Two balls collide bounce off each other as shown in the figure. The 1kg ball has a speed of 10cm/s after collision. The velocity (in cm/s) of the 0.5kg ball will be ____ .

$\left[\mathrm{Take} \sqrt{3}=1.73 ; \sqrt{1.54}=1.24\right]$

Conserving momentum along vertical

$-1\times 10\sin 30+0.5\times \mathrm{Vsin}\theta =0$

$\mathrm{Vsin}\theta =10.....\left(1\right)$

Conserving momentum along horizontal

$1\times 20-0.5\times 30=1\times 10\cos 30-0.5\mathrm{Vcos}\theta$

$\mathrm{Vcos}\theta =10(\sqrt{3}-1).....\left(2\right)$

$(1{)}^2+(2{)}^2={\mathrm{V}}^2={10}^2+\left[10\right(\sqrt{3}-1){]}^2$

$\Rightarrow \mathrm{V}=12.4\mathrm{cm}/\mathrm{s}$