Question

A cage of mass $M$ hangs from a light spring of force constant $k$. A body of mass $m$ falls from height $h$ inside the case and sticks to its floor. The amplitude of oscillations of the cage will be ______.$(M=m)$

Answer 1

Velocity of the cage after striking the ball, through momentum conservation,

$m\sqrt{2gh}=(m+m)VV=\frac{m}{m+M}\sqrt{2gh}$

Now, let spring elongates through x then, by applying energy connservation.

$-\frac{1}{2}K{(x+\frac{mq}{K})}^2+\frac{1}{2}K{\left(\frac{mq}{K}\right)}^2=\frac{1}{2}(m+M)V^2$

where, $\frac{mq}{K}$ is initial elongation.

As $m=M,$

And putting all given information:

$-\frac{1}{2}K(x^2+\frac{2mgx}{K}+\frac{m^2{g}^2}{K^2})+\frac{K{m}^2{g}^2}{K^2}=\frac{1}{2}2m\times 2gh\frac{m^2}{4m^2}$

$-K{x}^2-2mgx=\frac{mgh}{2}x=\frac{2mg\pm \sqrt{4m^2{g}^2-\frac{4Kmgh}{2}}}{2K}$

Amplitute$\left(x\right)=\frac{2mg\pm \sqrt{4m^2{g}^2-\frac{4Kmgh}{2}}}{K}$