| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on outer surface of sphere |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring energy conservation and the condition for leaving the surface (N=0). Part (i) is a show-that requiring straightforward application of these principles, parts (ii) and (iii) follow directly from the result. Slightly easier than average due to the guided structure and standard techniques. |
| Spec | 6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of energy equation at \(A\) and \(B\) | M1 | 3 terms needed; allow sign error, missing \(m/g/r\) |
| \(mg0.6\cos\frac{\pi}{6} = mg0.6\cos\theta + \frac{1}{2}mv^2\) | A1 | Allow if \(\theta\) replaced by \(\varphi + \pi/6\) |
| \(F = ma\) radially | M1 | Allow sign error, missing \(m/g\) |
| \(mg\cos\theta - R = \frac{mv^2}{0.6}\) | A1 | |
| Use of \(R = 0\) | M1 | May be incorporated in previous step |
| \(\cos TOB = \frac{\sqrt{3}}{3}\) AG | A1 | Completely correct; not given if decimals used for angle |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of \(\frac{\sqrt{3}}{3}\) in 'correct' equation in (i) | M1 | \(mg0.6\cos\frac{\pi}{6} = mg0.6 \times \frac{\sqrt{3}}{3} + \frac{1}{2}mv^2\) or \(mg\frac{\sqrt{3}}{3} = \frac{mv^2}{0.6}\); equation must have gained M1 in (i) but allow restart here |
| \(1.84\ \text{ms}^{-1}\) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of \(F = ma\) tangentially | M1 | \(mg\sin\theta = ma\) seen; allow missing \(m/g\), \(-\) sign; allow M1 if angular accel found |
| \(8.00\ \text{ms}^{-2}\) | A1 | |
| [2] |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of energy equation at $A$ and $B$ | M1 | 3 terms needed; allow sign error, missing $m/g/r$ |
| $mg0.6\cos\frac{\pi}{6} = mg0.6\cos\theta + \frac{1}{2}mv^2$ | A1 | Allow if $\theta$ replaced by $\varphi + \pi/6$ |
| $F = ma$ radially | M1 | Allow sign error, missing $m/g$ |
| $mg\cos\theta - R = \frac{mv^2}{0.6}$ | A1 | |
| Use of $R = 0$ | M1 | May be incorporated in previous step |
| $\cos TOB = \frac{\sqrt{3}}{3}$ AG | A1 | Completely correct; not given if decimals used for angle |
| **[6]** | | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $\frac{\sqrt{3}}{3}$ in 'correct' equation in (i) | M1 | $mg0.6\cos\frac{\pi}{6} = mg0.6 \times \frac{\sqrt{3}}{3} + \frac{1}{2}mv^2$ or $mg\frac{\sqrt{3}}{3} = \frac{mv^2}{0.6}$; equation must have gained M1 in (i) but allow restart here |
| $1.84\ \text{ms}^{-1}$ | A1 | |
| **[2]** | | |
## Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $F = ma$ tangentially | M1 | $mg\sin\theta = ma$ seen; allow missing $m/g$, $-$ sign; allow M1 if angular accel found |
| $8.00\ \text{ms}^{-2}$ | A1 | |
| **[2]** | | |5\\
\includegraphics[max width=\textwidth, alt={}, center]{3e8248ca-74f1-443f-a5db-d7da532d2815-3_449_442_1281_794}
A fixed smooth sphere of radius 0.6 m has centre $O$ and highest point $T$. A particle of mass $m \mathrm {~kg}$ is released from rest at a point $A$ on the sphere, such that angle $T O A$ is $\frac { \pi } { 6 }$ radians. The particle leaves the surface of the sphere at $B$ (see diagram).\\
(i) Show that $\cos T O B = \frac { \sqrt { 3 } } { 3 }$.\\
(ii) Find the speed of the particle at $B$.\\
(iii) Find the transverse acceleration of the particle at $B$.
\hfill \mbox{\textit{OCR M3 2013 Q5 [10]}}