Two particles which are initially at rest move towards each other under the action of their mutual attraction. If their speeds are $v$ and $2v$ at any instant,then the speed of the center of mass of the system is,

Two bodies of masses $M$ and $4M$ initially at rest,start moving towards each other due to their mutual attraction. The velocity of their centre of mass when the first body attains a velocity $v_0$ is

$A$ child is standing at one end of a long trolley moving with a speed $v$ on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed $u$,the centre of mass of the system (trolley + child) will move with a speed

Two blocks of masses $2 \text{ kg}$ and $1 \text{ kg}$ respectively,are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in $2 \text{ s}$ is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$)

The equation $M\vec{A} = \vec{F}_{ext}$ denotes what?

Two blocks,each of mass $m$,are connected to each other by a massless string and placed on two inclined planes as shown in the figure. After releasing the system from rest,find the magnitude of the acceleration of the centre of mass of the two blocks. (Take $g = 10 \, m/s^2$)