A2 October 2020 Q1
1. A particle \(P\) of mass 0.5 kg is moving with velocity \((4\mathbf{i} + 3\mathbf{j})\ \text{m s}^{-1}\) when it receives an impulse \(\mathbf{J}\) N s. Immediately after receiving the impulse, \(P\) is moving with velocity \((-\mathbf{i} + 6\mathbf{j})\ \text{m s}^{-1}\).
The angle between the direction of the impulse and the direction of motion of \(P\) immediately before receiving the impulse is \(\alpha^\circ\)
| Scheme | Marks | AO |
|---|---|---|
| Impulse-momentum equation | M1 | 3.1a |
| \(\mathbf{J} = 0.5(-\mathbf{i} + 6\mathbf{j} - 4\mathbf{i} - 3\mathbf{j})\) \(\big(\mathbf{J} = 0.5(-5\mathbf{i} + 3\mathbf{j})\big)\) | A1 | 1.1b |
| Find magnitude of \(\mathbf{J}\): | M1 | 1.1b |
| \(|\mathbf{J}|^2 = \dfrac{1}{4}(25 + 9)\), \(|\mathbf{J}| = \dfrac{\sqrt{34}}{2}\ (\text{N s})\) | A1 | 1.1b |
| (4) |
Notes
M1: Dimensionally correct. Must be subtracting, but condone subtracting in the wrong order.
A1: Correct unsimplified equation
M1: Correct application of Pythagoras to find the magnitude. (from \(\pm\mathbf{J}\))
A1: 2.9 or better (2.9154….) (from \(\pm\mathbf{J}\))
| Scheme | Marks | AO |
|---|---|---|
 | ||
| Correct use of trig | M1 | 3.1a |
| \(\alpha^\circ = 180^\circ - \tan^{-1}\dfrac{3}{4} - \tan^{-1}\dfrac{3}{5}\) or \(\alpha^\circ = \tan^{-1}\dfrac{4}{3} + \tan^{-1}\dfrac{5}{3}\) | A1ft | 1.1b |
| \(\alpha = 112\) | A1 | 1.1b |
| (3) | ||
| (7 marks) |
Notes
M1: Correct use of trig to find a relevant angle using \(4\mathbf{i} + 3\mathbf{j}\) and their \(\mathbf{J}\)
i.e. \(\alpha^\circ\) or \(180^\circ - \alpha^\circ\) Allow \(\left|\dfrac{\mathbf{a}.\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\right|\)
A1ft: Correct unsimplified expression for the required angle. Follow their \(\mathbf{J}\) A0 for \(\left|\dfrac{\mathbf{a}.\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\right|\)
Do not ISW
A1: 110 or better (112.166…..) or accept \(247.8\ldots^\circ\)
Alternative (b)
| Scheme | Marks | AO |
|---|---|---|
| Use scalar product of \(\mu\mathbf{J}\) and \(4\mathbf{i} + 3\mathbf{j}\) to find the angle | M1 | 3.1a |
| \(\cos\alpha^\circ = \dfrac{-20 + 9}{\sqrt{34} \times 5}\) | A1ft | 1.1b |
| \(\alpha = 112\) | A1 | 1.1b |
| (3) |
Alternative (b) 2
| Scheme | Marks | AO |
|---|---|---|
| Use of cosine rule in triangle of momenta or equivalent | M1 | 3.1a |
| \(\alpha^\circ = 180^\circ - \cos^{-1}\left(\dfrac{34 + 25 - 37}{2 \times 5 \times \sqrt{34}}\right)\) | A1ft | 1.1b |
| \(\alpha = 112\) | A1 | 1.1b |
| (3) |