Physics
posted by Ashley .
A 0.0600 kg tennis ball, moving with a speed of 2.50 m/s, has a headon collision with a 0.0900 kg ball initially moving away from it at a speed of 1.00 m/s. Assuming a perfectly elastic collision, what is the speed and direction of each ball after the collision?
I was given that the answer for the speeds of the balls are .70m/s and 2.2m/s.
equations:
m1v1+m2v2 = m1v1'+m2v2'
V1V2=V2'  V1'
My attempt:
with the given, I know that
m1= .06 kg
m2=.09 kg
v1= 2.5 m/s
v2= 1.0 m/s
after this, I got to .135=m1v1'+m2v2'
and 1.5= V2'V1'

I agree with this
m1v1+m2v2 = m1v1'+m2v2'
I disagree with this:
V1V2=V2'V1'
where in the world did that come from? 
I asked someone else for help on this problem and they told me to use that equation. That's even what they wrote down for me. I agree it doesn't make sense.
Anyways the initial and final mass will stay the same right? So I will be at...
135=(.06)v1'+(.09)v2' 
Now you have one other equation you can use, the magic equation conservation of energy is true
Intial KE= final KE
You know initial KE. Final ke, 1/2 m1*V1'^2 + 1/2 m2 v2'^2 is equal to that initial.
Now, in your 135=(.06)v1'+(.09)v2'
solve for v2' in terms of all the other stuff, then put that v2' into the conservationof energy equation (yes square v2') It is messy algebra, you will get a quadratic you can solve (quadratic equation).