If a bomb is thrown at a certain angle with the horizontal and after exploding on the way the different fragments move in different directions,then the centre of mass

$A$ body falling vertically downwards under gravity breaks into two parts of unequal masses. The centre of mass of the two parts taken together

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 . . . . . . $\text{cm}$.

$A$ projectile of mass $3m$ explodes at its highest point into three equal fragments. One fragment returns to the point of projection. The second fragment comes to rest. The distance of the third fragment from the point of projection when it lands on the ground is ......... $m$. (The range of the projectile without explosion was $100 \ m$)

What are some of the notable points in the derivation of $M\vec{A} = \vec{F}_{ext}$?

Two blocks of masses $2 \,kg$ and $1 \,kg$ are tied to the ends of a string which passes over a light frictionless pulley. The blocks are held at rest at the same horizontal level and then released suddenly. The distance traversed by their centre of mass in $2 \,s$ is (acceleration due to gravity $g = 10 \,ms^{-2}$) (in $\,m$)