Further Mathematics Questions and Answers

\(\text{Net force = Upward force - weight}\)

\(ma = F - mg  \implies F = ma + mg \)

\(F = (80 \times 10) + (80 \times 0.5) = 800 + 40 = 840N\)

\(F = F\cos \theta + F\sin \theta\) (Resolving into its components)

\(10N = 10 \cos 60, 10 \sin 60\)

\(6N = 6 \cos 330, 6 \sin 330\)

\(R = F_{1} + F_{2} = (10 \cos 60, 10 \sin 60) + (6 \cos 330, 6 \sin 330)\)

The y- component : \(10 \sin 60 + 6 \sin 330 = 10 \times \frac{\sqrt{3}}{2} + 6 \times \frac{-1}{2}\)

= \((5\sqrt{3} - 3)N\)

Momentum of bodies with common velocity = \(m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v\)

\((10 \times 5) + (15 \times 2)  = (10 + 15)v\)

\(50 + 30 = 25v \implies v = \frac{80}{25} = 3.2ms^{-1}\)

The unit vector, \(\hat{n}\) is given by \(\hat{n} = \frac{\overrightarrow{r}}{|\overrightarrow{r}|}\)

= \(\frac{(-2i + 5j)}{\sqrt{(-2)^{2} + 5^{2}}} = \frac{(-2i + 5j)}{\sqrt{29}}\)

= \(\frac{1}{\sqrt{29}}(-2i + 5j)\) 

\(\overrightarrow{r} . \overrightarrow{t} = |\overrightarrow{r}||\overrightarrow{t}|\cos \theta\)

\(\overrightarrow{r} . \overrightarrow{t} = (3i + 4j) . (-5i + 12j) = -15 + 48 = 33\)

\(|\overrightarrow{r}| = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5\)

\(|\overrightarrow{t}| = \sqrt{(-5)^{2} + 12^{2}| = \sqrt{169} = 13\)

\(\cos \theta = \frac{\overrightarrow{r} . \overrightarrow{t}}{|\overrightarrow{r}||\overrightarrow{t}|}\)

 \(\cos \theta = \frac{33}{5 \times 13} = \frac{33}{65}\)

\(\theta = \cos^{-1} {\frac{33}{65}} \approxeq 59.5°\)