2 dimensional collision confusion

[NKC]

Homework Statement

A 2 kg mass with initial velocity v = (5i + 7j) collides perfectly elastically with a 3 kg
mass with initial velocity v = (-1i -3j) After the collision, the 2kg mass has a speed of
√50 m/s and the 3 kg mass is traveling at an angle of 329.77Θ as measured from the positive x
axis. Determine the speed of the 3 kg mass after the collision and the angle of the 2 kg mass
after the collision (as measured from the positive x axis).
i and j are the x and y-axis respectively.

Homework Equations

P = MV
KE = 0.5MV^2
A*B = XYCosΘ

The Attempt at a Solution

Initial:
KEm1 = 0.5(2)(5i + 7j)^2
KEm2 = 0.5(3(-1i - 3j)^2
Pm1 = 2(5i + 7j)
Pm2 = 3(-1i - 3j)
Final:
KEm1 = 0.5(2)(√50)^2 Cosø
KEm2 = 0.5(3)V^2 Cos(329.77)
Pm1 = 2(√50)
Pm2 = 3V
I know that initial momentum and KE of the system should equal the final momentum and KE of the system, but I can't figure out how to set that up with a 2 dimensional collision. Normally it would be KEi = KEf and Pi = Pf, but with i and j coordinators I can't figure out how to make that work. Should KE and P for the y coordinate and the x coordinate be calculated separately?

 

[Qube]

Should KE and P for the y coordinate and the x coordinate be calculated separately?

Yes.

 

[NKC]

Yes.

Yes to both KE and Momentum? I thought kinetic energy didn't have a direction being a scalar.

 

[BvU]

So you answer your own question. "Should KE and P for the y coordinate and the x coordinate be calculated separately?": Yes on three counts: KE, px and Py.

 

[NKC]

So you answer your own question. "Should KE and P for the y coordinate and the x coordinate be calculated separately?": Yes on three counts: KE, px and Py.

Just to be clear, that leaves me with 3 equations for initial values, and 3 for final values right?

 

[BvU]

Yes. The initial ones have no unknowns, the final ones 2.

You will have to do something about the final ones: where do the cosines in the KE come from ? And why aren't the P vectors instead of numbers ?