Question 117 of 130

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 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:

A$m\left(\dfrac{{\theta }_{0}-{\theta }_{1}}{{\theta }_{0}+{\theta }_{1}}\right)$B$m\left(\dfrac{{\theta }_{0}+{\theta }_{1}}{{\theta }_{0}-{\theta }_{1}}\right)$C$\dfrac{m}{2}\left(\dfrac{{\theta }_{0}+{\theta }_{1}}{{\theta }_{0}-{\theta }_{1}}\right)$D$\dfrac{m}{2}\left(\dfrac{{\theta }_{0}-{\theta }_{1}}{{\theta }_{0}+{\theta }_{1}}\right)$

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Answer: (B) $m\left(\dfrac{{\theta }_{0}+{\theta }_{1}}{{\theta }_{0}-{\theta }_{1}}\right)$

$u=\sqrt{2 gl\left(1-cos {\theta }_{0}\right)}$ .......(i)

$v=$ velocity of ball after collision

$v=\left(\dfrac{m-M}{m+M}\right)u$

Since ball rises up to angle ${\theta }_{1}$

$v=\sqrt{2gl\left(1-cos {\theta }_{1}\right)}=\left(\dfrac{m-M}{m+M}\right)u$ ......(ii)

From (i) and (ii)

$\dfrac{m-M}{m+M}=\sqrt{\dfrac{1-cos {\theta }_{1}}{1-cos {\theta }_{0}}}=\dfrac{sin\left(\dfrac{{\theta }_{1}}{2}\right)}{sin\left(\dfrac{{\theta }_{0}}{2}\right)}$

$\Rightarrow \dfrac{M}{m}=\dfrac{{\theta }_{0}-{\theta }_{1}}{{\theta }_{0}+{\theta }_{1}}\Rightarrow M=\left(\dfrac{{\theta }_{0}-{\theta }_{1}}{{\theta }_{0}+{\theta }_{1}}\right)m$