AQA M3 — Question 6

Exam BoardAQA
ModuleM3 (Mechanics 3)
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeOblique collision, find velocities/angles
DifficultyStandard +0.8 This M3 oblique collision question requires resolving velocities in two perpendicular directions (along and perpendicular to line of centres), applying conservation of momentum and Newton's restitution law, then recombining components. While systematic, it involves multiple steps with trigonometry and careful bookkeeping across 4-5 equations, making it moderately challenging but still a standard M3 exercise.
Spec6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact
6 Two smooth billiard balls \(A\) and \(B\), of identical size and equal mass, move towards each other on a horizontal surface and collide. Just before the collision, \(A\) has velocity \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a direction inclined at \(30 ^ { \circ }\) to the line of centres of the balls, and \(B\) has velocity \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a direction inclined at \(60 ^ { \circ }\) to the line of centres, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{fc5bfc4b-68bb-4a23-874b-87e9558dc990-05_508_1420_532_294} The coefficient of restitution between the balls is \(\frac { 1 } { 2 }\).
  1. Find the speed of \(B\) immediately after the collision.
  2. Find the angle between the velocity of \(B\) and the line of centres of the balls immediately after the collision.
Question 6:
Part (a)(i):
AnswerMarks Guidance
Working/AnswerMark Guidance
Using sine rule in velocity triangle: \(\frac{50}{\sin\phi} = \frac{35}{\sin\theta}\) where geometry from bearing 120°M1 A1 Setting up triangle correctly
Angle at frigate between ship's velocity and joining lineM1
Two bearings foundA1 A1 One mark each
Part (a)(ii):
AnswerMarks Guidance
Working/AnswerMark Guidance
Find time using shorter intercept directionM1 M1
Answer in hoursA1 A1 A1
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
Minimum speed when frigate's velocity perpendicular to relative velocityM1 A1
Minimum speed \(= 50\sin(30°) = 25 \text{ km h}^{-1}\)A1 
# Question 6:

## Part (a)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Using sine rule in velocity triangle: $\frac{50}{\sin\phi} = \frac{35}{\sin\theta}$ where geometry from bearing 120° | M1 A1 | Setting up triangle correctly |
| Angle at frigate between ship's velocity and joining line | M1 | |
| Two bearings found | A1 A1 | One mark each |

## Part (a)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Find time using shorter intercept direction | M1 M1 | |
| Answer in hours | A1 A1 A1 | |

## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Minimum speed when frigate's velocity perpendicular to relative velocity | M1 A1 | |
| Minimum speed $= 50\sin(30°) = 25 \text{ km h}^{-1}$ | A1 | |

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6 Two smooth billiard balls $A$ and $B$, of identical size and equal mass, move towards each other on a horizontal surface and collide. Just before the collision, $A$ has velocity $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a direction inclined at $30 ^ { \circ }$ to the line of centres of the balls, and $B$ has velocity $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a direction inclined at $60 ^ { \circ }$ to the line of centres, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{fc5bfc4b-68bb-4a23-874b-87e9558dc990-05_508_1420_532_294}

The coefficient of restitution between the balls is $\frac { 1 } { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $B$ immediately after the collision.
\item Find the angle between the velocity of $B$ and the line of centres of the balls immediately after the collision.
\end{enumerate}

\hfill \mbox{\textit{AQA M3  Q6}}
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