How Far Does the Fragment Land After a Mid-Air Explosion?

[cb]

I need some help with the problem.

A shell is shot with an initial velocity v0 of 20 m/s, at an angle of 60 degrees with the horizontal. At the top of the trajectory, the shell explodeds into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and the air drag is negligible


First, I labled three parts of the trajectory.

1. At the initial position

K= (1/2) m * v^2
U = 0

2. At the max height of the trajectory

K = (1/2) (m/2) * v^2 ----> Because the shell exploded in half
U = (m/2) * g * h

3. At the end of the the trajectory

K = 0
U = 0


Then using the conservation of energy from points 1 to 2 I found the velocity at point 2 to be 22.36 m/s.

Now I have to finish the problem and find the total distance traveled. Can someone check my work up to now and give me some advice on how to find the total distance.

 

[Ba]

the kinetic energy does not equal 0 at the end of the trajectory unless you are willing to calculate the work done by the Earth in slowing the piece to rest. Consider the moment that it touches the earth, not the impact.

 

[cb]

True, but now I still have to solve for the total distance traveled.

I was thinking about finding the height with the equation h= v0^2* sin^2(alpha) / (2*g). But since the mass is cut in half, will that make the equation invalid?

If that is true I can find the distance for the first half. But I'm not sure how I would find the distance for the second half.

 

[Soveraign]

Use conservation of momentum to find out the new velocity of the half of the shell still moving horizontally. Keep in mind you started out with some P=Mv (where v is the horizontal component of the velocity). Now you have P = mv1 + mv2, where m = M/2. P must be the same in both situations and you already know v1 is zero (at the instant of the explosion). Find v2.

So now you have the new velocity of the shell piece and, treating this now as a typical projectile motion problem, you should be able to find the additional distance traveled.