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.
m1v1+m2v2 = m1v1'+m2v2'
V1-V2=V2' - V1'
with the given, I know that
m1= .06 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:
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...
- 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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