| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – horizontal circle in free space (no surface) |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem requiring resolution of forces and application of circular motion formula (T cos θ = mg, T sin θ = mrω²). It's slightly easier than average as it's a textbook setup with straightforward two-part structure, though it does require coordinating multiple equations and understanding of uniform circular motion concepts. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(OG(\text{arc}) = 1.8\sin(\pi/2)/(\pi/2)\) | B1 | \(1.1459..\) or \(3.6/\pi\) |
| \(OX(1.8 \times 2 + \pi \times 1.8) = 1.1459 \times \pi \times 1.8\) | M1 | \(0.70017..\) |
| \(OX = 0.7(00) \text{ m}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(OY = 1.8\tan22\) | B1 | C of M solid \(= 0.727247..\text{m}\) from O |
| \(OG(\text{lamina}) = 2 \times 1.8\sin(\pi/2)/(3\pi/2)\) | B1 | C of M lamina \(= 0.763943..\) or \(2.4/\pi\) |
| \(1.8\tan22 \times (W + 27.5) =\) | M1 | \(27.5\left[4 \times 1.8/(3\pi) - 0.727247\right] =\) |
| \(0.7W + 0.763943 \times 27.5\) | A1 | \(W(0.727247 - 0.70017)\) |
| \(W = 37(.3)\text{N}\) | A1 | Accept to 2sf as sensitive to rounding error |
## Question 5:
### Part (i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $OG(\text{arc}) = 1.8\sin(\pi/2)/(\pi/2)$ | B1 | $1.1459..$ or $3.6/\pi$ |
| $OX(1.8 \times 2 + \pi \times 1.8) = 1.1459 \times \pi \times 1.8$ | M1 | $0.70017..$ |
| $OX = 0.7(00) \text{ m}$ | A1 | |
**Total: 3 marks**
### Part (ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $OY = 1.8\tan22$ | B1 | C of M solid $= 0.727247..\text{m}$ from O |
| $OG(\text{lamina}) = 2 \times 1.8\sin(\pi/2)/(3\pi/2)$ | B1 | C of M lamina $= 0.763943..$ or $2.4/\pi$ |
| $1.8\tan22 \times (W + 27.5) =$ | M1 | $27.5\left[4 \times 1.8/(3\pi) - 0.727247\right] =$ |
| $0.7W + 0.763943 \times 27.5$ | A1 | $W(0.727247 - 0.70017)$ |
| $W = 37(.3)\text{N}$ | A1 | Accept to 2sf as sensitive to rounding error |
**Total: 5 marks**
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{5998f4b1-21da-4c25-8b09-91a1cb1eee42-3_365_679_264_733}
A uniform metal frame $O A B C$ is made from a semicircular $\operatorname { arc } A B C$ of radius 1.8 m , and a straight $\operatorname { rod } A O C$ with $A O = O C = 1.8 \mathrm {~m}$ (see diagram).\\
(i) Calculate the distance of the centre of mass of the frame from $O$.
A uniform semicircular lamina of radius 1.8 m has weight 27.5 N . A non-uniform object is formed by attaching the frame $O A B C$ around the perimeter of the lamina. The object is freely suspended from a fixed point at $A$ and hangs in equilibrium. The diameter $A O C$ of the object makes an angle of $22 ^ { \circ }$ with the vertical.\\
(ii) Calculate the weight of the frame.
\hfill \mbox{\textit{CAIE M2 2014 Q5}}