S4 Physics Practice Paper: Force, Motion and Energy (12 questions, 100 marks)
- 2026-11-08
S4 Physics: Force, Motion and Energy
S4 Physics paper: acceleration, momentum, work and power, energy efficiency, Hooke law, friction and circular motion. 12 questions, 100 marks, 75 minutes, full answers.
Download Question paper PDF | Answer key PDF
Paper details Level: S4 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Force and Motion (4 questions, 33 marks)
Uniform acceleration and Newton second law: list the given data first.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A car starting from rest accelerates uniformly and reaches 25 m/s in 10 s. Find (a) the acceleration; (b) the distance travelled in these 10 s. | Answer: (a) a = 25 / 10 = 2.5 m/s squared; (b) s = 0.5 x 2.5 x 100 = 125 m. Marking: acceleration 3 marks, distance 5 marks, 1 mark deducted for a missing unit. |
| 2 | A 2.0 kg block on a frictionless horizontal surface is pushed by a constant horizontal force of 6.0 N, starting from rest. Find (a) the acceleration; (b) the speed after 4.0 s; (c) the displacement in 4.0 s. | Answer: (a) a = 6.0 / 2.0 = 3.0 m/s squared; (b) v = 3.0 x 4.0 = 12 m/s; (c) s = 0.5 x 3.0 x 16 = 24 m. Marking: 2 marks each, 2 marks for stating that the surface is frictionless. |
| 3 | A small ball is thrown vertically upwards at 15 m/s. Take g = 10 m/s squared and ignore air resistance. Find (a) the greatest height above the hand; (b) the total time from the throw to the return to the same point. | Answer: (a) h = 15 squared / (2 x 10) = 11.25 m; (b) t = 2 x 15 / 10 = 3.0 s. Marking: zero velocity at the top 2 marks, working 3 marks, total time 3 marks. |
| 4 | A 3.0 kg trolley moving at 4.0 m/s collides with a stationary 1.0 kg trolley and the two move together after the collision. Find (a) the common velocity; (b) the loss of kinetic energy. | Answer: (a) 3.0 x 4.0 = (3.0 + 1.0) x v, so v = 3.0 m/s; (b) kinetic energy before 24 J, after 18 J, loss 6.0 J. Marking: momentum equation 4 marks, velocity 2 marks, energy loss 3 marks. |
Section B: Work, Energy and Power (4 questions, 33 marks)
Conservation of energy and efficiency: quote energy in joules.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A worker raises a 20 kg box through 5.0 m at constant speed. Take g = 10 m/s squared. (a) Find the work done by the worker. (b) If the process takes 4.0 s, find the average power. | Answer: (a) W = 20 x 10 x 5.0 = 1000 J; (b) P = 1000 / 4.0 = 250 W. Marking: work 4 marks, power 4 marks, 1 mark deducted for a missing joule or watt unit. |
| 6 | A 0.50 kg ball falls freely from a height of 20 m. Take g = 10 m/s squared and ignore air resistance. Find (a) the kinetic energy just before it reaches the ground; (b) its speed there. | Answer: (a) KE = 0.50 x 10 x 20 = 100 J; (b) v = sqrt(2 x 10 x 20) = 20 m/s. Marking: energy conservation 3 marks, kinetic energy 2 marks, speed 3 marks. |
| 7 | A 12 kg box slides down a frictionless slope of height 3.2 m. Take g = 10 m/s squared. Find (a) the loss of gravitational potential energy; (b) the speed at the bottom of the slope. | Answer: (a) change in potential energy = 12 x 10 x 3.2 = 384 J; (b) v = sqrt(2 x 10 x 3.2) = 8.0 m/s. Marking: potential energy 4 marks, speed 4 marks, the conservation statement is required. |
| 8 | An electric hoist with an input power of 500 W raises a 15 kg load through 4.0 m in 5.0 s. Take g = 10 m/s squared. Find (a) the useful work output; (b) the efficiency of the hoist. | Answer: (a) useful work W = 15 x 10 x 4.0 = 600 J; (b) input energy 2500 J, efficiency = 600 / 2500 x 100 per cent = 24.0 per cent. Marking: useful work 4 marks, input energy 2 marks, efficiency 3 marks. |
Section C: Applied Mechanics (4 questions, 34 marks)
Hooke law, friction and circular motion: mind direction and units.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A spring of stiffness 200 N/m is stretched by 0.15 m. Find the tension in the spring. | Answer: F = kx = 200 x 0.15 = 30 N. Marking: formula 4 marks, substitution and answer 3 marks, unit 1 mark. |
| 10 | An 8.0 kg block slides at 10 m/s on a horizontal floor and the coefficient of friction between them is 0.25. Take g = 10 m/s squared. Find (a) the friction force; (b) the deceleration; (c) the time and the distance taken to stop. | Answer: (a) f = 0.25 x 8.0 x 10 = 20 N; (b) a = 20 / 8.0 = 2.5 m/s squared; (c) t = 10 / 2.5 = 4.0 s and s = 100 / (2 x 2.5) = 20 m. Marking: 2 marks each, 2 marks for stating the direction. |
| 11 | A 0.40 kg ball moves on a horizontal plane at 4.0 m/s along a circle of radius 0.50 m. Find the magnitude and the direction of the centripetal force. | Answer: F = m v squared / r = 0.40 x 16 / 0.50 = 12.8 N, directed towards the centre of the circle. Marking: formula 4 marks, substitution 2 marks, direction 2 marks. |
| 12 | Two perpendicular forces of 6.0 N and 8.0 N act together on a particle of mass 2.0 kg. Find (a) the magnitude of the resultant force; (b) the acceleration of the particle. | Answer: (a) R = sqrt(36 + 64) = 10 N; (b) a = 10 / 2.0 = 5.0 m/s squared. Marking: vector diagram or Pythagoras 4 marks, resultant 3 marks, acceleration 3 marks. |
Find a tutor — freeTutor Registration →
A TutorZone practice paper; original questions and passages, not publisher material.


