Question

A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with speed v each in mutually perpendicular directions. The total energy released in the process of the explosion is-
(A) $\dfrac{2}{3}m{{v}^{2}}$
(B) \[\dfrac{3}{2}m{{v}^{2}}\]
(C) \[\dfrac{4}{3}m{{v}^{2}}\]
(D) \[\dfrac{3}{4}m{{v}^{2}}\]

Answer 1

Hint: Apply the law of conservation of momentum and form equation. Find the speed of the third fragment which has a mass 2m. The total energy released is the difference of final kinetic energy and initial kinetic energy.

Complete step by step answer:
Let the speed of the third fragment 2m mass be u.
Let the direction of one fragment be x-axis then the other fragment direction will be y-axis since both are perpendicular.
Momentum before collision=Momentum after collision
$4m\times 0=m\times \left( v\hat{i} \right)+m\times \left( v\hat{\pm } \right)+2m\times u$
By solving, we get:
$u=\dfrac{v}{\sqrt{2}}$
Now we will find the initial and final kinetic energy of the system.
Initial Kinetic Energy $K{{E}_{i}}=\dfrac{1}{2}\left( 4m \right)\left( 0 \right)=0$
Final Kinetic Energy \[K{{E}_{f}}=\dfrac{1}{2}\left( m{{v}^{2}}+m{{v}^{2}}+2m\dfrac{{{v}^{2}}}{2} \right)=\dfrac{3}{2}m{{v}^{2}}\]
Therefore, Kinetic energy released is
\[\begin{align}
& =K{{E}_{f}}-K{{E}_{i}} \\
& =\dfrac{3}{2}m{{v}^{2}}-0 \\
& =\dfrac{3}{2}m{{v}^{2}} \\
\end{align}\]
Hence, option (B) is the correct answer.

Note:Remember that we can apply the principle of conservation of momentum only when no external force is applied on the system.Principle of conservation of momentum said that the total momentum of the system remains constant provided that no external force acts on it.