A cannon fires a shell with a velocity V at an angle relative to the horizontal direction. Upon reaching the highest point of its trajectory, the shell explodes into two pieces of equal mass. One of the pieces retraces its path back to the cannon. What is the speed of the other piece immediately following the explosion?

Step-by-step solution

At the highest point of the trajectory, the vertical component of the velocity of the shell is zero. The velocity of the shell is purely horizontal and is given by v = V . The momentum of the shell just before the explosion is M V , where M is the mass of the shell. After the explosion, the shell breaks into two equal pieces, each of mass M 2 . Since one of the pieces retraces its path back to the cannon, its velocity must be exactly reversed. Therefore, the velocity of this piece just after the explosion is -V . L