1 A particle \(P\) is projected with speed \(u\) at an angle of \(30 ^ { \circ }\) above the horizontal from a point \(O\) on a horizontal plane and moves freely under gravity. The particle reaches its greatest height at time \(T\) after projection. Find, in terms of \(u\), the speed of \(P\) at time \(\frac { 2 } { 3 } T\) after projection. \includegraphics[max width=\textwidth, alt={}, center]{7251b13f-1fae-4138-80ea-e6b8091c94ab-04_362_750_258_653} A light inextensible string of length \(a\) is threaded through a fixed smooth ring \(R\). One end of the string is attached to a particle \(A\) of mass \(3 m\). The other end of the string is attached to a particle \(B\) of mass \(m\). The particle \(A\) hangs in equilibrium at a distance \(x\) vertically below the ring. The angle between \(A R\) and \(B R\) is \(\theta\) (see diagram). The particle \(B\) moves in a horizontal circle with constant angular speed \(2 \sqrt { \frac { \mathrm {~g} } { \mathrm { a } } }\). Show that \(\cos \theta = \frac { 1 } { 3 }\) and find \(x\) in terms of \(a\).

**Question 1:**

| Answer | Mark | Guidance |
|--------|------|----------|
| For greatest height, $T = \dfrac{u}{2g}$ | B1 | |
| At $t = \dfrac{2T}{3}$, $\uparrow v_v = \dfrac{u}{2} - \dfrac{2Tg}{3} = \dfrac{u}{6}$ | M1 | |
| $\rightarrow v_h = \dfrac{u\sqrt{3}}{2}$ | A1 | |
| Speed $= \sqrt{v_v^2 + v_h^2} = \sqrt{\dfrac{u^2}{36} + \dfrac{3u^2}{4}}$ | M1 | |
| $= \dfrac{\sqrt{7}}{3}u$ | A1 | |
| **Total** | **5** | |

1 A particle $P$ is projected with speed $u$ at an angle of $30 ^ { \circ }$ above the horizontal from a point $O$ on a horizontal plane and moves freely under gravity. The particle reaches its greatest height at time $T$ after projection.

Find, in terms of $u$, the speed of $P$ at time $\frac { 2 } { 3 } T$ after projection.\\

\includegraphics[max width=\textwidth, alt={}, center]{7251b13f-1fae-4138-80ea-e6b8091c94ab-04_362_750_258_653}

A light inextensible string of length $a$ is threaded through a fixed smooth ring $R$. One end of the string is attached to a particle $A$ of mass $3 m$. The other end of the string is attached to a particle $B$ of mass $m$. The particle $A$ hangs in equilibrium at a distance $x$ vertically below the ring. The angle between $A R$ and $B R$ is $\theta$ (see diagram). The particle $B$ moves in a horizontal circle with constant angular speed $2 \sqrt { \frac { \mathrm {~g} } { \mathrm { a } } }$.

Show that $\cos \theta = \frac { 1 } { 3 }$ and find $x$ in terms of $a$.\\

\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q1 [5]}}