A Level Physics 9702 Paper 21, May/June 2025

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Question paper

9702/21 · May/June 2025

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1
Figure 1 of 2 for question 1
Figure 2 of 2 for question 1
a

Define acceleration.

[1]
b(i)

In an experiment, two objects A and B are released from the side of a building, as shown in Fig. 1.1. Object A is released from rest at a height of 14 m above the horizontal ground. Object B is released with an initial upwards vertical velocity u at a height of 3.6 m above the ground. Both objects take the same time to reach the ground and they do not collide with each other. Air resistance is negligible. Calculate the time taken for object A to reach the ground.

[2]
b(ii)

Use your answer in (b)(i) to calculate u.

[2]
c(i)

In a second experiment, object B is released from the same height and given the same initial speed as in (b) but at a release angle θ to the vertical, as shown in Fig. 1.2. State and explain whether the time taken for object B to reach the ground is less than, the same as or greater than the time in (b)(i).

[2]
c(ii)

By considering energy, state and explain the effect of the change in release angle on the speed at which B reaches the ground.

[2]
2
Figure 1 of 2 for question 2
Figure 2 of 2 for question 2
a

State the principle of moments.

[2]
b

Three objects A, B and C are placed on a horizontal beam. The beam is in equilibrium, as shown in Fig. 2.1. The beam is uniform and has length 6.0 m. The pivot is at the midpoint of the beam. Object A has mass 60 kg and is at one end of the beam. Object B has mass 45 kg and is at a distance x from the pivot. Object C has mass 80 kg and is at the other end of the beam. Calculate x.

[3]
c(i)

The beam is 0.80 m above horizontal ground. Object A is removed and replaced by a spring connected to the ground and the beam, as shown in Fig. 2.2. After the change, the beam is again horizontal and in equilibrium. The positions of B and C are unchanged. The spring has an unstretched length of 0.59 m and obeys Hooke's law. Calculate the spring constant of the spring.

[3]
c(ii)

Calculate the elastic potential energy of the spring.

[2]
3
Figure 1 of 2 for question 3
Figure 2 of 2 for question 3
a

Define power.

[1]
b(i)

An electric car is powered by a motor. The car is travelling at a constant speed of 35 m s−1 along a straight horizontal road, as shown in Fig. 3.1. There is a total resistive force of 1750 N acting on the car. Calculate the power transmitted to the wheels of the car by the motor.

[2]
b(ii)

Calculate the useful work done by the motor when the car travels a distance of 17 km.

[2]
b(iii)

The potential difference (p.d.) across the motor has a constant value of 600 V and the motor has an efficiency of 85%. Calculate the current in the motor.

[3]
c(i)

The car in (b) now reaches a slope, as shown in Fig. 3.2. The car continues down the slope at the same speed as in (b). State and explain the effect, if any, of the slope on: the air resistance acting on the car

[1]
c(ii)

the current in the motor.

[1]
4
Figure 1 of 3 for question 4
Figure 2 of 3 for question 4
Figure 3 of 3 for question 4
a

State the principle of superposition.

[2]
b(i)

Light of wavelength 7.2×10−7 m is incident normally on a double slit, as shown in Fig. 4.1. A screen is at a distance D from the double slit. The double slit and the screen are parallel. The separation of the slits in the double-slit arrangement is 0.16 mm. The resulting interference pattern on the screen contains nine dark fringes, as shown in Fig. 4.2. The distance between the centres of the first and ninth dark fringes is 3.2 cm. Calculate D.

[3]
b(ii)

The slit separation is now gradually decreased from 0.16 mm to 0.04 mm. The distance between the centres of adjacent dark fringes is x. On Fig. 4.3, sketch the variation of x with slit separation.

[3]
5
Figure 1 of 2 for question 5
Figure 2 of 2 for question 5
a

Use the definitions of speed v, frequency f and wavelength λ to derive the wave equation v=fλ.

[2]
b(i)

A source of sound waves of frequency 236 Hz is travelling at a constant velocity of 20 m s−1. A stationary observer has a microphone connected to a cathode-ray oscilloscope (CRO). The microphone detects the sound waves as the source moves directly towards the observer. The resulting trace on the CRO is shown in Fig. 5.1. The time-base on the CRO is set to 1.0 ms div−1. Calculate the frequency of the sound waves detected by the microphone.

[2]
b(ii)

Determine the speed of the sound in air.

[2]
6

A nichrome wire X of length 45 cm and cross-sectional area 4.7×10−7 m2 is connected into the circuit shown in Fig. 6.1. The resistance of X is 1.1 Ω. The cell has electromotive force (e.m.f.) 1.3 V and negligible internal resistance.

Figure 1 of 2 for question 6
Figure 2 of 2 for question 6
a(i)

Calculate the current in X.

[1]
a(ii)

The number density of charge carriers (electrons) in nichrome is 8.5×1028 m−3. Calculate the average drift speed of the charge carriers in X.

[2]
a(iii)

Calculate the resistivity of the nichrome.

[3]
b(i)

Wire Y is identical to wire X. Wire Y is added to the circuit in parallel with wire X, as shown in Fig. 6.2. State and explain the effect, if any, of this change on: the reading on the ammeter

[2]
b(ii)

the average drift speed of the charge carriers in X.

[1]
7
a

Nitrogen-12 (712N) is an unstable isotope of nitrogen that decays by the emission of radiation to carbon-12 (612C). Complete the full nuclear equation for the decay, including all the particles involved. 712N→

[3]
b(i)

Hydrogen-1 (11H) is an isotope of hydrogen. An atom of hydrogen-1 comprises a proton with an orbiting electron. An antiparticle equivalent of hydrogen-1 comprises an antiproton with an orbiting positron. The antiquarks in the antiproton are the antiparticles of the quarks in a proton. State the charge on the positron in terms of the elementary charge e.

[1]
b(ii)

State the group (class) of fundamental particle to which the positron belongs.

[1]
b(iii)

In Table 7.1, state the flavour and charge of the three antiquarks that comprise the antiproton. Table 7.1

[3]

Mark scheme

9702/21 · May/June 2025

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1a · 1 mark

  • B1

    rate of change of velocity

1b(i) · 2 marks

  • C1

    s = ut + ½at2 and u = 0 or s = ½at2 t = (2  14 / 9.81)

  • A1

    t = 1.7 s

1b(ii) · 2 marks

  • C1

    s = ut + ½at2 u = (3.6 – ½  9.81  1.72) / 1.7

  • A1

    u = (–) 6.2 m s–1

1c(i) · 2 marks

  • B1

    Less time (as it reaches a lower height)

  • B1

    (because) the initial vertical (component of the) velocity is smaller (than in part (b))

1c(ii) · 2 marks

  • B1

    The (total) initial energy is the same (as in part (b))

  • B1

    change in gravitational potential energy is same, so speed is the same

2a · 2 marks

  • B1

    in (rotational) equilibrium

  • B1

    sum / total of clockwise moments about a point = sum / total of anticlockwise moments about the (same) point.

2b · 3 marks

  • C1

    80  9.81  3 or 60  9.81  3 or 45  9.81  x

  • C1

    80  9.81  3 = (60  9.81  3) + (45  9.81  x)

  • A1

    x = 1.3 m

2c(i) · 3 marks

  • C1

    k = F / x

  • C1

    x = 0.80 – 0.59 = 0.21 m

  • A1

    k = (60  9.81) / 0.21 = 2800 N m−1

2c(ii) · 2 marks

  • C1

    E = ½ kx2 or E = ½ Fx or E = ½ F2/k = ½  2800  0.212 or = ½  60  9.81  0.21 or = ½  (60  9.81)2 / 2800

  • A1

    = 62 J

3a · 1 mark

  • B1

    work done per unit time

3b(i) · 2 marks

  • C1

    P = F  v = 1750  35

  • A1

    = 6.1  104 W

3b(ii) · 2 marks

  • C1

    W = Fs = 1750  17 000

  • A1

    = 3.0  107 J

3b(iii) · 3 marks

  • C1

    P = V  I Power in = 600  I

  • C1

    Efficiency = useful power output / (total) power input

  • A1

    0.85 = 6.1  104 / (600  I) I = 120 A

3c(i) · 1 mark

  • B1

    Air resistance is the same, as the speed is the same

3c(ii) · 1 mark

  • B1

    The motor is producing less power (because of gravitational force / conversion of gravitational potential energy to kinetic energy) so the current will be smaller.

4a · 2 marks

  • B1

    (when two or more) waves meet/overlap (at a point)

  • B1

    (resultant) displacement is sum of the individual displacements

4b(i) · 3 marks

  • C1

    Fringe width, x = 3.2  10−2 / 8 = 4.0  10−3 (m)

  • C1

    D = ax /  = (4.0  10−3  0.16  10−3) / 7.2  10−7

  • A1

    = 0.89 m

4b(ii) · 3 marks

  • B1

    Curved line with a negative gradient of decreasing magnitude throughout, from slit separation 0.04 mm to 0.16 mm

  • B1

    Line of negative gradient ending at (0.16, 0.4), from slit separation 0.04 mm

  • B1

    Line of negative gradient passing through (0.08, 0.8) and (0.04, 1.6)

5a · 2 marks

  • B1

    wavelength: - wavelength = distance between successive / adjacent in phase points / wavefronts / crests / troughs -  = d / (number of) oscillations frequency: - frequency = (number of) oscillations / cycles crests / troughs / wavefronts (passing a point) per unit time - f = (number of) oscillations / t One correct point from either list

  • B1

    One correct point from both lists and speed = distance / time and one of: - wavelength  frequency (= distance per unit time) = speed - [(number of) oscillations / t ]  [d / (number of) oscillations] = f - v (= d / t) =  / (1/f) = f or v (= d / t) =  / T = f

5b(i) · 2 marks

  • C1

    T = 4  10−3 f = 1/T = 1 / 0.004

  • A1

    f = 250 Hz

5b(ii) · 2 marks

  • C1

    fo = fs v / (v – vs) 250 = 236  v / (v – 20) v = (250  20) / (250 – 236)

  • A1

    = 360 m s−1

6a(i) · 1 mark

  • A1

    I = 1.3 / 1.1 = 1.2 A

6a(ii) · 2 marks

  • C1

    v = I / nqA = 1.2 / (8.5  1028  1.60  10−19  4.7  10−7)

  • A1

    = 1.9  10−4 m s−1

6a(iii) · 3 marks

  • C1

     = RA / L

  • C1

    = (1.1  4.7  10−7) / 0.45

  • A1

    = 1.1  10−6  m

6b(i) · 2 marks

  • M1

    (Total) resistance decreases (and the potential difference stays the same)

  • A1

    (so the reading on the ammeter) increases

6b(ii) · 1 mark

  • B1

    (The average drift speed will be) the same because the current is the same (in X).

7a · 3 marks

  • B1

    beta-plus shown

  • B1

    neutrino shown

  • B1

    symbols, nucleon numbers and proton numbers all correct

7b(i) · 1 mark

  • B1

    +1

7b(ii) · 1 mark

  • B1

    Lepton(s)

7b(iii) · 3 marks

  • B1

    3 correct quark flavours

  • B1

    Charge on anti-up quark –⅔(e)

  • B1

    Charge on anti-down quark (+)⅓(e)

Questions

  1. 1
    Question 19 marks
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  2. 2
    Question 210 marks
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  3. 3
    Question 310 marks
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  4. 4
    Question 48 marks
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  5. 5
    Question 56 marks
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  6. 6
    Question 69 marks
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  7. 7
    Question 78 marks
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