A bomb of mass 3.0 ~kg explodes in air into two pieces of masses...

Velocity of second piece can be find out using conservation of momentum. $$\begin{aligned}& \mathrm{O}=m_1 v_1+m_2 v_2=2 \timesv_1+1 \times80 \\ \Rightarrow \quad & v_1=-40 \mathrm{~m}/ \mathrm{s} . \end{aligned}$$Negative sign showing that particle is moving in opposite direction of other particle direction. Energy imparted to the fragments are converted into their kinetic energy. $$\begin{aligned} \therefore \quad\text{ Total energy }& =\frac{1}{2}m_1 v_1^2+\frac{1}{2}m_2 v_2^2 \\ & =\frac{1}{2} \times2 \times₁₆₀₀⁺\frac{1}{2} \times1 \times6400 \\ & =4800 \mathrm{~J}=4.8 \mathrm{~kJ} . \end{aligned}$$