A simple pendulum, made of a string of length | vedclass

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Question

Just before collision speed of $m$

            $v = \sqrt {2gL\left( {1 - \cos {	heta _ \circ }} ight)} $

Vedclass · Accepted Answer

$m\left( {\frac{{{	heta _0} + {	heta _1}}}{{{	heta _0} - {	heta _1}}}} ight)$

Vedclass · Answer

Just before collision speed of $m$

            $v = \sqrt {2gL\left( {1 - \cos {	heta _ \circ }} ight)} $

Just after collision speed os $ M $

           ${v_1} = \sqrt {2gL\left( {1 - \cos {	heta _1}} ight)} $

$\begin{array}{l}
And\,{v_1} = \left( {\frac{{M - m}}{{M + m}}} ight)v\,\,;\,\,\frac{{{v_1}}}{v} = \frac{{M - m}}{{M + m}}\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt {\frac{{1 - \cos {	heta _1}}}{{1 - \cos {	heta _0}}}}  = \frac{{M - m}}{{M + m}}\
\frac{{\sin \left( {{	heta _1}/2} ight)}}{{\sin \left( {{	heta _0}/2} ight)}} = \frac{{M - m}}{{M + m}}\,\,\,\,\,\left[ {1 - \cos 2	heta  = 2{{\sin }^2}	heta } ight]\
\frac{{{	heta _1}}}{{{	heta _0}}} = \frac{{M - m}}{{M + m}}
\end{array}$

$\begin{array}{l}
M{	heta _1} + m{	heta _1} = M{	heta _0} - m{	heta _0}\
M = m\left[ {\frac{{{	heta _1} + {	heta _0}}}{{{	heta _0} - {	heta _1}}}} ight]
\end{array}$

A simple pendulum, made of a string of length $l$ and a bob of mass $m$ , is released from a small angle $\theta_0$. It strikes a block of mass $M$, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle $\theta_1$. Then $M$ is given by

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