| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2002 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Two-sphere oblique collision |
| Difficulty | Challenging +1.2 This is a standard M4 oblique collision problem requiring systematic application of momentum conservation (perpendicular and parallel to line of centres) and Newton's restitution law. Part (a) involves algebraic manipulation to show a given result, while part (b) adds energy considerations. The perpendicularity condition provides the key constraint, but the solution follows a well-established method taught in M4 courses. More challenging than basic mechanics due to the oblique collision setup and multi-variable algebra, but still a textbook-style question without requiring novel insight. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| CLM: \(mu \cos \theta = kmv\) | M1A1 | |
| NIL: \(eu \cos \theta = v\) | M1A1 | |
| Eliminating \(\theta\) | M1 | |
| \(e = \frac{1}{k}\) | A1 | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}m v_e^2 + \frac{1}{3}(2M)(\frac{1}{2}u\cos\theta)^2 = \frac{3}{4}\frac{1}{2}mu\dot{2}\) (or equivalent) | M1A1 | |
| \(\frac{1}{2}m(u\sin\theta)^2 + \frac{1}{3}(2M)(\frac{1}{2}u\cos\theta)^2 = \frac{3}{4}\frac{1}{2}m\dot{u}^2\) | M1A1√ | |
| [4 sin²\(\theta\) + 2 cos²\(\theta\) = 3] | (M1) | |
| \(4\sin^2\theta + 2(1 - \sin^2\theta) = 3\) | M1 | |
| \(\sin^2\theta = \frac{1}{2}\) | A1 | (6 marks) |
| \(\theta = 45°\) | A1 |
## Part (a)
CLM: $mu \cos \theta = kmv$ | M1A1
NIL: $eu \cos \theta = v$ | M1A1
Eliminating $\theta$ | M1
$e = \frac{1}{k}$ | A1 | (6 marks)
## Part (b)
$\frac{1}{2}m v_e^2 + \frac{1}{3}(2M)(\frac{1}{2}u\cos\theta)^2 = \frac{3}{4}\frac{1}{2}mu\dot{2}$ (or equivalent) | M1A1
$\frac{1}{2}m(u\sin\theta)^2 + \frac{1}{3}(2M)(\frac{1}{2}u\cos\theta)^2 = \frac{3}{4}\frac{1}{2}m\dot{u}^2$ | M1A1√
[4 sin²$\theta$ + 2 cos²$\theta$ = 3] | (M1)
$4\sin^2\theta + 2(1 - \sin^2\theta) = 3$ | M1
$\sin^2\theta = \frac{1}{2}$ | A1 | (6 marks)
$\theta = 45°$ | A1
[$\frac{1}{3}m(u\cos\theta)^2 - \frac{1}{3}(2M)(\frac{1}{2}u\cos\theta)^2 = \frac{3}{4}\frac{1}{2}m(u\cos\theta)^2$, then max M1]
---\includegraphics{figure_1}
A smooth uniform sphere $S$ of mass $m$ is moving on a smooth horizontal table. The sphere $S$ collides with another smooth uniform sphere $T$, of the same radius as $S$ but of mass $km$, $k > 1$, which is at rest on the table. The coefficient of restitution between the spheres is $e$. Immediately before the spheres collide the direction of motion of $S$ makes an angle $\theta$ with the line joining their centres, as shown in Fig. 1.
Immediately after the collision the directions of motion of $S$ and $T$ are perpendicular.
\begin{enumerate}[label=(\alph*)]
\item Show that $e = \frac{1}{k}$. [6]
\end{enumerate}
Given that $k = 2$ and that the kinetic energy lost in the collision is one quarter of the initial kinetic energy,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the value of $\theta$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2002 Q5 [12]}}