| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.8 This is a multi-part circular motion problem requiring 3D geometry (finding radius from vertical distance), resolving forces in two directions (radial and vertical), and then extending to a coupled system with string tension. Part (iii) requires setting up simultaneous equations with the hanging mass equilibrium. More demanding than standard horizontal circle problems but uses established M2 techniques without requiring novel insight. |
| Spec | 3.03d Newton's second law: 2D vectors6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos\theta = 3/5\) or \(\sin\theta = 4/5\) or \(\tan\theta = 4/3\) or \(\theta = 53.1°\) | B1 | \(\theta =\) angle to vertical |
| \(R\cos\theta = 0.2\times9.8\) | M1 | |
| \(R = 3.27\ \text{N}\) or \(49/15\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = 4\) | B1 | |
| \(R\sin\theta = 0.2\times4\times\omega^2\) | M1 | |
| A1 | ||
| \(\omega = 1.81\ \text{rad s}^{-1}\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\varphi = 26.6°\) or \(\sin\varphi = \frac{1}{\sqrt{5}}\) or \(\cos\varphi = \frac{2}{\sqrt{5}}\) or \(\tan\varphi = 0.5\) | B1 | \(\varphi =\) angle to horizontal |
| \(T = 0.98\) or \(0.1g\) | B1 | |
| \(N\cos\theta = T\sin\varphi + 0.2\times9.8\) | M1 | Vertically, 3 terms |
| \(N\times3/5 = 0.438 + 1.96\) | A1 | |
| \(N = 4.00\) | A1 | may be implied |
| \(N\sin\theta + T\cos\varphi = 0.2\times4\times\omega^2\) | M1 | Horizontally, 3 terms |
| \(4\times4/5 + 0.98\cos26.6° = 0.8\omega^2\) | A1 | |
| \(\omega = 2.26\ \text{rad s}^{-1}\) | A1 | [8] [15] |
# Question 7(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\theta = 3/5$ or $\sin\theta = 4/5$ or $\tan\theta = 4/3$ or $\theta = 53.1°$ | B1 | $\theta =$ angle to vertical |
| $R\cos\theta = 0.2\times9.8$ | M1 | |
| $R = 3.27\ \text{N}$ or $49/15$ | A1 | **[3]** |
---
# Question 7(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = 4$ | B1 | |
| $R\sin\theta = 0.2\times4\times\omega^2$ | M1 | |
| | A1 | |
| $\omega = 1.81\ \text{rad s}^{-1}$ | A1 | **[4]** |
---
# Question 7(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\varphi = 26.6°$ or $\sin\varphi = \frac{1}{\sqrt{5}}$ or $\cos\varphi = \frac{2}{\sqrt{5}}$ or $\tan\varphi = 0.5$ | B1 | $\varphi =$ angle to horizontal |
| $T = 0.98$ or $0.1g$ | B1 | |
| $N\cos\theta = T\sin\varphi + 0.2\times9.8$ | M1 | Vertically, 3 terms |
| $N\times3/5 = 0.438 + 1.96$ | A1 | |
| $N = 4.00$ | A1 | may be implied |
| $N\sin\theta + T\cos\varphi = 0.2\times4\times\omega^2$ | M1 | Horizontally, 3 terms |
| $4\times4/5 + 0.98\cos26.6° = 0.8\omega^2$ | A1 | |
| $\omega = 2.26\ \text{rad s}^{-1}$ | A1 | **[8] [15]** |7
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8e1225a2-cb98-4b71-a4af-0150f093f852-4_444_771_258_687}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
A particle $P$ of mass 0.2 kg is moving on the smooth inner surface of a fixed hollow hemisphere which has centre $O$ and radius $5 \mathrm {~m} . P$ moves with constant angular speed $\omega$ in a horizontal circle at a vertical distance of 3 m below the level of $O$ (see Fig.1).\\
(i) Calculate the magnitude of the force exerted by the hemisphere on $P$.\\
(ii) Calculate $\omega$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8e1225a2-cb98-4b71-a4af-0150f093f852-4_592_773_1231_687}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A light inextensible string is now attached to $P$. The string passes through a small smooth hole at the lowest point of the hemisphere and a particle of mass 0.1 kg hangs in equilibrium at the end of the string. $P$ moves in the same horizontal circle as before (see Fig. 2).\\
(iii) Calculate the new angular speed of $P$.
\hfill \mbox{\textit{OCR M2 2010 Q7 [15]}}