A Level Physics 9702 Paper 23, May/June 2026

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Question paper

9702/23 · May/June 2026

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

Define momentum.

[1]
b(i)

A ball X of mass 0.15 kg moves along a horizontal frictionless surface. The ball collides elastically with ball Y that is initially stationary. Balls X and Y have equal diameters but different masses. Figure 1.1 shows the variation with time of the velocity of ball X before, during and after the collision. Determine the magnitude of the acceleration of ball X during the collision. acceleration = ................................................ ms−2

[2]
b(ii)

Determine the magnitude of the change in momentum of ball X due to the collision. change in momentum = ............................................ kg ms−1

[2]
b(iii)

Determine the total distance travelled by ball X during the collision. distance = ......................................................m

[2]
c(i)

By considering relative speeds, explain why the velocity of ball Y after the collision must be +0.80 ms−1.

[2]
c(ii)

On Figure 1.1, sketch the variation of the velocity of ball Y with time between time=0 and time=0.20 s.

[2]
c(iii)

Calculate the mass of ball Y. mass = ..................................................... kg

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

Define the Young modulus of a material.

[1]
b(i)

A long, thin thread of high-density polythene has a uniform cross-section. The thread is suspended from one end so that it hangs vertically. The thread has a mass of 0.018 kg and a cross-sectional area of 2.6×10−6 m2. High-density polythene has a density of 960 kg m−3 and a Young modulus of 8.0×108 Pa. Show that the length of the thread is 7.2 m.

[2]
b(ii)

Show that the stress at the top of the thread due to its own weight is 6.8×104 Pa.

[2]
b(iii)

Figure 2.1 shows the variation with distance from the top of the thread of the stress in the thread due to its own weight. Suggest an explanation for the fact that the variation shown in Figure 2.1 is linear.

[2]
c(i)

The thread in 2(b) is now laid horizontally on a frictionless bench. A tensile force is applied to the ends of the thread so that there is a uniform stress throughout the thread of 6.8×104 Pa. Calculate the extension of the thread. extension = ......................................................m

[2]
c(ii)

Calculate the elastic potential energy of the thread. elastic potential energy = ....................................................... J

[2]
d

Suggest, with a reason, whether the total extension of the vertically suspended thread in 2(b) is less than, equal to or greater than the answer in 2(c)(i).

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

State the principle of superposition.

[2]
b(i)

Parallel rays of coherent light of wavelength 560 nm are incident normally in free space on two narrow slits as shown in Figure 3.1. Light passing through the double slit reaches a screen that is parallel to the plane of the slits. The distance between the centres of the slits is 1.4×10−5 m. The distance between the slits and the screen is 0.43 m. State what is meant by coherent.

[1]
b(ii)

State what happens to the light at the point where it passes through the slits.

[1]
b(iii)

At point X on the screen, there is a central bright fringe. Explain how this bright fringe is formed.

[3]
b(iv)

At point Y on the screen, the dark fringe closest to point X is formed. Calculate the distance XY. distance = ......................................................m

[3]
c(i)

On the screen in Figure 3.1: draw a cross (×) labelled with the letter B at the position of another bright fringe

[1]
c(ii)

draw a cross (×) labelled with the letter D at the position of another dark fringe.

[1]
d

Calculate the frequency of the light in 3(b). frequency = .................................................... Hz

[2]
4

A light-dependent resistor (LDR) is connected across the terminals of a cell that has internal resistance. A voltmeter is connected across the LDR.

a

Draw a circuit diagram to show this arrangement.

[3]
b(i)

The cell has an electromotive force (e.m.f.) of 4.90 V. At a particular light intensity, the LDR has a resistance of 230 Ω and a potential difference (p.d.) across it of 4.78 V. Identify one possible source of systematic error in measuring the p.d. across the LDR using a digital voltmeter.

[1]
b(ii)

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

[2]
b(iii)

Determine the internal resistance of the cell. internal resistance = ...................................................... Ω

[2]
b(iv)

The intensity of the light illuminating the LDR is gradually decreased. By placing one tick (✓) in each row, complete Table 4.1 to show what happens to the quantities indicated as the light intensity decreases. Table 4.1

decreasesunchangedincreases
resistance of LDR
current in cell
p.d. across LDR
[3]
5
a

State what is meant by a meson.

[1]
b

The elementary charge is e. Explain, in terms of quark composition, how a baryon may have an overall charge of: • −1e .................................................................................................................................... ........................................................................................................................................... • +2e .................................................................................................................................... ...........................................................................................................................................

[3]
c(i)

The nuclide 818X undergoes β+ decay to produce nuclide QPY and one other particle Z, as shown in the equation 818X→ QPY+ SRβ++ 00Z State the values of P, Q, R and S. P = ............................................................... Q = ............................................................... R = ............................................................... S = ...............................................................

[2]
c(ii)

State the name of particle Z.

[1]
c(iii)

State the net change in quark composition as nuclide X turns into nuclide Y.

[1]
c(iv)

Explain why the emission of particle Z results in the β+ particles from the decays of different nuclei of 818X having a continuous range of kinetic energies.

[2]

Mark scheme

9702/23 · May/June 2026

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1a · 1 mark

  • B1

    product of mass and velocity

1b(i) · 2 marks

  • C1

    a = Δv / (Δ)t or =(v – u) / (Δ)t or = gradient of slope

  • A1

    acceleration = ( –0.8 – 1.6) / (0.12 – 0.09) = (–)80 m s–2

1b(ii) · 2 marks

  • C1

    0.15  1.6 or 0.15  0.8

  • A1

    change in momentum = 0.15  ( –0.8 – 1.6) = (–)0.36 kg m s–1

1b(iii) · 2 marks

  • C1

    ½1.60  0.020 or 1.62 / (2  80) or 1.6  0.02 – ½  80  0.022 or ½  80  0.022 =0.016

  • A1

    distance = 0.016 + 0.004 = 0.020 m

1c(i) · 2 marks

  • B1

    relative speed of approach = relative speed of separation

  • B1

    (speed of separation = 1.6 m s–1 so) vY – (–0.80) = 1.6, so vY = 0.80 m s–1

1c(ii) · 2 marks

  • B1

    a straight horizontal line from (0, 0) to (0.09, 0) or a straight horizontal line from (0.12, 0.8) to (0.20, 0.8)

  • B1

    Both horizontal lines correct and straight diagonal line from (0.09, 0) to (0.12, 0.8)

1c(iii) · 2 marks

  • C1

    (0.15  1.6) = (0.15  (–0.80)) + 0.80m 0.36 = 0.8m

  • A1

    mass = 0.45 kg

2a · 1 mark

  • B1

    stress per unit strain

2b(i) · 2 marks

  • C1

     = m / V V = 0.018 / 960 = 1.875  10–5

  • A1

    (length =) 0.018 / (960  2.6  10–6) = 7.2 (m)

2b(ii) · 2 marks

  • C1

    ( =) F / A

  • A1

    = (0.018  9.81) / (2.6  10–6) = 6.8  104 (Pa)

2b(iii) · 2 marks

  • B1

    weight (of thread supported at a point) decreases as distance (from top of thread) increases

  • B1

    (uniform cross-section so) stress (at a point in the thread) is proportional to the weight (of thread below the point)

2c(i) · 2 marks

  • C1

     = x / L or E =  /  or x = L / E

  • A1

    extension = 7.2  (6.8  104) / (8.0  108) = 6.1  10–4 m

2c(ii) · 2 marks

  • C1

    E = ½Fx or E = ½kx2 or E = ½F2 / k = ½  6.8  104  2.6  10–6  6.1  10–4 or = ½  0.018  9.81  6.1  10–4 or = ½  (6.8  104  2.6  10–6/6.1  10–4 )  (6.1  10–4)2 or = ½  (0.018  9.81/6.1  10–4 )  (6.1  10–4)2 or = ½  (6.8  104  2.6  10–6)2 / (6.8  104  2.6  10–6/ 6.1  10–4) or = ½  (0.018  9.81)2 / (0.018  9.81/ 6.1  10–4)

  • A1

    = 5.4  10–5 J

2d · 1 mark

  • B1

    (extension in 2(b) is) less (than in 2(c)(i)) because stress (in vertical thread) is less (than 6.8  104 Pa everywhere except the top)

3a · 2 marks

  • B1

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

  • B1

    (resultant) displacement is sum of the individual displacements

3b(i) · 1 mark

  • B1

    constant phase difference

3b(ii) · 1 mark

  • B1

    (the light) diffracts

3b(iii) · 3 marks

  • B1

    waves (from each slit) meet / superpose (at the screen)

  • B1

    waves have 0 path difference / 0 phase difference

  • B1

    constructive interference occurs / (resultant wave has) maximum amplitude

3b(iv) · 3 marks

  • C1

     = ax / D

  • C1

    x = (560  10–9  0.43) / (1.4  10–5) = 0.017

  • A1

    distance = 0.017 / 2 = 0.0086 m

3c(i) · 1 mark

  • B1

    either: B marked above Y at same distance above Y as XY or: B marked below 0.43 m arrow, at a distance from X that is 2XY

3c(ii) · 1 mark

  • B1

    D marked below X at distance below X equal to XY

3d · 2 marks

  • C1

    c = f

  • A1

    frequency = (3.00  108) / (560  10–9) = 5.4  1014 Hz

4a · 3 marks

  • B1

    correct circuit symbol used for cell with internal resistance

  • B1

    correct symbol for LDR connected to terminals of cell

  • B1

    correct symbol for voltmeter connected in parallel with LDR

4b(i) · 1 mark

  • B1

    zero error / calibration error

4b(ii) · 2 marks

  • C1

    P = V2 / R or P = VI and V = IR or P = I2R and V = IR = 4.782 / 230 or = 4.78  (4.78 / 230) or = (4.78 / 230)2  230

  • A1

    = 0.099 W

4b(iii) · 2 marks

  • C1

    E = V + Ir r = (4.90 – 4.78)  230 / 4.78 or 4.78 / 4.90 = 230 / (230 + r) or (4.90 – 4.78) / 4.90 = r / (230 + r) or (4.90 – 4.78) / 4.78 = r / 230 or 4.90 / (4.78 / 230) = r + 230

  • A1

    internal resistance = 5.8 

4b(iv) · 3 marks

  • B1

    resistance of LDR increases

  • B1

    current in cell decreases

  • B1

    p.d. across LDR increases

5a · 1 mark

  • B1

    (a hadron consisting of) one quark and one antiquark

5b · 3 marks

  • C1

    quark charges have magnitudes of either ⅔(e) or ⅓(e)

  • A1

    –1e : either: –⅓(e), –⅓(e) and –⅓(e) or: –⅔(e), –⅔(e) and +⅓(e)

  • A1

    +2e: +⅔(e), +⅔(e) and +⅔(e)

5c(i) · 2 marks

  • A1

    P = 18 and R = 0

  • A1

    Q = 7 and S = (+)1

5c(ii) · 1 mark

  • B1

    (electron) neutrino

5c(iii) · 1 mark

  • B1

    up to down

5c(iv) · 2 marks

  • B1

    total energy emitted (in each decay) is constant

  • B1

    particle Z takes some of the energy

  • B1

    the energy (emitted) is shared between Z and the + particle

  • B1

    the way the total energy is distributed is different for each decay

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