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Cars colliding into each other.

  1. Jun 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Cars B and C are at rest with their brakes off. Car A plows into B at high speed, pushing B into C. Assume the collisions are completely inelastic. What fraction of the initial energy is dissipated in the collision involving car C?
    Data: MA = 1010 kg; MB = 1250 kg; MC = 1470 kg.



    2. Relevant equations
    Pinitial=Pfinal
    KE=.5mv^2

    3. The attempt at a solution
    Pi=Pf
    m_a*Vo=(m_a+m_b+m_c)*Vo'
    Q+.5m_aVo^2=.5(m_a+m_b+m_c)Vo'^2; where Q is the kinetic energy lost

    I'm pretty sure these equations are correct but I don't know where to go from here?
     
  2. jcsd
  3. Jun 20, 2011 #2

    gneill

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    Staff: Mentor

    Treat the problem as two separate collisions. Prior to any collision, car A has all the available kinetic energy. A collides with B and they stick together and move with a new velocity with momentum is conserved. The pair then have a collision with C, resulting in another new velocity for the trio (KE final).
     
  4. Jun 21, 2011 #3
    Ok, so how would I be able to find the fraction of the first one?
     
  5. Jun 21, 2011 #4

    gneill

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    Staff: Mentor

    Find the KE after the first collision, then the remaining KE after the second collision. The difference between these two is the KE lost in the second collision. Compare this with what you started with before any collisions.
     
  6. Jun 21, 2011 #5
    Ok sorry, I'm struggling with this one. So I have:
    for the first collision:
    MA*Vo=(MA+MB)*Vo'
    then I solved for Vo'=((MA/(MA+MB))Vo

    then plugged it into:
    .5MAVo^2=.5(MA+MB)Vo'^2=.5((MA^2)/(MA+MB))-Q; I assumed Vo=1
    then solved for Q, but then I am unsure of:
    1. if this makes any sense
    2. if .5MAVo^2=Eo then going forth from here can I do the same for the second collision solving for Q2 then add them together and end up with:
    (Eo-(Q+Q2))/Eo for the fraction?
     
  7. Jun 21, 2011 #6

    gneill

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    Staff: Mentor

    Kinetic energy isn't conserved in an inelastic collision. So don't try to write conservation of kinetic energy equations :smile:

    You found the velocity Vo' of the car A+B tangle, and that will yield the kinetic energy of the pair as it moves towards car C. Also, there's no need to assume that Vo=1. Just keep Vo as Vo... it'll cancel out in the end when you form your ratio of energies.

    The KE of cars A+B is (1/2)(MA+MB)*Vo'^2. You can plug in the expression for Vo'.

    You've got Vo' and the mass of the A+B pair, so determine the final velocity after the A+B+C collision using conservation of momentum. Use that to find the expression for the final KE.
     
  8. Jun 23, 2011 #7
    I'm not sure I follow because if i use consveration of momentum again ill just end up with another variable Vo'' because if i do:
    (Ma+Mb)Vo'=(Ma+Mb+Mc)Vo'' it doesnt work out.
    so i am confused on how to find final KE
     
  9. Jun 23, 2011 #8

    gneill

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    Staff: Mentor

    The first car, A alone, starts off with some velocity Vo. It collides with and 'sticks' to car B, and together they end up with some velocity Vo' that is a calculable fraction of Vo. The pair also retains the original momentum.

    This pair then collides with car C, resulting in yet another velocity, say Vo''. It will be a fraction of the the velocity Vo'. Vo' and Vo'' can therefore be written in terms of the original Vo by substitution.
     
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