Three particles of masses 1 kg, 2kg and 3 kg are situated at the corners of an equilateral triangle move at speed 6ms−1,3ms−1 and 2ms−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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3 ms-1

5 ms-1

6 ms-1

zero

answer is D.

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Detailed Solution

v→cm=m1v→1+m2v→2+m3v→3m1+m2+m3⇒ v→cm= Total momentum Total mass Here total momentum of system is zero,because momenfum of each particle is same in magnitude and they are symmetrically oriented as shown. So p→1+p→2+p→3=0So, velocity of CM of the system will be zero

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