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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 for time $t$ . If displacement of smaller block is $x$ , then displacement of the heavier block at this moment would be

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$\frac{x}{2}$

$\frac{F t^{2}}{6 m} + \frac{x}{3}$

$\frac{x}{3}$

$\frac{F t^{2}}{4 m} - \frac{x}{2}$

answer is D.

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Detailed Solution

a_com = $\frac{F}{m + 2 m} = \frac{F}{3 m}$
X_com = $\frac{1}{2}$ a_comt²= $\frac{F t^{2}}{6 m}$
X_com= $\frac{m (x) + 2 m (x')}{m + 2 m}$
x' = $\frac{F t^{2}}{4 m} - \frac{x}{2}$

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