| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| 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 restitution. Part (a) is a 'show that' using standard projectile equations in rotated axes. Parts (b) and (c) involve resolving velocities and applying coefficient of restitution. While requiring multiple steps and careful angle work, it follows a well-established template for this topic with no novel insight needed. |
| Spec | 3.02i Projectile motion: constant acceleration model6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y = 10t\sin 40° - \dfrac{1}{2}gt^2\cos 30°\) | M1A1 | |
| \(y = 0 \Rightarrow t = \dfrac{20\sin 40°}{g\cos 30°}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\dot{x} = 10\cos 40° + g\sin 30°\left(\dfrac{20\sin 40°}{g\cos 30°}\right)\) | M1 | |
| \(\dot{x} = 15.08 \text{ ms}^{-1}\) | A1 | |
| \(\dot{y} = 10\sin 40° - g\cos 30°\left(\dfrac{20\sin 40°}{g\cos 30°}\right)\) | M1 | |
| \(\dot{y} = -6.427 \text{ ms}^{-1}\) | A1 | Allow 3 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\dot{x}\) will be unchanged | B1 | |
| Rebound \(\dot{y} = 6.427 \times 0.5 = 3.214\) | M1 | Allow using 3 sf |
| Rebound speed \(= \sqrt{15.08^2 + 3.214^2}\) | m1 | |
| \(= 15.4 \text{ ms}^{-1}\) | A1F |
## Question 7:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $y = 10t\sin 40° - \dfrac{1}{2}gt^2\cos 30°$ | M1A1 | |
| $y = 0 \Rightarrow t = \dfrac{20\sin 40°}{g\cos 30°}$ | A1 | AG |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\dot{x} = 10\cos 40° + g\sin 30°\left(\dfrac{20\sin 40°}{g\cos 30°}\right)$ | M1 | |
| $\dot{x} = 15.08 \text{ ms}^{-1}$ | A1 | |
| $\dot{y} = 10\sin 40° - g\cos 30°\left(\dfrac{20\sin 40°}{g\cos 30°}\right)$ | M1 | |
| $\dot{y} = -6.427 \text{ ms}^{-1}$ | A1 | Allow 3 sf |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\dot{x}$ will be unchanged | B1 | |
| Rebound $\dot{y} = 6.427 \times 0.5 = 3.214$ | M1 | Allow using 3 sf |
| Rebound speed $= \sqrt{15.08^2 + 3.214^2}$ | m1 | |
| $= 15.4 \text{ ms}^{-1}$ | A1F | |7 A particle is projected from a point $O$ on a smooth plane which is inclined at $30 ^ { \circ }$ to the horizontal. The particle is projected down the plane with velocity $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $40 ^ { \circ }$ above the plane and first strikes it at a point $A$. The motion of the particle is in a vertical plane containing a line of greatest slope of the inclined plane.\\
\includegraphics[max width=\textwidth, alt={}, center]{719b82f7-2ab5-48db-9b2a-98284096a78a-6_449_963_488_529}
\begin{enumerate}[label=(\alph*)]
\item Show that the time taken by the particle to travel from $O$ to $A$ is
$$\frac { 20 \sin 40 ^ { \circ } } { g \cos 30 ^ { \circ } }$$
\item Find the components of the velocity of the particle parallel to and perpendicular to the slope as it hits the slope at $A$.
\item The coefficient of restitution between the slope and the particle is 0.5 . Find the speed of the particle as it rebounds from the slope.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2009 Q7 [11]}}