question_answer

A particle of mass 4 m which is at rest explodes into three fragments. Two of the fragments each of mass 'm' are found to move with a speed of 'v' each in mutually perpendicular directions. The total energy released in this process is ??..

Answer:

The resultant momentum of the particles of masses m and \[m=\sqrt{2}mv.\] Let x be the velocity of particle of mass 2 m. Applying the law of conservation of momentum \[:0=\sqrt{2}mv+2mx\Rightarrow x=-\frac{V}{\sqrt{2}}\] \[\therefore \]KE of three particles = \[\frac{1}{2}m{{v}^{2}}+\frac{1}{2}m{{v}^{2}}+\frac{1}{2}(2m)\left( \frac{v}{\sqrt{2}} \right)=\frac{3m{{v}^{2}}}{2}\]