M2 June 2006 Q8
8. Two particles \(A\) and \(B\) move on a smooth horizontal table. The mass of \(A\) is \(m\), and the mass of \(B\) is \(4m\). Initially \(A\) is moving with speed \(u\) when it collides directly with \(B\), which is at rest on the table. As a result of the collision, the direction of motion of \(A\) is reversed. The coefficient of restitution between the particles is \(e\).
(a) Find expressions for the speed of \(A\) and the speed of \(B\) immediately after the collision. (7)
In the subsequent motion, \(B\) strikes a smooth vertical wall and rebounds. The wall is perpendicular to the direction of motion of \(B\). The coefficient of restitution between \(B\) and the wall is \(\tfrac{4}{5}\). Given that there is a second collision between \(A\) and \(B\),
(b) show that \(\tfrac{1}{4} < e < \tfrac{9}{16}\). (5)
Given that \(e = \tfrac{1}{2}\),
(c) find the total kinetic energy lost in the first collision between \(A\) and \(B\). (3)
| Scheme | Marks |
|---|---|
| \(mu = 4mw - mv\) | M1 A1 |
| \(eu = w + v\) | M1 A1 |
| \(\Rightarrow\ w = \left(\dfrac{1 + e}{5}\right)u,\ \ v = \left(\dfrac{4e - 1}{5}\right)u\) indep | M1 A1 A1 |
| (7) |
| Scheme | Marks |
|---|---|
| \(w^{\prime} = \left(\dfrac{4 + 4e}{25}\right)u\) | B1 f.t. |
| Second collision \(\Rightarrow w^{\prime} > v\) | |
| \(\Rightarrow\ \ \dfrac{4 + 4e}{25} > \dfrac{4e - 1}{5}\) | M1 |
| \(\Rightarrow\ \ e < 9/16\) dep | M1 A1 |
| Also \(v > 0\ \Rightarrow\ e > 1/4\) Hence result (*) | B1 |
| (5) |
| Scheme | Marks |
|---|---|
| KE lost \(= \tfrac{1}{2}mu^2 - \left[\tfrac{1}{2}.4m\{(u/5)(1 + e)\}^2 + \tfrac{1}{2}m\{(u/5)(4e - 1)\}^2\right]\) | M1 A1 f.t. |
| \(= \dfrac{3}{10}mu^2\) | A1 cao |
| (3) | |
| (15 marks) |