Two spherical objects have masses $M$ and $5M$ and radii $R$ and $2R$ respectively. They are released in free space with their centers at a distance of $12R$. If they attract each other only due to gravitational force,the distance traveled by the smaller object before collision is: (in $R$)

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 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 particles of mass $m$ each are fixed at the opposite ends of a massless rod of length $5 \ m$ which is oriented vertically on a smooth horizontal surface and released. Find the displacement of the lower mass on the ground when the rod makes an angle of $37^o$ with the vertical. (in $m$)

Two men $A$ and $B$ are standing on a plank. $B$ is at the middle of the plank and $A$ is at the left end of the plank. The bottom surface of the plank is smooth. The system is initially at rest and the masses are as shown in the figure. $A$ and $B$ start moving such that the position of $B$ remains fixed with respect to the ground and $A$ meets $B$. Then the point where $A$ meets $B$ is located at

In a system,two particles of masses $m_1 = 3 \text{ kg}$ and $m_2 = 2 \text{ kg}$ are placed at a certain distance from each other. The particle of mass $m_1$ is moved towards the center of mass of the system through a distance of $2 \text{ cm}$. In order to keep the center of mass of the system at the original position,the particle of mass $m_2$ should move towards the center of mass by a distance of . . . . . . $\text{cm}$.