| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on table with string above |
| Difficulty | Standard +0.3 This is a standard M2 circular motion problem requiring resolution of forces and application of F=mrω². Part (i) involves straightforward calculation using Pythagoras, resolving vertically and horizontally. Part (ii) requires recognizing that maximum speed occurs when normal reaction becomes zero. Slightly above average difficulty due to the 3D geometry setup and the conceptual understanding needed in part (ii), but follows a well-practiced template for this topic. |
| Spec | 3.03i Normal reaction force3.03j Smooth contact model: and limitations6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin\theta = \frac{1}{2}\) or \(\theta = 30\) | B1 | \(\theta\) is angle with horizontal. May have angle with vertical |
| \(T\cos\theta = 0.2 \times 1.2\cos\theta \times 2.5^2\) | M1 | Attempt at resolving horizontally |
| A1 | \(cv(r)\) but not \(r = 1.2\) | |
| \(T = 1.5\) N | A1 | Rounding to 1.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R + T\sin\theta = 0.2g\) | M1 | Attempt at resolving vertically |
| A1 FT | FT on \(cv(T)\) | |
| \(R = 1.21\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r = \sqrt{1.2^2 - 0.6^2} = 1.2\cos\theta\) | B1 | May have been seen in (i), must be used here |
| \(R = 0\) | B1 | May be implied |
| \(T_1\sin\theta = 0.2g\) | B1 | |
| \(T_1\cos\theta = 0.2 \times v^2/r\) or \(0.2 \times r\omega^2\) | M1 | Attempt at resolving |
| \(v = 4.2 \text{ ms}^{-1}\) | A1 |
## Question 4(i)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin\theta = \frac{1}{2}$ or $\theta = 30$ | B1 | $\theta$ is angle with horizontal. May have angle with vertical |
| $T\cos\theta = 0.2 \times 1.2\cos\theta \times 2.5^2$ | M1 | Attempt at resolving horizontally |
| | A1 | $cv(r)$ but not $r = 1.2$ |
| $T = 1.5$ N | A1 | Rounding to 1.5 |
---
## Question 4(i)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R + T\sin\theta = 0.2g$ | M1 | Attempt at resolving vertically |
| | A1 FT | FT on $cv(T)$ |
| $R = 1.21$ N | A1 | |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = \sqrt{1.2^2 - 0.6^2} = 1.2\cos\theta$ | B1 | May have been seen in (i), must be used here |
| $R = 0$ | B1 | May be implied |
| $T_1\sin\theta = 0.2g$ | B1 | |
| $T_1\cos\theta = 0.2 \times v^2/r$ or $0.2 \times r\omega^2$ | M1 | Attempt at resolving |
| $v = 4.2 \text{ ms}^{-1}$ | A1 | |
---4 A particle $P$ of mass 0.2 kg is attached to one end of a light inextensible string of length 1.2 m . The other end of the string is fixed at a point $A$ which is 0.6 m above a smooth horizontal table. $P$ moves on the table in a circular path whose centre $O$ is vertically below $A$.\\
(i) Given that the angular speed of $P$ is $2.5 \mathrm { rad } \mathrm { s } ^ { - 1 }$, find
\begin{enumerate}[label=(\alph*)]
\item the tension in the string,
\item the normal reaction between the particle and the table.\\
(ii) Find the greatest possible speed of $P$, given that the particle remains in contact with the table.
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2012 Q4 [12]}}