$A$ body of mass $20 \ kg$ is moving with a velocity of $2v$ and another body of mass $10 \ kg$ is moving with velocity $v$ along the same direction. The velocity of their centre of mass is

Two particles of equal mass have velocities $2\,\hat{i}\,ms^{-1}$ and $2\,\hat{j}\,ms^{-1}$. The first particle has an acceleration $(3\,\hat{i} + 3\,\hat{j})\,ms^{-2}$ while the acceleration of the second particle is zero. The centre of mass of the two particles moves in

Show that the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of the centre of mass.

Two particles $A$ and $B$ initially at rest move towards each other under a mutual force of attraction. At the instant when the velocity of $A$ is $v$ and that of $B$ is $2v$,the velocity of the centre of mass of the system is:

$A$ wooden plank of mass $90 \ kg$ and length $3.3 \ m$ is floating on still water. $A$ girl of mass $20 \ kg$ walks from one end to the other end of the plank. The distance through which the plank moves is (in $cm$)

$M$ mass of a man stands at one end of a plank of length $L$ which lies at rest on a frictionless surface. The man walks to the other end of the plank. If the mass of the plank is $\frac{M}{3}$,then the distance that the man moves relative to the ground is: