Q. A shell is fired from cannon with a velocity $v\ {ms}^{- 1}$ at an angle of $\theta$ with the horizontal direction. At the highest point in its path, it explodes into 2 pieces of equal masses. One of the pieces retraces its path to the cannon. The speed of the other piece immediately after the explosion is
A) $\sqrt{3/2}\ v\cos\theta$
B) $v\cos\theta$
C) $2v\cos\theta$
D) $3v\cos\theta$
Explanation
(d) At the highest point, velocity of shell before explosion
$= v\cos\theta$
$\therefore\ \ Momentum\ $of shell before explosion
$= mv\cos\theta\ $
After explosion momentum
$= - \frac{m}{2}v\cos{\theta + \frac{m}{2}V}$
According to the law of conservation of momentum
$mv\cos{\theta = - \frac{mv\cos\theta}{2} + \frac{m}{2}V}$
$\therefore\ \ \ V = 3v\cos\theta$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.