Cambridge AS & A Level Physics 9702/22 is the May/June 2026 Paper 2 (AS Level Structured Questions), variant 2. It is an AS component worth 60 marks, it lasts 1 hour 15 minutes, and candidates answer all six questions on the question paper. The biggest question was Question 5, a 12-mark circuits question on parallel resistors, drift speed and an LDR. More than half the marks were for calculations. 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.

9702/22 at a glance

Question	Topic	Marks	Main command word
1	Kinematics: scalars and vectors, a car’s velocity-time graph, displacement sketch	11	Calculate, sketch
2	Equilibrium of a plank and block, Hooke’s law, elastic deformation	8	State, show that
3	Progressive waves, two-source interference, energy of a floating ball	10	Calculate, explain
4	Work, energy and power: an electric train on a slope	10	Calculate, determine
5	Parallel resistors, drift speed, an LDR	12	Derive, calculate
6	Alpha-particle scattering, momentum, uncertainty, quarks	9	Show that, calculate
	Total	60

Question-by-question walkthrough

Question 1: the car’s velocity-time graph (11 marks)

The graph showed a car speeding up to 12 m s⁻¹ in 10 s, travelling at constant velocity until 35 s, stopping at 40 s, then moving backwards with a steady negative acceleration to −10 m s⁻¹ at 80 s.

• Scalars and vectors (2 marks): amplitude and power are scalars (1 mark); force and momentum are vectors (1 mark).
• Distance from the graph (1 + 1 marks): the area under the line gives the distance. From 0 to 40 s it is ½ × 10 × 12 + 25 × 12 + ½ × 5 × 12 = 390 m.
• Maximum acceleration (2 marks): the steepest section is the braking from 35 s to 40 s: (0 − 12) ÷ 5 gives a magnitude of 2.4 m s⁻². A minus sign was allowed. One mark was for a = Δv ÷ t or the gradient with numbers in.
• Speed back at point P (2 marks): at 40 s the car is 150 m past P and at rest, then accelerates back at 0.25 m s⁻². v² = 2 × 0.25 × 150 = 75, so the speed is 8.7 m s⁻¹. The mark scheme gave no mark for a symbol equation on its own; the first mark was for the full substitution. Four other routes (finding t = 34.6 s first) were accepted, but rounding t to 35 s and getting 8.8 or 8.6 m s⁻¹ lost the answer mark.
• Displacement sketch (3 marks): one mark for a line that rises and then falls without crossing the time axis, one for the maximum at t = 40 s, and one for the right shape: gradient increasing from 0 to 10 s, constant from 10 to 35 s, decreasing from 35 to 40 s and decreasing from 40 to 80 s. A step change in displacement scored nothing. The line stays above the axis because the car comes back only 200 m of the 390 m it travelled.

Question 2: the plank and the block (8 marks)

A block sat on the centre of a plank. Ropes from the block passed over two pulleys and down to the ends of the plank, all at angle θ to the vertical.

• Conditions for equilibrium (2 marks): resultant force is zero (1 mark) and resultant torque, or moment, about any point is zero (1 mark). “Forces balance”, “forces cancel”, “moments balance” and “no moments” were ignored, and writing “momentum” for “moment” scored nothing.
• Show that R + W_P = W_B − R (2 marks): write one full force equation, either 2T cos θ = R + W_P for the plank or 2T cos θ + R = W_B for the block (first mark). The second mark needs both equations and then equating them. Using sin θ instead of cos θ throughout earned 1 of the 2 marks.
• Calculate R (2 marks): 2R = 8.1g − 4.5g = 3.6 × 9.81, so R = 18 N. Using g = 10 was allowed.
• Name of the law (1 mark): Hooke’s law.
• Elastic deformation (1 mark): when the tension is removed, the rope returns to its original length (the extension is zero). “Mass” was ignored as a stand-in for tension.

Question 3: interference of water waves (10 marks)

Two dippers, in phase at 3.0 Hz, sent out circular waves. Point P was 3.9 m from dipper A and 2.4 m from dipper B.

• Progressive wave (1 mark): a wave that transfers energy. References to the medium were ignored.
• Period (1 mark): T = 1/f = 0.33 s.
• Wave speed (3 marks): the path difference at P is 3.9 − 2.4 = 1.5 m, which is one wavelength (1 mark). Then v = fλ = 3.0 × 1.5 = 4.5 m s⁻¹ (1 mark for the equation, 1 for the answer). An error in the period was carried forward.
• Change in potential energy of a floating ball (3 marks): the ball moves between −15 cm and +15 cm, so the height difference is twice the amplitude, 0.30 m. ΔE = mgΔh = 0.44 × 9.81 × 0.30 = 1.3 J. Using 0.15 m earned only the first mark.
• Raising the frequency to 4.0 Hz (2 marks): the wavelength changes, so the path difference at P is no longer a whole number of wavelengths and the waves arriving at P are not in phase (1 mark). The amplitude of the ball decreases (1 mark). “Out of phase” or “not complete constructive interference” was accepted for the first mark, but “180° out of phase” or “antiphase” was not.

Question 4: the electric train (10 marks)

A train of mass 1.5 × 10⁴ kg climbed a slope at constant speed, its potential energy rising from 7.4 MJ to 13.0 MJ in 52 s. Its kinetic energy was 1.3 × 10⁵ J.

• Speed (3 marks): ½mv² = 1.3 × 10⁵, so v² = 2 × 1.3 × 10⁵ ÷ 1.5 × 10⁴ and v = 4.2 m s⁻¹. This is where the symbol equation pays: if you wrote E = ½mv² first and then slipped to v instead of v², you could still score 2 out of 3. With no symbol equation and the same slip, the scheme gave 0.
• Define power (1 mark): work done per unit time. “Rate of doing work” and “energy transferred per unit time” were also accepted. “Work in unit time” and “energy/time” alone scored nothing.
• Resistive force (3 marks): the motor’s useful work goes partly into potential energy and partly against the resistive force. Useful work = 1.1 × 10⁵ × 52 = 5.72 × 10⁶ J; minus the 5.6 × 10⁶ J potential energy gain leaves 1.2 × 10⁵ J. The distance is 4.2 × 52 m, so F = 1.2 × 10⁵ ÷ (4.2 × 52) = 550 N. Two power-based methods reaching the same value were also credited. The scheme accepted 555 N but no other value that rounds to 560 N, and failing to convert both MJ values to joules counted as one power-of-ten error.
• Efficiency (3 marks): input power = IV = 95 × 1400 = 1.33 × 10⁵ W (1 mark). Efficiency = useful output ÷ input × 100 = 1.1 × 10⁵ ÷ 1.33 × 10⁵ × 100 = 83%. Taking the useful output as the potential energy gain per second (1.08 × 10⁵ W) earned the two method marks but gave 81%, which lost the answer mark.

Question 5: parallel resistors, drift speed and an LDR (12 marks)

• Derive the parallel formula (2 marks): state that I_T = I₁ + I₂ + I₃ (1 mark), then substitute I = V/R with the same V across each resistor and divide by V to reach 1/R_T = 1/R₁ + 1/R₂ + 1/R₃ (1 mark). If you used separate symbols V₁, V₂, V₃, you had to say they are equal. A correct derivation for only two resistors earned 1 mark.
• Current in the battery (2 marks): two 2.8 Ω wires in parallel give 1.4 Ω, so I = 12 ÷ 1.4 = 8.6 A. Adding 4.3 A + 4.3 A also works. An answer of 4.3 A with no working earned 1 mark.
• Drift speed (2 marks): each wire carries half the current, 4.3 A. v = I ÷ (Anq) = 4.3 ÷ (6.1 × 10⁻⁶ × 8.5 × 10²⁸ × 1.6 × 10⁻¹⁹) = 5.2 × 10⁻⁵ m s⁻¹. Writing the symbol formula alone scored nothing, because it is on the formula sheet. Using the full 8.6 A in one wire (giving 1.0 × 10⁻⁴ m s⁻¹) earned 1 mark with working shown.
• Name of component X (1 mark): light-dependent resistor (LDR).
• Resistance of X (3 marks): the bare wire still takes 12 ÷ 2.8 = 4.3 A, so the branch with X carries 5.0 − 4.3 = 0.71 A (1 mark). That branch has resistance 12 ÷ 0.71 = 16.8 Ω (1 mark), and X = 16.8 − 2.8 = 14 Ω. Two other routes, through the p.d. across the wire or the total resistance of 2.4 Ω, reached the same answer.
• How the environment changed (2 marks): the battery current fell, so the resistance of X increased (1 mark, which must be seen). That means the light intensity on X decreased (1 mark). The scheme also allowed a fall in temperature.

Question 6: alpha particles and quarks (9 marks)

An alpha particle at 5.60 × 10⁶ m s⁻¹ bounced straight back at 5.38 × 10⁶ m s⁻¹ from a gold nucleus of mass 197u.

• Evidence that the atom is mostly empty space (1 mark): most alpha particles passed straight through, or were deviated by only very small angles. “Pass through” without “straight”, and “some” instead of “most”, were ignored.
• Show that the momentum is 3.72 × 10⁻²⁰ kg m s⁻¹ (2 marks): p = mv = 4 × 1.66 × 10⁻²⁷ × 5.60 × 10⁶ = 3.72 × 10⁻²⁰ kg m s⁻¹. Every step had to be seen, and using 1.67 × 10⁻²⁷ kg lost the second mark.
• Speed of the gold nucleus (2 marks): momentum before = momentum after, with the alpha particle’s final velocity negative: 4u × 5.60 × 10⁶ = 197u × v − 4u × 5.38 × 10⁶. This gives v = 2.2 × 10⁵ m s⁻¹. Treating the rebounding alpha particle’s velocity as positive gives 4500 m s⁻¹, which earned 1 mark with working shown. Assuming an elastic collision (speed of approach = speed of separation) was also credited.
• Percentage uncertainty (2 marks): the change in velocity is 5.60 × 10⁶ + 5.38 × 10⁶ = 1.098 × 10⁷ m s⁻¹, because the direction reverses. Its absolute uncertainty is 5 × 10⁴ + 5 × 10⁴ = 1 × 10⁵ m s⁻¹ (1 mark), so the percentage uncertainty is 0.9%.
• Up quarks in a hydrogen-2 nucleus (1 mark): one proton (uud) and one neutron (udd), so 3.
• Quark composition of a meson (1 mark): one quark and one antiquark. “Two quarks” was ignored, and naming particular quarks was not needed.

Where the marks are on 9702/22

Type of task	Parts	Marks
Calculate or determine	1(b)(ii)–(iv), 2(b)(ii), 3(b), 3(c)(i), 4(a), 4(b)(ii)–(iii), 5(b), 5(c)(ii), 6(b)(ii)–(iii), 6(c)	35
State or define	1(a), 1(b)(i), 2(a), 2(c), 3(a), 4(b)(i), 5(c)(i), 6(a), 6(d)	12
Show that or derive	2(b)(i), 5(a), 6(b)(i)	6
Explain	3(c)(ii), 5(c)(iii)	4
Sketch	1(b)(v)	3
Total		60

• Calculations carried 35 of the 60 marks, so writing the equation, then the substitution, then the answer with its unit is the habit that protects the most marks.
• Definitions and recall were worth 12 marks: scalars and vectors, equilibrium, Hooke’s law, elastic deformation, progressive wave, power, the LDR, the scattering observation and the meson.
• The two “explain” parts, 3(c)(ii) and 5(c)(iii), each needed a reason and a conclusion. In 5(c)(iii), the reason mark was mandatory: without “the resistance of X increased”, the light-intensity mark could not be given.

How to self-mark 9702/22 with the mark scheme

1. Learn the mark types first. B marks are independent. M marks must be seen, and the A mark that follows depends on them. C marks are method marks that you get automatically if the final answer is right. A marks are for the answer.
2. Check the final answer, then work backwards. If your answer is right, take all the C marks. If it is wrong, look for each C mark in your working: the equation, the substitution, the intermediate value.
3. Check units and significant figures. A missing or wrong unit normally costs the final mark. Answers are rounded to the scheme’s significant figures, and over-rounding mid-calculation (as with t = 35 s in 1(b)(iv)) can push you off the accepted value.
4. Read the guidance column for “Ignore” and “Not”. That is where “forces balance”, “work in unit time” and “180° out of phase” are ruled out.
5. Apply ecf where the scheme allows it. Your period from 3(b)(i) and your speed from 4(a) carried forward. Your current from 5(b)(i) also carried into 5(b)(ii) and 5(c)(ii).
6. Treat “show that” strictly. In 2(b)(i) and 6(b)(i), the given result earns nothing unless every step that leads to it is written down.
7. Log each lost mark as missing knowledge, a definition worded wrongly or a calculation slip, and revise in that order.

Sit 9702/22 as a timed mock

The paper allows 75 minutes for 60 marks, a little over one minute per mark. A workable plan:

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

Question 1 gets extra time because the displacement sketch needs care: mark the maximum at 40 s first, then draw each section’s gradient.

Other May/June 2026 9702 papers

• 9702 Paper 12 (Multiple Choice)
• 9702 Paper 32 (Advanced Practical Skills)
• 9702 Paper 42 (A Level Structured Questions)
• 9702 Paper 52 (Planning, Analysis and Evaluation)

Practise and download

• Download 9702/22 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

Need help with a 9702 topic? Book a free trial lesson with an A Level Physics tutor →

Frequently asked questions

What topics were on 9702 Paper 22 May/June 2026? +

Six questions: a car's velocity-time graph; equilibrium of a plank and block hung from ropes, with Hooke's law; interference of water waves from two dippers; the power and efficiency of an electric train on a slope; parallel resistors, drift speed and an LDR; and alpha-particle scattering, momentum and quarks.

How many marks is 9702/22 and how long is it? +

It is out of 60 marks and lasts 1 hour 15 minutes. All six questions are compulsory, and Question 5, on circuits, is the largest at 12 marks.

What was the answer to the resistive force question on 9702/22? +

550 N. The motor does 1.1 × 10⁵ × 52 = 5.72 × 10⁶ J of useful work, 5.6 × 10⁶ J of which becomes potential energy. The remaining 1.2 × 10⁵ J is done against friction over 4.2 × 52 m, giving F = 550 N. The mark scheme accepted 555 N but not 560 N.

What was the efficiency of the train's motor in 9702/22? +

83%. The input power is 95 × 1400 = 1.33 × 10⁵ W, and the useful output is 1.1 × 10⁵ W. Using the potential energy gain per second as the output instead gives 81%, which loses the answer mark.

Is 9702/22 an AS or A Level paper? +

It is Paper 2, the AS Level structured questions paper. AS candidates sit it with Papers 1 and 3, and full A Level candidates also sit Papers 4 and 5.

Where can I download 9702/22 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.

Ready to Excel in Your Studies?

Get personalised help from Tutopiya's expert tutors. Whether it's IGCSE, IB, A-Levels, or any other curriculum — we match you with the perfect tutor and your first session is free.

Book Your Free Trial