| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | September |
| Marks | 16 |
| Topic | Impulse and momentum (advanced) |
| Type | Multiple successive collisions |
| Difficulty | Challenging +1.8 This is a complex multi-stage mechanics problem requiring energy conservation, circular motion analysis, collision dynamics, and momentum considerations across multiple collisions. While the individual techniques are A-level standard (pendulum motion, circular motion, collisions), the extended multi-step reasoning, careful tracking of energy through multiple stages, and the conceptual understanding required for parts (iii) and (iv) elevate this significantly above routine exercises. The derivation in part (i) alone requires combining several conservation principles across two different collision events and motion types. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}mu^2 = mgh(1 - \cos \theta)\) | M1 | Conservation of energy for A |
| \(mu = 2mv_B\) | M1 | Conservation of momentum; \(v_B\) is velocity of B after first collision |
| \(e = \frac{v_B}{u}\) | M1 | Restitution |
| \(v_B = \frac{1}{2}u\) and \(e = \frac{1}{2}\) | A1 | |
| \(2m \times \frac{1}{2}u + m \times (-\frac{1}{2}u) = mv_A + 2mv_B\) | M1 | Conservation of momentum for 2nd collision with unchanged \(v_B\) |
| \(\frac{1}{2} = \frac{v_A - v_B}{\frac{1}{2}u}\) | M1 | Restitution for 2nd collision |
| \(v_A = \frac{1}{2}u\) | A1 | |
| \(\frac{1}{2}m(\frac{1}{2}u)^2 = mgh(1 - \cos \phi)\) | M1 | Conservation of energy for A |
| \(\frac{1}{8}mu^2 = \frac{1}{4}mgh(1 - \cos \theta) = mgh(1 - \cos \phi)\) | A1 | AG so elimination of \(\frac{1}{2}mu^2\) oe must be clearly shown |
| \(\Rightarrow \cos \phi = \frac{3 + \cos \theta}{4}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(V_B = \frac{1}{4}u\) | B1 | |
| \(2m \times \frac{1}{4}u + m \times (-\frac{1}{2}u) = mv_A + 2mv_B\) | M1 | Conservation of momentum for 3rd collision; Or with signs reversed, if different 'positive' direction chosen |
| \(\frac{1}{2} = \frac{v_A - v_B}{\frac{1}{4}u - (-\frac{1}{2}u)}\) | M1 | Restitution for 3rd collision |
| \(v_B = -\frac{1}{8}u\) and \(v_A = \frac{1}{4}u\) | A1 | |
| KE loss \(= \frac{\frac{1}{2} \times (\frac{1}{4}u)^2 + \frac{1}{2} \times 2m \times (-\frac{1}{8}u)^2}{\frac{1}{2}mu^2} = \frac{3}{32} = 9.4\%\) | A1 | |
| So percentage of KE remaining is 90.6% | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| During and between the 1st and 2nd collisions there is no external horizontal force acting on \(A\) and while there is an external force acting on \(B\) its velocity and hence momentum is the same just before the 2nd collision as it is just after the 1st collision | E1 | Identifying the unchanged velocity for \(B\) in the two collisions |
| Answer | Marks | Guidance |
|---|---|---|
| Between the 2nd and 3rd collisions a component of the tension in the string \(OA\) is horizontal and so the momentum of the system is not conserved | E1 | Identifying the relevant external force acting |
## (i)
$\frac{1}{2}mu^2 = mgh(1 - \cos \theta)$ | M1 | Conservation of energy for A
$mu = 2mv_B$ | M1 | Conservation of momentum; $v_B$ is velocity of B after first collision
$e = \frac{v_B}{u}$ | M1 | Restitution
$v_B = \frac{1}{2}u$ and $e = \frac{1}{2}$ | A1 |
$2m \times \frac{1}{2}u + m \times (-\frac{1}{2}u) = mv_A + 2mv_B$ | M1 | Conservation of momentum for 2nd collision with unchanged $v_B$
$\frac{1}{2} = \frac{v_A - v_B}{\frac{1}{2}u}$ | M1 | Restitution for 2nd collision
$v_A = \frac{1}{2}u$ | A1 |
$\frac{1}{2}m(\frac{1}{2}u)^2 = mgh(1 - \cos \phi)$ | M1 | Conservation of energy for A
$\frac{1}{8}mu^2 = \frac{1}{4}mgh(1 - \cos \theta) = mgh(1 - \cos \phi)$ | A1 | AG so elimination of $\frac{1}{2}mu^2$ oe must be clearly shown
$\Rightarrow \cos \phi = \frac{3 + \cos \theta}{4}$ | A1 |
## (ii)
$V_B = \frac{1}{4}u$ | B1 |
$2m \times \frac{1}{4}u + m \times (-\frac{1}{2}u) = mv_A + 2mv_B$ | M1 | Conservation of momentum for 3rd collision; Or with signs reversed, if different 'positive' direction chosen
$\frac{1}{2} = \frac{v_A - v_B}{\frac{1}{4}u - (-\frac{1}{2}u)}$ | M1 | Restitution for 3rd collision
$v_B = -\frac{1}{8}u$ and $v_A = \frac{1}{4}u$ | A1 |
KE loss $= \frac{\frac{1}{2} \times (\frac{1}{4}u)^2 + \frac{1}{2} \times 2m \times (-\frac{1}{8}u)^2}{\frac{1}{2}mu^2} = \frac{3}{32} = 9.4\%$ | A1 |
So percentage of KE remaining is 90.6% | A1 |
## (iii)
During and between the 1st and 2nd collisions there is no external horizontal force acting on $A$ and while there is an external force acting on $B$ its velocity and hence momentum is the same just before the 2nd collision as it is just after the 1st collision | E1 | Identifying the unchanged velocity for $B$ in the two collisions
## (iv)
Between the 2nd and 3rd collisions a component of the tension in the string $OA$ is horizontal and so the momentum of the system is not conserved | E1 | Identifying the relevant external force acting
---A point $O$ is situated a distance $h$ above a smooth horizontal plane, and a particle $A$ of mass $m$ is attached to $O$ by a light inextensible string of length $h$. A particle $B$ of mass $2m$ is at rest on the plane, directly below $O$, and is attached to a point $C$ on the plane, where $BC = l$, by a light inextensible string of length $l$. $A$ is released from rest with the string $OA$ taut and making an acute angle $\theta$ with the downward vertical (see diagram).
\includegraphics{figure_8}
$A$ moves in a vertical plane perpendicular to $CB$ and collides directly with $B$. As a result of this collision, $A$ is brought to rest and $B$ moves on the plane in a horizontal circle with centre $C$. After $B$ has made one complete revolution the particles collide again.
\begin{enumerate}[label=(\roman*)]
\item Show that, on the next occasion that $A$ comes to rest, the string $OA$ makes an angle $\phi$ with the downward vertical through $O$, where $\cos \phi = \frac{3 + \cos \theta}{4}$. [9]
\end{enumerate}
$A$ and $B$ collide again when $AO$ is next vertical.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the percentage of the original energy of the system that remains immediately after this collision. [5]
\item Explain why the total momentum of the particles immediately before the first collision is the same as the total momentum of the particles immediately after the second collision. [1]
\item Explain why the total momentum of the particles immediately before the first collision is different from the total momentum of the particles immediately after the third collision. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q8 [16]}}