| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Three-particle sequential collisions |
| Difficulty | Standard +0.8 This is a multi-stage collision problem requiring systematic application of conservation of momentum and coefficient of restitution equations across two separate collisions. Part (i) is a standard 'show that' verification, parts (ii-iii) require algebraic manipulation with a parameter m, and part (iv) demands inequality reasoning about direction reversal. The sequential nature and parametric analysis elevate this above routine A-level mechanics questions, but the techniques are standard for Further Mechanics. |
| 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 | Marks | Guidance |
| \(1.2 \times 4 = 1.2v_A + 1.8v_B\) | M1* | Conservation of momentum. Allow one minor slip, e.g. transpose masses |
| \(\dfrac{v_B - v_A}{4} = \dfrac{3}{4}\) | M1* | Restitution. Allow sign error |
| Attempt to solve for \(v_A\) and \(v_B\) | M1dep | |
| \(v_B = 2.8\) | A1 (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v_A = -0.2\) | B1 | 0.2 in opposite direction. Allow "away from B" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.8 \times 2.8 = 1.8V_B + mV_C\) | M1* | Conservation of momentum. Allow 1 minor slip. NB \(v_C > v_B\) |
| \(\dfrac{V_C - V_B}{2.8} = \dfrac{3}{4}\) | M1* | Restitution. Allow sign error |
| Attempt to solve for \(V_B\) in terms of \(m\) | M1dep | \(V_C\) must be eliminated |
| \(V_B = \dfrac{5.04 - 2.1m}{1.8 + m}\) oe | A1 | \(\dfrac{8.82}{1.8+m} - 2.1\) or \(\dfrac{25.2 - 10.5m}{5m + 9}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Direction reversed \(\Rightarrow V_B < 0\) | M1 | Seen or implied by eg \(\dfrac{5.04 - 2.1m}{1.8 + m} < 0\). If \(V_c\) found in error, \(V_c < 2.1\) or \(\dfrac{8.82}{1.8+m} < 2.1\) |
| \(m > 2.4\) | A1 | Must be from an inequality |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.2 \times 4 = 1.2v_A + 1.8v_B$ | M1* | Conservation of momentum. Allow one minor slip, e.g. transpose masses |
| $\dfrac{v_B - v_A}{4} = \dfrac{3}{4}$ | M1* | Restitution. Allow sign error |
| Attempt to solve for $v_A$ and $v_B$ | M1dep | |
| $v_B = 2.8$ | A1 (AG) | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v_A = -0.2$ | B1 | 0.2 in opposite direction. Allow "away from B" |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.8 \times 2.8 = 1.8V_B + mV_C$ | M1* | Conservation of momentum. Allow 1 minor slip. NB $v_C > v_B$ |
| $\dfrac{V_C - V_B}{2.8} = \dfrac{3}{4}$ | M1* | Restitution. Allow sign error |
| Attempt to solve for $V_B$ in terms of $m$ | M1dep | $V_C$ must be eliminated |
| $V_B = \dfrac{5.04 - 2.1m}{1.8 + m}$ oe | A1 | $\dfrac{8.82}{1.8+m} - 2.1$ or $\dfrac{25.2 - 10.5m}{5m + 9}$ |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Direction reversed $\Rightarrow V_B < 0$ | M1 | Seen or implied by eg $\dfrac{5.04 - 2.1m}{1.8 + m} < 0$. If $V_c$ found in error, $V_c < 2.1$ or $\dfrac{8.82}{1.8+m} < 2.1$ |
| $m > 2.4$ | A1 | Must be from an inequality |
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\includegraphics[max width=\textwidth, alt={}, center]{5960a9cf-2c51-4c07-9973-c29604762df7-3_218_1335_251_367}
Three particles $A$, $B$ and $C$ are free to move in the same straight line on a large smooth horizontal surface. Their masses are $1.2 \mathrm {~kg} , 1.8 \mathrm {~kg}$ and $m \mathrm {~kg}$ respectively (see diagram). The coefficient of restitution in collisions between any two of them is $\frac { 3 } { 4 }$. Initially, $B$ and $C$ are at rest and $A$ is moving with a velocity of $4.0 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ towards $B$.\\
(i) Show that immediately after the collision between $A$ and $B$ the speed of $B$ is $2.8 \mathrm {~ms} ^ { - 1 }$.\\
(ii) Find the velocity of $A$ immediately after this collision.\\
$B$ subsequently collides with $C$.\\
(iii) Find, in terms of $m$, the velocity of $B$ after its collision with $C$.\\
(iv) Given that the direction of motion of $B$ is reversed by the collision with $C$, find the range of possible values of $m$.
\hfill \mbox{\textit{OCR Further Mechanics AS 2018 Q4 [11]}}