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F(avg) and Fmax during collision

  1. Dec 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Superball Hits Wall Figure 7-24 shows an approximate plot of force magnitude versus time during the collision of a 60 g Superball with a wall. The initial velocity of the ball is 32 m/s perpendicular to the wall. It rebounds directly back with approximately the same speed, also perpendicular to the wall. What is Fmax, the maximum magnitude of the force on the ball during the collision?

    http://i234.photobucket.com/albums/ee9/locowise/10_32.gif


    2. Relevant equations
    change in p= p(final)-p(initial)=Favg*change in time


    3. The attempt at a solution
    F(avg)= 657
    But no idea how to find Fmax!
     
  2. jcsd
  3. Dec 11, 2007 #2

    malty

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    Just curious how did you calculate F (avg)?
     
  4. Dec 11, 2007 #3
    change in momentum=f(avg)*change in time

    .058kg(-34-34)=(avg)*change in time

    -3.944=f(avg)*.006s

    im not sure about the negatives
     
  5. Dec 11, 2007 #4

    malty

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    Ok, I'd got 640N but that's because we both took different masses? Now all I'm going to ask is this, if you were to get the average Force from the graph only (i.e let's pretend you could only work it out from the graph, and the F values were given) how would you do it?
     
  6. Dec 11, 2007 #5
    sum all the forces and divide by the number of forces (integral?)
     
  7. Dec 11, 2007 #6

    malty

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    Yep, but that is unnecessary here? What are you doing when you sum all those forces and divide by the number of forces?
     
  8. Dec 11, 2007 #7
    taking the average
     
  9. Dec 11, 2007 #8

    malty

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    Lol, yeah :rofl: should have said (other than taking the average)

    What are you doing when you're intergrating something?
     
  10. Dec 11, 2007 #9
    taking the area under curve
     
  11. Dec 11, 2007 #10

    malty

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    Correct!

    When you're taking the average of any function what you are essentially doing is finding a rectangle of equal area that represents the area under the curve of that function.

    The "height" of this rectangle will give the average value.. does this help?
     
  12. Dec 12, 2007 #11
    ok but how do i find the area if i dont have the height
     
  13. Dec 12, 2007 #12
    You have the area (the integral) and you have one of the sides (the time duration of the impulse). The F_avg is easily found with this info.
     
  14. Dec 13, 2007 #13
    Im not trying to find the Favg
     
  15. Dec 13, 2007 #14

    malty

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    The area of you're rectangle is equal to some height on the y-axis multiplied by the some length l on the x-axis.

    If I told you the length of your rectangle was 6 (seconds).


    Now look at your graph what is the height equal to?


    What I'm trying to get you to realise here is that the area, can be redrawn as a single rectangle, the value of the height of this rectangle will give you the average.

    So what if you were to convert the region under curve into a rectangle of equal area?
    Hopefully you would notice that the height of this rectangle is exactly half the height of highest point of the curve in your given question.
    Just mess around with at bit and you should see it.


    Now, this means that Fmax is equal to twice the height of a rectangle of equal area .. ..
     
  16. Dec 13, 2007 #15
    Sorry, I read the wrong post. It's simple geometry to find Fmax.... Since you know what the impulse is (Favg * dt) all you have to do is break down the area under the Fmax curve into expressions involving Fmax. The left and right triangles are obviously 1/3dt * 1/2 * Fmax and the central block is a rectangle with base 1/3dt and height Fmax. Just solve for Fmax...
     
  17. Apr 15, 2008 #16
    Ans=160N
     
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