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 \(A\) with the string taut. It is given that \(OA\) makes an angle \(\theta\) with the downward vertical through \(O\), where \(\tan \theta = \frac{3}{4}\). The particle \(P\) is projected perpendicular to \(OA\) in an upwards direction with speed \(\sqrt{5ag}\), and it starts to move along a circular path in a vertical plane. When \(P\) is at the point \(B\), where angle \(AOB\) is a right angle, the tension in the string is \(T\). Find \(T\) in terms of \(m\) and \(g\). [5]

Question 2:
2 | mv2
At B, T+mgsin=
a | B1
1 1
Energy A to B: mu2 − mv2 =mga(cos+sin)
2 2 | M1A1
Substitute for u and  to find T:
 4 3
T =mg5−2 −3 
 5 5 | M1
8
T = mg
5 | A1
5
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 $A$ with the string taut. It is given that $OA$ makes an angle $\theta$ with the downward vertical through $O$, where $\tan \theta = \frac{3}{4}$. The particle $P$ is projected perpendicular to $OA$ in an upwards direction with speed $\sqrt{5ag}$, and it starts to move along a circular path in a vertical plane. When $P$ is at the point $B$, where angle $AOB$ is a right angle, the tension in the string is $T$.

Find $T$ in terms of $m$ and $g$. [5]

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q2 [5]}}