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Two cars driving 30 and 40 mi/hr is like hitting a brick wall at 70 mi/hr

  1. Aug 5, 2009 #1
    I hear this from people sometimes. Upon hearing about a bad head on collision, I will hear a person promptly add the two speeds together and claim the wreck for the drivers is as bad as slamming into a brick wall with that new speed (S = s1+s2). Is this really a good analogy? I'm not so sure it is.

    When two cars collide head on mv = F[tex]\Delta[/tex]t applies, and the time it takes for the two cars to stop when they hit might be longer than slamming into a brick wall, so the force (which is what I think people mean by "is the same as running into...") could be smaller. I realize I'm not being extremely specific about the crash so there are a lot of things to nitpick over, but lets say we're talking about an inelastic accordion-like wreck on a frictioney (made up word) surface.

    Am I wrong or does this analogy to a brick wall fall apart?
  2. jcsd
  3. Aug 5, 2009 #2


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    Only if there is a huge difference in mass, like a car going 30 mph head on into a train coming at 40 mph. If the masses are the same, then it's like hitting a solid wall at 35 mph, assuming it's a direct head on and not a partial head on where the cars spin.
  4. Aug 5, 2009 #3


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    Hi Pupil! :smile:

    A car "gives" a lot more than a brick wall does.

    So it would be better to say it's the same as slamming into the front half of a stationary car up against a brick wall.

    EDIT: mgb_phys has changed my mind … see below. :smile:
    Last edited: Aug 6, 2009
  5. Aug 5, 2009 #4
    LOL! Well, yeah, I suppose that would be a more apt analogy. With the impulse equation that was what I was getting at when you say it 'gives' more. The brick wall also accelerate much less than the car when hit by another car, so won't that slow the car's acceleration as well?
  6. Aug 5, 2009 #5


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    Also imagine two identical cars hitting each other head on, both going at 30mph.
    Would that be any different from one car hitting a wall at 30mph?
  7. Aug 5, 2009 #6
    I would think so, since the [tex]\Delta[/tex]t in mv = F([tex]\Delta[/tex]t) is probably larger for the car collision than for the wall, which would result in a smaller average force.

    However both cars would stop dead as if they ran into a brick wall, since their total momentum would be 0 (assuming perfect in-elasticity).

    Right? :shy:
  8. Aug 6, 2009 #7
    So if one car travels at 30 and one at 40, they both hit head on, and one car takes "70 mph" of force. What does the other take... 0? lol..
  9. Aug 6, 2009 #8


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    Yes, if both cars have the same speed, then it makes no difference whether the wall is there or not.

    I prefer mgb_phys's :smile: reasoning to mine …

    ignore my previous post! :biggrin:
  10. Aug 6, 2009 #9
    Since I've been following this thread, I just have to ask. What is it with the brick wall thing? I know that it's common for people to say this, but It's a bad analogy any way you look at it. As others have already pointed out, there's a lot of difference between a brick wall and a car. If you're going to make analogies why not say that two cars, one traveling 30mph and the other traveling 40mph and colliding with each other head on, is similar to one car traveling 70mph and colliding head on with a parked car?
    Yes, because a brick wall is not similar to a car. Let's phrase the last sentence a different way. "Would that be any different from one car hitting a beach ball at 30mph?"
    No, they both experience the same force.
  11. Aug 6, 2009 #10


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    If two identical cars hit each other head on they are going to both stop exactly over the midpoint of the collision. Therefore it is equivalent to hitting an immovable object at that same point.
  12. Aug 6, 2009 #11
    Yes, that is true. I was not considering the brick wall to be immovable.
  13. Aug 6, 2009 #12
    So I've changed my mind. You can make an analogy with the brick wall if you consider the brick wall to be immovable. But the OPs example is still incorrect when saying that two cars, one traveling 30mph and the other traveling 40mph and hitting each other head on, is similar to one car hitting a brick wall at 70mph. It is actually similar to one car hitting a brick wall at 35mph.
  14. Aug 6, 2009 #13
    How so?
  15. Aug 6, 2009 #14
    I would suggest someone calculate or reference something. This is getting to be a bit of a ridiculousness thread....
  16. Aug 6, 2009 #15
    By calculating the average speed of the two cars. As "mgb phys" has already pointed out:
    However, he neglected to mention that both cars must be traveling at the same speed for this to be true. If one car is traveling at 30mph and the other at 40mph it would be the same if both cars were traveling at 35mph. 30+40=70, 35+35=70. If one car was traveling at 40mph and the other car at 60mph then it would be the same as hitting an "immovable" brick wall at 50mph.
  17. Aug 7, 2009 #16
    Well, I'm not very good with math but I'm going to give this a try. The solution stated in my previous post works just fine as long as both cars have the same mass. But what if they have different mass and are traveling at different speeds?

    For example, suppose we have Car1 which weighs 2500 lbs and is traveling at 30mph, and we have Car2 which weighs 3500 lbs and is traveling at 40mph. Car1 and Car2 crash head on.

    Now, I want to crash Car1 into an immovable brick wall with the same force as when it collided with Car2. How fast does Car1 have to be traveling? If I do the same with Car2, how fast does it have to be traveling?

    Here is my solution and I need the much wiser PF members to tell me if I'm on the right track or not. Since I know the following to be true when both cars have the same mass and same speed:

    (M1 * V12) / 2 = (M2 * V22) / 2

    All I have to do is solve for V1 or V2.

    Speed of Car1 = sqrt((M2 * V22) / M1) = 47.33 mph
    Speed of Car2 = sqrt((M1 * V12) / M2) = 25.35 mph

    The numbers look about right but I'm not sure if it's correct or not.
  18. Aug 8, 2009 #17


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    This is only true in an elastic collision where kinetic energy is conserved. I would assume an inelastic collision where the cars front ends collapse and the cars remain attached to each other. In this case you just need to conserve momentum so that the momentum of the combined cars moving as the same speed is the same as the momemtum was before the collision to determine the final speed:

    (m1 + m2) v = m1 v1 + m2 v2

    v = (m1 v1 + m2 v2) / (m1 + m2)

    Then the impact for car1 is related to v1-v and for car2 it's related to v2-v.
  19. Aug 8, 2009 #18
    There we go. That's what I'm talking about. Now instead of debating what if's, we can talk in absolutes. :biggrin:
  20. Aug 8, 2009 #19
    Ok well I'm from Aus so i'll use km/h instead :P

    if you think of it in terms of change in momentum (as has been previously stated), for the cars the change in momentum will be different for either car. Assuming that both cars weigh 1,000kg, car1 is travelling west at 80km/h and car2 is travelling east at 100km/h.

    total initial momentum = mv
    = (1,000 x 100/3.6) - (1,000 x 80/3.6)
    = 27777.78 - 22222.22
    = 5555.6 Ns-1 east

    initial momentum = final momentum
    5555.6 = mv where m is now the combined mass of the cars
    v= 2.7 ms-1 East(10 km/h)

    now we also know that [tex]\Delta[/tex]p = [tex]\Delta[/tex]Ft

    assuming that the crash takes 0.5 sec to happen (including all crumpling etc).... i have just plucked this number out of thin air btw

    [tex]\Delta[/tex]p = 22222.2 Ns-1 east

    22222.2 Ns-1 east = F x 0.5
    therefore Fav = 44444.4 N

    to find the force if the car had hit an imovable brick wall....

    t = 0.5 sec
    a = (v - u)/t
    F = ma

    from this we can solve to see what speed the car would have to be travelling at to experience the same force

    44444.4 = 1000 x (-u)/0.5
    -u = 22.22 ms-1
    = 80 km/h

    thus is can be shown that two cars hitting head on can be shown to be the same as one car hitting a brick wall at the average of their speeds
  21. Aug 8, 2009 #20
    If M1 = M2 and V1 = V2 then it has to be true (before the collision). I was originally bebating whether I should use momentum or kinetic energy.

    nooma, thanks for the detailed proof that if the cars are of the same mass we need only calculate their average speed.

    So, I'm assuming that the current consensus is that my method is wrong? I guess I'll have to give it some more thought.

    To refresh:
    Car1 initial speed = 30mph weight = 2500lbs
    Car2 initial speed = 40mph weight = 3500lbs
    If the cars crash head on, at what speed must the cars travel in order to hit an immovable wall with the same force? Anyone have answers?
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