Two blocks of masses 2 kg and 1 kg respectively are tied to the ends of a string which passes over a light frictionless pulley . The masses are held at rest at the same horizontal level and then released . The distance traversed by centre of mass in 2 s is ( g = 10 `m//s^(2)`)

Options

A. 1.42 m
B. 2.22 m
C. 3.12 m
D. 3.33 m

Explanation

`V_(cm)=a_(cm)t=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)g t =(10xx2)/(g)=2.22 m`