| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2003 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Two-sphere oblique collision |
| Difficulty | Challenging +1.8 This is a challenging M4 mechanics problem requiring multiple collision analysis with both ball-ball and ball-wall interactions, coefficient of restitution calculations, and geometric reasoning across five interconnected parts. While the individual techniques (momentum conservation, Newton's law of restitution, kinematics) are standard M4 content, the extended multi-stage nature with two collisions and careful tracking of both balls' positions demands sustained problem-solving over many steps. The final part requiring a distance ratio involves synthesizing all previous results. This is harder than typical M4 questions but not exceptionally so for this advanced module. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.02d Constant acceleration: SUVAT formulae6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| P before: \(\rightarrow \frac{13u}{12}\cos\alpha = u\), \(\uparrow \frac{13u}{12}\sin\alpha = \frac{5u}{12}\) | B1, B1 | |
| \(\rightarrow u\) | \(\rightarrow 0\) | |
| \(m\) | \(2m\) | |
| PCLM (\(\rightarrow\)): \(mu = mv + 2m\frac{3u}{5} \Rightarrow v = -\frac{u}{5}\), i.e. \(\frac{u}{5}\) // CB | M1 A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| NLI \(\rightarrow eu = v_2 - v_1 \Rightarrow eu = \frac{3u}{5} - (-\frac{u}{5})\), i.e. \(e = \frac{4}{5}\) | M1, A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| \(O \rightarrow C\): \(t_1 = \frac{d_1}{3u/5} = \frac{5d_1}{3u}\) | B1 | |
| P travels \(\frac{u}{5} \times \frac{5d_1}{3u} = \frac{d_1}{3}\) in direction CB | M1 | |
| \(\therefore\) P is \(d_1 + \frac{d_1}{3} = \frac{4d_1}{3}\) from \(w\) | A1 cso | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| After hitting \(w\), Q has speed \(\frac{3u}{10}\) in direction CB | B1 | |
| Velocity of Q relative to P in direction CB is \(\frac{u}{10}\) | M1 | |
| Time for Q to travel \(\frac{4}{3}d_1\) is: \(\frac{4d_1/3}{u/10} = \frac{40d_1}{3u}\) | A1 | |
| Total time between collisions is: \(\frac{40d_1}{3u} + \frac{5d_1}{3u} = \frac{15d_1}{u}\) | A1 cso | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| For collision to occur P must travel \(\uparrow d_2\) and \(\downarrow d_2\) in time \(\frac{15d_1}{u}\) | ||
| \(d_2 \uparrow\): \(t_2 = \frac{d_2}{5u/12} = \frac{12d_2}{5u}\) | B1 | |
| \(\downarrow d_2\) velocity \(\downarrow\) is \(\frac{5u}{24}\), \(\therefore t_3 = \frac{d_2}{5u/24} = \frac{24d_2}{5u}\) | B1, B1 | |
| Total time is: \(\frac{36d_2}{5u} = \frac{15d_1}{u}\) | M1 | |
| \(\therefore 12d_2 = 25d_1\), i.e. \(d_1:d_2 = 12:25\) | A1 | (5) |
| (18 marks) |
## Part (a)
| Content | Marks | Notes |
|---------|-------|-------|
| P before: $\rightarrow \frac{13u}{12}\cos\alpha = u$, $\uparrow \frac{13u}{12}\sin\alpha = \frac{5u}{12}$ | B1, B1 | |
| | | |
| $\rightarrow u$ | $\rightarrow 0$ | | $\rightarrow v$ | $\rightarrow \frac{3u}{5}$ |
| $m$ | $2m$ | | $m$ | $2m$ |
| | | |
| PCLM ($\rightarrow$): $mu = mv + 2m\frac{3u}{5} \Rightarrow v = -\frac{u}{5}$, i.e. $\frac{u}{5}$ // CB | M1 A1 | (4) |
## Part (b)
| Content | Marks | Notes |
|---------|-------|-------|
| NLI $\rightarrow eu = v_2 - v_1 \Rightarrow eu = \frac{3u}{5} - (-\frac{u}{5})$, i.e. $e = \frac{4}{5}$ | M1, A1 | (2) |
## Part (c)
| Content | Marks | Notes |
|---------|-------|-------|
| $O \rightarrow C$: $t_1 = \frac{d_1}{3u/5} = \frac{5d_1}{3u}$ | B1 | |
| P travels $\frac{u}{5} \times \frac{5d_1}{3u} = \frac{d_1}{3}$ in direction CB | M1 | |
| $\therefore$ P is $d_1 + \frac{d_1}{3} = \frac{4d_1}{3}$ from $w$ | A1 cso | (3) |
## Part (d)
| Content | Marks | Notes |
|---------|-------|-------|
| After hitting $w$, Q has speed $\frac{3u}{10}$ in direction CB | B1 | |
| Velocity of Q relative to P in direction CB is $\frac{u}{10}$ | M1 | |
| Time for Q to travel $\frac{4}{3}d_1$ is: $\frac{4d_1/3}{u/10} = \frac{40d_1}{3u}$ | A1 | |
| Total time between collisions is: $\frac{40d_1}{3u} + \frac{5d_1}{3u} = \frac{15d_1}{u}$ | A1 cso | (4) |
## Part (e)
| Content | Marks | Notes |
|---------|-------|-------|
| For collision to occur P must travel $\uparrow d_2$ and $\downarrow d_2$ in time $\frac{15d_1}{u}$ | | |
| | | |
| $d_2 \uparrow$: $t_2 = \frac{d_2}{5u/12} = \frac{12d_2}{5u}$ | B1 | |
| | | |
| $\downarrow d_2$ velocity $\downarrow$ is $\frac{5u}{24}$, $\therefore t_3 = \frac{d_2}{5u/24} = \frac{24d_2}{5u}$ | B1, B1 | |
| | | |
| Total time is: $\frac{36d_2}{5u} = \frac{15d_1}{u}$ | M1 | |
| | | |
| $\therefore 12d_2 = 25d_1$, i.e. $d_1:d_2 = 12:25$ | A1 | (5) |
| | **(18 marks)** | |\includegraphics{figure_2}
A small ball $Q$ of mass $2m$ is at rest at the point $B$ on a smooth horizontal plane. A second small ball $P$ of mass $m$ is moving on the plane with speed $\frac{13}{12}u$ and collides with $Q$. Both the balls are smooth, uniform and of the same radius. The point $C$ is on a smooth vertical wall $W$ which is at a distance $d_1$ from $B$, and $BC$ is perpendicular to $W$. A second smooth vertical wall is perpendicular to $W$ and at a distance $d_2$ from $B$. Immediately before the collision occurs, the direction of motion of $P$ makes an angle $\alpha$ with $BC$, as shown in Fig. 2, where $\tan \alpha = \frac{5}{12}$. The line of centres of $P$ and $Q$ is parallel to $BC$. After the collision $Q$ moves towards $C$ with speed $\frac{5}{4}u$.
\begin{enumerate}[label=(\alph*)]
\item Show that, after the collision, the velocity components of $P$ parallel and perpendicular to $CB$ are $\frac{1}{4}u$ and $\frac{5}{12}u$ respectively.
[4]
\item Find the coefficient of restitution between $P$ and $Q$.
[2]
\item Show that when $Q$ reaches $C$, $P$ is at a distance $\frac{4}{5}d_1$ from $W$.
[3]
\end{enumerate}
For each collision between a ball and a wall the coefficient of restitution is $\frac{1}{2}$.
Given that the balls collide with each other again,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item show that the time between the two collisions of the balls is $\frac{15d_1}{u}$.
[4]
\item find the ratio $d_1 : d_2$.
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2003 Q6 [18]}}