Need help with 2D Collision Problem?

Question

So far I know this type of collision is somewhat inelastic. But, I'm not sure what equation I should use to approach this problem. Please help!

Collision between two pucks on an air-hockey table.
Puck A has a mass of 3.1 kg and is moving along the x axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 7.1 kg and is initially at rest. After the collision, the two pucks fly apart with the angles: Puck A= 65 degrees, Puck B= 37 degrees.

(a) Find the final speed of:
puck A=
puck B=

(b) Find the kinetic energy of the A+B system:
before the collision=
after the collision=

Answer 1

I've done so many of these that I wrote a program to save me some work. Here are the results:
GIVENS
masses m1, m2 = 3.1 7.1
pre-collision velocities v1i, v2i = 5.5 0
pre-collision angles Θ1i, Θ2i = 0 0 deg
post-collision angles Θ1f, Θ2f = 65 -37 deg
SETUP
Px = m1v1cos(Θ1i) + m2v2cos(Θ2i) = 17.05
Py = m1v1sin(Θ1i) + m2v2sin(Θ2i) = 0
P = sqrt(Px^2 + Py^2) = 17.05
ΘBias = arctan(Py/Px) = 0 deg
RESULTS
post-collision velocities v1f, v2f:
v1f = P / ((1/tan(Θ1f) - 1/tan(Θ2f))*m1*sin(Θ1f)) =
3.38392960822387
v2f = P / ((1/tan(Θ2f) - 1/tan(Θ1f))*m2*sin(Θ2f)) =
2.22503758849494
DETAILS
y momenta: final 1, 2, 1+2; initial 1+2 =
9.50733343962322 -9.50733343962322 0 0
x momenta: final 1, 2, 1+2; initial 1+2 =
4.43334239152943 12.6166576084706 17.05 17.05
kinetic energy: final 1, 2, 1+2; initial 1+2 =
17.7490183697919 17.5753125592645 35.3243309290564 46.8875
As you can see, the collision is partly elastic (final KE < initial KE)