M4 June 2018 Q2
2. A small ball \(B\), moving on a smooth horizontal plane, collides with a fixed smooth vertical wall. Immediately before the collision the angle between the direction of motion of \(B\) and the wall is \(\alpha\). The coefficient of restitution between \(B\) and the wall is \(\dfrac{3}{4}\). The kinetic energy of \(B\) immediately after the collision is 60% of its kinetic energy immediately before the collision.
Find, in degrees, the size of angle \(\alpha\). (8)
| Scheme | Marks |
|---|---|
| Velocity before & after: parallel to wall : \(u\) and \(u\) | B1 |
| Perpendicular to the wall : \(v\) and \(\dfrac{3}{4}v\) Allow with \(ev\) | B1 |
| Kinetic energy: \(\dfrac{1}{2}m\left(\dfrac{9}{16}v^2 + u^2\right) = 0.6\times\dfrac{1}{2}m\left(v^2 + u^2\right)\) | M1A2 |
| \(\dfrac{90}{16}v^2 + 10u^2 = 6v^2 + 6u^2\) | |
| \(4u^2 = \dfrac{6}{16}v^2\qquad u^2 = \dfrac{3}{32}v^2\) | |
| \(\tan\alpha = \dfrac{v}{u} = \sqrt{\dfrac{32}{3}}\) | M1A1 |
| \(\alpha = 73^\circ\) (or better 72.976..... ) | A1 |
| (8) | |
| (8 marks) |
2 Alt
| Velocity before & after: parallel to wall : \(u\cos\alpha\) and \(u\cos\alpha\) | B1 |
| Perpendicular to the wall : \(u\sin\alpha\) and \(\dfrac{3}{4}u\sin\alpha\) | B1 |
| Kinetic energy: \(\dfrac{1}{2}m\left(\dfrac{9}{16}(u\sin\alpha)^2 + (u\cos\alpha)^2\right) = 0.6\times\dfrac{1}{2}m\left((u\sin\alpha)^2 + (u\cos\alpha)^2\right)\) | M1A2 |
| \(\dfrac{9}{16}\sin^2\alpha + \cos^2\alpha = \dfrac{3}{5} = \dfrac{9}{16} + \dfrac{7}{16}\cos^2\alpha\) | M1 |
| \(\cos^2\alpha = \dfrac{3}{35},\quad \alpha = \cos^{-1}\sqrt{\dfrac{3}{35}} = 73.0^\circ\) (1.27 radians) | A1,A1 |