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 explosion is-

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

Answer 1

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

By Conservation of momentum, the net momentum must be 0
The total momentum of 2 m masses is $M=\sqrt{2m^2{v}^2}=\sqrt{2}mv$
$\therefore 2m{v}^{\prime }=\sqrt{2}mv⟹v^{\prime }=v/\sqrt{2}$
hence $KE=m{v}^2+m{v}^2/2=3m{v}^2/2$