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Golf ball hit by club (conservation problem)

  1. Jul 9, 2010 #1
    1. The problem statement, all variables and given/known data

    In a typical tee shot, a golf ball is hit by the 300 g head of a club moving at a speed of 40 m/s. The collision with the ball happens so fast that the collision can be treated as the collision of a 300 g mass with a stationary ball—the shaft of the club and the golfer can be ignored. The 46 g ball takes off with a speed of 70 m/s.
    A) What is the change in momentum of the ball?
    B)What is the speed of the club head immediately after the collision?
    C) What fraction of the club’s kinetic energy is transferred to the ball?
    D) What fraction of the club’s kinetic energy is “lost” to thermal energy?


    2. Relevant equations
    Pi=Pf
    KE=.5*m*v^2
    P=mV


    3. The attempt at a solution

    For A I did Pf-Pi=.046*70-0=3.22 Ns

    For B I did conservation of momentum: .3*40=.3*v+.046*70

    v=(.3*40-.046*70)/.3=29.3 m/s

    For C the initial KE of the club is .5*.3*40^2=240 J
    the final KE of the ball is 0.5*0.046*70*70=112.7 J

    112.7/240=.47

    for D

    KEi=240 J from before and the final is

    0.5*0.3*29.3*29.3 + 0.5*0.046*70*70=241.5 J

    which is where I am stuck because I am confused why the initial energy is less than the final energy. Help?
    [
     
  2. jcsd
  3. Jul 9, 2010 #2

    kuruman

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    It's a round off error. The final speed of the club is 29.666... If you use this value, your final energy will be slightly less than the initial energy. It helps to solve the problem symbolically (if you can) and plug in at the very end. This way, problems like this will be avoided.
     
  4. Jul 9, 2010 #3
    Yeah I went back and used exact values before. And when I do the calculation

    v=(.3*40-.046*70)/.3=29.2666 m/s=(439/15) m/s

    Also, if I plug in your value of 29.666 I get 244.7 J in the end.

    So I'm still stuck.
     
  5. Jul 9, 2010 #4

    kuruman

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    I did it in two different ways and I agree with you. The final kinetic energy is greater than the initial kinetic energy by about 1.2 J. Now what?

    1. We are both wrong and we need a third opinion.
    2. This is a poorly conceived problem.
    3. The golfer cheated by attaching a firecracker on his club and added extra speed to the ball.

    I vote for option 2. What do you think?
     
  6. Jul 9, 2010 #5
    You're right I think it is option 2. So I did the problem again "assuming" a completely elastic collision with a stationary target and if energy is conserved then the MOST the golf ball can go is 69.36 m/s. So I think whoever wrote this problem meant for the final speed of the ball to be less than 70 m/s. I think an elastic collision is when you get the most speed out of the target right?

    Thanks for your help.
     
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