M2 June 2015 Q3
3. A particle \(P\) of mass 0.75 kg is moving with velocity \(4\mathbf{i}\) m s\(^{-1}\) when it receives an impulse \((6\mathbf{i} + 6\mathbf{j})\) N s. The angle between the velocity of \(P\) before the impulse and the velocity of \(P\) after the impulse is \(\theta^\circ\).
Find
| Scheme | Marks |
|---|---|
| \(0.75\mathbf{v} = 6\mathbf{i} + 6\mathbf{j} + 0.75 \times 4\mathbf{i}\ \ (= 9\mathbf{i} + 6\mathbf{j})\) | M1 A1 |
| \(\mathbf{v} = 12\mathbf{i} + 8\mathbf{j}\) | A1 |
 | |
| \(\theta = \tan^{-1}\left(\dfrac{2}{3}\right)\) or \(\theta = \cos^{-1}\left(\dfrac{1 + 13 - 8}{2\sqrt{13}}\right)\), or equivalent | M1 |
| 33.7\(^\circ\) or 0.588 radians | A1 |
| (5) |
Notes
M1 Impulse momentum equation. Must be considering \(\pm m(\mathbf{v} - \mathbf{u})\)
A1 Correct unsimplified
A1 Award in (a) if seen or if (a) is completed correctly. Award in (b) if (a) is incomplete and this mark has not been awarded and correct \(\mathbf{v}\) seen for the first time in (b)
Could have a velocity triangle rather than momentum, in which case the vectors are \(4\mathbf{i}, 8\mathbf{i} + 8\mathbf{j}, 12\mathbf{i} + 8\mathbf{j}\)
M1 Correct trig to find the required angle
A1 Accept 34\(^\circ\) or better. Must be the final answer
3a alt
| \(\begin{pmatrix}6\\6\end{pmatrix} = 0.75\begin{pmatrix}v\cos\theta\\v\sin\theta\end{pmatrix} - 0.75\begin{pmatrix}4\\0\end{pmatrix}\) | M1A1 |
| \(\Rightarrow 0.75 \times v\cos\theta = 9,\ \ 0.75 \times v\sin\theta = 6\) | A1 |
| \(\Rightarrow \tan\theta = \dfrac{2}{3}\) | M1 |
| 33.7\(^\circ\) or 0.588 radians | A1 |
| Scheme | Marks |
|---|---|
| Change in KE \(= \dfrac{1}{2} \times \dfrac{3}{4}(144 + 64) - \dfrac{1}{2} \times \dfrac{3}{4}(16)\) | M1 A1ft |
| \(= 72\) (J) | A1 |
| (3) | |
| (8 marks) |
Notes
M1 Finding a difference between KE terms. Must use \(\dfrac{1}{2}mv^2\) in both terms (all of v, not just one component of it.)
A1ft follow their v. Allow \(\pm\)
A1 CAO