Two particles `A` and `B` of masses `2m` and `m` are placed on a smooth surface at separation `d`. They move towards each other due to the mutual attractive force.
(i) The particles will meet at distance `d//3` from the initial position of mass `2m`.
(ii) The speed of `A` will be half of speed of `B` until the particles collide (excluding the initial speed).
(iii) The distance covered by `B` is always double that covered by `A` until the particles collide.
(iv) The velocity of center of mass is always zero.

Options

A. `(i) ,(ii)`
B. `(i),(ii),(iii)`
C. `(ii),(iii),(iv)`
D. all options are correct

Answer

D

Explanation

(i) Since the location of center of mass is fixed
`2mx = m(d - x) rArr x = (d)/(3)`
(ii) Since the initial momentum of system is zero and it will always remain zero (since no external force)
`0 = 2m v_(A) - mv_(B) rArr v_(A) = (v_(B))/(2)`
(iii) `2 mx_(1) =mx_(2) rArr v_(A) = (v_(B))/(2)`
(iii) `2mx_(1) = mx_(2) rArr x_(2) = 2x_(1)`
(iv) `vec(v)_(c.m.) = 0 `= constant