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(D) Let the mass of the upper prism be $m$ and the mass of the lower prism be $3m$.Since the horizontal plane is smooth and there are no external hori

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

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(D) Let the mass of the upper prism be $m$ and the mass of the lower prism be $3m$.Since the horizontal plane is smooth and there are no external horizontal forces acting on the system of the two prisms,the horizontal position of the centre of mass of the system remains unchanged.Let the lower prism move a distance $k$ to the left. Then the upper prism must move a distance $(a-b-k)$ to the right relative to the ground to maintain the centre of mass position.Applying the principle of conservation of the centre of mass:$3m \cdot k = m \cdot (a - b - k)$Dividing both sides by $m$:$3k = a - b - k$$4k = a - b$$k = \frac{a - b}{4}$Thus,the distance moved by the lower prism is $\frac{a - b}{4}$.

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