One end of a light inextensible string of length \(a\) is attached to a fixed point \(O\). A particle of mass \(m\) is attached to the other end of the string. The particle is held at the point \(A\) with the string taut. The angle between \(OA\) and the downward vertical is equal to \(\alpha\), where \(\cos \alpha = \frac{4}{5}\). The particle is projected from \(A\), perpendicular to the string in an upwards direction, with a speed \(\sqrt{3ga}\). It then moves along a circular path in a vertical plane. The string first goes slack when it makes an angle \(\theta\) with the upward vertical through \(O\). Find the value of \(\cos \theta\). [5]

Question 2:
2 | 1 1
 mv2  m3gamgacoscos
2 2 | M1 A1 | Energy equation.
m
T mgcos v2
a | B1 | N2L
4
agcos3ga2gacos2ga
5 | M1 | Combine to find cos.
7
cos
15 | A1
5
Question | Answer | Marks | Guidance

One end of a light inextensible string of length $a$ is attached to a fixed point $O$. A particle of mass $m$ is attached to the other end of the string. The particle is held at the point $A$ with the string taut. The angle between $OA$ and the downward vertical is equal to $\alpha$, where $\cos \alpha = \frac{4}{5}$. The particle is projected from $A$, perpendicular to the string in an upwards direction, with a speed $\sqrt{3ga}$. It then moves along a circular path in a vertical plane. The string first goes slack when it makes an angle $\theta$ with the upward vertical through $O$.

Find the value of $\cos \theta$. [5]

\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q2 [5]}}