Physics bomb explosion problem... answer gets 10 points!

A bomb with 1.0x10^4 J explodes into three pieces.
The first piece is 1.0 kg and travels in the positive y direction with a velocity of 60 m/s.
The second piece is 4.0 kg and travels in the positive x direction at 40 m/s.
Find the third piece's mass, velocity, and angle.

Diagram:
http://www.flickr.com/photos/82417987@N08/7545812162/in/photostream

Some Equations:
momentum(p) = mass*velocity
p(before) = p(after)
Potential Energy = Kinetic Energy

Thanks in advance. Best, explained answer gets 10 points!
If you can start me off on the right foot that's really helpful too

You are going to need to solve equations simultaneously.

First conservation of energy. The KE of the pieces should equal the given potential energy.
KE = 1/2*m*v^2
1.0*10^4 = 1/2*(1)*(60)^2 + 1/2*(4)*(40)^2 + 1/2*(m)*(v)^2 <- where m and v are the unknown mass and velocity

For conservation of momentum, we assume the bomb was stationary before explosion:
Now apply conservation of momentum in the x-direction:
The first piece is only in y, so it is not included in the equation. The x-mom eqn becomes:
4*40 + m_x*v_x = 0

Now the y-mom equation:
1*60 +m_y*v_y = 0

At first it looks like there are too many unknowns for only 3 eqns. But consider that m_x=m_y because both are piece 3. And v^2 = v_x^2 + v_y^2. Simplifying we have:

10000 = (m)*(v)^2
-160 = m*v_x
-60 = m*v_y
v^2 = v_x^2 + v_y^2

4 eqns for 4 unknowns

solve for v_x and v_y and plug into the first eqn:
v_x = -160/m
v_y = -60/m
10000 = (m)*[(160/m)^2+(60/m)^2]
10000 = m*(25600/m^2 + 3600/m^2) = 29200/m
10000*m=29200 -> m=2.92kg

Plug this into eqn 1 to get the magnitude of the velocity.
10000 = (2.92)*(v)^2 -> v = 58.52m/s

With the known mass, you can get the components of the velocities.
v_x = -54.79m/s
v_y = -20.55m/s

The angle with respect to the horizontal axis is:
arctan(-20.55/-54.79) = 20.56 degrees below the horizontal (in quadrant III)