| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, vector velocity form |
| Difficulty | Standard +0.3 This is a standard M3 oblique collision question requiring application of conservation of momentum (in i-direction), Newton's law of restitution, and recognition that j-components are unchanged. The setup is straightforward with clear given values, and the solution follows a routine procedure taught in all M3 courses. While it requires multiple steps and careful vector work, it involves no novel insight or particularly challenging conceptual leaps—slightly easier than average A-level maths. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks |
|---|---|
| The impulse acts along the line of centres, which is parallel to \(\mathbf{i}\), so no impulse in \(\mathbf{j}\) direction | B1 |
| Answer | Marks |
|---|---|
| Only \(\mathbf{i}\) components change; \(\mathbf{j}\) components remain \(3\mathbf{j}\) and \(-2\mathbf{j}\) respectively | B1 |
| Conservation of momentum (i-component): \(1(2) + 2(-1) = 1(v_A) + 2(v_B)\) | M1 |
| \(v_A + 2v_B = 0\) | A1 |
| NEL: \(v_B - v_A = 0.5(2-(-1)) = 1.5\) | M1 |
| \(v_B - v_A = 1.5\) | A1 |
| \(v_A = -1, \quad v_B = 0.5\) | A1 |
| Velocity of \(A\): \((-\mathbf{i} + 3\mathbf{j})\) m s\(^{-1}\); Velocity of \(B\): \((0.5\mathbf{i} - 2\mathbf{j})\) m s\(^{-1}\) |
# Question 6:
## Part (a):
| The impulse acts along the line of centres, which is parallel to $\mathbf{i}$, so no impulse in $\mathbf{j}$ direction | B1 | |
## Part (b):
| Only $\mathbf{i}$ components change; $\mathbf{j}$ components remain $3\mathbf{j}$ and $-2\mathbf{j}$ respectively | B1 | |
| Conservation of momentum (i-component): $1(2) + 2(-1) = 1(v_A) + 2(v_B)$ | M1 | |
| $v_A + 2v_B = 0$ | A1 | |
| NEL: $v_B - v_A = 0.5(2-(-1)) = 1.5$ | M1 | |
| $v_B - v_A = 1.5$ | A1 | |
| $v_A = -1, \quad v_B = 0.5$ | A1 | |
| Velocity of $A$: $(-\mathbf{i} + 3\mathbf{j})$ m s$^{-1}$; Velocity of $B$: $(0.5\mathbf{i} - 2\mathbf{j})$ m s$^{-1}$ | | |6 Two smooth spheres, $A$ and $B$, have equal radii and masses 1 kg and 2 kg respectively.
The sphere $A$ is moving with velocity $( 2 \mathbf { i } + 3 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$ and the sphere $B$ is moving with velocity $( - \mathbf { i } - 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ on the same smooth horizontal surface.
The spheres collide when their line of centres is parallel to the unit vector $\mathbf { i }$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-16_456_1052_721_550}
\begin{enumerate}[label=(\alph*)]
\item Briefly state why the components of the velocities of $A$ and $B$ parallel to the unit vector $\mathbf { j }$ are not changed by the collision.
\item The coefficient of restitution between the spheres is 0.5 .
Find the velocities of $A$ and $B$ immediately after the collision.
\includegraphics[max width=\textwidth, alt={}, center]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-17_2484_1709_223_153}\\
$7 \quad$ A ball is projected from a point $O$ on a smooth plane which is inclined at an angle of $35 ^ { \circ }$ above the horizontal. The ball is projected with velocity $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ above the plane, as shown in the diagram. The motion of the ball is in a vertical plane containing a line of greatest slope of the inclined plane. The ball strikes the inclined plane at the point $A$.\\
\includegraphics[max width=\textwidth, alt={}, center]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-18_321_838_605_577}\\
(a) Find the components of the velocity of the ball, parallel and perpendicular to the plane, as it strikes the inclined plane at $A$.\\
(b) On striking the plane at $A$, the ball rebounds. The coefficient of restitution between the plane and the ball is $\frac { 4 } { 5 }$.
Show that the ball next strikes the plane at a point lower down than $A$.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-19_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2010 Q6 [7]}}