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 with `vecg` and then after some time accelerate down with g/3
C. initially accelerate with `vecg` and then the acceleration is 0
D. initially accelerate with `vecg` and then will accelerate down with g/9

Answer

D

Explanation

Before the string again becomes tight `a_(cm) = bar(g)` and after that
`a_(cm) = ((M - m)/(M + m))^(2) barg = (2/6)^(2) cdot bar(g) = barg//9`.