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The initial speed of two moving bodies given an increment of velocity.

  1. Jun 3, 2012 #1
    Car A has twice the mass of Car B but only half as much Ek (Kinetic Energy).
    When both cars increase their speed by 6.00m/s their resulting Ek's are equal

    What were the initial speeds of both cars?

    ANS 1= 4.24m/s
    ANS 2= 2.12 m/s





    Ke=1/2mV^2





    m(a)=2m(b)

    Ek(a)=1/2Ek(b)

    Ek(a)=1/2mV^2=1/2Ek(b)=1/2(1/2mV^2)
    Ek(a)=1/2mV^2=1/4mV^2
    Ek(a)=1/2mV^2= 1/4(2m)V^2
    1/2mV^2=1/2mV^2

    1/2m(vi+6.00m/s)^2=1/2m(vi+6.00m/s)^2

    Now, i realize that i must have made a mistake somewhere because this expression as i have demonstrated as equal.

    However this is not the answer.
    I am under the assumption that my mistake lies in how i made the expressions equal to one another.

    Can anyone find the flaw here?
    Merci.

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



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jun 3, 2012
  2. jcsd
  3. Jun 3, 2012 #2
    Hi LiamC, welcome to PF! :smile:

    There seems to be something wrong with the question...

    If, when both the cars increase their speed by 6m/s, their speed becomes same, then, their original speeds should have been the same too!! So you cannot have different speeds for both cars!! (I think you mis-typed kinetic energy as speed?)
     
  4. Jun 3, 2012 #3
    Yup, sorry my bad.

    It's kinetic energy in place of speed.
     
  5. Jun 3, 2012 #4
    Okay, so what you have done in your attempt is just loop back the given data on itself, and that leads you to nowhere. You also seem to have assumed their initial speeds are equal, which is wrong. :wink:

    Try framing the equations this way.

    [tex]m_A = 2m_B = 2m[/tex]

    And,

    [tex]\frac{1}{4}mv_B^2 = \frac{1}{2}(2m)v_A^2[/tex]

    This gives you a relation between [itex]v_A[/itex] and [itex]v_B[/itex]. Now use the second part of the problem which says the kinetic energies are equal if their speeds increase by 6m/s. How would you write equation for this?
     
  6. Jun 3, 2012 #5
    I added subscripts to your masses to make things more clear.


    You set up the equations wrong. When you replaced m with 2m on the right hand side, you were replacing mb with 2ma, when ma = 2mb.
     
  7. Jun 3, 2012 #6
    Would that be

    1/4m(vi(b)=6.00m/s)^2=1/2(2m)(vi(a)+6.00m/s)^2
     
  8. Jun 3, 2012 #7
    Noo! Why the 1/4?? What is the kinetic energy of a body in terms of m and v? The RHS is correct, by the way :smile:
     
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