| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with speed relationships |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring conservation of momentum and Newton's restitution law. The algebra is straightforward, and part (b) involves routine manipulation of inequalities. While it has multiple parts and some algebraic manipulation, it follows a well-practiced template with no novel insight required, making it slightly easier than average. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) Momentum: \(30mu + 15mu = 6mv + 5mkv\) | M1 A1 | |
| \(45u = (6 + 5k)v\) | M1 A1 | |
| \(v = \frac{45u}{5k + 6}\) | A1 | |
| (ii) Elasticity: \((kv - v)/(3u - 5u) = -e\) | M1 A1 | |
| \((k - 1)v = (-2u)(-e)\) | M1 A1 | |
| \(v = \frac{2eu}{k-1}\) | A1 | |
| (b) \(\frac{45u}{5k+6} = \frac{2eu}{k-1}\) | M1 A1 | |
| \(e = \frac{45(k-1)}{2(5k+6)}\) | M1 A1 | |
| \(0 \le e \le 1\), so \(0 \le 45k - 45 \le 10k + 12\) | M1 A1 A1 | |
| \(1 \le k \le \frac{57}{35}\) | A1 | (13 marks) |
(a)(i) Momentum: $30mu + 15mu = 6mv + 5mkv$ | M1 A1 |
$45u = (6 + 5k)v$ | M1 A1 |
$v = \frac{45u}{5k + 6}$ | A1 |
(ii) Elasticity: $(kv - v)/(3u - 5u) = -e$ | M1 A1 |
$(k - 1)v = (-2u)(-e)$ | M1 A1 |
$v = \frac{2eu}{k-1}$ | A1 |
(b) $\frac{45u}{5k+6} = \frac{2eu}{k-1}$ | M1 A1 |
$e = \frac{45(k-1)}{2(5k+6)}$ | M1 A1 |
$0 \le e \le 1$, so $0 \le 45k - 45 \le 10k + 12$ | M1 A1 A1 |
$1 \le k \le \frac{57}{35}$ | A1 | (13 marks)Two railway trucks $A$ and $B$, whose masses are $6m$ and $5m$ respectively, are moving in the same direction along a straight track with speeds $5u$ and $3u$ respectively, and collide directly. Immediately after this impact the speeds of $A$ and $B$ are $v$ and $kv$ respectively, in the same direction as before. The coefficient of restitution between $A$ and $B$ is $e$.
Modelling the trucks as particles,
\begin{enumerate}[label=(\alph*)]
\item show that \begin{enumerate}[label=(\roman*)] \item $v = \frac{45u}{5k + 6}$, \item $v = \frac{2eu}{k - 1}$. \end{enumerate} [8 marks]
\item Use the fact that $0 \leq e \leq 1$ to deduce the range of possible values of $k$. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q5 [13]}}