Pre-U Pre-U 9794/3 2012 June — Question 10 10 marks

Exam BoardPre-U
ModulePre-U 9794/3 (Pre-U Mathematics Paper 3)
Year2012
SessionJune
Marks10
TopicMomentum and Collisions 1
TypeThree-particle sequential collisions
DifficultyChallenging +1.2 This is a sequential collision problem requiring systematic application of conservation of momentum and Newton's restitution law across multiple impacts. While it involves three collisions and tracking particle positions, each collision follows standard A-level mechanics procedures. The multi-part structure and need to track positions between collisions elevates it above routine questions, but the techniques are all standard for Further Maths mechanics.
Spec6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact

10 \includegraphics[max width=\textwidth, alt={}, center]{f0c32e07-f3a0-4d58-bd00-c266177ceaac-4_81_949_1283_598} Three particles \(A , B\) and \(C\), having masses \(1 \mathrm {~kg} , 2 \mathrm {~kg}\) and 5 kg , respectively, are placed 1 metre apart in a straight line on a smooth horizontal plane (see diagram). The particles \(B\) and \(C\) are initially at rest and \(A\) is moving towards \(B\) with speed \(14 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The coefficient of restitution between each pair of particles is 0.5 .
  1. Find the velocity of \(B\) immediately after the first impact and show that \(A\) comes to rest.
  2. Show that \(B\) reverses direction after an impact with \(C\).
  3. Find the distance between \(B\) and \(C\) at the instant that \(B\) collides with \(A\) for the second time.

(i) COM: \(1 \times 14 + 2 \times 0 = U + 2V\) — B1
NEL: \(V - U = 0.5(14 - 0)\) — B1
Solution: \(U = 0\) ms\(^{-1}\), — B1
\(V = 7\) ms\(^{-1}\) — B1 [4]
(ii) For impact of \(B\) on \(C\):
COM: \(2 \times 7 + 5 \times 0 = 2U + 5V\) — B1
NEL: \(V - U = 0.5(7 - 0)\) — B1
Solution: \(U = -0.5\) ms\(^{-1}\) — B1 [3]
(iii) \(V = 3\) ms\(^{-1}\) — B1
\(B\) reaches \(A\) in 2 seconds. — B1
Distance between \(A\) and \(C\) is \(1 + 2 \times 3 = 7\) metres — B1 [3]
Total: [10]
*Depends on 2nd B1. SC If NEL is \(V + U = \ldots\) then max B1B0B0B1. May be seen/awarded in (ii). ft their \(U\). ft their \(V\).*
**(i)** COM: $1 \times 14 + 2 \times 0 = U + 2V$ — B1

NEL: $V - U = 0.5(14 - 0)$ — B1

Solution: $U = 0$ ms$^{-1}$, — B1

$V = 7$ ms$^{-1}$ — B1 [4]

**(ii)** For impact of $B$ on $C$:

COM: $2 \times 7 + 5 \times 0 = 2U + 5V$ — B1

NEL: $V - U = 0.5(7 - 0)$ — B1

Solution: $U = -0.5$ ms$^{-1}$ — B1 [3]

**(iii)** $V = 3$ ms$^{-1}$ — B1

$B$ reaches $A$ in 2 seconds. — B1

Distance between $A$ and $C$ is $1 + 2 \times 3 = 7$ metres — B1 [3]

**Total: [10]**

*Depends on 2nd B1. SC If NEL is $V + U = \ldots$ then max B1B0B0B1. May be seen/awarded in (ii). ft their $U$. ft their $V$.*
10\\
\includegraphics[max width=\textwidth, alt={}, center]{f0c32e07-f3a0-4d58-bd00-c266177ceaac-4_81_949_1283_598}

Three particles $A , B$ and $C$, having masses $1 \mathrm {~kg} , 2 \mathrm {~kg}$ and 5 kg , respectively, are placed 1 metre apart in a straight line on a smooth horizontal plane (see diagram). The particles $B$ and $C$ are initially at rest and $A$ is moving towards $B$ with speed $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The coefficient of restitution between each pair of particles is 0.5 .\\
(i) Find the velocity of $B$ immediately after the first impact and show that $A$ comes to rest.\\
(ii) Show that $B$ reverses direction after an impact with $C$.\\
(iii) Find the distance between $B$ and $C$ at the instant that $B$ collides with $A$ for the second time.

\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2012 Q10 [10]}}