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9702 w24 QP 41 Merged

This document is the Cambridge International AS & A Level Physics Paper 4 for October/November 2024, consisting of structured questions. It includes instructions for answering the exam, relevant physical constants, and various physics formulas. The paper covers topics such as Newton's law of gravitation, specific heat capacity, the Avogadro constant, and simple harmonic motion.

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0% found this document useful (0 votes)

79 views114 pages

9702 w24 QP 41 Merged

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Cambridge International AS & A Level

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PHYSICS 9702/41
Paper 4 A Level Structured Questions October/November 2024

2 hours

You must answer on the question paper.

No additional materials are needed.

INSTRUCTIONS
● Answer all questions.
● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs.
● Write your name, centre number and candidate number in the boxes at the top of the page.
● Write your answer to each question in the space provided.
● Do not use an erasable pen or correction fluid.
● Do not write on any bar codes.
● You may use a calculator.
● You should show all your working and use appropriate units.

INFORMATION
● The total mark for this paper is 100.
● The number of marks for each question or part question is shown in brackets [ ].

This document has 24 pages. Any blank pages are indicated.

DC (DE) 345927
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Data

acceleration of free fall g = 9.81 m s–2

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

elementary charge e = 1.60 × 10–19 C

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

rest mass of proton mp = 1.67 × 10–27 kg

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rest mass of electron me = 9.11 × 10–31 kg

Avogadro constant NA = 6.02 × 1023 mol–1

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

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

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

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

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1
( = 8.99 × 109 m F–1)
4rf0

Planck constant h = 6.63 × 10–34 J s

Stefan–Boltzmann constant σ = 5.67 × 10–8 W m–2 K–4

Formulae

uniformly accelerated motion s = ut + 12 at 2

v 2 = u 2 + 2as

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hydrostatic pressure ∆p = ρg∆h

upthrust F = ρgV

fs v
Doppler effect for sound waves fo = v ! v
s

electric current I = Anvq

resistors in series R = R1 + R2 + ...

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1 1 1
resistors in parallel = + + ...
R R1 R2

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GM
gravitational potential ϕ =– r

GMm
gravitational potential energy EP = – r

1 Nm
pressure of an ideal gas p = 3 V 〈c2〉

simple harmonic motion a = – ω 2x

velocity of particle in s.h.m. v = v0 cos ωt

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v = !ω (x 02 - x 2)

Q
electric potential V =
4rf0 r

Qq
electrical potential energy EP =
4rf0 r

1 1 1
capacitors in series = + + ...
C C1 C2

capacitors in parallel C = C1 + C2 + ...

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t
discharge of a capacitor x = x 0 e - RC

BI
Hall voltage VH =
ntq

alternating current/voltage x = x0 sin ωt

radioactive decay x = x0e–λt

0.693
decay constant λ =
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t1
2

IR (Z - Z 2) 2
intensity reflection coefficient = 1
I0 (Z 1 + Z 2) 2

Stefan–Boltzmann law L = 4πσr 2T 4

∆λ Df v
Doppler redshift á ác
λ f
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1 (a) State Newton’s law of gravitation.

...................................................................................................................................................

............................................................................................................................................. [2]

(b) A planet may be considered as a uniform sphere.

A satellite is in circular orbit of period T around the planet at a height h above the surface.

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The height of the orbit can be adjusted by use of the satellite’s rocket engines.
2
Fig. 1.1 shows the variation with h of T 3 .

1600

1200
2 2
T 3 / s3

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800

400

0
0 2 4 6 8 10 12
h / 106 m

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Fig. 1.1

(i) By reference to forces, explain why the orbit of the satellite is circular.

...........................................................................................................................................

..................................................................................................................................... [2]
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(ii) Use Newton’s law of gravitation to show that h and T are related by
GA 2
(h + B)3 = T
4π2
where G is the gravitational constant and A and B are constants that depend on the
properties of the planet.
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[3]

(iii) Use the gradient and intercept of the line in Fig. 1.1 to determine values for A and B.
Give units with your answers.
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A = ....................................... unit ..................

B = ....................................... unit ..................

[5]

[Total: 12]
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2 (a) Define specific heat capacity.

...................................................................................................................................................

............................................................................................................................................. [2]

(b) Two solid blocks X and Y are made from different metals. The blocks have different initial
temperatures. Block Y is initially at room temperature.

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The blocks are placed in direct thermal contact with each other at time t = 0. Fig. 2.1 shows
the variation with t of the temperatures of the two blocks.

100

75 X

temperature / °C

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25 Y

0
0 0.5 1.0 1.5 2.0 2.5 3.0
t / min

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Fig. 2.1

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(i) State three conclusions that may be drawn from Fig. 2.1. The conclusions may be
qualitative or quantitative.

1 ........................................................................................................................................

...........................................................................................................................................

2 ........................................................................................................................................

...........................................................................................................................................
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3 ........................................................................................................................................

...........................................................................................................................................
[3]

(ii) The ratio mass of block Y is equal to 1.3.

mass of block X
The metal in block Y has a specific heat capacity of 901 J kg–1 K–1.

Determine the specific heat capacity of the metal in block X.

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specific heat capacity = .......................................... J kg–1 K–1 [3]

[Total: 8]
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3 (a) (i) State what is meant by the Avogadro constant.

...........................................................................................................................................

..................................................................................................................................... [1]

(ii) State the relationship between the Avogadro constant NA, the molar gas constant R and
the Boltzmann constant k.

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[1]

(b) Two samples X and Y of ideal gases are both at thermodynamic temperature T.

Sample X has volume V and consists of N molecules, each of mass m.

Sample Y has volume 2V and consists of 2N molecules, each of mass 2m.

(i) Complete Table 3.1 by giving expressions, in terms of some or all of N, m, T, V and the

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constants in (a)(ii), for the quantities indicated.

Table 3.1

sample X sample Y

pressure

amount of
substance

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mean-square speed
of molecules

internal energy

[4]
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(ii) The temperature of sample X is now varied.

On Fig. 3.1, sketch the variation with thermodynamic temperature of the root-mean-
square (r.m.s.) speed of the molecules of the gas.

r.m.s. speed
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0
0
thermodynamic temperature

Fig. 3.1

[2]

[Total: 8]
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4 (a) State what is meant by simple harmonic motion.

...................................................................................................................................................

............................................................................................................................................. [2]

(b) A block is suspended from a spring, as shown in Fig. 4.1.

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spring

block

floor

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Fig. 4.1

The block is pulled down and released at time t = 0. It then oscillates vertically with simple
harmonic motion.

Fig. 4.2 shows the variation of the velocity v of the block with height h of the base of the block
above the floor.

v / cm s–1

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0 2 4 6 8 10 12
h / cm

–5
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–10

Fig. 4.2

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(i) Determine the amplitude, in cm, of the oscillations.

amplitude = .................................................... cm [1]

(ii) Show that the angular frequency of the oscillations is 3.2 rad s–1.
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[2]

(iii) Calculate the period T of the oscillations.

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T = ....................................................... s [2]

(iv) On Fig. 4.3, sketch the variation of h with time t from t = 0 to t = 6.0 s.

10.0

h / cm

7.5
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5.0

2.5

0
0 1 2 3 4 5 6
t/s
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Fig. 4.3

[4]

[Total: 11]

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5 (a) State the relationship between electric field and electric potential.

...................................................................................................................................................

............................................................................................................................................. [2]

(b) Two charged isolated insulating spheres X and Y are near to each other, as shown in Fig. 5.1.

X Y

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Fig. 5.1

P is a point on the line joining the centres of the spheres.

Explain why it is not possible for the total electric potential and the resultant electric field to
simultaneously be zero at point P.

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...................................................................................................................................................

............................................................................................................................................. [3]

(c) The magnitudes of the charges on spheres X and Y in Fig. 5.1 are Q and 2Q respectively.
The spheres may be considered as point charges at their centres.

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Point P is a distance x from the centre of sphere X.

The electric potential at point P is zero.

(i) Show that the distance y of point P from the centre of sphere Y is equal to 2x.

[2]
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(ii) State an expression, in terms of Q, x and the permittivity of free space ε0, for the electric
field strength EX at P due to sphere X.

EX = ......................................................... [1]

(iii) Determine an expression, in terms of Q, x and ε0, for the resultant electric field strength
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E at point P due to the two spheres.

E = ......................................................... [2]
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[Total: 10]
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6 (a) (i) State what is meant by rectification of an alternating voltage.

...........................................................................................................................................

..................................................................................................................................... [1]

(ii) State the difference between half-wave rectification and full-wave rectification.

...........................................................................................................................................

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..................................................................................................................................... [2]

(b) (i) Complete Fig. 6.1 to show a circuit that produces half-wave rectification of an alternating
input voltage VIN to produce output voltage VOUT across the resistor R.

VIN C R VOUT

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Fig. 6.1

[2]

(ii) State the purpose of the capacitor C in the circuit of Fig. 6.1.

...........................................................................................................................................

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..................................................................................................................................... [1]

+12

VIN / V

0
0 0.01 0.02 0.03 0.04
t/s
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–12

Fig. 6.2

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Fig. 6.3 shows the variation of VOUT with t.

VOUT / V

8
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0
0 0.01 0.02 0.03 0.04
t/s

Fig. 6.3

The maximum energy stored in the capacitor is 0.041 J.

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(i) Show that the capacitance of C is 570 μF.

[2]

(ii) Determine the resistance of R.

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resistance = ..................................................... Ω [3]

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[Total: 11]

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7 (a) Define magnetic flux density.

...................................................................................................................................................

............................................................................................................................................. [2]

(b) A long, straight wire carries a current into the page, as shown in Fig. 7.1.

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Fig. 7.1

On Fig. 7.1, draw four field lines to represent the magnetic field around the wire due to the
current in it. [3]

(c) Two identical wires X and Y are placed parallel to each other. The wires both carry current
into the page, as shown in Fig. 7.2.

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X Y

Fig. 7.2
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(i) Explain why the two wires exert a magnetic force on each other.

...........................................................................................................................................

..................................................................................................................................... [2]

(ii) On Fig. 7.2, draw an arrow to show the direction of the magnetic force exerted on wire X.
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Label your arrow F. [1]

(iii) The current in X is double the current in Y.

State how the magnetic force exerted on wire Y compares with the magnetic force
exerted on wire X.

...........................................................................................................................................

..................................................................................................................................... [2]
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(iv) The direction of the current in both wires is now reversed.

State, with a reason, the effect of this change on the direction of the force on wire X.

...........................................................................................................................................

..................................................................................................................................... [1]

[Total: 11]
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8 A polished sheet of magnesium in a vacuum emits electrons when it is illuminated by ultraviolet

radiation.

(a) State the name of this phenomenon.

............................................................................................................................................. [1]

(b) For emission of electrons to occur, the frequency of the ultraviolet radiation must be at least
8.8 × 1014 Hz.

(i) Calculate the work function energy of magnesium.

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work function energy = ...................................................... J [2]

(ii) For ultraviolet radiation with a frequency of 11 × 1014 Hz, calculate the maximum speed
of the emitted electrons.

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maximum speed = ................................................ m s–1 [3]

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(c) The frequency f of the ultraviolet radiation incident on the magnesium sheet is varied between
8.0 × 1014 Hz and 11 × 1014 Hz.

On Fig. 8.1, sketch the variation with f of the maximum kinetic energy EMAX of the emitted
electrons. Use the space below for any working that you need.
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2.0

1.5
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EMAX / 10–19 J

1.0

0.5

0
8.0 8.5 9.0 9.5 10.0 10.5 11.0
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f / 1014 Hz

Fig. 8.1

[3]

[Total: 9]
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9 Fluorine-18 ( 189F) decays by beta-plus (β+) emission with a half-life of 110 minutes.

(a) (i) State the name of the beta-plus particle.

..................................................................................................................................... [1]

(ii) Show that the decay constant of fluorine-18 is 1.05 × 10–4 s–1.
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[1]

(iii) Determine the activity of 2.1 × 10–12 kg of fluorine-18.

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activity = .................................................... Bq [3]

(b) A small sample of fluorine-18 injected into the body acts as a tracer for use in medical imaging.

(i) Describe how the interaction of a β+ particle with an electron in the body enables the
formation of an image.

...........................................................................................................................................
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...........................................................................................................................................

..................................................................................................................................... [3]

(ii) Suggest why 110 minutes is a suitable half-life for a nuclide used as a tracer in medical
diagnosis.

...........................................................................................................................................
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...........................................................................................................................................

..................................................................................................................................... [2]

[Total: 10]

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10 (a) Explain how redshift leads to the idea that the Universe is expanding.

...................................................................................................................................................

............................................................................................................................................. [3]

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(b) Stars in a distant galaxy emit radiation. The total luminosity of the stars in the galaxy is
1.90 × 1036 W.
The emission spectrum of the radiation contains a line X at a wavelength of 658 nm.

Radiation from the galaxy is observed on the Earth. The observed radiation has a radiant flux
intensity of 8.42 × 10–16 W m–2. In the observed emission spectrum, line X is at a wavelength
of 726 nm.

Determine:

(i) the distance d of the galaxy from the Earth

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d = ..................................................... m [2]

(ii) the speed v of the galaxy relative to the Earth.

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v = ................................................ m s–1 [2]

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(c) Observations of many galaxies, such as the one in (b), lead to many pairs of values of d and
v. Plotting these values reveals a trend.

(i) On Fig. 10.1, sketch the variation of v with d.

v
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0
0
d

Fig. 10.1

[2]

(ii) State the name of the quantity represented by the gradient of the line in Fig. 10.1.
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..................................................................................................................................... [1]

[Total: 10]
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To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge
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Cambridge International AS & A Level

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PHYSICS 9702/42
Paper 4 A Level Structured Questions October/November 2024

2 hours

You must answer on the question paper.

No additional materials are needed.

INFORMATION
● The total mark for this paper is 100.
● The number of marks for each question or part question is shown in brackets [ ].

This document has 24 pages. Any blank pages are indicated.

DC (PB/FC) 336256/3
© UCLES 2024 [Turn over
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2
, ,

Data

acceleration of free fall g = 9.81 m s–2

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

elementary charge e = 1.60 × 10–19 C

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

rest mass of proton mp = 1.67 × 10–27 kg

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rest mass of electron me = 9.11 × 10–31 kg

Avogadro constant NA = 6.02 × 1023 mol–1

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

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

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

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

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1
( = 8.99 × 109 m F–1)
4rf0

Planck constant h = 6.63 × 10–34 J s

Stefan–Boltzmann constant σ = 5.67 × 10–8 W m–2 K–4

Formulae

uniformly accelerated motion s = ut + 12 at 2

v 2 = u 2 + 2as

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hydrostatic pressure ∆p = ρg∆h

upthrust F = ρgV

fs v
Doppler effect for sound waves fo = v ! v
s

electric current I = Anvq

resistors in series R = R1 + R2 + ...

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1 1 1
resistors in parallel = + + ...
R R1 R2

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32
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GM
gravitational potential ϕ =– r

GMm
gravitational potential energy EP = – r

1 Nm
pressure of an ideal gas p = 3 V 〈c2〉

simple harmonic motion a = – ω 2x

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velocity of particle in s.h.m. v = v0 cos ωt

v = !ω (x 02 - x 2)

Q
electric potential V =
4rf0 r

Qq
electrical potential energy EP =
4rf0 r

1 1 1
capacitors in series = + + ...
C C1 C2

capacitors in parallel C = C1 + C2 + ...

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t
discharge of a capacitor x = x 0 e - RC

BI
Hall voltage VH =
ntq

alternating current/voltage x = x0 sin ωt

radioactive decay x = x0e–λt

0.693
decay constant λ =
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t1
2

IR (Z - Z 2) 2
intensity reflection coefficient = 1
I0 (Z 1 + Z 2) 2

Stefan–Boltzmann law L = 4πσr 2T 4

∆λ Df v
Doppler redshift á ác
λ f
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4
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1 A metal wheel consists of an axle A, eight spokes and a rim, as shown in Fig. 1.1.

spoke

axle A
rim

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Fig. 1.1

Point X is on the rim at the end of one of the spokes.

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The rim has a radius of 0.85 m.
The wheel is rotating clockwise with an angular speed of 140 rad s–1.

(a) For point X, determine:

(i) the speed

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speed = ................................................ m s–1 [2]

(ii) the centripetal acceleration.

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acceleration = ................................................ m s–2 [2]

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5
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(b) There is a uniform magnetic field of flux density 0.18 T into the plane of the page.

(i) State Lenz’s law of electromagnetic induction.

...........................................................................................................................................

..................................................................................................................................... [2]

(ii) Show that the time taken for point X to complete one revolution is 45 ms.
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[1]

(iii) Calculate the magnetic flux cut by spoke AX during one revolution of the wheel.
Give a unit with your answer.
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magnetic flux = .................................... unit ............ [3]

(iv) Determine the magnitude of the electromotive force (e.m.f.) induced across spoke AX.
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induced e.m.f. = ...................................................... V [2]

(v) Use Lenz’s law to explain whether the potential is higher at end A or end X of the spoke.

...........................................................................................................................................
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...........................................................................................................................................

..................................................................................................................................... [1]

[Total: 13]

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6
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2 The Sun may be considered as a uniform sphere with a mass of 1.99 × 1030 kg and a surface
temperature of 5780 K.

A probe with a mass of 2.63 kg moves in a straight line towards the Sun.
When it is at a distance x from the centre of the Sun, the probe measures the gravitational field
strength g due to the Sun and the radiant flux intensity F of radiation from the Sun.

(a) Define gravitational field.

...................................................................................................................................................

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............................................................................................................................................. [1]

(b) For the position of the probe where x = 1.47 × 1011 m:

(i) calculate g

g = ............................................... N kg–1 [2]

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(ii) determine the gravitational potential energy EP of the probe.

EP = ....................................................... J [2]

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4πGM
g= F
L
where M is the mass of the Sun, L is the luminosity of the Sun and G is the
gravitational constant.
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[3]

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(ii) Fig. 2.1 shows the variation of g with F.

g / 10–3 N kg–1

4
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0
0 0.5 1.0 1.5 2.0
F / 103 W m–2

Fig. 2.1

Determine a value for the luminosity L of the Sun. Give a unit with your answer.
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L = .................................... unit ............ [2]

(iii) Use your answer in (c)(ii) to determine the radius r of the Sun.
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r = ...................................................... m [2]

[Total: 12]
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8
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3 (a) Define specific latent heat.

...................................................................................................................................................

............................................................................................................................................. [2]

(b) A dish containing 7.2 × 10–5 m3 of a substance rests on a laboratory bench. The substance is
initially a liquid of density 710 kg m–3. Atmospheric pressure is 1.0 × 105 Pa.

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The liquid is heated at its boiling point so that it completely vaporises. The increase in the
internal energy of the substance during this process is 17.6 kJ. The final volume of the vapour
is 0.017 m3.

(i) Show that the magnitude of the work done on the substance when it vaporises is 1.7 kJ.

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[2]

(ii) Use the information in (b)(i) to calculate the thermal energy Q, in kJ, supplied to the
substance to cause it to vaporise.

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Q = ..................................................... kJ [2]

(iii) Use your answer in (b)(ii) to determine a value for the specific latent heat of vaporisation
LV, in kJ kg–1, of the substance.
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LV = .............................................. kJ kg–1 [2]

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Suggest and explain whether LF is likely to be less than, the same as, or greater than the
answer in (b)(iii).

...................................................................................................................................................

...................................................................................................................................................
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...................................................................................................................................................

............................................................................................................................................. [3]

[Total: 11]
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4 (a) State three of the basic assumptions of the kinetic theory of gases.

1 ................................................................................................................................................

...................................................................................................................................................

2 ................................................................................................................................................

...................................................................................................................................................

3 ................................................................................................................................................

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...................................................................................................................................................
[3]

(b) Explain how molecular movement causes the pressure exerted by a gas.

...................................................................................................................................................

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...................................................................................................................................................

............................................................................................................................................. [3]

〈c 2〉 / 106 m2 s–2 X

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Y
2

0
0 100 200 300 400
T/K

Fig. 4.1
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Fig. 4.2 shows the variation with T of the product pV for samples of the two gases, where p is
the pressure of the gas and V is the volume of the gas.

3
Y
pV / 103 J

2
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1 X

0
0 100 200 300 400
T/K

Fig. 4.2

State three conclusions about the gases and their samples that may be drawn from Fig. 4.1
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and Fig. 4.2. The conclusions may be qualitative or quantitative. Use the space below for any
working that you need.
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1 ................................................................................................................................................

...................................................................................................................................................

2 ................................................................................................................................................

...................................................................................................................................................

3 ................................................................................................................................................
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...................................................................................................................................................
[3]

[Total: 9]

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5 Fig. 5.1 shows a pendulum consisting of a metal sphere suspended by a thin string.

thin string

metal sphere

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oscillations

Fig. 5.1 (not to scale)

The sphere undergoes small oscillations about its equilibrium position. The oscillations may be
considered to be simple harmonic.

Fig. 5.2 shows the variation with time t of the displacement x of the sphere from its
equilibrium position.

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0.02
x /m

0.01

0
0 0.2 0.4 0.6 0.8 1.0 1.2
t/s

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–0.01

–0.02

Fig. 5.2

(a) On Fig. 5.1, draw an arrow, from the centre of the sphere, to represent the direction of the
resultant force acting on the sphere when it is in the position shown. [1]
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(b) The mass of the sphere is 0.15 kg.

(i) State the amplitude of the oscillations.

amplitude = ...................................................... m [1]

(ii) Determine the angular frequency of the oscillations.

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angular frequency = .............................................. rad s–1 [2]

(iii) Calculate the total energy of the oscillations.

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total energy = ....................................................... J [2]

6
EK / 10–3 J
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0
–0.02 –0.01 0 0.01 0.02
x /m
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Fig. 5.3
[3]

[Total: 9]

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6 (a) State Coulomb’s law.

...................................................................................................................................................

............................................................................................................................................. [2]

(b) Fig. 6.1 shows an isolated hollow conducting sphere that is positively charged.

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+
+ +
+ +
+ +
+

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Fig. 6.1

On Fig. 6.1, draw field lines to represent the electric field outside the sphere. [3]

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E / 105 N C–1

1
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0
0 2 4 6 8
x / cm

Fig. 6.2

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(i) Determine the radius, in cm, of the sphere.

radius = .................................................... cm [1]

(ii) Calculate the charge on the sphere.

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charge = ...................................................... C [3]

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(iii) Suggest an explanation for the fact that the electric field inside the sphere is zero.

...........................................................................................................................................

..................................................................................................................................... [1]

[Total: 10]
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7 (a) Define the capacitance of a parallel‑plate capacitor.

...................................................................................................................................................

............................................................................................................................................. [2]

(b) An initially uncharged capacitor X, of capacitance C, is gradually charged so that the final
potential difference (p.d.) between its plates is V and the final charge is Q.

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(i) On Fig. 7.1, sketch the variation of charge with p.d. for capacitor X as the p.d. increases
from 0 to V.

Q
charge

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0 V
p.d.

Fig. 7.1
[2]

(ii) Determine an expression, in terms of Q and V, for the work W done on capacitor X
during the charging process. Explain your reasoning.

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W = ......................................................... [2]
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, ,

(c) Another capacitor Y is initially uncharged. The fully charged capacitor X in (b) is now
connected to capacitor Y, as shown in Fig. 7.2.

Y
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Fig. 7.2

The capacitance of capacitor Y is 3C.

(i) Complete Table 7.1 to show expressions, in terms of Q and V, for the final p.d.s across,
and the final charges on, the two capacitors.
Use the space below for any working that you need.
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Table 7.1

X Y
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final p.d.

final charge

[3]

(ii) State whether the total energy stored in the two capacitors is less than, the same as, or
greater than the energy initially stored in capacitor X.

..................................................................................................................................... [1]
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[Total: 10]

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18
, ,

8 (a) State what is meant by the frequency of an alternating current.

...................................................................................................................................................

............................................................................................................................................. [1]

(b) An alternating current I in a resistor of resistance 680 Ω varies with time t according to

I = 3.5 sin (40πt)

where I is in A and t is in s.

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(i) Show that the period of the alternating current is 50 ms.

[1]

(ii) On Fig. 8.1, sketch the variation of I with t between t = 0 and t = 100 ms.

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I /A

0
0 25 50 75 100
t / ms

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–2

–4

Fig. 8.1
[3]
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19
, ,

(iii) Determine the root‑mean‑square (r.m.s.) current in the resistor.

r.m.s. current = ....................................................... A [1]

(c) Use data from (b), including your answer in (b)(iii), to show by calculation that the mean
power in the 680 Ω resistor is half of the peak power.
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[3]

[Total: 9]
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20
, ,

9 Electrons in a vacuum are accelerated from rest through a potential difference (p.d.) V to form a
beam. The electrons each have mass m and charge q.

The beam is incident on a graphite crystal that acts as a diffraction grating. After passing through
the crystal, the beam reaches a fluorescent screen. An interference pattern is observed on
this screen.

(a) Explain what this observation shows about the nature of electrons.

...................................................................................................................................................

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...................................................................................................................................................

............................................................................................................................................. [1]

(b) Determine an expression, in terms of m, q and V, for the momentum p of an electron in

the beam.

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p = ......................................................... [3]

Describe and explain the effect of this change on the interference pattern observed.

...................................................................................................................................................

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...................................................................................................................................................

............................................................................................................................................. [2]
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21
, ,

(d) The electrons are now accelerated through different values of V, resulting in pairs of
corresponding values for p and the de Broglie wavelength λ.

1
(i) On Fig. 9.1, sketch the variation of p with λ .

p
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0
0
1
λ

Fig. 9.1
[2]

(ii) State the name of the quantity represented by the gradient of the line in Fig. 9.1.
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..................................................................................................................................... [1]

[Total: 9]
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22

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23
, ,

10 (a) Radioactive decay is both random and spontaneous.

(i) State what is meant by random.

...........................................................................................................................................

..................................................................................................................................... [1]

(ii) State what is meant by spontaneous.

...........................................................................................................................................
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..................................................................................................................................... [1]

(iii) State one piece of evidence for the random nature of decay.

...........................................................................................................................................

..................................................................................................................................... [1]

(b) (i) Describe the differences between nuclear fission and nuclear fusion.

...........................................................................................................................................
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...........................................................................................................................................

..................................................................................................................................... [3]

(ii) Explain, with reference to the variation of binding energy per nucleon with nucleon
number, why the processes of nuclear fission and nuclear fusion both result in a release
of energy.
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...........................................................................................................................................

..................................................................................................................................... [2]

[Total: 8]
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Permission to reproduce items where third‑party owned material protected by copyright is included has been sought and cleared where possible. Every
reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the
publisher will be pleased to make amends at the earliest possible opportunity.

To avoid the issue of disclosure of answer‑related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge
Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download
at www.cambridgeinternational.org after the live examination series.

Cambridge Assessment International Education is part of Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge
Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge.
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, ,

Cambridge International AS & A Level

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PHYSICS 9702/51
Paper 5 Planning, Analysis and Evaluation October/November 2024

1 hour 15 minutes

You must answer on the question paper.

No additional materials are needed.

INFORMATION
● The total mark for this paper is 30.
● The number of marks for each question or part question is shown in brackets [ ].

This document has 8 pages.

DC (LO) 345928
© UCLES 2024 [Turn over
* 0000800000002 *

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2
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1 A thin cylindrical bar magnet of length L and cross-sectional area A is attached to a block.
An identical magnet is attached to a trolley, as shown in Fig. 1.1.

s D
L
N N
bench

P
block magnets trolley

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Fig. 1.1

The trolley is held so that the separation of the N poles of the two magnets is s.

Point P is a distance D from the N pole of the magnet on the stationary trolley.

The trolley is released. The speed v of the trolley at point P is determined using one light gate.

It is suggested that v is related to s by the relationship

mv 2 KA 2 B 2 L2
= -Q
2D s4

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where B is the magnetic flux density at the N pole of one of the magnets, m is the mass of the
trolley, and K and Q are constants.

Plan a laboratory experiment to test the relationship between v and s.

Draw a diagram showing the arrangement of your equipment.

Explain how the results could be used to determine values for K and Q.

In your plan you should include:

● the procedure to be followed

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● the measurements to be taken

● the control of variables

● the analysis of the data

● any safety precautions to be taken.

DO NOT WRITE IN THIS MARGIN

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9702/51/O/N/24
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Diagram
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.......................................................................................................................................................... [15]

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5
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2 A student investigates an electrical circuit. A power supply of electromotive force (e.m.f.) Es and
negligible internal resistance is connected in series to three resistors, each of resistance Z.

A cell, an ammeter and a resistor of resistance R are connected in parallel across one of these
resistors, as shown in Fig. 2.1.

+ Es –
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Z Z Z

A
R

Fig. 2.1

The current I is measured by the ammeter for different values of R.

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It is suggested that I and R are related by the equation

3E – Es = I(3R + 2Z)

where E is the e.m.f. of the cell.

1
(a) A graph is plotted of I on the y-axis against R on the x-axis.

Determine expressions for the gradient and y-intercept.

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gradient = ...............................................................

y-intercept = ...............................................................
[1]

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(b) Values of R and I are given in Table 2.1.

Table 2.1

R / kΩ I / μA 1 –1
/A
I

1.50 194 ± 2

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1.75 180 ± 2

1.92 172 ± 2

2.22 160 ± 2

2.48 150 ± 2

2.72 144 ± 2

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1
Calculate and record values of / A–1 in Table 2.1.
I
1
Include the absolute uncertainties in . [2]
I
1 1
(c) (i) Plot a graph of / A–1 against R / kΩ. Include error bars for . [2]
I I
(ii) Draw the straight line of best fit and a worst acceptable straight line on your graph. Label
both lines. [2]

(iii) Determine the gradient of the line of best fit. Include the absolute uncertainty in your
answer.

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gradient = ......................................................... [2]

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7200

7000
1 –1
/A
I

6800
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6600

6400

6200
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6000

5800

5600
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5400

5200
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5000
1.40 1.60 1.80 2.00 2.20 2.40 2.60 2.80
R / kΩ

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8
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(iv) Determine the y-intercept of the line of best fit. Include the absolute uncertainty in your
answer.

y-intercept = ......................................................... [2]

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(d) (i) Using your answers to (a), (c)(iii) and (c)(iv), determine the values of E and Z. Include
appropriate units.

Data: Es = (2.20 ± 0.05) V

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E = ...............................................................

Z = ...............................................................
[2]

(ii) Determine the absolute uncertainty in E.

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absolute uncertainty in E = ......................................................... [1]

(e) The experiment is repeated. Determine the resistance R that gives a value of I of 250 μA.
DO NOT WRITE IN THIS MARGIN

R = ...................................................... Ω [1]

[Total: 15]
To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge
Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download
at www.cambridgeinternational.org after the live examination series.
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9702/51/O/N/24
* 0000800000001 *

, ,

Cambridge International AS & A Level

¬Oz> 4mHuOªE]z5W
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¥55¥ueE¥EEU¥uuU
* 5 8 1 7 0 0 8 4 1 6 *

PHYSICS 9702/52
Paper 5 Planning, Analysis and Evaluation October/November 2024

1 hour 15 minutes

You must answer on the question paper.

No additional materials are needed.

INFORMATION
● The total mark for this paper is 30.
● The number of marks for each question or part question is shown in brackets [ ].

This document has 8 pages.

DC (DE/FC) 336258/2
© UCLES 2024 [Turn over
* 0000800000002 *

DO NOT WRITE IN THIS MARGIN

2
, ,

1 Fig. 1.1 shows a coil made from resistance wire.

Fig. 1.1

The coil is placed in cooking oil of mass m. The total length of the resistance wire in the oil is L.

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A potential difference V is applied to the coil. The temperature of the oil increases by Δθ in time t.

It is suggested that Δθ is related to L by the relationship

AtV 2
= mKΔθ + Z
L
where A is the cross-sectional area of the wire, and K and Z are constants.

Plan a laboratory experiment to test the relationship between Δθ and L.

Draw a diagram showing the arrangement of your equipment.

Explain how the results could be used to determine values for K and Z.

DO NOT WRITE IN THIS MARGIN

In your plan you should include:

● the procedure to be followed

● the measurements to be taken

● the control of variables

● the analysis of the data

● any safety precautions to be taken.

DO NOT WRITE IN THIS MARGIN

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9702/52/O/N/24
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3
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Diagram
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.......................................................................................................................................................... [15]
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5
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2 A student investigates the refraction of white light entering a transparent rectangular block. A
narrow beam of light enters the block at the midpoint of one of the shorter sides. The angle of
incidence θ is measured, as shown in Fig. 2.1.

block

beam of light

θ
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Fig. 2.1 (not to scale)

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The distance d between the corner of the block and the point where the beam of light touches the
boundary of the block is measured.

The experiment is repeated for different values of θ.

It is suggested that d and θ are related by the equation

B sin2 θ
=
B + d2 n2
where B and n are constants.
1
(a) A graph is plotted of d 2 on the y-axis against on the x-axis.
sin2 θ
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Determine expressions for the gradient and y-intercept.

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gradient = ...............................................................

y-intercept = ...............................................................
[1]
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1
(b) Values of θ, and d are given in Table 2.1.
sin2 θ
Table 2.1

1
θ /° d / cm d 2 / cm2
sin2 θ

28.5 4.39 24.8 ± 0.2

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33.5 3.28 21.4 ± 0.2

42.5 2.19 17.1 ± 0.2

50.0 1.70 14.7 ± 0.2

57.5 1.41 12.9 ± 0.2

63.5 1.25 11.8 ± 0.2

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Calculate and record values of d 2 / cm2 in Table 2.1.
Include the absolute uncertainties in d 2. [2]
1
(c) (i) Plot a graph of d 2 / cm2 against . Include error bars for d 2. [2]
sin2 θ
(ii) Draw the straight line of best fit and a worst acceptable straight line on your graph. Label
both lines. [2]

(iii) Determine the gradient of the line of best fit. Include the absolute uncertainty in your
answer.

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gradient = ......................................................... [2]

DO NOT WRITE IN THIS MARGIN

ĬÑú¾Ġ´íÈõÏĪÅĊßú¶þ×
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650

600

550
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d 2 / cm2

500

450

400
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350

300

250
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200

150

100
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1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

1
sin2 θ

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8
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(iv) Determine the y-intercept of the line of best fit. Include the absolute uncertainty in your
answer.

y-intercept = ......................................................... [2]

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(d) (i) Using your answers to (a), (c)(iii) and (c)(iv), determine the values of B and n. Include
appropriate units.

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B = ...............................................................

n = ...............................................................
[2]

(ii) Determine the percentage uncertainty in n.

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percentage uncertainty in n = ..................................................... % [1]

(e) The experiment is repeated. Determine the angle θ that gives a value of d of 30.0 cm. DO NOT WRITE IN THIS MARGIN

θ = ....................................................... ° [1]

[Total: 15]
To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge
Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download
at www.cambridgeinternational.org after the live examination series.
ĬÑú¾Ġ´íÈõÏĪÅĊÝú¶Ā×
© UCLES 2024 ĬàËòÑĤùĂÐîĆ¿¾đěÑĘĂ
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9702/52/O/N/24
Cambridge International AS & A Level

PHYSICS 9702/41
Paper 4 A Level Structured Questions October/November 2024
MARK SCHEME
Maximum Mark: 100

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the
details of the discussions that took place at an Examiners’ meeting before marking began, which would have
considered the acceptability of alternative answers.

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.

Cambridge International will not enter into discussions about these mark schemes.

Cambridge International is publishing the mark schemes for the October/November 2024 series for most
Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level
components.

This document consists of 15 printed pages.

© Cambridge University Press & Assessment 2024 [Turn over

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Generic Marking Principles

These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the
specific content of the mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these
marking principles.

GENERIC MARKING PRINCIPLE 1:

Marks must be awarded in line with:

• the specific content of the mark scheme or the generic level descriptors for the question
• the specific skills defined in the mark scheme or in the generic level descriptors for the question
• the standard of response required by a candidate as exemplified by the standardisation scripts.

GENERIC MARKING PRINCIPLE 2:

Marks awarded are always whole marks (not half marks, or other fractions).

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:

• marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond
the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate
• marks are awarded when candidates clearly demonstrate what they know and can do
• marks are not deducted for errors
• marks are not deducted for omissions
• answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the
question as indicated by the mark scheme. The meaning, however, should be unambiguous.

GENERIC MARKING PRINCIPLE 4:

Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level
descriptors.

© Cambridge University Press & Assessment 2024 Page 2 of 15

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
GENERIC MARKING PRINCIPLE 5:

Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may
be limited according to the quality of the candidate responses seen).

GENERIC MARKING PRINCIPLE 6:

Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or
grade descriptors in mind.

Science-Specific Marking Principles

1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks
should not be awarded if the keywords are used incorrectly.

2 The examiner should not choose between contradictory statements given in the same question part, and credit should not be awarded for
any correct statement that is contradicted within the same question part. Wrong science that is irrelevant to the question should be ignored.

3 Although spellings do not have to be correct, spellings of syllabus terms must allow for clear and unambiguous separation from other
syllabus terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection).

4 The error carried forward (ecf) principle should be applied, where appropriate. If an incorrect answer is subsequently used in a scientifically
correct way, the candidate should be awarded these subsequent marking points. Further guidance will be included in the mark scheme
where necessary and any exceptions to this general principle will be noted.

5 ‘List rule’ guidance

For questions that require n responses (e.g. State two reasons …):

• The response should be read as continuous prose, even when numbered answer spaces are provided.
• Any response marked ignore in the mark scheme should not count towards n.
• Incorrect responses should not be awarded credit but will still count towards n.
• Read the entire response to check for any responses that contradict those that would otherwise be credited. Credit should not be
awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this
should be treated as a single incorrect response.
• Non-contradictory responses after the first n responses may be ignored even if they include incorrect science.

© Cambridge University Press & Assessment 2024 Page 3 of 15

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
6 Calculation specific guidance

Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show
your working’.

For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded
by the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values.

For answers given in standard form (e.g. a  10n) in which the convention of restricting the value of the coefficient (a) to a value between 1
and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme.

Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded.
Exceptions to this general principle will be noted in the mark scheme.

7 Guidance for chemical equations

Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme.

State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

© Cambridge University Press & Assessment 2024 Page 4 of 15

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Abbreviations

/ Alternative and acceptable answers for the same marking point.

() Bracketed content indicates words which do not need to be explicitly seen to gain credit but which indicate the context for an
answer. The context does not need to be seen but if a context is given that is incorrect then the mark should not be awarded.

___ Underlined content must be present in answer to award the mark. This means either the exact word or another word that has the
same technical meaning.

Mark categories

B marks These are independent marks, which do not depend on other marks. For a B mark to be awarded, the point to which it refers must
be seen specifically in the candidate’s answer.

M marks These are mandatory marks upon which A marks later depend. For an M mark to be awarded, the point to which it refers must be
seen specifically in the candidate’s answer. If a candidate is not awarded an M mark, then the later A mark cannot be awarded
either.

C marks These are compensatory marks which can be awarded even if the points to which they refer are not written down by the candidate,
providing subsequent working gives evidence that they must have known them. For example, if an equation carries a C mark and
the candidate does not write down the actual equation but does correct working which shows the candidate knew the equation, then
the C mark is awarded.

If a correct answer is given to a numerical question, all of the preceding C marks are awarded automatically. It is only necessary to
consider each of the C marks in turn when the numerical answer is not correct.

A marks These are answer marks. They may depend on an M mark or allow a C mark to be awarded by implication.

© Cambridge University Press & Assessment 2024 Page 5 of 15

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Question Answer Marks

1(a) (gravitational) force is (directly) proportional to product of masses B1

force (between point masses) is inversely proportional to the square of their separation B1

1(b)(i) (gravitational) force acts perpendicular to direction of motion B1

gravitational force provides centripetal acceleration B1

1(b)(ii) (F =) GMm / x2 = mx2 and  = 2 / T M1

GMm / x2 = 42mx / T2

completion of algebra leading to x3 = GMT2 / 42 A1

clear indication that B = radius of planet and that A = mass (of planet) B1

1(b)(iii) gradient = 3√(42 / GA) C1

e.g. (1280 – 360) / (12  106) = 3√(42 / [6.67  10–11  A]) C1

A = 1.3  1024 kg A1

intercept = gradient  B C1

e.g. 360 = ((1280 – 360)  B) / (12  106) A1

B = 4.7  106 m

© Cambridge University Press & Assessment 2024 Page 6 of 15

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Question Answer Marks

2(a) (thermal) energy per unit mass (to change temperature) B1

(thermal) energy per unit change in temperature B1

2(b)(i) Any three bulleted points from: B3

• the blocks end up in thermal equilibrium

• heat capacity of Y is larger than heat capacity of X
• no heat loss to the surroundings

Up to 2 points from these six:

• initial temperature of X = 85 °C
• initial temperature of Y = 25 °C
• the temperature change of X = 45 °C
• the temperature change of Y = 15 °C
• the temperature change in X is three times that in Y
• final temperature of both = 40 °C

2(b)(ii)  = 45 °C for X and 15 °C for Y C1

mc  45 = 1.3  m  901  15 C1

c = 390 J kg K–1 A1

© Cambridge University Press & Assessment 2024 Page 7 of 15

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Question Answer Marks

3(a)(i) number of particles per unit amount of substance B1

3(a)(ii) NA = R / k B1

3(b)(i) X pressure and Y pressure both = NkT / V B1

X amount = N / NA and Y amount = 2N / NA B1

X mean-square speed = 3kT / m and Y mean-square speed = 3kT / 2m B1

X internal energy = 3NkT / 2 and Y internal energy = 3NkT B1

3(b)(ii) line passing through the origin and not returning to either axis B1

curve with positive decreasing gradient B1

© Cambridge University Press & Assessment 2024 Page 8 of 15

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Question Answer Marks

4(a) (motion in which) acceleration is (directly) proportional to displacement B1

(motion in which): B1
acceleration is (always) in the opposite direction to displacement
or
acceleration is (always) directed towards a fixed point

4(b)(i) amplitude = (9.5 – 3.5) / 2 A1

= 3.0 cm

4(b)(ii)  = v0 / x0 C1

= 9.5 / 3.0 = 3.2 rad s–1 A1

4(b)(iii) T = 2 /  C1

= 2 / 3.2 A1

= 2.0 s

4(b)(iv) attempted sinusoidal curve starting with a minimum at t = 0 B1

sinusoidal curve of period 2.0 s from t = 0 to t = 6.0 s B1

all peaks shown at h = 9.5 cm B1

all troughs shown at h = 3.5 cm B1

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

5(a) (electric) field equals (electric) potential gradient M1

reference to minus sign A1

5(b) • for potential to be zero, one potential must be positive and the other potential must be negative B3
• for potential to be zero, the charges must have opposite sign
• for field to be zero, the fields (due to X and Y) must be in opposite directions
• for field to be zero, the charges must have the same sign
• the signs of the charges cannot (simultaneously) be both the same and opposite (so not possible)
Any three points, 1 mark each

5(c)(i) VX = (–) Q / 4ε0x and VY = (–) 2Q / 4ε0y C1

(VX + VY = 0 so) Q / 4ε0x = 2Q / 4ε0y leading to y = 2x A1

5(c)(ii) EX = Q / 4ε0x2 A1

5(c)(iii) EY = 2Q / 4ε0(2x)2 C1

( = Q / 8ε0x2)

(opposite charges so fields in same direction so magnitudes add): A1

E = (Q / 4ε0x2) + (Q / 8ε0x2)

= 3Q / 8ε0x2

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Question Answer Marks

6(a)(i) conversion (from a.c.) to d.c. B1

6(a)(ii) half-wave: voltage in one direction is removed B1

full-wave: voltage in one direction is reversed B1

6(b)(i) one gap connected by a single diode and other gap connected directly B1

diode drawn (in a circuit) with correct circuit symbol B1

6(b)(ii) smoothing B1

6(c)(i) E = ½CV2 C1

C = 2  0.041 / 122 = 5.7  10–4 F = 570 F A1

6(c)(ii) 8.0 = 12.0 exp (– 0.010 / RC) C1

ln (8.0 / 12.0) = – 0.010 / (R  5.7  10–4) C1

R = 43  A1

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

7(a) • force per unit length B2

• force per unit current
• length / current perpendicular to field
1 mark for any two points, 2 marks for all three points

7(b) concentric circles around the wire (at least two circles needed) B1

spacing between circles increases with distance from wire (at least four circles needed) B1

arrows showing direction of field is clockwise B1

7(c)(i) (each) wire sits in the (magnetic) field created by the other B1

current (in one wire) is perpendicular to (magnetic) field (due to other wire) so (magnetic) force acts (on wire) B1

7(c)(ii) arrow drawn, starting from X and pointing towards Y, labelled F B1

7(c)(iii) (forces have) equal magnitudes B1

(forces are in) opposite directions B1

7(c)(iv) no change (in the direction of the force) since both the current in X and the field due to Y have reversed B1

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

8(a) photoelectric effect B1

8(b)(i) E = hf C1

work function = 6.63  10–34  8.8  1014 A1

= 5.8  10–19 J

8(b)(ii) hf =  + ½ mvMAX2 C1

6.63  10–34  11  1014 = (5.8  10–19) + (½  9.11  10–31  vMAX2) C1

vMAX = 5.7  105 m s–1 A1

8(c) EMAX shown as zero from f = 8.0 to 8.8 and non-zero from f = 8.8 to 11 B1

all non-zero EMAX shown as a single straight line with a positive gradient B1

line passing through (11, 1.45) B1

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

9(a)(i) positron B1

9(a)(ii)  = ln 2 / (110  60) = 1.05  10–4 s–1 A1

9(a)(iii) N = M / (18 u) or (M in grams  NA / 18) C1

N = (2.1  10–12) / (18  1.66  10–27) or (2.1  10–9  6.02  1023) / 18

( = 7.0  1013)

A = N C1

= 1.05  10–4  7.0  1013 A1

= 7.4  109 Bq

9(b)(i) • (pair) annihilation occurs B3

• the mass of the two particles is converted into energy

• two gamma photons are formed and travel in opposite directions

or
two gamma photons are formed and leave the body

• difference in arrival times of photons (at detector) is processed

Any three points, 1 mark each

9(b)(ii) with a shorter half-life: sample would (almost) fully decay before the test is complete B1

a longer half-life: exposes patient to harmful/ionising radiation unnecessarily B1

or
with a longer half-life: a larger dose (of tracer) needed to produce detectable activity

9702/41 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

10(a) • redshift is the increase in observed wavelength / decrease in observed frequency (caused by Doppler effect) B3
• radiation from distant galaxies is observed to be redshifted
• redshift provides evidence that galaxies are moving apart
• galaxies moving apart means Universe must be expanding
Any three points, 1 mark each

10(b)(i) F = L / 4d2 C1

d = √(1.90  1036 / [4  8.42  10–16]) A1

= 1.34  1025 m

10(b)(ii)  /  = v / c C1

(726 – 658) / 658 = v / (3.00  108)

v = 3.1  107 m s–1 A1

10(c)(i) line with positive gradient passing through the origin B1

straight line with positive gradient B1

10(c)(ii) Hubble constant B1

Cambridge International AS & A Level

PHYSICS 9702/42
Paper 4 A Level Structured Questions October/November 2024
MARK SCHEME
Maximum Mark: 100

Published

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.

Cambridge International will not enter into discussions about these mark schemes.

This document consists of 14 printed pages.

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Generic Marking Principles

GENERIC MARKING PRINCIPLE 1:

Marks must be awarded in line with:

GENERIC MARKING PRINCIPLE 2:

Marks awarded are always whole marks (not half marks, or other fractions).

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:

GENERIC MARKING PRINCIPLE 4:

Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level
descriptors.

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
GENERIC MARKING PRINCIPLE 5:

GENERIC MARKING PRINCIPLE 6:

Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or
grade descriptors in mind.

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Abbreviations

/ Alternative and acceptable answers for the same marking point.

___ Underlined content must be present in answer to award the mark. This means either the exact word or another word that has the
same technical meaning.

Mark categories

B marks These are independent marks, which do not depend on other marks. For a B mark to be awarded, the point to which it refers must
be seen specifically in the candidate’s answer.

A marks These are answer marks. They may depend on an M mark or allow a C mark to be awarded by implication.

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED

Question Answer Marks

1(a)(i) v = r C1

= 0.85  140 A1

= 120 m s–1

1(a)(ii) a = r2 or a = v2 / r C1

a = 0.85  1402 or 1202 / 0.85 A1

= 1.7  104 m s–2

1(b)(i) direction of (induced) e.m.f. M1

is such as to (produce effects that) oppose the change that caused it A1

1(b)(ii) T = 2 /  A1

= 2 / 140 = 0.045 s = 45 ms

1(b)(iii)  = BA C1

= 0.18    0.852 C1

= 0.41 Wb A1

1(b)(iv) E = /t C1

= 0.41 / 0.045 A1

= 9.1 V

1(b)(v) force (on spoke) must be anticlockwise, so current is from A to X (by Fleming’s left hand rule), so X is at the higher potential B1

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

2(a) force per unit mass B1

2(b)(i) g = GM / x2 C1

= (6.67  10–11  1.99  1030) / (1.47  1011)2 A1

= 6.14  10–3 N kg– 1

2(b)(ii) EP = – GMm / x C1

= – (6.67  10–11  1.99  1030  2.63) / (1.47  1011)

= – 2.37  109 J A1

2(c)(i) F = L / 4x2 C1

(g = GM / x2 and so) x2 = GM / g M1

and

x2 = L / 4F

elimination of x and subsequent algebra shown leading to g = 4GMF / L A1

2(c)(ii) correct read-off of pair of values of g and F and full substitution of values of g, G, M and F into equation C1
e.g. L = (4  6.67  10–11  1.99  1030  1.83  103) / (8.0  10–3)

L = 3.8  1026 W A1

2(c)(iii) L = 4 r2T4 C1

3.8  1026 = (4  5.67  10–8  57804)  r2

r = 6.9  108 m A1

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

3(a) (thermal) energy per unit mass (to cause change of state) B1

(thermal) energy to change state at constant temperature B1

3(b)(i) W = pV C1

= 1.0  105  0.017 = 1700 J = 1.7 kJ A1

3(b)(ii) U = Q + W C1

Q = 17.6 + 1.7 A1

= 19.3 kJ

3(b)(iii) mass = 710  7.2  10–5 C1

( = 0.051 kg)

L = 19.3 / 0.051 A1

= 380 kJ kg–1

3(c) fusion involves (much) smaller volume change (than vaporisation) B1

smaller change in intermolecular spacing so smaller change in internal energy B1

negligible work done (by substance during fusion) so LF is less (than LV) B1

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

4(a) • molecules are in (constant) random motion B3

• (all) collisions between molecules are (perfectly) elastic
• no forces between molecules (except during collisions)
• volume of molecules is negligible (compared with volume of gas)
• collisions involving molecules are instantaneous
Any three points, 1 mark each

4(b) • molecules collide with (walls of) container B3

• momentum of molecule changes during collision (with walls)
• change in momentum is caused by force on molecule by wall
• molecule experiences force from wall so molecule exerts force on wall
• many molecules exerting force across the area of the wall leads to pressure (on the wall)
Any three points, 1 mark each

4(c) Any three bulleted points from: B3

• both gases are ideal

Up to 2 points from:
• mass of one molecule of gas X is 3.3  10–27 kg
• mass of one molecule of gas Y is 6.6  10–27 kg
• mass of one molecule of gas Y is double mass of one molecule of gas X

Up to 2 points from:
• sample of X contains 0.27 mol / 1.6  1023 molecules
• sample of Y contains 0.81 mol / 4.9  1023 molecules
• sample of Y contains treble the amount of gas / number of molecules as sample of X

Up to 2 points from:
• mass of gas X is 5.4  10–4 kg
• mass of gas Y is 3.2  10–3 kg
• mass of gas Y is six times mass of gas X

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

5(a) arrow from sphere, perpendicular to string, pointing left and down B1

5(b)(i) amplitude = 0.016 m A1

5(b)(ii) angular frequency = 2 / T C1

= 2 / 0.40 A1

= 16 rad s–1

5(b)(iii) total energy = ½m2x02 C1

= ½  0.15  15.72  0.0162 A1

= 4.7  10–3 J

5(c) dome-shaped curve starting and ending on the x-axis, with peak at x = 0 B1

maximum EK shown as 4.7  10–3 J B1

minimum x shown as –0.016 m and maximum x shown as +0.016 m at the ends of the line B1

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

6(a) (electric) force is (directly) proportional to product of charges B1

force (between point charges) is inversely proportional to the square of their separation B1

6(b) at least four straight, radial lines to/from surface of sphere B1

at least four straight radial lines drawn, approximately equally spaced B1

arrows pointing away from the surface of the sphere B1

6(c)(i) radius = 3.2 cm A1

6(c)(ii) E = Q / (40x2) C1

Q = e.g. 2.2  105  4  8.85  10–12  0.0322 C1

= 2.5  10–8 C A1

6(c)(iii) • the (positive) charge is all the way around the surface B1
• a charge placed inside the sphere is pulled equally in all directions
• if the field was not zero, the charges would move (until field is zero)
• electric field lines go from positive charge to negative charge, and there are no negative charges inside the sphere
Any point, 1 mark

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

7(a) charge / potential (difference) M1

charge is charge on one plate, and potential is p.d. between the plates A1

7(b)(i) straight line starting at the origin B1

line with positive gradient ending at (V, Q) B1

7(b)(ii) work done is the area under the graph B1

W = ½QV A1

7(c)(i) final p.d. shown as V / 4 for both capacitors B1

final charges add together to give Q B1

charge on Y = 3  charge on X (and both charges shown as a multiple of Q) B1

Fully correct answer:

X Y

final p.d. V/4 V/4

final charge Q/4 3Q / 4

7(c)(ii) less than B1

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

8(a) number of cycles per unit time B1

8(b)(i) period = 2 / 40 = 0.050 s = 50 ms A1

8(b)(ii) sinusoidal curve, starting at (0, 0) and initially increasing from there B1

periodic line showing 2 cycles with period 50 ms from t = 0 to t = 100 ms B1

all peaks shown at I = +3.5 A and all troughs shown at I = –3.5 A B1

8(b)(iii) Ir.m.s = 3.5 / √2 A1

= 2.5 A

8(c) P = I2R C1

peak power = 3.52  680 (= 8330 W) M1

or
mean power = 2.472  680 (= 4170 W)

peak and mean powers both calculated correctly, with supporting working, and compared leading to conclusion that mean A1
power is half the peak power

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Question Answer Marks

9(a) diffraction is characteristic of wave behaviour so shows that electrons can behave like waves B1

9(b) qV = ½mv2 C1

p = mv C1

p = m  √(2qV / m) A1

= √(2qVm)

9(c) (electrons have) greater momentum so smaller (de Broglie) wavelength B1

fringes become closer together B1

9(d)(i) straight line with positive gradient B1

line with positive gradient passing through the origin B1

9(d)(ii) Planck constant B1

9702/42 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Question Answer Marks

10(a)(i) cannot predict when a particular nucleus will decay B1

or
cannot predict which nucleus will decay next

10(a)(ii) (decay is) not affected by external (environmental) factors B1

10(a)(iii) fluctuations in (measured) count rate B1

10(b)(i) • large nuclei undergo fission whereas small nuclei undergo fusion B3
• fission involves one nucleus splitting into two (or more) (smaller) nuclei
• fusion involves two nuclei joining together to form one (larger) nucleus
• fission is (usually) initiated by neutron bombardment
• fusion is (usually) initiated by (very) high temperatures
Any three points, 1 mark each

10(b)(ii) binding energy per nucleon is greatest for intermediate nucleon numbers B1
(may be shown on sketch graph with axes labelled ‘binding energy per nucleon’ and ‘nucleon number’)

both fusion and fission involve an increase in binding energy (per nucleon) B1

Cambridge International AS & A Level

PHYSICS 9702/51
Paper 5 Planning, Analysis and Evaluation October/November 2024
MARK SCHEME
Maximum Mark: 30

Published

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.

Cambridge International will not enter into discussions about these mark schemes.

This document consists of 11 printed pages.

9702/51 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Generic Marking Principles

GENERIC MARKING PRINCIPLE 1:

Marks must be awarded in line with:

GENERIC MARKING PRINCIPLE 2:

Marks awarded are always whole marks (not half marks, or other fractions).

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:

GENERIC MARKING PRINCIPLE 4:

Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level
descriptors.

9702/51 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
GENERIC MARKING PRINCIPLE 5:

GENERIC MARKING PRINCIPLE 6:

Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or
grade descriptors in mind.

Science-Specific Marking Principles

1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks
should not be awarded if the keywords are used incorrectly.

5 ‘List rule’ guidance

For questions that require n responses (e.g. State two reasons …):

9702/51 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
6 Calculation specific guidance

Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show
your working’.

7 Guidance for chemical equations

Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme.

State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

9702/51 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Question Answer Marks

1 Defining the problem

s is the independent variable and v is the dependent variable or vary s and measure v 1

keep D constant 1

Methods of data collection

labelled diagram of workable experiment including: 1

• light gate positioned at P
• light gate connected to timer / data logger
• labels for light gate and P and data logger / timer and at least one other label from block, magnet(s), trolley, s and D

measure D with a rule(r) and measure L with a rule(r) or calipers 1

description to determine v at P, e.g. (measure length of) card to interrupt beam 1

method to measure s, e.g. use calipers 1

9702/51 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Question Answer Marks

1 Method of Analysis

1 1 1
plot a graph of v2 against 4
or equivalent (e.g. 4 against v2)
s s

Do not accept logarithms.

m  gradient 1
K=
2DA2B 2L2

m 1
(or K = for 4 against v2)
2DA B L  gradient
2 2 2
s

m  y -intercept 1
Q=−
2D

m  y -intercept 1
(or Q = KA2B 2L2  y -intercept or Q = for 4 against v2)
2D  gradient s

9702/51 Cambridge International AS & A Level – Mark Scheme October/November 2024
PUBLISHED
Question Answer Marks

1 Additional detail including safety considerations 6

D1 method to stop the trolley (after passing point P), e.g. labelled block / buffer / cushion drawn after P
or
place a block / buffer / cushion after P to stop the trolley

D2 keep L, A, m and B constant

d 2
D3 use micrometer / calipers to measure diameter (d) of the magnet and A =
4

D4 method to secure block to bench, e.g. clamp block to bench or (heavy) mass on top of block
or
method to secure magnets, e.g. use glue to stick magnets to trolley / block

D5 method to increase the accuracy of measuring s or D, e.g. use a marker to left of the trolley

D6 measure B using a (calibrated) Hall probe and adjust / rotate probe until maximum value
or
measure B using Hall probe first in one direction, then in the opposite direction and average

D7 use a (top-pan) balance to measure m

D8 use of strong magnets to increase v

D9 repeat measurements of v for each value of s and average v

D10 relationship valid if a straight line is produced (passing through  − 2DQ  )

 m 
Do not accept line passing through the origin.

9702/51 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

2(a) gradient = 3 1
3E − E s

y-intercept = 2Z
3E − E s

2(b) 1
1
/ A−1
I

5150 or 5155

5560 or 5556

5810 or 5814

6250

6670 or 6667

6940 or 6944
Values correct as shown above.

1 1
Uncertainties in from 50 or 60 to 90 or 100.
I

2(c)(i) Six points from (b) plotted correctly. 1

Must be within half a small square. Diameter of points must be less than half a small square.

1 1
Error bars in plotted correctly.
I
All error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.

9702/51 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

2(c)(ii) Straight line of best fit drawn. 1

Do not accept line from top point to bottom point.
Points must be balanced.
Line must pass between (1.63, 5400) and (1.67, 5400) and between (2.58, 6800) and (2.62, 6800).

Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1
All error bars must be plotted.

2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1
Distance between data points must be greater than half the length of the drawn line.

Gradient determined of worst acceptable line with clear substitution of data points into y / x. 1

uncertainty = (gradient of line of best fit – gradient of worst acceptable line)

or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)

2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x into y = mx + c. 1

y-intercept of worst acceptable line determined by substitution into y = mx + c. 1

uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line

or
uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)

Do not accept ECF from false origin method.

9702/51 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

2(d)(i) E determined using gradient 1

and
E and Z given to 2 or 3 or 4 significant figures.

1 3  3 + gradient  Es 1 E
E=  + Es  = = + s
3  gradient  3  gradient gradient 3

1
E= + 0.733
gradient

Z determined using y-intercept 1

and
E and Z given with SI units with correct powers of ten.

Z=
( 3E − Es )  y -intercept or 3  y -intercept
Z=
2 2  gradient

Unit of E: V
Unit of Z: 

2(d)(ii) Absolute uncertainty in E with method shown. 1

 gradient 1  0.05
uncertainty =   +
 gradient gradient  3
or

correct substitution for max/min methods.

9702/51 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

2(e) Value of R determined to a minimum of two significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution and 1
correct use of power of ten.

1
−6
− y -intercept
R= 250  10
gradient

1 2Z
R= −
gradient  250  10 −6 3

3E − 2.2 2Z
R= −
3  250  10−6 3

Cambridge International AS & A Level

PHYSICS 9702/52
Paper 5 Planning, Analysis and Evaluation October/November 2024
MARK SCHEME
Maximum Mark: 30

Published

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.

Cambridge International will not enter into discussions about these mark schemes.

This document consists of 10 printed pages.

9702/52 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Generic Marking Principles

GENERIC MARKING PRINCIPLE 1:

Marks must be awarded in line with:

GENERIC MARKING PRINCIPLE 2:

Marks awarded are always whole marks (not half marks, or other fractions).

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:

GENERIC MARKING PRINCIPLE 4:

Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level
descriptors.

9702/52 Cambridge International AS & A Level – Mark Scheme October/November 2024
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GENERIC MARKING PRINCIPLE 5:

GENERIC MARKING PRINCIPLE 6:

Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or
grade descriptors in mind.

Science-Specific Marking Principles

1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks
should not be awarded if the keywords are used incorrectly.

5 ‘List rule’ guidance

For questions that require n responses (e.g. State two reasons …):

9702/52 Cambridge International AS & A Level – Mark Scheme October/November 2024
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6 Calculation specific guidance

Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show
your working’.

7 Guidance for chemical equations

Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme.

State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

9702/52 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

1 Defining the problem

L is the independent variable and  or temperature change/increase is the dependent variable 1

or
vary L and measure  or temperature change/increase

keep t constant 1

Methods of data collection

labelled diagram of workable experiment including: 1

• oil in a beaker/container (on a bench)
• coil fully submerged in oil
• (bulb of) thermometer in the oil
• at least three labels from thermometer, coil or resistance wire, oil, beaker/container, clamp/stand, bench
Do not accept other heating sources.

method to determine V – diagram of workable circuit including: 1

• power supply connected to wire
• voltmeter positioned to measure V across the coil

measure the initial and final temperature and find the difference  1

method to determine t, e.g. use stopwatch/timer 1

and
method to determine L e.g. use a rule(r) to measure L / length of wire or e.g. using number of turns and measure the
diameter of the coil with rule(r) / calipers

9702/52 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

1 Method of Analysis

1 1 1
plot a graph of  against or equivalent, e.g. against 
L L
Do not accept logarithms.

1
1 1
for  against for against 
L L

AtV 2 AtV 2  gradient

K= K=
m  gradient m

1
1 1
for  against for against 
L L

Z = −mK  y -intercept Z = AtV 2  y -intercept

or
AtV  y -intercept
2
Z=−
gradient

9702/52 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

1 Additional detail including safety considerations 6

D1 precaution linked to hot oil / beaker / wire, e.g. use of gloves to prevent burns from oil
or
precaution linked to spillage of oil, e.g. perform experiment in a tray

D2 keep A and m and V constant

d 2
D3 use a micrometer to measure the diameter (d) of the wire and A =
4

D4 repeat measurements of d along the wire and average

D5 method to reduce heat loss e.g. add insulation around the container / add a lid to the container

D6 method to keep V constant, e.g. adjust / change a variable resistor / power supply to keep V or voltmeter reading
constant

D7 use a balance to determine the mass of the oil

and
mass of oil = mass of (beaker + oil) − mass of beaker
or
place beaker on balance and zero balance, then add oil and read balance

D8 stir the oil for uniform temperature

or
keep the initial temperature (of oil) constant

D9 repeat the experiment for the same value of L and average  / average temperature change

 Z 
D10 relationship valid if a straight line is produced (passing through  − )
 mK 
Do not accept line passing through the origin.

D11 method to determine L accurately, e.g. measure length of unwound coil

9702/52 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

2(a) gradient = Bn2 1

y-intercept = − B

2(b) 1
d2 / cm2

615 or 615.0

458 or 458.0

292 or 292.4

216 or 216.1

166 or 166.4

139 or 139.2
Values correct as shown above.

Uncertainties in d2 decreasing from 10 to 4 or 5. 1

2(c)(i) Six points from (b) plotted correctly. 1

Must be within half a small square. Diameter of points must be less than half a small square.

Error bars in d2 plotted correctly. 1

All error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.

2(c)(ii) Straight line of best fit drawn. 1

Do not accept line from top point to bottom point.
Points must be balanced.
Line must pass between (1.90, 250) and (2.00, 250) and between (3.55, 500) and (3.65, 500).

Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1
All error bars must be plotted.

9702/52 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1
Distance between data points must be greater than half the length of the drawn line.

Gradient determined of worst acceptable line with clear substitution of data points into y / x. 1

uncertainty = (gradient of line of best fit – gradient of worst acceptable line)

or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)

2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x into y = mx + c. 1

y-intercept of worst acceptable line determined by substitution into y = mx + c. 1

uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line

or
uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)

Do not accept ECF from false origin method.

2(d)(i) B determined using y-intercept (B = – y-intercept) and B and n given to 2 or 3 or 4 significant figures. 1

n determined using gradient 1

and
B and n given with SI units with correct powers of ten.

gradient gradient
n= or n =
B −y -intercept

Unit for B: cm2

No unit for n.

9702/52 Cambridge International AS & A Level – Mark Scheme October/November 2024
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Question Answer Marks

2(d)(ii) Percentage uncertainty in n determined with method shown. 1

1  y -intercept gradient 
percentage uncertainty =  +   100
2  y -intercept gradient 
or

correct substitution for max/min methods.

2(e)  determined to a minimum of two significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution and correct 1
power of ten.

gradient
 = sin−1
−y -intercept + 900
or
n 2B
 = sin−1
B + 900

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