Two balls collide bounce off each other as shown in the figure. The 1kg ball has a speed of $\text{10}\text{cm/s}$ after collision. The velocity of the $\text{0}\text{.5}\text{kg}$ ball will be               

${\text{m}}_{\text{1}}=1=1000\text{g,}{\text{m}}_{\text{2}}=0.5\text{kg}=500\text{g}$

${\text{u}}_{\text{1}}=20\text{cm/s,}{\text{u}}_{\text{2}}=30\text{cm/s}$

${\text{v}}_{\text{1}}=10\text{cm/s,}{\text{v}}_{\text{2}}=?$

Along horizontal direction :-

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

$\Rightarrow 1000\left(20\right)+500\left[30\right]=100\times 10\times \frac{\sqrt{3}}{2}+{\text{500v}}_{\text{2}}\cos \theta$

$20+\left(-15\right)=5\sqrt{3}-0.5{\text{v}}_{\text{2}}\cos \theta$

$1=\sqrt{3}-0.1{\text{v}}_{\text{2}}\cos \theta$

${\text{v}}_{\text{2}}\cos \theta =7.32$

${\text{v}}_{\text{2}}^{\text{2}}{\cos }^2\theta =53.58\to \left(1\right)$

Along normal : -

$0+0={\text{m}}_{\text{2}}{\text{v}}_{\text{2}}\sin \theta +{\text{m}}_{\text{1}}{\text{v}}_{\text{1}}\sin 30°$

$0=0.5{\text{v}}_{\text{2}}\sin \theta +1\left(10\right)\frac{\text{v}}{\text{2}}$

${\text{v}}_{\text{2}}\sin \theta =10$

${\text{v}}_{\text{2}}^{\text{2}}{\sin }^2\theta =100\to \left(2\right)$

$\left(1\right)+\left(2\right)$

${\text{v}}_{\text{2}}^{\text{2}}=153.58$

${\text{v}}_{\text{2}}=12\text{cm/s}$