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S4 Physics momentum paper 4: ten common slips
S4 Physics momentum paper 4: ten common slips
Ten items rewritten from the usual slips: direction signs, unit conversion and forgetting to divide by force or time.
Direct answers Items 01 to 03 retry the basic formulae, 04 to 06 momentum change and average force, 07 to 09 energy, sticking and bullets, and 10 the minus sign for opposite motion.
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 0.2 kg ball moves right at 15 m/s. Find its momentum.
- FoundationA 500 g ball moves at 4 m/s. Find its momentum.
- FoundationA force of 12 N acts for 0.5 s. Find the impulse.
- IntermediateA 2 kg body moving at 6 m/s is stopped by a steady 4 N resistance. Find the time taken.
- IntermediateA 0.5 kg ball hits a wall at 8 m/s and rebounds at 6 m/s. Find the size of its momentum change.
- IntermediateA 1 200 kg car travelling at 20 m/s stops uniformly in 5 s. Find the average braking force.
- ChallengeA 3 kg body has momentum 18 kg m/s. Find its speed and kinetic energy.
- ChallengeA 4 kg trolley at 3 m/s sticks to a stationary 2 kg trolley. Find the common speed and the kinetic energy lost.
- ChallengeA 60 g bullet at 400 m/s embeds in a 5.94 kg stationary block. Find the speed of the block.
- Error redoA 3 kg ball at 4 m/s east meets a 1 kg ball at 2 m/s west and they stick. A student writes 14 kg m/s; find the correct momentum.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 3 kg m/s to the right | p = m v = 0.2 x 15 = 3 kg m/s; mass must be in kg and the direction follows the velocity. |
| 2 | 2 kg m/s | 500 g is 0.5 kg, so p = 0.5 x 4 = 2 kg m/s; putting 500 straight into the formula is the usual slip. |
| 3 | 6 N s | Impulse = F t = 12 x 0.5 = 6 N s, which is also the change in momentum, since N s and kg m/s are the same unit. |
| 4 | 3 s | The momentum change is 12 kg m/s, so t = 12 / 4 = 3 s; skipping the division gives the wrong 12 s. |
| 5 | 7 kg m/s | The change is m(u + v) = 0.5 x (8 + 6) = 7 kg m/s, because the rebound reverses the direction. |
| 6 | 4 800 N | The change is 1 200 x 20 = 24 000 kg m/s, so F = 24 000 / 5 = 4 800 N, directed against the motion. |
| 7 | 6 m/s and 54 J | v = 18 / 3 = 6 m/s, and the energy is p squared over 2m = 324 / 6 = 54 J. |
| 8 | 2 m/s and 6 J lost | Momentum 12 kg m/s over 6 kg gives 2 m/s; energy falls from 18 J to 12 J, so 6 J is lost. |
| 9 | 4 m/s | The bullet carries 24 kg m/s into 6 kg in total, so v = 24 / 6 = 4 m/s, with the mass written as 0.06 kg. |
| 10 | 10 kg m/s east, then 2.5 m/s | Opposite directions need a minus sign: 3 x 4 – 1 x 2 = 10 kg m/s, not 12 + 2, so the pair moves east at 2.5 m/s. |
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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