Question

A shell is hired from a cannon with velocity v m/s 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 the speed of the other piece immediately after explosion is

Answer 1

Let $2m$ be the mass of the cannon which is projected with velocity $v$ at an angle $\theta$ with the horizontal. The horizontal component of the velocity of the cannon is $v_x=v\cos \theta$. This component remains constant.

At the maximum height, the cannon has only this component as the vertical component is momentarily zero.

So, momentum before exploding $=2mv\cos \theta$

After exploding, one of the masses retraces the path of the cannon backwards. This is possible on if the part has velocity$-v\cos \theta$

Therefore, momentum after explosion$=-mv\cos \theta +mu$

So, by conservation of momentum,

$2mv\cos \theta =-mv\cos \theta +mu$

$u=\left(3v\cos \theta \right)m/s$