| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with two possible outcomes |
| Difficulty | Standard +0.3 This is a standard M1 momentum conservation question with straightforward application of the principle across three scenarios. Part (i) is routine (particles at rest), part (ii)(a) requires sign consideration for opposite directions, and part (ii)(b) involves coalescence—all are textbook exercises requiring methodical application rather than insight, making it slightly easier than average. |
| Spec | 6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| 4(i) | In Q4 right to left may be used as the positive sense throughout. \(0.18 × 2 − 3m = 0\) | M1 |
| 4(i) | \(m = 0.12\) | A1 |
| 4(i) | A1 | |
| 4(i) | [3] | |
| 4(ia) | Momentum after \(= -0.18 × 1.5 + 1.5m\) | B1 |
| 4(ia) | \(0.18 × 2 − 3m = -0.18 × 1.5 + 1.5m\) | M1 |
| 4(ia) | \(m = 0.14\) | A1 |
| 4(ia) | [3] | |
| 4(ib) | \(0.18 × 2 − 3m\) | B1ft |
| 4(ib) | \(= (0.18 + m)1.5\) | B1 |
| 4(ib) | \(m = 0.02\) | B1ft |
| 4(ib) | \(0.18 × 2 − 3m = -(0.18 + m)1.5\) | B1 |
| 4(ib) | \(m = 0.42\) |
4(i) | In Q4 right to left may be used as the positive sense throughout. $0.18 × 2 − 3m = 0$ | M1 | For focusing Momentum 'before' is zero
4(i) | $m = 0.12$ | A1 |
4(i) | | A1 |
4(i) | | [3] | 3 marks possible if g included consistently
4(ia) | Momentum after $= -0.18 × 1.5 + 1.5m$ | B1 |
4(ia) | $0.18 × 2 − 3m = -0.18 × 1.5 + 1.5m$ | M1 | For using conservation of momentum
4(ia) | $m = 0.14$ | A1 |
4(ia) | | [3] | 3 marks possible if g included consistently
4(ib) | $0.18 × 2 − 3m$ | B1ft | It wrong momentum 'before'
4(ib) | $= (0.18 + m)1.5$ | B1 |
4(ib) | $m = 0.02$ | B1ft |
4(ib) | $0.18 × 2 − 3m = -(0.18 + m)1.5$ | B1 |
4(ib) | $m = 0.42$ | | [4] | 0 marks if g included4\\
\includegraphics[max width=\textwidth, alt={}, center]{ae5d1e27-5853-48aa-9046-86ce1c1a154a-3_149_606_1626_772}
Two particles of masses 0.18 kg and m kg move on a smooth horizontal plane. They are moving towards each other in the same straight line when they collide. Immediately before the impact the speeds of the particles are $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively (see diagram).\\
(i) Given that the particles are brought to rest by the impact, find m .\\
(ii) Given instead that the particles move with equal speeds of $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ after the impact, find
\begin{enumerate}[label=(\alph*)]
\item the value of m , assuming that the particles move in opposite directions after the impact,
\item the two possible values of m , assuming that the particles coalesce.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2007 Q4 [10]}}