Question

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 'v' each in mutually perpendicular directions. The minimum energy released in the process of explosion is

• A. $\left(2/3\right)m{v}^2$
• B. $\left(3/2\right)m{v}^2$
• C. $\left(4/3\right)m{v}^2$
• D. $\left(3/4\right)m{v}^2$

Answer 1

The correct option is B $\left(3/2\right)m{v}^2$

From law of conservation of momentum the velocity v of the third fragment of mass 2m is given by

$\left(2m\right)V=\sqrt{(mv{)}^2+(m{v}^2)}$ $\therefore V=\frac{v}{\sqrt{2}}$

Hence,energy released $\Delta E=\frac{1}{2}\left(2m\right)V^2+\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2=\frac{3}{2}m{v}^2$