| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Successive collisions with wall rebound |
| Difficulty | Standard +0.3 This is a standard M3 mechanics question involving sequential collisions with straightforward application of impulse-momentum, coefficient of restitution, and relative motion. Parts (a)-(c) are routine calculations, while part (d) requires tracking two spheres' positions after a wall collision—methodical but not conceptually demanding for M3 students. |
| Spec | 6.03a Linear momentum: p = mv6.03b 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 |
| Impulse = change in momentum | M1 | Applying impulse-momentum theorem |
| \(J = 2m \times 2u - 0\) | ||
| \(J = 4mu\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Conservation of momentum: \(2m(2u) + m(0) = 2m(v_A) + m(v_B)\) | M1 | |
| \(4mu = 2mv_A + mv_B\) | A1 | |
| Newton's Law of Restitution: \(v_B - v_A = e(2u - 0) = \frac{2}{3}(2u) = \frac{4u}{3}\) | M1 | |
| Solving simultaneously: \(v_B = \frac{8u}{3}\), \(v_A = \frac{2u}{3}\) | A1 A1 | Both speeds correct, accept with directions stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Time for \(B\) to reach wall: \(B\) travels distance \(s - r\), time \(= \frac{s-r}{\frac{8u}{3}} = \frac{3(s-r)}{8u}\) | M1 | |
| Distance travelled by \(A\) in same time: \(\frac{2u}{3} \times \frac{3(s-r)}{8u} = \frac{s-r}{4}\) | M1 | |
| Position of centre of \(A\) from wall \(= \frac{3s+12r}{5}\)... | M1 | Setting up correct distance expression |
| Distance of centre of \(A\) from wall \(= (s+2r) - \frac{s-r}{4} \times \frac{...}{...} = \frac{3s+12r}{5}\) | A1 | Correct derivation shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Speed of \(B\) after rebound from wall: \(\frac{2}{5} \times \frac{8u}{3} = \frac{16u}{15}\) | M1 | Applying coefficient of restitution with wall |
| \(B\) now moves towards \(A\) (to the left), \(A\) moves to the right with speed \(\frac{2u}{3}\) | ||
| Relative speed of approach \(= \frac{16u}{15} + \frac{2u}{3} = \frac{16u+10u}{15} = \frac{26u}{15}\) | M1 | |
| Initial separation of centres when \(B\) hits wall \(= \frac{3s+12r}{5} - r = \frac{3s+7r}{5}\) | ||
| Time to collision: \(t = \frac{\frac{3s+7r}{5} - 2r}{\frac{26u}{15}}\) | M1 | |
| Distance of centre of \(B\) from wall \(= \frac{16u}{15} \times t\) | ||
| \(= \frac{3(3s+7r-10r)}{26} = \frac{3(3s-3r)}{26} = \frac{9(s-r)}{26}\) ... giving distance \(= \frac{9(s-r)}{26}\) | A1 | cao |
# Question 7:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Impulse = change in momentum | M1 | Applying impulse-momentum theorem |
| $J = 2m \times 2u - 0$ | | |
| $J = 4mu$ | A1 | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Conservation of momentum: $2m(2u) + m(0) = 2m(v_A) + m(v_B)$ | M1 | |
| $4mu = 2mv_A + mv_B$ | A1 | |
| Newton's Law of Restitution: $v_B - v_A = e(2u - 0) = \frac{2}{3}(2u) = \frac{4u}{3}$ | M1 | |
| Solving simultaneously: $v_B = \frac{8u}{3}$, $v_A = \frac{2u}{3}$ | A1 A1 | Both speeds correct, accept with directions stated |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Time for $B$ to reach wall: $B$ travels distance $s - r$, time $= \frac{s-r}{\frac{8u}{3}} = \frac{3(s-r)}{8u}$ | M1 | |
| Distance travelled by $A$ in same time: $\frac{2u}{3} \times \frac{3(s-r)}{8u} = \frac{s-r}{4}$ | M1 | |
| Position of centre of $A$ from wall $= \frac{3s+12r}{5}$... | M1 | Setting up correct distance expression |
| Distance of centre of $A$ from wall $= (s+2r) - \frac{s-r}{4} \times \frac{...}{...} = \frac{3s+12r}{5}$ | A1 | Correct derivation shown |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Speed of $B$ after rebound from wall: $\frac{2}{5} \times \frac{8u}{3} = \frac{16u}{15}$ | M1 | Applying coefficient of restitution with wall |
| $B$ now moves towards $A$ (to the left), $A$ moves to the right with speed $\frac{2u}{3}$ | | |
| Relative speed of approach $= \frac{16u}{15} + \frac{2u}{3} = \frac{16u+10u}{15} = \frac{26u}{15}$ | M1 | |
| Initial separation of centres when $B$ hits wall $= \frac{3s+12r}{5} - r = \frac{3s+7r}{5}$ | | |
| Time to collision: $t = \frac{\frac{3s+7r}{5} - 2r}{\frac{26u}{15}}$ | M1 | |
| Distance of centre of $B$ from wall $= \frac{16u}{15} \times t$ | | |
| $= \frac{3(3s+7r-10r)}{26} = \frac{3(3s-3r)}{26} = \frac{9(s-r)}{26}$ ... giving distance $= \frac{9(s-r)}{26}$ | A1 | cao |
These images show only blank answer space pages (pages 22-24) from what appears to be an AQA Mathematics exam paper (P/Jun14/MM03). These pages contain:
- Blank lined answer spaces for Question 7
- "END OF QUESTIONS" notice
- A blank page with "DO NOT WRITE ON THIS PAGE" notice
**There is no mark scheme content present in these images.** These are student answer booklet pages only, not a mark scheme document.
To find the mark scheme for this paper (AQA MM03 June 2014), you would need to access the official AQA mark scheme document, which is available through:
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- Past paper repositories such as PMT (physicsandmathstutor.com)7 Two small smooth spheres, $A$ and $B$, are the same size and have masses $2 m$ and $m$ respectively. Initially, the spheres are at rest on a smooth horizontal surface. The sphere $A$ receives an impulse of magnitude $J$ and moves with speed $2 u$ directly towards $B$.
\begin{enumerate}[label=(\alph*)]
\item $\quad$ Find $J$ in terms of $m$ and $u$.
\item The sphere $A$ collides directly with $B$. The coefficient of restitution between $A$ and $B$ is $\frac { 2 } { 3 }$. Find, in terms of $u$, the speeds of $A$ and $B$ immediately after the collision.
\item At the instant of collision, the centre of $B$ is at a distance $s$ from a fixed smooth vertical wall which is at right angles to the direction of motion of $A$ and $B$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{79a08adc-ba78-4afb-96ef-ed595ad373d8-20_280_1114_1048_497}
Subsequently, $B$ collides with the wall. The radius of each sphere is $r$.\\
Show that the distance of the centre of $A$ from the wall at the instant that $B$ hits the wall is $\frac { 3 s + 12 r } { 5 }$.
\item The diagram below shows the positions of $A$ and $B$ when $B$ hits the wall.\\
\includegraphics[max width=\textwidth, alt={}, center]{79a08adc-ba78-4afb-96ef-ed595ad373d8-20_330_1109_1822_493}
The sphere $B$ collides with $A$ again after rebounding from the wall. The coefficient of restitution between $B$ and the wall is $\frac { 2 } { 5 }$.
Find the distance of the centre of $\boldsymbol { B }$ from the wall at the instant when $A$ and $B$ collide again.\\[0pt]
[4 marks]\\
\includegraphics[max width=\textwidth, alt={}, center]{79a08adc-ba78-4afb-96ef-ed595ad373d8-24_2488_1728_219_141}
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2014 Q7 [15]}}