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Homework Help: Impulse Momentum (et al.) test

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data
    I took a test today and I am pretty sure I got everything right except for this one problem that was on there which I am unsure of. If I remember correctly, it went like this:

    It said there was a particle of mass 2 kg moving with a speed of 5 m/s to the right. Then there was a graph that more or less looked like this one I just made:


    The question asked what the final velocity was after the impulse.

    2. Relevant equations

    I believe this is correct and relevant.
    [tex]\int F dt = J = ΔP = mv_{f} - mv_{i}[/tex]

    3. The attempt at a solution

    I wagered that since the impulse was equal to the area under the curve on a Force v. Time graph, that the problem could be done in this way:

    [tex]J = \frac{1}{2}(base)(height)[/tex]
    [tex]J = \frac{1}{2}(0.10)(10)[/tex]
    [tex]J = 0.5[/tex]

    [tex]J = mv_{f} - mv_{i}[/tex]
    [tex]0.5 = (2)v_{f} - (2)(5)[/tex]
    [tex]10.5 = (2)v_{f}[/tex]
    [tex]5.25 = v_{f}[/tex]

    So I said the velocity afterwards was 5.25 m/s (and part b asked the direction, which I claimed was to the right).

    It's unlike any practice/hw/quiz problems we have had before and there was some equation she wrote on the board:
    [tex]F_{avg}Δt = mv_{f} - mv_{i}[/tex]

    Which I believe is basically what I did, I just didn't use that equation.

    Did I mess up?
  2. jcsd
  3. Nov 2, 2011 #2

    Doc Al

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    Staff: Mentor

    Assuming that the impulse was also to the right, you did fine.

    FYI: You could have used your instructor's formula; In this case the average force is 10/2 = 5 N. Same answer, of course. (Since it's really the same thing.)
  4. Nov 2, 2011 #3
    Whew, thank you.

    Another question that I am a little sketchy about is one where they showed some function U(x) which represented potential energy with respect to distance.The mass of the thing was m kg The question(s) were:

    a.)What is force:
    I simply took the derivative of U, and my final answer was U'(x). Sorry I can't remember the function, but it was something that had the general shape of y = 1/x.

    b.)If the object was released from it's origin, what is it's speed at x = 2?

    I claimed that mgU(0) was the total energy on the system, and so, after the thing fell or whatever, the final velocity could be found via:

    [tex]mgU(0) = mgU(2) + \frac{1}{2}mv^{2}[/tex]
    Solving for v obviously.

  5. Nov 2, 2011 #4

    Doc Al

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    Staff: Mentor

    Almost. Given a potential energy function, the associated force is given by -dU/dx. The minus sign is important.

    U(x) is the potential energy, not mgU(x). (You're confusing this with a gravitational PE problem, perhaps?)

    You want:
    [tex]U(0) = U(2) + \frac{1}{2}mv^{2}[/tex]
  6. Nov 2, 2011 #5
    Oh ok, I got confused trying to replicate the problem from memory into latex.

    Yes I used [itex]F = -\frac{dU}{dx}[/itex]
    and I did use U(x) for potential, not mgU(x).
  7. Nov 2, 2011 #6

    Doc Al

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    Staff: Mentor

    Good! :cool:
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