AS June 2018 Q4
4. A particle \(P\) of mass \(3m\) is moving in a straight line on a smooth horizontal floor. A particle \(Q\) of mass \(5m\) is moving in the opposite direction to \(P\) along the same straight line.
The particles collide directly.
Immediately before the collision, the speed of \(P\) is \(2u\) and the speed of \(Q\) is \(u\).
The coefficient of restitution between \(P\) and \(Q\) is \(e\).
When \(P\) and \(Q\) collide they are at a distance \(d\) from a smooth fixed vertical wall, which is perpendicular to their direction of motion. After the collision with \(P\), particle \(Q\) collides directly with the wall and rebounds so that there is a second collision between \(P\) and \(Q\). This second collision takes place at a distance \(x\) from the wall.
Given that \(e = \dfrac{1}{18}\) and the coefficient of restitution between \(Q\) and the wall is \(\dfrac{1}{3}\)
| Scheme | Marks | AO |
|---|---|---|
| Complete strategy to find speed of \(Q\) | M1 | 3.1b |
 | ||
| Use of CLM | M1 | 3.1a |
| \(6mu - 5mu\ (= mu) = 3mv + 5mw\) | A1 | 1.1b |
| Use of impact law | M1 | 3.1a |
| \(w - v = 3ue\) | A1 | 1.1b |
| \(\left.\begin{array}{l} 3v + 5w = u \\ 3w - 3v = 9ue \end{array}\right\} \Rightarrow 8w = u + 9ue, \quad w = \dfrac{u}{8}(9e + 1)\) * | A1* | 2.1 |
| (6) |
Notes
M1: Complete strategy e.g. use of CLM, impact law and solution of simultaneous equations.
M1: CLM equation. Requires all terms and dimensionally correct. Condone sign errors.
A1: Correct unsimplified equation
M1: Impact law. Condone sign error. Must be used the right way round.
A1: Correct unsimplified equation
Signs consistent with CLM equation.
A1*: Obtain given answer from correct working
| Scheme | Marks | AO |
|---|---|---|
| \(v = w - 3ue = \dfrac{u}{8}(1 - 15e)\) and \(v \gt 0\) | M1 | 3.1b |
| \(\Rightarrow (0 \leqslant)\ e \lt \dfrac{1}{15}\) | A1 | 1.1b |
| (2) |
Notes
M1: Find speed of \(P\) and form correct inequality consistent with their directions.
A1: Correct solution. Need not mention the lower limit.
| Scheme | Marks | AO |
|---|---|---|
| Complete strategy to find time for \(Q\) to get to second collision | M1 | 3.1a |
| Speed of \(Q\) after impact with wall \(= \dfrac{u}{16}\) | B1 | 1.1b |
 | ||
| Time for Q: \(\dfrac{16d}{3u} + \dfrac{16x}{u}\) follow their \(\dfrac{u}{16}\) and \(\dfrac{16d}{3u}\) | A1ft | 1.1b |
| Complete strategy to find time for \(P\) to get to second collision \(= \dfrac{48(d - x)}{u}\) | B1ft | 1.1b |
| Use both at the same place at the same | M1 | 2.1 |
| \(x = \dfrac{128d}{192} = \dfrac{2d}{3}\) | A1 | 1.1b |
| (6) | ||
| (14 marks) |
Notes
M1: Complete strategy e.g. find time to wall and back again
B1: Correct use of impact law
A1ft: Correct unsimplified equation using \(\text{time} = \dfrac{\text{distance}}{\text{speed}}\) and following their \(\dfrac{u}{16}\) and \(\dfrac{16d}{3u}\)
B1ft: Correct use of \(\text{time} = \dfrac{\text{distance}}{\text{speed}}\) Follow their \(\dfrac{u}{48}\)
M1: find \(x\) by putting both particles in the same place at the same time. Must be valid expressions for the times.
A1: Correct answer or exact equivalent
Alternative (c)
| Scheme | Marks | AO |
|---|---|---|
| Complete strategy to find position of second collision | M1 | 3.1a |
| Speed of \(Q\) after impact with wall \(= \dfrac{u}{16}\) | B1 | 1.1b |
| Distance apart when Q strikes the wall \(= \dfrac{8d}{9}\) | B1ft | 1.1b |
| Gap closing at \(\dfrac{u}{16} + \dfrac{u}{48}\) | A1ft | 1.1b |
| \(t = \dfrac{\dfrac{8d}{9}}{\dfrac{u}{16} + \dfrac{u}{48}} \quad \left(= \dfrac{32d}{3u}\right)\) | M1 | 2.1 |
| \(x = \dfrac{u}{16} \times \dfrac{32d}{3u} = \dfrac{2d}{3}\) | A1 | 1.1b |
| (6) |
M1: e.g. by considering distances and relative velocities
B1: Correct use of impact law
B1ft: Follow their \(\dfrac{u}{48}\) and \(\dfrac{3u}{16}\)
A1ft: Follow their \(\dfrac{u}{16}\) and \(\dfrac{u}{48}\)
M1: Correct use of \(\text{time} = \dfrac{\text{distance}}{\text{speed}}\)
A1: Correct answer
Alternative (c)
| Scheme | Marks | AO |
|---|---|---|
| Complete strategy to find position of second collision | M1 | 3.1a |
| Speed of \(Q\) after impact with wall \(= \dfrac{u}{16}\) | B1 | 1.1b |
| Distance apart when \(Q\) strikes the wall \(= \dfrac{8d}{9}\) | B1ft | 1.1b |
| Ratio of speeds: \(v_Q : v_P = 3 : 1\) | A1ft | 1.1b |
| Distance travelled by \(Q\) \(= \dfrac{3}{4} \times \dfrac{8d}{9}\) | M1 | 2.1 |
| \(x = \dfrac{2d}{3}\) | A1 | 1.1b |
| (6) |
M1: e.g. by considering distances and relative velocities
B1: Correct use of impact law
B1ft: Follow their \(\dfrac{u}{48}\) and \(\dfrac{3u}{16}\)
A1ft: Follow their \(\dfrac{u}{16}\) and \(\dfrac{u}{48}\)
M1: Correct use of ratio to find \(x\)
A1: Correct answer