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Height of a bounce

  1. Nov 15, 2009 #1
    I could use some help starting and solving this problem. It is due tonight. Any help would be appreciated.

    A 190 ball is dropped from a height of 1.8 and bounces on a hard floor. The force on the ball from the floor is shown in the figure (use the link). How high does the ball rebound?

    http://session.masteringphysics.com/problemAsset/1013630/13/jfk.Figure.P09.40.jpg
     
  2. jcsd
  3. Nov 15, 2009 #2

    cepheid

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    Dark Visitor,

    I've noticed you're posting the problems without using the template provided for homework help. This template is provided for a reason, namely that listing all of the given information in a problem, and the relevant equations and physical concepts that apply to it, is a very useful method of approaching a physics problem.

    The third part of the template (your attempt at a solution) is also critical. You must make an effort, and should not expect us to do your homework for you. It would gain you nothing if we did. You wouldn't learn anything. We already know how to do your homework problems, so it wouldn't gain us anything either. Please read about this further at the forum rules here

    https://www.physicsforums.com/showthread.php?t=94379

    before continuing to post.

    What have you done so far? What are your thoughts on how to approach the problem?
     
  4. Nov 15, 2009 #3
    I know what you are saying, and I am sorry for posting the way I do, but if I don't know where to begin, then I don't know what to put on a post for that.

    I know that we are given the weight of the ball, the height it is dropped from, the amount of time it took for the bounce, and the force from the floor on the ball. What I don't understand is how we get an h (height) out of that information. I don't know any equations that would work.
     
  5. Nov 15, 2009 #4

    cepheid

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    Hey Dark Visitor,

    Sorry I couldn't reply sooner. Hopefully you'll be back. Are you aware of the impulse-momentum theorem? Basically the impulse supplied to the ball by the floor (which is equal to the area underneath the force vs. time graph) is equal to the ball's change in momentum. If you know the ball's change in momentum, then you know its upward initial velocity and can easily figure out how high it will go either by conservation of energy or using basic kinematics.
     
  6. Nov 15, 2009 #5
    I see what you are saying. But how do I figure out the area under that section of the graph? And how does that connect me to the initial velocity or the height?
     
  7. Nov 15, 2009 #6

    cepheid

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    It's a triangle, and there is a formula to calculate the area of a triangle in terms of the lengths of its base and its height.

    p = mv
     
  8. Nov 15, 2009 #7
    So if I use the formula A = b*h, does that mean I multiply 2(2*1000)?
     
  9. Nov 15, 2009 #8

    cepheid

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    That's not quite the right formula for the area of the triangle.

    Also:

    The base of the triangle is the time interval during which the force is non-zero (hint, it is not 2 ms).

    The height of the triangle is the maximum force reached, i.e. the value of the force at the vertex, which is indeed 1000 N.
     
  10. Nov 15, 2009 #9
    Excuse me. I meant to type A = 1/2(b*h). My mistake.

    I know. It is 4 ms, but since we are making 2 right triangles, I figured we just multiply by 2. Is that a mistake?
     
  11. Nov 15, 2009 #10

    cepheid

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    It's a mistake because A = 1/2b*h applies to any triangle with base b and height h. So you don't have to break it up into individual right triangles, and even if you did, each of those would have only half of the base of the original, and you would get the same answer (you would have to, right? I mean, you're calculating the same area with both methods).
     
  12. Nov 15, 2009 #11
    I see. But does it make a difference that one measurement is in Newtons and the other is in milliseconds? Or do I multiply them as it is?
     
  13. Nov 15, 2009 #12

    cepheid

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    By definition, impulse has units of N*s = m*kg/s which is the same units as momentum. So once you calculate the area in N*s, you know what the change in momentum is. So, so long as you express the time values in seconds, the units are correct for the quantity you are calculating.
     
  14. Nov 15, 2009 #13
    So I can calculate it like this:

    1/2(1000 N * .004 s) = 2 m*kg/s
     
  15. Nov 15, 2009 #14

    cepheid

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    That looks right to me.
     
  16. Nov 15, 2009 #15
    So is that the answer? Or is there something else I need to do?
     
  17. Nov 15, 2009 #16

    ideasrule

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    That's the impulse imparted on the ball. You still have to calculate the final momentum, which is just initial momentum + impulse (be careful about the sign, though). Since p=mv and you know m, you can calculate v. Once you calculate v, you can figure out the final height that the ball reaches.
     
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