| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse on inclined plane |
| Difficulty | Challenging +1.2 This is a multi-part mechanics question requiring energy conservation, impulse-momentum theorem, and vector resolution on an inclined plane. While it involves several steps and careful geometric reasoning (particularly recognizing why θ=15°), the techniques are standard M3 material with clear structure. The geometric insight needed is moderate rather than novel, and the calculations follow established methods once the setup is understood. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Since plane is smooth, impulse is perpendicular to plane (so \(\theta = 15\)) | B1 | |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of \(v^2 = (u^2) + 2 \times g \times 2.5\) | M1 | |
| \(v = 7 \text{ ms}^{-1}\) | A1 | |
| Speed parallel to plane is \(7\sin 15°\) | B1 | 1.81(173...) |
| \(u = 7\sin 15° / \cos 60°\) | M1 | Allow sin/cos errors |
| \(u = 3.62\) | A1 | |
| \(I = 0.45(7\cos 15° + u\sin 60°)\) | M1 | Allow sin/cos errors or \(I = 0.45(7\cos 15° + 7\sin 15°\tan 60°)\) |
| \(I = 4.45\) | A1 | 4.45477... May see \(e = 0.464\) |
| [7] | ||
| *Or* using triangle with sides 3.15, 0.45 \(\times\) 7, \(I\) and 0.45 \(\times u\), correct angles 135°, 15°, 30° | M1 | Need 2 correct sides and 1 correct angle; All correct |
| Use of sin or cos rule (correct) | M1 | \(I\cos 15° = 3.15 + 0.45u\cos 45°\) M1; \(I\sin 15° = mu\cos 45°\) B1 |
| \(u = 3.62\) | A1 | Solve simultaneous equations — M1, dep attempt at two components of \(I\) |
| \(I = 4.45\) | A1 | Answers — A1A1 |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Since plane is smooth, impulse is perpendicular to plane (so $\theta = 15$) | B1 | |
| | **[1]** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $v^2 = (u^2) + 2 \times g \times 2.5$ | M1 | |
| $v = 7 \text{ ms}^{-1}$ | A1 | |
| Speed parallel to plane is $7\sin 15°$ | B1 | 1.81(173...) |
| $u = 7\sin 15° / \cos 60°$ | M1 | Allow sin/cos errors |
| $u = 3.62$ | A1 | |
| $I = 0.45(7\cos 15° + u\sin 60°)$ | M1 | Allow sin/cos errors or $I = 0.45(7\cos 15° + 7\sin 15°\tan 60°)$ |
| $I = 4.45$ | A1 | 4.45477... May see $e = 0.464$ |
| | **[7]** | |
| *Or* using triangle with sides 3.15, 0.45 $\times$ 7, $I$ and 0.45 $\times u$, correct angles 135°, 15°, 30° | M1 | Need 2 correct sides and 1 correct angle; All correct |
| Use of sin or cos rule (correct) | M1 | $I\cos 15° = 3.15 + 0.45u\cos 45°$ M1; $I\sin 15° = mu\cos 45°$ B1 |
| $u = 3.62$ | A1 | Solve simultaneous equations — M1, dep attempt at two components of $I$ |
| $I = 4.45$ | A1 | Answers — A1A1 |
---2\\
\includegraphics[max width=\textwidth, alt={}, center]{cc74a925-f1c8-4f59-a421-b46444cae5ec-3_442_636_255_715}\\
$B$ is a point on a smooth plane surface inclined at an angle of $15 ^ { \circ }$ to the horizontal. A particle $P$ of mass 0.45 kg is released from rest at the point $A$ which is 2.5 m vertically above $B$. The particle $P$ rebounds from the surface at an angle of $60 ^ { \circ }$ to the line of greatest slope through $B$, with a speed of $u \mathrm {~ms} ^ { - 1 }$. The impulse exerted on $P$ by the surface has magnitude $I$ Ns and is in a direction making an angle of $\theta ^ { \circ }$ with the upward vertical through $B$ (see diagram).\\
(i) Explain why $\theta = 15$.\\
(ii) Find the values of $u$ and $I$.
\hfill \mbox{\textit{OCR M3 2012 Q2 [8]}}