M3 June 2012 Q5
5. A fixed smooth sphere has centre \(O\) and radius \(a\). A particle \(P\) is placed on the surface of the sphere at the point \(A\), where \(OA\) makes an angle \(\alpha\) with the upward vertical through \(O\). The particle is released from rest at \(A\). When \(OP\) makes an angle \(\theta\) to the upward vertical through \(O\), \(P\) is on the surface of the sphere and the speed of \(P\) is \(v\).
Given that \(\cos\alpha = \dfrac{3}{5}\)
(a) show that \[v^2 = \frac{2ga}{5}(3 - 5\cos\theta)\] (4)
(b) find the speed of \(P\) at the instant when it loses contact with the sphere. (8)
| Scheme | Marks |
|---|---|
| Conservation of energy : Loss in GPE = gain in KE | M1 |
| \(mga\left(\cos\alpha - \cos\theta\right) = \dfrac{1}{2}mv^2\) | A2,1,0 |
| Substitute for \(\cos\alpha\) and rearrange to given answer: | A1 |
| \(v^2 = \dfrac{2mga}{m}\left(\dfrac{3}{5} - \cos\theta\right) = \dfrac{2ga}{5}\left(3 - 5\cos\theta\right)\) * | |
| (4) |
| Scheme | Marks |
|---|---|
| Considering the acceleration towards the centre of the hemisphere: | M1 |
| \(mg\cos\theta - R = \dfrac{mv^2}{a}\) | A2,1,0 |
| Substitute for \(v^2\) to form expression for \(R\): | DM1 |
| \(R = mg\cos\theta - \dfrac{mv^2}{a} = mg\left(3\cos\theta - 2\cos\alpha\right)\left(= mg\left(3\cos\theta - \dfrac{6}{5}\right)\right)\) | A1 |
| Loses contact with the surface when \(R = 0\) | M1 |
| \(\cos\theta = \dfrac{2}{5}\) | A1 |
| \(v^2 = \dfrac{2ga}{5},\ \ \ v = \sqrt{\dfrac{2ga}{5}}\) | A1 |
| (8) | |
| (12 marks) |
Alt:
| \(R = 0 \Rightarrow mg\cos\theta = \dfrac{mv^2}{a}\) | DM1 |
| \(\cos\theta = \dfrac{v^2}{ga}\) | A1 |
| Substitute in given (a) \(v^2 = \dfrac{2ga}{5}\left(3 - 5\dfrac{v^2}{ga}\right)\) | M1 |
| \(v^2 = \dfrac{6ga}{5} - 2v^2,\ \ \ 3v^2 = \dfrac{6ga}{5}\) | A1 |
| \(v = \sqrt{\dfrac{2ga}{5}}\) | A1 |