Paper 1 Dynamics Answers
| Q | Ans | Brief reason |
|---|
| 1 | B | ∑Fext=dp/dt. |
| 2 | B | a=F/m=12/3.0=4.0 ms−2. |
| 3 | A | Constant velocity means zero acceleration. |
| 4 | C | Same interaction, forces on different bodies. |
| 5 | B | p=mv=0.20(15)=3.0 kgms−1. |
| 6 | B | Impulse equals change in momentum. |
| 7 | B | Area under F-t gives impulse. |
| 8 | C | Momentum conservation requires negligible resultant external impulse. |
| 9 | A | v=0.50(2.0)/(2.00)=0.50 ms−1. |
| 10 | C | Elastic collisions conserve both. |
| 11 | C | N−mg=ma, so N>mg. |
| 12 | C | Apparent weight is normal contact force. |
| 13 | B | Internal forces cancel for the combined system. |
| 14 | B | Resultant force and acceleration oppose velocity. |
| 15 | A | N=kgms−2. |
| 16 | A | FavgΔt=Δp. |
| 17 | B | Momentum may still be conserved; kinetic energy is not. |
| 18 | C | Third-law partners are equal and opposite. |
| 19 | B | Momentum is vector mass times velocity. |
| 20 | B | Reversal makes Δp=pf−pi large. |