Bomb explodes at rest, conservation of motion problem

[T531]

Homework Statement

A bomb at rest 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 attached, but link just in case:
http://www.flickr.com/photos/82417987@N08/7545812162/in/photostream

Homework Equations

momentum(p) = mass*velocity
p(before) = p(after)
Potential Energy = Kinetic Energy
KE =1/2mv^2

The Attempt at a Solution

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

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:
4.0kg*40m/s + m(ofx)*v(of x) = 0

Now the y-mom equation:
1.0kg*60m/s +m(of y)*v(of y) = 0

I know: m(of x)=m(of y) because both are piece 3. And v^2 = v(of x)^2 + v(of y)^2.

Not sure how to find v(of x) and v(of y)

 

[nasu]

Solve the last two equations for vx and vy. Then plug into the energy conservation and solve for m.

 

[T531]

Okay, so
v(of x) = -160/m
v(of y) = -60/m
1.0x10^4 J = (m)*[(160/m)^2+(60/m)^2]
1.0x10^4 J = m*(25600/m^2 + 3600/m^2) = 29200/m
1.0x10^4 J*m=29200
m=2.92kg

then do I use KE = mv^2 ?
if so 1.0x10^4 J = (2.92kg)v^2
therefore v = 58.5 m/s

then v(of x) = -160/m = -160/(2.92 kg) = -54.8m m/s
and v(of y) = -60/m = -60/(2.92 kg) = -20.5 m/s
so theta = inverse tan(-20.5/-54.8) = 20.5 degrees

So:
Mass = 2.92 kg
Velocity = 58.5 m/s
Theta = 20.5 degrees

Can someone verify that?
Thanks