| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Conical pendulum (horizontal circle) |
| Difficulty | Moderate -0.8 This is a standard conical pendulum problem requiring resolution of forces (vertical equilibrium to find tension, horizontal component for centripetal force). The calculation is straightforward with given values, involving basic trigonometry and the centripetal force formula. The 'show that' part removes problem-solving challenge, making this easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03m Equilibrium: sum of resolved forces = 06.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\sin 30° = 0.4g\), \(T = 8N\) | M1, A1 | Resolves vertically [2] |
| (ii) \(Tcos 30° = 0.4v^2 / 0.2 (= 0.4\omega^2 \times 0.2)\), \(v = 1.86 \text{ ms}^{-1}\) | M1, A1ft | Newton's Second Law radially; It only on T from part (i) [2] |
**(i)** $\sin 30° = 0.4g$, $T = 8N$ | M1, A1 | Resolves vertically [2]
**(ii)** $Tcos 30° = 0.4v^2 / 0.2 (= 0.4\omega^2 \times 0.2)$, $v = 1.86 \text{ ms}^{-1}$ | M1, A1ft | Newton's Second Law radially; It only on T from part (i) [2]
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\includegraphics[max width=\textwidth, alt={}, center]{9d377c95-09b8-4893-b29f-8517a5016e8b-2_381_1079_255_534}
A particle $P$ of mass 0.4 kg is attached to a fixed point $A$ by a light inextensible string. The string is inclined at $60 ^ { \circ }$ to the vertical. $P$ moves with constant speed in a horizontal circle of radius 0.2 m . The centre of the circle is vertically below $A$ (see diagram).\\
(i) Show that the tension in the string is 8 N .\\
(ii) Calculate the speed of the particle.
\hfill \mbox{\textit{CAIE M2 2011 Q1 [4]}}