| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2003 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, find velocities/angles |
| Difficulty | Challenging +1.2 This is a standard M4 oblique collision problem requiring systematic application of conservation of momentum and Newton's law of restitution in the line of centres direction. Parts (a) and (b) involve routine coordinate transformations and algebraic manipulation to show given results. Part (c) requires tracking both spheres' trajectories to the wall using basic kinematics. While multi-step and requiring careful bookkeeping, it follows a well-established template for this topic with no novel insights needed. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| No impulse perpendicular to line of centres \(\Rightarrow\) velocity perpendicular to line of centres unchanged \(= U\cos\alpha\) | B1 | (*) |
| (CLM) \(U\sin\alpha = v + w\) | M1 A1 | |
| NLI: \(eU\sin\alpha = w - v\) | M1 A1 | |
| \(\Rightarrow v = \frac{1}{2}U\sin\alpha(1-e)\) | M1 A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Component perpendicular to wall \(= v\sin\alpha + U\cos\alpha\cos\alpha\) | M1 | |
| \(= \frac{1}{2}U\sin^2\alpha(1-e) + U\cos^2\alpha\) | ||
| \(= \frac{1}{2}U(\sin^2\alpha - e\sin^2\alpha + 2 - 2\sin^2\alpha)\) | M1 | |
| \(= \frac{1}{2}U[2 - \sin^2\alpha(1+e)]\) | A1 | (*) |
| Component parallel to wall \(= U\cos\alpha\sin\alpha - v\cos\alpha\) | M1 | |
| \(= U\cos\alpha\sin\alpha - \frac{1}{2}U\sin\alpha\cos\alpha(1-e)\) | A1 | |
| \(= \frac{1}{2}U\cos\alpha\sin\alpha(1+e)\) | A1 | (6) (*) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(e = \frac{2}{3}\), \(\tan\alpha = \frac{3}{4}\) | ||
| Component perpendicular to wall \(= \frac{1}{2}U\!\left(2 - \frac{9}{25}\times\frac{5}{3}\right) = \frac{7u}{10}\) | B1 | |
| Component parallel to wall \(= \frac{1}{2}U\times\frac{4}{5}\times\frac{3}{12}\times\frac{5}{3} = \frac{2u}{5}\) | ||
| Distance of \(A\) from \(X\): \(d\cot\theta = \frac{4d}{3}\) | B1 | |
| \(BX = d\cot\theta\) | M1 | |
| \(\cot\theta = \frac{2u}{5}\times\frac{7u}{10} = \frac{4}{7}\) | A1 | |
| Total distance \(AB = \frac{4d}{3} + \frac{4d}{7} = \frac{40d}{21}\) | A1 | (5) |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| No impulse perpendicular to line of centres $\Rightarrow$ velocity perpendicular to line of centres unchanged $= U\cos\alpha$ | B1 | (*) |
| (CLM) $U\sin\alpha = v + w$ | M1 A1 | |
| NLI: $eU\sin\alpha = w - v$ | M1 A1 | |
| $\Rightarrow v = \frac{1}{2}U\sin\alpha(1-e)$ | M1 A1 | (7) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Component perpendicular to wall $= v\sin\alpha + U\cos\alpha\cos\alpha$ | M1 | |
| $= \frac{1}{2}U\sin^2\alpha(1-e) + U\cos^2\alpha$ | | |
| $= \frac{1}{2}U(\sin^2\alpha - e\sin^2\alpha + 2 - 2\sin^2\alpha)$ | M1 | |
| $= \frac{1}{2}U[2 - \sin^2\alpha(1+e)]$ | A1 | (*) |
| Component parallel to wall $= U\cos\alpha\sin\alpha - v\cos\alpha$ | M1 | |
| $= U\cos\alpha\sin\alpha - \frac{1}{2}U\sin\alpha\cos\alpha(1-e)$ | A1 | |
| $= \frac{1}{2}U\cos\alpha\sin\alpha(1+e)$ | A1 | (6) (*) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $e = \frac{2}{3}$, $\tan\alpha = \frac{3}{4}$ | | |
| Component perpendicular to wall $= \frac{1}{2}U\!\left(2 - \frac{9}{25}\times\frac{5}{3}\right) = \frac{7u}{10}$ | B1 | |
| Component parallel to wall $= \frac{1}{2}U\times\frac{4}{5}\times\frac{3}{12}\times\frac{5}{3} = \frac{2u}{5}$ | | |
| Distance of $A$ from $X$: $d\cot\theta = \frac{4d}{3}$ | B1 | |
| $BX = d\cot\theta$ | M1 | |
| $\cot\theta = \frac{2u}{5}\times\frac{7u}{10} = \frac{4}{7}$ | A1 | |
| Total distance $AB = \frac{4d}{3} + \frac{4d}{7} = \frac{40d}{21}$ | A1 | (5) |6.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{47e1d96b-4582-4324-a946-66989a2c66fc-5_652_725_296_620}
\end{center}
\end{figure}
A small smooth uniform sphere $S$ is at rest on a smooth horizontal floor at a distance $d$ from a straight vertical wall. An identical sphere $T$ is projected along the floor with speed $U$ towards $S$ and in a direction which is perpendicular to the wall. At the instant when $T$ strikes $S$ the line joining their centres makes an angle $\alpha$ with the wall, as shown in Fig. 3.
Each sphere is modelled as having negligible diameter in comparison with $d$. The coefficient of restitution between the spheres is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that the components of the velocity of $T$ after the impact, parallel and perpendicular to the line of centres, are $\frac { 1 } { 2 } U ( 1 - e ) \sin \alpha$ and $U \cos \alpha$ respectively.
\item Show that the components of the velocity of $T$ after the impact, parallel and perpendicular to the wall, are $\frac { 1 } { 2 } U ( 1 + e ) \cos \alpha \sin \alpha$ and $\frac { 1 } { 2 } U \left[ 2 - ( 1 + e ) \sin ^ { 2 } \alpha \right]$ respectively.
The spheres $S$ and $T$ strike the wall at the points $A$ and $B$ respectively.\\
Given that $e = \frac { 2 } { 3 }$ and $\tan \alpha = \frac { 3 } { 4 }$,
\item find, in terms of $d$, the distance $A B$.
\section*{END}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2003 Q6 [18]}}