Q. A body of mass $M$ at rest explodes into three pieces, two of which of mass $M/4$ each are thrown off in mutually perpendicular directions with speeds of 3$ms\ ^{- 1}$ and 4$ms^{- 1}$ respectively. Then the third piece will be thrown off with a speed of
A) 1.5 $ms^{- 1}$
B) 2$\ ms^{- 1}$
C) 2.5$\ ms^{- 1}$
D) 3.0$\ ms^{- 1}$
Explanation
(c) As there is no external force, hence
$\overrightarrow{p} = {\overrightarrow{p}}_{1} + {\overrightarrow{p}}_{2} + {\overrightarrow{p}}_{3} =$ constant
$\Rightarrow \ \ \left| {\overrightarrow{p}}_{3} \right| = \left| {\overrightarrow{p}}_{1} + {\overrightarrow{p}}_{2} \right| = \frac{m}{4}\sqrt{(3)^{2} + (4)^{2}} = \frac{5m}{4}$
[since $v_{1}$ and $v_{2}$ are mutually perpendicular]
$\therefore\ p_{3} = \frac{m}{2}v_{3} = \frac{5m}{4}$
$\Rightarrow \ \ v_{3} = \frac{5}{2} = 2.5ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.