Q. A shell is fired from a cannon with velocity $v\ ms^{- 1}$ 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/s of the piece immediately after the explosion is
A) $3v\cos\theta$
B) 2$v\cos\theta$
C) $\frac{3v}{2}\cos\theta$
D) $\frac{\sqrt{3}v\cos\theta}{2}$
Explanation
(a) In Case of projectile motion at the highest point
$(v)_{vertical} = 0$ and $(v)_{horizontal} = v\cos\theta$
The initial linear momentum of the system will be $mv\cos\theta.$ Now as force of blasting is internal and force of gravity is vertical
So linear momentum of the system along horizontal is conserved
$p_{1} + p_{2} = mv\cos\theta$
$m_{1}v_{1} + m_{2}v_{2} = mv\cos\theta$
But it is given that $m_{1} = m_{2} = \frac{m}{2}$ and as one part retraces its path,
$v_{1} = - v\cos\theta$
$\therefore$ $\frac{1}{2}m\left( - v\cos\theta \right) + \frac{1}{2}mv^{2} = mv\cos\theta$
or $v_{2} = 3v\cos\theta$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.