| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Multiple wall bounces or returns |
| Difficulty | Standard +0.3 This is a standard M2 mechanics question involving coefficient of restitution and impulse-momentum theorem. Part (a) requires straightforward application of impulse = change in momentum with the restitution formula. Part (b) involves setting up distance equations for the two journeys (O to wall, wall to O) using the reduced speed after impact. While it requires careful bookkeeping of the speeds and distances, it follows a well-practiced template for collision problems with no novel insight required. Slightly easier than average due to the structured setup and standard techniques. |
| Spec | 6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Speed after impact \(= \frac{2}{3}u\) | B1 | Allow for velocity \(= -\frac{2u}{3}\) |
| Impulse \(=\) change in momentum \(= \pm\left(m \cdot \frac{2}{3}u - m(-u)\right) = \frac{5}{3}mu\) | M1 | Need to consider momentum before and after collision and use change of direction |
| \(\lambda = \frac{5}{3}\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Speed after second collision \(= e^2u = \frac{4}{9}u\) | B1 | Allow negative |
| Total time taken \(= \frac{1}{u} + \frac{2}{eu} + \frac{1}{e^2u} = \frac{1}{u} + \frac{3}{u} + \frac{9}{4u}\) | M1 | Use of time \(= \frac{\text{distance}}{\text{speed}}\) to find total time in terms of \(u\). At least one term dealt with correctly |
| A2 | \(-1\) each error | |
| Alt M1A2: Ratio of times \(\frac{1}{3}:1:\frac{3}{4}\) | M1, A2 | Or equivalent. \(-1\) each error |
| \(\frac{4}{u} + \frac{9}{4u} = \frac{25}{4u} = 3\), \(\quad u = \frac{25}{12}\) o.e. | DM1 | Use total time \(= 3\) and solve for \(u\) |
| A1 | Accept 2.08 and 2.1 (or better) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Speed after impact $= \frac{2}{3}u$ | B1 | Allow for velocity $= -\frac{2u}{3}$ |
| Impulse $=$ change in momentum $= \pm\left(m \cdot \frac{2}{3}u - m(-u)\right) = \frac{5}{3}mu$ | M1 | Need to consider momentum before and after collision and use change of direction |
| $\lambda = \frac{5}{3}$ | A1 | cso |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Speed after second collision $= e^2u = \frac{4}{9}u$ | B1 | Allow negative |
| Total time taken $= \frac{1}{u} + \frac{2}{eu} + \frac{1}{e^2u} = \frac{1}{u} + \frac{3}{u} + \frac{9}{4u}$ | M1 | Use of time $= \frac{\text{distance}}{\text{speed}}$ to find total time in terms of $u$. At least one term dealt with correctly |
| | A2 | $-1$ each error |
| **Alt M1A2:** Ratio of times $\frac{1}{3}:1:\frac{3}{4}$ | M1, A2 | Or equivalent. $-1$ each error |
| $\frac{4}{u} + \frac{9}{4u} = \frac{25}{4u} = 3$, $\quad u = \frac{25}{12}$ o.e. | DM1 | Use total time $= 3$ and solve for $u$ |
| | A1 | Accept 2.08 and 2.1 (or better) |
---5. A particle of mass $m \mathrm {~kg}$ lies on a smooth horizontal surface. Initially the particle is at rest at a point $O$ midway between a pair of fixed parallel vertical walls. The walls are 2 m apart. At time $t = 0$ the particle is projected from $O$ with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a direction perpendicular to the walls. The coefficient of restitution between the particle and each wall is $\frac { 2 } { 3 }$. The magnitude of the impulse on the particle due to the first impact with a wall is $\lambda m u \mathrm {~N} \mathrm {~s}$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\lambda$.
The particle returns to $O$, having bounced off each wall once, at time $t = 3$ seconds.
\item Find the value of $u$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q5 [9]}}