Two masses $m_1$ and $m_2$ start to move towards each other due to mutual gravitational force. If the distance covered by $m_1$ is $x$,then the distance covered by $m_2$ is:

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

Two particles of equal mass have velocities $\overrightarrow{v}_1=4 \hat{i} \text{ ms}^{-1}$ and $\overrightarrow{v}_2=4 \hat{j} \text{ ms}^{-1}$. The first particle has an acceleration $\overrightarrow{a}_1=(5 \hat{i}+5 \hat{j}) \text{ ms}^{-2}$ while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of

Two masses $m_1$ and $m_2$ $(m_1 > m_2)$ are connected by a massless,flexible,and inextensible string passed over a massless and frictionless pulley. The acceleration of the center of mass of the system is

$A$ small ball $B$ of mass $m$ is suspended with a light inelastic string of length $L$ from a block $A$ of the same mass $m$,which can move on a smooth horizontal surface as shown in the figure. The ball is displaced by an angle $\theta$ from the equilibrium position and then released. The displacement of the centre of mass of the $A+B$ system until the string becomes vertical is

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: