| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Collision with unchanged direction |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring systematic application of conservation of momentum and Newton's restitution law across two collisions. While it has multiple parts and requires careful algebraic manipulation, the techniques are routine for M2 students and the constraint that A's direction is unchanged provides helpful guidance. The 'show that' part (c) requires comparing velocities but follows directly from the algebra without requiring novel insight. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| \(mu = mv_1 + 2mv_2\) | M1 A1 | |
| \(eu = -v_1 + v_2\) | M1 A1 | |
| \(v_1 = \frac{u}{3}(1 - 2e); \quad v_2 = \frac{u}{3}(1 + e)\) | M1 A1 A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_1 > 0 \Rightarrow \frac{u}{3}(1 - 2e) > 0 \Rightarrow e < \frac{1}{2}\) | M1 A1 | (2) |
| Answer | Marks |
|---|---|
| \(2mv_2 = 2mv_3 + 4mv_4\) | M1 |
| \(ev_2 = -v_3 + v_4\) | M1 A1 |
| \(v_3 = \frac{v_2}{3}(1 - 2e) = \frac{u}{9}(1 - 2e)(1 + e)\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| i.e. if \(\frac{u}{3}(1 - 2e) > \frac{u}{9}(1 - 2e)(1 + e)\) | M1 | |
| i.e. if \(3 > 1 + e\) (as \((1 - 2e) > 0\)) | M1 | |
| i.e. if \(2 > e\) | M1 | |
| which is always true, so further collision occurs | A1 cso | (6) |
## Part (a)
$mu = mv_1 + 2mv_2$ | M1 A1
$eu = -v_1 + v_2$ | M1 A1
$v_1 = \frac{u}{3}(1 - 2e); \quad v_2 = \frac{u}{3}(1 + e)$ | M1 A1 A1 | (7)
## Part (b)
$v_1 > 0 \Rightarrow \frac{u}{3}(1 - 2e) > 0 \Rightarrow e < \frac{1}{2}$ | M1 A1 | (2)
## Part (c)
$v_2 \rightarrow \quad \rightarrow 0$
$2mv_2 = 2mv_3 + 4mv_4$ | M1
$ev_2 = -v_3 + v_4$ | M1 A1
$v_3 = \frac{v_2}{3}(1 - 2e) = \frac{u}{9}(1 - 2e)(1 + e)$ | M1 A1
Further collision if $v_1 > v_3$
i.e. if $\frac{u}{3}(1 - 2e) > \frac{u}{9}(1 - 2e)(1 + e)$ | M1
i.e. if $3 > 1 + e$ (as $(1 - 2e) > 0$) | M1
i.e. if $2 > e$ | M1
which is always true, so further collision occurs | A1 cso | (6)
**Total: (15 marks)**A uniform sphere $A$ of mass $m$ is moving with speed $u$ on a smooth horizontal table when it collides directly with another uniform sphere $B$ of mass $2m$ which is at rest on the table. The spheres are of equal radius and the coefficient of restitution between them is $e$. The direction of motion of $A$ is unchanged by the collision.
\begin{enumerate}[label=(\alph*)]
\item Find the speeds of $A$ and $B$ immediately after the collision.
[7]
\item Find the range of possible values of $e$.
[2]
\end{enumerate}
After being struck by $A$, the sphere $B$ collides directly with another sphere $C$, of mass $4m$ and of the same size as $B$. The sphere $C$ is at rest on the table immediately before being struck by $B$. The coefficient of restitution between $B$ and $C$ is also $e$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show that, after $B$ has struck $C$, there will be a further collision between $A$ and $B$.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2003 Q7 [15]}}