| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Sphere rebounds off fixed wall obliquely |
| Difficulty | Standard +0.3 This is a standard Further Maths mechanics question on oblique collisions with a wall. Part (a) requires applying Newton's experimental law (e = separation speed / approach speed) in the perpendicular direction, involving straightforward trigonometry. Part (b) tests understanding that a smooth wall means no impulse parallel to the wall, requiring comparison of parallel velocity components. While it's Further Maths content, the problem-solving is methodical and follows standard procedures taught in the specification, making it slightly easier than average overall. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(7\sin 40° \times e = 5\sin 26°\) | M1 | Forms equation for coefficient of restitution. 7 must be seen with \(\sin/\cos 40°\) and 5 must be seen with \(\sin/\cos 26°\) |
| Obtains correct equation | A1 | |
| \(e = \frac{5\sin 26°}{7\sin 40°} = 0.49\) | A1 | Answer to 2 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(7\cos 40° = 5.36\), \(5\cos 26° = 4.49\) | M1 | Obtains correct components parallel to wall before and after collision |
| \(5.36 \neq 4.49\) | A1 | Deduces components not equal or reduced, by valid numerical comparison or stating \(7\cos 40° \neq 5\cos 26°\) |
| As components of velocity parallel to wall are not equal, the wall cannot be smooth | R1 | Completes rigorous mathematical argument. Numerical values must be stated |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $7\sin 40° \times e = 5\sin 26°$ | M1 | Forms equation for coefficient of restitution. 7 must be seen with $\sin/\cos 40°$ and 5 must be seen with $\sin/\cos 26°$ |
| Obtains correct equation | A1 | |
| $e = \frac{5\sin 26°}{7\sin 40°} = 0.49$ | A1 | Answer to 2 sf |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $7\cos 40° = 5.36$, $5\cos 26° = 4.49$ | M1 | Obtains correct components parallel to wall before and after collision |
| $5.36 \neq 4.49$ | A1 | Deduces components not equal or reduced, by valid numerical comparison or stating $7\cos 40° \neq 5\cos 26°$ |
| As components of velocity parallel to wall are not equal, the wall cannot be smooth | R1 | Completes rigorous mathematical argument. Numerical values must be stated |
---6 A ball moving on a smooth horizontal surface collides with a fixed vertical wall. Before the collision, the ball moves with speed $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and at an angle of $40 ^ { \circ }$ to the wall.
After the collision, the ball moves with speed $5 \mathrm {~ms} ^ { - 1 }$ and at an angle of $26 ^ { \circ }$ to the wall.
Model the ball as a particle.\\
6
\begin{enumerate}[label=(\alph*)]
\item Find the coefficient of restitution between the ball and the wall, giving your answer correct to two significant figures.\\
6
\item Determine whether or not the wall is smooth.
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2019 Q6 [6]}}