A semi-cylinder is cut out of a block and the block is kept on the horizontal surface. A small body of mass m is released at the top of the cylinder shaped hole. (Friction is negligible.) If the block is able to move on the horizontal surface without friction, then the force exerted on the block by the small body at the lowest point is 7mg/2. Find the ratio of M/m?

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answer is 4.

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Detailed Solution

Momentum conservation along x axis $0 = m v_{2} - M v_{1}$

Work energy theorem,

$m g R = \frac{1}{2} m v_{2}^{2} + \frac{1}{2} M v_{1}^{2}$

from (i) and (ii)

$\Rightarrow 2 mgR = {mv}_{2}^{2} + \frac{m^{2}}{M} v_{2}^{2}$

$\sqrt{\frac{2 gRM}{M + m}} = v_{2} and v_{1} = \frac{m}{M} \sqrt{\frac{2 gRM}{M + m}}$

Making FBD of the particle w.r.t wedge at the lowest point.

$\frac{7 mg}{2} - mg = \frac{m {(v_{1} + v_{2})}^{2}}{R}$

$Substituting the values, we get: \frac{M}{m} = 4$ X

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