| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Successive collisions with wall rebound |
| Difficulty | Challenging +1.2 This is a multi-stage collision problem requiring systematic application of conservation of momentum and restitution equations across three collisions. While it involves several steps and careful bookkeeping of velocities, the techniques are standard for Further Mechanics AS. Part (a) is a 'show that' requiring methodical calculation through two collisions. Part (b) adds a distance-based constraint requiring kinematic reasoning to relate the coefficient of restitution to the given distance ratio. The problem is more involved than typical A-level mechanics but uses only standard methods without requiring novel insight. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 1st collision for \(A\) & \(B\): \(2mu = 2mv_A + mv_B\) | M1 | 3.1b – Conservation of momentum |
| \(\frac{1}{2} = \frac{v_B - v_A}{u}\) | M1 | 1.1a – Restitution |
| \(v_A = \frac{1}{2}u\) | A1 | 1.1 |
| 2nd collision for \(A\) & \(B\): \(2m\times\frac{1}{2}u + mU_B = 2mV_A + mV_B\) | M1 | 1.1 – Conservation of momentum. May see \(-U_B\) or \(\pm eu\). Do not allow assumed value of \(U_B\) e.g. \(\frac{1}{2}u\) or \(u\). |
| \(\frac{1}{2} = \frac{V_B - V_A}{\frac{1}{2}u - U_B}\) | M1 | 1.1 – Restitution. Do not allow assumed value of \(U_B\) e.g. \(\frac{1}{2}u\) or \(u\). SC1 if assumed value for \(V_B\) has been used (giving M0M0), provided \( |
| \(u + U_B = 2V_A + V_B\) and \(u - 2U_B = 4V_B - 4V_A\) | ||
| \(\Rightarrow 3u = 6V_B \Rightarrow V_B = \frac{1}{2}u\) | A1 | 2.1 – AG Intermediate work towards cancellation must be seen |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v_B = u\) | B1 | 1.1 – Award if seen in (a) |
| Collision for \(B\) & wall: \(e = \pm\frac{U_B}{u}\) or \(U_B = \pm eu\) | M1 | 3.1b – Restitution. May see \(V_{2B}\) or similar instead of \(\pm eu\) with use of restitution at the end. Award if seen in (a) |
| \(\frac{\frac{4}{5}d}{\frac{1}{2}u} = \frac{d}{u} + \frac{\frac{1}{5}d}{eu}\) | M1 | 3.1b – Seeing that \(A\) travels \(\frac{4}{5}d\) at \(\frac{1}{2}u\) in the same time as \(B\) travels \(d\) at \(u\) and \(\frac{1}{5}d\) at \(eu\). Do not allow assumed rebound speed. |
| \(\frac{3}{5} = \frac{1}{5e}\) | M1 | 1.1 – Correctly cancelling \(d\) and \(u\) and simplifying their 3 term equation including \(e\) in the denominator |
| So coefficient of restitution between \(B\) and wall is \(\frac{1}{3}\) | A1 | 3.2a |
| [5] |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 1st collision for $A$ & $B$: $2mu = 2mv_A + mv_B$ | M1 | 3.1b – Conservation of momentum |
| $\frac{1}{2} = \frac{v_B - v_A}{u}$ | M1 | 1.1a – Restitution |
| $v_A = \frac{1}{2}u$ | A1 | 1.1 |
| 2nd collision for $A$ & $B$: $2m\times\frac{1}{2}u + mU_B = 2mV_A + mV_B$ | M1 | 1.1 – Conservation of momentum. May see $-U_B$ or $\pm eu$. Do not allow assumed value of $U_B$ e.g. $\frac{1}{2}u$ or $u$. |
| $\frac{1}{2} = \frac{V_B - V_A}{\frac{1}{2}u - U_B}$ | M1 | 1.1 – Restitution. Do not allow assumed value of $U_B$ e.g. $\frac{1}{2}u$ or $u$. SC1 if assumed value for $V_B$ has been used (giving M0M0), provided $|U_B| \leq u$, direction of travel is towards A and equations are otherwise correct. |
| $u + U_B = 2V_A + V_B$ and $u - 2U_B = 4V_B - 4V_A$ | | |
| $\Rightarrow 3u = 6V_B \Rightarrow V_B = \frac{1}{2}u$ | A1 | 2.1 – **AG** Intermediate work towards cancellation must be seen |
| **[6]** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_B = u$ | B1 | 1.1 – Award if seen in (a) |
| Collision for $B$ & wall: $e = \pm\frac{U_B}{u}$ or $U_B = \pm eu$ | M1 | 3.1b – Restitution. May see $V_{2B}$ or similar instead of $\pm eu$ with use of restitution at the end. Award if seen in (a) |
| $\frac{\frac{4}{5}d}{\frac{1}{2}u} = \frac{d}{u} + \frac{\frac{1}{5}d}{eu}$ | M1 | 3.1b – Seeing that $A$ travels $\frac{4}{5}d$ at $\frac{1}{2}u$ in the same time as $B$ travels $d$ at $u$ and $\frac{1}{5}d$ at $eu$. Do not allow assumed rebound speed. |
| $\frac{3}{5} = \frac{1}{5e}$ | M1 | 1.1 – Correctly cancelling $d$ and $u$ and simplifying their 3 term equation including $e$ in the denominator |
| So coefficient of restitution between $B$ and wall is $\frac{1}{3}$ | A1 | 3.2a |
| **[5]** | | |6 Particles $A$ of mass $2 m$ and $B$ of mass $m$ are on a smooth horizontal floor. $A$ is moving with speed $u$ directly towards a vertical wall, and $B$ is at rest between $A$ and the wall (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{74bada9e-60cf-4ed4-abd0-4be155b7cf81-5_224_828_354_244}
A collides directly with $B$. The coefficient of restitution in this collision is $\frac { 1 } { 2 }$.\\
$B$ then collides with the wall, rebounds, and collides with $A$ for a second time.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $B$ after its second collision with $A$ is $\frac { 1 } { 2 } u$.
The first collision between $A$ and $B$ occurs at a distance $d$ from the wall. The second collision between $A$ and $B$ occurs at a distance $\frac { 1 } { 5 } d$ from the wall.
\item Find the coefficient of restitution for the collision between $B$ and the wall.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2019 Q6 [11]}}