| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.8 This M3 projectile question involves inclined plane motion with restitution, requiring coordinate transformation, multi-step kinematics, and algebraic manipulation. Part (a)(ii) requires proving a symmetry property, and part (b) involves analyzing motion after an elastic collision with coefficient of restitution e. While the techniques are standard for M3, the extended reasoning across multiple bounces and the need to work in tilted coordinates makes this significantly harder than routine projectile questions. |
| Spec | 3.02i Projectile motion: constant acceleration model6.03i Coefficient of restitution: e |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Taking axes along and perpendicular to slope | M1 | |
| Perpendicular to slope: \(0 = u\sin(\alpha-\theta)t - \frac{1}{2}g\cos\theta \cdot t^2\) | M1 A1 | |
| \(t = \frac{2u\sin(\alpha-\theta)}{g\cos\theta}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Horizontal velocity = 0 at A: \(u\cos\alpha - g\sin\theta \cdot t = 0\) | M1 | |
| \(t = \frac{u\cos\alpha}{g\sin\theta}\) | A1 | |
| Equating: \(\frac{2u\sin(\alpha-\theta)}{g\cos\theta} = \frac{u\cos\alpha}{g\sin\theta}\) | M1 | |
| \(2\sin(\alpha-\theta)\sin\theta = \cos\alpha\cos\theta\) | A1 | |
| Expanding: \(2(\sin\alpha\cos\theta - \cos\alpha\sin\theta)\sin\theta = \cos\alpha\cos\theta\) | M1 | |
| \(2\sin\alpha\sin\theta\cos\theta = \cos\alpha(\cos^2\theta + 2\sin^2\theta)\)... leading to \(\tan\alpha = \frac{\cos^2\theta+2\sin^2\theta}{2\sin\theta\cos\theta}\)... \(k=2\) giving \(\tan\alpha = 2\tan\theta\) | A1 | \(k=2\) |
# Question 7:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Taking axes along and perpendicular to slope | M1 | |
| Perpendicular to slope: $0 = u\sin(\alpha-\theta)t - \frac{1}{2}g\cos\theta \cdot t^2$ | M1 A1 | |
| $t = \frac{2u\sin(\alpha-\theta)}{g\cos\theta}$ | A1 | |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Horizontal velocity = 0 at A: $u\cos\alpha - g\sin\theta \cdot t = 0$ | M1 | |
| $t = \frac{u\cos\alpha}{g\sin\theta}$ | A1 | |
| Equating: $\frac{2u\sin(\alpha-\theta)}{g\cos\theta} = \frac{u\cos\alpha}{g\sin\theta}$ | M1 | |
| $2\sin(\alpha-\theta)\sin\theta = \cos\alpha\cos\theta$ | A1 | |
| Expanding: $2(\sin\alpha\cos\theta - \cos\alpha\sin\theta)\sin\theta = \cos\alpha\cos\theta$ | M1 | |
| $2\sin\alpha\sin\theta\cos\theta = \cos\alpha(\cos^2\theta + 2\sin^2\theta)$... leading to $\tan\alpha = \frac{\cos^2\theta+2\sin^2\theta}{2\sin\theta\cos\theta}$... $k=2$ giving $\tan\alpha = 2\tan\theta$ | A1 | $k=2$ |7 A projectile is fired from a point $O$ on the slope of a hill which is inclined at an angle $\alpha$ to the horizontal. The projectile is fired up the hill with velocity $U$ at an angle $\theta$ above the hill and first strikes it at a point $A$. The projectile is modelled as a particle and the hill is modelled as a plane with $O A$ as a line of greatest slope.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find, in terms of $U , g , \alpha$ and $\theta$, the time taken by the projectile to travel from $O$ to $A$.
\item Hence, or otherwise, show that the magnitude of the component of the velocity of the projectile perpendicular to the hill, when it strikes the hill at the point $A$, is the same as it was initially at $O$.
\end{enumerate}\item The projectile rebounds and strikes the hill again at a point $B$. The hill is smooth and the coefficient of restitution between the projectile and the hill is $e$.\\
\includegraphics[max width=\textwidth, alt={}, center]{fc5bfc4b-68bb-4a23-874b-87e9558dc990-06_428_1332_1023_338}
Find the ratio of the time of flight from $O$ to $A$ to the time of flight from $A$ to $B$. Give your answer in its simplest form.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 Q7}}