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Two cars traveling toward each other question

  1. Dec 6, 2007 #1
    An ongoing argument that i can't seem to settle.

    Say two identical cars with the same total weight are traveling at each other at 100km/h and hit head on.

    Would this be the same as one vehicle hitting a brick wall at 100km/h or would the speeds combine and come to 200km/h?

    Assume that the brick wall will "give" or crush to the same degree as the crumple zone of the cars (or simpler, assume in the second example one vehicle strikes another one that is backed up against an immovable brick wall).


    Thanks for your input!
     
  2. jcsd
  3. Dec 6, 2007 #2
    Nice question !
    It's the same, because of the conservation of momentum, the two cars will be at rest right on, exactly similar to the case a car hit the brick wall.

    From this example, we can see that in a head on crash, the heavier the car, the less risk for the driver because the final velocity will still be on the same direction with the heavier car. The smaller car will be smashed back.
     
  4. Dec 6, 2007 #3
    soo they will just hit each other and the energy won't become like a car running into a brick wall at 200km/h
     
  5. Dec 6, 2007 #4

    dst

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    In opposing directions? It's the equivalent of hitting a brick wall at 200kmh. Let's change the reference point in space to one of the cars, and let's say this car is stationary relative to everything else. Then, the other car is already approaching at 100kmh, and also, because of the shift in reference point, the speed of the newly stationary car has to be added onto the opposite side.
     
  6. Dec 6, 2007 #5

    Danger

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    Let's put it this way: if some idiot driver does something to cause a collision, I'd sure rather T-bone him than hit head-on.
     
  7. Dec 7, 2007 #6
    The velocity of one car in the other car's reference is 200km/h, but the impact is just similar to the 100km/h smash into a brick wall because the 2nd car is not a firmly standstill object like the wall.
     
  8. Dec 11, 2007 #7
    In the case where you assume a perfectly inelastic collision, there is no difference between hitting a solidly built, unyielding, brick wall or an identical car travelling toward you at the same speed.

    If the brick wall yields a bit, again assuming perfect inelasticity, you're better off hitting the brick wall.

    In the real world, collisions are always at least slightly elastic.
     
  9. Nov 14, 2010 #8
  10. Nov 14, 2010 #9

    Danger

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    I saw that Mythbusters, Chug. Cool work-up to the final act. I am retracting my previous statement because I missed out on a serious factor in the initial post. It specified a head-on collision, which I somehow overlooked. I stand by my opinion that T-boning a car is better than hitting it head-on, simply because it will crumple differently, but that wasn't part of the original question. My apologies if I misled anyone.
     
  11. Nov 14, 2010 #10
    It was pretty interesting. And I thought of your orginal comment as a funny statement, so no harm done. haha
     
  12. Nov 14, 2010 #11

    Danger

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    That is my ideal world, where stupidity is mistaken for humour. :biggrin:
     
  13. Nov 15, 2010 #12
    Only if the wall ends up traveling at 100 km/h in the original direction of the 200 km/h car. Make sure you do a complete Galilean transformation, not just half of one. It's the change in speed that determines how much energy is dissipated in the collision.
     
  14. Nov 15, 2010 #13
    Both cars are moving in opposite directions (ie towards each other). Lets say that one car is CAR A and the other car is CAR B, now according to te rules of finding the relative velocity of any object against any other object we find that the relative velocity of car A to car B is 200 Km/h and vice versa. now , for the same effect (ie same amount of damage TO ONE OF THE CARS (say car A) will be equal to a variable "X". But the total damage of the whole system (ie to both of the cars) will be 2X.

    Agreed ?

    Now , as the wall is stationary therefore the speed of the car should be 200Km/h so that the damage is equal to the TOTAL DAMAGE IN THE SYSTEM (ie previously equal to both cars ). But if you just want that the car should be damaged EQUAL TO ONE OF THE CARS ( ie car A) then car A should move at 100Km/h.

    Now this explaination is only valid in the case that all other influences are kept constant. For eg : there should be no wind, no water molecules to disrupb the experiment. ETC.
     
  15. Nov 15, 2010 #14
    Not only is it the same force, but its the same situation. Neither reference is more valid than the other. One car traveling 100km/h hitting another car traveling 100km/h and one car traveling 200km/h hitting another car at "rest" are two different descriptions of the exact same event.
     
  16. Nov 15, 2010 #15

    Danger

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    Rethink, pal. I know that it is counter-intuitive and hard to accept, but a car hitting another car at 100 km/h has the same reaction whether that other car is stationary or closing at an additional 100 km/h. I sure as hell don't like it, but that's how it is.
     
  17. Nov 15, 2010 #16
    No. KE = 1/2 mv^2; if a single car hits a wall at 100 km/h, the KE dissipated is 1X, if it hits the wall at 200 km/h, the KE is 4X. To get the same total KE as the two car collision, the single car would have to hit the wall at 141.4 km/h.
     
  18. Nov 15, 2010 #17
    Yes. The key is the speed change during the event; if a car traveling at 200 km/h hits another at rest, the final speed of both cars is 100 km/h, which gives a difference in speed of 100 km/h for both cars and the same energy dissipated as in the two cars at 100 km/h each.
     
  19. Nov 15, 2010 #18
    Danger, you don't need to like it 'cause it ain't true! If the single car hits a stationary car, the final speed of the conglomeration is 50 km/h.
     
    Last edited: Nov 15, 2010
  20. Nov 15, 2010 #19
    @mender.....if u would have read my post properly X does NOT REPRESENT ENERGY....it represents the amount of damage..........DAMAGE
     
  21. Nov 15, 2010 #20

    Danger

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    I must be missing something here. By my reckoning, the final speed will be zero in any measurement system. :confused:
     
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