Explanation

(b) The linear momentum of exploding part will remain conserved.
Applying conservation of linear momentum, We write,
$m_{1}u_{1} = m_{2}u_{2}$
Here,$m_{1} = {18kg,m}_{2} = 12kg$
$u_{1} = 6ms^{- 1},u_{2} = ?$
$\therefore 18 \times 6 = 12\ u_{2}$
$\Longrightarrow {\ u}_{2} = \frac{18 \times 6}{12}\ \ 9ms^{- 1}\ $
Thus, kinetic energy of 12 kg mass
$k_{2} = \frac{1}{2}m_{2}u_{2}^{2}$
$= \frac{1}{2} \times 12 \times (9)^{2}$
=$6 \times 81\ $
=486 J