| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Conical pendulum (horizontal circle) |
| Difficulty | Standard +0.3 Standard conical pendulum problem requiring resolution of forces and circular motion equations. Part (i) uses T cos 60° = mg and T sin 60° = mv²/r with given angle. Part (ii) reverses this to find angle from given angular speed. Straightforward application of standard mechanics formulas with no novel insight required, slightly above average due to two-part structure and algebraic manipulation. |
| 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 |
| \(T\cos 60 = 0.2g\) | M1 | Resolve vertically for P |
| \(T = 4\text{N}\) | A1 | |
| \(T\sin 60 = 0.2v^2/(2.4\sin 60)\) | M1 | Newton's Second Law with \(a = v^2/r\) |
| \(v = 6 \text{ ms}^{-1}\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(T\sin\theta = 0.2 \times 4^2 \times 1.2\sin\theta\) | M1 | Newton's Second Law with \(a = \omega^2 r\) |
| \(T = 3.84\) | A1 | |
| \(3.84\cos\theta = 0.2g\) | M1 | Resolve vertically with new tension |
| \(\theta = 58.6°\) | A1 | [4] |
## Question 4:
### Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $T\cos 60 = 0.2g$ | M1 | Resolve vertically for P |
| $T = 4\text{N}$ | A1 | |
| $T\sin 60 = 0.2v^2/(2.4\sin 60)$ | M1 | Newton's Second Law with $a = v^2/r$ |
| $v = 6 \text{ ms}^{-1}$ | A1 | [4] |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $T\sin\theta = 0.2 \times 4^2 \times 1.2\sin\theta$ | M1 | Newton's Second Law with $a = \omega^2 r$ |
| $T = 3.84$ | A1 | |
| $3.84\cos\theta = 0.2g$ | M1 | Resolve vertically with new tension |
| $\theta = 58.6°$ | A1 | [4] |
---4 One end of a light inextensible string of length 2.4 m is attached to a fixed point $A$. The other end of the string is attached to a particle $P$ of mass $0.2 \mathrm {~kg} . P$ moves with constant speed in a horizontal circle which has its centre vertically below $A$, with the string taut and making an angle of $60 ^ { \circ }$ with the vertical.\\
(i) Find the speed of $P$.
The string of length 2.4 m is removed, and $P$ is now connected to $A$ by a light inextensible string of length 1.2 m . The particle $P$ moves with angular speed $4 \mathrm { rad } \mathrm { s } ^ { - 1 }$ in a horizontal circle with its centre vertically below $A$.\\
(ii) Calculate the angle between the string and the vertical.
\hfill \mbox{\textit{CAIE M2 2014 Q4 [8]}}