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

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zero

6 $m s^{- 1}$

3 $m s^{- 1}$

5 $m s^{- 1}$

answer is D.

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Detailed Solution

$\begin{array}{l} {\vec{v}}_{cm} = \frac{m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2} + m_{3} {\vec{v}}_{3}}{m_{1} + m_{2} + m_{3}} \\ \Rightarrow {\vec{v}}_{cm} = \frac{Total momentum}{Total mass} \end{array}$

Here total momentum of system is zero,because momenfum of each particle is same in magnitude and they are symmetrically oriented as shown.

$So {\vec{p}}_{1} + {\vec{p}}_{2} + {\vec{p}}_{3} = 0$

So, velocity of CM of the system will be zero

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