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Linear Impulse - Jackhamer Rebounds

  1. Apr 9, 2009 #1
    1. The problem statement, all variables and given/known data
    During Operation the jack hammer develops on the concrete surface develops a force @ t=0 f=0, @ t=.2, F=90, @ t=.4, F=0. Graph looks like a triangle
    To achieve this the 2lb spike S, is fired from rest into the surface @ 200ft/s. Determine the speed of the spike just after rebounding


    2. Relevant equations

    mv + E Fdt=mv

    3. The attempt at a solution

    Well it seems to me that the initial velocity is 0, and the final is 200ft. How do i know the rebound velocity.
     
  2. jcsd
  3. Apr 9, 2009 #2
    impulse = F *(change in T)

    impulse = mv

    mv=F *(change in T)
     
  4. Apr 9, 2009 #3

    LowlyPion

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    First of all convert to SI units before things get out of whack.
    200 f/s = 60.96 m/s
    90 lbs force = 400.5 N
    2 lbs = .907 kg

    Now what is your impulse?

    Δmv = F*Δt

    But your time varies according to the graph. So your total impulse is the area under the function of Force(t) that they give you.

    With total Impulse then you have your Δmv.
     
  5. Apr 13, 2009 #4
    It is still not clicking

    mass .907

    is my initial velocity from rest, or is 60.96

    My force is 400500 N at its peak,

    i do not understand which t's to use.
     
  6. Apr 13, 2009 #5

    LowlyPion

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    Just curious how is your force so high? Are there units to your problem that you haven't supplied? Is the peak force 90lbs or some other value? Your original statement doesn't really say.
     
  7. Apr 13, 2009 #6
    90 lbs force = 400.5 N

    the Forve is 10^3 N = 400500 N
     
  8. Apr 13, 2009 #7

    LowlyPion

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    And where does the multiplier come in?

    Don't they give you the force as 90 already at t = .2 sec?

    Because it looks to me like the area under the Force as a function of time that they give, is the area of a triangle, over the range of t = 0 to t = .4.
    So ...

    I = 1/2*(400.5 N) * .4sec = 80.1 N-s

    That impulse is available to change the momentum of the .907 kg hammer.

    Figure then the original momentum and subtract the impulse, and that should yield the return momentum and hence the return velocity shouldn't it?
     
  9. Apr 15, 2009 #8
    Momentum = P

    P=mv = .907*60.96 = 55.301

    P-I = -24.799 <-- Pretty sure i'm wrong but im gonna keep going

    Return Momentum = RP

    RP = mv = -24.799 = .907v therefore v = -27.341 m/s
     
  10. Apr 15, 2009 #9

    LowlyPion

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    You probably want to multiply by 3.281 to get back to feet / sec. don't you?
     
  11. Apr 15, 2009 #10
    yes, i would, so i take it i am correct
     
  12. Apr 15, 2009 #11
    Ok, one more question

    For the graph, the units of force is 10^3 N

    Which makes F = 400500 N

    Do I use this number, it give a totaly different answer

    I = 80100 N-s
    P=55.301
    V=-88234.17601 m/s
     
  13. Apr 15, 2009 #12

    LowlyPion

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    Is there more to this problem than you have provided? Is there a mass to the jackhammer that is more than just 2 lb? Is there a cross section of the tip to consider? These numbers don't seem reasonable to me.
     
  14. Apr 15, 2009 #13
    nope thats it

    on the graph the yaxis is the force and the x axis is t(ms)

    am i correct based on the info i have provided
     
  15. Apr 15, 2009 #14

    LowlyPion

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    Oh and time is ms. Well that makes the difference. Your F at 103 and your time at 10-3 then effectively cancel in magnitude and yield the same thing as though you were calculating with 90 lbs and .4 sec. The earlier answer is what I got.

    Is it correct? I guess your grader is the judge.
     
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