Momentum Physics Problem: Two astronauts throw billiard balls that collide

My attempt is shown below.

Here is a better photo:

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The collision can be inelastic even if the balls don't stick together. Inelastic simply means that total KE is not conserved.

If the balls stick together, the collision is often called "completely inelastic".

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Okay, thank you so much for telling me!
I just tried to write LaTeX, I clicked preview and nothing happened to the text so I don't believe I'm writing it correct.
here is my work though:
pi=pf
mv1(initial) + mv2(initial) = mv1 (final) + mv2 (final)
so substitute the values im given:
4.2-3.5=3.5cos35 + v2cosx (i think it's subtraction since Erin's ball in this case is being thrown in the opposite direction)
In an elastic collision the KE is conserved so KEi=KEf
1/2mv1^2 (initial) + 1/2mv2^2 (initial) = 1/2mv1^2 (final) + 1/2mv2^2 (final)
plugging all the information in I find that v2 is 4.2 m/s.
i substitute the value i found for v2 into the first equation and i get that 0.7=3.4cos35 + 4.2cosx
so therefore I use the inverse cosine function and i calculate x to be an angle of 119.7 degrees.
i'm not sure if this is right or not because i was trying to imagine the scenario in my head but 119 degrees doesn't seem right.

in33dphysicshelp said:

In an elastic collision the KE is conserved so KEi=KEf
1/2mv1^2 (initial) + 1/2mv2^2 (initial) = 1/2mv1^2 (final) + 1/2mv2^2 (final)
plugging all the information in I find that v2 is 4.2 m/s.

You are assuming the collision is elastic. This is not justified.

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In addition to what has already been said about enrgy conservation, your momentum conservation is incomplete. This is a two-dimensional collision which means that you need to write two momentum conservation equations, one in the direction of the original motion and one perpendicular to it.

Okay this is what I updated.
I split the momentum into x and y components.
i did 4.2-3.5=3.4cos35 + v2fcosx
0=3.4sin35 + v2f sinx
i did v2fsinx/v2fcosx and got tanx = 2.09/1.95 so i did the inverse of tan, got that the angle was about 46.98 degrees.
then to calculate for v2f i just plugged in 46.98 degrees into one of the momentum equations and i got that v2f is -3.06 m/s.

in33dphysicshelp said:

Okay this is what I updated.
I split the momentum into x and y components.
i did 4.2-3.5=3.4cos35 + v2fcosx
0=3.4sin35 + v2f sinx

OK

in33dphysicshelp said:

i did v2fsinx/v2fcosx and got tanx = 2.09/1.95

Check this. Are the numbers in the right place on the right side?

in33dphysicshelp said:

Homework Statement: Two astronauts, George and Erin, throw two identical billiard balls at each other in space, far from any gravitational influence.

Why do so many physics teachers think there is no gravity in space?