| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, find velocities/angles |
| Difficulty | Standard +0.8 This is a multi-part oblique collision problem requiring conservation of momentum in two dimensions, geometric reasoning about equal time to boundaries, and distance calculations. While it involves several steps and coordinate resolution, the techniques are standard for Further Maths mechanics with clear guidance from the question structure. The conceptual demand is moderate—higher than typical A-level but not requiring novel insight. |
| Spec | 6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum |
| Answer | Marks | Guidance |
|---|---|---|
| The perpendicular components of momentum will be equal; as balls have equal masses the perpendicular velocity components will be equal; as both balls are the same distance from the sides they will both hit the sides at the same time | M1 | Recognises magnitudes of perpendicular components of momentum will be the same; condone missing reference to magnitudes |
| Uses equal masses to argue perpendicular velocity components are equal, and because balls are same distance from sides, balls hit the side at the same time | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(m\begin{bmatrix}2\\0\end{bmatrix} = m\begin{bmatrix}0.8\cos 30°\\0.8\sin 30°\end{bmatrix} + m\begin{bmatrix}v\cos\theta\\-v\sin\theta\end{bmatrix}\) | M1 | Forms conservation of momentum equations for both components |
| \(v\cos\theta = 2 - 0.8\cos 30° = 2 - \frac{2\sqrt{3}}{5}\), \(v\sin\theta = 0.8\sin 30° = 0.4\) | A1 | Obtains correct pair of equations |
| \(\tan\theta = \dfrac{0.4}{2 - \dfrac{2\sqrt{3}}{5}} = 0.3060...\) | M1 | Forms equation for \(\tan\theta\) from their pair of equations |
| \(\theta = 17.014... = 17.0°\) to 1 dp | A1 | Must see \(\theta = 17.01\) or better; accept 17; AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = \dfrac{0.8\sin 30°}{\sin 17.0°}\) | M1 | Uses momentum equation or Pythagoras' theorem to find \(v\) |
| \(v = 1.37\) | A1 | Correct value for \(v\), AWRT 1.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.8\cos 30° = 0.693\); \(1.37\cos 17.0° = 1.31\); Parallel component for red ball is greater, so red ball has greater speed and travels greater distance as both balls take same time to reach edges | M1 | Uses values and given information to create argument with two comparable quantities. Eg calculates components of velocity parallel to sides of table. Argument based on comparative size of angles, or based on speed and time. Not necessary to see numeric values in argument. |
| Completes reasoned argument to conclude red ball travels greater distance | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Include the rotational motion of the balls in the model | B1 | Suggests including air resistance or taking account of rotational motion of balls. Must be a refinement and not a criticism. |
## Question 7(a):
The perpendicular components of momentum will be equal; as balls have equal masses the perpendicular velocity components will be equal; as both balls are the same distance from the sides they will both hit the sides at the same time | M1 | Recognises magnitudes of perpendicular components of momentum will be the same; condone missing reference to magnitudes
Uses equal masses to argue perpendicular velocity components are equal, and because balls are same distance from sides, balls hit the side at the same time | R1 |
---
## Question 7(b):
$m\begin{bmatrix}2\\0\end{bmatrix} = m\begin{bmatrix}0.8\cos 30°\\0.8\sin 30°\end{bmatrix} + m\begin{bmatrix}v\cos\theta\\-v\sin\theta\end{bmatrix}$ | M1 | Forms conservation of momentum equations for both components
$v\cos\theta = 2 - 0.8\cos 30° = 2 - \frac{2\sqrt{3}}{5}$, $v\sin\theta = 0.8\sin 30° = 0.4$ | A1 | Obtains correct pair of equations
$\tan\theta = \dfrac{0.4}{2 - \dfrac{2\sqrt{3}}{5}} = 0.3060...$ | M1 | Forms equation for $\tan\theta$ from their pair of equations
$\theta = 17.014... = 17.0°$ to 1 dp | A1 | Must see $\theta = 17.01$ or better; accept 17; AG
## Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = \dfrac{0.8\sin 30°}{\sin 17.0°}$ | M1 | Uses momentum equation or Pythagoras' theorem to find $v$ |
| $v = 1.37$ | A1 | Correct value for $v$, AWRT 1.4 |
**Total: 2 marks**
---
## Question 7(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.8\cos 30° = 0.693$; $1.37\cos 17.0° = 1.31$; Parallel component for red ball is greater, so red ball has greater speed and travels greater distance as both balls take same time to reach edges | M1 | Uses values and given information to create argument with two comparable quantities. Eg calculates components of velocity parallel to sides of table. Argument based on comparative size of angles, or based on speed and time. Not necessary to see numeric values in argument. |
| Completes reasoned argument to conclude red ball travels greater distance | R1 | |
**Total: 2 marks**
---
## Question 7(e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Include the rotational motion of the balls in the model | B1 | Suggests including air resistance or taking account of rotational motion of balls. Must be a refinement and not a criticism. |
**Total: 1 mark**
---7 Two snooker balls, one white and one red, have equal mass.
The balls are on a horizontal table $A B C D$\\
The white ball is struck so that it moves at a speed of $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ parallel to $A B$\\
The white ball hits a stationary red ball.\\
After the collision, the white ball moves at a speed of $0.8 \mathrm {~ms} ^ { - 1 }$ and at an angle of $30 ^ { \circ }$ to $A B$
After the collision, the red ball moves at a speed $v \mathrm {~ms} ^ { - 1 }$ and at an angle $\theta$ to $C D$\\
When the collision takes place, the white ball is the same distance from $A B$ as the distance the red ball is from CD
The diagram below shows the table and the velocities of the balls after the collision.\\
\includegraphics[max width=\textwidth, alt={}, center]{0afe3ff2-0af5-4aeb-98c5-1346fa803388-08_595_1370_1121_335}
Not to scale
After the collision, the white ball hits $A B$ and the red ball hits $C D$\\
Model the balls as particles that do not experience any air resistance.\\
7
\begin{enumerate}[label=(\alph*)]
\item Explain why the two balls hit the sides of the table at the same time.\\
7
\item Show that $\theta = 17.0 ^ { \circ }$ correct to one decimal place.\\
7
\item $\quad$ Find $v$\\
7
\item Determine which ball travels the greater distance after the collision and before hitting the side of the table.
Fully justify your answer.\\
7
\item State one possible refinement to the model that you have used.\\
$8 \quad$ In this question use $g$ as $9.8 \mathrm {~ms} ^ { - 2 }$
A rope is used to pull a crate, of mass 60 kg , along a rough horizontal surface.\\
The coefficient of friction between the crate and the surface is 0.4
The crate is at rest when the rope starts to pull on it.\\
The tension in the rope is 240 N and the rope makes an angle of $30 ^ { \circ }$ to the horizontal.\\
When the crate has moved 5 metres, the rope becomes detached from the crate.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2022 Q7 [11]}}