A Level Physics 9702 Paper 22, Oct/Nov 2025

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9702/22 · Oct/Nov 2025

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1

Scientists are investigating the variation in air pressure at different locations on a mountain.

a

The scientists take measurements of several physical quantities at each location. Complete Table 1.1 by stating the SI base unit for each quantity and identifying with a tick (✓) whether each quantity is a scalar or a vector. Use the space for any working. Table 1.1

quantity measuredSI base unitscalarvector
air temperature
air pressure
[2]
b(i)

At one location, the density of the air is 1.1 kg m−3. A spherical weather balloon is filled with a gas and released from rest. The balloon has radius 0.90 m. Calculate the upthrust acting on the balloon when it is released. upthrust = ...................................................... N

[2]
b(ii)

Explain why an upthrust acts on the balloon.

[2]
b(iii)

The balloon has weight 19 N. Calculate the magnitude of the initial acceleration of the balloon. acceleration = ................................................ ms−2

[3]
c

A quantity c relating to the motion of the balloon is calculated from three measured quantities k, F and v using the formula c=2kFv2. The percentage uncertainties in the measured quantities are given in Table 1.2. Table 1.2

measured quantitypercentage uncertainty
k5%
F3%
v4%

The calculated value of c is 1.8. Determine the absolute uncertainty in c. absolute uncertainty = .........................................................

[2]
2

A spacecraft in deep space uses jets of hot gas from its thrusters to change its velocity. Fig. 2.1 shows a side view of the spacecraft and some of its thrusters. Thruster A is a distance of 1.6 m leftwards from the centre of gravity of the spacecraft. Thruster C is a distance of 0.40 m upwards from the centre of gravity of the spacecraft. Thrusters A and B can produce forces on the spacecraft in the upwards direction only. Thruster C can produce a force on the spacecraft in the leftwards direction only. All the thrusters shown produce forces entirely in the same plane as the centre of gravity.

Figure 1 of 3 for question 2
Figure 2 of 3 for question 2
Figure 3 of 3 for question 2
a(i)

Thruster A is activated, producing a force of 60 N upwards on the spacecraft. Thruster C is also activated, producing a force of 220 N in the leftwards direction on the spacecraft. Calculate the resultant moment due to these forces about the centre of gravity. resultant moment = ...................................................Nm

[2]
a(ii)

State and explain whether the forces from A and C are a couple.

[1]
b

Thrusters A and C are now switched off and the spacecraft is stationary. Thruster B is activated at time t1, producing a constant force on the spacecraft until the fuel runs out at time t2. As the fuel is used, the total mass of the spacecraft decreases. On Fig. 2.2, sketch the variation of speed of the spacecraft with time from t1 to t2.

[2]
c(i)

The spacecraft now splits apart into a carrier and a payload as shown in Fig. 2.3. During the split, an average force of 5500 N acts on the payload for a time of 0.36 s. The velocity of the payload increases by 8.5 ms−1 in the upwards direction. The combined mass of the carrier and payload is 2.5×103 kg. State the principle of conservation of momentum.

[2]
c(ii)

Show that the mass of the payload is 230 kg.

[2]
c(iii)

Calculate the magnitude of the change in velocity of the carrier. change in velocity = ................................................ ms−1

[3]
3

A spring is fixed at one end and attached to the frame of a pulley at the other end. A cable is passed around the wheel of the pulley. The spring is stretched to a fixed length using the cable and pulley. Fig. 3.1 shows the view from above of the spring, cable and pulley. The spring obeys Hooke’s law and has a spring constant k of 250 Nm−1. A force F acts on the spring. The tension in the cable is T. The pulley is in equilibrium.

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

On Fig. 3.2, draw labelled arrows to show the directions of the forces acting on the pulley.

[2]
b(i)

The force F is 110 N. Determine T. T = ...................................................... N

[1]
b(ii)

Calculate the extension of the spring. extension = ......................................................m

[2]
c(i)

A second identical spring with the same spring constant of 250 Nm−1 is now also connected to the pulley, as shown in Fig. 3.3. The tension in the cable is kept the same. The pulley is again in equilibrium. Determine the extension of the springs. extension = ..................................................... m

[2]
c(ii)

The elastic potential energy stored in the spring in Fig. 3.1 is E1. The total elastic potential energy stored in the two springs in Fig. 3.3 is E2. Calculate the ratio E1E2. ratio = .........................................................

[2]
4

A laser emits visible light of a single frequency in a vacuum. The light is incident normally on a double slit and then forms a pattern of bright and dark fringes on a screen, as shown in Fig. 4.1. The separation of the slits is 1.0×10−3 m. The distance from the slits to the screen is 4.8 m. The distance between the centres of adjacent dark fringes on the screen is 3.3 mm.

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

Explain how the pattern of bright and dark fringes is formed.

[3]
b

Calculate the frequency of the light emitted by the laser. frequency = .................................................... Hz

[4]
c

The double slit is removed. A second laser is placed beside the first laser. The second laser produces visible light of a different frequency from that of the first laser. The beams of light from the two lasers overlap on the screen. Explain why a steady pattern of bright and dark fringes is not formed on the screen.

[1]
5

Fig. 5.1 shows a circuit containing a battery, two fixed resistors X and Y, and a light-dependent resistor (LDR) Z. The battery has electromotive force (e.m.f.) 5.0 V and internal resistance 4.7 Ω. The current in X is I1 and the current in Y is I2. The resistance of X is 100 Ω. The resistance of Z varies with the intensity of light incident on it as shown in Fig. 5.2.

Figure 1 of 2 for question 5
Figure 2 of 2 for question 5
a

State Kirchhoff’s first law.

[1]
b(i)

The intensity of light incident on Z is 130 Wm−2. The current in the battery is 38 mA. Show that the terminal potential difference of the battery is 4.8 V.

[2]
b(ii)

Calculate the current I2 in Y. I2 = ...................................................... A

[3]
b(iii)

Calculate the power dissipated in Y. power = ..................................................... W

[2]
b(iv)

The intensity of the light incident on Z decreases. State and explain the effect on the terminal potential difference of the battery.

[3]
6
a

State what is meant by a fundamental particle.

[1]
b(i)

Particle Q is a meson with a charge of 0. Determine a possible quark composition for Q.

[2]
b(ii)

Particle Q has a mass of 0.67 u and a kinetic energy of 2.1×10−16 J. Calculate the speed of particle Q. speed = ................................................ ms−1

[3]
c(i)

Radium-228 (88228Ra) is a radioactive nuclide. State the number of electrons in a neutral atom of radium-228. number of electrons = .........................................................

[1]
c(ii)

A nucleus of radium-228 undergoes a series of decays to form nucleus X. During the process, 5 α-particles and 4 β− particles are emitted. Determine the number of protons and the number of neutrons in nucleus X. number of protons = ............................................................... number of neutrons = ...............................................................

[2]

Mark scheme

9702/22 · Oct/Nov 2025

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1a · 2 marks

  • B1

    air temperature: K and air pressure: kg m–1 s–2

  • B1

    scalar only ticked for both air temperature and air pressure

1b(i) · 2 marks

  • C1

    upthrust = 1.1  9.81  (4  0.90³ / 3)

  • A1

    = 33 N

1b(ii) · 2 marks

  • B1

    (due to difference in height / depth there is a) difference in pressure between top and bottom (of balloon)

  • B1

    (due to pressure difference, upwards) force on bottom of balloon is greater (than downwards force on top of balloon, so resultant force is upwards)

1b(iii) · 3 marks

  • C1

    ()F = 33 – 19 (= 14 N)

  • C1

    m = 19 / 9.81 (= 1.94 kg)

  • A1

    a = (33 – 19) / (19 / 9.81) = 7.2 m s–2

1c · 2 marks

  • C1

    5 + 3 + (2  4) (= 16%)

  • A1

    absolute uncertainty in c = 1.8  0.16 = (±) 0.3

2a(i) · 2 marks

  • C1

    60  1.6 (= 96 N m) or 220  0.40 (= 88 N m)

  • A1

    resultant moment = 96 – 88 = 8.0 N m

2a(ii) · 1 mark

  • B1

    the resultant force (of A and C) is not zero or forces (from A and C) are not parallel or forces (from A and C) are not equal (magnitude) or forces (from A and C) are not opposite (direction) or forces (from A and C) act through the same point so (they are) not a couple

2b · 2 marks

  • B1

    line with a positive gradient from (t1, 0) to t2

  • B1

    line with increasing positive gradient from t1 to t2

2c(i) · 2 marks

  • M1

    sum / total momentum before = sum / total momentum after or sum / total momentum (of a system of objects) is constant

  • A1

    if no (resultant) external force / for an isolated system

2c(ii) · 2 marks

  • C1

    ()p = F()t or F = ma and a = ()v / ()t

  • A1

    m = Ft / ()v (m =) 5500 × 0.36 / 8.5 = 230 (kg) or (m =) 5500 / (8.5 / 0.36) = 230 (kg)

2c(iii) · 3 marks

  • C1

    (p =) 5500  0.36 (= 1980 N s) or (p =) 230  8.5 (= 1955 N s) or a = 5500 / (2.5  10³ – 230) (= 2.42 m s–2)

  • C1

    1980 = (2.5  10³ – 230)v or 1955 = (2.5  10³ – 230)v or v = 2.42  0.36

  • A1

    v = 0.87 m s–1 or 0.86 m s–1

3a · 2 marks

  • B1

    an arrow horizontally on the page to the left labelled F

  • B1

    two arrows horizontally on the page to the right each labelled T

3b(i) · 1 mark

  • A1

    T = 110 / 2 = 55 N

3b(ii) · 2 marks

  • C1

    x = F / k

  • A1

    = 110 / 250 = 0.44 m

3c(i) · 2 marks

  • C1

    extension = 55 / 250 or 110 / (2  250)

  • A1

    extension = 0.22 m

3c(ii) · 2 marks

  • C1

    E = 1/2 Fx or E = 1/2 kx² or E = 1/2 F² / k

  • A1

    E1 / E2 = (1/2  110  0.44) / (2  (1/2  55  10.22)) or E1 / E2 = (1/2 k  0.44²) / (2  (1/2 k  0.22²)) or E1 / E2 = (1/2  110² / k) / (2  (1/2  55² / k)) E1 / E2 = 2.0 (no ECF from 3(b)(ii) and 3(c)(i))

4a · 3 marks

  • B1

    light diffracts / spreads (at slit(s))

  • B1

    light (from each slit) superposes / interferes (at screen)

  • B1

    when phase difference is 0 / path difference is n (where n is an integer) a bright fringe is formed or when phase difference is 180(°) / path difference is (n + 1) / 2 (where n is an integer) a dark fringe is formed

4b · 4 marks

  • C1

     = ax / D

  • C1

     = 1.0  10–3  3.3  10–3 / 4.8 (= 6.875  10–7 m)

  • C1

    f = v / 

  • A1

    = 3.0  10⁸ / 6.875  10–7 = 4.4  10¹⁴ Hz

4c · 1 mark

  • B1

    the light / beams (have different frequencies so) are not coherent / do not have constant phase difference

5a · 1 mark

  • B1

    sum of current(s) into junction = sum of current(s) out junction or (algebraic) sum of current(s) at a junction is zero

5b(i) · 2 marks

  • C1

    (V =) E – Ir

  • A1

    (V =) 5.0 – (38  10–3  4.7) = 4.8 (V)

5b(ii) · 3 marks

  • C1

    R(Z) = 120 

  • C1

    I2 = 38 × 10–3 – (4.8 / (100 + 120)) or R(EXT) = 4.8 / 38 × 10–3 (= 126 ) 1 / R(Y) = 1 / 126 – 1 / (120 + 100) (R(Y)= 297 ) I2 = 4.8 / 297

  • A1

    I2 = 0.016 A

5b(iii) · 2 marks

  • C1

    P = IV or P = I²R or P = V² / R

  • A1

    P = 0.016  4.8 or P = 0.016²  (4.8 / 0.016) or P = 4.8² / (4.8 / 0.016) P = 0.077 W

5b(iv) · 3 marks

  • B1

    resistance of the LDR / Z increases (as light intensity decreases)

  • M1

    total resistance (of the circuit) increases or current in the battery decreases

  • A1

    (potential difference across the internal resistor decreases so) the terminal potential difference increases

6a · 1 mark

  • B1

    (a particle that) cannot be divided / subdivided (into smaller particles)

6b(i) · 2 marks

  • C1

    any combination of one quark and one antiquark

  • A1

    up / charm / top and antiup / anticharm / antitop or down / strange / bottom and antidown / antistrange / antibottom

6b(ii) · 3 marks

  • C1

    E(K) = 1/2 mv²

  • C1

    2.1  10–16 = 1/2  0.67  1.66  10–27  v²

  • A1

    v = 6.1  10⁵ m s–1

6c(i) · 1 mark

  • A1

    number of electrons = 88

6c(ii) · 2 marks

  • A1

    (number of nucleons = 228 – 5 × 4 = 208) number of protons = 88 – (5 × 2) + 4 = 82

  • A1

    number of neutrons= 208 – 82 = 126 = 126

Questions

  1. 1
    Question 111 marks
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  2. 2
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  3. 3
    Question 39 marks
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  4. 4
    Question 48 marks
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  5. 5
    Question 511 marks
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  6. 6
    Question 69 marks
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