Question

A shell is fired from a cannon with velocity $vm{s}^{-1}$ 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 $m{s}^{-1}$ of the other piece immediately after the explosion is :

• A. $3v\cos \theta$
• B. $2v\cos \theta$
• C. $\frac{3}{2}\cdot v\cos \theta$
• D. $\sqrt{\frac{3}{2}\cdot v\cos \theta }$

Answer 1

The correct option is A $3v\cos \theta$

velocity of shell at highest point is $vcos\theta$
since first half got back to canon on the same path it came by. so its velocity after explosion was $vcos\theta$ backwards.
conserving momentum.
$mv\cos \theta =\left(m/2\right)(-v\cos \theta )+\left(m/2\right)v_1$
$v_1=3v\cos \theta$