| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Ball between two walls, successive rebounds |
| Difficulty | Standard +0.8 This is a two-stage collision problem requiring decomposition of velocities at two perpendicular walls with different restitution coefficients, then comparing kinetic energies. While methodical, it demands careful vector resolution, sequential application of collision laws, and algebraic manipulation to reach the 35% result. More challenging than standard single-collision questions but follows established FM1 techniques without requiring novel insight. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy to find the kinetic energy after the second impact | M1 | |
| Parallel to \(AB\) after first collision: \(u\cos 60°\) | M1 | Use of CLM parallel to wall; condone sin/cos confusion |
| Perpendicular to \(AB\) after first collision: \(\frac{1}{\sqrt{3}}u\sin 60°\) | M1 | Use NEL as model to find speed perpendicular to wall; condone sin/cos confusion |
| Components of velocity after first impact: \(\frac{u}{2},\ \frac{u}{2}\) | A1 | Both components correct with trig substituted (seen or implied) |
| Parallel to \(BC\) after collision: \(\frac{u}{2}\ \left(u\times\frac{1}{\sqrt{3}}\sin 60°\right)\) | M1 | Use of CLM parallel to wall; condone sin/cos confusion |
| Perpendicular to \(BC\) after collision: \(\sqrt{\frac{2}{5}}\times\frac{u}{2}\left(= \frac{1}{\sqrt{10}}u\right)\), \(\left(\sqrt{\frac{2}{5}}\times u\cos 60°\right)\) | M1 | Use NEL as model; condone sin/cos confusion |
| Components of velocity after second impact: \(\frac{u}{2},\ \frac{u}{\sqrt{10}}\) | A1 | Both components correct with trig substituted (seen or implied) |
| Final \(\text{KE} = \frac{1}{2}m\left(\frac{u^2}{4}+\frac{u^2}{10}\right)\left(= \frac{mu^2}{2}\times\frac{7}{20}\right)\) | M1 | Correct expression for total KE using their components after 2nd collision |
| Fraction of initial KE \(= \dfrac{\frac{mu^2}{2}\times\frac{7}{20}}{\frac{mu^2}{2}} = \frac{7}{20} = 35\%\) | A1* | Obtain given answer with sufficient working to justify it |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The answer is too large — rough surface means resistance so final speed will be lower | B1 | Clear explanation of how the modelling assumption has affected the outcome |
## Question 4:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy to find the kinetic energy after the second impact | M1 | |
| Parallel to $AB$ after first collision: $u\cos 60°$ | M1 | Use of CLM parallel to wall; condone sin/cos confusion |
| Perpendicular to $AB$ after first collision: $\frac{1}{\sqrt{3}}u\sin 60°$ | M1 | Use NEL as model to find speed perpendicular to wall; condone sin/cos confusion |
| Components of velocity after first impact: $\frac{u}{2},\ \frac{u}{2}$ | A1 | Both components correct with trig substituted (seen or implied) |
| Parallel to $BC$ after collision: $\frac{u}{2}\ \left(u\times\frac{1}{\sqrt{3}}\sin 60°\right)$ | M1 | Use of CLM parallel to wall; condone sin/cos confusion |
| Perpendicular to $BC$ after collision: $\sqrt{\frac{2}{5}}\times\frac{u}{2}\left(= \frac{1}{\sqrt{10}}u\right)$, $\left(\sqrt{\frac{2}{5}}\times u\cos 60°\right)$ | M1 | Use NEL as model; condone sin/cos confusion |
| Components of velocity after second impact: $\frac{u}{2},\ \frac{u}{\sqrt{10}}$ | A1 | Both components correct with trig substituted (seen or implied) |
| Final $\text{KE} = \frac{1}{2}m\left(\frac{u^2}{4}+\frac{u^2}{10}\right)\left(= \frac{mu^2}{2}\times\frac{7}{20}\right)$ | M1 | Correct expression for total KE using their components after 2nd collision |
| Fraction of initial KE $= \dfrac{\frac{mu^2}{2}\times\frac{7}{20}}{\frac{mu^2}{2}} = \frac{7}{20} = 35\%$ | A1* | Obtain **given answer** with sufficient working to justify it |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| The answer is too large — rough surface means resistance so final speed will be lower | B1 | Clear explanation of how the modelling assumption has affected the outcome |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{681833ac-b266-4ac8-881e-46ede398ce58-08_513_807_303_630}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 represents the plan view of part of a horizontal floor, where $A B$ and $B C$ are perpendicular vertical walls.
The floor and the walls are modelled as smooth.\\
A ball is projected along the floor towards $A B$ with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a path at an angle of $60 ^ { \circ }$ to $A B$. The ball hits $A B$ and then hits $B C$.
The ball is modelled as a particle.\\
The coefficient of restitution between the ball and wall $A B$ is $\frac { 1 } { \sqrt { 3 } }$\\
The coefficient of restitution between the ball and wall $B C$ is $\sqrt { \frac { 2 } { 5 } }$
\begin{enumerate}[label=(\alph*)]
\item Show that, using this model, the final kinetic energy of the ball is $35 \%$ of the initial kinetic energy of the ball.
\item In reality the floor and the walls may not be smooth. What effect will the model have had on the calculation of the percentage of kinetic energy remaining?
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM1 Q4 [9]}}