Question

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 :

• A. $m\left(\frac{{\theta }_o+{\theta }_1}{{\theta }_o-{\theta }_1}\right)$
• B. $\frac{m}{2}\left(\frac{{\theta }_o-{\theta }_1}{{\theta }_o+{\theta }_1}\right)$
• C. $\frac{m}{2}\left(\frac{{\theta }_o+{\theta }_1}{{\theta }_o-{\theta }_1}\right)$
• D. $m\left(\frac{{\theta }_o-{\theta }_1}{{\theta }_o+{\theta }_1}\right)$

Answer 1

The correct option is D $m\left(\frac{{\theta }_o-{\theta }_1}{{\theta }_o+{\theta }_1}\right)$

$u=\sqrt{2gl(1-cos{\theta }_o)}$ ........ (1)

$v$ = velocity of ball after collision

$v=\frac{m-M}{m+M}u$

Since ball rises upto height ${\theta }_1$

$v=\sqrt{2gl(1-cos{\theta }_1)}=\frac{m-M}{m+M}u$ ......... (2)

From (1) and (2) $\frac{m-M}{m+M}=\frac{1-cos{\theta }_1}{1-cos{\theta }_0}=\frac{sin\frac{{\theta }_1}{2}}{sin\frac{{\theta }_0}{2}}$

$\frac{M}{m}=\frac{{\theta }_0-{\theta }_1}{{\theta }_0+{\theta }_1}$

$M=\left(\frac{{\theta }_0-{\theta }_1}{{\theta }_0+{\theta }_1}\right)m$