Zero momentum reference frame and an inelastic collision

[Cutetys]

Homework Statement

In my reference frame, I see two carts collide. Both cart A and Cart B have a mass of 2 kg. Cart A has initial velocity v= 3 m/s i-hat + 2 m/s j-hat, cart B has initial velocity v= 3 m/s I-hat + 4 m/s j-hat. 12 J are released in the collision.

In previous parts of this question I found:
The velocity of the zero momentum reference frame relative to me is: 3 m/s I-hat +3 m/s j-hat.
And the two velocities of the carts in the zero momentum reference frame are: Cart A = 0 m/s I-hat - 1 m/s j-hat, Cart B = 0 m/s I-hat + 1 m/s j-hat.

Q: Determine the final velocity of the two carts relative to the zero momentum reference frame.

Homework Equations

ΔK= Kf - Ki
K= 1/2*m*v²
m*v1i + m*v2i = m*v1f + m*v2i
vAo=vAB + vBo maybe?

The Attempt at a Solution

So I know change in ΔK is involved as that it is given in the the question and it isn’t relevant for any questions to follow. I’ve tried combining the momentum and kinetic energy equations but that proved to be useless. (EDIT: I did not originally show my work for this. Sorry about that, here it is :
va= 3² m/s i-hat + 2² m/s j-hat = √13 m/s
vb= 3² m/s I-hat + 4² m/s j-hat = √25 m/s
-12 J = Kf - (1/2*2kg(√(25m/s)² + (√13m/s)²)
-12 = v1f + v2f - 38 J
26 J - v2f = v1f
2 kg (3 m/s i-hat + 2 m/s j-hat) + 2 kg (3 m/s I-hat + 4 m/s j-hat) = 2 kg (v1f + v2f)
2 kg (3 m/s i-hat + 2 m/s j-hat) + 2 kg (3 m/s I-hat + 4 m/s j-hat) = 2 kg (26 J - v2f) + 2 kg(v2f)
2 kg (3 m/s i-hat + 2 m/s j-hat) + 2 kg (3 m/s I-hat + 4 m/s j-hat) = 52 kg*J - 2 kg(v2f) + 2 kg(v2f))

I know how to attempt this question if the collision were elastic (v12i = -v12f) but as this collision is inelastic I haven’t found a way to use it.This collision does not appear to be totally inelastic either so I cannot find the velocities through that route.

At this point I have no clue what to try next and I really could use an idea to bounce off of. Any help is greatly appreciated.

 

[haruspex]

I’ve tried combining the momentum and kinetic energy equations but that proved to be useless.

That should be the way. please post your working.

 

[Cutetys]

That should be the way. please post your working.

Sorry bout that, here’s my work:
va= 3 m/s i-hat + 2 m/s j-hat = 13^1\2 m/s
vb= 3 m/s I-hat + 4 m/s j-hat = 25^1/2 m/s
-12 J = Kf - (1/2*2((25^1\2)^2 + (13^1/2)*2)
-12 = v1f + v2f - 38 J
26 J - v2f = v1f
2 kg (3 m/s i-hat + 2 m/s j-hat) + 2 kg (3 m/s I-hat + 4 m/s j-hat) = 2 kg (v1f + v2f)
2 kg (3 m/s i-hat + 2 m/s j-hat) + 2 kg (3 m/s I-hat + 4 m/s j-hat) = 2 kg (26 J - v2f) + 2 kg(v2f)
2 kg (3 m/s i-hat + 2 m/s j-hat) + 2 kg (3 m/s I-hat + 4 m/s j-hat) = 52 kg*J - 2 kg(v2f) + 2 kg(v2f)

And that’s where I come to to a problem. - 2 kg(v2f) + 2 kg(v2f) = 0, so from this equation I can not find v2f or v1f.

 

[haruspex]

3 m/s i-hat + 2 m/s j-hat = 13^1\2 m/s

I assume you mean the magnitude of that vector is √13.

-12 = v1f + v2f - 38 J

v1f and v2f are not energies.

Your work would be much easier to read if you were to use the subscript and superscript buttons (X₂, X²). Better still, LaTeX.

 

[Cutetys]

I assume you mean the magnitude of that vector is √13.

v1f and v2f are not energies.

Your work would be much easier to read if you were to use the subscript and superscript buttons (X₂, X²). Better still, LaTeX.

Sorry, I'm new to the forum so I didn't realize those option were there. I've edited my original post so hopefully it's easier to read.
You are also correct, I meant the magnitude of the vector. I also edited it to make that more clear.
These edits probably don't matter too much as I think I know where I'm going wrong now. Thank you so much!