Comprehension # 7
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 forceF 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 :-

Explanation

a. The acceleration of the centre of mass is given by
`a_(CM) =F/M=F/(2M) `
The position of the centre of mass at time to is
`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`. As the centre of mass is at x, we should have
`x=(mx_1+mx_2)/(2m)`
`or, (Ft^2)/(4m)=(x_1+x_2)/2`
`or, x_1+x_2=(Ft^2)/(2m)` .......i
the extension of th spring is `x_2-x_1`. therefore
`x_2-x_1=x_0` ......ii ltbr `from i and ii `x_1=1/2((Ft^2)/(2m)-x_0)`
and `x_2=1/2((Ft^2)/(2m)+x_0)`.