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CIE A-Level Physics 9702 Paper 2 Practice Paper (AS Structured) – 60 marks, 1 hour 15 minutes, answers and mark scheme included
TutorZone original practice paper – CIE A-Level Physics 9702 Paper 2 (AS structured)
6 questions, 23 parts – 60 marks – 1 hour 15 minutes – calculator allowed
Paper structure
| Paper | Time | Marks | Weighting |
|---|---|---|---|
| Paper 2 (AS structured) | 1 h 15 min | 60 marks | 46% of AS |
| Paper 1 (multiple choice) | 1 h 15 min | 40 marks | 31% of AS |
| Paper 3 (AS practical skills) | 2 h | 40 marks | 23% of AS |
Questions
Total: 60 marks · 1 hour 15 minutes · calculator allowed
Q 1 · Kinematics [10]
(a) Distinguish between a scalar quantity and a vector quantity, giving one physics example of each. [2]
(b) A car accelerates uniformly from 8.0 m/s to 26.0 m/s in 6.0 s. Calculate its acceleration and the distance travelled. [3]
(c) A ball is released from rest and falls 45 m (g = 9.81 m/s2, air resistance negligible). Calculate the time of fall and the speed on landing. [3]
(d) A velocity-time graph rises uniformly from 0 to 12 m/s in 4.0 s, then stays at 12 m/s until t = 10.0 s. Find the total displacement. [2]
Q 2 · Forces and momentum [10]
(a) State Newton’s second law and give the condition under which F = ma applies. [2]
(b) A car of mass 1200 kg experiences a resultant force of 3.6 kN. Calculate its acceleration. [2]
(c) Trolley A (2.0 kg) moving at 3.0 m/s collides with stationary trolley B (1.0 kg) and they move off together. Find the common velocity and the kinetic energy lost. [4]
(d) Explain why kinetic energy is not conserved in part (c). [2]
Q 3 · Work, energy and materials [10]
(a) A pump raises 250 kg of water through 12 m in 20 s. If the efficiency is 70%, calculate the output power and the input power (g = 9.81 m/s2). [4]
(b) A spring extends 0.060 m under a force of 12 N (within the elastic limit). Find its spring constant and the energy stored. [3]
(c) A wire of length 2.50 m and diameter 0.40 mm extends by 1.2 mm under a load of 8.0 N. Calculate the Young modulus. [3]
Q 4 · Waves [10]
(a) A sound wave has frequency 512 Hz and speed 340 m/s. Calculate its wavelength. [2]
(b) Describe the difference between a transverse wave and a longitudinal wave, with one example of each. [2]
(c) Two coherent microwave sources are 0.12 m apart, wavelength 0.030 m, screen 2.0 m away. Calculate the fringe spacing. [3]
(d) A string of length 0.80 m has a fundamental frequency of 250 Hz. Find the frequency and wavelength of the third harmonic. [3]
Q 5 · Electricity and d.c. circuits [12]
(a) Distinguish between electromotive force (e.m.f.) and potential difference (p.d.). [2]
(b) A wire of length 1.5 m and cross-sectional area 2.0×10^-7 m2 has resistance 0.90 ohm. Calculate its resistivity. [2]
(c) A cell of e.m.f. 6.0 V and internal resistance 0.50 ohm is connected to a 2.5 ohm resistor. Find the current, terminal p.d., power in the external resistor and power dissipated internally. [4]
(d) A 12 V supply is in series with a 4.0 kohm fixed resistor and a thermistor; the output is taken across the thermistor. At 20 C the thermistor is 2.0 kohm. Find the output voltage and state how it changes when temperature rises. [4]
Q 6 · Particle physics [8]
(a) Give the quark composition of a proton and of a neutron. [2]
(b) Write the equation for beta-minus decay and state the quark change involved. [2]
(c) Show that C-14 decaying to N-14 plus an electron and an antineutrino conserves charge and nucleon number. [2]
(d) Calculate the energy of a photon of wavelength 520 nm in J and in eV. h = 6.63×10^-34 J s, c = 3.00×10^8 m/s. [2]
Answers and mark scheme
| Part | Answer | Marking points | marks |
|---|---|---|---|
| 1(a) | A scalar has magnitude only; a vector has both magnitude and direction. e.g. speed (scalar), velocity (vector). | 1 mark: scalar magnitude only, vector magnitude and direction 1 mark: correct example pair such as speed/velocity |
2 |
| 1(b) | a = (26.0 – 8.0)/6.0 = 3.0 m/s2; s = (8.0 + 26.0)/2 x 6.0 = 102 m | 1 mark: a = 3.0 m/s2 1 mark: uses mean velocity 17.0 m/s 1 mark: s = 102 m |
3 |
| 1(c) | t = sqrt(2 x 45/9.81) = 3.03 s; v = 9.81 x 3.03 = 29.7 m/s | 1 mark: t = sqrt(2s/g) 1 mark: t = 3.03 s 1 mark: v = 29.7 m/s |
3 |
| 1(d) | Displacement = 0.5 x 4.0 x 12 + 6.0 x 12 = 24 + 72 = 96 m (area under the graph) | 1 mark: area split into two sections 1 mark: 96 m |
2 |
| 2(a) | The resultant force on a body equals its rate of change of momentum; when mass is constant this gives F = ma. | 1 mark: resultant force equals rate of change of momentum 1 mark: mass constant so F = ma |
2 |
| 2(b) | a = F/m = 3600/1200 = 3.0 m/s2 | 1 mark: 3.6 kN = 3600 N 1 mark: a = 3.0 m/s2 |
2 |
| 2(c) | Momentum: 2.0 x 3.0 = 3.0 x v so v = 2.0 m/s; KE before = 0.5 x 2.0 x 3.0^2 = 9.0 J; KE after = 0.5 x 3.0 x 2.0^2 = 6.0 J; loss = 3.0 J | 1 mark: 6.0 = 3.0v 1 mark: v = 2.0 m/s 1 mark: KE 9.0 J before, 6.0 J after 1 mark: loss 3.0 J |
4 |
| 2(d) | The collision is inelastic: some kinetic energy is transferred to internal energy (heat) and sound, though momentum is still conserved. | 1 mark: inelastic collision 1 mark: KE converted to heat/sound while momentum conserved |
2 |
| 3(a) | W = 250 x 9.81 x 12 = 29,430 J; P_out = 29,430/20 = 1,471.5 W = 1.5 kW; P_in = 1,471.5/0.70 = 2,102 W = 2.1 kW | 1 mark: W = 29,430 J 1 mark: P_out = 1.5 kW 1 mark: divide by 0.70, not multiply 1 mark: P_in = 2.1 kW |
4 |
| 3(b) | k = 12/0.060 = 200 N/m; E = 0.5 x 12 x 0.060 = 0.36 J | 1 mark: k = 200 N/m 1 mark: E = 0.5Fx 1 mark: E = 0.36 J |
3 |
| 3(c) | A = pi(0.20×10^-3)^2 = 1.26×10^-7 m2; stress = 8.0/1.26×10^-7 = 6.4×10^7 Pa; strain = 1.2×10^-3/2.50 = 4.8×10^-4; E = 1.3×10^11 Pa | 1 mark: A = 1.26×10^-7 m2 (use radius) 1 mark: stress 6.4×10^7 Pa, strain 4.8×10^-4 1 mark: E = 1.3×10^11 Pa |
3 |
| 4(a) | lambda = v/f = 340/512 = 0.664 m | 1 mark: lambda = v/f 1 mark: 0.664 m |
2 |
| 4(b) | In a transverse wave the oscillation is perpendicular to the direction of travel (e.g. light); in a longitudinal wave it is parallel (e.g. sound). | 1 mark: oscillation perpendicular / parallel to travel 1 mark: correct examples |
2 |
| 4(c) | x = lambda D/a = 0.030 x 2.0/0.12 = 0.50 m | 1 mark: x = lambda D/a 1 mark: substitution 1 mark: 0.50 m |
3 |
| 4(d) | f3 = 3 x 250 = 750 Hz; the third harmonic has 3 antinodes, lambda = 2L/3 = 1.60/3 = 0.533 m | 1 mark: f3 = 750 Hz 1 mark: lambda = 2L/3 1 mark: 0.533 m |
3 |
| 5(a) | e.m.f. is the energy converted from other forms to electrical energy per unit charge in the source; p.d. is the energy converted from electrical to other forms per unit charge across a component. | 1 mark: e.m.f. energy gained per unit charge in the source 1 mark: p.d. energy lost per unit charge across a component |
2 |
| 5(b) | rho = RA/L = 0.90 x 2.0×10^-7/1.5 = 1.2×10^-7 ohm m | 1 mark: rho = RA/L 1 mark: 1.2×10^-7 ohm m |
2 |
| 5(c) | I = 6.0/(0.50+2.5) = 2.0 A; V = 6.0 – 2.0×0.50 = 5.0 V; P_ext = 2.0^2 x 2.5 = 10 W; P_int = 2.0^2 x 0.50 = 2.0 W | 1 mark: I = 2.0 A (total 3.0 ohm) 1 mark: V = 5.0 V 1 mark: P_ext = 10 W 1 mark: P_int = 2.0 W |
4 |
| 5(d) | V_out = 12 x 2.0/(4.0+2.0) = 4.0 V; as temperature rises the thermistor resistance falls, so its share of the voltage falls and V_out decreases. | 1 mark: divider 12 x 2.0/6.0 1 mark: 4.0 V 1 mark: resistance falls as temperature rises 1 mark: V_out decreases |
4 |
| 6(a) | Proton uud (charge +1); neutron udd (charge 0). | 1 mark: uud 1 mark: udd |
2 |
| 6(b) | n -> p + e- + antineutrino; a d quark changes into a u quark. | 1 mark: n -> p + e- + antineutrino 1 mark: d -> u |
2 |
| 6(c) | Nucleon number 14 = 14 + 0 + 0; charge 6 = 7 + (-1) + 0, so both are conserved. | 1 mark: nucleon number 14 = 14 1 mark: charge 6 = 7 – 1 |
2 |
| 6(d) | E = hc/lambda = 6.63×10^-34 x 3.00×10^8/520×10^-9 = 3.83×10^-19 J = 2.39 eV | 1 mark: 3.83×10^-19 J 1 mark: 2.39 eV |
2 |
Topic map
| Question | Topic | Marks |
|---|---|---|
| 1 | Kinematics | 10 |
| 2 | Forces and momentum | 10 |
| 3 | Work, energy and materials | 10 |
| 4 | Waves | 10 |
| 5 | Electricity and d.c. circuits | 12 |
| 6 | Particle physics | 8 |
Common lost marks
- Finding displacement with v = u + at instead of mean velocity or area.
- Multiplying by 70% instead of dividing by 0.70 when finding input power.
- Using the diameter in A = pi r^2, so the stress and Young modulus are 4x too small.
- Quoting the terminal p.d. as the e.m.f. and ignoring the Ir drop.
- Leaving out the antineutrino in beta-minus decay or reversing d -> u.
FAQ
Who is this paper for?
AS students preparing for CIE A-Level Physics 9702 who want focused practice on Paper 2 structured calculation and explanation questions.
Does the answer booklet include a mark scheme?
Yes. Every part lists its marking points so you can mark yourself or a tutor can mark point by point.
Can I print it?
Yes, both the question paper and the answer booklet are A4 PDFs ready to print.
This is original TutorZone practice material, not a past examination paper.
Format follows the CIE 9702 Physics syllabus (2025-2027) Paper 2: 1 h 15 min, 60 marks, 46% of AS.
