| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, vector velocity form |
| Difficulty | Challenging +1.2 This is a standard M3 oblique collision problem requiring conservation of momentum, impulse-momentum theorem, Newton's experimental law, and kinematics. While it involves multiple parts and vector calculations, each step follows routine procedures taught in mechanics modules. The multi-part structure and need to work with 2D vectors elevates it slightly above average difficulty, but no novel insight is required—it's a textbook application of standard collision mechanics principles. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Conservation of momentum: \(2m(3\mathbf{i}+\mathbf{j}) + m(2\mathbf{i}-5\mathbf{j}) = 2m\mathbf{v}_A + m(2\mathbf{i}+\mathbf{j})\) | M1 | Using conservation of momentum with all terms |
| \(8\mathbf{i} - 3\mathbf{j} = 2\mathbf{v}_A + 2\mathbf{i} + \mathbf{j}\) | A1 | Correct equation |
| \(\mathbf{v}_A = (3\mathbf{i} - 2\mathbf{j})\ \text{m s}^{-1}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Impulse \(= m[(2\mathbf{i}+\mathbf{j}) - (2\mathbf{i}-5\mathbf{j})]\) | M1 | \(m(\mathbf{v}_B - \mathbf{u}_B)\) |
| \(= m[6\mathbf{j}]\) | A1 | Correct subtraction |
| \(= 6m\mathbf{j}\ \text{Ns}\) | A1 | Correctly shown (given answer) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Line of centres is in direction of impulse, i.e. \(\mathbf{j}\) direction | B1 | Identifying line of centres as \(\mathbf{j}\) direction |
| Component of relative velocity of approach along \(\mathbf{j}\): \((-5) - (1) = -6\), speed of approach \(= 6\) | M1 | Using components along line of centres |
| Component of relative velocity of separation along \(\mathbf{j}\): \((1) - (-2) = 3\), speed of separation \(= 3\) | A1 | Correct separation speed |
| \(e = \frac{3}{6} = \frac{1}{2}\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| After collision: \(\mathbf{v}_A = (3\mathbf{i}-2\mathbf{j})\), \(\mathbf{v}_B = (2\mathbf{i}+\mathbf{j})\) | B1 | Both velocities stated |
| Relative velocity of A with respect to B: \((\mathbf{i}-3\mathbf{j})\) | M1 | Finding relative velocity vector |
| At collision centres are \(0.1\ \text{m}\) apart (sum of radii) | B1 | Initial separation = \(0.1\ \text{m}\) |
| Need to find when separation \(= 1.10\ \text{m}\), so relative displacement \(= 1.00\ \text{m}\) | M1 | Setting up equation for required relative displacement |
| \( | \mathbf{i}-3\mathbf{j} | = \sqrt{1+9} = \sqrt{10}\), so \(t = \frac{1.00}{\sqrt{10}}\) |
## Question 7:
**(a) Find the velocity of A immediately after the collision**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Conservation of momentum: $2m(3\mathbf{i}+\mathbf{j}) + m(2\mathbf{i}-5\mathbf{j}) = 2m\mathbf{v}_A + m(2\mathbf{i}+\mathbf{j})$ | M1 | Using conservation of momentum with all terms |
| $8\mathbf{i} - 3\mathbf{j} = 2\mathbf{v}_A + 2\mathbf{i} + \mathbf{j}$ | A1 | Correct equation |
| $\mathbf{v}_A = (3\mathbf{i} - 2\mathbf{j})\ \text{m s}^{-1}$ | A1 | Correct answer |
**(b) Show that impulse on B is $6m\mathbf{j}$ Ns**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Impulse $= m[(2\mathbf{i}+\mathbf{j}) - (2\mathbf{i}-5\mathbf{j})]$ | M1 | $m(\mathbf{v}_B - \mathbf{u}_B)$ |
| $= m[6\mathbf{j}]$ | A1 | Correct subtraction |
| $= 6m\mathbf{j}\ \text{Ns}$ | A1 | Correctly shown (given answer) |
**(c) Find the coefficient of restitution**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Line of centres is in direction of impulse, i.e. $\mathbf{j}$ direction | B1 | Identifying line of centres as $\mathbf{j}$ direction |
| Component of relative velocity of approach along $\mathbf{j}$: $(-5) - (1) = -6$, speed of approach $= 6$ | M1 | Using components along line of centres |
| Component of relative velocity of separation along $\mathbf{j}$: $(1) - (-2) = 3$, speed of separation $= 3$ | A1 | Correct separation speed |
| $e = \frac{3}{6} = \frac{1}{2}$ | A1 | Correct value |
**(d) Time until centres are 1.10 m apart**
| Answer/Working | Marks | Guidance |
|---|---|---|
| After collision: $\mathbf{v}_A = (3\mathbf{i}-2\mathbf{j})$, $\mathbf{v}_B = (2\mathbf{i}+\mathbf{j})$ | B1 | Both velocities stated |
| Relative velocity of A with respect to B: $(\mathbf{i}-3\mathbf{j})$ | M1 | Finding relative velocity vector |
| At collision centres are $0.1\ \text{m}$ apart (sum of radii) | B1 | Initial separation = $0.1\ \text{m}$ |
| Need to find when separation $= 1.10\ \text{m}$, so relative displacement $= 1.00\ \text{m}$ | M1 | Setting up equation for required relative displacement |
| $|\mathbf{i}-3\mathbf{j}| = \sqrt{1+9} = \sqrt{10}$, so $t = \frac{1.00}{\sqrt{10}}$ | A1 | $t = \frac{1}{\sqrt{10}} \approx 0.316\ \text{s}$ |
These images show only blank answer space pages (pages 22, 23, and 24) from an AQA exam paper (P47914/Jun12/MM03). They contain no mark scheme content, question text, answers, or marking guidance — only ruled lines for student responses and the standard "END OF QUESTIONS" / "DO NOT WRITE ON THIS PAGE" notices.
There is no mark scheme content to extract from these pages.7 Two smooth spheres, $A$ and $B$, have equal radii and masses $2 m \mathrm {~kg}$ and $m \mathrm {~kg}$ respectively. The spheres are moving on a smooth horizontal plane. The sphere $A$ has velocity $( 3 \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ when it collides with the sphere $B$, which has velocity $( 2 \mathbf { i } - 5 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. Immediately after the collision, the velocity of the sphere $B$ is $( 2 \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of $A$ immediately after the collision.
\item Show that the impulse exerted on $B$ in the collision is $( 6 m \mathbf { j } )$ Ns.
\item Find the coefficient of restitution between the two spheres.
\item After the collision, each sphere moves in a straight line with constant speed. Given that the radius of each sphere is 0.05 m , find the time taken, from the collision, until the centres of the spheres are 1.10 m apart.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2012 Q7 [15]}}