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Impulsive force question

  1. Dec 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose that on the time interval [0,h] we have a force acting along the x-axis with magnitude [itex] 0 \leq f(t) [/itex] at time t, and f(0) = f(h) = 0. For a particle of mass m starting at rest, so that its distance x(t) from 0 satisfies 0 = x(0) = x'(0), show that [itex]x'(h) = \int_0^h f(t) dt[/itex].


    3. The attempt at a solution
    This is one of those problems where I swear I am right and the book is wrong. :biggrin:

    Let P denote the particle. Then on the interval [0,h], the force f(t) acts on P along the x-axis so henceforth I will write f(t) = (f(t),0) = f(t). The second law tells me that mx'' = f(t). Hence
    [tex]\int_0^h f(t) dt = \int_0^h m x'' dt = \int_0^h mv' dt = mv(h) - mv(0) = mv(h)[/tex]

    since v(0) = x'(0) = 0. So this seems to contradict what I was suppose to show except in the special case that m=1. I haven't really used my assumption that f(0) = 0 = f(h), but I can't think of how that would tie into this problem other than tell me that there is no net force acting on P at both t=0 and t=h. Would someone mind giving me some pointers on this one? :smile:

    - Sam
     
  2. jcsd
  3. Dec 28, 2011 #2

    Redbelly98

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    Homework Helper

    You are correct. It's likely this is a simple typo in the book, omitting the m from the solution, especially since they went to the trouble of saying "a particle of mass m".

    EDIT added:
    When in doubt it's often helpful to check units in an equation, which in this case proves the equation in the book must be wrong. Note that the "dt" inside the integral contributes a factor of seconds to the resulting units.
     
    Last edited: Dec 28, 2011
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