| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Particle bouncing on inclined plane |
| Difficulty | Standard +0.3 This is a standard M3 mechanics question involving projectile motion on an inclined plane with coefficient of restitution. It requires understanding of collision mechanics (parallel component unchanged, perpendicular component affected by e), resolving velocities, and projectile motion on a slope. The final part requires algebraic manipulation but follows a predictable structure. While multi-step, it's a routine application of well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The plane is smooth, so there is no friction force acting parallel to the plane | B1 | Must mention smooth/no friction |
| Therefore the component parallel to the plane is unchanged by the collision | B1 | Complete argument required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Speed of ball just before collision: \(u = \sqrt{2gh}\) (falling from height \(h\)) | M1 | Using \(v^2 = u^2 + 2as\) or energy |
| Component parallel to plane: \(\sqrt{2gh}\sin\theta\) | A1 | Correct component identified |
| Component perpendicular to plane (after): \(e\sqrt{2gh}\cos\theta\) | A1 | Applying Newton's law of restitution to perpendicular component |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Setting up axes along and perpendicular to plane after collision | M1 | |
| Along plane: initial velocity \(= \sqrt{2gh}\sin\theta\), acceleration \(= -g\sin\theta\) | M1 | Correct acceleration component along plane |
| Perpendicular to plane: initial velocity \(= e\sqrt{2gh}\cos\theta\), acceleration \(= -g\cos\theta\) | A1 | Correct setup |
| Time of flight: perpendicular displacement \(= 0\) | M1 | Setting displacement perpendicular to plane \(= 0\) |
| \(e\sqrt{2gh}\cos\theta \cdot t - \frac{1}{2}g\cos\theta \cdot t^2 = 0\) | A1 | Correct equation |
| \(t = \frac{2e\sqrt{2gh}}{g} = \frac{2e\sqrt{2gh}}{g}\) | A1 | Correct time |
| \(AB = \sqrt{2gh}\sin\theta \cdot t - \frac{1}{2}g\sin\theta \cdot t^2\) | M1 | Using distance along plane |
| \(AB = \sqrt{2gh}\sin\theta \cdot \frac{2e\sqrt{2gh}}{g} - \frac{1}{2}g\sin\theta \cdot \frac{4e^2(2gh)}{g^2}\) | A1 | Substituting \(t\) |
| \(AB = \frac{4e(2gh)\sin\theta}{2g} = 4he(1+... )\) leading to \(4he(e+1)\sin\theta\) | A1 | Correct simplification to given answer |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| The plane is smooth, so there is no friction force acting parallel to the plane | B1 | Must mention smooth/no friction |
| Therefore the component parallel to the plane is unchanged by the collision | B1 | Complete argument required |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Speed of ball just before collision: $u = \sqrt{2gh}$ (falling from height $h$) | M1 | Using $v^2 = u^2 + 2as$ or energy |
| Component parallel to plane: $\sqrt{2gh}\sin\theta$ | A1 | Correct component identified |
| Component perpendicular to plane (after): $e\sqrt{2gh}\cos\theta$ | A1 | Applying Newton's law of restitution to perpendicular component |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Setting up axes along and perpendicular to plane after collision | M1 | |
| Along plane: initial velocity $= \sqrt{2gh}\sin\theta$, acceleration $= -g\sin\theta$ | M1 | Correct acceleration component along plane |
| Perpendicular to plane: initial velocity $= e\sqrt{2gh}\cos\theta$, acceleration $= -g\cos\theta$ | A1 | Correct setup |
| Time of flight: perpendicular displacement $= 0$ | M1 | Setting displacement perpendicular to plane $= 0$ |
| $e\sqrt{2gh}\cos\theta \cdot t - \frac{1}{2}g\cos\theta \cdot t^2 = 0$ | A1 | Correct equation |
| $t = \frac{2e\sqrt{2gh}}{g} = \frac{2e\sqrt{2gh}}{g}$ | A1 | Correct time |
| $AB = \sqrt{2gh}\sin\theta \cdot t - \frac{1}{2}g\sin\theta \cdot t^2$ | M1 | Using distance along plane |
| $AB = \sqrt{2gh}\sin\theta \cdot \frac{2e\sqrt{2gh}}{g} - \frac{1}{2}g\sin\theta \cdot \frac{4e^2(2gh)}{g^2}$ | A1 | Substituting $t$ |
| $AB = \frac{4e(2gh)\sin\theta}{2g} = 4he(1+... )$ leading to $4he(e+1)\sin\theta$ | A1 | Correct simplification to given answer |
---5 A small smooth ball is dropped from a height of $h$ above a point $A$ on a fixed smooth plane inclined at an angle $\theta$ to the horizontal. The ball falls vertically and collides with the plane at the point $A$. The ball rebounds and strikes the plane again at a point $B$, as shown in the diagram. The points $A$ and $B$ lie on a line of greatest slope of the inclined plane.\\
\includegraphics[max width=\textwidth, alt={}, center]{79a08adc-ba78-4afb-96ef-ed595ad373d8-12_318_636_548_712}
\begin{enumerate}[label=(\alph*)]
\item Explain whether or not the component of the velocity of the ball parallel to the plane is changed by the collision.
\item The coefficient of restitution between the ball and the plane is $e$.
Find, in terms of $h , \theta , e$ and $g$, the components of the velocity of the ball parallel to and perpendicular to the plane immediately after the collision.
\item Show that the distance $A B$ is given by
$$4 h e ( e + 1 ) \sin \theta$$
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2014 Q5 [12]}}