# Horizontal Speed of Fragment

1. Nov 3, 2008

### shimizua

1. The problem statement, all variables and given/known data
A 14.0-kg shell is fired from a gun with a muzzle velocity 160.0 m/s at 27.0o above the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. What is the horizontal speed of the other fragment?

2. Relevant equations
The velocity is 160/s at an angle of 27deg - velocity must always have a speed and direction.
That's equivalent to 260 sin(27) vertically and 160 cos(27) horizontally. You can see this if you draw a triangle.

A nice feature about physics is that you can treat forces at right angles independantly so we can ignore the horizontall bit fro now and just look at the vertical.
It starts off going up at 125sin(33) and slows down at 9.8m/s^2
v^2 = u^2 + 2 a s At the top when u=0 and a = -9.8
160sin() ^2 = 2 * 9.8 * s where s (the height) = 164.31

The kinetic energy is = 1/2 m v^2 and potential energy is = m g h
At the top of the curve you have used up some of the initial KE as PE - so we now work out how much KE is left.
KE = 1/2m v^2 - mgh = 1/2* M * 160^2 - M * 9.8 * 164.31 = M ( 1/2*160^2 - 9.8*164.31).

Now the shell is only moving horizontally at the top of the curve so all this kinetic energy is going to go into the horizontal velocity.
But the mass has just halved - so:

M * those numbers = 1/2 1/2 M V^2 where V is the new velocity.

3. The attempt at a solution
so i tried doing this and it came up wrong. i saw that someone had already asked this question before but i dont know what i am doing wrong. i also just tried doing m1v1=m2v2 and that was wrong also

2. Nov 3, 2008