nuclear volume ∝ A nuclear surface ∝ A2/ A=N λ

with N=nNA

λ = Decay constant

,A = activity of a radioisotope

m=m0 exp(-λt) n=n0 exp(-λt) f=1exp(-λt) %=100exp(-λt) f=frequency (hz) at t≠0 n=number of mole at the time t≠0 n0 =initial number of mole at the time t=0s m=mass of the particle at the time t≠0 m0 =initial mass at the time t=0s One mole of oxygen atoms contains 6.02214179×10^23 oxygen atoms. Also, one mole of nitrogen atoms contains 6.02214179×10^23 nitrogen atoms.

The number 6.02214179×10^23 is called Avogadro's number (NA) or Avogadro's constant, after the 19th century scientist Amedeo Avogadro.

Type

Range

Relative Strength

Characteristic time

Typical particles

Mediating particle

Strong

Fm

1

< 10 -22 s

n, p, 

Gluon

Electromagnetic



10-2

10-14 – 10-20

e, n, p, , 

Photon

Weak

10-3 fm

10-7

10-8 – 10-13

all

Weak boson

Gravitational



10-38

Years

all

Graviton

TUTORIAL: 1. Suppose the range for 5.0MeVα ray is known to be 2.0 mm in a certain material.

Does this mean that every 5.0MeVα a ray that strikes this material travels 2.0 mm, or does the range have an average value with some statistical fluctuations in the distances traveled? Explain.

2. What is the difference between γ rays and characteristic x rays? Is either necessarily more energetic than the other? Which can be the most energetic?

3. Ionizing radiation interacts with matter by scattering from electrons and nuclei in the substance. Based on the law of conservation of momentum and energy, explain why electrons tend to absorb more energy than nuclei in these interactions.

4. What characteristics of radioactivity show it to be nuclear in origin and not atomic?

5. What is the source of the energy emitted in radioactive decay? Identify an earlier conservation law, and describe how it was modified to take such processes into account.

6. Consider Figure. If an electric field is substituted for the magnetic field with positive charge instead of the north pole and negative charge instead of the south pole, in which directions will the α,β , and γ rays bend?

7. Explain how an α particle can have a larger range in air than a β particle with the same energy in lead.

8. Arrange the following according to their ability to act as radiation shields, with the best first and worst last. Explain your ordering in terms of how radiation loses its energy in matter. (a) A solid material with low density composed of low-mass atoms. (b) A gas composed of high-mass atoms. (c) A gas composed of low-mass atoms. (d) A solid with high density composed of high-mass atoms.

9. Often, when people have to work around radioactive materials spills, we see them wearing white coveralls (usually a plastic material). What types of radiation (if any) do you think these suits protect the worker from, and how?

10. Is it possible for light emitted by a scintillator to be too low in frequency to be used in a photomultiplier tube? Explain.

11. The weak and strong nuclear forces are basic to the structure of matter. Why we do not experience them directly?

12. Define and make clear distinctions between the terms neutron, nucleon, nucleus, nuclide, and neutrino.

13. What are isotopes? Why do different isotopes of the same element have similar chemistries?

14. Star Trek fans have often heard the term “antimatter drive.” Describe how you could use a magnetic field to trap antimatter, such as produced by nuclear decay, and later combine it with matter to produce energy. Be specific about the type of antimatter, the need for vacuum storage, and the fraction of matter converted into energy.

15. What conservation law requires an electron’s neutrino to be produced in electron capture? Note that the electron no longer exists after it is captured by the nucleus.

16. Neutrinos are experimentally determined to have an extremely small mass. Huge numbers of neutrinos are created in a supernova at the same time as massive amounts of light are first produced. When the 1987A supernova occurred in the

Large Magellanic Cloud, visible primarily in the Southern Hemisphere and some 100,000 light-years away from Earth, neutrinos from the explosion were observed at about the same time as the light from the blast. How could the relative arrival times of neutrinos and light be used to place limits on the mass of neutrinos?

17. What do the three types of beta decay have in common that is distinctly different from alpha decay?

18. In a 3×109-year-old rock that originally contained some 238U, which has a halflife of 4.5×109 years, we expect to find some 238U remaining in it. Why are 226Ra,222Rn, and 210Po also found in such a rock, even though they have much shorter half-lives (1600 years, 3.8 days, and 138 days, respectively)?

19. Does the number of radioactive nuclei in a sample decrease to exactly half its original value in one half-life? Explain in terms of the statistical nature of radioactive decay.

20. Radioactivity depends on the nucleus and not the atom or its chemical state.

Why, then, is one kilogram of uranium more radioactive than one kilogram of uranium hexafluoride?

21. Explain how a bound system can have less mass than its components. Why is this not observed classically, say for a building made of bricks?

22. Spontaneous radioactive decay occurs only when the decay products have less mass than the parent, and it tends to produce a daughter that is more stable than the parent. Explain how this is related to the fact that more tightly bound nuclei are more stable. (Consider the binding energy per nucleon.)

23. To obtain the most precise value of BE from the equation BE=[ZM(1H)+Nmn]c2−m(AX)c2, we should take into account the binding energy of the electrons in the neutral atoms. Will doing this produce a larger or smaller value for BE? Why is this effect usually negligible?

24. How does the finite range of the nuclear force relate to the fact that BE/A is greatest for A near 60?

25. Why is the number of neutrons greater than the number of protons in stable nuclei having A greater than about 40, and why is this effect more pronounced for the heaviest nuclei?

26. A physics student caught breaking conservation laws is imprisoned. She leans against the cell wall hoping to tunnel out quantum mechanically. Explain why her chances are negligible. (This is so in any classical situation.)

27. When a nucleus α decays, does the α particle move continuously from inside the nucleus to outside? That is, does it travel each point along an imaginary line from inside to out? Explain.

28. The energy of 30.0 eV is required to ionize a molecule of the gas inside a Geiger tube, thereby producing an ion pair. Suppose a particle of ionizing radiation deposits

0.500 MeV of energy in this Geiger tube. What maximum number of ion pairs can it create?

29. A particle of ionizing radiation creates 4000 ion pairs in the gas inside a Geiger tube as it passes through. What minimum energy was deposited, if 30.0 eV is required to create each ion pair?

30. (a) Repeat Exercise, and convert the energy to joules or calories. (b) If all of this energy is converted to thermal energy in the gas, what is its temperature increase, assuming 50.0cm3 of ideal gas at 0.250-atm pressure? (The small answer is consistent with the fact that the energy is large on a quantum mechanical scale but small on a macroscopic scale.)

31. Suppose a particle of ionizing radiation deposits 1.0 MeV in the gas of a Geiger tube, all of which goes to creating ion pairs. Each ion pair requires 30.0 eV of energy. (a) The applied voltage sweeps the ions out of the gas in 1.00μs. What is the current? (b) This current is smaller than the actual current since the applied voltage in the Geiger tube accelerates the separated ions, which then create other ion pairs in subsequent collisions. What is the current if this last effect multiplies the number of ion pairs by 900?

32. Find the radius of a 238Pu nucleus. 238Pu is a manufactured nuclide that is used as a power source on some space probes.

33. (a) Calculate the radius of 58Ni, one of the most tightly bound stable nuclei. (b) What is the ratio of the radius of 58Ni to that of 258Ha, one of the largest nuclei ever made? Note that the radius of the largest nucleus is still much smaller than the size of an atom.

34. When an electron and positron annihilate, both their masses are destroyed, creating two equal energy photons to preserve momentum. (a) Confirm that the annihilation equation e++e−→γ+γ conserves charge, electron family number, and total number of nucleons. To do this, identify the values of each before and after the annihilation. (b) Find the energy of each γ ray, assuming the electron and positron are initially nearly at rest. (c) Explain why the two γ rays travel in exactly opposite directions if the center of mass of the electron-positron system is initially at rest.

35. A rare decay mode has been observed in which

222

Ra emits a 14C nucleus.

(a) The decay equation is 222Ra→AX+14C. Identify the nuclide AX.

(b) Find the energy emitted in the decay. The mass of 222Ra is 222.015353 u.

36. (a) Write the complete α decay equation for 226Ra. (b) Find the energy released in the decay. (a) Write the complete α decay equation for

249

Cf.

(b) Find the energy released in the decay.

37. (a) Write the complete β− decay equation for the neutron. (b) Find the energy released in the decay.

38. (a) Write the complete β+ decay equation for 11C. (b) Calculate the energy released in the decay. The masses of 11C and 11B are 11.011433 and 11.009305 u, respectively.

39. (a) Calculate the energy released in the α decay of 238U. (b) What fraction of the mass of a single 238U is destroyed in the decay?

The mass of 234Th is 234.043593 u. (c) Although the fractional mass loss is large for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?

40. An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of 14C. Estimate the minimum age of the charcoal, noting that 210=1024.

41. A 60Co source is labeled 4.00 mCi, but its present activity is found to be 1.85×107 Bq. (a) What is the present activity in mCi? (b) How long ago did it actually have a 4.00-mCi activity?

42. (a) Calculate the activity R in curies of 1.00 g of 226Ra. (b) Discuss why your answer is not exactly 1.00 Ci, given that the curie was originally supposed to be exactly the activity of a gram of radium 43. Show that the activity of the

14

C in 1.00 g of 12C found in living tissue is 0.250 Bq.

44. Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long periods of time. Natural thorium is almost 100% 232Th, with a half-life of 1.405×1010y. If an average lantern mantle contains 300 mg of thorium, what is its activity?

45.What fraction of a radioactive sample decays in 80 yr if the half-life of the material is 140 yr? (0.672) 46.Measurements show that only 18 percent of a radioactive material remains after 12.0 h. What is the half-life of this material? (4.8505 h) 47. Iodine-131, M=130,906114 , is used to treat thyroid disorder because, when ingested, it localizes in the thyroid gland. Its half-life is 8.1 days. What is the activity of 0.85 g of Iodine? (4.24 x1015 Bq) 48. The isotope of lead has an atomic mass of 207.976 627 u. Calculate: (a) The mass defect in atomic mass units, and (b) The binding energy in joules. (2.635 x 10 16 J) Pb 208 82 49.Write the alpha-decay process for .

50. The half-life of is 1600 years. This isotope is used to make the dial of a watch glow in the dark. What mass of radium will disappear if the watch is in use for fifty years if initially 1 x 10-9 kg is used? (2.14 x 10 -11kg)