- JEE Mains 2019
- Hard
- MCQ
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
Answer & Solution
(C) \(m\left( {\frac{{{\theta _0} + {\theta _1}}}{{{\theta _0} - {\theta _1}}}} \right)\)
Step-by-step Solution
Detailed explanation
\(v = \sqrt {2gL\left( {1 - \cos {\theta _ \circ }} \right)} \)
Just after collision speed os \( M \)
\({v_1} = \sqrt {2gL\left( {1 - \cos {\theta _1}} \right)} \)
\(\begin{array}{l}
And\,{v_1} = \left( {\frac{{M - m}}{{M + m}}} \right)v\,\,;\,\,\frac{{{v_1}}}{v} = \frac{{M - m}}{{M + m}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt {\frac{{1 - \cos {\theta _1}}}{{1 - \cos {\theta _0}}}} = \frac{{M - m}}{{M + m}}\\
\frac{{\sin \left( {{\theta _1}/2} \right)}}{{\sin \left( {{\theta _0}/2} \right)}} = \frac{{M - m}}{{M + m}}\,\,\,\,\,\left[ {1 - \cos 2\theta = 2{{\sin }^2}\theta } \right]\\
\frac{{{\theta _1}}}{{{\theta _0}}} = \frac{{M - m}}{{M + m}}
\end{array}\)
\(\begin{array}{l}
M{\theta _1} + m{\theta _1} = M{\theta _0} - m{\theta _0}\\
M = m\left[ {\frac{{{\theta _1} + {\theta _0}}}{{{\theta _0} - {\theta _1}}}} \right]
\end{array}\)
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