AQA M1 2013 January — Question 5 6 marks

Exam BoardAQA
ModuleM1 (Mechanics 1)
Year2013
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeCollision with two possible outcomes
DifficultyModerate -0.3 This is a straightforward momentum conservation problem with a clear setup. Students must recognize that particle A can move in either direction after collision (hence two solutions), then apply conservation of momentum with careful attention to sign conventions. While requiring systematic thinking about directions, it's a standard M1 collision question with routine calculations and no conceptual surprises.
Spec6.03b Conservation of momentum: 1D two particles

5 Two particles, \(A\) and \(B\), are moving towards each other along the same straight horizontal line when they collide. Particle \(A\) has mass 5 kg and particle \(B\) has mass 4 kg . Just before the collision, the speed of \(A\) is \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the speed of \(B\) is \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). After the collision, the speed of \(A\) is \(0.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and both particles move on the same straight horizontal line. Find the two possible speeds of \(B\) after the collision.
(6 marks)

Question 5:
Two possible speeds of B after collision
AnswerMarks Guidance
Working/AnswerMark Notes
Define positive direction, e.g. direction of A's motionM1 Setting up momentum equation
Initial momentum: \(5(4) - 4(3) = 20 - 12 = 8 \text{ Ns}\)A1 Correct initial momentum
After collision A has speed \(0.6\ \text{ms}^{-1}\), two cases for directionM1 Considering both directions for A
Case 1: \(5(0.6) + 4v_B = 8 \Rightarrow 4v_B = 5 \Rightarrow v_B = 1.25\ \text{ms}^{-1}\)A1 First speed
Case 2: \(5(-0.6) + 4v_B = 8 \Rightarrow 4v_B = 11 \Rightarrow v_B = 2.75\ \text{ms}^{-1}\)A1 Second speed
Both values checked for validity (no particle passes through other)A1 Both valid
## Question 5:

**Two possible speeds of B after collision**

| Working/Answer | Mark | Notes |
|---|---|---|
| Define positive direction, e.g. direction of A's motion | M1 | Setting up momentum equation |
| Initial momentum: $5(4) - 4(3) = 20 - 12 = 8 \text{ Ns}$ | A1 | Correct initial momentum |
| After collision A has speed $0.6\ \text{ms}^{-1}$, two cases for direction | M1 | Considering both directions for A |
| Case 1: $5(0.6) + 4v_B = 8 \Rightarrow 4v_B = 5 \Rightarrow v_B = 1.25\ \text{ms}^{-1}$ | A1 | First speed |
| Case 2: $5(-0.6) + 4v_B = 8 \Rightarrow 4v_B = 11 \Rightarrow v_B = 2.75\ \text{ms}^{-1}$ | A1 | Second speed |
| Both values checked for validity (no particle passes through other) | A1 | Both valid |
5 Two particles, $A$ and $B$, are moving towards each other along the same straight horizontal line when they collide. Particle $A$ has mass 5 kg and particle $B$ has mass 4 kg . Just before the collision, the speed of $A$ is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the speed of $B$ is $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. After the collision, the speed of $A$ is $0.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and both particles move on the same straight horizontal line.

Find the two possible speeds of $B$ after the collision.\\
(6 marks)

\hfill \mbox{\textit{AQA M1 2013 Q5 [6]}}