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Each particle maintains a direction towards the particle at the next corner. Time when particles will meet each other will be $\frac{2a}{\sqrt n v}$. Find the value of n

MY SOLUTION

The particles at adjacent corners will move with speed Vi and Vj. Therefore, relative velocity of first particle wrt second will be $$V_r=\sqrt{V^2+V^2}$$ $$=\sqrt 2 V$$ Distance to be covered is ‘a’

Time=$$\frac{a}{\sqrt 2 V}$$

Equating this with the original given expression $$\frac{a}{\sqrt 2 v}=\frac{2a}{\sqrt n V}$$$$n=8$$ Answer given is 4, what have I don’t wrong?

Aditya
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1 Answers1

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Consider what has happened after a small interval of time, $\delta t$. The first particle will have travelled a distance of $v \delta t$ towards the second particle whereas the second particle has travelled a distance of $v \delta t$ perpendicular to the line joining them.

The distance between them is the hypotenuse of a right-angled triangle with the other sides having lengths $a-v \delta t$ and $v \delta t$. Ignoring distances as small as $ (\delta t)^2$, the distance between the two particles is just $a-v \delta t$.

Therefore the distance between two particles will have decreased by $v \delta t$. This will be the same throughout the motion and so the particles will meet when $a=vt$, i.e. after time $\frac {a}{v}$.

In your solution, the relative velocity should just be $v$ along the line joining the particles since the second particle is moving at right angles to the first at every instant.