A2 June 2023 Q1
1. A particle \(P\) of mass 2 kg is moving with velocity \((-4\mathbf{i} + 3\mathbf{j})\ \text{m s}^{-1}\) when it receives an impulse \((-6\mathbf{i} + 42\mathbf{j})\ \text{N s}\).
The angle through which the direction of motion of \(P\) has been deflected by the impulse is \(\alpha^\circ\)
| Scheme | Marks | AO |
|---|---|---|
| Impulse-momentum: | M1 | 3.1a |
| \((-6\mathbf{i} + 42\mathbf{j}) = 2\{\mathbf{v} - (-4\mathbf{i} + 3\mathbf{j})\}\) | A1 | 1.1b |
| Find magnitude of their v: \(\sqrt{(-7)^2 + 24^2}\) | M1 | 1.1b |
| \(25\ (\text{m s}^{-1})\) | A1 | 1.1b |
| (4) |
Notes
M1: Dimensionally correct, mass \(\times\) velocity. Must be subtracting momenta but condone subtracting in the wrong order. M0 if \(g\) is included.
A1: Correct unsimplified equation.
M1: Correct application of Pythagoras to find the magnitude of their \(v\).
M0 for an incorrect speed if there is no evidence of Pythagoras being used on their velocity.
A1: Correct answer following the correct velocity.
| Scheme | Marks | AO |
|---|---|---|
| Use scalar product \(\cos\alpha = \dfrac{(-4 \times -7) + (3 \times 24)}{\sqrt{(-4)^2 + 3^2} \times \sqrt{(-7)^2 + 24^2}}\) | M1 | 3.1a |
| \(\alpha = 37\) or better | A1 | 1.1b |
| (2) | ||
| (6 marks) |
Notes
M1: Complete method to find the required angle. Correct use of scalar product with their v. The formula must be correct, \(\cos\alpha = \dfrac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\) M0 if the fraction is up the wrong way. Do not ISW.
A1: cao in degrees
1(b) alt 1
| Scheme | Marks | AO |
|---|---|---|
| Use cosine rule in a vector triangle: \(\cos\alpha = \dfrac{\{(-4)^2 + 3^2\} + \{(-7)^2 + 24^2\} - (3^2 + (-21)^2)}{2 \times 5 \times \sqrt{(-7)^2 + 24^2}}\) | M1 | 3.1a |
| \(\alpha = 37\) or better | A1 | 1.1b |
| (2) |
M1: Complete method to find the required angle. Correct use of cosine rule on \((\mathbf{v} - \mathbf{u})\) or \((\mathbf{u} - \mathbf{v})\) vector triangle for their v. M0 if using \((\mathbf{v} + \mathbf{u})\). Do not ISW.
A1: cao in degrees
1(b) alt 2
| Scheme | Marks | AO |
|---|---|---|
| Use inverse tan: Eg \(\alpha = \tan^{-1}\left(\dfrac{24}{7}\right) - \tan^{-1}\left(\dfrac{3}{4}\right)\) \(\phantom{\text{Eg }}\alpha = 90 - \tan^{-1}\left(\dfrac{7}{24}\right) - \tan^{-1}\left(\dfrac{3}{4}\right)\) | M1 | 3.1a |
| \(\alpha = 37\) or better | A1 | 1.1b |
| (2) |
M1: Complete method to find the required angle. Correct use of inverse tan formulae for their v. Do not ISW.
M0 for \(\tan^{-1}\left(\dfrac{3}{4}\right)\) alone which also gives the value 36.869…
A1: cao in degrees