M3 June 2007 Q6
6. A particle \(P\) is free to move on the smooth inner surface of a fixed thin hollow sphere of internal radius \(a\) and centre \(O\). The particle passes through the lowest point of the spherical surface with speed \(U\). The particle loses contact with the surface when \(OP\) is inclined at an angle \(\alpha\) to the upward vertical.
(a) Show that \[U^2 = ag(2 + 3\cos\alpha).\] (7)
The particle has speed \(W\) as it passes through the level of \(O\). Given that \(\cos\alpha = \dfrac{1}{\sqrt{3}}\),
(b) show that \[W^2 = ag\sqrt{3}.\] (5)
| Scheme | Marks |
|---|---|
| Energy \(\dfrac{1}{2}m\left(U^2 - v^2\right) = mga(1 + \cos\alpha)\) | M1 A1=A1 |
| \(\swarrow\) \((T +)\ mg\cos\alpha = \dfrac{mv^2}{a}\) | M1 A1 |
| Leaves circle when \(T = 0\) | |
| \(g\cos\alpha = \dfrac{U^2 - 2ga - 2ga\cos\alpha}{a}\) Eliminating \(v\) | M1 |
| Leading to \(U^2 = ag(2 + 3\cos\alpha)\) * cso | A1 |
| (7) |
| Scheme | Marks |
|---|---|
| Using conservation of energy from the lowest point of the surface | |
| \(\dfrac{1}{2}m\left(U^2 - W^2\right) = mga\) | M1 A1=A1 |
| \(W^2 = U^2 - 2ag\) | |
| Using \(\cos\alpha = \dfrac{1}{\sqrt{3}}\), \(W^2 = ag\left(2 + \dfrac{3}{\sqrt{3}}\right) - 2ag\) | M1 |
| \(= ag\sqrt{3}\) * cso | A1 |
| (5) | |
| (12 marks) |
Alternative to part (b) using conservation of energy from the point where \(P\) loses contact with surface.
| \(\left(V^2 = ag\cos\alpha = \dfrac{ga}{\sqrt{3}}\right)\) | |
| Energy \(\dfrac{1}{2}m\left(W^2 - V^2\right) = mga\cos\alpha\) | M1 A1 |
| \(\dfrac{1}{2}m\left(W^2 - \dfrac{1}{\sqrt{3}}ag\right) = mga \times \dfrac{1}{\sqrt{3}}\) | A1 |
| Leading to \(W^2 = ag\sqrt{3}\) * cso | M1 A1 (5) |
Alternative to part (b) using projectile motion from the point where \(P\) loses contact with surface.
| \(V^2 = ag\cos\alpha = \dfrac{ga}{\sqrt{3}}\) | |
| \(\downarrow\) \(W_y^2 = V^2\sin^2\alpha + 2ga\cos\alpha\) | |
| \(= \dfrac{1}{\sqrt{3}}ag\left(1 - \dfrac{1}{3}\right) + 2ga \times \dfrac{1}{\sqrt{3}} = \dfrac{8\sqrt{3}}{9}ag\) | M1 A1 |
| \(\leftarrow\) \(V_x = V\cos\alpha\) | A1 |
| \(W^2 = W_y^2 + V_x^2 = \dfrac{8\sqrt{3}}{9}ag + \dfrac{1}{3}ag\sqrt{3} \times \dfrac{1}{3} = ag\sqrt{3}\) * cso | M1 A1 (5) |