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Mechanical Energy of Collision

  1. Jul 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A 12000 kg car traveling at 10 m/s strikes and couples with a 6000 kg car at rest. What is the loss of mechanical energy in the collision?
    Answer is 200,000 J.


    2. Relevant equations

    I want to know HOW to arrive at the correct answer above.

    3. The attempt at a solution

    Relevant equations tried:

    1. Speed at final combination = Vf=(m1/m1+m2)Vi
    =(12000/12000+6000)(10)
    = 6.7 m/s

    2. Kf-Ki
    K= 1/2 mv^2
    Kf= 1/2(12000)(6.7)^2 + 1/2(6000)(6.7)^2
    = 269340 + 134670
    = 404010
    Ki= 1/2(12000)(10)^2 + 1/2 (6000)(10)^2
    = 600,000 + 300,000
    = 900,000

    K=Kf-Ki
    =404010 - 900,000
    = -495990
    =Not even close to the right answer.

    I've done this problem over and over for two hours, not kidding, trying different combinations of equations, can someone please tell me where I am going wrong?

    Thank you very much!
    =

    K
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 6, 2009 #2

    turin

    User Avatar
    Homework Helper

    It looks like you just messed up the initial kinetic energy calculation. How fast is the 6000 kg car travelling, initially?
     
  4. Jul 6, 2009 #3
    Thank you for replying! The 6000 kg car is at rest.
     
  5. Jul 6, 2009 #4
    So would that be:

    Ki= 1/2(12000)(10)^2 + 1/2 (6000)(0)^2
    = 600,000 + 0
    = 600,000

    Kf-Ki=
    404010-600,000 = -195990, still incorrect. :(
     
  6. Jul 6, 2009 #5

    turin

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    Homework Helper

    If you round 20/3=6.7 to only two sig. figs., then you should round -195990 to only two sig. figs., and then it is correct. However, what you should really do is keep 20/3, instead of brutally rounding it off at the beginning. Without rounding, it is correct exactly as given.
     
  7. Jul 6, 2009 #6
    I apologize, I'm new to this. I don't understand what you mean by 20/3, which is not part of any of the equations. Should it be?

    But you are so right about the rounding! Thank you! I was going crazy trying to figure it out:

    Kf = 1/2(12000)(6.66666667)^2 + 1/2(6000)(6.66666667)^2
    = 266666.7 + 133333.3
    = 400,000

    and the revised (thanks!)
    Ki= 1/2(12000)(10)^2 + 1/2 (6000)(0)^2
    = 600,000 + 0
    = 600,000

    So Kf-Ki =
    400,000 - 600,000 = -200,000, negative because the energy is lost, correct?

    Thank you ever so much, you have very much made my day!
     
  8. Jul 6, 2009 #7

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    He meant, in the calculation of the speed after the collision:

    (12000/18000)(10 m/s) = (dividing both numerator and denominator by 6000) = (2/3)(10 m/s)

    = 20/3 m/s

    This is the exact final velocity. Keeping this until the end will prevent the propagation of rounding errors. Try to simplify everything as much as possible before whipping out the calculator.
     
  9. Jul 6, 2009 #8
    Ok, I get it, thank you very much!
     
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