Question

Two blocks of masses $m$ and $2m$ are kept on a smooth horizontal surface. They are connected by an ideal spring of force constant $k$. Initially the spring is unstretched. A constant force is applied to the heavier block in the direction shown in figure. Suppose at time $t$ displacement of smaller block is $x$, then displacement of the heavier block at this moment would be:

• A. $\frac{x}{2}$
• B. $\frac{F{t}^2}{6m}+\frac{x}{3}$
• C. $\frac{x}{3}$
• D. $\frac{F{t}^2}{4m}-\frac{x}{2}$

Answer 1

The correct option is D $\frac{F{t}^2}{4m}-\frac{x}{2}$

Acceleration of COM
$a_{com}=\frac{F}{m+2m}=\frac{F}{3m}$
$\therefore x_{com}=\frac{1}{2}{a}_{com}{t}^2=\frac{F^2}{6m}{t}^2$ ...(i)
Let displacement of the heavier block at $t\text{is}{x}^{\prime }$
Further $x_{com}^{\prime }=\frac{m\left(x\right)+2m\left(x^{\prime }\right)}{m+2m}$
$x_{com}^{\prime }=\frac{x+2x^{\prime }}{3}$ ...(ii)
from equation (i) and (ii).
$x_{com}=x_{com}^{\prime }$
$\Rightarrow x^{\prime }=\frac{F{t}^2}{4m}-\frac{x}{2}$
$\therefore \left(d\right)$