Q. A particle is projected with 200$\ ms^{- 1}$, at an angle of 60${^\circ}$. At the highest point it explodes into three particles of equal masses. One goes vertically upward with velocity 100$\ ms^{- 1},$ the second particle goes vertically downward with the same velocity as the first. Then what is the velocity of the third particle?
Explanation
(d) At the highest point momentum of particle before explosion $\overrightarrow{p}\text{=}mv\cos{60{^\circ}}$
$= m \times 200\frac{1}{2} = 100m$ horizontally
Now as three is no external force during explosion, hence
$\overrightarrow{p} = {\overrightarrow{p}}_{1} + {\overrightarrow{p}}_{2} + {\overrightarrow{p}}_{3}$
However, since velocities of two fragments, of masses $m/3$ each, are 100 $ms^{- 1}$ downward and $100ms^{- 1}$ upward.
hence, ${\overrightarrow{p}}_{1} = - {\overrightarrow{p}}_{2}$ or ${\overrightarrow{p}}_{1} + {\overrightarrow{p}}_{2}$
${\overrightarrow{p}}_{3} = \frac{m}{3}.v_{3} = {\overrightarrow{p}}_{\ } = 100$ m horizontally
$v_{3} = 300ms^{- 1}$ horizontally
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.