In the figure the heavy mass $m$ moves down the smooth surface of a wedge making an angle $α$ with the horizontal. The wedge at rest $t = 0$ is on a smooth surface. The mass of the wedge is $M$ . The direction of motion of the mass $m$ makes an angle $β$ with the horizontal, then $' Tan β^{'}$ is

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$(1 + \frac{m}{M}) \tan α$

$\frac{m}{M} \tan α$

$\frac{M}{m} \tan α$

$(1 + \frac{M}{m}) \tan α$

answer is C.

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Detailed Solution

From free body diagrams
$N \sin α = M a - - - (1)$
$N + m A \sin α = m g \cos α - - - (2)$
$m g \sin α + m A \cos α = m a - - - (3)$
on solving equations $(1) (2) (3)$
$A = \frac{m g \cos α \sin α}{M + m \sin^{2} α}; a = \frac{(M + m) g \sin α}{M + m \sin^{2} α}$
Now $\vec{a_{b l o c k}} = \vec{a_{block /wedge}} + \vec{a_{wegde}}$
$∴ {\vec{a}}_{block} = (a \cos α \hat{i} - a \sin α \hat{j}) - A \hat{i}$
$(a \cos α - A) \hat{i} - a \sin α \hat{j}$
$∴ \tan β = \frac{a \sin α}{a \cos α - A} = (1 + \frac{m}{M}) \tan α$