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Momentum of a snooker ball

  1. May 3, 2008 #1
    1. The problem statement, all variables and given/known data
    A snooker ball of mass 0.350kg hits the side of a snooker table at right angles, and bounces off also at right angles. If its speed before collision is 2.8ms^-1 and its speed after is 2.5ms^-1, calculate the change in its momentum.
    (The answer to the question is not 0.105kg ms^-1)

    2. Relevant equations
    p = mv (well, that's what I've tried answering it with)

    3. The attempt at a solution
    I'm assuming my method is flawed, but I thought it might be worth using pythagoras to work out the actual (?) velocity.
    I got 3.75 but if you do 3.75*0.35 i reach 1.3 which according to the answers in the back of the book are incorrect.

    Could someone enlighten me please? I'm really stumped.

  2. jcsd
  3. May 3, 2008 #2


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    Hi Crosshash! :smile:

    You're misreading the question … the ball goes forward and back along the same line! :smile:
  4. May 3, 2008 #3
    argh! you're kidding me? I've been stuck on this question for almost 30 minutes and it's that simple!?

    I thought it deflects off the side at a right angle :P

    I get the right answer now, thankyou!
  5. May 3, 2008 #4
    Now if you could just figure out what a snooker is :rofl:
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