Challenging +1.8 This is a Further Mechanics question requiring systematic application of conservation of momentum and Newton's experimental law across multiple collisions, tracking three particles through sequential impacts. Students must determine velocities after each collision and verify no third collision occurs by checking relative positions/velocities—requiring careful algebraic manipulation and physical insight beyond standard two-particle problems, but following established methods without requiring novel proof techniques.
3 Three particles, \(A , B\) and \(C\), of masses \(2 \mathrm {~kg} , 3 \mathrm {~kg}\) and 5 kg respectively, are at rest in a straight line on a smooth horizontal plane with \(B\) between \(A\) and \(C\).
Collisions between \(A\) and \(B\) are perfectly elastic. The coefficient of restitution for collisions between \(B\) and \(C\) is \(e\).
\(A\) is projected towards \(B\) with a speed of \(5 u \mathrm {~ms} ^ { - 1 }\) (see diagram).
\includegraphics[max width=\textwidth, alt={}, center]{493f11f4-e25c-4eeb-a0ab-20ec6d7a7a7d-2_186_903_2330_251}
Show that only two collisions occur.
oe; eg stating that the greatest leftwards speed of B occurs when \(e = 1\) and evaluating this
So A and B are both travelling in the negative direction, but \(
v_A
\geq
[8]
\(V_C = \frac{3u(1 + e)}{2}\), but this need not be stated; ignore any wrong value seen, provided not used further. Allow statement such as 'B is not catching up with A' in place of a formal inequality, provided correct working is seen
A colliding with B: $2 \times 5u = 2v_A + 3v_B$ | M1 [1.1] | Conservation of momentum
$\frac{v_B - v_A}{5u} = 1$ | M1 [1.1] | NEL
$v_A = -u$ and $v_B = 4u$ | A1 [1.1] | Both
B colliding with C: $3 \times 4u = 3v_B + 5V_C$ | M1 [1.1] | Conservation of momentum
$\frac{V_C - V_B}{4u} = e$ | M1 [1.1] | NEL
$v_B = \frac{u(3 - 5e)}{2}$ | A1 [1.1] | Correct $V_B$
$e \leq 1 \Rightarrow V_B \geq \frac{u(3-5)}{2} = -u$ | M1 [2.1] | oe; eg stating that the greatest leftwards speed of B occurs when $e = 1$ and evaluating this
So A and B are both travelling in the negative direction, but $|v_A| \geq |v_B|$ hence B and A do not collide again, ie there are only two collisions (AG) | E1 [2.2a] | Conclusion correctly drawn
| [8] | $V_C = \frac{3u(1 + e)}{2}$, but this need not be stated; ignore any wrong value seen, provided not used further. Allow statement such as 'B is not catching up with A' in place of a formal inequality, provided correct working is seen
---
3 Three particles, $A , B$ and $C$, of masses $2 \mathrm {~kg} , 3 \mathrm {~kg}$ and 5 kg respectively, are at rest in a straight line on a smooth horizontal plane with $B$ between $A$ and $C$.
Collisions between $A$ and $B$ are perfectly elastic. The coefficient of restitution for collisions between $B$ and $C$ is $e$.\\
$A$ is projected towards $B$ with a speed of $5 u \mathrm {~ms} ^ { - 1 }$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{493f11f4-e25c-4eeb-a0ab-20ec6d7a7a7d-2_186_903_2330_251}
Show that only two collisions occur.
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q3 [8]}}