Course Guide AS A2 OCR Physicsassets.pearsonschool.com/asset_mgr/versions/2012... · 1W = J s 1 att oer For ce Extension Gradient = k Area under line = W ork done W 1 Fx 2 W ork done

OCR Physics

In Exclusive Partnership

Course GuideAS A2

Heinemann is working exclusively with OCR to produce an exciting suite of resources tailored to the new OCR GCE Physics specification. Written by experienced examiners, Heinemann’s brand new resources provide you with tailored support for teaching the revised specification. Comprehensive support for AS and A2, and motivating exam preparation in our unique Exam Café, will give your students every chance of success.

Exciting resources developed in exclusive partnership to support the new GCE Physics specification

Course Structure

22

AS

AS Student Book with Exam Café

CD-ROM

Roger Hackett and Robert Hutchings

AS

Exclusively endorsed by OCR for GCE Physics A


AS Teacher Support CD-ROM

AS

01865 888118 Graham BoneDavid Styles


ISBN 978 0 435961 85 1

Teacher SupportExclusively endorsed by OCR for GCE Biology A

A2 Student Book with Exam Café

CD-ROM


A2



A2 Teacher Support CD-ROM

A2

AS Revision Guide

A2 Revision Guide

l OCR and Heinemann working in exclusive partnership to provide better support for you.

l Engaging resources written by experienced examiners and tailored to the new specification.

l A full ready-to-use teaching scheme that can be customised to give you total freedom and flexibility.

l Innovative Exam Café CD-ROM provided FREE in the back of every Student Book.

What can you expect from Heinemann’s OCR AS and A2 Physics?

3

Authors

Student Book and Exam Café CD-ROM

Exam Café CD-ROM

Teacher Support CD-ROM

Revision Guide

AS Roger Hackett Robert Hutchings

Graham Bone Colin Gregory David Styles David Webb

Roger Hackett David Webb Graham Bone David Styles

David Sang

A2 Roger Hackett Robert Hutchings

Graham Bone Colin Gregory David Styles David Webb

Roger Hackett David Webb Graham Bone

David Sang

Our authors have been specially selected because of their expertise and experience as examiners and practising teachers, and are dedicated to providing you with a set of resources that meet your needs in the classroom.

77

76

Work andenergy

P = Wt

W = Fx cos U

Joule

Elastic limit

Efficiency =Useful output energy

Total input energy 100

mgh = mv 212

Gravitational potential energy = mgh

Kinetic energy = mv 212

Chemical energyElectrical potential energyElectromagnetic wave energyGravitational potential energyInternal energy

Nuclear energySound energyKinetic energy

Forms ofenergy

Transfer andtransform

Create and destroy

Sankey diagrams

Conservation ofenergy

Ductile Polymeric Brittle

Stre

ss

Strain

Stre

ss

Strain

Stre

ss

Strain

Strain = ExtensionLength

Stress = FA

Young modulus = StressStrain

W = F Distance moved in the direction of the force

1W = 1Js – 1

Watt

Power

Forc

e

Extension

Gradient = k

Area under line =Work done

W = Fx12

Work done

F = kx

Hooke’s law

ElasticPlasticMaterials

Efficiency

1.3 Work and energy summary

1 The table shows how the braking distance (see spread1.2.8) for a car of mass 800 kg varies with the initial speedof the car when a constant braking force F is applied.Speed (m s–1) 0 10 20 30 40Distance (m) 0 6 24 x 96

(a) Calculate the kinetic energy in J of the car when it istravelling at 20 m s–1. (b) How much work in J is done by the braking force tobring the car to rest from 20 m s–1? (c) Calculate the value of the braking force in N. (d) Calculate the braking distance x in m of the car froman initial speed of 30 m s–1.

(e) Write down a general equation which will allow you tocalculate the braking distance of the car from any initialspeed. (f) One simplistic method of measuring the severity of acar crash is by the amount of kinetic energy whichmust be dissipated. Using this method, determinewhether a car hitting a wall at 20 m s–1 is a worse crashthan two identical cars each travelling at 10 m s–1 inopposite directions making a head-on collision.

2 A model carruns on anarrow flexibletrack. It hasbeen formedinto a verticalcircular loopas shown inFigure 1.

Assume that there are no frictional forces. The car is notdriven by any motive force. It just freewheels. (a) The car approaches the loop at speed u. Explain whythe speed v at C must be less than u. (b) Using the law of conservation of energy show that uand v are related by the equation u2 = v2 + 4gr where g is the acceleration due to gravity. (c) There is a minimum value of v necessary for the car toreach C. Unless u is greater or equal to

____5grthe carwill fall off the track before reaching C. Write down anexpression for the minimum value of v. (d) For a loop of radius 20 cm, find the minimum height towhich the end of the track, beyond the left-hand sideof the diagram, must be raised for the car to completethe circle, when released from rest at the end of thetrack.

(e) The boy playing with this toy only has enough track toraise the end 40 cm above the floor. Calculate theminimum speed at which he must release the car if it isto perform the stunt.

3 A student takes a single measurement to measure theYoung modulus of steel. He stretches a fine wire of length1.5 m and cross-sectional area 5.2 10–8 m2 using a forceof 20 N. He measures the extension to be 2.8 mm. (a) Calculate (i) the stress in the wire, (ii) the strain causedand (iii) the value of the Young modulus of steel fromthese data. (b) A very tall building requires a series of lifts to reach allfloors of the building. The reasons for this arrangementinclude convenience and logistics as well as physics.The following calculation suggests one of the physicalproblems. The cables of length 70 m supporting a liftconsist of two steel ropes each of 100 strands giving atotal cross-sectional area of 1.0 10–4 m2. Consider afull lift to carry 8 passengers of average mass 75 kg.Calculate by how much an empty lift movesdownwards when it is entered by 8 passengers.

4 The gravitational force between the Earth and the Moon is2.0 1020 N. To try to visualise the magnitude of this forceimagine that the force holding the Earth and Moon togetheris provided instead by a steel cable. (a) Calculate the minimum diameter of such a cable. Usethe ultimate tensile stress of steel, 1.5 109 N m–2. (b) Your answer to (a) should be about 400 km. Show thatthe mass of this cable would be greater than the massof the Moon which is 7 1022 kg. The distancebetween the Earth and the Moon is 4 108 m and thedensity of steel is 7.9 103 kg m–3.5 The ultimate tensile stress is not the value looked at byengineers when considering safety in their design of abuilding or aircraft or whatever. It is the yield stress atwhich permanent or plastic deformation starts. The basicrule is that the maximum stresses in the design should beno more than one quarter of the yield stress. (a) The yield stress of steel is 3 108 N m–2 and theultimate breaking stress is about 109 N m–2. How muchsmaller is the maximum stress used in design than theultimate breaking stress for steel? (b) A tug boat assists a tanker to dock at a quay using asteel cable in which the maximum safe tension is6 105 N. Calculate the diameter of this steel cable. (c) Explain why it is better to use a long cable to tow aboat or a car, rather than a short one. Hint: Firstconsider the extension and increase in strain whenthere are sudden changes in motion, e.g. a jerk. (d) Calculate the strain energy stored in 100 m of cable atthe maximum safe tension. Take the Young modulus ofsteel to be 2.1 1011 N m–2.

Practice questionsModule 3

Work and energyPractice questions

C

v

B

u

A

u

r

Figure 1

934 physics.U1 M3.indd 76-77

10/3/08 11:34:01 am

Student Books

Exclusively endorsed by OCR, these Student Books offer accessible and engaging material to help students understand the underlying principles of science. Careful explanations of key equations, plenty of worked examples, practice questions and exam-style questions all ensure that students have plenty of opportunities to improve their skills.

18

19

1.1 8 Constant acceleration equations

By the end of this spread, you should be able to . . .

✱Derive and use equations of motion for constant acceleration.

IntroductionMotion with constant acceleration is a process we encounter frequently. Drop an object

on to the floor, and you will see it falling with constant acceleration. Often it is possible to

make the assumption that, at least over a short time, the acceleration remains constant.

To deal with the calculations required to interpret such events, it is useful to be able to

work from an algebraic equation.

Equations of motionWhen dealing with constant acceleration over a given time, we always use the following

standard algebraic symbols, shown in Table 1

Symbol QuantityAlternative quantity SI unit

s distanced moved displacement metre

t time interval

second

a acceleration

m s–2

u speed at the start velocity at the start m s–1

v speed at the end velocity at the end m s–1

Table 1 Standard algebraic symbols for constant acceleration

Note: If you are not concerned with the direction of the motion you will be using the

scalar quantities distance and speed. If you need to use a direction, then displacement

and velocity are required. However, since we normally deal with motion in a straight line,

it makes no difference in practice whether the quantities are vectors or scalars.

The sketch graph Figure 1 shows the motion we are dealing with, using the symbols

given in Table 1. The distance which the moving object has travelled is the area under

the graph, and is shaded pink. There are several ways in which this area can be

calculated, so various other details have been added to the graph. These are:

• The area of the rectangle is ut.

• The height of the triangle is at.

• The area of the triangle is ½t × at =½at2.

This gives the distance as ut + ½at2 or vt – ½at2.

One other useful equation is that the distance is equal to the average speed time.

In algebraic symbols: s = u + v_____2 t

Like the equation defining acceleration, you may notice that each of these equations

misses out one of the five symbols we started with. The only symbol that is not missing

from all of them is the time t. There are, however, many practical examples where you

require a relationship between u, v, a and s. It is not too difficult to eliminate t from any

two of the equations so far deduced. Try Question 1 below.

Summary of the equations of motion for constant acceleration

v = u + at term not included: s

s = u + v_____2 t term not included: a

s = ut + ½at2 term not included: v

s = vt – ½at2 term not included: u

v2 = u2 + 2as term not included: t

Module 1Motion

Constant acceleration equations

Examiner tip

Algebra is simply a set of instructions

that allow us to solve calculations by

defining quantities or magnitudes, and

expressing the relationship between

them. The equations of motion

described below provide an ideal

opportunity, at the start of your AS

course, to improve your algebraic skills.

The sooner you become familiar with

algebra the better, since:

• you will often need to take an

algebraic approach to solving

problems

• using algebra will speed up your

work• in exams, the percentage of available

marks gained by students is much

higher for numerical questions than

for descriptive ones.

Figure 1 Velocity–time graph

Velo

city

vv u at

u

t0

0

Area ut

Area at2

Time

12

Examiner tip

You will find it very useful to be able to

quote these equations from memory.

The first is a basic definition, and your

specification requires you to know the

second, third and fifth anyway.

Questions1 Starting with the equations v = u + at and s = u

+ v_____2 t show that:

v2 = u2 + 2as

2 A lorry increases its speed at a steady rate from 22 m s–1 to 29 m s–1 in 70 s.

(a) What is its acceleration?

(b) How far does it travel while accelerating?

3 An object in free fall has a constant acceleration of 9.81 m s–2. It goes past two light

gates during its fall, and the time interval between passing the first and the second is

0.269 s. The distance between the light gates is 1.36 m.

(a) What is the object’s speed as it passes the first light gate?

(b) At what height above the first light gate was it dropped?

4 An electron in a TV tube starts from rest and accelerates uniformly until it reaches the

screen, a distance of 0.40 m, with a speed of 5.6 107 m s–1.

(a) What is the acceleration of the electron?

(b) How long does the electron take to travel through the tube?

5 At the scene of a traffic accident the police notice that a car has taken 28.0 m to stop.

The assumed acceleration of the car is –8.0 m s–2. What value does this assumption

give for the speed of the car just before the accident?

6 Explain why you could not use the equations of motion for a firework rocket’s ascent.

Figure 3 The firework rocket uses up its fuel as it rises

Figure 2 Acceleration of an object

in free fall, captured with multiple-

flash photography


10/3/08 11:32:14 am

4


AS



We listen to teachers’ needs...

Sample pages from OCR AS Physics Student Book.

Text is structured in line with the new OCR specification, by Unit and Module.

Learning objectives are taken from the specification to highlight what students need to know and understand.

Questions students should be able to answer after studying each spread.Examination tips.

77

76

Work andenergy

P = Wt

W = Fx cos U

Joule

Elastic limit

Efficiency =Useful output energy

Total input energy 100

mgh = mv 212

Gravitational potential energy = mgh

Kinetic energy = mv 212

Chemical energyElectrical potential energyElectromagnetic wave energyGravitational potential energyInternal energy

Nuclear energySound energyKinetic energy

Forms ofenergy

Transfer andtransform

Create and destroy

Sankey diagrams

Conservation ofenergy

Ductile Polymeric Brittle

Stre

ss

Strain

Stre

ss

Strain

Stre

ss

Strain

Strain = ExtensionLength

Stress = FA

Young modulus = StressStrain

W = F Distance moved in the direction of the force

1W = 1Js – 1

Watt

Power

Forc

e

Extension

Gradient = k

Area under line =Work done

W = Fx12

Work done

F = kx

Hooke’s law

ElasticPlasticMaterials

Efficiency

1.3 Work and energy summary

1 The table shows how the braking distance (see spread1.2.8) for a car of mass 800 kg varies with the initial speedof the car when a constant braking force F is applied.Speed (m s–1) 0 10 20 30 40Distance (m) 0 6 24 x 96

(a) Calculate the kinetic energy in J of the car when it istravelling at 20 m s–1. (b) How much work in J is done by the braking force tobring the car to rest from 20 m s–1? (c) Calculate the value of the braking force in N. (d) Calculate the braking distance x in m of the car froman initial speed of 30 m s–1.

(e) Write down a general equation which will allow you tocalculate the braking distance of the car from any initialspeed. (f) One simplistic method of measuring the severity of acar crash is by the amount of kinetic energy whichmust be dissipated. Using this method, determinewhether a car hitting a wall at 20 m s–1 is a worse crashthan two identical cars each travelling at 10 m s–1 inopposite directions making a head-on collision.

2 A model carruns on anarrow flexibletrack. It hasbeen formedinto a verticalcircular loopas shown inFigure 1.

Assume that there are no frictional forces. The car is notdriven by any motive force. It just freewheels. (a) The car approaches the loop at speed u. Explain whythe speed v at C must be less than u. (b) Using the law of conservation of energy show that uand v are related by the equation u2 = v2 + 4gr where g is the acceleration due to gravity. (c) There is a minimum value of v necessary for the car toreach C. Unless u is greater or equal to

____5grthe carwill fall off the track before reaching C. Write down anexpression for the minimum value of v. (d) For a loop of radius 20 cm, find the minimum height towhich the end of the track, beyond the left-hand sideof the diagram, must be raised for the car to completethe circle, when released from rest at the end of thetrack.

(e) The boy playing with this toy only has enough track toraise the end 40 cm above the floor. Calculate theminimum speed at which he must release the car if it isto perform the stunt.

3 A student takes a single measurement to measure theYoung modulus of steel. He stretches a fine wire of length1.5 m and cross-sectional area 5.2 10–8 m2 using a forceof 20 N. He measures the extension to be 2.8 mm. (a) Calculate (i) the stress in the wire, (ii) the strain causedand (iii) the value of the Young modulus of steel fromthese data. (b) A very tall building requires a series of lifts to reach allfloors of the building. The reasons for this arrangementinclude convenience and logistics as well as physics.The following calculation suggests one of the physicalproblems. The cables of length 70 m supporting a liftconsist of two steel ropes each of 100 strands giving atotal cross-sectional area of 1.0 10–4 m2. Consider afull lift to carry 8 passengers of average mass 75 kg.Calculate by how much an empty lift movesdownwards when it is entered by 8 passengers.

4 The gravitational force between the Earth and the Moon is2.0 1020 N. To try to visualise the magnitude of this forceimagine that the force holding the Earth and Moon togetheris provided instead by a steel cable. (a) Calculate the minimum diameter of such a cable. Usethe ultimate tensile stress of steel, 1.5 109 N m–2. (b) Your answer to (a) should be about 400 km. Show thatthe mass of this cable would be greater than the massof the Moon which is 7 1022 kg. The distancebetween the Earth and the Moon is 4 108 m and thedensity of steel is 7.9 103 kg m–3.5 The ultimate tensile stress is not the value looked at byengineers when considering safety in their design of abuilding or aircraft or whatever. It is the yield stress atwhich permanent or plastic deformation starts. The basicrule is that the maximum stresses in the design should beno more than one quarter of the yield stress. (a) The yield stress of steel is 3 108 N m–2 and theultimate breaking stress is about 109 N m–2. How muchsmaller is the maximum stress used in design than theultimate breaking stress for steel? (b) A tug boat assists a tanker to dock at a quay using asteel cable in which the maximum safe tension is6 105 N. Calculate the diameter of this steel cable. (c) Explain why it is better to use a long cable to tow aboat or a car, rather than a short one. Hint: Firstconsider the extension and increase in strain whenthere are sudden changes in motion, e.g. a jerk. (d) Calculate the strain energy stored in 100 m of cable atthe maximum safe tension. Take the Young modulus ofsteel to be 2.1 1011 N m–2.

Practice questionsModule 3

Work and energyPractice questions

C

v

B

u

A

u

r

Figure 1


10/3/08 11:34:01 am


Exam practice questions provided at the end of each module. Answers are in the back of the book.

18

19

1.1 8 Constant acceleration equations

By the end of this spread, you should be able to . . .

✱Derive and use equations of motion for constant acceleration.

IntroductionMotion with constant acceleration is a process we encounter frequently. Drop an object

on to the floor, and you will see it falling with constant acceleration. Often it is possible to

make the assumption that, at least over a short time, the acceleration remains constant.

To deal with the calculations required to interpret such events, it is useful to be able to

work from an algebraic equation.

Equations of motionWhen dealing with constant acceleration over a given time, we always use the following

standard algebraic symbols, shown in Table 1

Symbol QuantityAlternative quantity SI unit

s distanced moved displacement metre

t time interval

second

a acceleration

m s–2

u speed at the start velocity at the start m s–1

v speed at the end velocity at the end m s–1

Table 1 Standard algebraic symbols for constant acceleration

Note: If you are not concerned with the direction of the motion you will be using the

scalar quantities distance and speed. If you need to use a direction, then displacement

and velocity are required. However, since we normally deal with motion in a straight line,

it makes no difference in practice whether the quantities are vectors or scalars.

The sketch graph Figure 1 shows the motion we are dealing with, using the symbols

given in Table 1. The distance which the moving object has travelled is the area under

the graph, and is shaded pink. There are several ways in which this area can be

calculated, so various other details have been added to the graph. These are:

• The area of the rectangle is ut.

• The height of the triangle is at.

• The area of the triangle is ½t × at =½at2.

This gives the distance as ut + ½at2 or vt – ½at2.

One other useful equation is that the distance is equal to the average speed time.

In algebraic symbols: s = u + v_____2 t

Like the equation defining acceleration, you may notice that each of these equations

misses out one of the five symbols we started with. The only symbol that is not missing

from all of them is the time t. There are, however, many practical examples where you

require a relationship between u, v, a and s. It is not too difficult to eliminate t from any

two of the equations so far deduced. Try Question 1 below.

Summary of the equations of motion for constant acceleration

v = u + at term not included: s

s = u + v_____2 t term not included: a

s = ut + ½at2 term not included: v

s = vt – ½at2 term not included: u

v2 = u2 + 2as term not included: t

Module 1Motion

Constant acceleration equations

Examiner tip

Algebra is simply a set of instructions

that allow us to solve calculations by

defining quantities or magnitudes, and

expressing the relationship between

them. The equations of motion

described below provide an ideal

opportunity, at the start of your AS

course, to improve your algebraic skills.

The sooner you become familiar with

algebra the better, since:

• you will often need to take an

algebraic approach to solving

problems

• using algebra will speed up your

work• in exams, the percentage of available

marks gained by students is much

higher for numerical questions than

for descriptive ones.

Figure 1 Velocity–time graph

Velo

city

vv u at

u

t0

0

Area ut

Area at2

Time

12

Examiner tip

You will find it very useful to be able to

quote these equations from memory.

The first is a basic definition, and your

specification requires you to know the

second, third and fifth anyway.

Questions1 Starting with the equations v = u + at and s = u

+ v_____2 t show that:

v2 = u2 + 2as

2 A lorry increases its speed at a steady rate from 22 m s–1 to 29 m s–1 in 70 s.

(a) What is its acceleration?

(b) How far does it travel while accelerating?

3 An object in free fall has a constant acceleration of 9.81 m s–2. It goes past two light

gates during its fall, and the time interval between passing the first and the second is

0.269 s. The distance between the light gates is 1.36 m.

(a) What is the object’s speed as it passes the first light gate?

(b) At what height above the first light gate was it dropped?

4 An electron in a TV tube starts from rest and accelerates uniformly until it reaches the

screen, a distance of 0.40 m, with a speed of 5.6 107 m s–1.

(a) What is the acceleration of the electron?

(b) How long does the electron take to travel through the tube?

5 At the scene of a traffic accident the police notice that a car has taken 28.0 m to stop.

The assumed acceleration of the car is –8.0 m s–2. What value does this assumption

give for the speed of the car just before the accident?

6 Explain why you could not use the equations of motion for a firework rocket’s ascent.

Figure 3 The firework rocket uses up its fuel as it rises

Figure 2 Acceleration of an object

in free fall, captured with multiple-

flash photography


10/3/08 11:32:14 am

5


A2



Don’t forGet

our A2 resources coming in the Autumn term!

End-of-module summary pages help students link together all the topics within each module.

In our unique Exam Café, students will find lots of support to help them prepare for their exams. They can Relax and Prepare with handy revision advice, Refresh their memories by testing their understanding and Get That Result through practising exam-style questions, accompanied by plenty of hints and tips.

An Exam Café CD-ROM is included FREE in the back of every Student Book.

6

Sample screen from OCR AS Physics Exam Café CD-ROM.

Study and revision skills to support students making the transition from GCSE to A Level.

Links to how students can use New Scientist to reinforce their learning.

Sample answers to each module with tips on how to improve, examiner tips and advice to students on practical skills.

Questions to test understanding. Also vocabulary tests, a glossary, and revision flashcards.

“Three stages is a great idea – something you can work through.”

Sophie Wilson, Student, Headington School.

“A really great and original way of encouraging students to revise and study for exams.”Marie-Lise Tassoni, Student, Bexhill College.

“I think it’s an extremely positive idea to make students see their potential.”Sophie East, Student, Oxford.

7

Sample screen from OCR AS Physics Exam Café CD-ROM.

Links directly to the module and specification.

What do students think of Exam Café?

Three stages allow students to see questions against student answers and examiner feedback.

Student activities focus on improving subject knowledge.

Teacher Support CD-ROM

The AS and A2 Teacher Support CD-ROMs help you plan and deliver the new specification with confidence. Each provide you with:

l weekly teaching plans and guidance sheets to help you save time

l customisable student practical sheets with accompanying teacher and technician notes

l a media bank of all the diagrams and learning objectives in the Student Book, all ready to use in PowerPoint format.

8

Sample lesson plan from OCR AS Physics Teacher Support CD-ROM.


AS



9

Sample screen from OCR AS Physics Teacher Support CD-ROM.


A2



Revision Guides

l Clearly written and well designed to aid revision.

l Written by experienced examiners and tailored to the new specification.

l Packed with examiner tips.

l Targeted at ensuring understanding with quick-check questions on each topic.

10

Sample pages from OCR Revise AS Physics

10

11

Using vectors

1UNIT

Mod

ule

1M

odule 1

Using vectors

Vector quantities, such as displacement, velocity and acceleration, have both

magnitude and direction. They cannot simply be represented by a numerical value.

Instead, to include their direction, they can be represented by drawing vector diagrams.

In a vector diagram:

the length of a line represents the magnitude of the vector quantity

the direction of a line represents the direction of the vector quantity.

Vector diagrams are a type of scale drawing.

Adding two vectors

Draw a vector triangle as follows.

Choose a suitable scale. Draw a line (AB) to represent the first

vector quantity. Add an arrow to the line.

From the end of the first vector, draw a line (BC) to represent

the second. Add an arrow.

Return to the start of the first vector. Draw a line from this point to the end of

the second vector. This line (AC) represents the resultant of the two. Add two

arrowheads to show that this is the resultant of the other two vectors. This process is

called adding two vectors.

When the two vectors are perpendicular, the triangle is right-angled and the resultant is

the hypotenuse. Use Pythagoras’ theorem to calculate the resultant.

Resolving a vector

Sometimes it is useful to resolve (break down) a vector quantity

into two components at 90° to one another.

Imagine turning the vector V round to point in the direction of

interest. If you turn the vector through an angle , the component

of V in this direction is V cos .

A vector may be replaced by two perpendicular components

whose values are V cos and V sin . Notice that if these two

components are added (see page 10), the resultant is the original

vector.

The perpendicular components of a vector are independent of one

another. Changing one component has no effect on the other.

Hint

Reminder of

Pythagoras’ theorem:

for a right-angled

triangle, a2 + b2 = c2.

ac

b

Quick check 1✔

Hint

You can think of this as

‘Going from A to B to

C is the same as going

directly from A to C’.

Hint

It is always advisable

to check a calculation

like this by drawing a

scale diagram and

measuring both the

length of the resultant

and the angle.

B

C

A

Quick check 2✔

V

V

V

V cos u

V cos u

V cos u

V sin u

V sin u

Resolving

Two perpendicular

components

Adding perpendicular

components

u

u

u

WoRked exAMple

A car is travelling at 20 m s–1 at 30° W of N (see diagram). Calculate the components

of its velocity due N and due W.

20 m s 1 20 m s 1 20 m s 1N

W

Step 1Step 2

Step 3

3030

60

60

STep 1 Draw a diagram; mark the relevant angles.

STep 2 Calculate the component due N. (The angle here is 30°.)

Component due N = 20 m s–1 × cos 30° = 17.3 m s–1

STep 3 Calculate the component due W. (The angle here is 60°.)

Component due W = 20 m s–1 × cos 60° = 10.0 m s–1

Quick check 3✔

Hint

Note that we could

have calculated this

last result using sin 30°

rather than cos 60°.

Quick check 4✔

1 A plane flies NE for 200 km, then due E for 400 km. Draw a scale diagram of its journey

and use it to deduce its distance from the starting point.

2 A whale swims 1000 km due S and then 400 km due E. How far is it from its starting point,

and in what direction?

3 The Earth’s gravity makes objects fall with an acceleration of g = 9.81 m s–2. What will be the

acceleration of an object down a smooth slope, inclined at 45° to the horizontal?

4 A boy is speeding at 8 m s–1 down a water slide which is inclined at 35° to the horizontal.

Calculate the horizontal and vertical components of his velocity.

QUICk CHeCk QUeSTIoNS ?Hint

Calculate the

component of g at 45°

to the horizontal.

Key words

magnitude

directionresolvingcomponents

WoRked exAMple

A ship is sailing due north at 8 m s–1. A passenger walks across the deck at 4 m s–1, in

an easterly direction. What is her resultant velocity?

The diagram shows how the two velocities are added.

STep 1 Draw a vector to represent the velocity of the ship.

STep 2 Draw a vector to represent the passenger’s velocity across the deck.

STep 3 Draw the resultant vector.

STep 4 Measure or calculate the resultant. In this case, we

have a right-angled triangle, so we can calculate

the resultant using Pythagoras’ theorem.

v2 = (8 m s–1)2 + (4 m s –1)2 = (64 + 16) m2 s–2

= 80 m2 s–2

v = 80 m2 s–2 = 8.9 m s–1

We must also state the direction of the passenger’s

resultant velocity. We need to find the angle from the

diagram.

tan = opp____adj

= 4/8 = 0.5

= tan–1 0.5 = 26.6°

The passenger’s resultant velocity is thus 8.9 m s–1 at 26.6° east of north.

4 m s 1

4 m s 1

8 m s 1

8 m s 1or

v

v

u

u

Examiner tip

In an examination you

may sketch the

triangle and use

Pythagoras and

trigonometry or you

may use a scale

drawing and measure v

and . Choose a

suitable scale, e.g. 1 cm

= 1 m s–1. Make sure

you have a ruler and

protractor with you!

Hints and tips help students avoid common errors.

Provides students with lots of opportunities to see problems worked through and answers are provided step by step.Clearly linked to

the specifications.

Enables students to check their knowledge and understanding. Answers are provided at the back of the book.

iv

v

Introductionvi

UNIT 1 Mechanics (G481)Module 1 Motion 2

1 Physical quantities and units 4

2 Estimated physical quantities 6

3 Scalar and vector quantities 8

4 Vector calculations 10

5 Vector resolution 12

6 Definitions in kinematics 14

7 Graphs of motion 16

8 Constant acceleration equations 18

9 Free fall20

10 Measurement of g 22

Summary and practice questions 24

End-of-module examination questions 26

Module 2 Forces in action 28

1 Force and the newton 30

2 Motion with non-constant acceleration 32

3 Equilibrium34

4 Centre of gravity 36

5 Turning forces 38

Further questions A 40

6 Density42

7 Pressure44

8 Car stopping distances 46

9 Car safety48



Module 3 Work and energy 54

1 Work and the joule 56

2 The conservation of energy 58

3 Potential and kinetic energies 60

Further questions B 62

4 Power and the watt 64

5 Efficiency66

6 Deformation of materials 68

7 Hooke’s law70

8 The Young modulus 72

9 Categories of materials 74



UNIT 2 Electrons, waves and

photons (G482)Module 1 Electric current 80

1 Electric current and charge 82

2 Kirchhoff’s first law 84

3 Electron drift velocity 86



Module 2 Resistance 92

1 Electromotive force 94

2 Potential difference 96

3 Resistance and Ohm’s law 98

4 Resistance of circuit components 100

5 Resistivity102

6 The effect of temperature on resistivity 104

7 Electrical power 106

8 Domestic electrical supply 108

9 Charging for electrical energy 110



Module 3 DC circuits 116

1 Series circuits 118

2 Parallel circuits 120

3 Circuit analysis 1 122

4 Circuit analysis 2 124

5 The potential divider 126

Further questions C 128



Module 4 Waves 134

1 Wave motion 136

2 Wave terminology 138

3 Wave speed 140

4 Wave properties 142

Further questions D 144

5 Electromagnetic waves 146

6 Polarisation 148

7 Interference 150

8 The Young double-slit experiment 152

9 The diffraction grating 154

Further questions E 156

10 Stationary waves 158

11 Stationary wave experiments 160

12 Stationary longitudinal waves 162


End-of-module examination questions 1 166

End-of-module examination questions 2 168

Module 5 Quantum physics 170

1 The energy of a photon 172

2 The photoelectric effect 1 174

3 The photoelectric effect 2 176

4 Wave–particle duality 178

5 Energy levels in atoms 180

Further questions F 182

6 Spectra184



Appendix: Accuracy and significant

figures 190

Answers 192

Glossary 204

Index 210

Contents

Contents

934 physics.prelims.indd 4-5

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10

11

Using vectors

1UNIT

Mod

ule

1

Module 1

Using vectors

Vector quantities, such as displacement, velocity and acceleration, have both

magnitude and direction. They cannot simply be represented by a numerical value.

Instead, to include their direction, they can be represented by drawing vector diagrams.

In a vector diagram:

the length of a line represents the magnitude of the vector quantity

the direction of a line represents the direction of the vector quantity.

Vector diagrams are a type of scale drawing.

Adding two vectors

Draw a vector triangle as follows.

Choose a suitable scale. Draw a line (AB) to represent the first

vector quantity. Add an arrow to the line.

From the end of the first vector, draw a line (BC) to represent

the second. Add an arrow.

Return to the start of the first vector. Draw a line from this point to the end of

the second vector. This line (AC) represents the resultant of the two. Add two

arrowheads to show that this is the resultant of the other two vectors. This process is

called adding two vectors.

When the two vectors are perpendicular, the triangle is right-angled and the resultant is

the hypotenuse. Use Pythagoras’ theorem to calculate the resultant.

Resolving a vector

Sometimes it is useful to resolve (break down) a vector quantity

into two components at 90° to one another.

Imagine turning the vector V round to point in the direction of

interest. If you turn the vector through an angle , the component

of V in this direction is V cos .

A vector may be replaced by two perpendicular components

whose values are V cos and V sin . Notice that if these two

components are added (see page 10), the resultant is the original

vector.

The perpendicular components of a vector are independent of one

another. Changing one component has no effect on the other.

Hint

Reminder of

Pythagoras’ theorem:

for a right-angled

triangle, a2 + b2 = c2.

ac

b

Quick check 1✔

Hint

You can think of this as

‘Going from A to B to

C is the same as going

directly from A to C’.

Hint

It is always advisable

to check a calculation

like this by drawing a

scale diagram and

measuring both the

length of the resultant

and the angle.

B

C

A

Quick check 2✔

V

V

V

V cos u

V cos u

V cos u

V sin u

V sin u

Resolving

Two perpendicular

components

Adding perpendicular

components

u

u

u

WoRked exAMple

A car is travelling at 20 m s–1 at 30° W of N (see diagram). Calculate the components

of its velocity due N and due W.

20 m s 1 20 m s 1 20 m s 1N

W

Step 1Step 2

Step 3

3030

60

60

STep 1 Draw a diagram; mark the relevant angles.

STep 2 Calculate the component due N. (The angle here is 30°.)

Component due N = 20 m s–1 × cos 30° = 17.3 m s–1

STep 3 Calculate the component due W. (The angle here is 60°.)

Component due W = 20 m s–1 × cos 60° = 10.0 m s–1

Quick check 3✔

Hint

Note that we could

have calculated this

last result using sin 30°

rather than cos 60°.

Quick check 4✔

1 A plane flies NE for 200 km, then due E for 400 km. Draw a scale diagram of its journey

and use it to deduce its distance from the starting point.

2 A whale swims 1000 km due S and then 400 km due E. How far is it from its starting point,

and in what direction?

3 The Earth’s gravity makes objects fall with an acceleration of g = 9.81 m s–2. What will be the

acceleration of an object down a smooth slope, inclined at 45° to the horizontal?

4 A boy is speeding at 8 m s–1 down a water slide which is inclined at 35° to the horizontal.

Calculate the horizontal and vertical components of his velocity.

QUICk CHeCk QUeSTIoNS ?Hint

Calculate the

component of g at 45°

to the horizontal.

Key words

magnitude

directionresolvingcomponents

WoRked exAMple

A ship is sailing due north at 8 m s–1. A passenger walks across the deck at 4 m s–1, in

an easterly direction. What is her resultant velocity?

The diagram shows how the two velocities are added.

STep 1 Draw a vector to represent the velocity of the ship.

STep 2 Draw a vector to represent the passenger’s velocity across the deck.

STep 3 Draw the resultant vector.

STep 4 Measure or calculate the resultant. In this case, we

have a right-angled triangle, so we can calculate

the resultant using Pythagoras’ theorem.

v2 = (8 m s–1)2 + (4 m s –1)2 = (64 + 16) m2 s–2

= 80 m2 s–2

v = 80 m2 s–2 = 8.9 m s–1

We must also state the direction of the passenger’s

resultant velocity. We need to find the angle from the

diagram.

tan = opp____adj

= 4/8 = 0.5

= tan–1 0.5 = 26.6°

The passenger’s resultant velocity is thus 8.9 m s–1 at 26.6° east of north.

4 m s 1

4 m s 1

8 m s 1

8 m s 1or

v

v

u

u

Examiner tip

In an examination you

may sketch the

triangle and use

Pythagoras and

trigonometry or you

may use a scale

drawing and measure v

and . Choose a

suitable scale, e.g. 1 cm

= 1 m s–1. Make sure

you have a ruler and

protractor with you!


A2



Summary of contents

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Course Guide AS A2 OCR Physicsassets.pearsonschool.com/asset_mgr/versions/2012... · 1W = J s 1 att oer For ce Extension Gradient = k Area under line = W ork done W 1 Fx 2 W ork done