CAIE A Level 9702

Physics 2025 · May/June

View PDF Login to Download
27 Questions Q&A Available
AI Paper Analysis & Study Guide
Auto-generated · SkoolAccess AI

This CAIE A Level Physics paper covers a broad spread of core mechanics and electricity topics, with some waves and materials content as well. Students can expect questions on vectors and scalars, forces, moments, energy, power, momentum, wave superposition and coherence, Young modulus, and circuit analysis with resistors and potential difference.…

Jump to Questions

Questions & Model Answers 27 Qs

Q1a
Short Answer 2 marks
Table 1.1 lists some physical quantities. Identify with ticks (✓) which quantities are vectors and which are scalars. Table 1.1 quantity scalar vector acceleration displacement gravitational potential energy speed temperature
Acceleration: vector; displacement: vector; gravitational potential energy: scalar; speed: scalar; temperature: scalar.
Vectors have magnitude and direction; scalars have magnitude only.
Q1b(i)
Structured 3 marks
A constant resultant force F acts on a car of mass m. The car moves from rest with constant acceleration a along horizontal ground. When the car has displacement s, the speed of the car is v. Using the concept of work done on the car, show that the kinetic energy EK of the car is given by the equation EK = 1/2 mv^2.
Work done = Fs = mas. Using v^2 = 2as for motion from rest, EK = work done = mas = 1/2 mv^2.
The work done by the resultant force equals the gain in kinetic energy.
Q1b(ii)
Short Answer 1 mark
The mass of the car is 920 kg. At time t = 0, the car is at rest. At time t = 5.8 s, its velocity is 17 m s−1. Calculate the kinetic energy of the car at time t = 5.8 s.
1.33 × 10^5 J
EK = 1/2 mv^2 = 0.5 × 920 × 17^2 ≈ 1.33 × 10^5 J.
Q1b(iii)
Structured 3 marks
Between time t = 0 and time t = 5.8 s, the work done against resistive forces is 4.7 × 10^4 J. Determine the average output power of the car during this time.
Average output power = total work done / time = (1.33 × 10^5 + 4.7 × 10^4) / 5.8 ≈ 3.1 × 10^4 W.
The engine must supply the gain in kinetic energy plus the work done against resistive forces.
Q1b(iv)
Short Answer 1 mark
At time t = 5.8 s, the speed of the car becomes constant. State and explain whether the output power of the car is greater than, less than or the same as the output power just before t = 5.8 s.
Less than. When speed becomes constant, acceleration is zero so no power is needed to increase kinetic energy; only power to overcome resistive forces remains.
Before 5.8 s the engine supplies power for both acceleration and resistive forces.
Q2a
Short Answer 1 mark
Define the moment of a force about a point.
Moment = force × perpendicular distance from the point to the line of action of the force.
This is the standard definition of moment.
Q2b(i)
Structured 3 marks
A tree of mass 270 kg grows out of sloping ground and is supported by a post, as shown in Fig. 2.1. The ground applies a total force R on the tree at point Q. The centre of gravity of the tree is a horizontal distance of 1.2 m from Q. The post applies a force F of 1800 N perpendicular to the line PQ. The line of action of F passes through point P at an angle θ to the vertical. P is a horizontal distance of 1.6 m from Q. The tree is in equilibrium and all forces act on the tree in the same plane. By taking moments about point Q, show that θ is 25°.
Taking moments about Q: 1800 × 1.6 sin θ = 270g × 1.2. Hence sin θ = (270g × 1.2)/(1800 × 1.6) ≈ 0.42, so θ ≈ 25°.
Equating clockwise and anticlockwise moments gives the angle.
Q2b(ii)
Structured 2 marks
On Fig. 2.2, draw a labelled scale vector triangle to represent the forces acting on the tree. The weight of the tree has been drawn to scale.
A closed vector triangle with the weight vector, the force from the post F, and the reaction R drawn to scale and labelled.
For equilibrium, the three forces must form a closed triangle.
Q2b(iii)
Short Answer 2 marks
The tree exerts a pressure of 150 kPa on the top of the post. Determine the surface area of the tree in contact with the post.
Area = F / p = 1800 / (1.50 × 10^5) = 1.2 × 10^-2 m^2.
Pressure = force / area.
Q3a
Short Answer 1 mark
Two progressive water waves X and Y travel along a straight line from point A to point B. The variation of displacement of the waves with distance from A at an instant in time is shown in Fig. 3.1. State the amplitude of wave X.
20 cm
Amplitude is the maximum displacement from equilibrium.
Q3b(i)
Short Answer 2 marks
Both waves have frequency 16 Hz. Determine the speed of wave X.
0.80 m s^-1
Use v = fλ from the graph to find the wavelength of wave X.
Q3b(ii)
Short Answer 1 mark
State and explain whether X and Y are coherent.
They are not coherent because they do not have the same wavelength/frequency and/or constant phase difference.
Coherent waves must have the same frequency and a constant phase difference.
Q3c
Structured 2 marks
Wave X and wave Y superpose to form a resultant wave. On Fig. 3.2, sketch the variation of displacement of the resultant wave with distance from A at the instant of time shown in Fig. 3.1.
A sketch showing point-by-point addition of the two waves, giving a resultant with varying amplitude according to superposition.
The resultant displacement is the algebraic sum of the two displacements.
Q3d
Short Answer 2 marks
The intensity of wave X is IX. The intensity of wave Y is IY. Use Fig. 3.1 to determine the ratio IX/IY.
IX/IY = (20^2)/(10^2) = 4
Intensity is proportional to amplitude squared.
Q4a
Short Answer 1 mark
A small ball is dropped from rest from height h1 above the ground and falls vertically downwards. The ball collides with the ground and bounces back vertically upwards, reaching a maximum height h2. Fig. 4.1 shows the ball just before and just after hitting the ground. State and explain whether the collision is elastic or inelastic.
Inelastic, because the speed after the collision is less than before, so kinetic energy is not conserved.
An elastic collision conserves kinetic energy.
Q4b(i)
Short Answer 2 marks
Calculate the change in momentum of the ball during the collision with the ground.
Change in momentum = 0.25(3.6 - (-5.2)) = 2.2 kg m s^-1 upward.
Take upward as positive; momentum changes from downward to upward.
Q4b(ii)
Short Answer 2 marks
Determine the average force on the ball during the collision with the ground.
Average force = Δp / Δt = 2.2 / 0.18 ≈ 12 N upward.
Average force equals rate of change of momentum.
Q4c
Short Answer 3 marks
Calculate the ratio h2/h1.
h2/h1 = (3.6^2)/(5.2^2) ≈ 0.48
Using v^2 = 2gh, height is proportional to speed squared.
Q5a
Short Answer 1 mark
Define the Young modulus.
Young modulus = tensile stress / tensile strain, provided the limit of proportionality is not exceeded.
This is the definition of Young modulus.
Q5b(i)
Structured 3 marks
A wire of unstretched length 0.81 m is made of a metal with Young modulus 95 GPa. The wire obeys Hooke's law and has a constant cross-sectional area. Fig. 5.1 shows the force-extension graph for the wire. Determine the cross-sectional area of the wire.
Using gradient of graph, F/x = EA/L, so A = (L/E)(F/x). From the graph gradient ≈ 1.0 × 10^5 N m^-1, giving A ≈ 8.5 × 10^-7 m^2.
For a wire obeying Hooke's law, F = (EA/L)x.
Q5b(ii)
Structured 3 marks
The extension of the wire is initially 2.0 × 10^-3 m. Determine the work done to increase the extension of the wire to 3.0 × 10^-3 m.
Work done = area under force-extension graph between 2.0 × 10^-3 m and 3.0 × 10^-3 m = 0.5(F1 + F2)Δx ≈ 0.25 J.
The work done is the area under the force-extension graph.
Q6a
Short Answer 1 mark
Define electric potential difference across a component.
Energy transferred per unit charge across the component.
Potential difference is work done per unit charge.
Q6b(i)
Short Answer 2 marks
A circuit contains four resistors and a battery of electromotive force (e.m.f.) 8.0 V with negligible internal resistance. When the variable resistor has resistance R, the currents in the circuit are 0.030 A, I1 and I2, as shown in Fig. 6.1. Determine the charge passing through the battery in a time of 4.0 minutes.
Charge = It = 0.030 × 240 = 7.2 C.
Charge is current multiplied by time.
Q6b(ii)
Short Answer 2 marks
Calculate I1.
I1 = 8.0 / 430 ≈ 0.0186 A.
Use Ohm's law for the 430 Ω resistor.
Q6b(iii)
Short Answer 1 mark
Calculate I2.
I2 = 0.030 - I1 ≈ 0.011 A.
Apply current conservation at the junction.
Q6b(iv)
Short Answer 2 marks
Determine R.
R = 8.0 / I2 - 210 - 240 ≈ 270 Ω.
Use series resistance and the current in the branch containing R.
Q6c
Structured 2 marks
The variable resistor in (b) is fitted with a scale so that its resistance can be accurately determined. The resistor of resistance 240 Ω is now replaced by a new resistor X of unknown resistance. A galvanometer is connected as shown in Fig. 6.2. With reference to ratios of resistances, explain how this circuit can be used to determine the resistance of X.
Adjust the variable resistor until the galvanometer reads zero. Then no current flows through the galvanometer, so the two branches are balanced and the ratio of resistances in one branch equals the ratio in the other; hence X can be found from the known resistance ratio.
At balance, the potential differences across the two branches are equal.
Paper Details
Board CAIE
Level A Level
Subject Physics
Code 9702
Year 2025
Session May/June
Q&A 27 questions
Interactive Q&A View View PDF Login to Download
Quick Tips
  • Read each question carefully before answering
  • Allocate time based on marks (≈1 min/mark)
  • Attempt all questions — partial marks count