A simple pendulum of length 0.2m has bob of mass 5gm, it is pulled aside through an angle 60° from the vertical. A spherical body of mass 2.5 gm is placed at the lowest position of the bob. When the bob is released it strikes the spherical body and comes to rest. What is the velocity of the spherical body after collision?$\left(\text{g}=9.8{\text{ms}}^{-2}\right)$  (in$\text{m/s}$ )           

$\Delta \text{OAB}$

$\cos \theta =\frac{x}{l}$

$\text{OC}=l$

$\text{OA}=x$

$\text{AC}=\text{h}\Rightarrow l-x$

$\text{h}=l-l\cos \theta$

$\text{h}=l-l\cos \theta$

$=l\left(1-\cos \theta \right)$

$\text{h}=l\left(1-\frac{1}{2}\right)\left[\because \cos 60°=\frac{1}{2}\right]$

$=\frac{l}{2}$

Total energy at B = T.E at ‘C’

$\text{mgh}=\frac{1}{2}{\text{mu}}_{\text{C}}^{\text{2}}$

${\text{u}}_{\text{C}}=\sqrt{\text{2gh}}$

$=\sqrt{2\times g\times \frac{l}{2}}$

${\text{u}}_{\text{C}}=\sqrt{\text{g}l}$

${\text{u}}_{\text{bob}}=\sqrt{\text{g}l}$

${\text{u}}_{\text{mass}}=0$

${\text{m}}_{\text{1}}={\text{m}}_{\text{bob}}=5\times {10}^{-3}\text{kg}$

${\text{m}}_2={\text{m}}_{\text{mass}}=2.5\times {10}^{-3}\text{kg}$

${\text{v}}_1=0,{\text{v}}_{\text{2}}=?$

${\text{m}}_{\text{1}}{\text{u}}_{\text{1}}+{\text{m}}_{\text{2}}{\text{u}}_{\text{2}}={\text{m}}_{\text{1}}{\text{v}}_{\text{1}}+{\text{m}}_{\text{2}}{\text{v}}_{\text{2}}$

$5\times {10}^{-3}\times \sqrt{\text{g}l}+0=0+2.5\times {10}^{-3}{\text{v}}_{\text{2}}$

${\text{v}}_{\text{2}}=2\sqrt{\text{g}l}$

${\text{v}}_{\text{2}}=2\sqrt{9.8\times 0.2}$

$=2\sqrt{9.8\times 2\times {10}^{-2}}$

$=2\sqrt{49\times 2\times 2\times {10}^{-2}}$

$=2\times 7\times 2\times {10}^{-1}$

${\text{v}}_{\text{2}}=2.8$