| 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 | Moderate -0.3 This is a standard conical pendulum problem requiring resolution of forces (tension into horizontal and vertical components), application of Newton's second law for circular motion, and basic trigonometry. While it involves multiple parts and circular motion concepts, the solution follows a well-established method with no novel insight required, making it slightly easier than average for A-level mechanics. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(10\cos30 \times 1.2\sin\theta - 10\sin30 \times 1.2\cos\theta\) | M1 | Creating a 3 term solvable equation in \(\sin\theta\) and \(\cos\theta\) |
| \(= 6 \times 0.8\sin\theta\) | A1 | |
| \(5.5923..\sin\theta = 6\cos\theta\) | M1 | |
| \(\theta = 47(.0)\) | A1 | |
| OR | ||
| \(10 \times 1.2\sin(\theta - 30) = 6 \times 0.8\sin\theta\) | M1 | Creating a 3 term solvable equation in \(\sin\theta\) and \(\cos\theta\) |
| \(10 \times 1.2\cos(120 - \theta) = 6 \times 0.8\sin\theta\) | A1 | |
| \(5.5923..\sin\theta = 6\cos\theta\) | M1 | |
| \(\theta = 47(.0)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mu = (10\cos30 - 6)/(10\sin30)\) | M1 | For using \(F = \mu R\) with a reasonable attempt to find F and R |
| \(\mu = 0.532\) | A1 |
## Question 2:
### Part (i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $10\cos30 \times 1.2\sin\theta - 10\sin30 \times 1.2\cos\theta$ | M1 | Creating a 3 term solvable equation in $\sin\theta$ and $\cos\theta$ |
| $= 6 \times 0.8\sin\theta$ | A1 | |
| $5.5923..\sin\theta = 6\cos\theta$ | M1 | |
| $\theta = 47(.0)$ | A1 | |
| OR | | |
| $10 \times 1.2\sin(\theta - 30) = 6 \times 0.8\sin\theta$ | M1 | Creating a 3 term solvable equation in $\sin\theta$ and $\cos\theta$ |
| $10 \times 1.2\cos(120 - \theta) = 6 \times 0.8\sin\theta$ | A1 | |
| $5.5923..\sin\theta = 6\cos\theta$ | M1 | |
| $\theta = 47(.0)$ | A1 | |
**Total: 4 marks**
### Part (ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mu = (10\cos30 - 6)/(10\sin30)$ | M1 | For using $F = \mu R$ with a reasonable attempt to find F and R |
| $\mu = 0.532$ | A1 | |
**Total: 2 marks**
---2\\
\includegraphics[max width=\textwidth, alt={}, center]{5998f4b1-21da-4c25-8b09-91a1cb1eee42-2_565_549_438_797}
A non-uniform rod $A B$ of weight 6 N rests in limiting equilibrium with the end $A$ in contact with a rough vertical wall. $A B = 1.2 \mathrm {~m}$, the centre of mass of the rod is 0.8 m from $A$, and the angle between $A B$ and the downward vertical is $\theta ^ { \circ }$. A force of magnitude 10 N acting at an angle of $30 ^ { \circ }$ to the upwards vertical is applied to the rod at $B$ (see diagram). The rod and the line of action of the 10 N force lie in a vertical plane perpendicular to the wall. Calculate\\
(i) the value of $\theta$,\\
(ii) the coefficient of friction between the rod and the wall.
\hfill \mbox{\textit{CAIE M2 2014 Q2}}