Two block of equal mass m are connected by an unstretched spring and the sytem is kept at rest on a frictionless horizontal surface. A constant force F is applied on one of the blocks puling it away from the other as shown in figure. a.Find the positon of the centre of mass at time t. b.If the extansion of the spring is `x_0` at time t find the displacement of the two blocks at this instant.

Explanation

(a) The acceleration of the centre of mass is `a_(CM) = F/(2m)`
The displacement of the centre of mass at time t will be `x=1/2a_(CM).t^(2) = (Ft^(2))/(4m)`
(b) Suppose the displacement of the first block is `x_(1)` and that of the second is `x_(2)`. Then,
`x=(mx_(1) + mx_(2))/(2m)` or `(Ft^(2))/(4m) =(x_(1) + x_(2))/2` (or) `x_(1) + x_(2) =(Ft^(2))/(2m)`.........(i)
Further, the extension of the spring is `x_(1)-x_(2)`
Therefore, `x_(1)-x_(2)=x_(0)`............(ii)
From Eqs. (i) and (ii), `x_(1) = 1/2((Ft^(2))/(2m) +x_(0))` and `x_(2) = 1/2((Ft^(2))/(2m) -x_(0))`