Q. A canon ball is fired with a velocity $200\ m/sec$ at an angle of $60{^\circ}$ with the horizontal. At the highest point of its flight it explodes into 3 equal fragments, one going vertically upwards with a velocity $100\ m/sec$, the second one falling vertically downwards with a velocity $100\ m/sec$. The third fragment will be moving with a velocity
Explanation
(b) Momentum of ball (mass $m$) before explosion at the highest point $= mv\widehat{i} = mu\cos 60{^\circ}\widehat{i}$
$= m \times 200 \times \frac{1}{2}\widehat{i} = 100\ m\widehat{i}\ kgms^{- 1}$
Let the velocity of third part after explosion is $V$
After explosion momentum of system $= {\overrightarrow{P}}_{1} + {\overrightarrow{P}}_{2} + {\overrightarrow{P}}_{3}$
$= \frac{m}{3} \times 100\widehat{j} - \frac{m}{3} \times 100\widehat{j} + \frac{m}{3} \times V\widehat{i}$
By comparing momentum of system before and after the explosion
$\frac{m}{3} \times 100\widehat{j} - \frac{m}{3} \times 100\widehat{j} + \frac{m}{3}V\widehat{i} = 100mi \Rightarrow V = 300m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.