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Linear Momentum

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    A 5 g box moving to the right at 20 cm/s makes an elastic head on collision with a 10 g box initially at rest.

    Mass 1 = .005 kg
    initial Velocity 1 = .2 m/s
    Mass 2 = .01 kg
    initial Velocity 2 = 0 m/s

    a.) what velocity does each box have after the collision?
    b.) what fraction of the initial kinetic engergy is transferred to the 10 g box?

    2. Relevant equations
    - (initial velocity 1 - initial velocity 2) = velocity final 2 - velocity final 1
    Mass 1 (initial velocity 1) + Mass 2 (initial velocity 2) = Mass 1 (velocity final 1) + Mass 2 (velocity final 2)
    Kinetic energy = .5(mass)(velocity)2

    3. The attempt at a solution

    a.) for the final velocities...
    - (Vo1 - V02) = Vf2 - Vf1
    - (0 - .2) = Vf2 - Vf1
    .2 + Vf1 = Vf2

    for Vf1...
    M1V01 + M2V02 = M1Vf1 + M2Vf2
    (.005)(.2) + (.01)(0) = (.005)Vf1 + (.01)(.2 + Vf1)
    .001 = .015(Vf1) + .002
    -.001 = .015(Vf1)
    -.067 [rounded number] = Vf1

    plug into first problem
    .2 + Vf1 = Vf2
    .133 = Vf2

    b.) [tex]\sum[/tex] K0 = [tex]\sum[/tex] Kf [not sure if this is the correct method for solving this equation...]
    K01 + K02 = Kf1 + Kf2
    .5(.005)(.2)2 + .5(.01)(0)2 = .5(.005)(-.067)2 + .5(.01)(.133)2

    Im not sure if its K01/Kf2 or K01/ (Kf1 + Kf2)
     
  2. jcsd
  3. Dec 7, 2008 #2

    rl.bhat

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    Homework Helper

    a.) for the final velocities...
    - (Vo1 - V02) = Vf2 - Vf1
    - (0 - .2) = Vf2 - Vf1
    .2 + Vf1 = Vf2
    The above is true when the masses are equal.
     
  4. Dec 7, 2008 #3
    Oh I see. I'm unsure now how this would work since there are 2 variables that I need to find.
     
  5. Dec 8, 2008 #4

    rl.bhat

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    Homework Helper

    Use (.005)(.2) + (.01)(0) = (.005)Vf1 + (.01)(Vf2) ....... (1)
    and .5(.005)(.2)2 + .5(.01)(0)2 = .5(.005)(Vf1)2 + .5(.01)(Vf2)2 .......(2)
    Solve the above equations to get the final velocities.
     
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