M1 January 2006 Q4
4. Two forces \(\mathbf{P}\) and \(\mathbf{Q}\) act on a particle. The force \(\mathbf{P}\) has magnitude 7 N and acts due north. The resultant of \(\mathbf{P}\) and \(\mathbf{Q}\) is a force of magnitude 10 N acting in a direction with bearing 120\(^\circ\). Find
(i) the magnitude of \(\mathbf{Q}\),
(ii) the direction of \(\mathbf{Q}\), giving your answer as a bearing. (9)
| Scheme | Marks |
|---|---|
| \(\mathbf{R} = 10\sqrt{3}/2\,\mathbf{i} - 5\mathbf{j}\) | M1 A1 |
| Using \(\mathbf{P} = 7\mathbf{j}\) and \(\mathbf{Q} = \mathbf{R} - \mathbf{P}\) to obtain \(\mathbf{Q} = 5\sqrt{3}\,\mathbf{i} - 12\mathbf{j}\) | M1 A1 |
| Magnitude \(= \sqrt{[(5\sqrt{3})^2 + 12^2]} \approx 14.8\) N (AWRT) | M1 A1 |
| angle with \(\mathbf{i} = \arctan(12/5\sqrt{3}) \approx 54.2^\circ\) | M1 A1 |
| bearing \(\approx 144^\circ\) (AWRT) | A1 |
| (9) | |
| (9 marks) |
Notes
(Corrected from the printed mark scheme: the angle with \(\mathbf{i}\) is printed as 64.2\(^\circ\); \(\arctan(12/5\sqrt{3}) \approx 54.2^\circ\), giving the bearing 144\(^\circ\).)
Alternative method
| Scheme | Marks |
|---|---|
| Vector triangle correct | B1 |
| \(Q^2 = 10^2 + 7^2 + 2 \times 10 \times 7\cos 60\) | M1 A1 |
| \(Q \approx 14.8\) N (AWRT) | A1 |
| \(\dfrac{14.8}{\sin 120} = \dfrac{10}{\sin\theta}\) | M1 A1ft |
| \(\Rightarrow \theta = 35.8\), \(\Rightarrow\) bearing 144 (AWRT) | M1 A1, A1 |
| (9) |