A block of mass M is suspended from a light spring of force constant k. Another mass m moving upwards with velocity v hits the mass M and gets embedded in it. What will be the amplitude of oscillation of the combined mass ?

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$\frac{Mv}{\sqrt{(M - m) k}}$

$\frac{MV}{\sqrt{(M + m) k}}$

$\frac{mv}{\sqrt{(M + m) k}}$

$\frac{mv}{\sqrt{(M - m) k}}$

answer is B.

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Detailed Solution

$mv = (M + m) v^{'} or v^{'} = (\frac{m}{M + m}) v Further, maximum P . E . = maximum K . E . ∴ \frac{1}{2} {kA}^{2} = \frac{1}{2} (M + m) v^{' 2} or A = {(\frac{M + m}{k})}^{1 / 2} \times V^{'} = {(\frac{M + m}{k})}^{1 / 2} \times (\frac{m}{M + m}) \times V = \frac{mv}{\sqrt{(M + m) k}}$

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