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there is a picture with this problem - or diagram really. Player two is moving towards (0,0) at an angle theta from the x-axis, in the 3rd quadrant. Player one is moving towards (0,0) at the same angle theta in the 4th quadrant. The final angle is in the first quadrant. So it's like an upside down V that meets at (0,0) on a regular x-y graph.

momentum is conserved, so delta p = 0, and pf=pi. After the collision you can use work-energy stuff to find the work done by friction: -uk*(total mass)*98*1.69. I broke the momentum equations up into their components, and used the y-component equation to solve for theta - I got 30 degrees, which sounds probable.

I tried to solve for the initial velocity after the collision using delta K = Work done by friction. I'm not entirely sure that's right. I ended up with -1/2(m1+m2)Vi^2 = the above expression for work done by friction. I got Vi = 4.1m/s. I was kind of dismissed that because I didn't think it was reasonable and wasn't sure about setting up delta K to equal work done by friction - thoughts?

If 4.1m/s IS right, does that mean that the velocity in the x-direction after collision is 0.1m/s, because velocity in y-direction is 4.0m/s? And if I am on the wrong track there, what is a better relationship to use?

Is there some way to solve for m2 that I'm not seeing? I keep getting that they equal each other, so I think solving for the x-component of V is probably the next step.

Thanks