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Calculate Average Force by finding Impulse over Impact Time

  1. Nov 12, 2007 #1
    [SOLVED] Calculate Average Force by finding Impulse over Impact Time

    1. The problem statement, all variables and given/known data

    A movie depicting a 55-g golf ball hit with a club shows that the impulse time between the club and the ball is 0.2 ms, and the initial speed of the ball is 136 mph.

    a) What is the average force exerted by the club on the ball during impact?
    b) Estimate the distance over which the club and the ball are in contact during the swing.

    2. Relevant equations




    3. The attempt at a solution

    First converted everything to SI units...
    Mass of ball: 0.055kg
    Impact time: 2x10^-4 seconds
    Initial velocity of golf ball: 60.797 m/s

    (a)
    Fave = change in momentum (impulse) / impact time.
    = Pfinal - Pinitial / impact time = mVf - mVi / impact time = m(Vf - Vi)/impact time

    = [tex] (0.055kg)(0m/s - 60.797m/s) / 2x10^-4 s = -16719.175 N [/tex]

    I have a feeling this is wrong because the Force should not be negative. Where I am getting tripped up in is what the final velocity of the ball should be ... I just assumed it was 0 m/s (when the ball would eventually come to rest) but I have a few doubts about whether this is what I am to do.

    (b) Fave = M * Impact speed / 2 * change in collision distance / impact speed =

    M * Impact speed^2 / 2*change in collision distance

    [tex](0.055kg)(60.797m/s)^2/ 2deltah = -16719.175N[/tex]

    delta h = -0.0060797m ...
     
    Last edited: Nov 12, 2007
  2. jcsd
  3. Nov 12, 2007 #2
    I've thought about my question more and now I'm thinking maybe I misinterpreted one of the calculations. Perhaps change in momentum can't equate to m(Vf - Vi) ...
     
  4. Nov 12, 2007 #3

    hage567

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    Homework Helper

    Well, you have your Vi and Vf mixed up. The ball is initially at rest, so Vi=0. Vf = 60.797 m/s because that's the velocity the ball is going after the force has acted on it. Change that up and you will get a positive number.
     
  5. Nov 12, 2007 #4
    Throughout the equation you were (correctly) calculating from the point of view of the ball (using the ball's mass, velocities, etc.). The question asks what the average force exerted by the club onto the ball is. Re-read Newton's laws for a hint.
     
  6. Nov 12, 2007 #5
    Thanks. That makes sense and I feel silly for not realizing it. I think what threw me off was how the question worded "the initial speed of the ball is 136mph" - so I immediately went Vi = 136mph ...

    Now I realize that its final velocity (in regards to velocity post-impact) is 136mph (60.797 m/s) and that its initial velocity is infact 0 m/s since it is as at rest.
     
  7. Nov 12, 2007 #6
    I found the three laws of Newton on a neat website (and am going to print them out and post it on my wall!).

    My guess is you are referencing the third law: III. For every action there is an equal and opposite reaction.

    But this doesn't seem to make conceptual sense to me. Yes by the conservation of momentum if you add up the momentum of the club and the ball it will equal 0. The reason the ball goes much farther is because it has to have a much larger velocity to compensate for its smaller mass relative to the club. But does this relation really hold when you examine forces? The club applies 16719.175 Newtons of Force on the ball, (and my assumption from action-reaction pairs is that that the ball, via your hint, is that the ball will exert a force of -16719.175N on the club ...) but this is a large amount of force, and while I'm not a golfer, it seems like this is an excessive amount of force for the golfer to withstand...

    EDIT: I read this statement which sort of made it clear to me: The average force on an object during a collision is the change in the objects momentum divided by the collision time. The change of the club's momentum divided by its collision time would give me a negative force. But if I look at the change of momentum (positive) for the ball and divide it by the impact time ; then I would get a positive force (which I now do!)
     
    Last edited: Nov 12, 2007
  8. Nov 12, 2007 #7
    Bingo!
     
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