CAIE M2 2011 June — Question 3 8 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2011
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeParticle on cone surface – with string attached to vertex or fixed point
DifficultyStandard +0.3 This is a standard circular motion problem on a cone surface requiring resolution of forces and application of F=mrω². Part (i) involves setting up force equations with the constraint that tension equals normal reaction, leading to straightforward algebra. Part (ii) requires recognizing that maximum angular speed occurs when the normal reaction becomes zero. While it requires careful geometry and force resolution, these are well-practiced techniques in M2 with no novel insight needed.
Spec3.03d Newton's second law: 2D vectors6.05c Horizontal circles: conical pendulum, banked tracks

3 \includegraphics[max width=\textwidth, alt={}, center]{1d2e8f3a-dab6-4306-bc4a-d47805947cd2-3_385_1154_253_497} A particle \(P\) of mass 0.5 kg is attached to the vertex \(V\) of a fixed solid cone by a light inextensible string. \(P\) lies on the smooth curved surface of the cone and moves in a horizontal circle of radius 0.1 m with centre on the axis of the cone. The cone has semi-vertical angle \(60 ^ { \circ }\) (see diagram).
  1. Calculate the speed of \(P\), given that the tension in the string and the contact force between the cone and \(P\) have the same magnitude.
  2. Calculate the greatest angular speed at which \(P\) can move on the surface of the cone.

AnswerMarks Guidance
(i) \(R\cos30° + T\cos60° = 0.5g\); \(F = 0.5g/(\cos30° + \cos60°)\); \(T\sin60° - R\sin30° = 0.5v^2/0.1\); \(v = 0.518\) ms\(^{-1}\)M1, A1, M1, A1 [4] or with R = T = F; F = 3.660...; R = T; Newton's Second Law with radial acceleration
(ii) \(R = 0\); \(T\cos60° = 0.5g\); \(T\sin60° = 0.5 \times \omega^2 \times 0.1\); \(\omega = 13.2\) rads\(^{-1}\) OR \(R = 0\); \(mv^2\sin30°/r\) or \(m\omega^2\sin30° = mg\cos30°\); \(\omega = 13.2\) rad s\(^{-1}\)B1, M1, M1, A1 [4] Could be implied; T = 10 N; Newton's Second Law with radial acceleration; Could be implied
(i) $R\cos30° + T\cos60° = 0.5g$; $F = 0.5g/(\cos30° + \cos60°)$; $T\sin60° - R\sin30° = 0.5v^2/0.1$; $v = 0.518$ ms$^{-1}$ | M1, A1, M1, A1 [4] | or with R = T = F; F = 3.660...; R = T; Newton's Second Law with radial acceleration

(ii) $R = 0$; $T\cos60° = 0.5g$; $T\sin60° = 0.5 \times \omega^2 \times 0.1$; $\omega = 13.2$ rads$^{-1}$ OR $R = 0$; $mv^2\sin30°/r$ or $m\omega^2\sin30° = mg\cos30°$; $\omega = 13.2$ rad s$^{-1}$ | B1, M1, M1, A1 [4] | Could be implied; T = 10 N; Newton's Second Law with radial acceleration; Could be implied
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\includegraphics[max width=\textwidth, alt={}, center]{1d2e8f3a-dab6-4306-bc4a-d47805947cd2-3_385_1154_253_497}

A particle $P$ of mass 0.5 kg is attached to the vertex $V$ of a fixed solid cone by a light inextensible string. $P$ lies on the smooth curved surface of the cone and moves in a horizontal circle of radius 0.1 m with centre on the axis of the cone. The cone has semi-vertical angle $60 ^ { \circ }$ (see diagram).\\
(i) Calculate the speed of $P$, given that the tension in the string and the contact force between the cone and $P$ have the same magnitude.\\
(ii) Calculate the greatest angular speed at which $P$ can move on the surface of the cone.

\hfill \mbox{\textit{CAIE M2 2011 Q3 [8]}}