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Work done by impulsive force

  1. Oct 24, 2014 #1
    1. The problem statement, all variables and given/known data
    A force exerts an impulse J on a particle changing its speed from u to 2u. The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is--
    Ans: Ju/2

    2. Relevant equations
    Impulse= force*time
    impulse=change in momentum


    3. The attempt at a solution
    Since applied force and initial velocity are oppositely directed, J=m(2u-(-u))=3mu.
    Work=Force*displacement=Impulse*displacement/time=Impulse*velocity
    But velocity is not constant. I don't know what to do from here.
     
  2. jcsd
  3. Oct 24, 2014 #2

    vela

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    Hint: You can express the kinetic energy in terms of momentum: ##K = \frac{p^2}{2m}##.
     
  4. Oct 24, 2014 #3

    haruspex

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    There's a bit of a catch here. Consider two phases. In the first phase, velocity changes from u in one direction to u in the opposite direction. How much work has been done so far?
     
  5. Oct 24, 2014 #4
    Zero!oo) because change in kinetic energy is zero. Then change in ke when the velocity becomes 2u will be 1/2m(4u^2)-1/2mu^2=3/2mu^2. Impulse is 3mu so work is Ju/2!! Yay! Thanks for helping haruspex and vela. But now that I think about it, there is no need to consider two phases of the motion. All I can do is subtract initial ke from final ke and express it in the form of impulse and u.
     
  6. Oct 25, 2014 #5

    haruspex

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    Yes, but I thought you might fall into the trap of applying vela's equation to the change in momentum, giving K = ##\frac{J^2}{2m} = \frac J{6u}##. The point being that the difference of the squares is not the same as the square of the difference.
     
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