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Momentum of a System

  1. Mar 29, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose that the momentum of a system is 8 kg m/s. What is the kinetic energy of the system if it has a mass of 10kg? If this system is brought to rest with a constant force in 5 seconds, what is the magnitude of this force?

    2. Relevant equations
    Momentum > p = (m)(v)
    Kinetic Energy > KE = 0.5(m)(v)2

    3. The attempt at a solution
    Whenever I get an answer this easily, I am always very cautious, because it seems too easy and I usually get it wrong. So far, I applied the momentum equation, p = (m)(v) > 8 = 10v, and solved to get the fact that velocity is 4/5 m/s. Then, I plugged that into the formula for Kinetic Energy, KE = 1/2(m)(v)2, and I got that the Kinetic Energy is 16/5 Joules. As for finding the magnitude of the opposing force, I do not even know where to begin. Did I get the Kinetic Energy right, and how would I begin to find the opposing magnitude? Thanks!
     
  2. jcsd
  3. Mar 29, 2017 #2
    You are absolutely correct with kinetic energy. That is exactly how you are supposed to find it.

    As for the opposing force, it's not as hard as it seems. But first, let me give you some definitions and a new equation. The change in momentum (Δρ) is also called Impulse (some textbooks will use different variables for impulse, but I use the letter "J").

    J=Force*Δt
    Change in momentum=Force*change in time

    So to find the opposing force, all you do is plug in to that equation. If you need more help with this problem, let me know.
     
  4. Mar 29, 2017 #3

    Orodruin

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    Your solution for the kinetic energy is correct, although it may be easier in this case to directly use the relation ##E_k = p^2/(2m)##.

    For the stopping time, apply Newton's second law.
     
  5. Mar 29, 2017 #4
    It is possible that the problem is not referring to this force. It's hard to know what exactly your physics class is doing right now without being in it myself. Now that I think about it, if you have not learned this equation yet, you may want to solve it the way Orodruin suggested above.
     
  6. Mar 30, 2017 #5

    Orodruin

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    To be honest, there is really not much to "learn". The impulse equation is just the integrated version of Newton's second law, instead of ##dp/dt = F## you write
    $$
    \Delta p = \int F\, dt.
    $$
     
  7. Mar 30, 2017 #6

    CWatters

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    The question implies the stopping force is constant so the deceleration is constant. You can use the equations of motion to find the deceleration and Newton to find the force.
     
  8. Mar 30, 2017 #7

    Orodruin

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    The deceleration is not requested, you just need to apply Newton's second law, i.e., ##dp/dt = F## or - for a constant force - ##\Delta p/\Delta t = F##. I would also argue that Newton's second law is the equation of motion for the system.
     
  9. Mar 31, 2017 #8
    I used this strategy, and I got that the force would be equivalent to -(8/5) N.
     
  10. Apr 1, 2017 #9

    haruspex

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    Correct.
     
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