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AP Problem

  1. Dec 2, 2004 #1
    Hey im in 12 grade but this is advanced placement so its college level... Anyways i could use some help on this problem i have

    A 5kg ball initally rest at the edfe of a 2-meter long, 1.2 meter high frictionless tavle, as show above. A hard plastic cude of mass .5kg slides across the table at a speed of 26 m/s and strikes the ball, causing the ball to leave the table in the direction in which the cude was moving. The figure below shows a graph of the force exerted on the ball by the cude as a function of time

    graph is:

    as time goes from 0-4 seconds, the force rises o-2.0(x10^3 Newtons).
    from 4.0-6.0 seconds the F stays at 2.0 (x10^3 Newtons)
    6-10 seconds force falls to 0. so it looks like a table /**\ type thing...

    Anyways, it wants the total impulse given to the ball?
     
  2. jcsd
  3. Dec 2, 2004 #2
    i know impulse=change in potentional, or the intengral of dt? i havent taken calculas yet so i have no idea about derivatives and integrations
     
  4. Dec 2, 2004 #3
    Impulse is [tex] F \Delta t [/tex]

    You can easily determine this from the graph, just the same way you get displacement from a velocity vs time graph, or work from a Force vs displacement graph.
     
  5. Dec 2, 2004 #4
    Whats that symbol infront of time.... we havent learned displacement yet...
     
  6. Dec 2, 2004 #5
    nevermind change in time.... ok thanks
     
  7. Dec 2, 2004 #6
    [tex]F \Delta t [/tex] is read F delta t and the delta means change in (time for this case). I'm amazed you have learned anything about displacement (maybe you refer to it simply as distance which is a scalar quantity unlike displacement, a vector). But to concentrate on this problem how would you find impulse looking at the graph?
     
  8. Dec 2, 2004 #7
    so i did J=2000(6-4), J=4,000.. so basically area under the FxT graph is the impulse... sasme as area in VxT=distance... next part is to find the horizontal velocity of the ball, which so if i know mv of block, i can use J=change in momentum... so 4000=(26x.5)-(5xV). so Velocity= hm.... nevermind that doesnt seem right... i guess we dont know change in momentum
     
  9. Dec 2, 2004 #8
    o wait, wouldnt it be 4000=(.5x26)-(.5v), V=7974... that defiantly isnt right
     
  10. Dec 3, 2004 #9

    NateTG

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    Homework Helper

    Do you know any calculus? (oops, nevermind, just saw the inital post.)

    The impulse is going to be equal to the area under the force curve in the force-time graph. You can check to make sure that force times time gives units equal to momentum, and be sure to check the scales.
     
    Last edited: Dec 3, 2004
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