Question

A 0.0600 kg tennis ball, moving with a speed of 2.50 m/s, has a head-on 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'
V1-V2=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'

Answer 1

I agree with this
m1v1+m2v2 = m1v1'+m2v2'
I disagree with this:
V1-V2=V2'-V1'

where in the world did that come from?

Answer 2

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'

Answer 3

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).