Cambridge AS & A Level Physics 9702/42 is the May/June 2026 Paper 4 (A Level Structured Questions), variant 2. It lasts 2 hours, is worth 100 marks and is sat by every full A Level candidate. It had ten compulsory questions, and the largest, on the first law of thermodynamics, carried 14 marks. This guide goes through the paper question by question and shows what the official mark scheme gave marks for. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of all five 9702 papers from this series, see A Level Physics 9702 May/June 2026 past papers: what came up and how it was marked.

9702/42 at a glance

Question	Topic	Marks	Main command word
1	Circular motion: an aircraft banking in a horizontal circle	10	Explain, calculate, determine
2	Gravitational fields: field lines, mass of the Earth, a satellite’s potential energy	9	Define, calculate, determine
3	Thermal physics: a gas cooled at constant pressure, the first law, specific heat capacity	14	Show, explain, determine
4	Kinetic theory: one particle in a cube, building up pV = ⅓Nm⟨c²⟩	10	State, determine, explain
5	Simple harmonic motion: a block on a spring, amplitude, kinetic energy sketch	8	Show, calculate, sketch
6	Electric fields: potential of a charged sphere, a Van de Graaff dome as a capacitor	13	Define, sketch, calculate
7	The Hall probe: faces, symbols and the origin of the Hall voltage	8	Identify, state, explain
8	Alternating current: rectification and smoothing from two graphs	7	State, determine
9	Quantum physics: de Broglie wavelength, wave and particle evidence, an accelerated electron	10	State, complete, calculate
10	Nuclear physics and astronomy: binding energy, fusion in a star, radiant flux intensity	11	Define, show, determine
	Total	100

The data and formulae pages at the front of the paper give the constants used below, such as g = 9.81 m s⁻², G = 6.67 × 10⁻¹¹ N m² kg⁻² and 1 u = 1.66 × 10⁻²⁷ kg.

Question-by-question walkthrough

The mark scheme uses four kinds of mark. B marks are independent points. C marks are for working, and a correct final answer earns all of them automatically (AFC, automatic full credit). M marks must be seen, and the A mark that follows is lost without them. A marks are for the answer.

Question 1: the banking aircraft (10 marks)

An aircraft flies at 180 m s⁻¹ in a horizontal circle of radius 7200 m, with its wings tilted so that the lift force is not vertical.

• Describing uniform circular motion (2 marks): one mark for the acceleration being perpendicular to the velocity (or directed towards the centre, with the velocity tangential), and one for the velocity and acceleration having constant magnitude, or for both changing direction. “Constant speed” with no mention of magnitude was not enough.
• Why the lift must tilt (2 marks): the only other force, the weight, is vertical, so the vertical component of lift balances it (1); the lift therefore needs a horizontal component to provide the centripetal acceleration (1).
• Direction of the resultant force (1 mark): an arrow from the aircraft pointing horizontally to the left in the diagram (towards the centre of the circle), within 5°. Any other arrow through the aircraft counted as a contradiction unless the labels made clear which one was the resultant.
• Centripetal acceleration (2 marks): a = v²/r = 180² ÷ 7200 = 4.5 m s⁻².
• Angle of tilt (3 marks): one mark for the reasoning (vertical component of lift balances the weight, horizontal component gives the centripetal acceleration, or mg = L cos θ with ma = L sin θ), then tan θ = a/g = 4.5 ÷ 9.81, so θ ≈ 25° (24.6°). Your acceleration from the previous part carried forward.

Question 2: gravitational fields (9 marks)

• Definition of gravitational field (1 mark): force per unit mass. The ratio had to be clear.
• Field lines near the ground (2 marks): at least four straight, parallel lines perpendicular to the ground and roughly evenly spaced (1), with arrows pointing towards the ground (1). Every arrow drawn had to point down.
• Mass of the Earth (2 marks): from g = GM/R², M = 9.81 × (6.37 × 10⁶)² ÷ 6.67 × 10⁻¹¹ = 5.97 × 10²⁴ kg. The question asked for three significant figures, and the answer had to be given to three.
• Gravitational potential at the surface (2 marks): φ = −GM/R, or −gR, giving −6.25 × 10⁷ J kg⁻¹. The A mark needed three significant figures, the minus sign and a unit; N m kg⁻¹ and m² s⁻² were also accepted.
• Potential energy of a 96.0 kg satellite at a height of 1.93 × 10⁶ m (2 marks): E = −GMm/r with r = (6.37 + 1.93) × 10⁶ m, giving −4.61 × 10⁹ J. Using only the Earth’s radius or only the height as r lost the method mark.

Question 3: the first law of thermodynamics (14 marks)

An ideal gas of 6.00 × 10²³ molecules (each 6.64 × 10⁻²⁷ kg) at 1.01 × 10⁵ Pa, 0.0243 m³, 296 K and internal energy 3676 J is cooled at constant pressure until its volume is 80% of the start value.

• Definition of specific heat capacity (2 marks): thermal energy per unit mass (1) per unit change in temperature (1). “Thermal energy per unit (mass × temperature change)” earned both. Using “energy” without “thermal” in an otherwise correct answer earned 1.
• Show the work done is 491 J (2 marks): W = pΔV = 1.01 × 10⁵ × 0.20 × 0.0243 = 491 J. The full substitution had to be seen.
• Sign of the work done on the gas (1 mark): positive, because the volume decreases (the gas is compressed).
• Why the internal energy change is −735 J (2 marks): internal energy is proportional to thermodynamic temperature, and at constant pressure temperature is proportional to volume (1), so the internal energy falls to 80% of 3676 J, a change of −0.20 × 3676 = −735 J (1). The alternative route used U = (3/2)NkT with a temperature fall of 0.20 × 296 K. Both marks had to come from the same method.
• Thermal energy transferred (2 marks): from ΔU = Q + W, −735 = Q + 491, so Q = −1226 J. The answer had to be negative.
• Specific heat capacity for this change (2 marks): the temperature fall is 0.20 × 296 ≈ 59 K (1); c = 1226 ÷ (6.00 × 10²³ × 6.64 × 10⁻²⁷ × 59) ≈ 5200 J kg⁻¹ K⁻¹ (1). Using 296 K or 237 K as the temperature change lost the first mark.
• Heating back to 296 K at constant volume (3 marks): the temperature change is the same, so the change in internal energy is the same (1); at constant volume no work is done on or by the gas (1); so less thermal energy is needed for the same temperature change, and the specific heat capacity is less (1).

Question 4: kinetic theory (10 marks)

• Two assumptions about collisions (2 marks): collisions are instantaneous (or last a negligible time compared with the time between them), and collisions are elastic. The list rule applied.
• Single particle in a cube of side L (3 marks, 1 each): change in momentum at each collision 2mu; time between collisions with the same face 2L/u; mean force Δp/Δt = mu²/L.
• Pressure on a face (2 marks): one mark for using Newton’s third law, so the force of the particle on the face equals the force of the face on the particle; then p = F/A = (mu²/L)/L² = mu²/L³.
• Getting to pV = ⅓Nm⟨c²⟩ (3 marks): the volume of the cube L³ is V, so pV = mu² (1); the particles move in three dimensions, so u² becomes ⅓⟨c²⟩ (1); and there are N particles, so the expression is multiplied by N (1). Algebra alone did not score the second and third marks. You had to say why each change is made.

Question 5: simple harmonic motion (8 marks)

A 27 g block oscillates on a spring with a period of 1.7 s and a total energy of 13 μJ.

• Show ω = 3.7 rad s⁻¹ (1 mark): ω = 2π/T = 2π ÷ 1.7 = 3.7 rad s⁻¹, with the working shown.
• Meaning of amplitude (1 mark): the maximum displacement from the equilibrium position. “Mean position” was accepted, “rest position” was not.
• Amplitude (3 marks): E = ½mω²x₀², so 13 × 10⁻⁶ = ½ × 0.027 × 3.7² × x₀², giving x₀ = 8.4 × 10⁻³ m. Note the energy is in microjoules.
• Sketch of kinetic energy against displacement (3 marks): a single dome-shaped curve peaking at 13 μJ at x = 0 (1); symmetrical about the vertical axis, with the ends at zero kinetic energy at equal distances either side (1); and an x-axis scale that puts those ends at ±8.4 × 10⁻³ m, with at least one scale marking on each side (1).

Question 6: electric potential and a Van de Graaff dome (13 marks)

• Definition of electric potential (2 marks): work done per unit charge (1) in moving a positive charge from infinity to the point (1).
• Potential at the surface of a sphere (1 mark): V₀ = Q/4πε₀R, with ε₀ named as the permittivity of free space.
• Sketch from R to 3R (3 marks): a curve starting at V₀ whose magnitude and gradient both decrease (1); passing through ½V₀ at 2R and ⅓V₀ at 3R (1); and staying entirely in the positive region (1).
• Charge on the dome (2 marks): with R = 0.119 m and V = −4.80 × 10⁴ V, Q = 4πε₀RV = −6.35 × 10⁻⁷ C. The answer had to be negative.
• Why the dome acts as a capacitor (1 mark): it stores charge (or energy).
• Capacitance (2 marks): C = Q/V, or 4πε₀R, = 1.32 × 10⁻¹¹ F (13.2 pF), and the unit was needed for the A mark.
• Energy stored (2 marks): E = ½QV = ½ × 6.35 × 10⁻⁷ × 4.80 × 10⁴ = 0.015 J (0.0152 J).

Question 7: the Hall probe (8 marks)

Current passes through a slice from face WXYZ to face PQRS, with the magnetic field perpendicular to faces WXQP and ZYRS.

• The faces with the Hall voltage (1 mark): PSZW and QRYX. All four letters were needed for each face.
• Symbols in V_H = BI/ntq (3 marks): n is the number density of charge carriers (number per unit volume; “density” or “number” alone was rejected); q is the charge on each carrier (“elementary charge” alone was not enough); t is the thickness of the slice.
• Origin of the Hall voltage (3 marks, any three): the charge carriers feel a force from the magnetic field; the force is perpendicular to both I and B; the carriers are deflected towards one face; the separated charge creates a p.d. between the faces; a steady state is reached when the magnetic force is balanced by the force from the resulting electric field.
• Which face is at the higher potential (1 mark): the carriers (electrons) are deflected towards QRYX, so PSZW is at the higher potential. Saying that positive carriers would make QRYX higher was also accepted.

Question 8: rectification and smoothing (7 marks)

Two graphs showed a sinusoidal input voltage and a rectified, smoothed output over 0.06 s.

• Components (2 marks): a diode rectifies and a capacitor smooths.
• Peak input voltage (1 mark): 3.6 V.
• Period of the input (1 mark): 0.040 s.
• Three further conclusions (3 marks): any three of the rectification being half-wave; an input frequency of 25 Hz (or angular frequency of about 160 rad s⁻¹); an r.m.s. input voltage of 2.5 V (3.6 ÷ √2 = 2.546 V; 2.55 V was allowed but 2.6 V was not, because it is rounded twice); a time constant of about 0.058 s (0.056 to 0.059 s allowed); a ripple of 1.6 V or 44%. Numbers needed a unit and two significant figures. Simply describing the graphs or reading values off them scored nothing.

Question 9: wave-particle duality (10 marks)

• The de Broglie wavelength (1 mark): the wavelength associated with a moving particle.
• Table of evidence (4 marks, one per cell):
  • wave nature of matter: electron diffraction or interference;
  • particle nature of matter: for example Brownian motion, states of matter, quantisation of charge, radioactive decay, Rutherford scattering or matter–antimatter annihilation;
  • wave nature of electromagnetic radiation: diffraction, interference or polarisation of light (or the Doppler effect in light);
  • particle nature of electromagnetic radiation: the photoelectric effect (emission or absorption spectra and Compton scattering were also accepted).
• Speed of an electron accelerated through 3.5 kV (3 marks): eV = ½mv², so 1.60 × 10⁻¹⁹ × 3500 = ½ × 9.11 × 10⁻³¹ × v², giving v = 3.5 × 10⁷ m s⁻¹.
• Its de Broglie wavelength (2 marks): λ = h/mv = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 3.5 × 10⁷) = 2.1 × 10⁻¹¹ m.

Question 10: binding energy and a star (11 marks)

Hydrogen fuses to helium in a star, with the overall reaction producing two particles X. The mass defect of helium-4 is 0.030 377 u.

• Definition of binding energy (2 marks): the energy needed to separate all the nucleons of the nucleus (M1) to infinity (A1). Describing a single nucleon was rejected.
• Particle X (1 mark): a positron (beta-plus). “Beta” alone was incomplete.
• Show the binding energy is 4.54 × 10⁻¹² J (2 marks): E = Δmc² = 0.030 377 × 1.66 × 10⁻²⁷ × (3.00 × 10⁸)², with the full substitution seen.
• Radiant flux intensity at 8.15 × 10¹⁶ m (2 marks): F = L/4πd² = 7.98 × 10²⁷ ÷ (4π × (8.15 × 10¹⁶)²) = 9.56 × 10⁻⁸ W m⁻².
• Rate of loss of hydrogen (4 marks): three independent C marks, then the answer. Helium-4 nuclei formed per second = 7.98 × 10²⁷ ÷ 4.54 × 10⁻¹² ≈ 1.76 × 10³⁹ (1); hydrogen nuclei used = 4 × that ≈ 7.03 × 10³⁹ s⁻¹ (1); mass = 1.67 × 10⁻²⁷ kg × that number (1); so the rate is 1.17 × 10¹³ kg s⁻¹ (1). Including the mass defect in the mass calculation lost the mark.

Where the marks are on 9702/42

Type of task	Marks	Where
Calculations, including “show that” and graph read-offs	42	1(b)(iii)–(iv), 2(b), 3(b)(i), 3(b)(iv)–(v), 5(a)(i), 5(a)(iii), 6(c)(i), 6(c)(iii)–(iv), 8(b), 9(c), 10(b)(ii), 10(c)
Explanations	19	1(b)(i), 3(b)(ii)–(iii), 3(c), 4(c), 6(c)(ii), 7(b), 8(c)
Recall statements and identifications	15	1(a), 4(a), 7(a), 8(a), 9(b), 10(b)(i)
Definitions	9	2(a)(i), 3(a), 5(a)(ii), 6(a), 9(a), 10(a)
Sketches and drawings	9	1(b)(ii), 2(a)(ii), 5(b), 6(b)(ii)
Derivations in symbols	6	4(b), 6(b)(i)
Total	100

• Calculations carry 42 marks, so careless arithmetic, missing units and lost minus signs cost more than anything else.
• Definitions are 9 quick marks. Every one needed a clear ratio (“per unit mass”, “per unit charge”) or a specific condition (“to infinity”).
• Explanations need the reason, not just the result. In 4(c), the algebra alone did not earn the marks; each step needed a stated reason.

How to self-mark 9702/42 with the mark scheme

1. Learn the abbreviations first. A slash separates alternative answers, words in brackets are context that need not be written, and an underlined word must appear. AFC means a correct final answer gets all the C marks. ecf means error carried forward, and the scheme says where it is allowed (“Possible ECF from 3(b)(iv)”) and where it is not (“No ECF from 5(a)(i)”).
2. Check the sign before anything else. In 2(b)(ii), 2(b)(iii), 3(b)(iv) and 6(c)(i), the A mark needed a negative answer.
3. Check the significant figures. 2(b)(i) and 2(b)(ii) had to be to three; most other answers needed at least two or three, as stated in the guidance column.
4. Award C marks only from your working if your final answer is wrong. If you wrote no equation and slipped, you can’t claim them.
5. Mark “show that” parts strictly. In 3(b)(i), 5(a)(i) and 10(b)(ii), the full substitution had to be visible.
6. Apply the list rule to “any three” parts (7(b)(i) and 8(c)): mark only your first three points, and ignore points that just describe the graphs.
7. Mark sketches point by point, checking shape, key values and scale separately in 5(b) and 6(b)(ii).

Sit 9702/42 as a timed mock

The paper allows 120 minutes for 100 marks, a little over one minute per mark. A workable plan:

Block	Marks	Time
Question 1	10	12 minutes
Question 2	9	11 minutes
Question 3	14	16 minutes
Question 4	10	12 minutes
Question 5	8	10 minutes
Question 6	13	15 minutes
Question 7	8	9 minutes
Question 8	7	8 minutes
Question 9	10	12 minutes
Question 10	11	13 minutes
Check signs, units and significant figures	–	2 minutes

Write the symbol equation before every substitution, so that a slip still leaves you the C marks.

Other May/June 2026 9702 papers

• 9702 Paper 12 (Multiple Choice)
• 9702 Paper 22 (AS Level Structured Questions)
• 9702 Paper 32 (Advanced Practical Skills)
• 9702 Paper 52 (Planning, Analysis and Evaluation)

Practise and download

• Download 9702/42 and its mark scheme on the Learning Portal →
• All 9702 May/June 2026 papers: overview and marking patterns
• Try a free A Level Physics quiz

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Frequently asked questions

What topics were on 9702 Paper 42 May/June 2026? +

Ten questions: circular motion of a banking aircraft; gravitational fields and a satellite; specific heat capacity and the first law of thermodynamics; kinetic theory; simple harmonic motion of a block on a spring; electric potential and a Van de Graaff dome; the Hall probe; rectification and smoothing; the de Broglie wavelength; and binding energy and fusion in a star.

How many marks is 9702/42 and how long is it? +

It is out of 100 marks and lasts 2 hours. All ten questions are compulsory, and Question 3, on thermodynamics, is the biggest at 14 marks.

What was the specific heat capacity answer in 9702/42 Question 3? +

The internal energy fell by 735 J and 491 J of work was done on the gas, so the thermal energy transferred was Q = −735 − 491 = −1226 J. The temperature fell by 0.20 × 296 = 59 K, giving c = 1226 ÷ (6.00 × 10²³ × 6.64 × 10⁻²⁷ × 59) ≈ 5200 J kg⁻¹ K⁻¹.

What was the rate of loss of hydrogen in 9702/42 Question 10? +

Each helium-4 nucleus releases 4.54 × 10⁻¹² J, so the star makes 7.98 × 10²⁷ ÷ 4.54 × 10⁻¹² ≈ 1.76 × 10³⁹ nuclei per second. Four hydrogen nuclei are used for each one, so the mass lost is 4 × 1.67 × 10⁻²⁷ × 1.76 × 10³⁹ ≈ 1.17 × 10¹³ kg s⁻¹.

Which answers on 9702/42 had to be negative? +

The gravitational potential at the Earth's surface (−6.25 × 10⁷ J kg⁻¹), the satellite's potential energy (−4.61 × 10⁹ J), the thermal energy transferred to the gas (−1226 J) and the charge on the dome (−6.35 × 10⁻⁷ C). The A mark needed the minus sign in each case.

Where can I download 9702/42 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge AS & A Level Physics 9702 past papers page.

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