A bomb of mass 3m kg explodes into two pieces of mass m kg and 2m kg. If the velocity of m kg mass is 16 m/sec, the total kinetic energy released in the explosion is
A)192 mJ
B)96 mJ
C)384 mJ
D)768 mJ
Correct Answer:
A
Solution :
By the conservation of momentum, \[{{m}_{A}}{{\nu }_{A}}={{m}_{B}}{{\nu }_{B}}\] \[\Rightarrow \,\,\,\operatorname{m} \times 16=2m\,\,\times \,\,{{\nu }_{A}}\,\,\Rightarrow \,\,{{\nu }_{B}}=8 m/sec\] \[=\frac{1}{2}{{m}_{A}}{{\nu }_{A}}^{2}+\frac{1}{2}{{m}_{B}}{{\nu }_{B}}^{2}\] \[=\frac{1}{2}\times m\times {{(16)}^{2}}+\frac{1}{2}\times (2m)\times {{8}^{2}}=192\,\,mJ\]