| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, find velocities/angles |
| Difficulty | Standard +0.8 This is a standard M3 oblique collision problem requiring resolution of velocities parallel and perpendicular to the line of centres, application of Newton's experimental law (coefficient of restitution), and conservation of momentum. While it involves multiple steps and careful component work, it follows a well-established procedure taught in M3 with no novel insights required. The multi-part structure and need for systematic application of several principles places it above average difficulty but well within the standard M3 syllabus. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Component of velocity of \(A\) perpendicular to line of centres is unchanged: \(u\sin 30° = \frac{u}{2}\) | B1 | Perpendicular component preserved |
| Component of velocity of \(A\) along line of centres before collision: \(u\cos 30° = \frac{u\sqrt{3}}{2}\) | B1 | |
| Let \(v_A\) = component of \(A\) along line of centres after, \(v_B\) = velocity of \(B\) after | ||
| Conservation of momentum along line of centres: \(4m \cdot \frac{u\sqrt{3}}{2} = 4m \cdot v_A + 3m \cdot v_B\) | M1 A1 | |
| \(2u\sqrt{3} = 4v_A + 3v_B\) | A1 | |
| Newton's Law of Restitution along line of centres: \(v_B - v_A = \frac{5}{9} \cdot \frac{u\sqrt{3}}{2}\) | M1 A1 | |
| \(v_B - v_A = \frac{5u\sqrt{3}}{18}\) | ||
| Solving: \(v_A = \frac{u\sqrt{3}}{4}\), \(v_B = \frac{7u\sqrt{3}}{12}\) | DM1 A1 | |
| \(\tan\alpha = \frac{u/2}{u\sqrt{3}/4} = \frac{2}{\sqrt{3}}\) | M1 | |
| \(\alpha = \arctan\!\left(\frac{2}{\sqrt{3}}\right) \approx 49.1°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Impulse \(= 3m \cdot v_B = 3m \cdot \frac{7u\sqrt{3}}{12}\) | M1 A1 | |
| \(= \frac{7mu\sqrt{3}}{4}\) | A1 | Accept equivalent exact form |
# Question 7:
## Part (a): Find the value of α (10 marks)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Component of velocity of $A$ perpendicular to line of centres is unchanged: $u\sin 30° = \frac{u}{2}$ | B1 | Perpendicular component preserved |
| Component of velocity of $A$ along line of centres before collision: $u\cos 30° = \frac{u\sqrt{3}}{2}$ | B1 | |
| Let $v_A$ = component of $A$ along line of centres after, $v_B$ = velocity of $B$ after | | |
| Conservation of momentum along line of centres: $4m \cdot \frac{u\sqrt{3}}{2} = 4m \cdot v_A + 3m \cdot v_B$ | M1 A1 | |
| $2u\sqrt{3} = 4v_A + 3v_B$ | A1 | |
| Newton's Law of Restitution along line of centres: $v_B - v_A = \frac{5}{9} \cdot \frac{u\sqrt{3}}{2}$ | M1 A1 | |
| $v_B - v_A = \frac{5u\sqrt{3}}{18}$ | | |
| Solving: $v_A = \frac{u\sqrt{3}}{4}$, $v_B = \frac{7u\sqrt{3}}{12}$ | DM1 A1 | |
| $\tan\alpha = \frac{u/2}{u\sqrt{3}/4} = \frac{2}{\sqrt{3}}$ | M1 | |
| $\alpha = \arctan\!\left(\frac{2}{\sqrt{3}}\right) \approx 49.1°$ | A1 | |
## Part (b): Magnitude of impulse on B (3 marks)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Impulse $= 3m \cdot v_B = 3m \cdot \frac{7u\sqrt{3}}{12}$ | M1 A1 | |
| $= \frac{7mu\sqrt{3}}{4}$ | A1 | Accept equivalent exact form |
These images show blank answer pages (pages 22, 23, and 24) from an AQA exam paper (P38108/Jun11/MM03). They contain no mark scheme content — they are continuation answer sheets and a "no questions printed on this page" page.
There is no mark scheme content to extract from these pages.7 Two smooth spheres, $A$ and $B$, have equal radii and masses $4 m$ and $3 m$ respectively. The sphere $A$ is moving on a smooth horizontal surface and collides with the sphere $B$, which is stationary on the same surface.
Just before the collision, $A$ is moving with speed $u$ at an angle of $30 ^ { \circ }$ to the line of centres, as shown in the diagram below.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Before collision}
\includegraphics[alt={},max width=\textwidth]{0590950d-145c-4ae2-bc3c-f61a9433d158-20_362_933_664_450}
\end{center}
\end{figure}
Immediately after the collision, the direction of motion of $A$ makes an angle $\alpha$ with the line of centres, as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{0590950d-145c-4ae2-bc3c-f61a9433d158-20_449_927_1244_456}
The coefficient of restitution between the spheres is $\frac { 5 } { 9 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\alpha$.
\item Find, in terms of $m$ and $u$, the magnitude of the impulse exerted on $B$ during the collision.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0590950d-145c-4ae2-bc3c-f61a9433d158-23_2349_1707_221_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2011 Q7 [13]}}