M3 January 2012 Q6
6. A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(l\). The other end of the string is attached to a fixed point \(O\). The particle is hanging in equilibrium at the point \(A\), vertically below \(O\), when it is set in motion with a horizontal speed \(\tfrac{1}{2}\sqrt{(11gl)}\). When the string has turned through an angle \(\theta\) and the string is still taut, the tension in the string is \(T\).
(a) Show that \(T = 3mg\left(\cos\theta + \dfrac{1}{4}\right)\). (8)
At the instant when \(P\) reaches the point \(B\), the string becomes slack.
Find
(b) the speed of \(P\) at \(B\), (3)
(c) the maximum height above \(B\) reached by \(P\) before it starts to fall. (4)
| Scheme | Marks |
|---|---|
| GPE gained \(= mgl(1 - \cos\theta)\) Conservation of energy: \(\ \dfrac{1}{2}m\dfrac{11gl}{4} = mgl(1 - \cos\theta) + \dfrac{1}{2}mv^2\) | M1A1 A1 |
| \(v^2 = gl\left(\dfrac{11}{4} - 2 + 2\cos\theta\right) = gl\left(\dfrac{3}{4} + 2\cos\theta\right)\) | |
| Resolving towards the centre of the circle: | M1 |
| \(T - mg\cos\theta = \dfrac{mv^2}{l}\) | A1 A1 |
| \(T - mg\cos\theta = mg\left(\dfrac{3}{4} + 2\cos\theta\right)\) | M1 |
| \(T = mg\left(\dfrac{3}{4} + 3\cos\theta\right) = 3mg\left(\cos\theta + \dfrac{1}{4}\right)\) * | A1 |
| (8) |
| Scheme | Marks |
|---|---|
| \(T = 0 \Rightarrow \cos\theta = -\dfrac{1}{4}\) | M1 |
| \(v^2 = gl\left(\dfrac{3}{4} + 2\cos\theta\right) = \dfrac{gl}{4},\quad v = \sqrt{\dfrac{gl}{4}}\) | M1, A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Horizontal component of velocity at \(B\) \(= \sqrt{\dfrac{gl}{4}} \times \cos(180 - \theta) = \dfrac{1}{4}\sqrt{\dfrac{gl}{4}}\) | B1ft |
| Extra height \(h \Rightarrow mgh + \dfrac{1}{2}m\dfrac{gl}{64} = \dfrac{1}{2}m\dfrac{gl}{4}\) | M1 A1 |
| \(h = \left(\dfrac{1}{8} - \dfrac{1}{128}\right)l = \dfrac{15l}{128} \quad (0.117l)\) | A1 |
| (4) | |
| (15 marks) |
OR
Using \(h = \dfrac{v^2\sin^2\theta}{2g} = \dfrac{\frac{gl}{4} \times \frac{15}{16}}{2g} = \dfrac{15l}{128}\)
OR
Using \(v^2 = u^2 + 2as,\ \ 0 = \dfrac{15gl}{64} - 2gh,\ \ h = \dfrac{15l}{128}\)