Two cars head on at 50mph

  1. Its probably been done previously but a search didn't return any results for me.

    But anyway,

    Vehicle 'a' (Va) travels head-on in to Vechicle 'b' (Vb). Both are travelling at 50mph. Both vehicles are identical in weight, tyre grip, transmission, general design, road surface etc.

    Is the impact speed 50mph or 100mph?

    The police and news reports always say 100mph. (Probably because higher number usually gets more attention from the viewer).
    But if you look at each vehicle in turn, each one is going through a transition of 50mph down to 0mph in the space of a crumple zone. So each vehicle is surely experiencing an impact speed of 50mph. Is it?


    But then, Va is 'receiving' the crash energy from Vb (and vice versa), so the energy absorption is higher than if either Va or Vb were to crash in to a stationary wall for example.
    So under this view could it be argued that the impact speed is indeed 100mph?

    (I suppose the term 'impact speed' is open to interpretation?)
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,055
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    Hi AJ_2010! :smile:
    I think if you analysed it, you'd find that the crumple zone is shorter, so the acceleration (and damage and injury) is greater.

    But the simple answer is that if you use the frame of reference of either vehicle, the the other vehicle is approaching at 100 mph. :wink:
     
  4. tim - a shorter crumple zone assumes less impact speed no? Therefore acceleration (or experienced 'g-force') would be less?

    Approaching at 100mph yes.
    I think this is why I mentioned at the end of my post that 'impact speed' probably needs interpretation.

    But the main point of my question is that I hear the police and news readers say 'impact speed' all the time. And they simply add up the speeds of the approaching vehicles.

    I was just wondering if this is actually misleading at all?
     
  5. tiny-tim

    tiny-tim 26,055
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    ok, longer instead of shorter :smile:

    no, it's the opposite of misleading, it's exactly the parameter that should be given …

    when you're about to be in a crash, it very much is your frame of reference that is the most relevant to finding what's about to happen to you!
     
  6. Totally agree.

    But if I look at it this way:
    I am in car Va travelling at 50mph and Vb hit us at 50mph.
    The car I am in comes to a stop from 50mph to 0mph within a fraction of a second and within a larger* crumple zone.

    *compared to hitting a stationary wall.

    Is this the same as me travelling at 100mph and hitting a stationary wall? Whereby I decelerate from 100mph to 0mph within a crumple zone of whatever it is and within a time frame of whatever it is.

    As kinetic energy increases on the square of velocity there is MUCH more energy involved in the impact speed of a 100mph car hitting a wall than a 50mph hitting another 50mph car..is there not? I may be wrong as I have not done the actual calcs for this.

    :)

    The wording of 'impact speed' is assuming the car is having to stop from quoted speed to zero. In this case my line of thinking would be that a head-on collision in this example should be an impact speed of 50mph and not 100mph.

    Happy to be proved wrong of course. :)
     
  7. essentially, yes...per tinytims post #4...
    but of course cars crumple differently than a wall so the nature of the deceleration would likely be different....If the wall were a perfect solid and experienced no deformation, for example, likely the single car would crumple a LOT more than if it were only 1/2 of a pair of cars crumpling.....


    no the energy is pretty much the same but how it is disipatted would likely be quite different...damage would be very similar.....look at the situation from the FT = MV perspective....
    it gets to the heart of the matter.....the momentum (mv) is equal to the force (F) of impact times the duration of the impact (T)....so as the duration (T) gets bigger for some momentum, the force (F) gets smaller....the resulting effects are dependent on both force and duration of the force....

    If that isn't especially clear, think about the car not crashing but instead just rolling to a stop from normal friction...you'd hardly feel the small force inside but it would take a relatively long time for the car to come to a stop. And of course no energy, nor momentum, would be lost to deformation of the auto body....

    The above logic is why you may see plastic barrels of sand on highways...the sand gives way gradually when impacted by a car and so extends the duration (T) of energy disippation reducing the impact (F) to passengers...if you just double the duration of impact, you cut the force in half.
     
    Last edited: Dec 13, 2010
  8. Since you're comparing to a wall, I think you need to take two cases into consideration rather than one.

    1) You are moving at 50 mph and the wall is stationary.

    2) You are moving at 50 mph and the wall is also moving at 50 mph.

    In which case would there be a stronger impact?

    The second case, and to make sense of it, if the wall was stationary in this second case, it'd be as if you were moving at 100 mph.
     
  9. Would this be a similar analogy though?
    Is the car going from 50mph to zero in both cases? Or in the moving wall situation - would the car be going from 50mph to -50mph?


    Or to add some 'spice' to these cases : an HGV travelling at 50mph hits head-on a car travelling at 50mph.
    The car would most likely go from 50mph to -20mph (or whatever it happens to be). And the HGV would go from 50mph to 20mph (for example).

    In this case is the impact speed 100mph or the change in velocity of the car? ie.70mph (assuming you are sitting in the car)


    I thought I had it all in place in my head but the replies here have got me thinking harder :)
     
  10. Oh ok, after looking at Naty's post, I think that answers your question the best.

    As another way to think about it (and I guess a direct answer to your question since it's from an mass x acceleration standpoint versus a momentum one), F = ma, where a = dv/dt, the change of velocity over change in time. You're correct about the change in velocity, but the length of time will be shorter for an impact where both cars are moving at 50 mph, resulting in a larger acceleration and a larger force.
     
  11. Hang on, if two objects are spaced out and moving at constant speeds there shouldnt ever be a collision.
     
  12. I'm confused, isn't this the same nature as the question?
     
  13. Yes, you will. And, if you stay in the inertial frame of your initial motion, you'll find that at the end of the collision, you are moving backwards at 50 mph.

    Each vehicle undergoes a 50mph deceleration over the distance of its own crumple zone. So, as far as the vehicle is concerned, it's exactly the same as if you hit a stationary, immovable wall at 50mph.
     
  14. K^2

    K^2 2,470
    Science Advisor

    It depends on what you want to compare it to.

    Two 50mph vehicles head on are equivalent to 100mph vehicle hitting a stationary vehicle. From that perspective, collision speed is 100mph.

    But if you want to compare to collision into solid wall, a 50mph collision will do the same damage as two identical 50mph vehicles head on.

    Since vehicles that collide usually aren't close enough to identical, the second comparison tends to be flawed. The first holds regardless of vehicle masses. So that would probably make more sense.
     
  15. You stated in the case of a an object moving towards a wall (which is also moving) there would be a collision.

    This cant be true as the two particles traveling at constant and equal velocities cant collide.
     
  16. I said they had the same speed, not velocity.

    In that example, the magnitudes of their velocity were the same, the signs (AKA directions) were different.
     
  17. The TV show about myths addressed this issue. Two cars identical going at 50 mph crashing head on will each receive an impact equal to a single car at 50 mph hitting an immovable wall. The crumple distance & forces on the driver were indeed measured and affirmed.

    But the reaction time would be based on a relative speed of 100 mph. E.g., if you were not paying attention & suddenly realized you are headed towards a brick wall, then you hit the brakes. The relative speed is 50 mph between your car & the wall. But if another car is approaching at 50 mph, the relative speed is 100 mph, meaning you'll have less time to slow down before impact.

    It's not a simple cut & dried problem, but it has been proven that the damage to the car(s) & impact force on the occupants is equivalent for 2 cars at 50 mph vs. a car & immobile wall at 50 mph. A single car hitting an immobile wall at 100 mph incurs much greater crumple of the car body, & much greater forces on the occupants. This scenario is more dangerous. Make sense?

    Claude
     
  18. boneh3ad

    boneh3ad 1,609
    Gold Member

    The crash is effectively at 50 mph, not 100 mph. Think of it in terms of energy. If you have one car hitting a wall at 50 mph, you have the energy from that one car (let's call it E), and it is dissipated by that one car (the wall is stationary). If you have two cars moving at 50 mph, they both have energy E, but the mass is also doubled, so you are dissipating twice the energy but with twice the mass.

    You could look at it in terms of forces if you wanted as well... or momentum. No matter how you slice it, a head on crash with two cars going 50 mph is the same as hitting a wall at 50 mph assuming the cars have the same mass.

    This actually was tested on the Mythbusters and they got it wrong the first time, so they revisited it after all the fans complained and got it right the second time.



    This isn't true either. Two vehicles moving at 50 mph will again, have 2E energy distributed between the two cars. That means that each car dissipates E amount of energy. If you have a stationary car and a 100 mph car, the total energy is 4E and the mass is 2M, so each car is dissipating 2E. That is going to be a much worse crash. To compare the head on collision to someone hitting a stationary car, the moving car would have to be moving at about 70 mph (50[itex]$\sqrt{2}$[/itex]).

    Apologies if the LaTeX is messed up. This site always acts so goofy when it translates it...
     
    Last edited by a moderator: Sep 25, 2014
  19. tiny-tim

    tiny-tim 26,055
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    Yes, if two cars at 50 mph simultaneously hit opposite sides of the same brick wall, then that's obviously the same as if the wall wasn't there.

    Putting the wall back: if the wall doesn't move at all, the effect on one car of hitting it at 50 mph is the same as the effect (on that car) of two cars crashing head-on each at 50mph.
     
  20. Fredrik

    Fredrik 10,155
    Staff Emeritus
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    Only if the other guy doesn't hit the brakes too.

    As long as we're neglecting friction between the cars and the ground, it's pretty obvious that velocities 50 and -50 (relative to the ground) are equivalent to 0 and 100. In both cases, the velocities are 50 and -50 in the center-of-mass frame, where the cars will eventually come to rest.

    You need to refresh and resend after each preview. It's not supposed to be that way, but no one knows what's causing it or how to solve it.
     
    Last edited: Dec 13, 2010
  21. K^2

    K^2 2,470
    Science Advisor

    Wrong. Final velocity of a 100mph car in a collision with stationary one is 50mph, not 0mph. Both the energy dissipation and momentum transfer are identical to 50mph head-on collision.

    Not to mention the fact that this is a simple coordinate system transformation.
     
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