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2D Momentum Problem

  1. May 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Car 1 weighing 1000 kg crashes into the rear of parked car 2 weighing 1100 kg and stops. Car2 moves ahead 2 ms2. What is the velocity of car1 at impact?


    2. Relevant equations

    (m1v1+m2v2)=(m1v3+m2v4)


    3. The attempt at a solution

    (m2V4)/m1
    (1100*2)/1000
    2.2ms2= 2200/1000


    Okay, now I know this is right, but what I don't know is how do they arrive at the equation (m2v4)/m1 ????
     
  2. jcsd
  3. May 16, 2012 #2
    If you examine your original conservation of momentum equation, let [itex]v_1, v_3[/itex] be the initial and final speeds of car 1, respectively, and let [itex]v_2,v_4[/itex] be the initial and final speeds of car 2, respectively. From the context of the problem, what must be true about speeds [itex]v_2[/itex] and [itex]v_3[/itex]?
     
  4. May 17, 2012 #3
    (1000*2.2+1100*0)=(1000*0+1100*2)

    V2 and v3 have no velocity so they have no momentum...correct?
     
  5. May 17, 2012 #4
    Correct.
     
  6. May 17, 2012 #5
    So because they have no momentum they cancel out which means you rewrite the equation to solve for velocity1

    (1000*2.2+1100*0)=(1000*0+1100*2)

    (1000)=(1100*2)

    (m1)=(m2v4)

    (m2V4)/m1

    (1100*2)/1000

    2.2ms2= 2200/1000

    Is that how it is done?
     
  7. May 17, 2012 #6
    Yes, that's the idea. Conservation of momentum dictates [itex]m_1 v_1 = m_2 v_4[/itex], solving for the initial speed is just an algebra problem.
     
  8. May 17, 2012 #7
    Thanks Steely Dan... that wasn't as hard as I thought it would be... mind you that was an easy momentum problem for you, but difficult for me. I imagine it would be harder if you had 1 vehcile t-bone another and they both have pre and post impact velocities.
     
  9. May 17, 2012 #8
    I was just re reading your posts...

    ...when I re read this post the light started to flicker.

    Thanks again Steely Dan
     
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