Secondary 4 Physics Practice Paper: Kinematics, Forces and Newton Laws, Work Energy and Power, Momentum and Collisions, Heat and Specific Heat Capacity, Waves and Lenses (100 marks)

2026-11-14

Physics – Secondary 4 – Practice Paper

S4 Physics practice paper, 24 parts on kinematics, forces, work and power, momentum, specific heat capacity and lenses. All values computed in python.

Download　Question paper PDF　|　Answer key PDF

Paper details　S4 Physics practice paper with 24 parts on kinematics, forces, work and power, momentum, specific heat capacity, waves and lenses. Answers with marking points.

Section A: Kinematics and Graphs (16 marks)

A car starts from rest and accelerates uniformly at 2.0 m/s2 for 6 s, then travels at constant speed for 10 s, and finally decelerates uniformly to rest in 5 s.

#	Question	Answer and marking notes
1	(a) Find the speed of the car at the end of the acceleration stage.	Answer: v = 2.0 x 6 = 12 m/s in the direction of motion. (Marking: formula 1, substitution 1, unit 1)
2	(b) Find the total distance travelled in the whole 21 s.	Answer: acceleration stage 36 m, constant speed stage 120 m, deceleration stage 30 m, total 186 m. (Marking: 1 mark per stage, 1 for sum, 1 for unit)
3	(c) Find the average speed for the whole journey (3 significant figures).	Answer: average speed = 186 / 21 = 8.86 m/s. (Marking: total distance over total time 2, value 1, unit 1)
4	(d) Find the acceleration in the deceleration stage and state its direction.	Answer: a = (0 – 12) / 5 = -2.4 m/s2; the negative sign shows it opposes the motion, magnitude 2.4 m/s2. (Marking: formula 1, value 1, direction 2)

Section B: Forces and Newton Laws (16 marks)

A 5 kg block on a rough horizontal table is pulled from rest by a horizontal force of 40 N. The coefficient of kinetic friction is 0.35 and g = 10 m/s2.

#	Question	Answer and marking notes
5	(a) Find the friction force acting on the block.	Answer: f = 0.35 x 5 x 10 = 17.5 N opposing the motion. (Marking: formula 1, value 1, direction 1)
6	(b) Find the acceleration of the block.	Answer: net force 22.5 N; a = 22.5 / 5 = 4.5 m/s2. (Marking: net force 1, formula 1, value 2)
7	(c) Find the speed of the block after 4 s.	Answer: v = 4.5 x 4 = 18 m/s. (Marking: formula 1, substitution 2, unit 1)
8	(d) If the pulling force is removed at the end of 4 s, find the further distance slid.	Answer: deceleration 3.5 m/s2; s = 18^2 / (2 x 3.5) = 46.3 m. (Marking: deceleration 2, formula 1, value 2)

Section C: Work, Energy and Power (16 marks)

A 20 kg crate is pushed 5 m up a smooth incline measured along the incline. The incline makes 30 degrees with the horizontal and g = 10 m/s2.

#	Question	Answer and marking notes
9	(a) Find the vertical height gained by the crate.	Answer: h = 5 x sin 30 = 2.5 m. (Marking: trigonometry 2, value 1)
10	(b) Find the work done against gravity in the process.	Answer: W = mgh = 20 x 10 x 2.5 = 500 J. (Marking: formula 1, substitution 2, unit 1)
11	(c) If the whole process takes 4 s, find the average power of the pushing force.	Answer: P = 500 / 4 = 125 W. (Marking: formula 1, substitution 2, unit 1)
12	(d) The crate is released from rest at that height and slides down the smooth incline. Find its speed at the bottom.	Answer: by energy conservation v = sqrt(2 x 10 x 2.5) = 7.07 m/s. (Marking: energy conservation 2, formula 1, value 2)

Section D: Momentum and Collisions (16 marks)

Trolley A of mass 2 kg moving at 3 m/s collides with stationary trolley B of mass 4 kg. After the collision the two trolleys stick together.

#	Question	Answer and marking notes
13	(a) Find the common speed of the two trolleys after collision.	Answer: momentum conservation gives 6 = 6v, so v = 1 m/s. (Marking: equation 1, substitution 1, value 1)
14	(b) Find the total kinetic energy before the collision.	Answer: total KE = 0.5 x 2 x 9 = 9 J. (Marking: formula 1, substitution 2, unit 1)
15	(c) Find the kinetic energy lost in the collision.	Answer: KE after = 3 J, so loss = 6 J. (Marking: final KE 2, subtraction 2)
16	(d) If the collision were perfectly elastic, find the velocities of A and B after collision.	Answer: A = -1 m/s (rebounds), B = 2 m/s; momentum 6 and KE 9 are both conserved. (Marking: A 2, B 2, check 1)

Section E: Heat and Specific Heat Capacity (16 marks)

A 500 W heater heats 0.5 kg of water from 20 degC to 80 degC. The specific heat capacity of water is 4200 J/(kg K).

#	Question	Answer and marking notes
17	(a) Find the energy absorbed by the water.	Answer: Q = mc dT = 0.5 x 4200 x 60 = 126000 J. (Marking: formula 1, substitution 2, unit 1)
18	(b) Find the time needed in theory.	Answer: t = 126000 / 500 = 252 s. (Marking: formula 1, substitution 2, unit 1)
19	(c) If the heating actually takes 300 s, find the efficiency of the heater.	Answer: efficiency = 126000 / 150000 = 0.84, i.e. 84 per cent. (Marking: definition 2, calculation 2, percent 1)
20	(d) Explain why the actual time is longer than the theoretical value.	Answer: some heat is lost to the surroundings and the container, so less energy heats the water. (Marking: heat loss 2, explanation 1)

Section F: Waves and Lenses (20 marks)

A sound wave of frequency 250 Hz travels in air with wavelength 1.2 m. A convex lens of focal length 10 cm has an object placed 30 cm in front of it.

#	Question	Answer and marking notes
21	(a) Find the speed of the sound wave.	Answer: v = f x wavelength = 250 x 1.2 = 300 m/s. (Marking: formula 1, substitution 2, unit 1)
22	(b) Find the time for the sound wave to travel 510 m.	Answer: t = 510 / 300 = 1.7 s. (Marking: formula 1, substitution 2, unit 1)
23	(c) Find the image distance.	Answer: 1/v = 1/10 – 1/30 = 1/15, so v = 15 cm behind the lens. (Marking: formula 2, substitution 2, value 2)
24	(d) Find the magnification and state the nature of the image.	Answer: m = 15 / 30 = 0.5; the image is inverted, diminished and real. (Marking: magnification 2, inverted 1, diminished 1, real 2)

Find a tutor — free Tutor Registration →

A TutorZone practice paper; original questions and passages, not publisher material.