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Momentum Conservation - so hard!

  1. Jul 12, 2007 #1
    Momentum Conservation --- so hard!

    1. The problem statement, all variables and given/known data
    Two automobiles collide at an intersection. One car of mass 1.4 X 10^3 kg is traveling at 45km/h ; the other car of mass 1.3 X 10^3 kg is traveling at 39km/h [E]. If the cars have a completely inelastic collision, what is their velocity just after collision?


    2. Relevant equations
    v1m1 + v2m2 = v(m1 + m2)


    3. The attempt at a solution

    Given information:
    m1 = 1.4 X 10^3 kg
    m2 = 1.3 X 10^3 kg
    v1 = 45km/h --> 12.5 m/s
    v2 = 39 km/h [E] --> 10.83 m/s
    vf ' = ?

    v1m1 + v2m2 = v(m1 + m2)
    (12.5)(1.4 X 10^3) + (10.83)(1.3 X 10^3) = v (1.4 X 10^3 + 1.3 X 10^3 kg)
    v = 11.7 m/s --> 42 km/H

    But the answer that I get is wrong. So anyone please help me how to do this. Thanks.
    The correct answer is 30 km/h.
     
  2. jcsd
  3. Jul 12, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    vector addition

    Realize that momentum is a vector. Add the momentum vectors to get the total momentum of the cars after they collide and stick together.
     
  4. Jul 12, 2007 #3
    I am really confused.

    Do I do the following:[tex]\sqrt{10.8^2 + 12.5^2}[/tex]

    I am not sure what to do since there are so many ways to do this, and I am not sure where to begin.
     
  5. Jul 12, 2007 #4

    Doc Al

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    Staff: Mentor

    Not exactly. You added the squares of the velocities; you should have added the squares of the momenta = mass*velocity.
     
  6. Jul 12, 2007 #5
    Oohhh. ok.

    I did this:

    p1= m1v1
    = (1.4 X 10 ^3)(12.5)
    = 17500 kg.m/s

    p2= m2v2
    = (1.3 X 10 ^3)(10.8)
    = 14040 kg.m/s

    [tex]\sqrt{17500^2 + 14040^2}[/tex]
    = 2.24 X 10^4 kg.m/s

    Now I think I got it:
    2.24 X 10^4 = v (m1 + m2)
    v = (2.24 X 10^4)/(1.4 X 10 ^3 + 1.3 X 10 ^3)
    v= 8.296296296 m/s
    v = 29.86666667 km/h

    Therefore, the speed is 30 km/h.

    Wow. Thanks for this Doc Al. I appreciate your help very much. You are the best.
     
    Last edited: Jul 12, 2007
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