2 \includegraphics[max width=\textwidth, alt={}, center]{f6887893-66c5-40df-ba8d-9439a5c268eb-2_582_798_616_671} A particle of mass \(m\) is attached to the end \(B\) of a light inextensible string. The other end of the string is attached to a fixed point \(A\) which is at a distance \(a\) above the vertex \(V\) of a circular cone of semi-vertical angle \(60 ^ { \circ }\). The axis of the cone is vertical. The particle moves with constant speed \(u\) in a horizontal circle on the smooth surface of the cone. The string makes a constant angle of \(30 ^ { \circ }\) with the vertical (see diagram). The tension in the string and the magnitude of the normal force acting on the particle are denoted by \(T\) and \(R\) respectively. Show that $$T = \frac { m } { \sqrt { } 3 } \left( g + \frac { 2 u ^ { 2 } } { a } \right) ,$$ and find a similar expression for \(R\). Deduce that \(u ^ { 2 } \leqslant \frac { 1 } { 2 } g a\).

## Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| *EITHER:* Resolve vertically: $T\cos 30° + R\cos 30° = mg$ | M1 A1 | |
| $[T + R = 2mg/\sqrt{3}]$ | | |
| Resolve horizontally: $T\sin 30° - R\sin 30° = mu^2/a \sin 60°$ | M1 A1 | |
| $[T - R = 4mu^2/a\sqrt{3}]$ | | |
| *OR:* Any 2 other independent resolutions, e.g.: | | |
| Along $BV$ (gives $T$): $T\cos 30° - mg\cos 60° = (mu^2/a\sin 60°)\sin 60°$ | | |
| Normal to $BV$: $R + T\cos 60° - mg\cos 30° = -(mu^2/a\sin 60°)\cos 60°$ | | |
| Along $BA$: $T + R\cos 60° - mg\cos 30° = (mu^2/a\sin 60°)\cos 60°$ | | |
| Normal to $BA$: $R\cos 30° - mg\cos 60° = -(mu^2/a\sin 60°)\sin 60°$ | | |
| Solve for $T$: $T = \frac{m}{\sqrt{3}}(g + 2u^2/a)$ **A.G.** | A1 | |
| Solve for $R$: $R = \frac{m}{\sqrt{3}}(g - 2u^2/a)$ | A1* | |
| Valid reason for deducing $u^2 \leq \frac{1}{2}ga$ (dep A1*): $R \geq 0$ (A.E.F.) | B1 | **Total: 7** |

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\includegraphics[max width=\textwidth, alt={}, center]{f6887893-66c5-40df-ba8d-9439a5c268eb-2_582_798_616_671}

A particle of mass $m$ is attached to the end $B$ of a light inextensible string. The other end of the string is attached to a fixed point $A$ which is at a distance $a$ above the vertex $V$ of a circular cone of semi-vertical angle $60 ^ { \circ }$. The axis of the cone is vertical. The particle moves with constant speed $u$ in a horizontal circle on the smooth surface of the cone. The string makes a constant angle of $30 ^ { \circ }$ with the vertical (see diagram). The tension in the string and the magnitude of the normal force acting on the particle are denoted by $T$ and $R$ respectively. Show that

$$T = \frac { m } { \sqrt { } 3 } \left( g + \frac { 2 u ^ { 2 } } { a } \right) ,$$

and find a similar expression for $R$.

Deduce that $u ^ { 2 } \leqslant \frac { 1 } { 2 } g a$.

\hfill \mbox{\textit{CAIE FP2 2010 Q2 [7]}}