Show that in the absence of any external force, the velocity of CM remains constant.

Explanation

(i) When a rigid body moves, its center of mass will also move along with the body. For kinematic quantities like velocity `(v_(CM))` and acceleration `(a_(CM))` of the center of mass, we can differentiate the expression for position of center of mass with respect to time once and twice respectively. For simplicity, let us take the motion along X direction only.
`vec(v)_(CM) = (d vec(x)_(CM))/(dt) = (sum m_(i)((d vec(x)_(i))/(dt)))/(sum m_(i))= (sum m_(i) vec(v)_(i))/(sum m_(i))`
`vec(v)_(CM) = (sum m_(i) vec(v)_(i))/(sum m_(i))`
`vec(a)_(CM) = (d)/(dt) ((d vec(x)_(CM))/(dt))=((d vec(v)_(CM))/(dt))=(sum m_(i)((d vec(v)_(i))/(sum m_(i))))/(sum m_(i))`
`= (sum m_(i) vec(a)_(i))/(sum m_(i))`
(ii) In the absence of external force, i.e. `vec(F)_(ext) = 0` the individual rigid bodies of a system can move or shift only due to the internal forces.
(iii) This will not affect the position of the center of mass. This means that the center of mass will be in a state of rest or uniform motion. Hence, `vec(v)_(CM)` will be zero when center of mass is at rest and constant when center of mass has uniform motion `(vec(v)_(CM) = 0 or vec(v)_(CM)=` constant). There will be no acceleration of center of mass, `(vec(a)_(CM) = 0)`
From equation
`0 = (sum m_(i) vec(v)_(i))/(sum m_(i))" "(or)` constant,
`vec(v)_(CM) = (sum m_(i)vec(v)_(i))/(sum m_(i)), vec(a)_(CM) = 0`
(iv) Here, the individual particles may still move with their respective velocities and accelerations due to internal forces.