Two particles are shown in figure. At `t = 0` a constant force `F = 6 N` starts acting on `3 kg`. Find the velocity of circle of mass of these particle at `t = 5 s`.
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Options

A. `5 m//s`
B. `4 m//s`
C. `6 m//s`
D. `3 m//s`

Answer

C

Explanation

( c) `2 kg` particle remains at rest. So `v_1 = 0`
for `3 kg` particle : `a_2 = (F)/(m) =(6)/(3) = 2 m//s^2`
Velocity at `t = 5 s` : `v_2 = at = 2 xx 5 = 10 m//s`
`v_(cm) = (m_1 v_1 + m_2 v_2)/(m_1 + m_2) =(2 xx 0 + 3 xx 10)/(2 + 3) = 6 m//s`.