Hi there, I have a quick question i want to settle an argument the RTA website states that For example, a head-on crash where both cars are travelling at only 50 km/h gives a collision speed of 100 km/h Ã¢â‚¬â€œ equivalent to driving into a stationary object at 100 km/h! i have read elsewhere that this is bullsh!t can anyone clarify this for me my understanding was that if you have 2 "corolla's" that collide head on 50km/h there are 2 "corolla's" to absorb the impact where if a singular "corolla" hit an immovable object there would only be one to absorb the impact thus ending up with the same 'damage' from both scenarios please correct me if I'm wrong [edit spelling] oh $%$% i spelt it wrong in the poll as well

Yes, there's the same amount of moving energy to be dissipated, so the impact force is the same. Whether both cars crumple is irrelevant, brick walls 'crumple' too under certain circumstances.

The quick answer is yes, it would double the impact. In the case of the two Corollas head on at 50km/h each, it should be the equivalent of one Corolla doing 100km/h and stacking into the front of another stationary Corolla.

there is one minor diference. Two crumple zones disipating some energy. If you hit an object that is 100% "Rock" solid (Yes pun intended) then it has no crumple zone. But other than that... Yep 50 + 50 = 100. Oh and to put this in perspective. 60kph is approximately the speed you reach falling from 3 stories.

Of course the real physics are applied in the whole crumple thing. Running a car into a cliff at 40km/h (I've been a passenger in this one) hurts a lot more than hitting a car at the same speed which in turn would hurt more than running into a giant beanbag that somebody left laying around.

You're talking about two issues. One, how much energy is there? And two, how is that energy dissipated? quickly, e=mv^2 The 2nd problem isn't so simple. NB 100km/hr closing speed isnt "double". e=1000kg x 50^2 =/ 2(1000kg x 100^2)

Well, the results wouldn't be pretty either way. http://mathforum.org/library/drmath/view/60747.html has a good discussion on it.

Actually not quite. it would be the same as crashing into an anchored corolla. If they are heading at each other the resulltant vector is 0 if one hits an other that is stationary and not anchored it has a resultant vector of e=mv^2 where m has doubled (Minus the energy that has gone into turning the fronts of the cars into scrap mettal)

I don't think its right. The amount of energy is MORE than double, because the equation is exponential no? quote "Thus a head-on collision is only half as bad as hitting a *wall* at twice the speed." I was right. In the case of hitting a stationary moveable object ie another identical car, parked, at 100kph the collision results are the same as two cars head on at 50kph. But if the car at 100kph hits an imoveable stationary object, ie wall or tree or something of the sort, the results are twice as bad. So it seems it is worse than the RTA stated.

yep....the question was "does it double the impact (read force)" which is mass x accleration. No talk whatsoever in the question on crumple zones or brick walls or energy dissapation. so in my opinion the answer to that question is yes, it will double the force

You can think of one car being stationary, and no it is not anchored. Both cars are accelerating from 50 back to zero in the first example, or if you consider the second example, one is accelerating from 50 to xkm/hr and the other is accelerating from 0 to -xkm/hr. Anchoring it would not be correct.

it is e=mv^2energy = mass times the square of the velocity) So double the speed (relative speed of the two objects) and you quadrupal the energy that has to be disipated.

I don't think that is correct. The velocity would be exactly the same for both equations would it not?

The "Relative" Velocity would be equil. And this is what counts. How things look to the outside world are not what you need to measure, what you measure is the "relative" relationships between teh objects you are measuring. Note the real world has other factors such as drag and friction and so on, but at these sort of velocities these factors are small change and can be ignored.