$A$ 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 perpendicular to its length. The acceleration of the centre of mass of the rod is:

Two particles of masses $m_1$ and $m_2$ initially at rest start moving towards each other under their mutual force of attraction. The speed of the centre of mass at any time $t$,when they are at a distance $r$ apart,is

Two blocks of masses $10 \; kg$ and $30 \; kg$ are placed on the same straight line with coordinates $(0,0) \; cm$ and $(x, 0) \; cm$ respectively. The block of $10 \; kg$ is moved on the same line through a distance of $6 \; cm$ towards the other block. The distance through which the block of $30 \; kg$ must be moved to keep the position of the centre of mass of the system unchanged is

Two particles have masses $m_1$ and $m_2$. If the first particle $(m_1)$ is moved by a distance $d$ towards the center of mass,by what distance should the second particle $(m_2)$ be moved so that the center of mass of the system remains at the same position?

$A$ shell fired from a gun at an angle to the horizontal explodes in mid-air. Then the centre of mass of the shell fragments will move

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: When a firecracker (rocket) explodes in mid-air,its fragments fly in such a way that the centre of mass of the fragments continues to move along the same parabolic path that the firecracker would have followed had it not exploded.
Reason $(R)$: The explosion of a firecracker (rocket) occurs due to internal forces only,and no external force acts to cause this explosion.