M2 June 2013 (R) Q5
5. Two particles \(P\) and \(Q\), of masses \(2m\) and \(m\) respectively, are on a smooth horizontal table. Particle \(Q\) is at rest and particle \(P\) collides directly with it when moving with speed \(u\). After the collision the total kinetic energy of the two particles is \(\dfrac{3}{4}mu^2\). Find
| Scheme | Marks |
|---|---|
| \(2mu = 2mv_P + mv_Q\) | M1 A1 |
| \(\tfrac{3}{4}mu^2 = \tfrac{1}{2}2mv_P^{\,2} + \tfrac{1}{2}mv_Q^{\,2}\) | M1 A1 |
| \(3v_Q^{\,2} - 4uv_Q + u^2 = 0\) or \(12v_P^{\,2} - 16uv_P + 5u^2 = 0\) | M1 A1 |
| \(v_Q = \tfrac{u}{3},\ v_P = \tfrac{5u}{6}\) or \(v_Q = u,\ v_P = \tfrac{u}{2}\) | DM1 A1 |
| \(v_Q = u\) | DM1 |
| ………… since \(v_Q > v_P\) | A1 |
| (10) |
Notes
M1 CLM. Needs all 3 terms of corrwct form but condone sign slips
A1 Correct equation
M1 KE after impact. 3 terms of correct form
A1 Correct equation
M1 Use CLM equation to form quadratic in \(v_P\) or \(v_Q\)
A1 Correct equation
DM1 Solve for a value of \(v_Q\). Dependent on the previous M1.
A1 A \(v_Q, v_P\) pair correct or two correct values for \(v_Q\)
DM1 Select solution from a choice of two. Dependent on all 4 M marks.
A1 Correct justification
| Scheme | Marks |
|---|---|
| \(e = \dfrac{u - \tfrac{u}{2}}{u}\ \ \ \ \ \left(\dfrac{v_Q - v_P}{u}\right)\) | M1 A1 ft |
| \(= \dfrac{1}{2}\) | A1 |
| (3) | |
| (13 marks) |
Notes
M1 Impact law. Must be used correctly. Condone \(\pm e\) Follow their speeds from (a).
A1 ft Correct for their speeds