Posted by **Ashley** on Wednesday, February 3, 2010 at 9:35pm.

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'

- Physics -
**bobpursley**, Wednesday, February 3, 2010 at 10:03pm
I agree with this

m1v1+m2v2 = m1v1'+m2v2'

I disagree with this:

V1-V2=V2'-V1'

where in the world did that come from?

- Physics -
**Ashley**, Wednesday, February 3, 2010 at 10:08pm
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'

- Physics -
**bobpursley**, Wednesday, February 3, 2010 at 10:25pm
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).

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