| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Successive collisions, three particles in line |
| Difficulty | Standard +0.3 This is a standard M3 collision question involving successive impacts with given coefficient of restitution. Parts (a)-(b) require routine application of conservation of momentum and Newton's restitution formula. Part (c) involves a straightforward inequality from comparing velocities. Part (d) is direct impulse calculation. While multi-step, each component follows textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| CLM: \(1 \times 3u + 3 \times 0 = 1 \times v_A + 3 \times v_B\) | M1 | Conservation of momentum for A and B |
| \(3u = v_A + 3v_B\) | A1 | Correct equation |
| NEL: \(v_B - v_A = \frac{1}{3}(3u - 0)\) | M1 | Newton's experimental law applied correctly |
| \(v_B - v_A = u\) | A1 | Correct restitution equation |
| Solving: \(v_A = 0\), so A is brought to rest | A1 | Show \(v_A = 0\) |
| \(v_B = u\) | A1 | Speed of B after collision |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| CLM for B and C: \(3 \times u + x \times 0 = 3v_B' + x v_C\) | M1 | Conservation of momentum for B and C |
| \(3u = 3v_B' + xv_C\) | A1 | Correct momentum equation |
| NEL: \(v_C - v_B' = \frac{1}{3}(u - 0)= \frac{u}{3}\) | M1 | Restitution equation for B and C |
| \(v_C - v_B' = \frac{u}{3}\) | A1 | Correct restitution equation |
| Solving: \(v_C = \frac{4u}{3+x}\) | A1 | Shown (given result) |
| \(v_B' = \frac{4u}{3+x} - \frac{u}{3} = \frac{u(9-x)}{3(3+x)}\) | A1 | Speed of B after second collision |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| For B to collide with A again, B must be moving in negative direction (back towards A): \(v_B' < 0\) | M1 | Correct condition stated |
| \(\frac{u(9-x)}{3(3+x)} < 0 \Rightarrow 9 - x < 0\) | ||
| \(\Rightarrow x > 9\) | A1 | Conclusion shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| When \(x = 5\): speed of C \(= \frac{4u}{3+5} = \frac{u}{2}\) | ||
| Impulse \(= mv_C - 0 = 5 \times \frac{u}{2}\) | M1 | Impulse = change in momentum of C |
| \(= \frac{5u}{2}\) | A1 | Correct magnitude |
## Question 3:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| CLM: $1 \times 3u + 3 \times 0 = 1 \times v_A + 3 \times v_B$ | M1 | Conservation of momentum for A and B |
| $3u = v_A + 3v_B$ | A1 | Correct equation |
| NEL: $v_B - v_A = \frac{1}{3}(3u - 0)$ | M1 | Newton's experimental law applied correctly |
| $v_B - v_A = u$ | A1 | Correct restitution equation |
| Solving: $v_A = 0$, so A is brought to rest | A1 | Show $v_A = 0$ |
| $v_B = u$ | A1 | Speed of B after collision |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| CLM for B and C: $3 \times u + x \times 0 = 3v_B' + x v_C$ | M1 | Conservation of momentum for B and C |
| $3u = 3v_B' + xv_C$ | A1 | Correct momentum equation |
| NEL: $v_C - v_B' = \frac{1}{3}(u - 0)= \frac{u}{3}$ | M1 | Restitution equation for B and C |
| $v_C - v_B' = \frac{u}{3}$ | A1 | Correct restitution equation |
| Solving: $v_C = \frac{4u}{3+x}$ | A1 | Shown (given result) |
| $v_B' = \frac{4u}{3+x} - \frac{u}{3} = \frac{u(9-x)}{3(3+x)}$ | A1 | Speed of B after second collision |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| For B to collide with A again, B must be moving in negative direction (back towards A): $v_B' < 0$ | M1 | Correct condition stated |
| $\frac{u(9-x)}{3(3+x)} < 0 \Rightarrow 9 - x < 0$ | | |
| $\Rightarrow x > 9$ | A1 | Conclusion shown |
### Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| When $x = 5$: speed of C $= \frac{4u}{3+5} = \frac{u}{2}$ | | |
| Impulse $= mv_C - 0 = 5 \times \frac{u}{2}$ | M1 | Impulse = change in momentum of C |
| $= \frac{5u}{2}$ | A1 | Correct magnitude |3 Three smooth spheres, $A , B$ and $C$, of equal radii have masses $1 \mathrm {~kg} , 3 \mathrm {~kg}$ and $x \mathrm {~kg}$ respectively. The spheres lie at rest in a straight line on a smooth horizontal surface with $B$ between $A$ and $C$. The sphere $A$ is projected with speed $3 u$ directly towards $B$ and collides with it.\\
\includegraphics[max width=\textwidth, alt={}, center]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-08_250_835_511_605}
The coefficient of restitution between each pair of spheres is $\frac { 1 } { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $A$ is brought to rest by the impact and find the speed of $B$ immediately after the collision in terms of $u$.
\item Subsequently, $B$ collides with $C$.
Show that the speed of $C$ immediately after the collision is $\frac { 4 u } { 3 + x }$.\\
Find the speed of $B$ immediately after the collision in terms of $u$ and $x$.
\item Show that $B$ will collide with $A$ again if $x > 9$.
\item Given that $x = 5$, find the magnitude of the impulse exerted on $C$ by $B$ in terms of $u$.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-09_2484_1709_223_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-10_2484_1712_223_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-11_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2010 Q3 [16]}}