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S4 Physics momentum paper 2: conservation and collisions
S4 Physics momentum paper 2: conservation and collisions
Ten items start from conservation of total momentum and cover sticking, recoil, explosions and elastic or inelastic collisions.
Direct answers Items 01 to 03 cover sticking collisions and recoil, 04 to 06 final speeds and energy loss, 07 to 09 reverse work, explosions and bullets, and 10 the sign of direction.
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 3 kg cart at 4 m/s hits a stationary 1 kg cart and they stick together. Find the common speed after the collision.
- FoundationA 2 kg ball moves east at 5 m/s and a 3 kg ball west at 2 m/s; they stick together. Find the final velocity.
- FoundationA 4 kg gun fires a 0.02 kg bullet at 400 m/s. Find the recoil speed of the gun.
- IntermediateA 5 kg ball at 6 m/s hits a stationary 1 kg ball and continues at 4 m/s. Find the speed of the 1 kg ball.
- IntermediateA 0.5 kg lump of clay at 8 m/s sticks to a stationary 0.5 kg block. How much kinetic energy is lost?
- IntermediateTwo 0.4 kg balls approach each other at 5 m/s and 3 m/s and stick together. Find the final velocity.
- ChallengeA 2 kg ball at 3 m/s sticks to a stationary 4 kg ball. Find the common speed and the kinetic energy lost.
- ChallengeA stationary 1 kg bomb explodes into two pieces, the 0.4 kg piece flying off at 6 m/s. Find the speed of the other piece.
- ChallengeA 0.15 kg ball at 20 m/s hits a stationary 0.5 kg ball and stops. Find the speed of the 0.5 kg ball.
- Error redoA 3 kg ball east at 4 m/s hits a 1 kg ball west at 2 m/s and they stick. A student answers 3.5 m/s; find the correct velocity.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 3 m/s | Total momentum before is 3 x 4 = 12 kg m/s and the combined mass is 4 kg, so v = 12 / 4 = 3 m/s in the original direction. |
| 2 | 0.8 m/s east | Taking east as positive, total momentum is 10 – 6 = 4 kg m/s over 5 kg, so v = 0.8 m/s east. |
| 3 | 2 m/s | Momentum is conserved from rest: 0.02 x 400 + 4 v = 0 gives v = -2 m/s, the minus meaning opposite to the bullet. |
| 4 | 10 m/s | Before the collision the momentum is 30 kg m/s and after it is 20, so the 1 kg ball carries the remaining 10, giving 10 m/s. |
| 5 | 8 J | The common speed is 4 m/s, so kinetic energy falls from 16 J to 8 J, a loss of 8 J. |
| 6 | 1.0 m/s towards the faster ball | Taking the 5 m/s ball as positive, total momentum is 2.0 – 1.2 = 0.8 kg m/s over 0.8 kg, giving 1.0 m/s. |
| 7 | 1 m/s with 6 J lost | v = 6 / 6 = 1 m/s; kinetic energy drops from 9 J to 3 J, so 6 J is lost, as always happens in a perfectly inelastic collision. |
| 8 | 4 m/s in the opposite direction | Momentum was zero before the explosion, so 0.4 x 6 = 0.6 x v gives v = 4 m/s opposite to the smaller piece. |
| 9 | 6 m/s | The whole 3 kg m/s of momentum passes to the 0.5 kg ball, so v = 3 / 0.5 = 6 m/s; momentum being conserved says nothing about kinetic energy. |
| 10 | 2.5 m/s east | With east positive, 12 – 2 = 10 kg m/s over 4 kg gives 2.5 m/s; the slip was adding the westward momentum instead of subtracting it. |
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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