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  • Can someone explain - A simple pendulum , made of a string of length l and a bob of mass m , is released from a small angl - Work Energy and Power - JEE Main

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 upto angle \theta_{1}. Then M is given by : 

  • Option 1)

    \frac{m}{2}\left ( \frac{\theta_{0} + \theta_{1}}{\theta_{0} - \theta_{1}} \right )

  • Option 2)

    m\left ( \frac{\theta_{0} + \theta_{1}}{\theta_{0} - \theta_{1}} \right )

  • Option 3)

    \frac{m}{2}\left ( \frac{\theta_{0} - \theta_{1}}{\theta_{0} + \theta_{1}} \right )

  • Option 4)

    m\left ( \frac{\theta_{0} - \theta_{1}}{\theta_{0} + \theta_{1}} \right )

Answers (1)

best_answer

 

Coffecient of Restitution (e) -

e= \frac{v_{2}-v_{1}}{u_{2}-u_{1}}

- wherein

Ratio of relative velocity after collision to relative velocity before collision.

 

 

Perfectly Elastic Collision -

e=1

- wherein

e\: :\: coefficient \: of\: restitution

before Collision                                           After collision

                         

apply momentum conservation

mv=-mv_{1}+mv_{2}\, \, \, \, \, -\left ( 1 \right )

from energy balance

we get

v=\sqrt{29l\left ( 1-cos\Theta _{0} \right )}

& v_{1}=\sqrt{29l\left ( 1-cos\Theta _{1} \right )}

Put value of  v & v_{1}  in eq^{n}\, \, \, \left ( 1 \right )

m\times \sqrt{29l\left ( 1-cos\Theta _{0} \right )}=-m\sqrt{29l\left ( 1-cos\Theta _{1} \right )}+mv_{2}\, \, \, \, \left ( 2 \right )

For elastic collision

e=1=\frac{V_{2}-\left ( -V_{1} \right )}{V}=\frac{ \sqrt{29l\left ( 1-cos\Theta _{1} \right )}+V_{2}}{ \sqrt{29l\left ( 1-cos\Theta _{0} \right )}}

\Rightarrow V_{2}=\sqrt{29l}\left ( \sqrt{\left ( 1-cos\Theta _{0} \right )}-\sqrt{\left ( 1-cos\Theta _{1} \right )} \right )\rightarrow put in equation(2)

We get \frac{m}{m}=\frac{\sqrt{1-cos\Theta _{0}}+\sqrt{1-cos\Theta _{1}}}{\sqrt{1-cos\Theta _{0}}-\sqrt{1-cos\Theta _{1}}}

By componendo & dividendo

\frac{m-m}{m+m}=\frac{\sqrt{1-cos\Theta _{1}}}{\sqrt{1-cos\Theta _{0}}}=\frac{Sin\left ( \frac{\Theta _{1}}{2} \right )}{Sin\left ( \frac{\Theta _{0}}{2} \right )}

\frac{m}{m}=\frac{\Theta _{0}-\Theta _{1}}{\Theta _{0}+\Theta _{1}}

m=m=\left ( \frac{\Theta _{0}-\Theta _{1}}{\Theta _{0}+\Theta _{1}} \right )

 


Option 1)

\frac{m}{2}\left ( \frac{\theta_{0} + \theta_{1}}{\theta_{0} - \theta_{1}} \right )

Option 2)

m\left ( \frac{\theta_{0} + \theta_{1}}{\theta_{0} - \theta_{1}} \right )

Option 3)

\frac{m}{2}\left ( \frac{\theta_{0} - \theta_{1}}{\theta_{0} + \theta_{1}} \right )

Option 4)

m\left ( \frac{\theta_{0} - \theta_{1}}{\theta_{0} + \theta_{1}} \right )

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