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Three particles of masses 1 kg, 2 kg and...

Three particles of masses `1 kg, 2 kg` 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 particles at the next corners symmetrically. Find velocity of `CM` of the system at this instant

A

`3ms^(-1)`

B

`5ms^(-1)`

C

`6ms^(-1)`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
D
`vecv_(CM)=(m_1vecv_1+m_2vecv_2+m_3vecv_3)/(m_1+m_2+m_3)=("Total momentum")/("Total mass")`
Here total momentum of system is zero, beccause momentum of each particle is same is magnitude and they are symmetrically oriented as shown.

`So vecp_1+vecp_2+vecp_3=0`
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