For the system shown in Fig. the string is light and pulley is frictionless. The `4 kg` block is given an upward velocity of `1 m//s`. The centre of mass of the two blocks will [neglect the impulse duration]

Options

A. accelerate down with `g//3`
B. initially accelerate downwards with `g` and then after some time accelerate down with `g//3`.
C. initially accelerate with `g` and then the acceleration is `0 `
D. initially accelerate with `g` and then with accelerate with `g//3`.

Answer

D

Explanation

Initially both the blocks are under the gravity effect so
`a_(CM)=(4g+2g)/(4+2)=g`
But once the jerk in the sting occurs and string becomes taut, the `4 kg` block moves down with acceleration `g//3` and `2 kg` block moves up with acceleration `g//3` so,
`a_(CM)=(4xxg/3-(2g)/3)/6=g/9`