M2 June 2014 Q5
5. A particle of mass \(m\) kg lies on a smooth horizontal surface. Initially the particle is at rest at a point \(O\) midway between a pair of fixed parallel vertical walls. The walls are 2 m apart. At time \(t = 0\) the particle is projected from \(O\) with speed \(u\) m s\(^{-1}\) in a direction perpendicular to the walls. The coefficient of restitution between the particle and each wall is \(\dfrac{2}{3}\). The magnitude of the impulse on the particle due to the first impact with a wall is \(\lambda mu\) N s.
The particle returns to \(O\), having bounced off each wall once, at time \(t = 3\) seconds.
| Scheme | Marks |
|---|---|
| Speed after impact \(= \dfrac{2}{3}u\) | B1 |
| Impulse = change in momentum \(= \pm\left(m.\text{their}\dfrac{2}{3}u - m.(-u)\right)\ \left(= \dfrac{5}{3}mu\right)\) | M1 |
| \(\lambda = \dfrac{5}{3}\) | A1 |
| (3) |
Notes
B1 Allow for velocity \(= -\dfrac{2u}{3}\)
M1 Need to consider momentum before and after the collision and use change of direction.
A1 cso
| Scheme | Marks |
|---|---|
| Speed after second collision \(= e^2u = \dfrac{4}{9}u\) | B1 |
| Total time taken \(= \dfrac{1}{u} + \dfrac{2}{eu} + \dfrac{1}{e^2u}\ \left(= \dfrac{1}{u} + \dfrac{3}{u} + \dfrac{9}{4u}\right)\) | M1 A2 |
| \(\dfrac{4}{u} + \dfrac{9}{4u} = \dfrac{25}{4u} = 3,\ \ \ u = \dfrac{25}{12}\) o.e. | DM1 A1 |
| (6) | |
| (9 marks) |
Notes
B1 Allow negative
M1 Use of time \(= \dfrac{\text{distance}}{\text{speed}}\) to find the total time in terms of \(u\). (At least one term dealt with correctly)
A2 -1 each error
DM1 Use total time = 3 and solve for \(u\)
A1 Accept 2.08 and 2.1 (or better)
Alt for M1A2
| Ratio of times for the three sections is \(\dfrac{1}{3} : 1 : \dfrac{3}{4}\) | M1 A2 |
A2 Or equivalent. -1 each error
SC The candidate who only considers the first return to O can score the first M1 in (b) for \(\dfrac{1}{u} + \dfrac{1}{eu}\) giving 1/6 marks