Physics - Kinematics Question with Solution | TestHub

From the figure: ${\vec{r}}_A=-100\hat{i}\mathrm{m},{\vec{v}}_A=20\hat{i}m/s$; ${\vec{r}}_B=-100\sqrt{3}\hat{j}\mathrm{m},{\vec{v}}_B=\frac{20}{\sqrt{3}}\hat{j}m/s$.

Relative position of A w.r.t. B: ${\vec{r}}_{AB}=-100\hat{i}+100\sqrt{3}\hat{j}$, $|{\vec{r}}_{AB}|=200\mathrm{m}$. Relative velocity: ${\vec{v}}_{AB}=20\hat{i}-\frac{20}{\sqrt{3}}\hat{j}$, $|{\vec{v}}_{AB}|=\sqrt{400+\frac{400}{3}}=\frac{40}{\sqrt{3}}m/s$.

${\vec{r}}_{AB}\cdot {\vec{v}}_{AB}=-2000-2000=-4000$, so $\cos \varphi =\frac{-4000}{200\times 40/\sqrt{3}}=-\frac{\sqrt{3}}{2}$, $\varphi =150^{\circ }$.

Closest approach $r_{\min }=200\sin 150^{\circ }=100\mathrm{m}>0$, so they do not collide. Hence (B) and (C).