Question

If a bomb of mass 3m kg explodes into two pieces of mass m kg and 2m kg, the velocity of mass m kg is 16 $m/s$ then the total kinetic energy released in the explosion will be,
A) 96 mJ.
B) 116 mJ.
C) 192 mJ.
D) 384 mJ.

Answer 1

Hint
Since there is no external force acting on the system, hence we can conserve momentum. Once done, find out the change in kinetic energy because that is the total energy released in the explosion.

Complete step by step answer
Given that the velocity of m mass is ${\text{16m}}{{\text{s}}^{{\text{ - 1}}}}$ and let the final velocity of 2m kg mass be V.
Now initially the system was at res, so initial momentum of the system = 0. Since no external force is acting on the system, therefore the momentum must be conserved.
i.e. ${P_i} = {P_f}$
Putting the equations and their values we have,
$\Rightarrow 0 = 16 \times m - 2m \times V $
$\Rightarrow 2mV = 16m $
$\Rightarrow V = 8{\text{m}}{{\text{s}}^{{\text{ - 1}}}} $
Now since initially the system was at rest, hence the initial kinetic energy is 0. The final kinetic energy of the system is,
$K{E_f} = \dfrac{1}{2}m \times {(16)^2} + \dfrac{1}{2}(2m) \times {(8)^2} $
$\Rightarrow K{E_f} = 192{\text{m J}} $
Now energy released due to explosion = Change in kinetic energy of the system,
$\Rightarrow \Delta KE = K{E_f} - K{E_i} $
$\Rightarrow \Delta KE = 192{\text{m J}} - 0 $
$\Rightarrow \Delta KE = 192{\text{m J}} $
Therefore, the correct answer is option (C).

Note
In this type of question, always see if there is any external force acting on the system or not. If no such external force exists, then the momentum of the system should be conserved and then solved.