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Q.

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 as shown in figure. Suppose at time t, displacement of smaller block is x, then displacement of the heavier block at this moment would be

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a

$\frac{x}{2}$

b

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

c

$\frac{x}{3}$

d

$\frac{{Ft}^{2}}{4 m} - \frac{x}{2}$

answer is D.

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Detailed Solution

$X_{CM} = \frac{m x + 2 m x_{2 m}}{m + 2 m}$

$∴ \frac{1}{2} a_{CM} t^{2} = \frac{x + 2 x_{2 m}}{3}$

$∴ \frac{3}{2} (\frac{F}{3 m}) t^{2} = x + 2 x_{2 m}$ $[∵ F = {Ma}_{CM}]$

$∴ x_{2 m} = \frac{{Ft}^{2}}{4 m} - \frac{x}{2}$

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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 to the heavier block in the direction as shown in figure. Suppose at time t, displacement of smaller block is x, then displacement of the heavier block at this moment would be