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A wedge shaped block 'A' of mass M is at rest on a smooth horizontal surface. A small block 'B' of mass 'm' placed at the top edge of inclined plane of length 'L' as shown in the figure. By the time, the block 'B' reaches the bottom end, the wedge A moves a distance of

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$\frac{mL}{m + M}$

$\frac{mLcos θ}{m + M}$

$\frac{mL}{Mcos θ}$

Zero

answer is B.

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Detailed Solution

$\begin{array}{l} Applying conservation of linear momentum along horizontal direction as there is no force is acting \\ ∴ \frac{mLcos θ}{t} = \frac{(M + m) x}{t} \\ ∴ x = \frac{mLcos θ}{M + m} \end{array}$

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