$M$ mass of a man stands at one end of a plank of length $L$ which lies at rest on a frictionless surface. The man walks to the other end of the plank. If the mass of the plank is $\frac{M}{3}$,then the distance that the man moves relative to the ground is:

There are some passengers inside a stationary railway compartment. The track is frictionless. The centre of mass of the compartment itself (without the passengers) is $C_1$,while the centre of mass of the 'compartment plus passengers' system is $C_2$. If the passengers move about inside the compartment along the track,then:

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:

$A$ man of mass $M$ stands at one end of a plank of length $L$ which lies at rest on a frictionless surface. The man walks to the other end of the plank. If the mass of the plank is $3M$,the distance that the man moves relative to the ground is:

$A$ firecracker is thrown with a velocity of $30 \, ms^{-1}$ in a direction that makes an angle of $75^{\circ}$ with the vertical axis. At some point on its trajectory,the firecracker splits into two identical pieces in such a way that one piece falls $27 \, m$ far from the shooting point. Assuming that all trajectories are contained in the same plane,how far will the other piece fall from the shooting point? (Take $g = 10 \, ms^{-2}$ and neglect air resistance)

Two particles of masses $m_1$ and $m_2$ $(m_1 > m_2)$,initially at rest,move towards each other under an inverse square law force of attraction. Pick out the correct statement about the centre of mass $(CM)$ of the system.