question_answer

A body of mass \[(4\,w)\] is lying in \[xy-\]plane at rest. It suddenly explodes into three pieces. Two pieces each of mass (m) move perpendicular to each other with equal speeds \[(\upsilon )\]. The total kinetic energy generated due to explosion is [NEET 2014]

A) \[m{{v}^{2}}\]

B) \[\frac{3}{2}m{{v}^{2}}\]

C) \[2m{{v}^{2}}\]

D) \[4\,m{{v}^{2}}\]

Correct Answer: B

Solution :

According to question, the third part of mass 2 m will move as shown in the figure, because the total momentum of the system after explosion must remain zero. Let the velocity of third part is \[v'\].

From the conservation of momentum

\[\sqrt{2}(mv)=(2m)\times v'\Rightarrow v'=\frac{v}{\sqrt{2}}\]

\[\Rightarrow \]   So total kinetic energy generated by the explosion

\[=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}m{{v}^{2}}+\frac{1}{2}(2m)v{{'}^{2}}\]

\[=m{{v}^{2}}+m\times {{\left( \frac{v}{\sqrt{2}} \right)}^{2}}\]

\[=m{{v}^{2}}+\frac{m{{v}^{2}}}{2}=\frac{3}{2}m{{v}^{2}}\]