| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Year | 2020 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Sphere rebounds off fixed wall obliquely |
| Difficulty | Standard +0.8 This is a Further Maths mechanics question requiring integration of the impulse-momentum theorem with a time-varying force function, resolution of velocities at an angle, and application of the coefficient of restitution concept. While the individual techniques are standard (integration, trigonometry, momentum conservation), combining them in this context with the given polynomial force model requires solid problem-solving skills beyond routine A-level questions. |
| Spec | 6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I = \int_0^{0.1} kt^2(0.1-t)^2\, dt\) | M1 (AO 3.3) | Models impulse of force using integral of form \(\int F\, dt\) |
| Integrates \(t^2(0.1-t)^2\) | M1 (AO 1.1a) | — |
| \(I = k\int_0^{0.1}\left(\frac{t^2}{100} - \frac{t^3}{5} + t^4\right)dt = \frac{k}{3000000}\) | A1 (AO 1.1b) | Obtains \(\int_0^{0.1} kt^2(0.1-t)^2\,dt = \frac{k}{3000000}\) |
| \(I = \frac{k}{3000000}\) | R1 (AO 2.1) | Completes rigorous argument linking impulse to \(\frac{k}{3000000}\); units not required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For maximum impulse the ball rebounds with same magnitude of velocity perpendicular to wall (i.e. \(e \leq 1\) or \(e = 1\)) | E1 (AO 3.3) | Explains ball must rebound with same velocity component perpendicular to wall for maximum impulse |
| As ball rebounds, minimum impulse must be greater than impulse needed to bring perpendicular velocity to zero (\(0 < e\) or \(e = 0\)) | E1 (AO 3.3) | Explains impulse must be greater than impulse needed to reduce perpendicular component to zero |
| \(I_{max} = 2(0.3 \times 4\sin 30°) = 1.2\); \(I_{min} > 0.3 \times 4\sin 30° = 0.6\) | B1 (AO 1.1b) | Calculates maximum and minimum magnitudes for impulse or velocity |
| \(0.6 < \frac{k}{3000000} \leq 1.2\) | M1 (AO 2.2a) | Uses impulse to deduce inequality for \(k\) |
| \(1800000 < k \leq 3600000\) | R1 (AO 2.1) | Reaches required conclusion from complete argument |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Parallel component \(= 4\cos 30°\) | B1 (AO 3.4) | Obtains correct component of velocity parallel to wall |
| \(I = 0.8\) | M1 (AO 3.3) | Applies impulse to motion perpendicular to wall |
| \(0.3v - (-0.3 \times 4\sin 30°) = 0.8\), so \(v = \frac{2}{3}\) | A1 (AO 1.1b) | Obtains correct rebound velocity component perpendicular to wall |
| Speed \(= \sqrt{\left(\frac{2}{3}\right)^2 + (4\cos 30°)^2} = 3.53 \text{ m s}^{-1}\) | A1F (AO 1.1b) | Obtains correct speed using perpendicular and parallel components; condone missing units |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I = \int_0^{0.1} kt^2(0.1-t)^2\, dt$ | M1 (AO 3.3) | Models impulse of force using integral of form $\int F\, dt$ |
| Integrates $t^2(0.1-t)^2$ | M1 (AO 1.1a) | — |
| $I = k\int_0^{0.1}\left(\frac{t^2}{100} - \frac{t^3}{5} + t^4\right)dt = \frac{k}{3000000}$ | A1 (AO 1.1b) | Obtains $\int_0^{0.1} kt^2(0.1-t)^2\,dt = \frac{k}{3000000}$ |
| $I = \frac{k}{3000000}$ | R1 (AO 2.1) | Completes rigorous argument linking impulse to $\frac{k}{3000000}$; units not required |
---
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For maximum impulse the ball rebounds with same magnitude of velocity perpendicular to wall (i.e. $e \leq 1$ or $e = 1$) | E1 (AO 3.3) | Explains ball must rebound with same velocity component perpendicular to wall for maximum impulse |
| As ball rebounds, minimum impulse must be greater than impulse needed to bring perpendicular velocity to zero ($0 < e$ or $e = 0$) | E1 (AO 3.3) | Explains impulse must be greater than impulse needed to reduce perpendicular component to zero |
| $I_{max} = 2(0.3 \times 4\sin 30°) = 1.2$; $I_{min} > 0.3 \times 4\sin 30° = 0.6$ | B1 (AO 1.1b) | Calculates maximum and minimum magnitudes for impulse or velocity |
| $0.6 < \frac{k}{3000000} \leq 1.2$ | M1 (AO 2.2a) | Uses impulse to deduce inequality for $k$ |
| $1800000 < k \leq 3600000$ | R1 (AO 2.1) | Reaches required conclusion from complete argument |
---
## Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Parallel component $= 4\cos 30°$ | B1 (AO 3.4) | Obtains correct component of velocity parallel to wall |
| $I = 0.8$ | M1 (AO 3.3) | Applies impulse to motion perpendicular to wall |
| $0.3v - (-0.3 \times 4\sin 30°) = 0.8$, so $v = \frac{2}{3}$ | A1 (AO 1.1b) | Obtains correct rebound velocity component perpendicular to wall |
| Speed $= \sqrt{\left(\frac{2}{3}\right)^2 + (4\cos 30°)^2} = 3.53 \text{ m s}^{-1}$ | A1F (AO 1.1b) | Obtains correct speed using perpendicular and parallel components; condone missing units |
---5 A ball, of mass 0.3 kg , is moving on a smooth horizontal surface.
The ball collides with a smooth fixed vertical wall and rebounds.\\
Before the ball hits the wall, the ball is moving at $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ to the wall as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{b0d0c552-71cb-4e5a-b545-de8a9052def0-06_634_268_584_886}
The magnitude of the force, $F$ newtons, exerted on the ball by the wall at time $t$ seconds is modelled by
$$F = k t ^ { 2 } ( 0.1 - t ) ^ { 2 } \quad \text { for } \quad 0 \leq t \leq 0.1$$
where $k$ is a constant.
The ball is in contact with the wall for 0.1 seconds.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{b0d0c552-71cb-4e5a-b545-de8a9052def0-07_2484_1709_219_153}
\end{center}
5 (b) Explain why $1800000 < k \leq 3600000$
Fully justify your answer.\\
5 (c) Given that $k = 2400000$
Find the speed of the ball after the collision with the wall.\\[0pt]
[4 marks]\\
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2020 Q5 [17]}}