Q. A shell is fired from a cannon with velocity $\ v\ m/sec$ at an angle $\theta$ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed in $m/sec$ of the other piece immediately after the explosion is
A) $3v\cos\theta$
B) $2v\cos\theta$
C) $\frac{3}{2}v\cos\theta$
D) $\frac{\sqrt{3}}{2}v\cos\theta$
Explanation
(a) Shell is fired with velocity $v$ at an angle $\theta$ with the horizontal
So its velocity at the highest point
= horizontal component of velocity $= v\cos\theta$
So momentum of shell before explosion $= mv\cos\theta$
When it breaks into two equal pieces one piece retraces its path to the canon, then other part moves with velocity $V$
So momentum of two pieces after explosion
$= \frac{m}{2}\ \left( - v\cos\theta \right) + \frac{m}{2}V$
By the law of conservation of momentum
$mv\cos\theta = \frac{- m}{2}\ v\cos\theta + \frac{m}{2}\ V \Rightarrow V = 3v\cos\theta$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.