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Does 1 car hitting a wall at 120mph = 2 cars colliding, each going 60mph?

  1. Oct 21, 2011 #1
    I've been searching around the internet and there doesn't seem to be a straight answer for this. People just spit out an answer and don't back it up with any explanation.

    This is a complex situation in real life, so let's assume both cars are of the same mass and simplify the scenario by assuming the simplest collision with no fusion of the two cars into one mass etc.

    Although this is a simplification of the real life scenario, i am curious if the energy to stop a car going 120mph that hits an immovable wall is equivalent to the energy to stop one of the two cars (each going 60mph) that collide and both stop at the collision point. Can anyone show this using physics formulas?
     
    Last edited: Oct 21, 2011
  2. jcsd
  3. Oct 21, 2011 #2

    Nabeshin

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    No.

    The kinetic energy of an object is equal to 1/2mv^2. Thus, the situation with the car going 120mph actually has twice the kinetic energy of the two cars each going 60mph.
     
  4. Oct 21, 2011 #3
    The short answer is no. The 120mph collision is harder. The common sense reason is because a wall barrier has infinite mass and infinite stiffness.
    For the explanation you can take two approaches: conservation of momentum or conservation of energy. But you have to be careful how you define the systems you are analyzing.

    Energy is not conserved during these collisions (theoretically it could be conserved if you were able to account for all the energy dissipated in heat, sound, deformation...etc)

    The energy of the all system is:
    Total Energy=KE+PE = 1/2*M*V² + 0 (assuming ΔZ's are zero)
    If you define
    system 1 as: barrier wall + veh@120
    and system 2 as: veh1@60 +veh2@60
    You will notice that system1 has more energy than system 2.

    Momentum is conserved in the 2 vehicles system:
    eq 1) m1*V1+m2*V2=m1*V1'+m2*V2' (V's are the velocities after impact)
    eq 2) restitution= (V2'-V1') / (V1-V2)
    by solving both equations you will find formulas for V's for any given restitution coefficient.
    The following relationships should hold true for any collision:
    (closing speed)+(closing speed)*restitution=ΔVeh1+ΔVeh2 (Δ is change of velocities)
    m1/m2 = ΔVeh2/ΔVeh1

    The change of velocity and the deformation of the veh@120 will tell a different story than the other 2 cars.

    PS: Crash test engineers use Barrier Equivalent Velocity (BEV) to analyze and compare vehicle collisions. You can find good technical info about it on the SAE website. Also mithbusters did an episode on this, cause the cocky one with the mustache wanted to be convinced :) Regards,
    JP
     
  5. Oct 21, 2011 #4
    Both answers above seem right on.

    The "harder" wall means the single vehicle will likely deform more suddenly (faster) than when two vehicles are involved and both deform....so from an FT = MV perspective, the force of the collison will be spread over a shorter time T when the wall is involved, increasing F.
     
  6. Oct 21, 2011 #5
    All of the above is definitely true with one big caveat: it assumes a "sticky" collision. That is, your car is reduced to a thin layer of smashed matter on the wall or the front fender of the other car. And of course it also assumes equal masses of the cars and infinite mass of the wall.

    If you consider "bouncy" collision where a small plastic bubble of a car hits 18-wheel truck & trailer unit (a moving wall in effect ) AND is thrown back (rather than becoming an integral part of its fender), the situation changes drastically.

    Ignoring any loss of energy due to deformations etc, as a first rough approximation, the speed of your car changes from +V not to -V (sticky case), not even to -2V (as many people mistakenly think), it's -3V. Remembering the origanal speed too, the change of speed becomes 4V and the energy your broken body needs to absorb 16x the energy of the sticky collision in a car vs car scenario or 4x in the double speed vs stationary wall (still sticky too) scenario.

    And the bottom line? Dunno... don't try to prove your right of way to the large trucks perhaps?
     
  7. Oct 22, 2011 #6
    well! Its undoubtedly 'no!. The energy imparted to the wall by a car moving with velocity 120 mph doesnot happen to be same as that of two cars moving wid velocity of 60 mph provided that they have same mass...as simple as that!
     
  8. Oct 22, 2011 #7

    A.T.

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    What if the energy does happen to be the same in both cases:
    - 120mph against a wall
    - 84.852mph frontal against an identical car going at the same speed

    In which one would you rather be, and why?
     
  9. Oct 22, 2011 #8
    In one case, your speed changes from 120mph to zero. In the other - 84.852 to zero. Do I need to say more?
     
  10. Oct 22, 2011 #9
  11. Oct 22, 2011 #10

    A.T.

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    Yes please elaborate on your reasoning. Keep in mind that you can change the speed of a car from 120mph to zero without any damage. So the change in velocity cannot be the only factor, can it?

    My example was meant for kapil phyreak, who argued based on total energy.
     
  12. Oct 22, 2011 #11
    We always assume "everything else being equal", otherwise we'll have to go into such things as crumbling capacities of the car frame, bonnet, engine. Its size and geometry. In short, it becomes a serious, large and costly project to answer.

    But, everything being equal (or even not so equal but within reason) you are still stuck with all that extra energy that has to go somewhere.
     
  13. Oct 22, 2011 #12

    A.T.

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    But it is not everything else being equal, aside of the amount of energy. That "somewhere" where the energy is going to is different in the two cases. So one cannot argue based on total energy alone. That was the point of the example.
     
  14. Oct 22, 2011 #13
    Suppose I pick my reference such that the car is approaching me at 60 mph, and the wall is approaching me from the opposite direction at 60mph. From the perspective of the wall, the car approaches at 120mph, yet from mine, they are both going 60 mph.

    Yet somehow my calculations of the same event would be different, and that's no bueno. It's because of that damned nonlinear relationship in kinetic energy that this gets confusing... Momentum provides the clear picture here.
     
  15. Oct 22, 2011 #14
    Didnt Discovery Channel's Mythbusters did this?
     
    Last edited: Oct 22, 2011
  16. Oct 22, 2011 #15

    cmb

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    OK, here's an addendum for you to chew on!:

    If you drive your car at 10mph into a solid wall that does not deform at all, and come to zero mph in a 'sticky' collision over 10ms (say), then your car gets bent.

    If you drive at 50mph into the back of a very solid truck that does not deform at all (for the sake of the question, let us assume it is 'of infinite mass' and was travelling at 40mph both before and after), and end up decelerating in a 'sticky' collision to 40mph over the same 10ms period as the first, your car gets bent.

    Does your car get bent (absorbs more collision energy) more, or less, or the same between these two scenarios?

    Consider the energy expended, before versus after, in each collision. Which is greater?

    (PS, the latter happened to me when someone swung out in front of me and I hit them from behind. It took me a while to figure out why what happened to my car was what happened. It was not immediately obvious to me.)
     
    Last edited: Oct 22, 2011
  17. Oct 22, 2011 #16
    Yes you are missing the ΔT factor that the Momentum Analysis gives you. Both scenarios are pretty bad for the drivers, no doubt. But they won’t necessarily have the same deceleration rates during impact.
     
    Last edited: Oct 22, 2011
  18. Oct 22, 2011 #17
    Agreed! If you are trying to determine if two impact scenarios are equivalent or "similar" like the original post thread, then you need the momentum analysis also. Total energy only gives you one side of the picture. I can bring my car from 60mph to zero at the stop sign….the big difference during an impact is the ΔT factor jeje And that ΔT depends of what you are hitting or the "somewhere" where the energy is going like you said.
     
  19. Oct 23, 2011 #18

    cjl

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    The 84.852mph case, without a doubt. In that case, the collision energy is dissipated into two cars, while in the 120mph case, it all goes into a single car.
     
  20. Oct 23, 2011 #19

    A.T.

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    Yes, exactly.
     
  21. Oct 25, 2011 #20
    what I don't understand here is that why doesn't the energy dissipate into the wall as well?
     
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