M3 June 2009 Q5
5. One end of a light inextensible string of length \(l\) is attached to a fixed point \(A\). The other end is attached to a particle \(P\) of mass \(m\), which is held at a point \(B\) with the string taut and \(AP\) making an angle \(\arccos\dfrac{1}{4}\) with the downward vertical. The particle is released from rest. When \(AP\) makes an angle \(\theta\) with the downward vertical, the string is taut and the tension in the string is \(T\).
(a) Show that \[T = 3mg\cos\theta - \dfrac{mg}{2}.\] (6)
At an instant when \(AP\) makes an angle of 60\(^\circ\) to the downward vertical, \(P\) is moving upwards, as shown in Figure 3. At this instant the string breaks. At the highest point reached in the subsequent motion, \(P\) is at a distance \(d\) below the horizontal through \(A\).
(b) Find \(d\) in terms of \(l\). (5)
| Scheme | Marks |
|---|---|
| Energy: \(\left(\dfrac{1}{2}mu^2 +\right)mgl\left(\cos\theta - \dfrac{1}{4}\right) = \dfrac{1}{2}mv^2\) | M1A1 |
| Resolving: \(T - mg\cos\theta = \dfrac{mv^2}{l}\) | M1A1 |
| Eliminate \(v^2\): \(T = mg\cos\theta + \dfrac{1}{l}\left(2mgl\left(\cos\theta - \dfrac{1}{4}\right)\right)\) | M1 |
| \(T = 3mg\cos\theta - \dfrac{mg}{2}\) * | A1 |
| Scheme | Marks |
|---|---|
 \(\Rightarrow v^2 = \dfrac{gl}{2}\) | M1 |
| vertical motion under gravity: \(\updownarrow\) \(0 = (v\cos 30^\circ)^2 - 2gs\) | M1 |
| \(0 = \dfrac{gl}{2} \times \dfrac{3}{4} - 2gs \Rightarrow s = \dfrac{3l}{16}\) | A1 |
| Distance below A \(= \dfrac{l}{2} - \dfrac{3l}{16} = \dfrac{5l}{16}\) | M1A1 |
| (11 marks) |
Alternative for end of (b) using energy
 | M1A1 |
| \(\dfrac{gl}{4} - \dfrac{gl}{2} = \dfrac{gl}{4} \times \dfrac{1}{4} - gd\) | M1 |
| \(d = \dfrac{1 - 4 + 8}{16}l = \dfrac{5l}{16}\) | A1 |