| Exam Board | OCR |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2018 |
| Session | March |
| Marks | 11 |
| Topic | Momentum and Collisions 1 |
| Type | Collision followed by wall impact |
| Difficulty | Standard +0.8 This is a multi-stage collision problem requiring conservation of momentum, Newton's restitution law, and inequality analysis to determine when a second collision occurs. Part (i) is a standard 'show that' requiring algebraic manipulation. Part (ii) requires tracking velocities through two collisions and setting up an inequality condition, which demands careful reasoning about relative motion and is more sophisticated than typical A-level mechanics questions. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4u = 4v_A + 11v_B\) | M1 | Conservation of momentum |
| \(\frac{v_B - v_A}{u} = e\) | M1 | Restitution |
| \(4u = 4(v_B - eu) + 11v_B\) oe | M1 | Elimination of \(v_A\) attempted, oe |
| \(v_B = \frac{8}{15}u(1 + e)\) | A1 [4] | AG |
| \(v_B = -\frac{2}{15}eu(1 + e)\) | B1ft | \(-\frac{1}{2}e\) times their \(v_B\) |
| \(v_A = \frac{8}{15}u(4 - 11e)\) | M1 | From simultaneous equations in (i) |
| \(4 - 11e < 0 \Rightarrow e > \frac{4}{11}\) | A1 | Condone \(\geq\) |
| 2nd collision occurs if \(\frac{1}{5}u(4 - 11e) > -\frac{2}{15}eu(1 + e)\) | M1 | For their \(v_A >\) their \(v_B\) soi; inequality must be correct way round |
| \(2e^2 - 9e + 4 > 0\) | M1 | Reduction to 3 term quadratic |
| \(e < \frac{1}{2}\) | A1 | Condone \(\leq\); \(e > 4\) must be rejected, but this can be implied |
| \(\frac{4}{11} < e < \frac{1}{2}\) | A1 [7] | cao |
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4u = 4v_A + 11v_B$ | M1 | Conservation of momentum |
| $\frac{v_B - v_A}{u} = e$ | M1 | Restitution |
| $4u = 4(v_B - eu) + 11v_B$ oe | M1 | Elimination of $v_A$ attempted, oe |
| $v_B = \frac{8}{15}u(1 + e)$ | A1 [4] | AG |
| $v_B = -\frac{2}{15}eu(1 + e)$ | B1ft | $-\frac{1}{2}e$ times their $v_B$ |
| $v_A = \frac{8}{15}u(4 - 11e)$ | M1 | From simultaneous equations in (i) |
| $4 - 11e < 0 \Rightarrow e > \frac{4}{11}$ | A1 | Condone $\geq$ |
| 2nd collision occurs if $\frac{1}{5}u(4 - 11e) > -\frac{2}{15}eu(1 + e)$ | M1 | For their $v_A >$ their $v_B$ soi; inequality must be correct way round |
| $2e^2 - 9e + 4 > 0$ | M1 | Reduction to 3 term quadratic |
| $e < \frac{1}{2}$ | A1 | Condone $\leq$; $e > 4$ must be rejected, but this can be implied |
| $\frac{4}{11} < e < \frac{1}{2}$ | A1 [7] | cao |5 Two particles $A$ and $B$ are on a smooth horizontal floor with $B$ between $A$ and a vertical wall. The masses of $A$ and $B$ are 4 kg and 11 kg respectively. Initially, $B$ is at rest and $A$ is moving towards $B$ with a speed of $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see diagram). $A$ collides directly with $B$. The coefficient of restitution between $A$ and $B$ is $e$.\\
\includegraphics[max width=\textwidth, alt={}, center]{bf86ac88-0fd1-4d49-a705-9b8d06fbac2a-3_209_803_1658_630}\\
(i) Show that immediately after the collision the speed of $B$ is $\frac { 4 } { 15 } u ( 1 + e )$.
After the collision between $A$ and $B$ the direction of motion of $A$ is reversed. $B$ subsequently collides directly with the vertical wall. The coefficient of restitution between $B$ and the wall is $\frac { 1 } { 2 } e$.\\
(ii) Given that there is a second collision between $A$ and $B$, find the range of possible values of $e$.
\hfill \mbox{\textit{OCR FM1 AS 2018 Q5 [11]}}