| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, find velocities/angles |
| Difficulty | Standard +0.3 This is a standard M3 oblique collision question requiring resolution of velocities parallel and perpendicular to the line of centres, application of conservation of momentum and Newton's law of restitution. While it involves multiple steps, the method is routine and well-practiced at this level, making it slightly easier than average. |
| Spec | 6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| \(m \times 8\cos 30° + m \times 4\cos 60° = mv_A + mv_B\) | M1A1 | OE unsimplified |
| \(v_A + v_B = 8.93\) | m1 | Dependent on both M1s |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{v_B - v_A}{8\cos 30° - 4\cos 60°} = \frac{1}{2}\) | M1A1 | OE unsimplified |
| \(v_B - v_A = 2.46\) | m1 | Dependent on both M1s |
| \(v_B = 5.70\) | A1F | AWRT (or \(3\sqrt{3} + \frac{1}{2}\)) |
| Momentum of \(B\) perpendicular to the line of centres is unchanged | B1 | PI (can also be gained in part (b)) |
| Speed of \(B = \sqrt{u_B^2 + v_B^2} = \sqrt{(4\sin 60°)^2 + (5.70)^2} = 6.67\) | m1A1F | Dependent on both M1s |
| 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Direction of \(B = \tan^{-1}\frac{4\sin 60°}{5.70} = 31.3°\) | m1A1F | Dependent on both M1s and B1 |
| 2 | ||
| 11 | Total |
### Part (a)
Conservation of linear momentum along the line of centres:
$m \times 8\cos 30° + m \times 4\cos 60° = mv_A + mv_B$ | M1A1 | OE unsimplified
$v_A + v_B = 8.93$ | m1 | Dependent on both M1s
Law of restitution along the line of centre:
$\frac{v_B - v_A}{8\cos 30° - 4\cos 60°} = \frac{1}{2}$ | M1A1 | OE unsimplified
$v_B - v_A = 2.46$ | m1 | Dependent on both M1s
$v_B = 5.70$ | A1F | AWRT (or $3\sqrt{3} + \frac{1}{2}$)
Momentum of $B$ perpendicular to the line of centres is unchanged | B1 | PI (can also be gained in part (b))
Speed of $B = \sqrt{u_B^2 + v_B^2} = \sqrt{(4\sin 60°)^2 + (5.70)^2} = 6.67$ | m1A1F | Dependent on both M1s
| | **9** |
### Part (b)
Direction of $B = \tan^{-1}\frac{4\sin 60°}{5.70} = 31.3°$ | m1A1F | Dependent on both M1s and B1
| | **2** |
| | **11** | **Total** |6 Two smooth billiard balls $A$ and $B$, of identical size and equal mass, move towards each other on a horizontal surface and collide. Just before the collision, $A$ has velocity $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a direction inclined at $30 ^ { \circ }$ to the line of centres of the balls, and $B$ has velocity $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a direction inclined at $60 ^ { \circ }$ to the line of centres, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{f8c04360-f54b-4d08-aee9-fe28612918d0-4_508_1420_532_294}
The coefficient of restitution between the balls is $\frac { 1 } { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $B$ immediately after the collision.
\item Find the angle between the velocity of $B$ and the line of centres of the balls immediately after the collision.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2006 Q6 [11]}}