Q. A 5 kg stationary bomb is exploded in three parts having mass 1: 1: 3 respectively. Parts having same mass move in perpendicular directions with velocity 39$\ ms^{- 1}$, then the velocity of bigger part will be
A) $10\sqrt{2}\ ms^{- 1}$
B) $\frac{10}{\sqrt{2}\ }ms^{- 1}$
C) $13\sqrt{2}\ ms^{- 1}$
D) $\frac{15}{\sqrt{2}\ }ms^{- 1}$
Explanation
(c) As $m_{1}:m_{2}:m_{3} = 1:1:3$
and momentum is conserved,
$\therefore\ \sqrt{p_{1}^{2} + p_{2}^{2} + p_{3}^{2}} = 3v_{3}$
$\sqrt{1 \times 39^{2} + 1 \times 39^{2}} = 3v_{3}$
$39\sqrt{2} = 3v_{3}$
$v_{3} = \frac{39\sqrt{2}}{3} = 13\sqrt{2}ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.