| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with coefficient of restitution |
| Difficulty | Standard +0.3 This is a straightforward Further Mechanics question requiring application of coefficient of restitution and impulse-momentum theorem. Part (a) uses standard formulas (impulse = m(v_after - v_before)), and part (b) requires setting up a time equation using constant velocity motion between bounces. While it's FM1 content, the problem-solving is routine with no novel insights needed—slightly easier than average A-level difficulty. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Speed after first impact \(= eu = \frac{3}{4}u\cdot\frac{\pi}{2}\) | B1 | Using NEL as a model to find speed after first impact |
| Impulse \(= \lambda mu = mv - mu = \pm\left[\frac{3}{4}mu - (-mu)\right]\) | M1 | Must be a difference of two terms, taking account of change in direction of motion |
| \(\lambda = \frac{7}{4}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Speed after second impact \(= \frac{3}{4}\times\frac{3}{4}u = \frac{9}{16}u\) | B1 | Using NEL as a model to find speed after second impact |
| Use of \(s = vt\) to find total time | M1 | Needs to be used for at least one stage of the journey |
| \(7 = \frac{2}{u} + \frac{4}{\frac{3}{4}u} + \frac{2}{\frac{9}{16}u}\left(= \frac{2}{u} + \frac{16}{3u} + \frac{32}{9u}\right)\) | A1 | Ur equivalent |
| \(63u = 18 + 48 + 32\) | M1 | Solve their linear equation for \(u\) |
| \(u = \frac{98}{63} = \frac{14}{9}\ (=1.\dot{5})\) | A1 | Accept 1.56 or better |
## Question 3:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Speed after first impact $= eu = \frac{3}{4}u\cdot\frac{\pi}{2}$ | B1 | Using NEL as a model to find speed after first impact |
| Impulse $= \lambda mu = mv - mu = \pm\left[\frac{3}{4}mu - (-mu)\right]$ | M1 | Must be a difference of two terms, taking account of change in direction of motion |
| $\lambda = \frac{7}{4}$ | A1 | cao |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Speed after second impact $= \frac{3}{4}\times\frac{3}{4}u = \frac{9}{16}u$ | B1 | Using NEL as a model to find speed after second impact |
| Use of $s = vt$ to find total time | M1 | Needs to be used for at least one stage of the journey |
| $7 = \frac{2}{u} + \frac{4}{\frac{3}{4}u} + \frac{2}{\frac{9}{16}u}\left(= \frac{2}{u} + \frac{16}{3u} + \frac{32}{9u}\right)$ | A1 | Ur equivalent |
| $63u = 18 + 48 + 32$ | M1 | Solve their linear equation for $u$ |
| $u = \frac{98}{63} = \frac{14}{9}\ (=1.\dot{5})$ | A1 | Accept 1.56 or better |
---\begin{enumerate}
\item A particle of mass $m \mathrm {~kg}$ lies on a smooth horizontal surface.
\end{enumerate}
Initially the particle is at rest at a point $O$ between two fixed parallel vertical walls.\\
The point $O$ is equidistant from the two walls and the walls are 4 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 $\frac { 3 } { 4 }$\\
The magnitude of the impulse on the particle due to the first impact with a wall is $\lambda m u$ Ns.\\
(a) Find the value of $\lambda$.
The particle returns to $O$, having bounced off each wall once, at time $t = 7$ seconds.\\
(b) Find the value of $u$.
\hfill \mbox{\textit{Edexcel FM1 Q3 [8]}}