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Inelastic Collision with Friction

  1. Jul 2, 2007 #1
    1. The problem statement, all variables and given/known data

    You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 [tex]kg[/tex], was stopped at a red light when it was hit from behind by car A, of mass 1400 [tex]kg[/tex]. The cars locked bumpers during the collision and slid to a stop. Measurements of the skid marks left by the tires showed them to be 7.25 [tex] m [/tex] long, and inspection of the tire tread revealed that the coefficient of kinetic friction between the tires and the road was 0.650.

    What was the speed ([tex] v [/tex]) of car A ([tex]mph[/tex]) just before the collision?

    By how many [tex]mph[/tex] was car A exceeding the speed limit 35.0 [tex]mph[/tex]?

    2. Relevant equations

    I have tried
    1.) [tex]m_1 v_1i + m_2 v_2i = (m_1+m_2)v_f[/tex] (does not work)
    2.) [tex]v_f^2-v_o^2=2-U_k(g)D[/tex] (does not work)

    3. The attempt at a solution

    The first equation above I know will not work because it was designed for a perfect inelastic collision. I need to take into account the friction, that is why I used equation two.

    [tex]v_f = 0 m/s[/tex]
    [tex]v_0 = ???[/tex] (need to find for problem #1)
    [tex]U_k = 0.650[/tex]
    [tex]g = 9.8 m/s^2[/tex]
    [tex]D = 7.25m[/tex]

    Any ideas to what I could be doing wrong?
  2. jcsd
  3. Jul 2, 2007 #2

    Doc Al

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    Staff: Mentor

    You need both of those equations. First there is an inelastic collision (equation #1), then there is sliding (equation #2).

    Work backwards!

    Good. Solve for v_0, which is the speed of the stuck cars after the collision.
  4. Jul 2, 2007 #3
    Hmmm... now I understand. [tex]v_0[/tex] in problem 2 = [tex]v_f[/tex] in problem 1.

    Problem 2:
    [tex]v_0 = 6.647 m/s[/tex]

    Problem 1:
    [tex]m_1 v_1i + m_2 v_2i = (m_1+m_2)v_f[/tex]
    [tex]1400 v_1i + 2100*0 = (1400+2100)6.647[/tex]
    [tex]v_1i = 16.618 m/s[/tex]

    is 16.618 m/s right?
  5. Jul 2, 2007 #4

    Doc Al

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    Staff: Mentor


    Correct approach, but revisit your calculation. (I think you forgot to multiply by 2.)

    Also the correct approach. But redo with revised numbers.

    And when you're done you'll have to convert from m/s to mph in order to answer the questions.
  6. Jul 2, 2007 #5
    First I wanted to just say thanks for the help!

    When I perform the calculation for [tex]v_f^2-v_o^2=2-U_k(g)D[/tex] I always get 6.647 m/s ??

    [tex]v_f = 0 m/s[/tex]
    [tex]v_0 = ???[/tex]
    [tex]U_k = 0.650[/tex]
    [tex]g = 9.8 m/s^2[/tex]
    [tex]D = 7.25m[/tex]

    [tex]0^2-v_o^2=2-0.650*9.8*7.25[/tex] = 6.647 m/s ???

    Where do I multiply by 2???

    Also, thanks for the heads up on the unit conversion. I almost forgot.
  7. Jul 2, 2007 #6

    Doc Al

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    Staff: Mentor

    Right here:
    Spot the 2!
    Let me rewrite that equation more clearly:
    [tex]v_f^2-v_o^2= -2 \mu_k g D[/tex]
  8. Jul 2, 2007 #7
    Thanks for the help Doc Al!

    I was able to get the right answer of 53.748 mph for question 1 and 18.8 mph for question 2.
  9. Aug 9, 2011 #8
    Thx this hlp me solved similar question.But may i ask why we should consider the acceleration as 9.8 m/s^2 ?
  10. Aug 9, 2011 #9

    Doc Al

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    Staff: Mentor

    Who says the acceleration is 9.8 m/s^2? (That's g, a constant describing the earth's gravity.)
  11. Aug 9, 2011 #10
    Because the average gravity of Earth is approximately 9.8 m/s2, (9.80665). Oh well, the vertical acceleration. Therefore, it's might be considered as convenient. What do I know. ^_^
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