Challenging +1.2 This is a standard vertical circular motion problem requiring conservation of energy and the condition for string becoming slack (tension = 0). While it involves multiple steps (energy equation, tension equation, solving for angle), the approach is well-established and commonly practiced in Further Maths mechanics. The given initial angle and speed lead to straightforward algebra, making this a typical textbook exercise rather than requiring novel insight.
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 string is held taut with \(OP\) making an angle \(\alpha\) with the downward vertical, where \(\cos \alpha = \frac{2}{3}\). The particle \(P\) is projected perpendicular to \(OP\) in an upwards direction with speed \(\sqrt{3ag}\). It then starts to move along a circular path in a vertical plane.
Find the cosine of the angle between the string and the upward vertical when the string first becomes slack.
[4]
Question 1:
1 | m
When string goes slack, mgcos= v2, v2 =agcos
a | B1 | N2L
May include T, but B1 not awarded until T = 0.
1 1
m.3ag− mv2 =mg(acos+acos)
2 2 | B1 | Energy equation.
2
So u2 −agcos=2agcos+
3 | M1 | Combine.
4
u2 − ag
3 5
cos= =
3ag 9 | A1
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 string is held taut with $OP$ making an angle $\alpha$ with the downward vertical, where $\cos \alpha = \frac{2}{3}$. The particle $P$ is projected perpendicular to $OP$ in an upwards direction with speed $\sqrt{3ag}$. It then starts to move along a circular path in a vertical plane.
Find the cosine of the angle between the string and the upward vertical when the string first becomes slack.
[4]
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q1 [4]}}