Question

Two blocks of equal mass m are connected by an un-stretched spring and the system is kept at rest on α frictionless horizontal surface. A constant force F is applied on one of the blocks pulling it away from the other as shown in figure. If the extension of the spring is $x_0$ at time t, find the displacement of the two blocks at this instant.

• A. ,
• B. ,
• C. ,
• D. ,

Answer 1

The correct option is C

,

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 = $\frac{m{x}_1+m{x}_2}{2m}$

$\frac{F{t}^2}{4m}=\frac{x_1+x_2}{2}$

$x_1+x_2=\frac{F{t}^2}{2m}$ ....... (i)

The extension of the spring is $x_2-x_1$ Therefore,

$x_2-x_1=x_0$ ....... (ii)

from (i) and (ii) , $x_1=\frac{1}{2}(\frac{F{t}^2}{2m}-x_0)$

and $x_2=\frac{1}{2}(\frac{F{t}^2}{2m}+x_0)$ .