Two bodies of masses $m_1$ and $m_2$ $(m_1 > m_2)$ are tied to the ends of a string which passes over a light frictionless pulley. The masses are initially at rest and then released. The acceleration of the centre of mass is

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ horizontal force $F$ is applied at one end of the rod at an angle $\theta$ with the rod. The acceleration of the centre of mass of the rod is:

Two particles of equal masses have velocities $\overrightarrow{v_1} = 2\hat{i} \ m/s$ and $\overrightarrow{v_2} = 2\hat{j} \ m/s$. The first particle has an acceleration $\overrightarrow{a_1} = (3\hat{i} + 3\hat{j}) \ m/s^2$,while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a

Two particles of mass $m_1$ and $m_2$ are separated by a distance $d$. If the first particle is moved towards the centre of mass by a distance $x$,then the distance by which the second particle must be moved towards the centre of mass to keep the centre of mass at the same position is:

In the figure shown,the blocks have equal masses. Friction,mass of the string,and the mass of the pulley are negligible. The magnitude of the acceleration of the centre of mass of the two blocks is (Acceleration due to gravity $= g$).

On a smooth horizontal plane,a uniform string of mass $M$ and length $L$ is lying in a state of rest. $A$ man of the same mass $M$ is standing next to one end of the string. Now,the man starts collecting the string. Finally,the man collects all the string and puts it in his pocket. What is the displacement of the man with respect to the Earth in the process of collection?