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

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

Options

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

Answer

C

Explanation

The acelertion of the centre of mass is `a_(CM)=F/(2m)`
The displcement of the centre of mass at time `t` will be
`x=1/2a_(CM)t^(2)=(Ft^(2))/(4m)`