| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 6 |
| 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 and application of circular motion formulas. While it involves multiple parts and coordinate geometry, the solution follows a routine procedure: resolve tension vertically (T cos θ = mg) and horizontally (T sin θ = ma), then use given acceleration to find unknowns. The steps are straightforward with no novel insight required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.15g = T\cos\theta\) | B1 | |
| \(T\sin\theta = 0.15 \times 7\) | M1 | For using Newton's second law horizontally |
| \(\theta = 35\) | A1 | [3 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Tension is \(1.83\) N | B1 ft | [1 mark] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(a = v^2 \div r\) and \(r = 2\sin\theta\) | |
| Speed is \(2.83\) ms\(^{-1}\) | A1 ft | ft \(v = \sqrt{14\sin\theta}\) [2 marks] |
## Question 2:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.15g = T\cos\theta$ | B1 | |
| $T\sin\theta = 0.15 \times 7$ | M1 | For using Newton's second law horizontally |
| $\theta = 35$ | A1 | **[3 marks]** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Tension is $1.83$ N | B1 ft | **[1 mark]** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $a = v^2 \div r$ and $r = 2\sin\theta$ |
| Speed is $2.83$ ms$^{-1}$ | A1 ft | ft $v = \sqrt{14\sin\theta}$ **[2 marks]** |
---2\\
\includegraphics[max width=\textwidth, alt={}, center]{6fe2c5e0-0496-4fb4-95d2-354b90607b5b-2_561_570_1274_790}
A particle of mass 0.15 kg is attached to one end of a light inextensible string of length 2 m . The other end of the string is attached to a fixed point. The particle moves with constant speed in a horizontal circle. The magnitude of the acceleration of the particle is $7 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The string makes an angle of $\theta ^ { \circ }$ with the downward vertical, as shown in the diagram. Find\\
(i) the value of $\theta$ to the nearest whole number,\\
(ii) the tension in the string,\\
(iii) the speed of the particle.
\hfill \mbox{\textit{CAIE M2 2005 Q2 [6]}}