M1 June 2008 Q5
5.
Two forces \(\mathbf{P}\) and \(\mathbf{Q}\) act on a particle at a point \(O\). The force \(\mathbf{P}\) has magnitude 15 N and the force \(\mathbf{Q}\) has magnitude \(X\) newtons. The angle between \(\mathbf{P}\) and \(\mathbf{Q}\) is 150\(^\circ\), as shown in Figure 1. The resultant of \(\mathbf{P}\) and \(\mathbf{Q}\) is \(\mathbf{R}\).
Given that the angle between \(\mathbf{R}\) and \(\mathbf{Q}\) is 50\(^\circ\), find
(a) the magnitude of \(\mathbf{R}\), (4)
(b) the value of \(X\). (5)
| Scheme | Marks |
|---|---|
| \((\uparrow)\ \ \ 15\sin 30^\circ = R\sin 50^\circ\) | M1 A1 |
| \(R \approx 9.79\ \ (\text{N})\) | DM1 A1 |
| (4) |
Alternatives using sine rule in (a) or (b); cosine rule in (b)
| (a) \(\dfrac{R}{\sin 30^\circ} = \dfrac{15}{\sin 50^\circ}\) | M1 A1 |
| \(R \approx 9.79\ \ (\text{N})\) | DM1 A1 (4) |
| Scheme | Marks |
|---|---|
| \((\rightarrow)\ \ X - 15\cos 30^\circ = R\cos 50^\circ\) ft their \(R\) | M1 A2 ft |
| \(X \approx 19.3\ \ (\text{N})\) | DM1 A1 |
| (5) | |
| (9 marks) |
Alternatives using sine rule in (a) or (b); cosine rule in (b)
| (b) \(\dfrac{X}{\sin 100^\circ} = \dfrac{15}{\sin 50^\circ} = \dfrac{R}{\sin 30^\circ}\) | M1 A2 ft on \(R\) |
| \(X \approx 19.3\ \ (\text{N})\) | DM1 A1 (5) |
| OR: cosine rule; any of \(X^2 = R^2 + 15^2 - 2 \times 15 \times R\cos 100^\circ\) \(R^2 = X^2 + 15^2 - 2 \times 15 \times X\cos 30^\circ\) \(15^2 = R^2 + X^2 - 2 \times X \times R\cos 50^\circ\) | M1 A2 ft on \(R\) |
| \(X \approx 19.3\ \ (\text{N})\) | DM1 A1 (5) |