Question

A bob of mass $m$ attached with a string of length $l$ tied to a point on ceiling is released from a position when its string is horizontal. At the bottom most point of its motion, an identical mass $m$ gently stuck to it. Find the maximum angle from the vertical to which it rotates:

• A. $co{s}^{-1}\left(\frac{2}{5}\right)$
• B. $co{s}^{-1}\left(\frac{3}{4}\right)$
• C. $co{s}^{-1}\left(\frac{1}{4}\right)$
• D. ${60}^o$

Answer 1

The correct option is B $co{s}^{-1}\left(\frac{3}{4}\right)$

Using conservation of mechanical energy,

$mgl=\frac{1}{2}m{v}_m^2$

$v_m=\sqrt{2gl}$
From conservation of linear momentum,
$v_{2m}=\frac{v_m}{2}=\frac{\sqrt{2gl}}{2}$

According to the problem,

$v^{\prime }=0$
From diagram, $h=l(1-cos\theta )$

Again applying conservation of energy,

$\frac{1}{2}\left(2m\right)(v_{2m}{)}^2=2mgh$
$\therefore \frac{l}{4}=l(1-cos\theta )$
$\therefore \theta =co{s}^{-1}\left(\frac{3}{4}\right)$