| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2001 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Range of coefficient of restitution |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring conservation of momentum and Newton's restitution law. Parts (a) and (b) are routine calculations with given values. Part (c) requires setting up inequalities to prevent further collisions, which is a common M2 technique but involves slightly more problem-solving than pure recall, making it slightly easier than average overall. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\rightarrow v_1\) \(\rightarrow v_2\) | M1A1 | |
| \(\rightarrow 2u\) \(\rightarrow u\) | ||
| A: \(\circ\) 2m \(\quad\) B: \(\circ\) 4m | ||
| CoM: \(4mu + 4mu = 2m \cdot v_1 + 4m \cdot v_2\) \(\Rightarrow\) \(4u = v_1 + 2v_2\) | M1A1 | |
| NEL: \(\frac{1}{2}(2u - u) = v_2 - v_1\) | M1A1cso(6) | |
| Solving to find \(v_2\); \(v_2 = \frac{3u}{2}\) | M1A1 | (2) |
| (b) Substitute for \(v_2\) in one equation; \(v_1 = v_2 - \frac{1}{2}u = u\) | M1A1 | (2) |
| (c) \(\rightarrow w_1\) \(\rightarrow w_2\) | M1A1 | |
| \(\rightarrow \frac{2}{3}u\) \(\rightarrow 0\) | ||
| \(\circ\) B \(\quad\) \(\circ\) C | ||
| 4m \(\quad\) m | ||
| CoM: \(4m(\frac{2}{3}u) = 4m \cdot w_1 + m \cdot w_2\) \(\Rightarrow\) \(6u = 4w_1 + w_2\) | M1A1 | |
| NEL: \(e(\frac{2}{3}u) = w_2 - w_1\) | M1A1 | |
| Solving for \(w_1\) as f(e): \(w_1 = \frac{3u(4 - e)}{10}\) or e as f(\(w_1\)): \(e = \frac{2(6u - 5w_1)}{3u}\) | M1A1 | (8) |
| Answer | Marks |
|---|---|
| Note: If \(w_1\) or e not found (not asked for): Setting \(w_1 = v = u \Rightarrow w_2 = 2u \Rightarrow e = \frac{2}{3}\) is M1A1 but need to deal with inequality for final M1A1 |
**(a)** $\rightarrow v_1$ $\rightarrow v_2$ | M1A1 |
$\rightarrow 2u$ $\rightarrow u$ | |
A: $\circ$ 2m $\quad$ B: $\circ$ 4m | |
CoM: $4mu + 4mu = 2m \cdot v_1 + 4m \cdot v_2$ $\Rightarrow$ $4u = v_1 + 2v_2$ | M1A1 |
NEL: $\frac{1}{2}(2u - u) = v_2 - v_1$ | M1A1cso(6) |
Solving to find $v_2$; $v_2 = \frac{3u}{2}$ | M1A1 | (2)
**(b)** Substitute for $v_2$ in one equation; $v_1 = v_2 - \frac{1}{2}u = u$ | M1A1 | (2)
**(c)** $\rightarrow w_1$ $\rightarrow w_2$ | M1A1 |
$\rightarrow \frac{2}{3}u$ $\rightarrow 0$ | |
$\circ$ B $\quad$ $\circ$ C | |
4m $\quad$ m | |
CoM: $4m(\frac{2}{3}u) = 4m \cdot w_1 + m \cdot w_2$ $\Rightarrow$ $6u = 4w_1 + w_2$ | M1A1 |
NEL: $e(\frac{2}{3}u) = w_2 - w_1$ | M1A1 |
Solving for $w_1$ as f(e): $w_1 = \frac{3u(4 - e)}{10}$ or e as f($w_1$): $e = \frac{2(6u - 5w_1)}{3u}$ | M1A1 | (8)
Requirement is that $w_1 \geq$ candidate's $v_1 = u$; $\Rightarrow$ $e \leq \frac{2}{5}$
**Note:** If $w_1$ or e not found (not asked for): Setting $w_1 = v = u \Rightarrow w_2 = 2u \Rightarrow e = \frac{2}{3}$ is M1A1 but need to deal with inequality for final M1A1 | |
---A particle $A$ of mass $2m$ is moving with speed $2u$ on a smooth horizontal table. The particle collides directly with a particle $B$ of mass $4m$ moving with speed $u$ in the same direction as $A$. The coefficient of restitution between $A$ and $B$ is $\frac{1}{2}$.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $B$ after the collision is $\frac{5}{3}u$.
[6]
\item Find the speed of $A$ after the collision.
[2]
\end{enumerate}
Subsequently $B$ collides directly with a particle $C$ of mass $m$ which is at rest on the table. The coefficient of restitution between $B$ and $C$ is $e$. Given that there are no further collisions,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the range of possible values for $e$.
[8]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2001 Q6 [16]}}