Explanation

(d) From law of conservation of momentum, when no external force acts upon a system of two (or more) bodies, then the total momentum of the system remains constant.
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Momentum before explosion =momentum after explosion.
since bomb v at rest, its velocity is zero, hence,
$mv = m_{1}v_{1} + m_{2}v_{2}$
$3 \times 0 = 2v_{1} + 1 \times 80$
$or\ v_{1} = - \frac{80}{2} = - 40ms^{- 1}$
Total energy imparted is
$KE = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}$
$= \frac{1}{2} \times 2 \times ( - 40)^{2} + \frac{1}{2} \times 1 \times (80)^{2}$
=1600+3200=4800J
=4.8kJ