| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.3 This is a standard M3 circular motion question with two familiar scenarios: horizontal circle in a bowl and vertical circle motion. Parts (i)-(ii) involve routine resolution of forces and circular motion equations. Parts (iii)-(v) use energy conservation and the leaving condition (R=0), all standard textbook techniques requiring no novel insight. Slightly easier than average due to clear structure and well-practiced methods. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05a Angular velocity: definitions6.05c Horizontal circles: conical pendulum, banked tracks6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Resolving vertically: \(R\cos 60° = mg\) | M1 | |
| \(R = \frac{0.4 \times 9.8}{\cos 60°} = 7.84\ \text{N}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R\sin 60° = \frac{mv^2}{r}\), where \(r = 2.7\sin 60°\) | M1 A1 | |
| \(v^2 = \frac{R\sin 60° \cdot r}{m} = \frac{7.84\sin 60° \times 2.7\sin 60°}{0.4}\) | M1 | |
| \(v = \sqrt{2.7 \times 9.8\sin^2 60°/\cos 60°}\), \(v \approx 7.94\ \text{m s}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Energy conservation from bottom: \(\frac{1}{2}mv^2 = \frac{1}{2}m(9)^2 - mg\cdot 2.7(1-\cos\theta)\) | M1 A1 | |
| \(v^2 = 81 - 2\times 9.8 \times 2.7(1-\cos\theta)\) | A1 | |
| \(v^2 = 81 - 52.92 + 52.92\cos\theta\) | A1 | Accept equivalent forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Radial equation (inward positive): \(R + mg\cos\theta = \frac{mv^2}{2.7}\) | M1 A1 | |
| Substituting \(v^2\): \(R = \frac{0.4}{2.7}(81 - 52.92 + 52.92\cos\theta) - 0.4\times 9.8\cos\theta\) | M1 | |
| \(R = 4.16 - 11.76\cos\theta\) | A1 A1 | Full substitution and simplification shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = 0\): \(\cos\theta = \frac{4.16}{11.76}\) | M1 | |
| \(v^2 = 81 - 52.92 + 52.92 \times \frac{4.16}{11.76}\) | M1 | |
| \(v \approx 5.58\ \text{m s}^{-1}\) | A1 A1 |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolving vertically: $R\cos 60° = mg$ | M1 | |
| $R = \frac{0.4 \times 9.8}{\cos 60°} = 7.84\ \text{N}$ | A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R\sin 60° = \frac{mv^2}{r}$, where $r = 2.7\sin 60°$ | M1 A1 | |
| $v^2 = \frac{R\sin 60° \cdot r}{m} = \frac{7.84\sin 60° \times 2.7\sin 60°}{0.4}$ | M1 | |
| $v = \sqrt{2.7 \times 9.8\sin^2 60°/\cos 60°}$, $v \approx 7.94\ \text{m s}^{-1}$ | A1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Energy conservation from bottom: $\frac{1}{2}mv^2 = \frac{1}{2}m(9)^2 - mg\cdot 2.7(1-\cos\theta)$ | M1 A1 | |
| $v^2 = 81 - 2\times 9.8 \times 2.7(1-\cos\theta)$ | A1 | |
| $v^2 = 81 - 52.92 + 52.92\cos\theta$ | A1 | Accept equivalent forms |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Radial equation (inward positive): $R + mg\cos\theta = \frac{mv^2}{2.7}$ | M1 A1 | |
| Substituting $v^2$: $R = \frac{0.4}{2.7}(81 - 52.92 + 52.92\cos\theta) - 0.4\times 9.8\cos\theta$ | M1 | |
| $R = 4.16 - 11.76\cos\theta$ | A1 A1 | Full substitution and simplification shown |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = 0$: $\cos\theta = \frac{4.16}{11.76}$ | M1 | |
| $v^2 = 81 - 52.92 + 52.92 \times \frac{4.16}{11.76}$ | M1 | |
| $v \approx 5.58\ \text{m s}^{-1}$ | A1 A1 | |
---2 A fixed hollow sphere with centre O has an inside radius of 2.7 m . A particle P of mass 0.4 kg moves on the smooth inside surface of the sphere.
At first, P is moving in a horizontal circle with constant speed, and OP makes a constant angle of $60 ^ { \circ }$ with the vertical (see Fig. 2.1).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{39e14918-5017-43c0-9b74-7c68717ad5f3-3_655_666_488_696}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}
(i) Find the normal reaction acting on P .\\
(ii) Find the speed of P .
The particle P is now placed at the lowest point of the sphere and is given an initial horizontal speed of $9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It then moves in part of a vertical circle. When OP makes an angle $\theta$ with the upward vertical and P is still in contact with the sphere, the speed of P is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the normal reaction acting on P is $R \mathrm {~N}$ (see Fig. 2.2).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{39e14918-5017-43c0-9b74-7c68717ad5f3-3_716_778_1653_696}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{figure}
(iii) Find $v ^ { 2 }$ in terms of $\theta$.\\
(iv) Show that $R = 4.16 - 11.76 \cos \theta$.\\
(v) Find the speed of P at the instant when it leaves the surface of the sphere.
\hfill \mbox{\textit{OCR MEI M3 2007 Q2 [18]}}