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Momentum/Impusle problem [revised]

  1. Dec 16, 2007 #1
    1. The problem statement, all variables and given/known data

    So my problem says: "Car A (in a frictionless surface) travels at a velocity and weighs 1100 kg, Car B is parked and weighs 2200 KG, they collide in an inelastic collision (meaning they stick together) and they move a distance of 2.8 M together. The road has a coefficient of friction of 0.4"

    2. Relevant equations

    Find the velocity of the Car A before impact

    3. The attempt at a solution

    Last time I posted this question, our professor did not give us a coefficient of friction, which deemed the problem impossible to solve with mathematical values, but now that he realized his mistake, he gave us the problem again (an a re-test) so I just wanted to know if I'm along the right lines.


    [tex]F_{f}[/tex] = [tex]U_{k}[/tex]mg,


    F = [tex]F_{f}[/tex],

    we can asume that

    ma = [tex]U_{k}[/tex]mg

    and the masses would divide out, so

    a = [tex]U_{k}[/tex]g
    a = (0.4)(-9.82 [tex]m/s^{2}[/tex])
    a = -3.982 [tex]m/s^{2}[/tex]

    so then we can use the kinematic equation of

    [tex]V_{f}[/tex]^2 = [tex]V_{o}[/tex]^2 + 2ad

    where [tex]V_{f}[/tex] = 0 (this is the only part I'm iffy about), so

    [tex]V_{o}[/tex] = [tex]\sqrt{-2ad}[/tex]
    [tex]V_{o}[/tex] = 4.7 m/s

    Is this correct? thank for the help :)
  2. jcsd
  3. Dec 16, 2007 #2
    There will be a vf that is nonzero, unless car A accelerates to 0m/s before it collides with car B
  4. Dec 16, 2007 #3
    Wouldn't the fact that they crashed and there is a force of friction mean they'll eventually come to a rest?

    EDIT: but not over that distance, now I see what I did.

    So should I approach it some other way? maybe through work?
    Last edited: Dec 16, 2007
  5. Dec 16, 2007 #4
    Yes eventually but momentum has to be conserved and right after they collide they will still be moving because if the collided them car A had a momentum that was not 0 so then the momentum right after the crash will have to get a nonzero number as well.

    Edit: Work backwards, you can use the fact that the two cars went 2.8m after the collision to your advantage.
  6. Dec 16, 2007 #5
    Well since I have a distance, and an acceleration and a mass, I can get work.

    W = F[tex]\Delta[/tex]d
    W = (m1 + m2)ad
    W = (3300)(-3.982)(2.8)
    W = -36793.68 J

    and since W = KE

    W = 1/2mv^2
    v = [tex]\sqrt{2W/m}[/tex]
    v = 4.72 m/s

    what? ok now I'm lost.
  7. Dec 16, 2007 #6
    Am I missing something in my interpretation of work?
  8. Dec 16, 2007 #7
    The acceleration was the value before they collided, that wont be the acceleration after they collide.
  9. Dec 16, 2007 #8
    Hmmmm alright, so I'm guessing kinematic equations won't work... I'm lost, can I get a clue please?
  10. Dec 16, 2007 #9

    Doc Al

    User Avatar

    Staff: Mentor

    View the problem as having two parts:
    (1) The collision. Hint: What's conserved?
    (2) The coming to a stop due to friction (after the collision). What's the acceleration?

    Begin by looking at part two and figure out the initial speed of the stuck-together cars immediately after the collision.
  11. Dec 16, 2007 #10
    well I know that Momentum is conserved, however kinetic energy isn't (inelastic collisions, do'h)

    and the acceleration would be the the coefficient of friction times acceleration due to gravity (like shown above).


    [tex]m_{1}[/tex][tex]v_{1}[/tex] = ([tex]m_{1}[/tex]+[tex]m_{2}[/tex])[tex]V^{'}[/tex]

    hmmmm... Where can this get me?
    Last edited: Dec 16, 2007
  12. Dec 16, 2007 #11

    Doc Al

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    Staff: Mentor

    Good. (I see you've already calculated the initial speed of the stuck-together cars immediately after the collision.)

    Good. Use conservation of momentum to figure out the initial speed of car A.
  13. Dec 16, 2007 #12
    OHHHHHH how can I be so dumb!!! that speed I Calculated is V'!!! wow I'm dumb!

    So now it should be:

    [tex]m_{1}[/tex][tex]v_{1}[/tex] = ([tex]m_{1}[/tex]+[tex]m_{2}[/tex])[tex]V^{'}[/tex]
    [tex]v_{1}[/tex] = (([tex]m_{1}[/tex]+[tex]m_{2}[/tex])[tex]V^{'}[/tex]) / [tex]m_{1}[/tex]
    [tex]v_{1}[/tex] = ((3300)(4.72))/1100
    [tex]v_{1}[/tex] = 14.16 m/s

    I think this should be right, is it?
    Last edited: Dec 16, 2007
  14. Dec 17, 2007 #13

    Doc Al

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    Staff: Mentor

    Looks good!
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