There are two bodies of different masses of 2 kg and 4 kg moving with velocities 10m/s and 2 m/s towards each other due to mutual gravitational attraction. What is the velocity of their centre of mass?

Q. Two particles of mass

2 k g

and

4 k g

are approaching towards each other with acceleration

1 m / {s e c}^{2}

and

2 m / {s e c}^{2}

respectively on a smooth horizontal surface. Find the acceleration of centre of mass of the system.

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Two bodies of masses 4 kg and 2 kg are moving towards each other with 3 ms−1 and 2 ms−1 respectively. Then velocity of center of mass is

The correct option is B 2ms−1 towards 4 kgVan=m1v1+m2v2m1+m2A=4kgspeed=3m/sB=2kgspeed=2m/s⇒4×3−2×24⇒84=2m/s

Two bodies of masses $4 k g$ and $2 k g$ are moving towards each other with $3 m s^{- 1}$ and $2 m s^{- 1}$ respectively. Then velocity of center of mass is

The correct option is B $2 m s^{- 1}$ towards $4 k g$
$\begin{matrix} V_{a n} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} A = 4 k g s p e e d = 3 m / s B = 2 k g s p e e d = 2 m / s \Rightarrow \frac{4 \times 3 - 2 \times 2}{4} \Rightarrow \frac{8}{4} = 2 m / s \end{matrix}$