M2 June 2017 Q1
1. A particle \(P\) of mass 0.5 kg is moving with velocity \(4\mathbf{j}\) m s\(^{-1}\) when it receives an impulse \(\mathbf{I}\) N s. Immediately after \(P\) receives the impulse, the velocity of \(P\) is \((2\mathbf{i} + 3\mathbf{j})\) m s\(^{-1}\).
Find
(a) the magnitude of \(\mathbf{I}\), (4)
(b) the angle between \(\mathbf{I}\) and \(\mathbf{j}\). (2)
| Scheme | Marks |
|---|---|
| \(\mathbf{I} = 0.5(2\mathbf{i} + 3\mathbf{j}) - 0.5(4\mathbf{j})\) | M1 |
| \(\left(= 0.5(2\mathbf{i} - \mathbf{j})\right)\) | A1 |
| \(|\mathbf{I}| = \dfrac{1}{2}\sqrt{2^2 + 1^2}\) | M1 |
| \(= \dfrac{1}{2}\sqrt{5}\ (= 1.12)\) Ns | A1 |
| (4) |
Notes
M1 Impulse-momentum equation. Dimensionally correct. Condone subtraction in wrong order.
A1 Correct unsimplified
M1 Correct method for modulus. Follow their \(\mathbf{I}\)
A1 1.1 or better (from correct solution only)
| Scheme | Marks |
|---|---|
| \(\tan^{-1}(\pm 2)\) or \(\tan^{-1}\left(\pm\dfrac{1}{2}\right)\) or \(\tan\theta = \pm 2\) or \(\tan\theta = \pm\dfrac{1}{2}\) or equivalent | M1 |
| Required angle = 117\(^\circ\) (116.6\(^\circ\) or better) | A1 |
| (2) | |
| (6 marks) |
Notes
M1 Correct method for a relevant angle. Follow their \(\mathbf{I}\)
A1 Accept 243\(^\circ\) (2.03 rads)
1balt
 | |
| \(\cos\alpha = \dfrac{16 + 5 - 13}{2\sqrt{5}\sqrt{16}} = \dfrac{1}{\sqrt{5}}\) | M1 |
| Required angle = 117\(^\circ\) (116.6\(^\circ\)) | A1 |
A1 Accept 243\(^\circ\)