| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.2 This is a standard M3 projectile-on-inclined-plane question requiring coordinate system rotation and component resolution. Part (a) involves routine application of kinematic equations in tilted coordinates, with part (a)(ii) being a straightforward verification. Part (b) adds coefficient of restitution but follows a predictable pattern. While requiring multiple steps and careful bookkeeping, the techniques are standard for M3 and don't require novel insight—slightly above average due to the algebraic manipulation and restitution component. |
| Spec | 3.02i Projectile motion: constant acceleration model6.03i Coefficient of restitution: e |
| Answer | Marks | Guidance |
|---|---|---|
| The projectile hits the plane again when \((Ut\sin\theta - \frac{1}{2}gt^2\cos\alpha) = 0\) | M1A1 | |
| \(\therefore t = \frac{2U\sin\theta}{g\cos\alpha}\) | A1F | Need to be simplified |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Component of velocity perpendicular to plane \(= U\sin\theta - g\frac{2U\sin\theta}{g\cos\alpha}\cos\alpha = -U\sin\theta\) | M1A1F | |
| The initial magnitude | A1 | AG |
| 3 |
| Answer | Marks |
|---|---|
| \(u = eU\sin\theta\) | M1 |
| \(a = -g\cos\alpha\) | m1 |
| \(s = 0\) | m1 |
| \(0 = eU\sin\theta \cdot T - \frac{1}{2}g\cos\alpha \cdot T^2\) | M1A1 |
| \(T = \frac{2eU\sin\theta}{g\cos\alpha} = et\) | A1F |
| \(t:T = 1:e\) | A1F |
| 4 | |
| 10 | |
| 75 | TOTAL |
### Part (a)(i)
The projectile hits the plane again when $(Ut\sin\theta - \frac{1}{2}gt^2\cos\alpha) = 0$ | M1A1 |
$\therefore t = \frac{2U\sin\theta}{g\cos\alpha}$ | A1F | Need to be simplified
| | **3** |
### Part (a)(ii)
Component of velocity perpendicular to plane $= U\sin\theta - g\frac{2U\sin\theta}{g\cos\alpha}\cos\alpha = -U\sin\theta$ | M1A1F |
The initial magnitude | A1 | AG
| | **3** |
### Part (b)
Newton's law of restitution perpendicular to plane:
$u = eU\sin\theta$ | M1 |
$a = -g\cos\alpha$ | m1 |
$s = 0$ | m1 |
$0 = eU\sin\theta \cdot T - \frac{1}{2}g\cos\alpha \cdot T^2$ | M1A1 |
$T = \frac{2eU\sin\theta}{g\cos\alpha} = et$ | A1F |
$t:T = 1:e$ | A1F |
| | **4** |
| | **10** |
| | **75** | **TOTAL** |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]{f8c04360-f54b-4d08-aee9-fe28612918d0-5_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 2006 Q7 [13]}}