In an explosion, a body breaks up into two pieces of unequal masses. Which of the statement is correct?
A. Both parts will have numerically equal momentum
B. Lighter part will have more momentum
C. Heavier part will have less momentum
D. Both parts will have equal kinetic energy

Hint:Let’s have a look at momentum. The product of the mass of a particle and its velocity is called momentum. Since it has both magnitude and direction the momentum is a vector quantity. Newton's second law of motion also refers to the momentum and states that the rate of change of momentum is equal to the force acting on the particle i.e., \[\overrightarrow P = mv\]

Formula Used:
The formula to find the momentum is,
\[\overrightarrow P = mv\]
Where, m is mass of body and v is momentum.

Complete step by step solution:
Consider that we have a body of mass M which is at rest having the velocity of \[{v_0}\]. Now this body of mass M explodes and becomes two pieces of unequal masses of say \[{m_1}\] and \[{m_2}\] having the velocities \[{v_1}\] and \[{v_2}\]. Here in the explosion, the momentum of a body will be conserved, that is the initial momentum of a body is equal to the final momentum of the body.
\[\overrightarrow {{P_i}} = \overrightarrow {{P_f}} \\ \]
Since, we know that initial momentum is zero,
\[0 = \overrightarrow {{P_1}} + \overrightarrow {{P_2}} \]
\[\Rightarrow 0 = {m_1}{v_1} + {m_2}{v_2}\]
\[\therefore {m_1}{v_1} = - {m_2}{v_2}\]
This shows that the direction of momentum is different but the magnitude will remain the same. Therefore, both parts will have numerically equal momentum.

Hence, option A is the correct answer.

Note:When the initial momentum of a particle is equal to the final momentum of a particle, then we can say that the momentum is conserved. This is nothing but the law of conservation of momentum.