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Second Law of Motion

Three particl...

Question

Three particles of masses $1 k g$ , $2 k g$ and $3 k g$ are situated at the corners of an equilateral triangle move at speed $6 {m s}^{- 1}$ , $3 {m s}^{- 1}$ , and $2 {m s}^{- 1}$ respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of $C M$ of the system at this instant

A

3 {m s}^{- 1}

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B

5 {m s}^{- 1}

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C

6 {m s}^{- 1}

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D

Zero

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Solution

The correct option is D Zero
${\to a}_{C M} = \frac{m_{1} {\to v}_{1} + m_{2} {\to v}_{2} + m_{3} {\to v}_{3}}{m_{1} + m_{2} + m_{3}}$
$= \frac{T o t a l m o m e n t u m}{T o t a l m a s s}$

Here total momentum of system is zero, because momentum of each particle is same in magnitude and they are symmetrically oriented a shown.
So, ${\to p}_{1} + {\to p}_{2} + {\to p}_{3} = 0$
${\to a}_{C M} = 0$

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