| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on cone surface – with string attached to vertex or fixed point |
| Difficulty | Standard +0.3 This is a standard circular motion problem on a cone surface requiring resolution of forces in two directions (horizontal and vertical), application of F=mrω², and use of the cone geometry. While it involves multiple steps and careful force resolution, it follows a well-established method taught in M2 with no novel insight required—slightly easier than average due to the convenient 45° angle simplifying trigonometry. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Use \(a = \omega^2 r\) \([16 \times 1.2 \sin 45°]\) | M1 | |
| Obtains acceleration as 13.6m s\(^{-2}\) (13.57645) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Uses Newton's 2nd Law either horizontally or perpendicular to OP to obtain a 3 term equation \(T \sin 45° + N \cos 45° = 0.3 \times 13.576\) or \(N - 0.3g \sin 45° = 0.3 \times 13.576 \cos 45°\) | M1 | |
| A1 ft | ||
| Resolves forces vertically or uses Newton's 2nd Law along OP to obtain a 3 term equation \(N \sin 45° - T \cos 45° = 0.3g\) or \(T + 0.3g \cos 45° = 0.3 \times 13.576 \sin 45°\) | M1 | |
| A1 ft | ||
| Obtains tension as 0.759 N (0.75868) | A1 | |
| Obtains force exerted by the cone as 5.00 N (5.00132) | A1 | 6 marks |
**(i)**
| Use $a = \omega^2 r$ $[16 \times 1.2 \sin 45°]$ | M1 |
| Obtains acceleration as 13.6m s$^{-2}$ (13.57645) | A1 | 2 marks |
**(ii)**
| Uses Newton's 2nd Law either horizontally or perpendicular to OP to obtain a 3 term equation $T \sin 45° + N \cos 45° = 0.3 \times 13.576$ or $N - 0.3g \sin 45° = 0.3 \times 13.576 \cos 45°$ | M1 |
| | A1 ft |
| Resolves forces vertically or uses Newton's 2nd Law along OP to obtain a 3 term equation $N \sin 45° - T \cos 45° = 0.3g$ or $T + 0.3g \cos 45° = 0.3 \times 13.576 \sin 45°$ | M1 |
| | A1 ft |
| Obtains tension as 0.759 N (0.75868) | A1 |
| Obtains force exerted by the cone as 5.00 N (5.00132) | A1 | 6 marks |
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\includegraphics[max width=\textwidth, alt={}, center]{3e7472a8-df1e-45c4-81fb-e4397bddf5ad-3_576_826_258_662}
A hollow cone with semi-vertical angle $45 ^ { \circ }$ is fixed with its axis vertical and its vertex $O$ downwards. A particle $P$ of mass 0.3 kg moves in a horizontal circle on the inner surface of the cone, which is smooth. $P$ is attached to one end of a light inextensible string of length 1.2 m . The other end of the string is attached to the cone at $O$ (see diagram). The string is taut and rotates at a constant angular speed of $4 \mathrm { rad } \mathrm { s } ^ { - 1 }$.\\
(i) Find the acceleration of $P$.\\
(ii) Find the tension in the string and the force exerted on $P$ by the cone.
\hfill \mbox{\textit{CAIE M2 2002 Q4 [8]}}