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Physics Help Needed! Plz

  1. Mar 22, 2005 #1
    I need help on these 2 questions.
    1) What impulse if needed to stop a 100kg mass traveling at a velocity of -25m/s?


    3) A force with a magnitude of 430 N is used to stop an object with a mass of 45kg moving at a velocity of 900m/s. How long will it take to bring the object to a full stop?

    -Any help on formulas and how to solve this would be appreciated.. Thanx!
  2. jcsd
  3. Mar 22, 2005 #2


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    it's been a while since I've done impulse, but if I remeber right, impulse is just the change in momentum. So 1) isn't really that bad. Just think, what is the initial momentum? What is the final momentum? the difference is the impulse.

    Also, from what i remember, impulse is also the force exerted over a certain time (the force times the change in time). So for 3) you just have to solve for the change in time. It's good to remember to for a change in momentum, the mass remains constant and the speed is the only thing that changes.

    Hope that helps, and check to make sure the equations I said were right.
  4. Mar 22, 2005 #3
    How would you set up the problems though?
  5. Mar 22, 2005 #4
    Initial & Final mass = 100kg
    Initial Velocity = 25m/s
    Final Velocity = 0m/s

    Impulse is any change in momentum. Momentum = Mass x Velocity.
    Once you have stopped the object from moving, there is no velocity. You can see that the momentum would be 0 kgm/s. The change in momentum of this operation is the impulse.

    Impulse = MVfinal - MVinitial

    For #2:
    Remember that F*(delta)T = m*(delta)V
  6. Mar 22, 2005 #5


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    Above utilizes {IMPULSE} = {Change in Momentum} = {Force}*{Time}.

  7. Mar 23, 2005 #6
    Thanxs.. did I set this one up right?
    A Metal sphere with a mass of 40kg rolls along a frictionless surface at 40m/s and strikes a stationary sphere having a mass of 400kg. The firsts sphere stops completely. At what speed does the second sphere move away from the point of impact?
    Va2=(-MbVb2/Ma) = (-400kg)(40m/s)/40kg
  8. Mar 23, 2005 #7
    You should try to be more specific with your equation, but the highlights are, its a closed system, so momentum is conserved. Your initial momentum [tex] \rho_i = m v [/tex] is equal to your final momentum [tex] \rho_f [/tex].

    [tex] \rho_i = \rho_f. [/tex]

    [tex] m_i v_i = m_f v_f [/tex]

    Solving for v_f will give you your answer.

    [tex] v_f = \frac {m_i v_i } { m_f } [/tex]
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