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S4 Physics Practice Paper: Newton’s Laws, Momentum and Energy (12 questions, 100 marks, 75 minutes)
S4 Physics: Newton’s Laws, Momentum and Energy
S4 physics paper: forces, momentum, work and energy, elastic collisions, projectiles and inclined planes, with g = 9.81 m/s². Mark scheme included.
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Paper details Level: S4 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Force and Motion (30 marks)
Section A
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 1. A 2.0 kg trolley starts from rest and accelerates uniformly to 6.0 m/s in 3.0 s. Find (a) the acceleration; (b) the resultant force on the trolley. | 1. (a) a = v/t = 6.0/3.0 = 2.0 m/s²; (b) F = ma = 2.0 × 2.0 = 4.0 N. Marking: formula or substitution required for full marks. |
| 2 | 2. A 5.0 kg box on a rough horizontal floor is pulled by a horizontal force of 40 N against a friction force of 15 N. Starting from rest, find (a) its acceleration; (b) the distance travelled in 4.0 s. | 2. (a) Net force = 40 − 15 = 25 N, so a = 25/5.0 = 5.0 m/s²; (b) s = ½at² = ½ × 5.0 × 4.0² = 40 m. Marking: no marks if friction is ignored. |
| 3 | 3. A 0.50 kg ball hits a wall perpendicularly at 8.0 m/s and rebounds at 6.0 m/s. The contact time is 0.020 s. Find (a) the magnitude of the change in momentum; (b) the average force exerted by the wall on the ball. | 3. Taking the rebound direction as positive: Δp = m(v − u) = 0.50 × (6.0 + 8.0) = 7.0 kg·m/s; F = Δp/Δt = 7.0/0.020 = 350 N. Marking: sign error costs 1 mark. |
| 4 | 4. A 3.0 kg block slides from rest down a frictionless 30° incline from a height of 1.5 m. Find (a) its speed at the bottom; (b) its kinetic energy at the bottom. | 4. (a) mgh = ½mv² → v = √(2 × 9.81 × 1.5) = 5.4 m/s; (b) KE = mgh = 3.0 × 9.81 × 1.5 = 44 J. |
Section B: Momentum and Energy (30 marks)
Section B
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | 5. A 4.0 kg object moving at 3.0 m/s on a smooth horizontal surface collides with a stationary 2.0 kg object and they stick together. Find (a) the common velocity after the collision; (b) the kinetic energy lost. | 5. (a) Momentum: 4.0 × 3.0 = 6.0v → v = 2.0 m/s; (b) initial KE = 18 J, final KE = 12 J, so 6.0 J is lost. |
| 6 | 6. A 0.15 kg ball is released from rest at a height of 1.8 m (ignore air resistance). Find (a) its speed just before hitting the ground; (b) the time of fall. | 6. (a) v = √(2gh) = √(2 × 9.81 × 1.8) = 5.9 m/s; (b) t = √(2h/g) = 0.61 s. |
| 7 | 7. A 60 kg passenger stands on a scale inside a lift that accelerates downwards at 2.0 m/s². Find the normal reaction from the scale on the passenger. | 7. Taking downwards as positive: mg − N = ma → N = m(g − a) = 60 × (9.81 − 2.0) = 469 N. Marking: must state that the normal reaction is less than the weight. |
| 8 | 8. A 1200 kg car travelling at 20 m/s brakes uniformly to rest over 40 m. Find (a) the deceleration; (b) the braking force; (c) the braking time. | 8. (a) v² = u² − 2as → a = 20²/(2 × 40) = 5.0 m/s²; (b) F = ma = 6000 N; (c) t = u/a = 4.0 s. |
Section C: Integrated Problems (40 marks)
Section C
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | 9. A 2.0 kg object moving at 5.0 m/s makes a one-dimensional elastic collision with a stationary 3.0 kg object. Find the velocities after the collision and verify with momentum and kinetic energy. | 9. Taking the initial direction as positive: v₁ = (m₁ − m₂)u₁/(m₁ + m₂) = −1.0 m/s; v₂ = 2m₁u₁/(m₁ + m₂) = 4.0 m/s. Check: momentum 2.0 × (−1.0) + 3.0 × 4.0 = 10; KE 1.0 + 24 = 25 = ½ × 2.0 × 5.0². |
| 10 | 10. A small ball is projected horizontally at 12 m/s from a height of 20 m (ignore air resistance). Find (a) the time of flight; (b) the horizontal range; (c) the speed on landing. | 10. (a) t = √(2h/g) = 2.0 s; (b) range = 12 × 2.0 = 24 m; (c) vy = 19.6 m/s, v = √(12² + 19.6²) = 23 m/s. |
| 11 | 11. Two blocks of mass 3.0 kg and 2.0 kg are connected by a light string over a smooth fixed pulley (Atwood machine). Find (a) the acceleration of the system; (b) the tension in the string. | 11. (a) a = (3.0 − 2.0) × 9.81/5.0 = 1.96 m/s²; (b) for the 2.0 kg block: T = m(g + a) = 2.0 × (9.81 + 1.96) = 23.5 N. Check with the 3.0 kg block: T = 3.0 × (9.81 − 1.96) = 23.5 N. |
| 12 | 12. A 5.0 kg crate is pulled from rest 4.0 m up a 25° incline by a force of 60 N along the incline, against a friction force of 8.0 N. Find (a) the work done by the pulling force; (b) the work done against gravity; (c) the net work done; (d) the final speed of the crate. | 12. (a) W = 60 × 4.0 = 240 J; (b) Wg = 5.0 × 9.81 × 4.0 × sin25° = 82.9 J; (c) net work = 240 − 82.9 − 32 = 125 J; (d) ½mv² = 125 → v = 7.1 m/s. |
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