| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, direction deflected given angle |
| Difficulty | Challenging +1.2 This is a standard M4 oblique collision problem requiring resolution of velocities along and perpendicular to the line of centers, application of conservation of momentum and Newton's restitution law, then finding the deflection angle. While it involves multiple steps and careful trigonometry, it follows a well-established method taught in M4 with no novel insight required. The 10 marks reflect the algebraic manipulation needed, but this is a textbook-style question slightly above average difficulty due to the algebraic complexity of reaching the given answer. |
| Spec | 6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v_1 + v_2 = u\cos 30°\) | M1 A1 | Conservation of momentum (horizontal) |
| \(-v_1 + v_2 = eu\cos 30°\) | M1 A1 | Newton's law of restitution |
| Subtracting: \(v_1 = \frac{u\sqrt{3}}{4}(1-e)\) | A1 | |
| \(\tan\theta = \tan(\alpha - 30°) = \frac{\tan\alpha - \tan 30°}{1 + \tan\alpha \tan 30°}\) | M1 | |
| \(\tan\alpha = \frac{u\sin 30°}{v_1} = \frac{2}{\sqrt{3}(1-e)}\) | M1 A1 | |
| \(\tan\theta = \dfrac{\dfrac{2}{\sqrt{3}(1-e)} - \dfrac{1}{\sqrt{3}}}{1 + \dfrac{2}{\sqrt{3}(1-e)}\cdot\dfrac{1}{\sqrt{3}}}\) | M1 | |
| \(= \dfrac{(1+e)\sqrt{3}}{5-3e}\) (*) | A1 | (10) |
# Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v_1 + v_2 = u\cos 30°$ | M1 A1 | Conservation of momentum (horizontal) |
| $-v_1 + v_2 = eu\cos 30°$ | M1 A1 | Newton's law of restitution |
| Subtracting: $v_1 = \frac{u\sqrt{3}}{4}(1-e)$ | A1 | |
| $\tan\theta = \tan(\alpha - 30°) = \frac{\tan\alpha - \tan 30°}{1 + \tan\alpha \tan 30°}$ | M1 | |
| $\tan\alpha = \frac{u\sin 30°}{v_1} = \frac{2}{\sqrt{3}(1-e)}$ | M1 A1 | |
| $\tan\theta = \dfrac{\dfrac{2}{\sqrt{3}(1-e)} - \dfrac{1}{\sqrt{3}}}{1 + \dfrac{2}{\sqrt{3}(1-e)}\cdot\dfrac{1}{\sqrt{3}}}$ | M1 | |
| $= \dfrac{(1+e)\sqrt{3}}{5-3e}$ (*) | A1 | **(10)** |
---3.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{d57ea92a-4d6a-46bf-a6aa-bbd5083e8726-3_469_1163_1217_443}
\end{center}
\end{figure}
A smooth uniform sphere $A$, moving on a smooth horizontal table, collides with a second identical sphere $B$ which is at rest on the table. When the spheres collide the line joining their centres makes an angle of $30 ^ { \circ }$ with the direction of motion of $A$, as shown in Fig. 1. The coefficient of restitution between the spheres is $e$. The direction of motion of $A$ is deflected through an angle $\theta$ by the collision.
Show that $\tan \theta = \frac { ( 1 + e ) \sqrt { 3 } } { 5 - 3 e }$.\\
(10 marks)\\
\hfill \mbox{\textit{Edexcel M4 Q3 [10]}}