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Three particles of masses 1 kg, 2 kg and 3 kg 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 CM of the system at this instant is _____ ms^-1

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Detailed Solution

${\vec{v}}_{a m} = \frac{m_{1} \vec{v} + m_{2} {\vec{v}}_{2} + m_{3} {\vec{v}}_{3}}{m_{1} + m_{2} + m_{3}}$

$\Rightarrow {\vec{v}}_{c m} = \frac{Total momentum}{Total mass}$

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

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

So, velocity of CM of the system will be zero

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