| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Sphere rebounds off fixed wall obliquely |
| Difficulty | Standard +0.8 This is a Further Maths FM1 question requiring sequential wall collisions with coefficient of restitution calculations. Students must resolve velocities in perpendicular directions, apply Newton's law of restitution twice, and work with trigonometric ratios. While systematic, it requires careful component resolution and understanding that tangential components are preserved while normal components follow the restitution law. Part (a) is a 'show that' requiring precise working, and the two-collision sequence with different restitution coefficients elevates this above standard single-collision problems. The modeling critique in (c) is routine for mechanics questions. |
| Spec | 6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| After hit \(AB\): \(\rightarrow 6\cos\alpha\ (= v\cos\beta)\ (= 3.6)\) | B1 | Use model to find component parallel to the wall |
| Use of impact law | M1 | Use model and impact law perpendicular to the wall |
| \(\uparrow 6e\sin\alpha\ (= v\sin\beta)\left(= \frac{24e}{5}\right)(= 4.8e)\) | A1 | Correct perpendicular component |
| \(\tan\beta = \frac{1}{3} = \frac{6e\sin\alpha}{6\cos\alpha}\left(= \frac{24e}{5} \div \frac{18}{5}\right)\) | M1 | Use \(\frac{1}{3}\) and their components to form equation in \(e\). \(\left(v = \frac{6\sqrt{10}}{5} = 3.79\right)\) |
| \(e = \frac{18}{3 \times 24} = \frac{1}{4}\) | A1* | Correct answer from correct exact working. If only see \(e\tan\alpha = \tan\beta\) with no explanation, score 0/5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| After hit \(BC\): \(\uparrow \frac{6}{5}\) | B1 | First component correct |
| \(\rightarrow \frac{1}{2} \times \frac{18}{5}\left(= \frac{9}{5}\right)\) | B1 | Second component correct. Alternative: B1 for speed of impact with \(BC = 3.79\)..., B1 for path on leaving \(BC\) at \(56.3...°\) to \(BC\) |
| Speed \(= \frac{3}{5}\sqrt{2^2 + 3^2}\) | M1 | Use Pythagoras' theorem or trigonometry to find the speed |
| \(= \frac{3\sqrt{13}}{5}\ \text{(m s}^{-1}\text{)}\) | A1 | Any equivalent form. 2.2 or better \((2.1633...)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| An appropriate refinement | B1 | Two independent refinements relating to the modelling e.g. include friction between floor and ball; include friction between ball and walls; give the ball dimensions; consider air resistance; spin/rotation. Do not accept: mass/gravity/levels/perpendicularity |
| A second independent appropriate refinement and no incorrect refinements | B1 | (same list as above) |
# Question 2:
## Part 2a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| After hit $AB$: $\rightarrow 6\cos\alpha\ (= v\cos\beta)\ (= 3.6)$ | B1 | Use model to find component parallel to the wall |
| Use of impact law | M1 | Use model and impact law perpendicular to the wall |
| $\uparrow 6e\sin\alpha\ (= v\sin\beta)\left(= \frac{24e}{5}\right)(= 4.8e)$ | A1 | Correct perpendicular component |
| $\tan\beta = \frac{1}{3} = \frac{6e\sin\alpha}{6\cos\alpha}\left(= \frac{24e}{5} \div \frac{18}{5}\right)$ | M1 | Use $\frac{1}{3}$ and their components to form equation in $e$. $\left(v = \frac{6\sqrt{10}}{5} = 3.79\right)$ |
| $e = \frac{18}{3 \times 24} = \frac{1}{4}$ | A1* | Correct answer from correct exact working. If only see $e\tan\alpha = \tan\beta$ with no explanation, score 0/5 |
**(5 marks)**
## Part 2b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| After hit $BC$: $\uparrow \frac{6}{5}$ | B1 | First component correct |
| $\rightarrow \frac{1}{2} \times \frac{18}{5}\left(= \frac{9}{5}\right)$ | B1 | Second component correct. Alternative: B1 for speed of impact with $BC = 3.79$..., B1 for path on leaving $BC$ at $56.3...°$ to $BC$ |
| Speed $= \frac{3}{5}\sqrt{2^2 + 3^2}$ | M1 | Use Pythagoras' theorem or trigonometry to find the speed |
| $= \frac{3\sqrt{13}}{5}\ \text{(m s}^{-1}\text{)}$ | A1 | Any equivalent form. 2.2 or better $(2.1633...)$ |
**(4 marks)**
## Part 2c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| An appropriate refinement | B1 | Two independent refinements relating to the modelling e.g. include friction between floor and ball; include friction between ball and walls; give the ball dimensions; consider air resistance; spin/rotation. Do **not** accept: mass/gravity/levels/perpendicularity |
| A second independent appropriate refinement and no incorrect refinements | B1 | (same list as above) |
**(2 marks)**
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a871044a-17c5-440d-8d8f-886939603dd4-06_524_638_255_717}
\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 $B C$ are fixed vertical walls with $A B$ perpendicular to $B C$.
A small ball is projected along the floor towards $A B$ with speed $6 \mathrm {~ms} ^ { - 1 }$ on a path that makes an angle $\alpha$ with $A B$, where $\tan \alpha = \frac { 4 } { 3 }$. The ball hits $A B$ and then hits $B C$.\\
Immediately after hitting $A B$, the ball is moving at an angle $\beta$ to $A B$, where $\tan \beta = \frac { 1 } { 3 }$\\
The coefficient of restitution between the ball and $A B$ is $e$.\\
The coefficient of restitution between the ball and $B C$ is $\frac { 1 } { 2 }$\\
By modelling the ball as a particle and the floor and walls as being smooth,
\begin{enumerate}[label=(\alph*)]
\item show that the value of $e = \frac { 1 } { 4 }$
\item find the speed of the ball immediately after it hits $B C$.
\item Suggest two ways in which the model could be refined to make it more realistic.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM1 2019 Q2 [11]}}