Three particles of equal masses are placed at the corners of an equilateral triangle, as shown in fig. Now particle A starts with a velocity V₁ towards line AB, particle B starts with a velocity V₂ towards line BC and particle C starts with velocity V₃ towards line CA. The displacement of C.M of three particle A, B and C after time ‘t’ will be (Assuming that V₁, V₂ and V₃ have the same magnitude)

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$\frac{v_{1} + \frac{\sqrt{3}}{2} v_{2} + \frac{v_{3}}{2}}{3}$

$\frac{v_{1} + v_{2} + v_{3}}{4}$

$\frac{v_{1} + v_{2} + v_{3}}{3}$

Zero

answer is A.

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Detailed Solution

Since internal force do not effect the position of cm.
$∴$ Displacement of com= 0

Vector sum of all the momentum is equal to zero

hence $V_{cm} = \frac{\vec{p_{1}} + \vec{p_{2}} + \vec{p_{3}}}{3 m} = 0$

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