Pre-U Pre-U 9794/3 2016 Specimen — Question 9 10 marks

Exam BoardPre-U
ModulePre-U 9794/3 (Pre-U Mathematics Paper 3)
Year2016
SessionSpecimen
Marks10
TopicMomentum and Collisions 1
TypeThree-particle sequential collisions
DifficultyStandard +0.8 This is a multi-stage collision problem requiring systematic application of conservation of momentum and Newton's restitution law across three sequential collisions, tracking positions and velocities throughout. While the individual collision calculations are standard A-level mechanics, the extended chain of reasoning, careful bookkeeping of multiple particles' states, and the geometric position calculation in part (iii) elevate this above typical textbook exercises. It's more demanding than standard two-particle collision questions but doesn't require novel mathematical insight beyond methodical application of principles.
Spec6.03a 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 impact

9 \includegraphics[max width=\textwidth, alt={}, center]{b18b1bc5-bf26-4161-b5a5-764b00e97bea-5_118_851_1265_607} Three particles \(A , B\) and \(C\), having masses of \(1 \mathrm {~kg} , 2 \mathrm {~kg}\) 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 \(14 \mathrm {~ms} ^ { - 1 }\). 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 \(B\) reversed direction after the impact with \(C\).
  3. Find the distances 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
\(U = 0\) ms\(^{-1}\) B1
\(V = 7\) ms\(^{-1}\) B1
(ii) COM: \(2 \times 7 + 5 \times 0 = 2U + 5V\) B1
NEL: \(V - U = 0.5(7 - 0)\) B1
\(U = -0.5\) ms\(^{-1}\) B1
(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
**(i)** COM: $1 \times 14 + 2 \times 0 = U + 2V$ **B1**

NEL: $V - U = 0.5(14 - 0)$ **B1**

$U = 0$ ms$^{-1}$ **B1**

$V = 7$ ms$^{-1}$ **B1**

**(ii)** COM: $2 \times 7 + 5 \times 0 = 2U + 5V$ **B1**

NEL: $V - U = 0.5(7 - 0)$ **B1**

$U = -0.5$ ms$^{-1}$ **B1**

**(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**
9\\
\includegraphics[max width=\textwidth, alt={}, center]{b18b1bc5-bf26-4161-b5a5-764b00e97bea-5_118_851_1265_607}

Three particles $A , B$ and $C$, having masses of $1 \mathrm {~kg} , 2 \mathrm {~kg}$ 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 $14 \mathrm {~ms} ^ { - 1 }$. 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 $B$ reversed direction after the impact with $C$.\\
(iii) Find the distances 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 2016 Q9 [10]}}