| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Projectile with plane collision |
| Difficulty | Standard +0.8 This M4 question requires resolving velocities in directions perpendicular and parallel to an inclined plane, applying Newton's experimental law with coefficient of restitution, and using the constraint that the final velocity is horizontal. It demands careful vector decomposition, simultaneous equation solving, and trigonometric manipulation—significantly more sophisticated than routine mechanics problems but standard for M4 level. |
| Spec | 6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Component of velocity along plane (before): \(u\sin\alpha\) | B1 | |
| Component of velocity perpendicular to plane (before): \(u\cos\alpha\) | B1 | |
| Tangential component unchanged: \(u\sin\alpha\) | M1 | No friction so tangential component preserved |
| Normal component after: \(e \cdot u\cos\alpha = \frac{1}{3}u\cos\alpha\) | M1 | Applying Newton's law of restitution |
| After impact motion is horizontal; resultant velocity makes angle \(\alpha\) below horizontal with the plane | M1 | Using condition that motion after is horizontal |
| \(\tan\alpha = \frac{\frac{1}{3}u\cos\alpha}{u\sin\alpha}\) | M1 | Forming equation using horizontal condition |
| \(\tan^2\alpha = \frac{1}{3}\) | A1 | |
| \(\alpha = \arctan\left(\frac{1}{\sqrt{3}}\right) = 30°\) | A1 |
# Question 2:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Component of velocity along plane (before): $u\sin\alpha$ | B1 | |
| Component of velocity perpendicular to plane (before): $u\cos\alpha$ | B1 | |
| Tangential component unchanged: $u\sin\alpha$ | M1 | No friction so tangential component preserved |
| Normal component after: $e \cdot u\cos\alpha = \frac{1}{3}u\cos\alpha$ | M1 | Applying Newton's law of restitution |
| After impact motion is horizontal; resultant velocity makes angle $\alpha$ below horizontal with the plane | M1 | Using condition that motion after is horizontal |
| $\tan\alpha = \frac{\frac{1}{3}u\cos\alpha}{u\sin\alpha}$ | M1 | Forming equation using horizontal condition |
| $\tan^2\alpha = \frac{1}{3}$ | A1 | |
| $\alpha = \arctan\left(\frac{1}{\sqrt{3}}\right) = 30°$ | A1 | |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2a3ae838-b58e-4957-8d98-f7d8a65df99a-03_604_741_123_605}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A smooth fixed plane is inclined at an angle $\alpha$ to the horizontal. A smooth ball $B$ falls vertically and hits the plane. Immediately before the impact the speed of $B$ is $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$, as shown in Figure 1. Immediately after the impact the direction of motion of $B$ is horizontal. The coefficient of restitution between $B$ and the plane is $\frac { 1 } { 3 }$.
Find the size of angle $\alpha$.
\hfill \mbox{\textit{Edexcel M4 2013 Q2 [6]}}