Two blocks of equal mass `m` are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force `F` is applied on the first block pulling away from the other as shown in Fig.

Then the displacement of the centre of mass in at time `t` is

Options

A. `(Ft^(2))/(2m)`
B. `(Ft^(2))/(3m)`
C. `(Ft^(2))/(4m)`
D. `(Ft^(2))/(m)`

Answer

C

Explanation

As `ma_(cm) = f rArr a_(cm) = (f)/(2m) " so " s_(cm) = (1)/(2) at^(2) = ((f)/(4m))t^(2)`