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Momentum Question Involving Elastic Collision

  1. Jun 18, 2013 #1
    1. The problem statement, all variables and given/known data
    A 67.6g tennis ball with an initial speed of 28.8m/s hits a wall and rebounds with the same speed. The figure below shows the force of the wall on the ball during the collision. What is the value of Fmax, the maximum value of the contact force during the collision, if the force is applied for ti=19.1ms?



    2. Relevant equations
    I= pf-pi
    I= area under the curvehttp://s1329.photobucket.com/user/GarrettSPB/media/momentum_zps3611f797.png.html


    3. The attempt at a solution
    I was able to find the impulse as the difference between the initial and final momentum. This ended up being 3.89 N*s. From that I know how to find Favg, but I'm stuck on how to find Fmax. Any suggestions would be appreciated.
     
  2. jcsd
  3. Jun 18, 2013 #2
    IWhere is the figure?
     
  4. Jun 18, 2013 #3
    http://s1329.photobucket.com/user/GarrettSPB/media/momentum_zps3611f797.png.html

    Can you see that?
     
  5. Jun 18, 2013 #4
    I cannot see a figure. Is it a graph of Force vs time or what? If you don't have some characteristics of the tennis ball other than it being elastic, I wouldn't know how to find the max force. You probably know the average force. Or does the ball hit a wall at an angle?
     
  6. Jun 18, 2013 #5
  7. Jun 18, 2013 #6
    You must integrate the curve which amounts to finding the area. The integral of the curve is the imnpulse that you have figured out. Is this enough help?
     
  8. Jun 18, 2013 #7
    Unfortunately not. I don't know how knowing the area under there can help me find Fmax. If I could break up the the graph into geometric shapes then I would have a better idea how to find Fmax. Is there some equation I'm missing?
     
  9. Jun 18, 2013 #8
    Look at the graph. You know what ti is, given to be 19.1 msec. Basically, the area under the curve is the impulse. You can break up the curve into three parts and find the area as a function of Fmax. Then equate this to what you have determined the impulse to be.
     
  10. Jun 18, 2013 #9
    I've tried doing that and it doesn't work. I choose the rectangle in the middle of the graph (t=19.1/3) and then evaluate for 3.89 N*s= 6.37msec* Fmax; Fmax=611N. And thats not correct
     
  11. Jun 18, 2013 #10
    The right side of the equation is incorrect. You have three geometric shapes, two triangles and 1 rectangle. The Area of the first triangle is (1/2)base X height = (1/2) * 0.00637 sec * Fmax. Now do the other triangel and the rectangle and add them.
     
  12. Jun 18, 2013 #11
    Oh wow. Thank you so much
     
  13. Jun 18, 2013 #12
    Your time is 19.1 not 19.1/3 if this makes a slight difference
     
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