Center of mass problem involving shell

[mmattson07]

Homework Statement

A shell is shot with an initial velocity 0 of 23 m/s, at an angle of θ0 = 54° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (Fig. 9-42). 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 that air drag is negligible?

http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c09/fig09_46new.gif

Homework Equations

consv momentum
M v = m1 u1 + m2 u2

The Attempt at a Solution

Here is what I tried

Init vertical speed= 23sin54= 18.61m/s
Init horiz speed= 23cos54=13.53m/s

Time to reach top of trajectory= 18.61/9.8= 1.899s
horiz distance to top= 13.52*1.899= 25.67m

The other shell will now have the horiz speed of 23 m/s and fall for the same time it took to rise so it goes another 23*1.899=43.677m

I added both these to get 69.347m but it isn't correct

 

[BigJDubs]

. What am I doing wrong?You have to consider the conservation of momentum. The initial momentum before the explosion is Mv. After the explosion, the final momentum is m1u1 + m2u2. Since the mass of each fragment is the same, m1 = m2. Therefore, m1u1 + m2u2 = Mv => m1(u1 - v) + m2(u2 - v) = 0 => u1 - v = -(u2 - v) => u1 = 2v - u2 From this equation, you can find the velocity of the second fragment once it leaves the explosion and then use kinematics equations to calculate the distance it travels.