A Bomb of mass 16kg at rest explodes into two pieces of masses 4kg & 12kg.  The velocity of the mass 12kg is 4 m/s.   The K.E. of the other mass is

Linear momentum is conserved.

Apply principle of conservation of linear momentum
 $\therefore 0=m_1{v}_1+m_2{v}_2$

$\mathrm{here} {\mathrm{m}}_1=4\mathrm{kg};{\mathrm{m}}_2=12\mathrm{kg};{\mathrm{v}}_2=4\mathrm{m}/\mathrm{s}$

On substituting 
 $$\begin{gathered} \Rightarrow \hspace{0.17em}4\times v_1+12\times 4=0 \\ \Rightarrow \hspace{0.17em}{v}_1=-12m/s \end{gathered}$$     
  $\therefore K.E_1=\frac{1}{2}{m}_1{v_1}^2$ 

On substituting 
 $$\begin{gathered} K.E_1=\frac{1}{2}\left(4\right)\times {(-12)}^2 \\ K.E_1=288 joule \end{gathered}$$