ERRATUM NOTICE

Advanced Extension Award Physics(H7651)

Tuesday 24 June, Afternoon

Notice to Invigilator

Before the start of the examination the following should be read to candidates.

1. Go to the Question Paper. (Pause)

2. Turn to page 15 of the Question Paper. (Pause)

3. Go to the second line of text below fi gure 5.2. (Pause)

4. The sentence reads “The corresponding tensions in the strings . . .”. (Pause)

5. This should read “The corresponding tensions in the wires . . .”. (Pause)

Repeat

1. Go to the Question Paper. (Pause)

2. Turn to page 15 of the Question Paper. (Pause)

3. Go to the second line of text below fi gure 5.2. (Pause)

4. The sentence reads “The corresponding tensions in the strings . . .”. (Pause)

5. This should read “The corresponding tensions in the wires . . .”. (Pause)

Please make this change in your question paper now.

This is the end of the announcement.

TIME

3 hours.

INSTRUCTIONS TO CANDIDATES

In the boxes on the answer booklet, write your centre number, candidate number and subject

title. Answer all seven questions.

INFORMATION FOR CANDIDATES

The total mark for this paper is 100.

The mark for each part of a question is shown in brackets and the total mark is given at the end

of the question.

The values of physical constants and relationships you may require are given on a separate

information leaflet which is an insert to the paper. Any additional data are given in the

appropriate question.

The approximate time which you should spend on any one question is given at the start of the

question.

You are reminded of the need to organise and present your answers clearly and logically and to

use specialist vocabulary where appropriate.

Materials required for examination – Answer booklet, Graph paper, Calculator

Items included with the question paper – Information Leaflet (insert)

4691

Advanced Extension Award2008

Physics

Advanced Extension Award

H7651

TUESDAY 24 JUNE, AFTERNOON

H76

51

ADVICE TO CANDIDATES

In calculations you are advised to show all the steps in your working, giving your answers at

each stage. Give final answers to a justifiable number of significant figures.

4691 2 [Turn over

1 (You are advised to spend about 30 minutes on this question.)

Read the passage and then answer the questions which follow.

Fermat’s principle of least time

(Adapted from The Feynman Lectures, by R B Feynman, R P Leighton and M Sands,

Addison-Wesley, Reading, Mass.)

The first way of thinking that clarified the law about the behaviour of light was 1 discovered by Fermat in about 1650, and it is called the principle of least time, or

Fermat’s principle. His idea is this: that out of all possible paths that it might take from

one point to another, light takes the path which requires the shortest time.

Originally, the statement that light travels along the path of shortest possible distance 5was made by Hero of Alexandria. This can be seen to hold true for understanding, for

example, the manner in which light is reflected off the surface of a mirror. It was this

that inspired Fermat to suggest to himself that perhaps refraction operated on a similar

basis. But in the case of refraction, light obviously does not use the path of shortest

distance, so Fermat tried the idea that it takes the shortest time. 10

In Fig. 1.1, our problem is to go from A to B in the shortest time.

Fig. 1.1

CE

A

B

XF

beach

water

4691 3 [Turn over

To illustrate that the best thing to do is not just go in a straight line, let us imagine

that someone has fallen out of a boat, and is screaming in the water at point B.

We are at point A on land, and we see the accident, and can run and can also swim.

But we can run faster than we can swim. What do we do? Do we go in a straight 15 line? By using a little more intelligence we would realise that it would be advantageous

to travel a little greater distance on land in order to decrease the distance in the

water, because we go much slower in the water. Let us try to show that the final solution

is the path ACB, and that this path takes the shortest time. If this is the path of least

time, then if we take any other, it will take longer. However, another path very close to 20 ACB will take almost the same time. So we consider a nearby point X on the shoreline,

and we calculate how long it would take to go from A to B by the two paths AXB and

ACB, and compare the new time with the old time. We want the difference, of course,

to be nearly zero if the distance XC is short. First look at the path AXB on land. If we

draw a perpendicular XE, we see that the path AX is shorter than the path AC by an

amount approximately equal to EC. Let us say we gain by not having to go that extra 25 distance. On the other hand, in the water, by drawing a corresponding perpendicular

CF, we see that we have to go an extra distance approximately equal to XF, and that is what

we lose. Or, in time, we gain the time it would have taken to go the distance EC, but we

lose the time it would have taken to go the distance XF. These times must be equal since, in

the first approximation, there is to be no change in time. 30

Of course we must realise that this is an illustration, not a proof, of Fermat’s principle.

For one thing it deals with someone running and swimming, not with light passing

from one medium to another. But, if we accept the principle and apply it to light,

we shall find it a useful method of arriving at the laws of reflection and refraction, and a

means of explaining other optical phenomena. 35

4691 4 [Turn over

(a) The passage states that the law of reflection relating the angles of incidence and

reflection at a plane mirror is consistent with both Hero’s principle of least distance

(lines 5–7) and Fermat’s principle of least time (lines 3–4).

Consider Fig. 1.2.

Fig. 1.2

A ray of light is to go from A to B by a path involving reflection at the plane mirror PQ.

ACB is a hypothetical path; ADB is the observed path. AO is the perpendicular from

A to the mirror. AO is extended to A′ so that AO = OA′. The straight line from A′ to B

meets the mirror at D.

Assume Hero’s principle and show that light follows the path ADB. Hence show that

the law of reflection follows. Explain why Fermat’s principle gives the same result as

Hero’s. [5]

(b) Explain the statement “But in the case of refraction, light obviously does not use the

path of shortest distance” (line 9). Illustrate your answer with a diagram. [2]

(c) Suppose that the speed at which the rescuer can swim is 1/n times the speed at which

he can run. Consider the logic described in lines 18–30 of the extract, and refer to

Fig. 1.1. ACB is to be the route of shortest time. Hence explain why

XC sin EXC = n XC sin XCF

Use this relation to show that the path of minimum time to get from A to B is

consistent with Snell’s law of refraction. [5]

A

A′

D

CO

B

QP

4691 5 [Turn over

(d) Fig. 1.3 shows the path that light sometimes takes in hot countries between an object

(here, a palm tree) and the eye of an observer a long way away.

Fig. 1.3 (not to scale)

Although the real path of the light is curved, our brains interpret light as having

travelled in a straight line from the source.

(i) Explain with the aid of a ray diagram how this mental interpretation may be

responsible for the observation of a mirage, suggesting the presence of a pool of

water at the tree. [2]

(ii) Suggest circumstances under which the path of light from the top of the tree might

follow such a curved path. Explain why the path is curved and why such a path is

consistent with Fermat’s principle. [4]

[Total 18 marks]

light path

4691 6 [Turn over

2 (You are advised to spend about 25 minutes on this question.)

In a simple kinetic model of an ideal gas, the molecules are assumed to move in random

directions, all with the same speed. However, in a more advanced model, the molecules

are thought of as moving with a spread of speeds. Fig. 2.1 shows part of this distribution

of speeds c for the molecules, each of mass m, in a sample of an ideal gas at a kelvin

temperature T. (A complete graph would require an extension of the c-axis towards

infinity.)

Fig. 2.1

The values of n on the vertical axis are proportional to the number of molecules moving at a

given speed c. The speeds c are shown on the horizontal axis.

0

1.0

2.0

3.0

0 105 2015 25 30 35 40

n

c/102 m s–1

4691 7 [Turn over

(a) (i) Copy Table 2.1. Read off from the graph in Fig. 2.1 the values of n corresponding to

the stated values of c, and enter them in your table.

[2]

Table 2.1

c/102 m s–1 n

0 0

5

10

15

20

25

30

35

40

(ii) Calculate the arithmetical mean speed of these molecules. Show your working

clearly, using the right-hand column for any processed data. [4]

(iii) Your answer to (a)(ii) is a good approximation to the arithmetical mean

speed c (or <c>) of all the molecules in the distribution. Suggest how an even better

approximation could be obtained. [1]

(iv) The root-mean-square speed crms of the molecules is given by the square root of the

arithmetic mean of the squares of their speeds. For this distribution, the value of

crms is 2130 m s–1. Find the ratio of your answer in (a)(ii) to this value of crms. [1]

(v) Deduce from Fig. 2.1 the value of cp, the most probable speed of the molecules in

the distribution. [1]

(vi) The median speed cm is the speed at which half of all the molecules have speeds

below cm, and half of them above. Describe how you would obtain an approximate

value for cm from Fig. 2.1. [1]

(vii) For the distribution in Fig. 2.1 the values of <c>, crms, cp and cm are all different.

This is because the distribution is asymmetrical. For a set of readings of the

count rate of a long-lived radioactive sample, the shape of the distribution is

symmetrical. For such a distribution, comment on the relative values of the mean

count rate, the root-mean-square count rate, the most probable count rate and the

median count rate. [2]

4691 8 [Turn over

(b) The values of c corresponding to maximum and minimum values of n in the distribution

of Fig. 2.1 are given by the solutions of the equation

Equation 2.1

In Equation 2.1, k is the Boltzmann constant.

Use Equation 2.1 or Fig. 2.1 to determine the value of c corresponding to a minimum

value of n in the distribution, together with the corresponding value of n. Use a solution

to Equation 2.1 to find also an expression for the value of c corresponding to the

maximum value of n in the distribution. [3]

[Total 15 marks]

ckT

mc kT12

222

0– e /mc⎛

⎝⎜⎞

⎠⎟=−

BLANK PAGE

(Questions continue overleaf)

4691 9 [Turn over

4691 10 [Turn over

3 (You are advised to spend about 25 minutes on this question.)

(a) This part of the question is about two quantities, conductivity and current density,

used in current electricity.

The conductivity σ of a conductor is the reciprocal of the resistivity.

The current density J in a conductor is the current per unit cross-sectional area of the

conductor.

A potential difference V is applied between the ends of a metal rod of length L and

cross-sectional area A, so as to produce a uniform electric field E between the ends of

the rod. The resistivity of the material of the rod is ρ. The current in the rod is I.

Starting with expressions for the resistance of the rod, the current density and the

electric field between the ends of the rod, use the defining equation for resistivity

to obtain an expression for the electrical conductivity σ of the metal in terms of the

uniform applied electric field E and the current density J. [3]

(b) In one model of electrical conduction in a metal, the free electrons in it are supposed

to move like the molecules of a kinetic-theory gas. Each electron is accelerated by the

applied electric field E until it collides with a lattice ion. In the collision the electron’s

velocity is reduced to zero, and the process of acceleration and collision is repeated. The

free electrons thus acquire an average drift velocity v towards the positive end of the

metal rod.

(i) By first considering the acceleration of a free electron by the electric field, obtain

an expression for the drift velocity v in terms of E, e, me and t, where t is the mean

time between electron-ion collisions and e and me are the electron charge and mass

respectively. [3]

(ii) Another expression for the drift velocity is

v = J

ne

where J is the current density in the metal rod, n is the number of free electrons

in the metal per unit volume of the metal and e is the electron charge. Use this

expression, and your answers to (a) and (b)(i), to find an expression for the mean

time t between collisions of the electrons with the ions in terms of e, me, σ and n.

[3]

4691 11 [Turn over

(iii) Besides attaining their drift velocity, the free electrons in this model also move with

random thermal motion, just like the molecules of an ideal gas.

State an expression for the root-mean-square speed of the electrons at temperature T

in terms of the Boltzmann constant k and the electron mass me.

Hence show that the mean distance λ travelled by the electrons between collisions

with lattice ions is given by

λ σ= 23

2nem kTe .

[3]

(iv) Copper contains 8.3 × 1028 free electrons per cubic metre. At room temperature, the

electrical conductivity of copper is 5.6 × 107 Ω–1 m–1.

Use the relation in (b)(iii) to deduce the mean distance travelled by the free

electrons in copper between collisions with lattice ions.

State a typical value for the interionic distance in a metal lattice. Compare your

answer with this value, and discuss whether, on the basis of the comparison, the

model is credible. [4]

[Total 16 marks]

4691 12 [Turn over

4 (You are advised to spend about 20 minutes on this question.)

A student is investigating the conservation of energy applied to radioactive decay.

(a) The student starts with alpha decay. He knows that alpha particles are helium nuclei,

each consisting of two protons and two neutrons. He wonders why these nucleons are

always emitted as a single alpha particle, rather than as four separate particles.

Explain why this is the case. Illustrate your answer with reference to the alpha decay of 232U to 228Th. Make use of the following data. [6]

Masses of neutral atoms 232U 232.03713 u

228Th 228.02872 u

Mass of alpha particle 4.00260 u

Proton mass 1.00728 u

Neutron mass 1.00867 u

(b) The student then reads about beta+ decay. Beta+ decay is decay by the emission of a

positron, a particle with mass me and charge +e.

The student looks up data for the beta+ decay of Nitrogen-13 to Carbon-13. He uses the

following values from a textbook.

Masses of neutral atoms 13N 13.005738 u 7

13C 13.003355 u 6

Electron or positron mass me 0.000549 u

He calculates the difference Δm in mass between the nitrogen atom and the carbon atom

and a positron. Using E = Δmc2, he calculates the total energy released in the beta+

decay of one Nitrogen-13 nucleus as 2.74 × 10–13 J. He is surprised to find that the

textbook gives a different value for the energy released.

Explain the physics error the student has made. Calculate the textbook value. [6]

[Total 12 marks]

BLANK PAGE

(Questions continue overleaf)

4691 13 [Turn over

4691 14 [Turn over

5 (You are advised to spend about 25 minutes on this question.)

Imagine that the first international space station, with artificially engineered ‘gravity’,

has been built. It is in the shape of a hollow cylinder of diameter 30 m. The floors of the

astronauts’ bedrooms are just inside the outer surface of the cylinder. The artificial gravity

has been produced by spinning the entire space station about its cylindrical axis, so that an

acceleration of 9.81 m s–2 is produced at the floor of an astronaut’s room. The arrangement

is sketched in Fig. 5.1.

Fig. 5.1

(a) (i) Calculate the angular velocity of rotation of the space station. [2]

(ii) Obtain a general expression, in terms of the radius R of the space station and its

angular velocity of rotation ω, for the acceleration a due to the artificial gravity at a

height h above the floor of an astronaut’s room. [1]

(b) An astronaut wakes up in his bedroom in a panic. He cannot remember whether he is

on the space station, or whether he is in the exact replica of a bedroom that was set up

in a simulator on Earth. He is desperate to know where he is, but it is the middle of the

night, there are no windows and the door is locked. On his bedside table is a slinky (a

long metal spring of very low spring constant but of finite mass), and a ruler. He decides

that by hanging the spring from his outstretched arm, and looking at the manner in

which the tension of the spring varies with distance from his hand, he will be able to tell

where he is.

Model the suspended spring as a series of five thin discs, each of mass m, connected

together by five identical massless, inextensible wires, each of length l, as shown in

Fig. 5.2.

30 m

axis

astronaut

floor of astronaut’sroom

ω

4691 15 [Turn over

Fig. 5.2

The wires are numbered from the one at the top (the one hanging from the astronaut’s hand)

as 5, 4, 3, 2, 1. The corresponding tensions in the strings are T5, T4, …, T1. The wires are

held so that the disc at the bottom is just above the floor of the room.

(i) Suppose that the astronaut is in the room in the simulator on Earth. Deduce, in terms

of the acceleration of free fall on Earth g and the mass m of each disc, the tensions T1,

T2 … T5 in each of the five wires. On graph paper, draw a graph of the tension T as a

function of height h above the floor. Your h-axis should extend from h = 0 to h = 5l. Add

suitable graduations and values to the T-axis. Label your graph E. [3]

(ii) Now suppose that the astronaut is on the space station rather than on Earth. Deduce, in

terms of l, m, R and ω, expressions for each of the tensions T1, T2, … T5 in the wires.

Remembering that the station has been designed to produce an artificial gravity with the

acceleration of free fall at the floor of a room equal to g, indicate on your graph in (b)(i) how, if at all, the tensions will be different from the case on Earth. Label your new

graph S. [5]

(iii) Having carried out this modelling exercise, the astronaut holds the real spring from his

hand.

(1) Sketch the general appearance of the coils of the spring. [1]

(2) Suggest how the astronaut might obtain an indication of how the tension in the

spring varies. [1]

(3) From his observations, how could he tell where he was? [1]

[Total 14 marks]

hand

wire

disc

5T5

T5

T4

T4

T3

T3

T2

T2

T1

T1

4

3

2

1floor

4l

5l

3l

2l

l

0

h

4691 16 [Turn over

6 (You are advised to spend about 20 minutes on this question.)

A metal rod rests on a pair of long, horizontal, parallel metal rails along which it can slide

without friction. The whole arrangement is in a region of uniform vertical magnetic field

perpendicular to, and into, the plane of the page. The rails are connected through a switch

to a fixed resistor and a cell of constant e.m.f. and negligible internal resistance. The

resistances of the rod and of the rails are negligible. The circuit is shown in Fig. 6.1.

Fig. 6.1 (plan view)

(a) Describe and explain what happens after the switch is connected, and during the

subsequent motion of the rod. [8]

(b) The following is a list of the values of relevant quantities in this arrangement.

Mass of rod = 25 g

Separation of rails = 45 cm

Flux density of magnetic field = 0.82 T, vertically downwards

E.m.f. of cell = 1.5 V

Resistance of fixed resistor = 2.2 Ω

Use these data to calculate the values of any quantities referred to in (a). [4]

[Total 12 marks]

rod long rails

7 (You are advised to spend about 25 minutes on this question.)

The year is 2028, and society has changed. The economy of the United Kingdom is now

based almost entirely on tourism and the production of television programmes, in particular

game shows and virtual reality series. Consequently, the Secondary school curriculum

concentrates on Hospitality Studies and Media Studies. You are now in your late 30s, and

are a successful member of the management team of a large hotel group. Although you

studied Physics up to AEA standard at school back in 2008 and followed this with a First

Class Honours degree in Physics, you were among the last to follow such a course. Science

has not been taught beyond Primary school level for a decade. Only a few universities have

Science Faculties, and these are very small. Physics is treated as being irrelevant, as are the

great pioneers such as Newton and Einstein.

You are very disturbed at the way things are going. Your daughter is about to go to

Secondary school. Write a letter to the Board of Governors of your daughter’s prospective

school emphasising the importance of continuing the study of Physics, at least in the early

years of Secondary school. Illustrate your letter by outlining Newton’s laws of motion and

pointing out practical applications of the laws which your daughter may have experienced at

her Primary school, perhaps without realising how they related to the laws. Remember that it

is likely that no member of the Board has your own background in Physics; most will have

stopped studying Science at the end of their Primary school careers.

In addition to the content of your answer, you will be assessed on the quality of your written communication.

[Total 13 marks]

4691 17

THIS IS THE END OF THE QUESTION PAPER

530-092-1

4691.02 1

ADVANCED EXTENSION AWARD

PHYSICS

Information Leaflet

The following may be of use in answering some of the questions.

Values of constantsspeed of light in free space c = 3.00 × 108 m s–1

permeability of free space μ0 = 4π × 10– 7 H m–1

permittivity of free space ε0 = 8.85 × 10–12 F m–1

1 —— = 9 × 109 F–1 m ( 4πε

0 )

elementary charge e = 1.60 × 10–19 C

the Planck constant h = 6.63 × 10–34 J s

unified atomic mass unit 1 u = 1.66 × 10–27 kg

electron mass me = 9.11 × 10–31 kg

proton mass mp = 1.673 × 10–27 kg

neutron mass mn = 1.675 × 10–27 kg

molar gas constant R = 8.31 J K–1 mol–1

the Avogadro constant NA

= 6.02 × 1023 mol–1

the Boltzmann constant k = 1.38 × 10–23 J K–1

gravitational constant G = 6.67 × 10–11 N m2 kg–2

acceleration of free fall on

the Earth’s surface g = 9.81 m s–2

normal atmospheric pressure patm

= 1.01 × 105 Pa

electron volt 1 eV = 1.60 × 10–19 J

H7651

4691.02 2 [Turn over

Formulae

The following equations may be useful in answering some of the questions in the examination:

Mechanics

equations for uniformly accelerated motion v = u + at

s = 1–2(u + v)t

s = ut + 1–2 at2

v2 = u2 + 2as

Momentum and Energy Δ(mv)force = rate of change of momentum F = ——– Δt

power P = Fv

Kinetic Theory

kinetic theory of gases pV = 1–3 Nmc2

––

average kinetic 3RTenergy of a molecule 1–

2 mc2

–– = 3–

2 kT = ——

2NA

Electricity

terminal potential difference Vload

= � – Ir

discharge of capacitor Q = Q0e–t/RC

time constant τ = RC

Atomic and Nuclear physics ΔNradioactive decay ––– = –λN Δt

N = N0e–λt

ln 2half-life T1–

2 = —–

λ

4691.02 3

Energy

mass-energy relationship E = mc2

Quantum Physics

energy-frequency relationship for

photons E = hf

hde Broglie equation λ = – p

Waves and Oscillations

two-slit interference λ = ay/d or λ = xs/D

simple harmonic motion a = –(2πf)2x

x = A cos 2πft x = A sin 2πft

Fields Fgravitational fields g = –– m

GM g = ––– r2

Felectric fields E = –– q

1 q E = —— –– 4πε

0 r2

V E = –– d

Magnetic effect of currents

force on a current-carrying conductor F = BIl

force on a moving charge F = Bqv

magnetic flux Φ = BA

d(NΦ)induced e.m.f. � = – ——— dt

4691.02 4 [Turn over

Mathematical equations

areas and volumes area of circle = πr2

surface area of cylinder = 2πrh + 2πr2

volume of cylinder = πr2h

surface area of sphere = 4πr2

volume of sphere = 4–3 πr3

radians arc = rθ

sin θ ≈ tan θ ≈ θ and cos θ ≈ 1 for small θ

logarithms ln(xn) = n ln x ln(ekx) = kx