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

Options

A. `3ms^(-1)`
B. `5ms^(-1)`
C. `6ms^(-1)`
D. zero

Answer

D

Explanation

`vecv_(cm)=(m_(1)vecv_(1)+m_(2)vecv_(2)+m_(3)vecv_(3))/(m_(1)+m_(2)+m_(3))`
`implies vecv_(cm)=("Total momentum")/("Total mass")`

here total momentum of system is zero beause momentum of each particle is same in magnitude and they are symmetrically oriented as shown.
So `vecp_(1)+vecp_(2)+vecp_(3)=0`