advanced physics for you blad spreads

7/25/2019 Advanced Physics for You Blad Spreads

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PHYSICSKeith Johnson

Simmone Hewet

Sue Hol

John Mille

Advanced

for You

2NDEditio

2

Sample

material


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troduction (for Students)

dvanced Physics for Youis designed to help andpport you during your advanced Physics course.o matter which exam course you are following, thisok will help you to make the transition to A-level.

he book is carefully laid out s o that each new idea isroduced and developed on a single page or on two

cing pages. Words have been kept to a minimumd as straightforward as possible, with clear diagrams,d a cartoon character called Phiz.ges with a red trianglein the top corner are theore difficult pages and can be left at first.

ch important fact or new formula is clearly printedheavy typeor is in a coloured box.

here is a summary of important facts at the end ofch chapter, to help you with revision.

orked examples are a very useful way of seeing howtackle problems in Physics. In this book there areer 200 worked examples to help you to learn how to

ckle each kind of problem.the back of the book there are extra sections givingu valuable advice on study skill s, practical work,vision and examination techniques, as well as morelp with mathematics.

Throughout the book there are Physics at Work pages.These show you how the ideas that you learn in Physicsare used in a wide range of interesting applications.

There is also a useful analysis of how the book coversthe different examination syllabuses, with full details ofwhich pages you need to study, on the web-site at :www.oxfordsecondary.co.uk/advancedforyouAt the end of each chapter there are a number ofquestions for you to practise your Physics and so gainin confidence. They range from simple fill-in-a-missing-word sentences (useful for doing quick revision) to moredifficult questions that will need more thought.

At the end of each main topic you will find a section offurther questions, mostly taken from actual advancedlevel examination papers.For all these questions, a Hints and Answers section atthe back of the book gives you helpful hints if you needthem, as well as the answers.

We hope that reading this book will make Physicsmore interesting for you and easier to understand.Above all, we hope that it will help you to make goodprogress in your studies, and that you will enjoy usingAdvanced Physics for You.

Keith Johnson Simmone Hewett Sue Holt John Miller

Introduction (for Teachers planning for September)

In considering the implementation of the new AS / A-level specifications forSeptember there are a number of things for you to consider:

Coverage of the new Specification A good text-book is a valuable backup and safety-net for your teaching, for

students to use as a follow-up for homework and after any absences.

For this revised edition the extensive new coverage is shown by: - An outline of the Contents, on the next page. - A short Summary of the specification coverage, see the inside of the back cover. - A detailed Map of the coverage of your particular specification, as a free PDF,

at: www.oxfordsecondary.co.uk/advancedforyou

Accessibility & Readability If your students are going to use the book as a backup to your teaching, then they

need to be able to study and understand independently.

Accessibility, readability, layout and clarity of presentation are all vital here. Judge for yourself by looking at the sample pages weve included. And perhaps ask your students what they think?

Support for your students, across the full ability range Your students will be supported not only by the clear layout and the hundreds of

worked examples ...but also by the expanded s upport for Maths (now 15 pages),and sections on Practical work and Study Skills.

All of the hundreds of questions have been analysed by Senior Examiners. Theyhave selected a large number of past-paper questions that are appropriate for thenew specifications, including Synoptic questions and questions on Practical work.

The Hints & Answers section has also been expanded.

In updating and expanding the very successful first edition, we have taken great careto provide a quality text-book. One that is clearly written, strongly supportive, andwith a slight touch of humour to present a friendly face of Physics.

We hope you will find it a useful and significant support to boost your teaching andenhance your results. Keith Johnson Simmone Hewett Sue Holt John Miller

On the following pages we have included all of Chapter 1, followed by samples fromother parts of the book, so that you can see the range of support that we provide.


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Waves

10. Wave Motion 122 Wavelength, period, frequency Amplitude, energy, phase Electromagnetic radiation, speed of light Longitudinal, transverse, polarisation Physics of hearing, decibels Physics at work : Earthquakes

11. Reflection and Refraction 138 Particles or waves, Snells Law, refractive index Total internal reflection, optical fibres Lenses and curved mirrors, ray diagrams Human eye, defects, spectacles Physics at work : Optical fibres and Telescopes

12. Interference and Diffraction 158 Superposition, phase, path difference

Stationary waves, nodes and anti-nodes Diffraction, resolution Physics at work : Radio astronomy Interference, Youngs slits, diffraction grating Physics at work : Holograms and Soap bubbles

Further Questions on Waves 180

Matter and Molecules

13. Materials 184 Density, crystal structure, size of atoms Hookes law, stress and strain Young modulus, properties of materials Physics at work : Composite materials

14.*Thermodynamics 198 Internal energy, measuring temperature Specific heat capacity, specific latent heat Physics at work : Marathons and Heat pumps

15.*Gases and Kinetic Theory 209 Pressure in gases, Boyles Law Absolute zero, Avogadro constant, ideal gases

Physics at work: Scuba diving and Ballooning

Further Questions on Matter & Molecules 222

Chapters marked *are usually not in AS, only in A-level.

For more details of which sections you need to studyfor your particular examination, see page 506, anddownload the relevant Specification Map fromwww.oxfordsecondary.co.uk/advancedforyou

Mechanics

. Basic Ideas 6Units, checking equationsSignificant figuresVectors and scalarsAdding and resolving vectors

. Looking at Forces 18Weight and other forcesFriction, drag, viscosity, Stokes LawLift, pressure, upthrust, Archimedes principleFree-body force diagramsPhysics at work : Friction and Racing cars

. Turning Effects of Forces 32Moments, conditions for equilibriumCouples, centre of mass

. Describing Motion 38Displacementtime graphsVelocitytime graphsEquations of motionFalling under gravity, g, projectiles

. Newtons Laws and Momentum 52First law, inertiaSecond law, momentumThird law, pairs of forcesImpulse, conservation of momentumPhysics at work : Car safety

. Work, Energy and Power 68Kinetic energy, potential energyConservation of energyEfficiency; kinetic energy and momentumElastic and inelastic collisions

.* Circular Motion 76Centripetal force, radiansHorizontal and vertical circlesPhysics at work : Bobsleighs and Space stations

.* Gravitational Forces and Fields 86Gravitational force, inverse-square law

Gravitational field, potential energyPhysics at work : Satellites and Space-probes

.* Simple Harmonic Motion 100Oscillations, period, frequencySimple pendulum, mass on a springDamping, resonancePhysics at work : Falling bridges and Cooking

urther Questions on Mechanics 114

ContentsElectricity and Magnetism

16. Current and Charge 226 Conduction of electricity, flow of charge Energy transfer and p.d. Resistance and resistivity Physics at work : Shocks and Electric cars

17 Electric circuits 244 Resistors in series and parallel Potential divider, potentiometer E.m.f. and internal resistance Kirchhoffs Law

18.*Magnetic Fields 256 Magnetic flux density Coils, solenoids, electromagnets Magnetic force on a wire; moving charges Physics at work : Magnetic trains

19.*Electromagnetic Induction 270 Magnetic flux Faradays Law, Lenzs Law A.c. generator Physics at work : Metal detectors

20.*Alternating Current 280 r.m.s. values, cathode-ray oscilloscope Transformer, national grid,

Rectification, diodes

21.*Electric Fields 288 Electric charges, charging by friction Electric fields, electric potential Physics at work : Crop-spraying, Photocopiers22.*Capacitors 300 Charge and discharge Capacitance, time constant Physics at work : Designing capacitors

Further Questions on Electricity 312

Quantum Physics

23. Electrons and Photons Thermionic emission; Thomson

Photo-electric effect, Plancks eq

24.*Spectra and Energy Levels Black-body radiation, Wiens Law Continuous spectrum and line sp

Further Questions on Quantum Phy

Nuclear Physics

25.*Radioactivity Atomic structure, isotopes, detec , , radiation, decay, half-life

26.*Nuclear Energy E= m c2, binding energy, fission Physics at work : Nuclear waste

27. Particle Physics Quarks & leptons; Feynman diag Physics at work : Identifying pa

Further Questions on Nuclear Phys

Modern Physics

28.*Special Relativity Time dilation, length contraction, r

29.*Astrophysics Classifying stars, Doppler, Hubble,

30.*Medical Imaging Techniques X-ray imaging, tracers, ultrasoun

Further Questions on Modern Phys

Extra Sections

Synoptic (cross-topic) Questions 446including Practical Questions

Study Skills 456

Making Progress in Physics

Revision Technique, Examination Technique

Doing Your Practical Work 466

Check Your Maths

Graphs, equations, exponentials, spre

Hints and Answers

Examination Syllabus Coverage

Index


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ow fast does light travel ? How much do you weigh ?hat is the radius of the Earth ?hat temperature does ice melt at ?

e can find the answers to all of these questions by measurement.peed, mass, length and temperature are all examples ofphysicalantities.easurement of physical quantities is an essential part of Physics.

this chapter you will learn :the difference between base and derived units,how you can use units to check equations,how to use significant figures,how to deal with vectors.

ating measurements

measurement requires a system of units.r example : How far is a distance of 12 ?

ithout a unit this is a meaningless question. You must alwaysve a measurement as a number multiplied by a unit.

r example :

2 m means 12 multiplied by the length of one metre.kg means 9 multiplied by the mass of one kilogram.

ut what do we mean by one metre and one kilogram ?etres and kilograms are two of the seven internationally agreedse units.

ase quantities and units

he Systme International (SI) is a system ofeasurement that has been agreed internationally.defines 7 base quantities and units, butu only need six of them at A-level.

he 7 base quantities and their units are listedthe table :

heir definitions are based on specific physicaleasurements that can be reproduced, verycurately, in laboratories around the world.

he only exception is the kilogram. This is theass of a particular metal cylinder, known as theototype kilogram, which is kept in Paris.

Derived units

Of course, we use far more physical quantities in Physics thanjust the 7 base ones. All other physical quantities are known asderived quantities. Both the quantity and its unit are derivedfrom a combination of base units, using a defining equation :

1 Basic Ideas

Physical quantity Defined as Unit

density mass (kg) volume (m3) kg m3

momentum mass (kg) velocity (m s1) kg m s1

force mass (kg) acceleration (m s2) kg m s2

pressure force (kg m s2or N) area (m2) kg m1s2(N m2)

work (energy) force (kg m s2or N) distance (m) kg m2s2(N m)

power work (kg m2s2or J ) time (s) kg m2s3(J s1)

electrical charge current (A) time (s) A s

potential difference energy (kg m2s2or J ) charge (A s or C) kg m2A1s3(J C1)

res istance potential difference ( kg m2A1s3or V ) current (A) kg m2A2s3( V A1)

Example 1Velocityis defined by the equation :

velocity =distance travelled in a given direction (m)

time taken (s)

Both distance (ie. length) and time are base quantities.The unit of distance is the metre and the unit of timeis the second. So, from the defining equation,the derived unit of velocity is metres per second,written m/sor ms1.

Example 3Forceis defined by the equation :

force =mass (kg) acceleration (m s2) (see page 55)

The derived unit of force is therefore :kilogram metres per second squared or kgms2.This is given a special name : the newton(symbol N).

Example 2Accelerationis defined by the equatio

acceleration =change in velocity (m

time taken (s)

Again, combining the units in the defigives us the derived unit of acceleratioThis is metres per second per secometres per second squared, written

What other units have you come across in addition to these

base units and base unit combinations ?Newtons, watts, joules, volts and ohms are just a few thatyou may remember.These are special names that are given to particular combinationsof base units.

The table below lists some common derived quantities andunits for you to refer to.

Some of the combinations are quite complicated. You can seewhy we give them special names !

Measuring temperature

and time

and weight.

Base quantity Base unit

Name Symbol Name Symbol

time t second s

length l metre m

mass m kilogram kg

temperature T, kelvin K

electric current I ampere A

amount of substance n mole mol

luminous intensity (Not used at A-level) candela cd


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Homogeneity of equations

e have seen that all units are derived from base units using equations.his means that in any correct equation the base units of each partust be the same. When this is true, the equation is said to bemogeneous. Homogeneous means composed of identical parts.

Prefixes

r very large or very small numbers, we can use standard prefixesth the base units.

he main prefixes that you need to know are shown in the table :

The importance of significant figures

What is the difference between lengths of 5 m, 5.0 m and 5.00 m ?

Writing 5.00 m implies that we have measured the length moreprecisely than if we write 5 m.

Writing 5.00 m tells us that the length is accurate to the nearestcentimetre.A figure of 5 m may have been rounded to the nearest metre.The actual length could be anywhere between 4

1

2m and 5

1

2m.

The number 5.00 is given to three significant figures(or 3 s.f.).

To find the number of significant figures you must count up thetotal number of digits, starting at the first non-zero digit,reading from left to right.The table gives you some examples :

It shows you how a number in the first column (where it is given to3 s.f.) would be rounded to 2 significant figures or 1 significant figure.

In the last example in the table, why did we change the number to

standard form ?Writing 1.7 102instead of 170 makes it clear that we are givingthe number to two significant figures, not three.(If you need more help on standard form look at the Check YourMathssection on page 473.)

Significant figures and calculations

How many significant figures should you give in your answersto calculations ?This depends on the precision of the numbers you use in thecalculation . Your answer cannot be any more precise than the datayou use. This means that you should round your answer to thesame number of significant figures as those used in the calculation.

If some of the figures are given less precisely than others, then roundup to the lowestnumber of significant figures.Example 7 explains this.

Make sure you get into the habit of rounding all your answers to thecorrect number of significant figures. You may lose marks in anexamination if you dont !

Example 4how that the following equation is homogeneous : kinetic energy =

1

2mass velocity2

rom the table on page 7 :

Unit of kinetic energy =joule =kg m2s2

Unit of1

2mass velocity2=kg (m s1)2=kg m2s2 (Note :

1

2is a pure number and so has no unit.)

The units on each side are the same and so the equation is homogeneous.

Example 7The swimmer in the photograph covers a distance of 100.0 m in 68 s.Calculate her average speed.

speed = distance travelledtime taken

= 100.0 m68 s

= 1.470 588 2 m s1

This is the answer according to your calculator.How many significant figures should we round to ?

The distance was given to 4 significant figures . But the time wasgiven to only 2 significant figures . Our answer cannot be any moreprecise than this, so we need to round to 2 significant figures.

Our answer should be stated as : 1.5 m s1 (2 s.f.)

xample 5hiz is trying to calculate the power Pof a resistor when

e is given its resistance Rand the current Iflowing through it.He cannot remember if the formula is : P=I2R or P=I2R.

y checking for homogeneity, we can work out which equation is correct :

Using the table on page 7 : Units of P=watts (W) =kg m2s3

Units of I2=A2

Units of R=ohms () =kg m2A2s3

Multiplyingtogether the units of I2and Rwould give ushe units of power. So the first equation is correct.

xample 6

) Energy stored in a chocolate bar =1 000 000 J=1 106J=1 megajoule=1 MJ

) Wavelength of an X-ray =0.000 000 001 m=1 109m=1 nanometre=1 nm

his is a useful way of checking an equation. It can be particularlyeful after you have rearranged an equation :

ne word of warning. This method shows that an equation couldcorrect but it doesnt prove that it is correct !

an you see why not ? Example 4 above is a good illustration.he equation for kinetic energy would still be homogeneous even ife had accidentally omitted the

1

2. Prefix Symbol Multiplier

tera T 10 12

giga G 109

mega M 106

kilo k 103

deci d 101

centi c 102

milli m 103

micro 106

nano n 109

pico p 1012

femto f 1015

3 s.f. 2 s.f.

4.62 4.6

0.005 01 0.0050

3.40 108 3.4 1

169 1.7 1


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0

Vectors and scalars

hrowing the javelin requires a force. If you want to throw it a longtance, what two things are important about the force you use ?

he javelins path will depend on both the sizeand the directionthe force you apply.

rce is an example of a vectorquantity.ectors have both size (magnitude) and direction.

her examples of vectors include : velocity, acceleration andomentum. They each have a size and a direction.

uantities that have size (magnitude) but nodirectione called scalars. Examples of scalars include : temperature,ass, time, work and energy.

he table shows some of the more common vectors and scalarsat you will use in your A-level Physics course :

ok back at the table of base quantities on page 6.e these vectors or scalars ? Most of the base quantities are scalars.

an you spot the odd one out ?

Vector addition

What is 4 kg plus 4 kg ? Adding two masses of 4 kg alwaysgivesthe answer 8 kg. Mass is a scalar. You combine scalars usingsimple arithmetic.

What about 4 N plus 4 N ? Adding two forces of 4 N can giveany answer between 8 N and 0 N.Why do you think this is ?Its because force is a vector. When we combine vectors we alsoneed to take account of their direction.

Often in Physics we will come across situations where two or morevectors are acting together. The overall effect of these vectors iscalled the resultant. This is the single vector that would have thesame effect.To find the resultant we must use the directions of the 2 vectors :

Vectors acting along the same straight line

Two vectors acting in the samedirection can simply be addedtogether :

Resultant F1F2

If the vectors act in oppositedirections, we need to take onedirection as positive, and the other as negative, before addingthem :

Resultant F1(F2) F1F2

Representing vectors

ectors can be represented in diagrams by arrows.

The length of the arrow represents the magnitude of the vector.

The direction of the arrow represents the direction of the vector.

ere are some examples :

orce vectors

horizontal force of 20 N : A vertical force of 10 N :

Using the same scale, how would you draw the vector forforce of 15 N at 20 to the horizontal ?

Example 8Phiz is standing on a moving walkway in an airport.The walkway is moving at a steady velocity of 1.50 m s1.a) Phiz starts to walk forwards along the walkway at 2.00 m s1.

What is his resultant velocity ?

Both velocity vectors are acting in the same direction.

Resultant velocity =1.50 m s1+2.00 m s1

=3.50 m s1 in the direction of the walkway.

b) Phiz then decides he is going the wrong way. He turns round

and starts to run at 3.40 m s1

in the opposite direction to themotion of the walkway. What is his new resultant velocity ?

The velocity vectors now act in opposite directions.Taking motion in the direction of the walkway to be positive :

Resultant velocity =+1.50 m s1 3.40 m s1

=1.90 m s1 (3 s.f.)

As this is negative, the resultant velocity acts in the oppositedirection to the motion of the walkway. He moves to the left.

Velocity vectors

velocity of 30 m s1in a NE direction : A velocity of 20 m s1due west :

Using the same scale, how would you draw the vector forvelocity of 15 m s1in a NW direction ?

N

E

45

W

S

Scalars Vectors

d is tance d isplacemen t

speed velocity

mass weight

pressure force

energy momentum

temperature acceleration

volume e lect ric cur rent

density torque

Scalars are simply added

4 kg 4 kg

resultant

F1

F2

resultant

F

F2

F


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2

Vector subtraction

The diagram shows the speed and direction ofa trampolinist at two points during a bounce :

What is the trampolinists change in speedfrom A to B ?

Change in speed =new speed old speed

=10 m s16 m s1

=4 m s1

What about his change invelocity?To find the change in a vector quantitywe use vector subtraction :

Change in velocity =new velocity old velocity

=10 m s1up6 m s1down

Remember, with vectors we must take account of the direction.In this example let us take the upward direction to be positive,

and the downward direction to be negative.We can then rewrite our equation as :

Change in velocity =+10 m s1(6 m s1)

=+10 m s1+6 m s1

=+16 m s1

So the change in velocity is 16 m s1in an upward direction.

Can you see that subtracting 6 m s1downwards is the same asadding 6 m s1acting upwards ?Vector subtraction is the same as the addition of a vector ofthe same size acting in the opposite direction.

Combining perpendicular vectors

o find the resultant of two vectors (X, Y ) acting at 90 to eachher, we draw the vectors as adjacent sides of a rectangle :

he resultant is the diagonalof the rectangle, as shown here :

he magnitude(size) of the resultant vector Rcan be found usingthagoras theorem :

R 2=X2+Y2

he directionof the resultant is given by the angle :

tan =opposite

adjacent =

Y

X =tan1(YX)

ou can also find the resultant using an accurate scale drawing.he magnitude of the resultant can be found by measuring itsngth with a ruler. The direction can be measured with aotractor. For more on scale drawing, see page 14.

resultan

tR

Y

X

xample 10A man tries to row directly across a river.He rows at a velocity of 3.0 m s 1.

he river has a current of velocity 4.0 m s1parallel to the banks.Calculate the resultant velocity of the boat.

he diagram shows the two velocity vectors.

We can find their resultant using Pythagoras theorem :

ize of resultant = (3.0 m s1)2+(4.0 m s

1)2

25 m s1

=5.0 m s1

=

Direction of resultant : tan=opposite

adjacent =

3.0

4.0 =tan1(3.04.0)=37

o the resultant velocity is 5.0 m s1at 37 to the bank.

xample 9wo tugs are pulling a ship into harbour. One tug pulls in a SEirection. The other pulls in a SW direction. Each tug pulls withforce of 8.0 104N. What is the resultant force on the ship ?

he two forces act at 90 to each other. Using Pythagoras theorem :

Magnitude of resultant = (8.0 104N)

2+(8.0 10

4N)

2

= 1.28 10 N10

=1.1 105N (2 s.f.)

ince both tugs pull with the same force, the vectors form adjacent sides of a square.he resultant is the diagonal of the square. So it acts at 45 to each vector.he resultant force must therefore act due south.

N

EW

S

resultant

45458.0

104N

8.0

104N

resu

ltant

4.0 m s1

3.0 m s1

Example 11A boy kicks a ball against a wall with a horizontal velocity of 4.5 m s1.The ball rebounds horizontally at the same speed.What is the balls change in velocity ?

Although the speed is the same, the velocity has changed . Why ?

Velocity is a vector, so a change in direction means a change invelocity.

Let us take motion towards the wall as positive,and motion away from the wall as negative.

Change in velocity = new velocity old velocity

= (4.5 m s1) (+4.5 m s1)

= 4.5 m s14.5 m s1

= 9.0 m s1 (2 s.f.)

So the change in velocity is 9.0 m s1in a direction away from the wall.

before impact

+4.5 m s

after impact

4.

6 m s1

10 m s1

A

B


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4

Resolving vectors

We have seen how to combine two vectors that are acting at 90,to give a single resultant. Now lets look at the reverse process.

We can resolvea vector into two componentsacting at rightangles to each other. The component of a vector tells you theeffect of the vector in that direction.

So how do we calculate the 2 components ?Look at the vector Vin this diagram :

We can resolve this vector into two components, V1and V2, atright angles to each other : V1acts at an angle 1to the original vector.

V2acts at an angle 2to the original vector. (1+2=90)

To find the size of V1and V2we need to use trigonometry :

cos1=adjacent

hypotenuse =

V1

V Rearranging this gives : V

1Vcos1

cos2=

adjacent

hypotenuse =

V2

V Rearranging this gives : V

2Vcos

2

So to find the component of a vector in any direction you need tomultiply by the cosine of the angle between the vector and thecomponent direction.

Combining vectors using scale drawing

more than one vector acts on an object we can find thesultant vectorby drawing each vector consecutivelyad to tail. The resultant Rwill be a line joining the startthe final end point. This works for any number ofctors acting at a point.

each vector is drawn carefully to scale, using a protractorensure the correct angles, then the resultant can be foundtaking measurements from the finished diagram.

otice that it doesnt matter in which order you choose todraw the vectors. You always end up with the samesultant. Two of the six possible arrangements for vectorsb, care shown here :y out the other combinations for yourself.

hat if, after drawing each vector head to tail, you end up backhere you s tarted ?this happens then the resultant is zero. This will be the

se for a system of forces acting on an object in equilibriume pages 32 and 36).

ector triangles

e can redraw the diagram on page 12 for tworpendicular vectors using the head-to-tail technique :

an you see that this gives the same resultant ?

ny two vectors can be combined in this way.he resultant becomes the third side of avector triangle.his is also a useful technique where the vectors are notrpendicular to each other.

the vectors lengths and directions are drawn accurately to scale,e resultant can be found by measuring the length of the resultantthe diagram with a ruler.

he direction is measured with a protractor.

he method also works forvector subtraction.gain the vectors are drawn head to tail, but the vectorbe subtracted is drawn in the reverse direction.

he diagrams show the resultant of the addition andbtraction of two vectors Pand Q:

you prefer a more mathematical solution, lengths and anglesn be calculated using trigonometry. For a right-angled triangle,thagoras theorem and sine, cosine or tangent will give youe magnitude and direction of the resultant (see also page 474).her triangles can be solved using the si ne or cosine rule.

though a mathematical solution does not require aale diagram, you should always draw a sketch using thead-to-tail method to ensure you have your triangle theht way round.

a

b

bb

c

cc

a

R

R

a

This system of vectors:

can be redrawn as:

or

X

X

X

This:

is the same as:

or

Y Y

Y

R

R

R

AddingP Q: SubtractingP Q :

P

P

P

QR

PQ R

PQ

Q

Q

Example 12A tennis player hits a ball at 10 m s1at an angle of 30to the ground.What are the initial horizontal and vertical componentsof velocity of the ball ?

Horizontalcomponent : vH=10 cos 30 =8.7 m s1 (2 s.f.)

The angles in a right angle add up to 90.So the angle between the balls path and the vertical=90 30 =60

Verticalcomponent : vV=10 cos 60 =5.0 m s1(2 s.f.)

vH

vV

10ms

1

30

Example 13A water-skier is pulled up a rampby the tension in the tow-rope.This is a force of 300 N actinghorizontally. The ramp is angledat 20.0 to the horizontal.

What are the components of the forcefrom the rope acting (a) parallel toand (b) perpendicular to the slope ?

The angle between the rope and the parallel slope direction =20So, the angle between the rope and the perpendicular direction =90 20 =70

a) Component of forceparallelto ramp Fpara

=300 N cos 20 =282 N (3 s.f.)

b) Component of forceperpendicularto ramp Fperp=300 N cos 70 =103 N (3 s.f.)

V2

1

2

20

20

300 N

Fpara

Fperp


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6

Combining non-perpendicular vectors by calculation

n the previous page you lear nt how to resolve a vector intoo perpendicular components. You can also use the samechnique to combine multiple vectors without the need forale drawings.

he trick is to first resolve each vector into its components.ten its convenient to resolve horizontally and vertically (butu can use any convenient mutually perpendicular directions).

the horizontal components can then be added togetherking account of their direction).milarly, all the vertical components can be added.he resultant horizontal and vertical vectors can then bembined using Pythagoras theorem.

e how this works in the following example :

xample 14hree forces act on a single point, as shown in the diagram :

ind the magnitude and direction of their resultant force.

tep 1 : Resolve the 3.0 N force into its horizontal and vertical components :

Horizontal component = 3.0 cos 30 =2.6 NVertical component = 3.0 cos 60 = 1.5 N

tep 2 : Replace the 3.0 N force with its components in the diagram (shown in black) :

tep 3 : Find the total force in the horizontal and vertical directions :

Total horizontal force = 6.0 2.6 = 3.4 N (to the left)Total vertical force = 2.0 +1.5 = 3.5 N (downwards)

tep 4 : Finally, combine the total horizontal and vertical forces into a single resultant force :

Using Pythagoras theorem :

Magnitude of resultant R = 3.42+3.52 =4.9 N (2 s.f.)

tan=opposite

adjacent =

3.5

3.4 =tan1 (3.53.4)=46

So the resultant force is 4.9 N acting at 46 below the original 6.0 N force.

Summary

There are 6 base quantities that you need for A-level(time, length, mass, temperature, electric current, andamount of substance).All other quantities are derived from these.

For an equation to be correct it must be homogeneous.This means that all the terms have the same units.But remember, a homogeneous equation may not beentirely right !

You should always give your numerical answers to thecorrect number of significant figures.

Scalarshave size (magnitude) only.

Vectorshave size (magnitude) and direction.Vectors can be represented by arrows.

When vectors are added together to finwe must take account of their direction

For vectors acting along the same straigone direction as positive and the other

To add two perpendicular vectorsXanPythagoras theorem : R2=X2+Y2

A single vector can be resolved to find iperpendicular directions.The component of a vector in any direcmultiplying the vector by the cosineof between the vector and the required dir

Subtracting a vector is the same as addthe same size, acting in the opposite dir

Questions

1.Can you complete the following sentences ?a) Measurements must be given as a number

multiplied by a .......b) Seconds, metres and kilograms are all ....... units.

Units made up of combinations of base units areknown as ....... units.

c) Vector quantities have both ....... and .......Scalars have only .......

d) The single vector that has the same effect as twovectors acting together is called the .......

e) The effect of a vector in a given direction is calledthe ....... in that direction.

f) The ....... of a vector in any direction is found bymutiplying the ....... by the ....... of the anglebetween the vector and the required direction.

2.Rewrite each of the following quantities using asuitable prefix :a) 2 000 000 000 Jb) 5900 gc) 0.005 sd) 345 000 Ne) 0.000 02 m

3.The drag force Fon a moving vehicle depends on its

cross-sectional areaA, its velocity vand the densityof the air, .a) What are the base units for each of these four

variables ?b) By checking for homogeneity, work out which of

these equations correctly links the variables :i) F =kA2vii) F =kA2viii) F =kAv2

(The constant khas no units.)

4.In a tug-of-war one team pulls to thof 600 N. The other team pulls to force of 475 N.a) Draw a vector diagram to show b) What is the magnitude and direc

resultant force ?

5.Two ropes are tied to a large bouldeis pulled with a force of 400 N due rope is pulled with a force of 300 Na) Draw a vector diagram to show b) What is the magnitude and direc

resultant force on the boulder ?

6.A javelin is thrown at 20 m s1at athe horizontal.a) What is the vertical component ob) What is the horizontal componen

7.A ball is kicked with a force of 120 horizontal.a) Calculate the horizontal componb) Calculate the vertical componen

8.Find the resultant of the following sy

forces (not drawn to scale) :a) by scale drawing,b) by calculation.

1.8 N

3.0

1.4 N45

Further questions on page 114.

6.0 N3.0 N

602.0 N

2.6 N

1.5 N

2.0 N

6.0 N

3.4 N

R 3.5 N


10/27

8

e live in a world full of movement. Humans, animals and theany forms of transport we use are obvious examples of objectssigned for movement. This chapter is about the Physics of motion.

this chapter you will learn :how to describe motion in terms of distance, displacement,speed, velocity, acceleration and time,how to use equations that link these quantities,how to draw and interpret graphs representing motion.

Distance and displacement

stance and displacement are both ways of measuring how far anject has moved. So what is the difference ?

stance is a scalar. Displacement is a vectorquantity (see page 10).splacement is the distance moved in a particular direction.

he snail in the picture moves from A to B along an irregular path :he distancetravelled is the total length of the dashed line.

ut what is the snails displacement?he magnitude of the displacement is the length of theaight line AB. The direction of the displacement is along this line.

Speed and velocity

he speed of an object tells you the distance moved per second,the rate of change ofdistance :

Acceleration

Acceleration is the rate of change of velocity :

4 Describing Motion

Military jet450 m s1

Racing car60 m s1

Cheetah27 m s1

Sprinter10 m s1

Tortoise0.060 m s1

Example 1The boat in the diagram sails 150 m due south and then 150 m due east.This takes a total time of 45 s.Calculate (a) the boats average speed, and (b) the boats average velocity.

a) average speed =distance travelled

time taken =

150 m +150 m

45 s =

300 m

45 s =6.7 m s1

b) average velocity =displacement

time taken =

212 m

45 s = 4.7 m s1in a SE direction

Example 2A ball hits a wall horizontally at 6.0 m s1. It rebounds horizontally at 4.4 m s1.The ball is in contact with the wall for 0.040 s . What is the acceleration of the ball ?

Taking motion towardsthe wall as positive :

change in velocity = new velocity old velocity

= (4.4 m s1) (+6.0 m s1)

= 10.4 m s1

acceleration =change in velocity

time taken =

10.4 m s1

0.040 s = 260 m s2 Negative, therefore in

away fromthe wall.

The runner is going ra constant speed.So how can he also His velocity is changhis direction is chang

peed is a scalar quantity, but velocity is a vector.

elocity measures the rate of change of displacement:

average speeddistance travelled(m)

time taken(s)

oth speed and velocity are measured in metres per second, written

/sor m s

1

. With velocity you also need to state the direction.

ing these equations you can find the averagespeed and theeragevelocity for a car journey.speedometer shows the actual or instantaneousspeed of the car.

his varies throughout the journey as you accelerate and decelerate.

how can we find the instantaneous speed or velocity at any point ?he answer is to find the distance moved, or the displacement, oververy small time interval. The smaller the time interval, the closere get to an instantaneous value (see also page 40).

average velocitydisplacement(m)

time taken(s)

Acceleration is measured in metres per second per second, ormetres per second squared, written m/s2or m s2.

It is a vector quantity, acting in a particular direction.

The change in velocity may be a change in speed,or direction, or both.If an object is slowing down, its change in velocity is negative.This gives a negative acceleration or deceleration.

Indicating direction

It is important to state the direction of vectors such as displacement,velocity and acceleration. In most motion problems you will be dealingwith motion in a straight line (linear motion).In this case you can use +and signs to indicate direction.For example, with horizontal motion, if you take motion to the rightas

positive, then :

3 m means a displacement of 3 m to the left

+8 m s1 means a velocity of 8 m s1to the right

4 m s2 means an acceleration of 4 m s2to the left(ora deceleration of an object moving towards the right)

The sign convention you choose is entirely up to you.In one question it may be easier to take upas positive whereas inanother you might use downas positive. It doesnt matter as long asyou keep to the sameconvention for the entire calculation.

acceleration change in velocity (m s1)

time taken(s)

45

N

150 m

before imp

+6

4.4 m

after impac


11/27

0

Displacementtime graphs

he diagram shows a graph of displacement against time, for a car :

hat type of motion does this straight line represent ?he displacement increases by equal amounts in equal times. Soe object is moving at constant velocity.

ou can calculate the velocity from the graph :

elocity from 0 to A =displacement

time taken

= gradient of line 0A

e steeper the gradient, the greater the velocity.

elocity is a vector so the graph also needs to indicate its direction.ositive gradients (sloping upwards) indicate velocity in one direction.egative gradients (sloping downwards) indicate velocity in theposite direction.

ow would you draw a displacementtime graph for a stationaryject ? If the velocity is zero, the gradient of the graph must also bero. So your graph would be a horizontal line.

his second graph is a curve. How is the velocity changing here ?he gradient of the graph is gradually increasing. This shows thate velocity is increasing. So the object is accelerating.

e could find the averagevelocity for this motion by dividinge total displacement sby the time taken t (=s/t).

ut how do we find the actual (instantaneous) velocity at any point ?he instantaneous velocity is given by the gradient at that point.his is found by drawing a tangent to the curve and calculating itsadient (=s/t; see the labels on the graph; see page 476).

hat would the gradient of a distancetime graph represent ?this case the gradient would give the scalar quantity, speed.

Velocitytime graphs

his graph shows the motion of a car travelling in a straight line :

starts at rest, speeds up from 0 to A, travels at constant velocityom A to B), and then slows down to a stop (B to C).

hat does the gradient tell us this time ?

adient of line 0A =change in velocity

time taken = acceleration

he steeper the line, the greater the acceleration.

positive gradient indicates acceleration. A negative gradientg. BC) indicates a negative acceleration (deceleration).

raight lines indicate that the acceleration is constant or uniform.

he graph is curved, the acceleration at any point is given by theadient of the tangent at that point.

The areaunder a velocitytime graph also gives usinformation. First lets calculate the displacementof the car during its motion.

From 0 to A :displacement = average velocity time taken

= (1

220 m s1) 10 s

= 100 mFrom A to B :displacement = average velocity time taken = 20 m s1 20 s = 400 mFrom B to C :displacement = average velocity time taken

= (1

220 m s1) 5 s

= 50 m

Example 3The velocitytime graph represents the motion of a stockcarstarting a race, crashing into another car and then reversing.

a) Describe the motion of the car during each labelled section.b) What is the maximum velocity of the car ?c) At which point does the car crash ?d) Does the car reverse all the way back to the start point ?

a) 0 to A : The car accelerates.A to B : The car is moving at constant velocity .B to C : The car rapidly decelerates to a standstill.C to D : The car is not moving.D to E : The velocity is increasing again but the values

are negative. Why ? The car is starting to reverse. E to F : The car is now reversing at constant velocity.

F to G : The car decelerates and stops.

b) From the graph, maximum velocity =15 m s1.

c) The car crashes at point B. This causes the rapid deceleration.

d) The area under the positive part of the velocitytime graph(shaded blue) gives the forward displacement of the car.The negative area (shaded red) gives the distance that the car reversed.As the red area is smaller than the blue area, we can see thatthe car did not reverse all the way back to the start point.

Compare these values with the areas under the velocitytime graph.(They are marked on the diagram.) What do you notice ?In each case the area under the graph gives the displacement.

This also works for non-linear velocitytime graphs :Area of shaded strip =vt =average velocity time interval

=displacement in this interval

Adding up the total area under the curve would give us the totaldisplacement during the motion.

What does the area under a speedtime graph represent ?This gives us the distancemoved.

200

40

A B

C D

10

5

5

15

60

velocity /m s1

8

50

10

A B

area = l h

10

20

velocity /m s1

15 20 25 30

area = bh1

2

= 10 s 20 m s1 = 20 s 20 m s 1

= 400 m

1

2

= 100 m

are

A

displacement

timet

s

0

The gradient of a displacementtime graphgives us the velocity.

displacement

time

tange

ntatX

s

t

t

X

s

0

velocity

time

A B

C

t0

v

The gradient of a velocitytime graphgives us the acceleration.

velocity

v

t0


12/27

2

s displacement(m)

u initial velocity(m s1)

v final velocity (m s1)

a constant acceleration (m s2)

t time interval(s)

Equations of motion

hese are 4 equations that you can use whenever an object movesth constant, uniform accelerationin a straight line.he equations are written in terms of the 5 symbols in the box :

hey are derived from our basic definitions of acceleration andlocity.heck your syllabus to see if you need to lear n these derivations,whether you only need to know how to use the equations tolve problems.

riting this in symbols : a =v u

t

So a t = v u which we can rearrange to give

om page 38, average velocity =displacement

time taken

the acceleration is constant, the average velocity during the motionll be halfway between uand v. This is equal to

1

2(u+v).

riting our equation for velocity in symbols :

1

2(u+v) =

s

t which we can rearrange to give

ing equation (1) to replace vin equation (2) :

s=1

2(u+u+a t) t

s=1

2(2u+a t) t which we can multiply out to give

om equation (1), t =v u

a

ing this to replace tin equation (2) :

s =1

2(u+v)

(v u)

a

2a s= (u+v) (vu)

2a s= v2u2 which we can rearrange to give

ote :

You can use these equations only if the acceleration is constant.

Notice that each equation contains only 4 of our five s u v a t variables.So if we know any 3 of the variables we can use these equations to find the other two.

erivations

om page 39, acceleration =change in velocity

time taken =

final velocity initial velocity

time taken

Example 4A cheetah starts from rest and accelerates at 2.0 m s 2due east for 10 s.Calculate a) the cheetahs final velocity, b) the distance the cheetah covers in this 10 s.

First, list what you know : s = ? u = 0 ( =from rest) v = ? a = 2.0 m s2

t = 10 s

a) Using equation (1) : vua t

v=0 +(2.0 m s210 s) =20 m s1 due east

b) Using equation (2) : s12(uv)t

s=1

2(0 +20 m s1) 10 s =100 m due east

Or you could find the displacement by plotting a velocitytime graphfor this motion. The magnitude of the displacement is equal tothe area under the graph. Check this for yourself.

Example 5An athlete accelerates out of her blocks at 5.0 m s2.a) How long does it take her to run the first 10 m ?b) What is her velocity at this point ?

First, list what you know : s = 10 m u = 0 v = ? a = 5.0 m s2

t = ?

a) Using equation (3) : s u t12a t2

10 m = 0 + (1

25.0 m s2 t2)

b) Using equation ( 1) : vua t

v=0 +(5.0 m s22.0 s) = 10 m s1 (2 s.f.)

Example 6A bicycles brakes can produce a deceleration of 2.5 m s2.How far will the bicycle travel before stopping, if it is movingat 10 m s1when the brakes are applied ?

First, list what you know : s = ? u = 10 m s1

v = 0 a = 2.5 m s2 (negative for deceleration)

Using equation (4) : v2u22a s

0 =(10 m s1)2+(2 2.5 m s2 s)

0 =100 m2s2(5.0 m s2 s) So s =100 m2s2

5.0 m s2

So t2=10 m

2.5 m s2 = 4.0 s2

v u a t . . . . . (1)

s 12(u v) t . . . . . (2)

s u t 12a t2 . . . . . (3)

v2 u2 2a s . . . . . (4)

20 m s1

v


13/27

49

Physics at work : Speed and stopping distance

Stopping distancesWhen you learn to drive you need to learn the Highway Code to pass your Driving Theory Test.The Highway Code includes a table of typical stopping distances for cars travelling at different speeds.A section of this is shown below. Notice that the total stopping distance is made up of two parts.The thinking distanceis the distance travelled before the driver reacts and starts to brake.You move a surprising distance after spotting a hazard before you even press the brake pedal.The braking distanceis the distance travelled while decelerating to a stop once youve pressed the brake.What deceleration and reaction time do these figures assume ? We can find out using our equations of motion.Using the 70 mph (31 m s1) figures :

v2=u2 + 2as0 =(31 m s1)2+(2 a75 m) So a=

(31 m s1)2

(2 75 m)=6.4 m s2(negative as it is a deceleration)

Reaction time = thinking distanceaverage speed

= 21 m31 m s1

=0.68 s

So the key to high speed is minimising drag by using streamlined shapes, as well as maximising engine power.This has been taken to extremes in vehicles such as Thrust SSC. SSC stands for supersonic car !Thrust SSC holds the official land speed record for a wheeled vehicle, achieving an incredible 763.035 mphusing a mix of car and aircraft technology. Thrusts driver, Andy Green, is now involved in designingBloodhound SSC, a rocket-powered car being designed to break the 1000 mph barrier !

Fast carsn 2014, the Hennessey Venom became the worlds fastest productionoad car, reaching 270.49 mph on the Space Shuttle landing strip in

Florida. This beat the previous record holder, the Bugatti Veyron,by just 0.63 mph !

So what limits the top speed of a car ? On page 45 you saw howa falling object will accelerate until it reaches its maximum speed, orerminal velocity. This is due to the increase in the drag force as the

object speeds up. Exactly the same effect limits the speed of a car.Even with the accelerator pressed to the floor, there is a maximum drivingorce available from the engine. This will accelerate the car rapidly at first

but as the speed increases the drag force on it will als o increase.Eventually the drag force becomes large enough to bala nce the motiveorce and so it is no longer possible to accelerate. The graph shows thehape of the velocitytime graph for a car reaching its maximum speed :

Note that these are only typical values. In practice theactual distances would depend on factors such as yourattention, road surface, weather and the vehicles condition.

max speed(acceleration0)

increasing drag at higher speedreduces the acceleration

t

v

Braking distance

14 m

Thinking distance

70 mph

50 mph

30 mph

75 m

15 m

9 m

21 m

38 m

Thrust SSC.

Bloodhound SSC.

The Hennessey Venom.

Physics at work : Car safety

Airbags

Airbags are designed to provide a cushion between your upper bodyand the steering wheel or dashboard. This can reduce the pressureexerted on you by more than 80% in a collision.

Without an airbag, your head would hit the steering wheel about80 milliseconds after an impact. To be of any use, the crash mustbe detected and the airbag inflated in l ess than 50 ms !

So the bag is inflated explosively. This could be dangerous if itinflated accidentally under normal driving conditions.

So what triggers the airbag ? Why does it inflate during a collisionand not whenever you brake hard ?

In a collision you are brought to rest very rapidly, say in 0.10 s.Even at low speeds this produces high deceleration. For example,at 20 mph (9 m s1) :

acceleration =v

t =

0 9 m s1

0.10 s =90 m s2=9.2g (whereg=acceleration d

This means that your seat belt must be able to exert a force over 9 times your body weight withou

Compare this with the deceleration during an emergency stop, even at high speed.On page 49 we used data from the Highway Code to calculate the typical deceleration of a car frThe answer of 6.4 m s2(0.65g) is 14 times smaller than the deceleration in even a low-speed cAn acceleration sensor can easily detect the difference between the two, and only activate the airbag

Sections of a car are designed to cin an accident.

Crash detection, inflation andthe airbag (to allow the drivemust all take place in less timyou to blink !

force

exerted

on

driver

/N

6000

chest hits steering wheel

without seat beltwith seat belt

400

80

4000

2000

Seat belts and crumple zones

Seat belts save lives ! In a collision with a stationary object, the front of yourcar stops almost instantly. But what about the passengers ?

Unfortunately they will obey Newtons first law and continue moving forwardsat constant velocity until a force changes their motion. Without a seat belt,this force will be provided by an impact with the steering wheel or windscreen.This can cause you serious injury, even at low speeds.A seat belt allows you to decelerate in a more controlled way, reducing the forceson your body. In an accident it is designed to stretch about 25 cm.This allows the restraining force to act over a longer time.

Newtons second law (F=(m v)/t, see page 55)tells us that the longer the time ttaken to reducea passengers momentum (m v), the smaller theforce Fneeded to stop them.

Look at the graph. It compares the forces actingon a driver involved in a collision with, andwithout, a seat belt :

Newtons second law also explains why modern

cars are designed with crumple zones.The passenger cell is designed as a rigid cage tomaximise passenger protection by transmittingforces away around the roof and floor of the car.But crumple zones are deliberately built into the cars bodywork.By crumpling, the car takes a longer time to come to rest.Again this means a lower rate of change of momentum and asmaller force acting on the passengers.So, although cars suffer significant damage in even a minorcollision, you are far more likely to walk away from an accident.


14/27

30

Electromagnetic waves

o you recognise this well-known experiment ?hite light is a mixture of several colours and can be splita prism to show a spectrum. Light travels as a transverse

ave and each colour of light has a different wavelength.ed light has the longest wavelength, violet the shortest.

he visible spectrum is a tiny part of a much wider spectrum,th wavelengths that range from 1015m to more than 1 km.

The regions of the electromagnetic spectrum

he diagram above shows how the spectrum is l abelled, but

ere are no sudden boundaries between the different regions.amma radiation and X-rays

amma radiation is emitted by radioactive nuclei (see page 360).rays are produced when high-speed electrons decelerate quickly.gh-energy X-rays have shorter wavelengths than low-energymma rays. They are given different names only because of theay that they are produced.

traviolet (UV)

hy is the arc welder in the photograph wearing eye protection ?ectric arcs, such as sparks and lightning, produce ultraviolet.s also given out by our Sun. Small amounts of UV are goodr us, producing vitamin D in our skin. Large amounts of high-ergy UV can damage living tissue.

ckily for us the Ozone Layer, high in the Earths atmosphere,sorbs UV rays with wavelengths less than 300 nm (3 107m).

ut its recent thinning increases the risk of skin cancer.

sible light

uman eyes can detect wavelengths from 400 nm (violet light) to00 nm (red light). Other animals have eyes that are sensitive toferent ranges. Bees, for example, can see ultraviolet radiation.

frared (IR)

ery object that has a temperature above absolute zero gives

t infrared waves. Rescue workers can use infrared viewerssearch for survivors trapped below collapsed buildings.was the first invisible p art of the spectrum to be discovered

y the astronomer William Herschel in 1800).

adio waves

adio waves range in wavelength from millimetres to tens ofometres. Microwaves are the shortest of the radio waves andey are used for mobile phone and satellite communication.nger wavelengths are used for radio transmissions.

The nature of electromagnetic radiation

All electromagnetic waves can travel through a vacuum at3.00 108m s1. This value is called the speed of light.But what are electromagnetic waves ?

Electromagnetic waves consist of electric and magnetic fields.Can you see that the fields oscillate in phase, and are at 90to each other and to the direction of travel of the wave ?Each field vibrates at the wave frequency.

Who discovered these ideas ?By the early 19thcentury it was known that an electric currentalways produces a magnetic field. Michael Faraday then showedthat changing a magnetic field produces an electric current.

In 1862 James Clerk Maxwell saw the connection :If a changing magnetic field produces a changing electric field,the electric field must create a changing magnetic field.The two oscillating fields are linked !

Maxwell predicted that an oscillating electric charge should

radiate an electromagnetic wave. He also derived an equationfor the speed cof the wave in free space, using the magneticfield constant oand the electric field constant o:

Using the constants in this equation gives the speed of light !Maxwell had shown that light waves are electromagnetic waves.He thought that light was just one part of a wider spectrum.In 1887, Heinrich Hertz proved that Maxwells ideas werecorrect when he discovered radio waves (see page 167).

Measuring the speed of light

Galileo had tried to measure the speed of light in 1638.Can you suggest why he did not s ucceed ?

The first successful direct measurement of the speed of lightwas achieved by Fizeau in 1849. His apparatus consistedof a rotating toothed wheel with Nteeth and a mirror.

A beam of light passes through a gap A between two teeth.It travels to the distant mirror and is reflected back towards

the wheel. The wheel spins, going faster and faster.If light leaving through one gap returns to the wheel as thenext tooth B takes the place of the gap, the light is cut off.

We now know the time tfor the light to travel to the mirrorand back, a distance 2d.The wheel spinsftimes per second and in time tit makes

1

2N of a turn while a cog replaces a gap. So t=

1

2N f.

This method gave Fizeau a value for cof 313 000 km s1.

c1

oo

Maxwells calculation for the value

ois the permeability of free space

ois the permittivity of free space

c=410

78.85 10

12

1

c=3.00 108m s1

Fizeaus calculation for the value o

The wheel had 720 teeth.The frequencyfwas 12.6 Hz.Distance 2dwas 17.26 km =1.72

Speed =distance

time, so c=

2d

1/(2N f

c = 1.726 104m 2 720= 3.13 108m s1

nature ofwave

velengthetres)

radio waves

102101 1031101102103104105106107108109101010111012

microwavesinfraredultraviolet

visible light

x-raysgamma

wavelength

m

A

B

toothed

wheel

d

Welders must protect themselves against UV.

Photographs taken with visible light and IR.


15/27

62

Standing waves and resonance

ou can trap waves on a string by attaching a vibrationnerator to a long cord, fastened firmly a t the other end :

he vibrator sends waves along the cord. The waves reflect backom the fixed end, and superpose with the incident waves frome vibration generator, to form a standing wave.

he output of the signal generator can be varied so that thenerator forces the cord to vibrate at different frequencies.certain frequencies, the string vibrates with a large

mplitude. These are the resonantfrequencies of the system.

esonance occurs when the frequency that is driving the systemom the vibration generator) matches a natural frequency ofe system (see also page 110).

he photograph shows what you might see. At a node the cordes not move at all, and you can see the string quite distinctly.the anti-nodes the string vibrates with a maximum ampl itude,d so it appears as a blur.

ationary waves in string instruments

he guitar and violin are just two types of string instrument.hen one of the strings is plucked or bowed in the middle,ansverse waves travel along the string in opposite directions.

the ends of the string the waves are reflected back and aationary wave, with nodes Nand anti-nodesA, is set up.

he string can only vibrate with certain frequencies. These aree resonant frequencies of the string; we call them harmonics.hy are only certain frequencies allowed ?

his is because you must have a whole number of stationaryave loops fitting into the length of the string.

he simplest way the string can vibrate is with just one loop.he wavelength is twice the length Lof the string.an the stationary wave have a longer wavelength than this ?o, one loop produces a sound of the lowest possible frequency.his is thefundamental frequency orfirst harmonic,f1.

the second diagram, the string vibrates with two loops.he wavelength is now equal to the length of the string.his is half as long as it was for one loop.he frequencyf2of this second harmonicmust be twice theequency of the fundamental frequency.

he third harmonic is at three times the fundamental frequency,cause the wavelength is 1

3of its value for one l oop.

an you predict the pattern for the rest of the harmonics ?

The first harmonic of a stretched string

Have you noticed that guitar strings have different thicknesses ?Each string has a different mass per unit length,.

If you tighten any of the strings, increasing its tension,T,why does the frequency of the note from the wire increase ?

When you strum the guitar string, waves move in both directionsalong the string and superpose to give a stationary wave.Both the tension andthe mass per unit length affect the speed cof these waves on the string.

From the previous page we know that when the string vibrates inits first harmonic mode, the wavelength of the stationary waveis twice the length Lof the string.

For any wave cf (page 125)and so the first harmonic frequencyf1is given by :

f1

c

or f1

T1

2L

Number of loops 1 2 3 4 5

Wavelength 2L2L2

2L3

2L4

2L5

Frequency f1 f2=2 f1 f3=3 f1 f4=4 f1 f5=5 f1

Using a sonometer

The factors that affect the first harmonic frequency of a stringcan be investigated using a sonometer, as shown :

One experiment is to see how the length Lof the wire affectsthe frequency of the note it produces. You can adjust the lengthby altering the position of the moveable bridge.Or you can investigate how the tension Tin the wire affects

the frequency, by adjusting the weights.How do you measure the frequency of the s ound from the wire ?This is done by adjusting the wire until the sound it producesmatches that of the tuning fork.

Unless you have a keen ear for music, you may find it difficult tomatch the sounds ! The paper rider will help.

If the tuning fork is placed as shown, it forces the wire to vibrate.When the paper rider vibrates vigorously, the wire is resonating.So then the frequency of the wire equals the frequency of the fork.

stretched rubbervibrationgenerator

signal generator(10100 Hz)

A

N

N

A

A A

length L

AA A

A AA A

N

N N

N NN

N

N NNN

N N

3rd harmonic

4th harmonic

2nd harmonic

1st harmonic

2

woodensonometer

wire

L

paperrider

vibrating tuning fork

In fac t : speed, c(m s1)Ttension, (N)

mass per unit length, (kg m )1

or cT

Example 3A guitar string has a mass per unit length of 7.95 104kg m1

The string is 0.600 m long and is under a tension of 78.5 N.What value is the first harmonic frequency of the guitar string ?

1f

Tc 1

2L

1

2 0.600 m

78.5 N

7.95 10 kg m262 Hz (3 s.f.)

4 = =

=

1

0.6

The string vibfirst harmonic

How could you increase thefundamental frequency of a string ?You could press down on the stringwith your finger to shortenits length. Or you could tighten thestring, to increase its tension.


16/27

06

Example 6In the circuit shown, the capacitor is charged to 6.0 V.The switch is now closed. Calculate :a) the time constant,b) the p.d. across the capacitor after 20 s,c) the charge on the capacitor after 20 s.

a) C R b) CRt

Vt V

0e c) Qt C Vt

=500 106F 100 103 = 6.0 e20 s 50 s

= 500 106

=50 s (2 s.f.) = 6.0 e0.4 Qt = 2.0 103C

Vt= 4.0 V (2 s.f.)

Exponential discharge

How does the charge on the capacitor plates vary as it discharges ?

The diagram shows our discharge circuit again :

At this instant, the charge on the capacitor is Q, the dischargecurrent is Iand the p.d. across the resistor a nd the capacitor is V.

In a short time t, a small charge Qflows off the capacitor plates.From page 228 we know that : Q=It

But for the resistor, I=V

R, and for the capacitor, V=

Q

C, so I=

Q

C R.

Substituting for Iin the equation for Q: Q=Q

C R t or

(The minus sign tells us that Qdecreases as time passes.)

This equation tells us that the rate of flow of charge off thecapacitor plates, Q/t, depends on the time constant C Randon the charge Qon the capacitor plates. As time passes, Qfallsand the capacitor discharges more and more slowly.

Lets look in detail at the graph of charge Qagainst time t:After a time C R, the charge has fallen to 1/e(about 1/3) of Q0.

After a further time C R, the charge has again fallen by 1/e.Now the charge is 1/e1/e(roughly 1/9) of its original value.

A further C Rseconds and the charge is 1/e1/e1/eof Q0.

The charge continues to fall by 1/eevery C Rseconds.We say that the charge falls exponentially with time, and

Qt Q

0e CRt

Q0 is the initial charge

Qt is the charge after time t

CRis the time constant

Q=C V, and so Vis proportional to Q. The p.d. Vmust also fall exponentially.

V=I R, and so Iis proportional to V. The currentImust fall exponentially.

Discharging a capacitor

hat factors affect the timetaken for a capacitor to discharge ?ou can investigate this using the apparatus shown :

he capacitor charges up when the switch is in position A.ou can then move the switch to position B and dischargee capacitor through the resistor.

he voltmeter records the p.d. across the capacitor.an you see that this is also the p.d. across the resistor ?hy will the voltmeter read 6 V at the start of the discharge ?

ou can record the p.d. every 10 s as the capacitor discharges.

he graph shows the results that you would get usingferent values for the capacitance Cand the resistance R:

hy does the discharge take longer as Cand Rincrease ?he larger the resistance, the more it resists the flow of charge.he charge moves more slowly around the circuit.

he greater the capacitance, the greater the charge stored.takes longer for the charge to flow off the capacitor plates.

hy does the p.d. fall more and more slowly with time ?ok at the circuit diagrams :

otice that the smaller the p.d., the smaller the current :other words, the smaller the rate of flow of charge.

the p.d. falls, the charge flows off the plates more slowlyd the time for the p.d. to drop takes longer and longer.

he time constant

hetime constantgives us information about the time itkes for a capacitor to discharge :

xample 5What is the time constant for the circuit s hown ?

Time constant, capacitance, C resistance,R

= 500 106F 56 103

= 28 seconds

CR

1

e

1

e

Q0

Q0

Q0

Q

1

e

2CRCR 3CR

t

V01

e

1

e

V0t

CReVtV0

V

V01

e

1

e

1

e

I0

1

e I0

I0

CR

I

Time constant, capacitance, C resistance,R (seconds) (farads) (ohms)

fact, the time constant is the time it takes for the p.d. tol to 1/e of its original value V0.

is one of those special mathematical numbers, rather like ).he value of e is 2.72 (3 s.f.) and so 1/eequals 0.37(2 s.f.).

ter a time CR, the p.d. Vacross the capacitor equals 0.37 V0.is now 37% (very roughly 1/3) of its original value.

ter a time 5CR, the p.d. will have fallen to about 0.7% of V0d we can consider the capacitor to be effectively discharged.

6 V

A B

R

C

V

time /s

p.

d.

/V

6 V

40 800

0

120 160 200 240

3 V

C=1

000

F

R=100

k

C=1000F

R=56k

C=

500

F

R=56k

I I I I

I= = = 60 A

6 V

6 VV

R 100 k I= = = 40 A

4 V

4 VV

R 100 k

100 k 100 k

500 F

56 k

S

R

C

Q = CVV

6.O V

10500 F

Q

t

Q

C R


17/27

08

Charging a capacitor : Exponential growth

hat factors affect the time taken for a capacitor to charge up ?ou can investigate this us ing the apparatus shown :

ere we are using a 6 V supply.he voltmeter is recording the p.d. across the capacitor.this also the p.d. across the resistor ?o, because the capacitor and resistor sharethe supply p.d.

hen the switch S is closed, the capacitor starts to charge up,d you can record the p.d. across the capacitor every 10 s.

he graph shows the results that you would get usingferent values for the capacitance Cand resistance R:

otice that, as the capacitor charges, the p.d. increases everore slowly, eventually reaching the supply p.d. of 6.0 V.

an you see that charging takes l onger as Cand Rincrease ?he time constant affects the charging process.

nce the charge on the capacitor is proportional to the p.d.ross its plates, the graph of charge Qagainst time mustthe same shape as the graph of p.d. against time.

ts look in more detail at the graph of charge against time :ote that the charge on the fully charged capacitor is Q0.

ter a time C R, the charge has risen to (1 1/e) of Q0.his is about 2/3 of the final charge Q0.

ter time 2 C R, the charge has risen to (1 (1/e1/e))(1 1/e2) of Q0. This is about 8/9 of Q0.

an you see a pattern ?

e say the charge increases exponentiallywith time, and :Q0is the final chargeQtis the charge after time tC Ris the time constant

e can write a similar equation for the p.d. across thepacitor as it charges up, but what about the current ?

ok at the circuit diagram again.

Cis the p.d. across the capacitor, VRthe p.d. across the resistor.an you see that VC+VRmust alwaysequal the supply voltage ?

as VCrises exponentially, VRmust fall exponentially.he current is proportional to the p.d. VRacross the resistor, soe charging currentfallsexponentially as the capacitor charges.

V

S

R

C

6 V

VC

VR

I

I

C

=500F

R

=56k

C=

1000F

R

=56k

C=

100

0F

R

=100

k

6 V

400 80 120 160 200

p.d. V /V

3 V

time /s

CR0 2CR 3CR

Q

t

(1 1/e3) Q0

(1 1/e2) Q0

(1 1/e) Q0

CRt

Qt Q

0(1 e )

Vt V

0(1 e )CR

t

It I

0e CRt

xample 7

he switch is closed in the circuit shown. Calculate :250 F

100 k6.0 V) the time constant,

) the p.d across the capacitor after 40 s, the charging current after 40 s.

) CR b) CRt

VtV

0(1 e ) c) The p.d. across R=6.0 V 4.8 V =1.2 V

=250 106F 100 103 =6.0 (1 e )e40 25 s

Using V =IR = 1.2 V :

=25 s (2 s.f.) =6.0 (1 e1.6) 1.2 V =I100 103

Vt=4.8 V (2 s.f.) I= 1.2 10

5A (2 s.f.)

Physics at work : Capacitors

Physics at work : Earthing

The capacitance of charged spheres

Have you seen this experiment being carried out ?

The student stands on a block of polystyrene and touches thedome of the Van de Graaff generator. When the Van de Graaffis switched on, the dome and the student become charged.Her hair demonstrates that like charges repel !

You need to take care when using a Van de Graaff generator.As the dome becomes highly charged, an el ectric field is createdat its surface. The electric field strength can be so large that theinsulation of the air breaks down, and sparks pass from the dometo any nearby object that is connected to Earth (see page 294).You may get an unexpected electric shock !

The dome of the Van de Graaff is storing electric charge.It is acting as a capacitor. Can we calculate its capacitance C?

The diagram shows an isolated sphere of radius rs, in a vacuum :The dome of the Van de Graaff is (almost) a sphere.

The charge on the sphere is +Q. From page 296 we know that

the electric potentialV

at the surface of the sphere is :V =

k Q

rs

, where k =1

4O

For any capacitor C =Q

V, and so : C = Q

Q

4Ors

C = Q4Ors

Q,

When a charged capacitor is connected to an uncharged capacitor (as in Example 3),electrons flow between the capacitors until the p.d. across both capacitors is the same.The capacitors share the charge, but not equally.

If the uncharged capacitor has a much larger capacitance than the charged capacitor,the charged capacitor loses almost all of its charge to the uncharged capacitor, and itis effectively discharged. Example 8 above shows that the Earth has a large capacitance.So when a charged conductor is connected to the Earth, it is (effectively) discharged.

The Earth has such a large capacitance that any loss or gain of charge causes a verysmall change in its potential. Therefore, we can treat the Earth as a convenient zeroof electrical potential energy, as we s aw in Chapter 21.

electric

field lines

+

+

+

+

+

Example 8Treating the Earth as an i solated sphere, what is its capacitance ?The radius of the Earth is 6.4 106m.

(From page 290, the value of Ois 8.85 1012F m1.)

C4OrsC=48.85 1012F m1 6.4 106m =7.1 104F, or 710 F (2 s.f.)

the s

for e

neu

C 4Ors


18/27

22

Thomsons measurement ofQ

mfor electron

1897 J J Thomson found a way to measuree ratio of charge Qto mass mfor the electron.

he diagram shows an electron beam moving through acuum tube called an electron deflection tube.

he beam can be deflected in 2 ways :

by an electricfield between the plates Y1and Y2,by a magneticfield produced by the 2 coilscarrying a current.

n electric field between Y1and Y

2can push electrons downwards

a constant force Q E(see page 292), as shown in the diagram :

he value of the electric field Eis found by measuring the p.d. Vtween the plates Y1and Y2and the separation dbetween those

ates, as we know that EV

d(see page 293).

current in coils 1 and 2 provides a magnetic field Bacrosse electron deflection tube at right angles to the electric field.

he coils are connected in series so they have the same current.he magnetic field gives an upwardsforce B Q von the electronam (see page 264).

he current through the coils is adjusted until the beam ofectrons is a horizontal line. At this point the upward magneticrce balances the downward electric force. The current is noted.

he electron deflection tube is removed, and the magnetic field Bdway between the two coils is measured using a sensor such asHall probe (see page 257) while the coils carry the same currentbefore.

ombining the electric and magnetic forces

e dont know the electron velocity v, so we need to eliminate itom our calculations. The kinetic energy gained by the electronom the electron gun is QVgun(see page 321).

electron gun coil 2

Y1

Y2

3000V

coil 1

fluorescentscreen

force FBQ v

force FQE

V

Balanced forces on the moving electron.

1

2m v 2Q Vgun v

2=2Q Vgun

m

ut B Q vQ E v=E

B and so v2

E2

B22Q Vgun

m=

E2

B2 and dividing both sides by 2V

gun

Q

m

E2

2Vgun B2

xample 2n this experiment, the electrons are accelerated by a p.d. of 2900 V. The same p.d. Vgunis applied across the

lates Y1and Y2, which are 10 cm apart. The magnetic field Bneeded to balance the forces on the electroneam is 9.1 104tesla.

Calculate the value ofQ

mgiven by this experiment.

EV

gun

d=

2900 V

0.10 m=29 000 V m1(or 29 000 N C1). Note that you must use din metres !

Q

m

E2

2VgunB2=

(29 000 V m1)2

2 2900 V (9.1 104T )2=1.8 1011C kg1 (2 s.f.)

Physics at work : Particle accelerators

A linear accelera

A synchrotron.

1, 2, 3 are acceleration e

1

2

particlesource

beam

Particle physicists can create new particles by colliding other particlestogether. To do this, the colliding particles need to be given largeamounts of energy by an accelerator. Accelerators use electric fieldsto accelerate particles such as protons to energies much greater thanthe 2 000 eV of the electron gun on page 321.

Linear accelerator or linacThis accelerates particles in a straight line :The cylindrical electrodes are connected to an alternating supplyso that they are alternately positive and negative. The frequencyof the p.d. is set so that as the particles emerge from eachelectrode they are accelerated across the next gap.Why do the electrodes get progressively longer ? To keep in stepwith the alternating p.d., the particles must take the same time totravel through each electrode. As the particles get faster the tubesmust get longer. The accelerator may be 3 km long !A linear accelerator can accelerate electrons to about 50 GeV.

CyclotronA cyclotron consists of two semicircular metal dees separated by

a small gap. When a charged particle enters the cyclotron,a perpendicular magnetic field makes it move along a circularpath at a steady speed. Each time the particle reaches the gapbetween the dees, an alternating p.d. accelerates it across.

The force on a moving charged particle in a magnetic field isgiven byFB Q v(see page 264). Since this provides thecentripetal force (F m v2/r , see page 80), we can write :

m v2

r = B Q v v =

B Q r

m

This shows that the velocity is proportional to the radius, so as theparticles get faster they spiral outwards. The time spent in eachdee stays the same (see the coloured box) :

So the alternating p.d. must reverse everym

B Qseconds.

Cyclotrons are used to accelerate heavy particles such as protonsand -particle s. These can reach energies of about 25 MeV.

SynchrotronIn modern accelerators, particles travel at speeds close to thespeed of light. At these very high speeds, Einsteins theory ofspecial relativity(see Chapter 28) is needed to explain theway in which particles move. As particles approach the speedof light, a constant force does not produce a constantacceleration you cannot use Ek=

1

2m v2in spe cial relativity !

Synchrotrons contain electromagnets which keep the particlesmoving in a circle. Regions of electric field at various pointsaround the loop give the particles extra energy, in pushessynchronised with the arrival of pulses of particles.

As the particles gain energy, the magnetic field is increased tokeep them moving in a circle of constant radius.

Synchrotrons can accelerate particles such as protons to energiesof more than 1000 GeV.

alternating supply

D

c

magnetic field B

v

B Q v

How long does a particle takeits semicircular path within eac

time =distance

speed=

r

B Q r/m

This is independent of the speSee also page 265.

particles extracted

evacuated pipe

accelerating cavity

d

focusing

magnet

particles

injected

linear

accelerator


19/27

84

Annihilation

Why does antimatter not last long ? As soon as an antiparticlemeets its particle, the two destroy each other. Their massis converted to energy. This is called annihilation.

For example, when an electron and a positron collide,they annihilate, producing two -ray photons of energy :

Why are two photons produced ? One photon could conservecharge and mass/energy. But momentummust also be conserved,and for that to happen, two photons are needed.If an electron and positron with the same speed collide head-on,the total momentum before the collision is zero.For zero momentum after the collision, we must have two identicalphotons moving in opposite directions.

When sufficient energy is available, annihilation can produceshort-lived particles. The energy is converted back into matter.This is how new particles are created in accelerator experiments.

hat is everything made of ? What holds everything together ?rticle physicists are continuing to search for answers to theseestions. This chapter introduces you to some of the key ideas.

this chapter you will learn :what the building blocks of all matter are,what antimatter is and how it is created,about the fundamental forces that hold everything in theUniverse together.

Matter and antimatter

ntimatter is not just science fiction ! Every particle hasequivalent antiparticle for example, antiprotons and

tineutrons. These are real particles.ntimatter is a bit like a mirror image. A particle and itstiparticle have the same mass. They carry equal but oppositearge and they spin in opposite directions.

me antiparticles have special names and symbols, butost are represented by a bar over the particle symbol.r example, p(pbar) represents an antiproton.

he existence of antimatter was predicted mathematically byitish physicist Paul Dirac in 1928. Diracs equation links themplex theories of special relativity and quantum mechanics.

he equation describes both negative electrons and an equivalentsitiveparticle. These anti-electrons or positrons(e+or +)

ere discovered experimentally by Carl Anderson in 1932.

ntiprotons and antineutrons were first observed in acceleratorperiments in the mid-1950s. Since then antiparticle s haveen observed or detected for all known particles.any antiparticles occur naturally. They are created by high-ergy collisions of cosmic rays with the molecules in our

mosphere. Positrons are also created in +-decay (page 366).his reaction is routinely used in hospitals for medical imaging

positron emission tomography (PET) (see page 432).

an antiparticles join together to make anti-atoms and realtimatter ? In theory, yes. In 1995, scientists at the Europeanboratory for Particle Physics (CERN) created their first atomsantihydrogen by joining positrons with antiprotons.

nly 9 atoms were made in total. They lasted just 1010s !

y 2011, several hundred antihydrogen atoms could be maded stored, but only for around 17 minutes.current production rates it would take us 100 billion yearsproduce just 1 g of antihydrogen !

antiproton + positron antihydrogen

p + e+ H

Paul Dirac predicted theexistence of antimatter in 1928.

An antiproton (blue) collidebubble chamber liquid. Th4 positive (red) and 4 nega

Example 1An electron and positron with negligible kinetic energy annihilate and produce two identical -ray pCalculate a) the energy released and b) the frequency of the -photons.

Rest mass of an electron =9.11 1031kg h=6.63 1034J s c=3.00 108m s1

(a) Using Einsteins equation (see page 375),E m c2 = (2 9.11 1031kg) (3.00 108m s1)2=1.64 1013J

(b) Using Plancks equation Eh f (see page 328),

Energy of each photon =1

21.64 1013J f =

E

h =

1

21.64 1013J

6.63 1034J s =1.24

Pair production

Pair production is the opposite of annihilation. High-energy photonscan vanish, creating particleantiparticle pairs in their place.For example, a -ray can produce an electronpositron pair :

A third particle such as an atomic nucleus or electron is often involvedindirectly. This recoils, carrying away some of the photon energy.

The photograph shows the creation of electronpositron pairs at X and Yin a bubble chamber (see page 359). The tracks are curved due to amagnetic field. They are shown in green (electrons) and red (positrons).Notice that the -ray photons leave no tracks as they are uncharged.

In event X, a third particle is involved. (This is an electron ejected fromthe bubble chamber liquid during the interaction.)

Why are the tracks at X more curved than at Y ?In event X the ejected electron carries off some of the energy.This leaves less energy for the electronpositron pair.The lower the energy (and speed) of the particles produced, the more theyare deflected in the magnetic field. This is why their paths curve more.

0

1e 0

1e 2

00

00

0

1e 0

1e

Y

e

27 Particle Physics

mirror image

particle

+

anti-particle


20/27

08

29 Astrophysics

tronomers have made huge strides in observing the Universe.oday we use all ranges of the electromagnetic spectrum tove us insights into the processes in distant stars and galaxies.

this chapter you will learn :how reflecting and refracting telescopes are used,how stars are classified, and how they change,how Doppler shift and cosmological red-shift are used.

Observing the sky

aked-eye astronomy

rly astronomers observed the sky with the naked eye. They sawat the fixed stars appeared to rotate around the Earth, taking ale less than a day to do this. They also saw things which didt move with the stars : the Sun, the Moon and the planetsercury, Venus, Mars, Jupiter and Saturn.hey saw other mysteries :

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