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

Two persons of mass $m_1$ and $m_2$ are standing at the two ends $A$ and $B$ respectively,of a trolley of mass $M$ as shown. Choose the incorrect statement,if $m_1 = m_2 = m$ and both the persons jump one by one,then

Two particles $X$ and $Y$,initially at rest,move towards each other under mutual attraction. If at any instant the velocity of particle $X$ is $V$ and the velocity of particle $Y$ is $2V$,then the velocity of their center of mass will be:

If one observes from the reference frame in which the centre of mass appears to be at rest,what will be the velocity of the centre of mass?

$A$ system consists of two particles of masses $m_1$ and $m_2$. If the particle of mass $m_1$ is moved towards the centre of mass through a distance $d$,then the distance the second particle should be moved,so as to keep the centre of mass at the same position is

$A$ particle of mass $3m$ is projected from the ground at some angle with the horizontal. The horizontal range is $R$. At the highest point of its path,it breaks into two pieces of mass $m$ and $2m$. The smaller mass comes to rest,and the larger mass finally falls at a distance $x$ from the point of projection. The value of $x$ is: