| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Ball between two walls, successive rebounds |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics collision problem requiring systematic application of restitution coefficients and component resolution. While it involves two collisions and requires algebraic manipulation to reach tan(α/2) = 1/3, the method is straightforward: resolve velocities parallel and perpendicular to walls, apply e = 2/3 twice, and use the given final angle condition. Part (b) is routine energy calculation. More mechanical than insightful, but above average difficulty due to the two-stage collision sequence and algebraic work required. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03i Coefficient of restitution: e6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Component \(v\cos\alpha\) (horizontal, unchanged for first ball) | B1 | One mark for each component correct |
| Component \(\frac{2}{3}v\sin\alpha\) (vertical, after first impact) | B1 | One mark for each component correct |
| Component \(\frac{4}{9}v\sin\alpha\) (vertical, after second impact) | B1 | One mark for each component correct |
| Component \(v\cos\alpha\) (horizontal, after second impact) | B1 | One mark for each component correct |
| \(\tan\dfrac{\alpha}{2} = \dfrac{\frac{4}{9}v\sin\alpha}{v\cos\alpha} = \dfrac{4}{9}\tan\alpha\) | M1 | Form expression for \(\tan\frac{\alpha}{2}\) in terms of \(\tan\alpha\) |
| \(t = \tan\dfrac{\alpha}{2} \Rightarrow t = \dfrac{4 \times 2t}{9(1-t^2)}\) | M1 | |
| \(1 - t^2 = \dfrac{8}{9},\quad t = \dfrac{1}{3}\) | A1* | Obtain given result from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\alpha = \dfrac{\frac{2}{3}}{1 - \frac{1}{9}} = \dfrac{3}{4}\) | B1 | |
| Change in KE: \(\dfrac{1}{2}mv^2 - \dfrac{1}{2}m\left(v^2\cos^2\alpha + \left(\dfrac{4}{9}v\right)^2\sin^2\alpha\right)\) | M1 | |
| \(\%\ \text{of KE lost} = 100\left(1 - \dfrac{\frac{1}{2}mv^2\left(\frac{16}{25} + \frac{16}{81} \times \frac{9}{25}\right)}{\frac{1}{2}mv^2}\right)\) | M1 | |
| \(= 28.888\ldots\ (\%)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Form and solve equation in \(\tan\frac{\alpha}{2}\) | M1 | |
| Obtain given answer from correct working | A1\* | NB: This is a "Show that .." question. A candidate who assumes, without proof, that \(\tan\frac{\alpha}{2} = e^2\tan\alpha\) can only score the last two marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Correct use of \(t = \frac{1}{3}\) | (b)B1 | Must be seen / used in part (b) |
| Dimensionally correct expression for change in KE | M1 | NB note that they may not show component parallel to the wall |
| Dimensionally correct expression for the percentage of KE lost | M1 | |
| Accept \(29(\%)\) or better. Accept \(\frac{260}{9}\) | A1 |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Component $v\cos\alpha$ (horizontal, unchanged for first ball) | B1 | One mark for each component correct |
| Component $\frac{2}{3}v\sin\alpha$ (vertical, after first impact) | B1 | One mark for each component correct |
| Component $\frac{4}{9}v\sin\alpha$ (vertical, after second impact) | B1 | One mark for each component correct |
| Component $v\cos\alpha$ (horizontal, after second impact) | B1 | One mark for each component correct |
| $\tan\dfrac{\alpha}{2} = \dfrac{\frac{4}{9}v\sin\alpha}{v\cos\alpha} = \dfrac{4}{9}\tan\alpha$ | M1 | Form expression for $\tan\frac{\alpha}{2}$ in terms of $\tan\alpha$ |
| $t = \tan\dfrac{\alpha}{2} \Rightarrow t = \dfrac{4 \times 2t}{9(1-t^2)}$ | M1 | |
| $1 - t^2 = \dfrac{8}{9},\quad t = \dfrac{1}{3}$ | A1* | Obtain given result from correct working |
---
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\alpha = \dfrac{\frac{2}{3}}{1 - \frac{1}{9}} = \dfrac{3}{4}$ | B1 | |
| Change in KE: $\dfrac{1}{2}mv^2 - \dfrac{1}{2}m\left(v^2\cos^2\alpha + \left(\dfrac{4}{9}v\right)^2\sin^2\alpha\right)$ | M1 | |
| $\%\ \text{of KE lost} = 100\left(1 - \dfrac{\frac{1}{2}mv^2\left(\frac{16}{25} + \frac{16}{81} \times \frac{9}{25}\right)}{\frac{1}{2}mv^2}\right)$ | M1 | |
| $= 28.888\ldots\ (\%)$ | A1 | |
## Part (a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Form and solve equation in $\tan\frac{\alpha}{2}$ | **M1** | |
| Obtain given answer from correct working | **A1\*** | NB: This is a "Show that .." question. A candidate who assumes, without proof, that $\tan\frac{\alpha}{2} = e^2\tan\alpha$ can only score the last two marks. |
## Part (b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Correct use of $t = \frac{1}{3}$ | **(b)B1** | Must be seen / used in part (b) |
| Dimensionally correct expression for change in KE | **M1** | NB note that they may not show component parallel to the wall |
| Dimensionally correct expression for the percentage of KE lost | **M1** | |
| Accept $29(\%)$ or better. Accept $\frac{260}{9}$ | **A1** | |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{361d263e-0ee1-47e9-8fc2-0f127f1c2d7e-24_553_951_258_557}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 represents the plan view of part of a horizontal floor, where $A B$ and $C D$ represent fixed vertical walls, with $A B$ parallel to $C D$.
A small ball is projected along the floor towards wall $A B$. Immediately before hitting wall $A B$, the ball is moving with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\alpha$ to $A B$, where $0 < \alpha < \frac { \pi } { 2 }$
The ball hits wall $A B$ and then hits wall $C D$.\\
After the impact with wall $C D$, the ball is moving at angle $\frac { 1 } { 2 } \alpha$ to $C D$.\\
The coefficient of restitution between the ball and wall $A B$ is $\frac { 2 } { 3 }$\\
The coefficient of restitution between the ball and wall $C D$ is also $\frac { 2 } { 3 }$\\
The floor and the walls are modelled as being smooth. The ball is modelled as a particle.
\begin{enumerate}[label=(\alph*)]
\item Show that $\tan \left( \frac { 1 } { 2 } \alpha \right) = \frac { 1 } { 3 }$
\item Find the percentage of the initial kinetic energy of the ball that is lost as a result of the two impacts.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM1 2020 Q7 [11]}}