M4 June 2014 (R) Q4
4. A smooth uniform sphere \(S\) is moving on a smooth horizontal plane when it collides obliquely with an identical sphere \(T\) which is at rest on the plane. Immediately before the collision \(S\) is moving with speed \(U\) in a direction which makes an angle of 60\(^\circ\) with the line joining the centres of the spheres. The coefficient of restitution between the spheres is \(e\).
(a) Find, in terms of \(e\) and \(U\) where necessary,
(i) the speed and direction of motion of \(S\) immediately after the collision,
(ii) the speed and direction of motion of \(T\) immediately after the collision. (12)
The angle through which the direction of motion of \(S\) is deflected is \(\delta^\circ\).
(b) Find
(i) the value of \(e\) for which \(\delta\) takes the largest possible value,
(ii) the value of \(\delta\) in this case. (3)
| Scheme | Marks |
|---|---|
| \(mv_1 + mv_2 = mu\cos 60^\circ\) Momentum | M1 A1 |
| \(-v_1 + v_2 = eu\cos 60^\circ\) Impact law | M1 A1 |
| \(v_1 = \dfrac{u(1 - e)}{4}\) | |
| Speed of \(S = \sqrt{\dfrac{u^2(1 - e)^2}{16} + \dfrac{3u^2}{4}}\) Solve for \(v_1\) and find speed | M1 A1 |
| \(= \dfrac{u}{4}\sqrt{e^2 - 2e + 13}\) | A1 |
| \(\tan\theta = \dfrac{u\sqrt{3}}{2v_1} = \dfrac{2\sqrt{3}}{(1 - e)}\) Use components to find dirn | M1 A1 |
| \(S\) moves at \(\arctan\dfrac{2\sqrt{3}}{(1 - e)}\) to the line of centres | M1 A1 |
| \(v_2 = \dfrac{u(1 + e)}{4}\) \(v_2\) in terms of \(u,\ e\) | B1 |
| \(T\) has speed \(\dfrac{u(1 + e)}{4}\) along the line of centres Conclusion | |
| (12) |
| Scheme | Marks |
|---|---|
| \(\theta\) is a max when \(e = 1\) then \(\theta = 90^\circ\) | M1 A1 |
| then deflection angle is \(90^\circ - 60^\circ = 30^\circ\) \(\delta = 30\) | A1 |
| (3) | |
| (15 marks) |