Edexcel M4 2008 June — Question 4 8 marks

Exam BoardEdexcel
ModuleM4 (Mechanics 4)
Year2008
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeSphere rebounds off fixed wall obliquely
DifficultyStandard +0.8 This M4 oblique collision problem requires resolving velocities parallel and perpendicular to the wall, applying the coefficient of restitution correctly to the normal component only, then using the geometric constraint that angles change from 2θ to θ. It demands careful vector resolution, understanding of restitution in 2D collisions, and algebraic manipulation involving trigonometric identities—significantly above average difficulty but standard for M4 level.
Spec6.03k Newton's experimental law: direct impact

4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{376d12ab-022c-4070-a1e0-88eacc2fe48e-2_451_357_1672_852} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} A small smooth ball \(B\), moving on a horizontal plane, collides with a fixed vertical wall. Immediately before the collision the angle between the direction of motion of \(B\) and the wall is \(2 \theta\), where \(0 ^ { \circ } < \theta < 45 ^ { \circ }\). Immediately after the collision the angle between the direction of motion of \(B\) and the wall is \(\theta\), as shown in Figure 1. Given that the coefficient of restitution between \(B\) and the wall is \(\frac { 3 } { 8 }\), find the value of \(\tan \theta\).
(8)

Question 4:
AnswerMarks Guidance
Working/AnswerMarks Notes
\(u\cos2\theta = v\cos\theta\)M1 A1
\(\frac{3}{8}u\sin2\theta = v\sin\theta\)M1 A1
\(3\tan2\theta = 8\tan\theta\)M1
\(\dfrac{6\tan\theta}{1-\tan^2\theta} = 8\tan\theta\)M1
\(\tan^2\theta = \frac{1}{4} \quad (\tan\theta \neq 0)\)
\(\tan\theta = \frac{1}{2}\)M1 A1 Total: 8
## Question 4:

| Working/Answer | Marks | Notes |
|---|---|---|
| $u\cos2\theta = v\cos\theta$ | M1 A1 | |
| $\frac{3}{8}u\sin2\theta = v\sin\theta$ | M1 A1 | |
| $3\tan2\theta = 8\tan\theta$ | M1 | |
| $\dfrac{6\tan\theta}{1-\tan^2\theta} = 8\tan\theta$ | M1 | |
| $\tan^2\theta = \frac{1}{4} \quad (\tan\theta \neq 0)$ | | |
| $\tan\theta = \frac{1}{2}$ | M1 A1 | **Total: 8** |

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4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{376d12ab-022c-4070-a1e0-88eacc2fe48e-2_451_357_1672_852}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

A small smooth ball $B$, moving on a horizontal plane, collides with a fixed vertical wall. Immediately before the collision the angle between the direction of motion of $B$ and the wall is $2 \theta$, where $0 ^ { \circ } < \theta < 45 ^ { \circ }$. Immediately after the collision the angle between the direction of motion of $B$ and the wall is $\theta$, as shown in Figure 1.

Given that the coefficient of restitution between $B$ and the wall is $\frac { 3 } { 8 }$, find the value of $\tan \theta$.\\
(8)\\

\hfill \mbox{\textit{Edexcel M4 2008 Q4 [8]}}