| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Particle bouncing on inclined plane |
| Difficulty | Standard +0.8 This M3 question requires resolving motion in a tilted coordinate system (inclined plane at 25°), finding time of flight using projectile equations, then applying coefficient of restitution to find rebound speed. It combines multiple mechanics topics (projectile motion, inclined planes, collisions) with careful vector resolution, making it moderately challenging but still within standard M3 scope. |
| 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Resolve perpendicular to plane: initial velocity = \(15\sin 30°\) | M1 | Setting up equation of motion perpendicular to plane |
| Acceleration perpendicular to plane = \(-g\cos 25°\) | ||
| \(s = ut + \frac{1}{2}at^2\) perpendicular to plane, \(s = 0\) at A | M1 | Using \(s=0\) for displacement perpendicular to plane |
| \(0 = 15\sin 30° \cdot t - \frac{1}{2}g\cos 25° \cdot t^2\) | A1 | Correct equation |
| \(t = \frac{2 \times 15\sin 30°}{g\cos 25°} = \frac{15}{9.8\cos 25°}\) | ||
| \(t \approx 1.69\) s | A1 | Accept \(1.68\) to \(1.69\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Velocity component along plane at A: \(v_{\parallel} = 15\cos 30° - g\sin 25° \cdot t\) | M1 | Using \(v = u + at\) along plane |
| \(v_{\parallel} = 15\cos 30° - 9.8\sin 25° \times 1.688...\) | A1 | Correct expression |
| \(v_{\parallel} \approx 12.990... - 6.994... \approx 5.996\) ms\(^{-1}\) (down the plane) | A1 | |
| Velocity component perpendicular to plane at A: \(v_{\perp} = 15\sin 30° - g\cos 25° \cdot t\) | M1 | |
| \(v_{\perp} = 7.5 - 9.8\cos 25° \times 1.688... \approx -7.5\) ms\(^{-1}\) (into plane) | A1 | |
| After rebound, perpendicular component = \(e \times v_{\perp} = \frac{2}{3} \times 7.5 = 5\) ms\(^{-1}\) | M1 | Applying Newton's law of restitution to perpendicular component |
| Along plane component unchanged = \(5.996\) ms\(^{-1}\) | A1 | |
| Speed \(= \sqrt{5.996^2 + 5^2}\) | M1 | Combining components |
| Speed \(\approx 7.81\) ms\(^{-1}\) | A1 |
# Question 5:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Resolve perpendicular to plane: initial velocity = $15\sin 30°$ | M1 | Setting up equation of motion perpendicular to plane |
| Acceleration perpendicular to plane = $-g\cos 25°$ | | |
| $s = ut + \frac{1}{2}at^2$ perpendicular to plane, $s = 0$ at A | M1 | Using $s=0$ for displacement perpendicular to plane |
| $0 = 15\sin 30° \cdot t - \frac{1}{2}g\cos 25° \cdot t^2$ | A1 | Correct equation |
| $t = \frac{2 \times 15\sin 30°}{g\cos 25°} = \frac{15}{9.8\cos 25°}$ | | |
| $t \approx 1.69$ s | A1 | Accept $1.68$ to $1.69$ |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Velocity component along plane at A: $v_{\parallel} = 15\cos 30° - g\sin 25° \cdot t$ | M1 | Using $v = u + at$ along plane |
| $v_{\parallel} = 15\cos 30° - 9.8\sin 25° \times 1.688...$ | A1 | Correct expression |
| $v_{\parallel} \approx 12.990... - 6.994... \approx 5.996$ ms$^{-1}$ (down the plane) | A1 | |
| Velocity component perpendicular to plane at A: $v_{\perp} = 15\sin 30° - g\cos 25° \cdot t$ | M1 | |
| $v_{\perp} = 7.5 - 9.8\cos 25° \times 1.688... \approx -7.5$ ms$^{-1}$ (into plane) | A1 | |
| After rebound, perpendicular component = $e \times v_{\perp} = \frac{2}{3} \times 7.5 = 5$ ms$^{-1}$ | M1 | Applying Newton's law of restitution to perpendicular component |
| Along plane component unchanged = $5.996$ ms$^{-1}$ | A1 | |
| Speed $= \sqrt{5.996^2 + 5^2}$ | M1 | Combining components |
| Speed $\approx 7.81$ ms$^{-1}$ | A1 | |
---5 A particle is projected from a point $O$ on a smooth plane, which is inclined at $25 ^ { \circ }$ to the horizontal. The particle is projected up the plane with velocity $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $30 ^ { \circ }$ above the plane. The particle strikes the plane for the first time 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]{a90a2de3-5cc0-4e87-b29a-2562f86eee17-12_518_839_552_630}
\begin{enumerate}[label=(\alph*)]
\item Find the time taken by the particle to travel from $O$ to $A$.
\item The coefficient of restitution between the particle and the inclined plane is $\frac { 2 } { 3 }$.
Find the speed of the particle as it rebounds from the inclined plane at $A$. (8 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2012 Q5 [12]}}