Explanation

(b) Applying law of conservation of liner momentum, $ie,$
$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$
$\ Here,\ m_{1} = 10\ g = 10^{- 2}\ kg,\ m_{2} = 5\ kg$
$u_{1} = 300\ {ms}^{- 1},\ u_{2} = 0$
$v_{1} = 0,\ v_{2} = ?$
$\therefore\ \ 10^{- 2} \times 300 + 5 \times 0 = 10^{- 2} \times 0{+ 5v}_{2}$
$or\ \ {5v}_{2} = 3$
$or\ \ v_{2} = \frac{3}{2}\ {ms}^{- 1}$
$= 60\ c{ms}^{- 1}$