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S4 Physics momentum paper 1: p = mv and F t = change in p
S4 Physics momentum paper 1: p = mv and F t = change in p
Ten items start from p = m v and impulse F t; the units are interchangeable and direction must be fixed first.
Direct answers Items 01 to 03 cover basic momentum and impulse, 04 to 06 change in momentum and average force, 07 to 09 reverse work and rebound, and 10 using the final momentum only.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S4 Physics Momentum and Impulse |
| Level | S4 |
02 Questions (10 items)
- FoundationA 2 kg ball moves at 3 m/s. Find its momentum.
- FoundationA 0.5 kg ball moves at 12 m/s. Find its momentum.
- FoundationA 20 N force pushes a box for 3 s. Find the impulse.
- IntermediateA 4 kg cart travelling at 5 m/s is braked to rest. Find the change in momentum.
- IntermediateA 0.2 kg ball speeds up from 10 m/s to 25 m/s in 0.5 s. Find the average force.
- IntermediateA 1 500 kg car slows uniformly from 20 m/s to rest in 5 s. Find the size of the braking force.
- ChallengeAn object has momentum 12 kg m/s and mass 3 kg. Find its speed.
- ChallengeAn impulse of 150 N s acts on a board for 0.02 s. Find the average force.
- ChallengeA 0.06 kg tennis ball hits a wall at 30 m/s and rebounds at 30 m/s. Find the size of its momentum change.
- Error redoA 2 kg ball goes from 4 m/s to 10 m/s in 3 s. A student answers 20 N; find the correct average force.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 6 kg m/s | p = m v = 2 x 3 = 6 kg m/s, pointing the same way as the velocity. |
| 2 | 6 kg m/s | p = 0.5 x 12 = 6 kg m/s; a light fast ball can carry the same momentum as a heavy slow one. |
| 3 | 60 N s | Impulse = F t = 20 x 3 = 60 N s, and a newton second is the same unit as kg m/s. |
| 4 | 20 kg m/s opposite to the motion | Change is 0 – 4 x 5 = -20 kg m/s, magnitude 20, and the braking force points against the motion. |
| 5 | 6 N | The momentum change is 0.2 x 15 = 3 kg m/s, so the force is 3 / 0.5 = 6 N. |
| 6 | 6 000.0 N | Momentum change is 1 500 x 20 = 30 000 kg m/s, so F = 30 000 / 5 = 6 000.0 N against the motion. |
| 7 | 4 m/s | v = p / m = 12 / 3 = 4 m/s; momentum points along the velocity, so a straight division works. |
| 8 | 7 500.0 N | F = impulse / time = 150 / 0.02 = 7 500.0 N; a short contact time means a very large force, which is why crashes hurt. |
| 9 | 3.6 kg m/s | Before +1.8 and after -1.8 kg m/s, so the change is 0.06 x 60 = 3.6 kg m/s; reversing direction means adding the two speeds. |
| 10 | 4 N | F = m(v – u) / t = 2 x 6 / 3 = 4 N; the slip was using the final momentum only instead of the change in momentum. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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