| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with two possible outcomes |
| Difficulty | Moderate -0.3 This is a straightforward application of conservation of momentum in two scenarios. Students must set up momentum equations (0.5×6 = 0.5v_P + m×4) for each case, with the only variation being the sign of one velocity. The algebra is simple and the question clearly guides students through both possibilities, making it slightly easier than average but still requiring proper understanding of vector nature of momentum. |
| Spec | 6.03b Conservation of momentum: 1D two particles |
1\\
\includegraphics[max width=\textwidth, alt={}, center]{470e70de-66ba-4dcc-a205-0c92f29471b1-2_227_878_269_635}
A particle $P$ of mass 0.5 kg is travelling with speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a smooth horizontal plane towards a stationary particle $Q$ of mass $m \mathrm {~kg}$ (see diagram). The particles collide, and immediately after the collision $P$ has speed $0.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $Q$ has speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Given that both particles are moving in the same direction after the collision, calculate $m$.\\
(ii) Given instead that the particles are moving in opposite directions after the collision, calculate $m$.
\hfill \mbox{\textit{OCR M1 2009 Q1 [6]}}