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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 plane, the distance moved by the lower prism is
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Verified Answer
The correct answer is:
$\frac{a-b}{4}$
If lower prism moves through a horizontal distance $\mathrm{k}$ since horizontal position of centre of mass of the two prims system remains unchanged
$\begin{aligned}
& 3 m k=m[(a-b)-k] \\
& \Rightarrow 3 k=(a-b) \\
& \Rightarrow 4 k=a-b \\
& \text { or } k=\frac{a-b}{4}
\end{aligned}$
$\begin{aligned}
& 3 m k=m[(a-b)-k] \\
& \Rightarrow 3 k=(a-b) \\
& \Rightarrow 4 k=a-b \\
& \text { or } k=\frac{a-b}{4}
\end{aligned}$
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