Two particles of masses `2m` and `3m` are placed at separation `d` on a smooth surface. They move towards each other due to mutual attractive force. Find (a) acceleration of c.m. (b) Velocity of c.m. when separation between particles becomes `d//3`. ( c) At what distance from the initial position of mass `2m`, the particles collide.

Explanation

(a) Since `vec(F)_(ext) = 0 , vec(a)_(c.m.) = 0`.
(b) Initially the particles are at rest `v_(1) = v_(2) = 0`, therefore , `v_(c.m.) = 0`. Since `vec(F)_(ext) = 0` , the velocity of c.m. is constant and hence `v_(c.m.)` is always zero whenever the separation becomes `d//2 , d//3` , etc.
( c) First , locate the c.m.
`x_(c.m.) = ( 2m xx 0 + 3 m xx d)/( 2m + 3m) = (3d)/(5)`
Since the c.m. is at rest , its position is fixed , hence particles will meet at c.m. i.e. at distance `3d//5 from A`.