The wedge shown can slide without friction on a horizontal floor. Mass of the wedge is $M$ and its angle of inclination is $θ = 30^{0}$ . A block of mass ‘ $m$ ’ slides down the wedge without friction when released on its inclined face. If path of the block relative to the ground makes an angle of $ϕ = 60^{0}$ with the horizontal, find the ratio of mass of the block to that of the wedge $(\frac{m}{M})$

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Detailed Solution

$\frac{a_{w}}{\sin (ϕ - ϕ)} = \frac{a_{B}}{\sin θ} \Rightarrow \frac{a_{w}}{a_{B}} = \frac{\sin (ϕ - θ)}{\sin θ} m a_{B} \cos ϕ = M a_{w} \Rightarrow \frac{m}{M} = \frac{a_{w}}{a_{B} \cos ϕ} = \frac{\sin (ϕ - θ)}{\sin θ . \cos ϕ} = \frac{1 \times 2 \times 2}{2 \times 1 \times 1} = 2$

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