Question

Two particles start simultaneously from points $A$ and $B$ as shown in the figure. Then what is the minimum distance between them?

• A. $5m$
• B. $10m$
• C. $15m$
• D. $20m$

Answer 1

The correct option is B $10m$

Let $\overrightarrow{V_{AB}}$ be the relative velocity of $A$ w.r.t $B$.
$\overrightarrow{V_A}=\sqrt{10}\cos {45}^{\circ }\hat{i}+\sqrt{10}\sin {45}^{\circ }\hat{j}$
$⟹\overrightarrow{V_A}=\sqrt{5}\hat{i}+\sqrt{5}\hat{j}$
$\overrightarrow{V_B}=-\sqrt{5}\hat{j}$
Hence, $\overrightarrow{V_{AB}}=\overrightarrow{V_A}-\overrightarrow{V_B}=\sqrt{5}\hat{i}+2\sqrt{5}\hat{j}$

Let $\alpha$ be the angle that $\overrightarrow{V_{AB}}$ makes with the line $AB$.
$\therefore \tan \alpha =\frac{2\sqrt{5}}{\sqrt{5}}=2$
Let $d_{min}$ be the minimum distance between the particles.
From the figure,
$d_{min}=5\sqrt{5}\sin \alpha =5\sqrt{5}\times \frac{2}{\sqrt{5}}=10m$