Question

Two blocks of equal mass $m$ each 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 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 $t$, The displacement of the left and right blocks at this instant is $x_1$ and $x_2$ respectively, then

• A. $x_1=\frac{1}{2}(\frac{F{t}^2}{2m}+x_0),x_2=\frac{1}{2}(\frac{F{t}^2}{2m}-x_0)$
• B. $x_1=\frac{1}{2}(\frac{F{t}^2}{2m}-x_0),x_2=\frac{1}{2}(\frac{F{t}^2}{2m}+x_0)$
• C. $x_1=\frac{1}{2}(\frac{F{t}^2}{2m}-x_0),x_2=\frac{1}{2}(\frac{F{t}^2}{2m}-x_0)$
• D. $x_1=\frac{1}{2}(\frac{F{t}^2}{2m}+x_0),x_2=\frac{1}{2}(\frac{F{t}^2}{2m}+x_0)$

Answer 1

The correct option is A $x_1=\frac{1}{2}(\frac{F{t}^2}{2m}+x_0),x_2=\frac{1}{2}(\frac{F{t}^2}{2m}-x_0)$


The acceleration of the center of mass of the system is,

$a_{COM}=\frac{F}{2m}$

The displacement of the center of mass at time $t$ will be,

$x=u_{COM}t+\frac{1}{2}{a}_{COM}{t}^2$

Since, the system is at rest initially.

$\therefore x=\frac{F{t}^2}{4m}.....\left(1\right)$

Given, at time $t$ extension in the spring is $x_0$, displacement of the first block is $x_1$ and that of the second is $x_2$. Then the position of COM will be,

$x=\frac{m{x}_1+m{x}_2}{2m}.....\left(2\right)$

From eqs. $\left(1\right)$ and $\left(2\right)$ we get,

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

$x_1+x_2=\frac{F{t}^2}{2m}.....\left(3\right)$

Now, the extension of the spring is,

$x_1-x_2=x_0.....\left(4\right)$

From eqs. $\left(3\right)$ and $\left(4\right)$ we get,


$x_1=\frac{1}{2}(\frac{F{t}^2}{2m}+x_0)\text{and}{x}_2=\frac{1}{2}(\frac{F{t}^2}{2m}-x_0)$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(A\right)$ is the correct answer.