| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | String through hole – hanging particle in equilibrium below table |
| Difficulty | Standard +0.3 This is a standard M2 circular motion problem with coupled particles through a hole. Part (i) involves straightforward application of T=mg for the hanging particle and T=mrω² for circular motion. Parts (ii-iii) reverse the setup but use identical principles with given values. The problem requires systematic application of Newton's laws but no novel insight or complex multi-step reasoning beyond typical M2 exercises. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(T = 4.9\) N | B1 | B0 for \(0.5g\) |
| \(T = 0.3 \times 0.2 \times \omega^2\) | M1 | or \(0.3v^2/0.2\) and \(\omega = v/0.2\) |
| \(\omega = 9.04\) rads\(^{-1}\) | A1, A1 | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\cos\theta = \sqrt{0.6}/0.8\ (0.968)\) | B1 | \((\theta=14.5°)\) angle to vertical or equiv.; angle consistent with diagram; can be their angle |
| \(T\cos\theta = 0.5 \times 9.8\) | M1, A1 | |
| \(T = 5.06\) N | A1 | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(T\sin\theta = 0.5 \times v^2/0.2\) | M1 | must be a component of \(T\) |
| \(v = 0.711\) ms\(^{-1}\) | A1, A1 | 3 marks total; \((\sin\theta = \frac{1}{4})\) can be their angle; 11 marks total |
# Question 6:
## Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $T = 4.9$ N | B1 | B0 for $0.5g$ |
| $T = 0.3 \times 0.2 \times \omega^2$ | M1 | or $0.3v^2/0.2$ and $\omega = v/0.2$ |
| $\omega = 9.04$ rads$^{-1}$ | A1, A1 | 4 marks total |
## Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $\cos\theta = \sqrt{0.6}/0.8\ (0.968)$ | B1 | $(\theta=14.5°)$ angle to vertical or equiv.; angle consistent with diagram; can be their angle |
| $T\cos\theta = 0.5 \times 9.8$ | M1, A1 | |
| $T = 5.06$ N | A1 | 4 marks total |
## Part (iii):
| Working | Mark | Guidance |
|---------|------|----------|
| $T\sin\theta = 0.5 \times v^2/0.2$ | M1 | must be a component of $T$ |
| $v = 0.711$ ms$^{-1}$ | A1, A1 | 3 marks total; $(\sin\theta = \frac{1}{4})$ can be their angle; 11 marks total |
---(i) Calculate the tension in the string and hence find the angular speed of $Q$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d6d87705-be4b-407d-b699-69fb441d88a7-4_489_1358_1286_392}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
The particle $Q$ on the plane is now fixed to a point 0.2 m from the hole at $A$ and the particle $P$ rotates in a horizontal circle of radius 0.2 m (see Fig. 2).\\
(ii) Calculate the tension in the string.\\
(iii) Calculate the speed of $P$.
\hfill \mbox{\textit{OCR M2 2006 Q6 [11]}}