AS June 2019 Q2
2. Two particles, \(A\) and \(B\), of masses \(2m\) and \(3m\) respectively, are moving on a smooth horizontal plane. The particles are moving in opposite directions towards each other along the same straight line when they collide directly. Immediately before the collision the speed of \(A\) is \(2u\) and the speed of \(B\) is \(u\). In the collision the impulse of \(A\) on \(B\) has magnitude \(5mu\).
| Scheme | Marks | AO |
|---|---|---|
| Using the Impulse-momentum principle for \(B\) | M1 | 3.1a |
| \(5mu = 3m(v_B - -u)\) | A1 | 1.1b |
| \(v_B = \dfrac{2u}{3}\) | A1 | 1.1b |
| Use of conservation of momentum | M1 | 3.1a |
| \(4mu - 3mu = 2mv_A + 3mv_B \left(= 2mv_A + 3m.\dfrac{2u}{3}\right)\) | A1ft | 1.1b |
| \(v_A = -\dfrac{u}{2}\) | A1 | 1.1b |
| Use of NLR | M1 | 3.4 |
| \(e = \dfrac{v_B - v_A}{2u + u} \left(= \dfrac{\dfrac{u}{2} + \dfrac{2u}{3}}{2u + u}\right)\) | A1ft | 1.1b |
| \(e = \dfrac{7}{18} = 0.39\) or better | A1 | 1.1b |
 | ||
| (9) |
Notes
M1: Correct no. of terms and dimensionally correct but condone sign errors but must be a difference of momenta
A1: Correct unsimplified equation (corrected from the printed mark scheme: the scheme prints “\(5mu == 3m(v_B - -u)\)” with a doubled equals sign)
A1: Correct appropriate velocity
M1: Use of CLM with correct no. of terms and dimensionally correct but condone sign errors Alternative: Use Impulse - momentum for \(A\)
A1ft: Correct unsimplified CLM equation
Or: \(-5mu = 2m(v_A - 2u)\)
A1: Correct speed
M1: Use of NLR with \(e\) on the correct side
A1ft: Correct unsimplified equation
A1: Correct answer
ALT
Could find \(v_A\) before \(v_B\):
M1A1A1 for first velocity, M1A1A1 for second
M1A1A1 for \(e\) found correctly
Candidates are approaching this in many different ways.
They need
- two of momentum impulse equation for each particle and CLM
- impact law
M1A1 for each correct equation (in the order seen)
Of the remaining 3 A marks,
A1 for a correct expression for \(v_A\) or \(v_B\)
A1 for a correct expression in \(e\)
A1 for the correct answer
e.g.
| Scheme | Marks |
|---|---|
| CLM: \(4mu - 3mu = 2mv_A + 3mv_B\) | M1A1 |
| Impact: \(v_B - v_A = 3ue\) | M1A1 |
| \(v_B = \dfrac{u}{5}(1+6e) \quad \text{or} \quad v_A = \dfrac{u}{5}(1-9e)\) | A1 |
| \(5mu = 3m\big(v_B - (-u)\big) \quad \left(= 3m\left(\dfrac{u}{5}(1+6e) + u\right)\right)\) Or \(-5mu = 2m(v_A - 2u) \quad \left(= 2m\left(\dfrac{u}{5}(1-9e) - 2u\right)\right)\) | M1A1 |
| \(5 = 3\left(\dfrac{1}{5}(1+6e) + 1\right)\) or \(-5 = 2\left(\dfrac{1}{5}(1-9e) - 2\right)\) | A1 |
| \(e = \dfrac{7}{18} = 0.39\) or better | A1 |
| Scheme | Marks | AO |
|---|---|---|
| KE Loss = Initial KE \(-\) Final KE | M1 | 2.1 |
| \(= \dfrac{1}{2}.2m(2u)^2 + \dfrac{1}{2}.3mu^2 - \left(\dfrac{1}{2}.2m\left(-\dfrac{u}{2}\right)^2 + \dfrac{1}{2}.3m\left(\dfrac{2u}{3}\right)^2\right)\) | A1ft A1ft | 1.1b 1.1b |
| \(= \dfrac{55mu^2}{12}\) | A1 | 1.1b |
| (4) | ||
| (13 marks) |
Notes
M1: Correct no. of terms and must be a difference.
Must be dimensionally correct at the point when they state their expression for the loss (change) in KE
A1ft: Unsimplified expression in \(u\) with at most 1 error, ft on their speeds from (a)
A1ft: Correct unsimplified expression in \(u\). (These first 3 marks can be scored for a correct loss or gain in KE), ft on their speeds from (a)
A1: cso Accept \(4.58mu^2\) or \(4.6mu^2\)