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JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 10)

A simple pendulum, made of a string of length $$\ell $$ 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 :
$${m \over 2}\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)$$
$${m \over 2}\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)$$
$$m\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)$$
$$m\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)$$

Explanation

JEE Main 2019 (Online) 12th January Morning Slot Physics - Center of Mass and Collision Question 94 English Explanation

v = $$\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} $$

v1 = $$\sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)} $$

By momentum conservation

m$$\sqrt {2gl\left( {1 - \cos {\theta _0}} \right)} $$
$$= M{V_m} - m\sqrt {2g\left( {1 - \cos \theta } \right)} $$

$$ \Rightarrow $$$$m\sqrt {2g\ell } \left\{ {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right\}$$

$$ = $$ MVm

and  e = 1 = $${{{V_m} + \sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)} } \over {\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} }}$$

$$\sqrt {2g\ell } $$ $$\left( {\sqrt {1 - \cos {\theta _0}} - \sqrt {1 - \cos {\theta _1}} } \right) $$
$$= $$ Vm     . . .(I)

m$$\sqrt {2g\ell } \left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)$$
$$ = $$ MVM     . . .(II)

Dividing

$${{\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)} \over {\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)}} = {M \over m}$$

By componendo divided

$${{m - M} \over {m + M}}$$ = $${{\sqrt {1 - \cos {\theta _1}} } \over {\sqrt {1 - \cos {\theta _0}} }} = {{\sin \left( {{{{\theta _1}} \over 2}} \right)} \over {\sin \left( {{{{\theta _0}} \over 2}} \right)}}$$

$$ \Rightarrow $$  $${M \over m} = {{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}} \Rightarrow M = {{{\theta _0} - \theta 1} \over {{\theta _0} + {\theta _1}}}$$

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