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Cutetys

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## 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*v

^{2}

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= 3

^{2}m/s i-hat + 2

^{2}m/s j-hat = √13 m/s

vb= 3

^{2}m/s I-hat + 4

^{2}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.

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