Question

A body of mass 4 m at rest explodes into three pieces. Two of the pieces each of mass m move with a speed v each in mutually perpendicular directions. The total kinetic energy released is

• A. $\frac{1}{2}m{v}^2$
• B. $m{v}^2$
• C. $\frac{3}{2}m{v}^2$
• D. $\frac{5}{2}m{v}^2$

Answer 1

The correct option is C $\frac{3}{2}m{v}^2$

By momentum conservation, $2m{v}_x=mv$

$v_x=\frac{v}{2}$

$2m{v}_y=mv$

$v_y=\frac{v}{2}$

$u=\sqrt{v_x^2+v_y^2}=\sqrt{\frac{2v^2}{4}}$=$\frac{v}{\sqrt{2}}$

Kinetic energy release=$\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+2m{\left(\frac{v}{2}\right)}^2$=$\frac{3}{2}m{v}^2$