Verified Solution • 2025–26 Session moderate

QJEEM19Ph7865: JEE Main 2019 Physics - Class 11 JEEM19Ph7865 Solution: A simple pendulum, made of a string of length and a bob of mass …

A simple pendulum, made of a string of length ℓ\ell and a bob of mass m, is released from a small angle θ0{{\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 θ1{{\theta _1}} . Then M is given by :

Multiple Choice Options

  • A m2(θ0+θ1θ0−θ1){m \over 2}\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)
  • B m2(θ0−θ1θ0+θ1){m \over 2}\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)
  • C m(θ0+θ1θ0−θ1)m\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)
  • D m(θ0−θ1θ0+θ1)m\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)

Step-by-Step Verified Solution

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

v = 2gℓ(1−cos⁡θ0)\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)}

v1 = 2gℓ(1−cos⁡θ1)\sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)}

By momentum conservation

m 2gl(1−cos⁡θ0)\sqrt {2gl\left( {1 - \cos {\theta _0}} \right)}
=MVm−m2g(1−cos⁡θ)= M{V_m} - m\sqrt {2g\left( {1 - \cos \theta } \right)}

⇒\Rightarrow m2gℓ{1−cos⁡θ0+1−cos⁡θ1}m\sqrt {2g\ell } \left\{ {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right\}

== MVm

and  e = 1 = Vm+2gℓ(1−cos⁡θ1)2gℓ(1−cos⁡θ0){{{V_m} + \sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)} } \over {\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} }}

2gℓ\sqrt {2g\ell } (1−cos⁡θ0−1−cos⁡θ1)\left( {\sqrt {1 - \cos {\theta _0}} - \sqrt {1 - \cos {\theta _1}} } \right)
== Vm     . . .(I)

m 2gℓ(1−cos⁡θ0+1−cos⁡θ1)\sqrt {2g\ell } \left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)
== MVM     . . .(II)

Dividing

(1−cos⁡θ0+1−cos⁡θ1)(1−cos⁡θ0+1−cos⁡θ1)=Mm{{\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−Mm+M{{m - M} \over {m + M}} = 1−cos⁡θ11−cos⁡θ0=sin⁡(θ12)sin⁡(θ02){{\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    Mm=θ0−θ1θ0+θ1⇒M=θ0−θ1θ0+θ1{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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