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Speeds of tennis balls after collision

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

    m1v1+m2v2 = m1v1'+m2v2'

    3. The attempt at a solution

    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 don't know where to start?
     
  2. jcsd
  3. Feb 3, 2010 #2
    Hi balletgirl

    Plug all the data to the equation you wrote and you lack one more equation, which is V1-V2=V2' - V1'
     
  4. Feb 3, 2010 #3
    Okay so I do 2.5-1.0= V2'-V1'. I don't understand how you find those final numbers.
    &
    .06kg(2.5m/s) + (.09kg)(1.0m/s) = m1v1'+m2v2'
     
  5. Feb 3, 2010 #4
    You have two equations and two variables so just use elimination or substitution
     
  6. Feb 3, 2010 #5
    I don't understand the concept.
    I thought I was solving for four variables: m1' v1' m2' and v2'
    Anyways I got to .135=m1v1'+m2v2'
    1.5= V2'-V1'

    it didn't work out when I tried using elimination or substitution
     
  7. Feb 3, 2010 #6
    Okay just realized a dumb mistake. The mass will stay the same. This will make everything a lot easier for me.
     
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