2 \includegraphics[max width=\textwidth, alt={}, center]{7fcedc6d-8dc1-4159-8a72-be0f6a3f659b-2_698_737_484_703} A particle \(P\) travels on a smooth surface whose vertical cross-section is in the form of two arcs of circles. The first arc \(A B\) is a quarter of a circle of radius \(\frac { 1 } { 8 } a\) and centre \(O\). The second arc \(B C\) is a quarter of a circle of radius \(a\) and centre \(Q\). The two arcs are smoothly joined at \(B\). The point \(Q\) is vertically below \(O\) and the two arcs are in the same vertical plane. The particle \(P\) is projected vertically downwards from \(A\) with speed \(u\). When \(P\) is on the \(\operatorname { arc } B C\), angle \(B Q P\) is \(\theta\) (see diagram). Given that \(P\) loses contact with the surface when \(\cos \theta = \frac { 5 } { 6 }\), find \(u\) in terms of \(a\) and \(g\).

## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| State or imply reaction $R$ is zero when contact lost | M1 | |
| $mv^2/a = mg\cos\theta\ [-R]$ | M1 | Use $F = ma$ radially when contact lost |
| $v^2 = ag\cos\theta = 5ag/6$ | A1 | Use $\cos\theta = 5/6$ to find $v^2$ |
| $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mg\{a/8 + a(1-\cos\theta)\}$ | M1 A1 | Use conservation of energy at $\theta$; if found by $v^2 = u^2 + 2gh$ lose this A1 only |
| $= 7mag/24$ (A.E.F.) | A1 | |
| $u^2 = 5ag/6 - 2 \times 7ag/24$ | M1 | Combine to find $u$ |
| $u = \sqrt{\frac{1}{4}ag}$ or $\frac{1}{2}\sqrt{ag}$ | A1 | |

**Total: 8 marks**

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\includegraphics[max width=\textwidth, alt={}, center]{7fcedc6d-8dc1-4159-8a72-be0f6a3f659b-2_698_737_484_703}

A particle $P$ travels on a smooth surface whose vertical cross-section is in the form of two arcs of circles. The first arc $A B$ is a quarter of a circle of radius $\frac { 1 } { 8 } a$ and centre $O$. The second arc $B C$ is a quarter of a circle of radius $a$ and centre $Q$. The two arcs are smoothly joined at $B$. The point $Q$ is vertically below $O$ and the two arcs are in the same vertical plane. The particle $P$ is projected vertically downwards from $A$ with speed $u$. When $P$ is on the $\operatorname { arc } B C$, angle $B Q P$ is $\theta$ (see diagram). Given that $P$ loses contact with the surface when $\cos \theta = \frac { 5 } { 6 }$, find $u$ in terms of $a$ and $g$.

\hfill \mbox{\textit{CAIE FP2 2013 Q2 [8]}}