| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on outer surface of cylinder |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring energy conservation and Newton's second law in the radial direction. The steps are routine: apply conservation of energy to find v², use F=ma radially to find R, then set R=0 to find when the particle leaves the surface. While it involves multiple steps, the techniques are well-practiced at this level with no novel insight required, making it slightly easier than average. |
| Spec | 6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}mv^2 + mg(0.6)(1 - \cos\theta) = \frac{1}{2}m \cdot 4^2\) | M1 | For using the pce; condone sin/cos and sign errors; need KE before and after and difference in PE |
| \(v^2 = 4.24 + 11.76\cos\theta\) | A1 | AG |
| M1 | For using Newton's 2nd law; condone sin/cos and sign errors; 3 terms needed | |
| \(R - 0.45g\cos\theta = 0.45v^2/0.6\) | A1 | |
| \(R = 3.18 + 13.23\cos\theta\) | A1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos\theta = -3.18/13.23\) | M1, A1 FT | For using \(R = 0\); \(-0.24036...\) or \(-106/441\) or \(\theta = 103.9°\) ft from \(R = A + B\cos\theta\), where \(A, B \neq 0\) |
| \([v^2 = 4.24 - 11.76 \times 3.18/13.23]\) | M1 | For substituting for \(\cos\theta\) |
| Speed is 1.19 ms\(^{-1}\) | A1 | CAO without wrong working |
| [4] |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}mv^2 + mg(0.6)(1 - \cos\theta) = \frac{1}{2}m \cdot 4^2$ | M1 | For using the pce; condone sin/cos and sign errors; need KE before and after and difference in PE |
| $v^2 = 4.24 + 11.76\cos\theta$ | A1 | AG |
| | M1 | For using Newton's 2nd law; condone sin/cos and sign errors; 3 terms needed |
| $R - 0.45g\cos\theta = 0.45v^2/0.6$ | A1 | |
| $R = 3.18 + 13.23\cos\theta$ | A1 | |
| | **[6]** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\theta = -3.18/13.23$ | M1, A1 FT | For using $R = 0$; $-0.24036...$ or $-106/441$ or $\theta = 103.9°$ ft from $R = A + B\cos\theta$, where $A, B \neq 0$ |
| $[v^2 = 4.24 - 11.76 \times 3.18/13.23]$ | M1 | For substituting for $\cos\theta$ |
| Speed is 1.19 ms$^{-1}$ | A1 | CAO without wrong working |
| | **[4]** | |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{cc74a925-f1c8-4f59-a421-b46444cae5ec-4_524_611_255_703}
A hollow cylinder is fixed with its axis horizontal. The inner surface of the cylinder is smooth and has radius 0.6 m . A particle $P$ of mass 0.45 kg is projected horizontally with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from the lowest point of a vertical cross-section of the cylinder and moves in the plane of the cross-section, which is perpendicular to the axis of the cylinder. While $P$ remains in contact with the surface, its speed is $v \mathrm {~ms} ^ { - 1 }$ when $O P$ makes an angle $\theta$ with the downward vertical at $O$, where $O$ is the centre of the cross-section (see diagram). The force exerted on $P$ by the surface is $R \mathrm {~N}$.\\
(i) Show that $v ^ { 2 } = 4.24 + 11.76 \cos \theta$ and find an expression for $R$ in terms of $\theta$.\\
(ii) Find the speed of $P$ at the instant when it leaves the surface.
\hfill \mbox{\textit{OCR M3 2012 Q4 [10]}}