A two particles P and Q are separated by distance d apart. P and Q move with velocity v and u making an angle 60^o and q with line PQ the graph between their relative separation (s) when time t is shown in figure (2). The velocity v in terms of u

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$\frac{u}{\sqrt{2}}$

$\frac{u}{\sqrt{3}}$

$u$

$\sqrt{3} u$

answer is B.

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Detailed Solution

As the bodies meet in air, relative motion along Y axis = 0
$∴ Vsin 60 = usin θ \Rightarrow usin θ = \frac{\sqrt{3} V}{2}$
Relative displacement along X axis=d
$\begin{array}{l} ∴ t = \frac{d}{Vcos 60 + ucos θ} = \frac{\sqrt{3} d}{2 u} \Rightarrow ucos θ = \frac{2 u}{\sqrt{3}} - \frac{V}{2} \\ (usin θ)^{2} + (ucos θ)^{2} = {(\frac{\sqrt{3} V}{2})}^{2} + {(\frac{2 u}{\sqrt{3}} - \frac{V}{2})}^{2} \end{array}$
On solving $\Rightarrow \frac{V}{u} = \frac{1}{\sqrt{3}} and V = \frac{u}{\sqrt{3}}$