A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\). The particle \(P\) is held at the point \(A\), where \(OA\) makes an angle \(\alpha\) with the downward vertical through \(O\), and with the string taut. The particle \(P\) is projected perpendicular to \(OA\) in an upwards direction with speed \(\sqrt{3ag}\). It then starts to move along a circular path in a vertical plane. The string goes slack when \(P\) is at \(B\), where \(OB\) makes an angle \(\theta\) with the upward vertical. Given that \(\cos \alpha = \frac{3}{5}\), find the value of \(\cos \theta\). [4]

Question 1:
1 | 1 1
m.3ag mv2 mgacoscos
2 2 | M1 | Energy equation, 4 terms, dimensionally correct,
mass must be present, allow sign errors, allow sin in
both terms on RHS
mv2
mgcos
a | B1 | N2L, may include tension initially but not awarded
until tension = 0 used
3 1 4 
mag m.agcosmga cos
 
2 2 5 
3 7
cos
2 10 | M1 | Dependent on tension = 0 and on an energy
equation, eliminate v2.
7
cos
15 | A1 | If no m in energy equation and no further errors,
award SCB2 for correct final answer
4
Question | Answer | Marks | Guidance

A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$. The particle $P$ is held at the point $A$, where $OA$ makes an angle $\alpha$ with the downward vertical through $O$, and with the string taut. The particle $P$ is projected perpendicular to $OA$ in an upwards direction with speed $\sqrt{3ag}$. It then starts to move along a circular path in a vertical plane. The string goes slack when $P$ is at $B$, where $OB$ makes an angle $\theta$ with the upward vertical.

Given that $\cos \alpha = \frac{3}{5}$, find the value of $\cos \theta$. [4]

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q1 [4]}}