M1 June 2018 Q1
1. Two particles, \(P\) and \(Q\), have masses \(3m\) and \(m\) respectively. They are moving in opposite directions towards each other along the same straight line on a smooth horizontal plane and collide directly. The speeds of \(P\) and \(Q\) immediately before the collision are \(2u\) and \(4u\) respectively. The magnitude of the impulse received by each particle in the collision is \(\dfrac{21mu}{4}\).
| Scheme | Marks |
|---|---|
| For \(P\): \(\ -\dfrac{21mu}{4} = 3m(v_P - 2u)\) | M1A1 |
| \(v_P = \dfrac{u}{4}\) | A1 |
| (3) |
Notes
M1 for using Impulse = Change in Momentum of \(P\) (must have \(3m\) in both terms) (M0 if clearly adding momenta or if \(g\) is included) but condone sign errors.
First A1 for a correct equation. (N.B. Could have \(-v_P\) in place of \(v_P\))
Second A1 for \(\dfrac{u}{4}\) oe (must be positive)
N.B. If they try to find \(v_Q\) first and then use CLM to find \(v_P\), M1 for a complete method to find \(v_P\), A1 for correct equations, A1 for the answer for \(v_P\).
If an incorrect \(v_Q\) is then just stated in (b), award relevant marks if seen in working for (a).
If no attempt at (b), then no marks for (b).
| Scheme | Marks |
|---|---|
| For \(Q\): \(\ \dfrac{21mu}{4} = m(v_Q - -4u)\) | M1A1 |
| \(v_Q = \dfrac{5u}{4}\) | A1 |
| (3) | |
| (6 marks) |
Notes
M1 for using Impulse = Change in Momentum of \(Q\) (must have \(m\) in both terms) (M0 if clearly adding momenta or if \(g\) is included) but condone sign errors.
First A1 for a correct equation. (N.B. Could have \(-v_Q\) in place of \(v_Q\))
Second A1 for \(\dfrac{5u}{4}\) oe (must be positive)
OR:
M1 for CLM with correct no. of terms, condone missing \(m\)’s or extra \(g\)’s and sign errors
First A1 for a correct equation
Second A1 for \(\dfrac{5u}{4}\) oe (must be positive)
OR
| CLM: \(\ 3m \times 2u - m \times 4u = 3m \times \dfrac{u}{4} + mv_Q\) | M1 A1 |
| \(v_Q = \dfrac{5u}{4}\) | A1 |