Moderate -0.3 This is a standard M1 resultant force question using the cosine rule and sine rule. While it requires two steps and careful angle work, it's a routine textbook exercise with no conceptual difficulty—slightly easier than average for A-level due to its straightforward application of triangle laws.
\includegraphics{figure_2}
Forces of magnitudes 8 N and 5 N act on a particle. The angle between the directions of the two forces is \(30°\), as shown in Fig. 1. The resultant of the two forces has magnitude \(R\) N and acts at an angle \(\theta°\) to the force of magnitude 8 N, as shown in Fig. 2. Find \(R\) and \(\theta\). [7]
**Method 1:**
$R \cos \theta = 8 + 5 \cos 30°$ | M1 | For attempt at resolving || or $\perp$ to 8 N force
$R \sin \theta = 5 \sin 30°$ | A1 | For one completely correct equation
Hence $R^2 = (12.33...)^2 + 2.5^2$ | A1 | For a second correct equation
$R = 12.6$ | M1, A1 | For correct method for either unknown; For correct value
$\tan \theta = \frac{2.5}{12.33...}$ | M1 | For correct method for second unknown
$\theta = 11.5$ | A1 | For correct value
**Method 2 (Alternative):**
Triangle of forces has 5, 8, $R$ and $150°$ | M1, A1 | For considering any triangle with 5, 8, $R$; For correct triangle drawn or used
$R^2 = 8^2 + 5^2 - 2 \times 5 \times 8 \cos 150°$ | M1, A1 | For use of cosine formulae attempted; For correct expression for $R^2$
Hence $R = 12.6$ | A1 | For correct value
$\sin \theta = \frac{5 \sin 150°}{12.58...} = 0.1987...$ | M1 | For use of sine formula with numerical $R$
Hence $\theta = 11.5$ | A1 | For correct value
| | **Total: 7 marks**
\includegraphics{figure_2}
Forces of magnitudes 8 N and 5 N act on a particle. The angle between the directions of the two forces is $30°$, as shown in Fig. 1. The resultant of the two forces has magnitude $R$ N and acts at an angle $\theta°$ to the force of magnitude 8 N, as shown in Fig. 2. Find $R$ and $\theta$. [7]
\hfill \mbox{\textit{OCR M1 Q2 [7]}}