M2 January 2011 Q8
8. A particle \(P\) of mass \(m\) kg is moving with speed 6 m s\(^{-1}\) in a straight line on a smooth horizontal floor. The particle strikes a fixed smooth vertical wall at right angles and rebounds. The kinetic energy lost in the impact is 64 J. The coefficient of restitution between \(P\) and the wall is \(\frac{1}{3}\).
(a) Show that \(m = 4\). (6)
After rebounding from the wall, \(P\) collides directly with a particle \(Q\) which is moving towards \(P\) with speed 3 m s\(^{-1}\). The mass of \(Q\) is 2 kg and the coefficient of restitution between \(P\) and \(Q\) is \(\frac{1}{3}\).
(b) Show that there will be a second collision between \(P\) and the wall. (7)
| Scheme | Marks |
|---|---|
| KE lost: \(\ \dfrac{1}{2} \times m \times 36 - \dfrac{1}{2} \times m \times v^2 = 64\) | M1A1 |
| Restitution: \(\ v = 1/3 \times 6 = 2\) | M1A1 |
| Substitute and solve for m: \(\ \dfrac{1}{2} \times m \times 36 - \dfrac{1}{2} \times m \times 4 = 64 = 16m\) | DM1 |
| \(m = 4\) answer given | A1 |
| (6) |
| Scheme | Marks |
|---|---|
| Conservation of momentum: \(\ 6 - 8 = 4w - 2v\) their "2" | M1A1ft |
| Restitution: \(\ v + w = \frac{1}{3}(2 + 3)\) their "2" | M1A1ft |
| \(v = \dfrac{5}{3} - w\) | |
| Solve for \(w\): \(\ -2 = 4w - 2\left(\dfrac{5}{3} - w\right) = 6w - \dfrac{10}{3}\) | DM1 |
| \(\dfrac{4}{3} = 6w\) | A1 |
| \(\left(w = 4/18 = 2/9\ \text{m s}^{-1}\right)\) | |
| \(w > 0 \Rightarrow\) will collide with the wall again | A1 |
| (7) | |
| (13 marks) |