| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Successive collisions, three particles in line |
| Difficulty | Challenging +1.2 This is a multi-collision momentum problem requiring systematic application of conservation of momentum and restitution equations across three collisions. While it involves multiple steps and algebraic manipulation, the techniques are standard for Further Mechanics AS. Part (c) requires insight that a third collision occurs only if velocities satisfy certain inequalities, which elevates it slightly above routine exercises but remains within expected problem-solving for this module. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03d Conservation in 2D: vector momentum6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 1st Collision: \(3.3u = 3.3v_A + 2.2v_B\) | M1 | Conservation of momentum - Must be seen |
| 1st Collision: \(\pm e = \dfrac{v_B - v_A}{u}\) | M1 | NEL - Must be seen |
| \(3u = 3v_A + 2v_B\) and \(2eu = 2v_B - 2v_A\) \(\Rightarrow 5v_A = 3u - 2eu \Rightarrow v_A = \dfrac{u(3-2e)}{5}\) | A1 | AG |
| \(3u = 3v_A + 2v_B\) and \(3eu = 3v_B - 3v_A\) \(\Rightarrow 5v_B = 3u + 3eu \Rightarrow v_B = \dfrac{3u(1+e)}{5}\) | A1 | Find \(v_B\) by elimination or substitution - AEF, award if seen in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 2nd Collision: \(2.2 \times \dfrac{3u(1+e)}{5} = 2.2V_B + V_C\) | M1ft | Conservation of momentum (ft their \(V_B\)) - May be in terms of \(V_A\) |
| 2nd Collision: \(\pm e = \dfrac{V_C - V_B}{\left(\dfrac{3u(1+e)}{5}\right)}\) | M1ft | NEL (ft their \(V_B\)) |
| \(\dfrac{33}{25}u(1+e) = \dfrac{11}{5}V_B + V_C\) and \(\dfrac{3eu(1+e)}{5} = V_C - V_B\) \(\Rightarrow \dfrac{16}{5}V_B = \dfrac{33}{25}u(1+e) - \dfrac{3eu(1+e)}{5}\) | M1ft | Attempt to eliminate \(V_C\) - \(V_C = \dfrac{33u(1+e)^2}{80}\) |
| \(\Rightarrow V_B = \dfrac{3u(1+e)(11-5e)}{80}\) | A1 | oe - Must be in terms of \(e\) and \(u\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{u(3-2e)}{5} > \dfrac{3u(1+e)(11-5e)}{80}\) | M1ft | Correct condition for further collision (ft their \(V_B\) from (b)) |
| \(3e^2 - 10e + 3 > 0\) | M1 | Rearranging to 3 term quadratic inequality in \(e\) |
| \(A\) and \(B\) collide again \(\Rightarrow e \neq 0\) | B1 | 2.2a |
| \((3e-1)(e-3) > 0\) and \(0 \leq e \leq 1\) and \(e \neq 0\) \(\Rightarrow 0 < e < \dfrac{1}{3}\) | A1 | \(e < \dfrac{1}{3}\) is not sufficient for A1 - If B1 not awarded then award A1 for \(0 \leq e < \dfrac{1}{3}\) |
## Question 6:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 1st Collision: $3.3u = 3.3v_A + 2.2v_B$ | M1 | Conservation of momentum - Must be seen |
| 1st Collision: $\pm e = \dfrac{v_B - v_A}{u}$ | M1 | NEL - Must be seen |
| $3u = 3v_A + 2v_B$ and $2eu = 2v_B - 2v_A$ $\Rightarrow 5v_A = 3u - 2eu \Rightarrow v_A = \dfrac{u(3-2e)}{5}$ | A1 | AG |
| $3u = 3v_A + 2v_B$ and $3eu = 3v_B - 3v_A$ $\Rightarrow 5v_B = 3u + 3eu \Rightarrow v_B = \dfrac{3u(1+e)}{5}$ | A1 | Find $v_B$ by elimination or substitution - AEF, award if seen in (b) |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 2nd Collision: $2.2 \times \dfrac{3u(1+e)}{5} = 2.2V_B + V_C$ | M1ft | Conservation of momentum (ft their $V_B$) - May be in terms of $V_A$ |
| 2nd Collision: $\pm e = \dfrac{V_C - V_B}{\left(\dfrac{3u(1+e)}{5}\right)}$ | M1ft | NEL (ft their $V_B$) |
| $\dfrac{33}{25}u(1+e) = \dfrac{11}{5}V_B + V_C$ and $\dfrac{3eu(1+e)}{5} = V_C - V_B$ $\Rightarrow \dfrac{16}{5}V_B = \dfrac{33}{25}u(1+e) - \dfrac{3eu(1+e)}{5}$ | M1ft | Attempt to eliminate $V_C$ - $V_C = \dfrac{33u(1+e)^2}{80}$ |
| $\Rightarrow V_B = \dfrac{3u(1+e)(11-5e)}{80}$ | A1 | oe - Must be in terms of $e$ and $u$ only |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{u(3-2e)}{5} > \dfrac{3u(1+e)(11-5e)}{80}$ | M1ft | Correct condition for further collision (ft their $V_B$ from (b)) |
| $3e^2 - 10e + 3 > 0$ | M1 | Rearranging to 3 term quadratic inequality in $e$ |
| $A$ and $B$ collide again $\Rightarrow e \neq 0$ | B1 | 2.2a |
| $(3e-1)(e-3) > 0$ and $0 \leq e \leq 1$ and $e \neq 0$ $\Rightarrow 0 < e < \dfrac{1}{3}$ | A1 | $e < \dfrac{1}{3}$ is not sufficient for A1 - If B1 not awarded then award A1 for $0 \leq e < \dfrac{1}{3}$ |
---6 Three particles $A , B$ and $C$ are free to move in the same straight line on a large smooth horizontal surface. Their masses are $3.3 \mathrm {~kg} , 2.2 \mathrm {~kg}$ and 1 kg respectively. The coefficient of restitution in collisions between any two of them is $e$.
Initially, $B$ and $C$ are at rest and $A$ is moving towards $B$ with speed $u \mathrm {~ms} ^ { - 1 }$ (see diagram). $A$ collides directly with $B$ and $B$ then goes on to collide directly with $C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{0501e5a4-2137-4e7d-98ff-2ee81941cbf3-4_221_1342_552_246}
\begin{enumerate}[label=(\alph*)]
\item The velocities of $A$ and $B$ immediately after the first collision are denoted by $\mathrm { v } _ { \mathrm { A } } \mathrm { ms } ^ { - 1 }$ and $\mathrm { V } _ { \mathrm { B } } \mathrm { ms } ^ { - 1 }$ respectively.
\begin{itemize}
\item Show that $\mathrm { v } _ { \mathrm { A } } = \frac { \mathrm { u } ( 3 - 2 \mathrm { e } ) } { 5 }$.
\item Find an expression for $\mathrm { V } _ { \mathrm { B } }$ in terms of $u$ and $e$.
\item Find an expression in terms of $u$ and $e$ for the velocity of $B$ immediately after its collision with $C$.
\end{itemize}
After the collision between $B$ and $C$ there is a further collision between $A$ and $B$.
\item Determine the range of possible values of $e$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2020 Q6 [12]}}