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Two smooth and similar right angled prisms are arranged on a smooth horizontal plane as shown in the figure. The lower prism has a mass 3 times the upper prism. The prisms are held in an initial position as shown and are then released. As the upper prism touches the horizontal plan, the distance moved by the lower prism is

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$a - b$

$\frac{a - b}{3}$

$\frac{a - b}{4}$

$\frac{b - a}{2}$

answer is D.

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Detailed Solution

The system's centre of mass is not moved in a horizontal direction at all. This implies

$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} = 0$

If x represents the lower prism's displacement, then

$x_{cm} = \frac{m (a - b - x) + 3 m (- x)}{m + 3 m} = 0$

$a - b - x - 3 x = 0$

$x = \frac{a - b}{4}$

Hence the correct answer is $\frac{a - b}{4} .$

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