Question

A bomb of mass 2 kg which is at rest explodes into three fragments of equal masses. Two of the fragments are found to move with a speed of 1 m/s each in mutually perpendicular direction. The total energy released during explosion is then

• A. 1.5 J
• B. 2 J
• C. 4 J
• D. 5 J

Answer 1

The correct option is B 2 J

$ifthevelocityof3rdfragmentisv$

$then,$

$mv=\sqrt{{(m\times 1)}^2+{(m\times 1)}^2}$

$v=m{s}^{-1}$

$energyreleased$

$=\frac{1}{2}\left(1\right)(1{)}^2+\frac{1}{2}\left(1\right)(1{)}^2+\frac{1}{2}\left(1\right)(\sqrt{2}{)}^2$

$=2J$