Q. A shell is fired from a cannon with a velocity $v$ at an angle $\theta$ with the horizontal direction. At the highest point in its path, it explodes into two pieces, one retraces its path to the cannon and the speed of the other piece immediately after the explosion is
A) $3v\cos\theta$
B) $2v\cos\theta$
C) $\left( \frac{3}{2} \right)v\cos\theta$
D) $\left( \frac{\sqrt{3}}{2} \right)v\cos\theta$
Explanation
(a) As we know that at the highest point, the shell has only the horizontal component of velocity which is $v\cos\theta$. If $u$ be he velocity of second exploded piece, then applying conservation of linear momentum along $x$-axis
$\therefore\ 2mv\cos{\theta = - mv\cos{\theta + mu}}$
Or $u = 3v\cos\theta$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.