Question

A body of mass $4m$ 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 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

Answer: (C)

$\frac{3}{2}m{v}^2$

By momentum conservation

$2mv\text{'}=2–\sqrt{m{v}^{2}mv\text{'}}=2mV$

$v\text{'}=v^2–\sqrt{v\text{'}}=v^2$

$KE=2l^{2}KE=2l^{2}m{v}^2+12m{v}^2+12\left(2m\right)v^{2}2\left(2m\right)v^{2}2$

$=m{v}^2+m{v}^{2}2=m{v}^2+m{v}^{2}2$

$=\frac{3}{2}m{v}^2$