[Updated on: Thu, 24 March 2005 04:52]

ok, where to start...

right, so, if you cut it down to the basics, the impact of two exactly the same cars travelling in exactly opposite directions at 50km/h would be the same as if one car hit a solid wall at 100km/h.
but, of course, the wall is not the same as a car. the cars are made of easily shape changing metal (or plastic)

Quote:

As for the original question, if two cars travelling at 50km/h collide head-on the impact will be the same as one car travelling at 50km/h hitting a solid object (eg concrete wall). In other words, hitting another car at the same speed is the same as hitting a solid object.

this is definetely incorrect, have a think about the relative speeds involved for the cars: to car 1, car 2 is travelling at 100km/h

Quote:

Potential energy of the cars; PE = 0.5*m*V^2

i belive you mean kinetic energy, KE, here?
KE = 1/2m.v^2
PE = m.g.h



to solve this problem, we need to consider the velocity of one car relative to the other. by using vectors (i cant draw them here), we can conclude that car 2 is travelling at 100km/h relative to car 1, and vice versa. note that this is assuming they are travelling at exactly the same velocity (50km/h, opposite direction), ignoring air resistance, and that they are exactly the same car, with the same weight, and no acceleration until the moment of impact (no acceleration or deceleration or change of direction)
assuming the mass of each car is, say, 1000kg, both car's initial velocity is 14m/s (50km/h = 13.9m/s) in opposite directions, the final velocity is 0m/s for both cars, and the time taken for the whole process is 0.5s, we can use newton's 2nd law to find the force exerted on each car:
F = (m.v - m.u)/t
= (1000x14 - 1000x0)/0.5
= 14000/0.5
= 28000 N

for the impact with the wall (assuming it will not move at all), however, the time will be much less, due to the fact that only the car has collapsible parts, where the wall does not, whereas in the collision between two cars, both objects are able to change shape. therefore, we will assume the time for the collision with the wall to be 0.1s. so the force on the car:
F = (m.v - m.u)/t
= (1000x14 - 1000x0)/0.1
= 14000/0.1
= 140000 N

as you can see, the force for the impact with the concrete wall would be much higher than for a collision between two cars


jeez, that would have to be my longest post ever! i hope to never have to do it again

[Updated on: Thu, 24 March 2005 06:29]

[Updated on: Thu, 24 March 2005 06:05]

Quote:

What does hitting a stationary car have to do with anything??? Of course the "damage" will be less when hitting a stationary car, you will be pushing the other car along, coming to a steady stop.

ah, sorry, i thought you said hitting a car at 50km/h, but you actually said a brick wall. but in that less damage bit, i was assuming the car would stay put and not crumple, much like the brick wall...hmmm...


anyway, this is the way i like to put it:

a car travelling towards a concrete wall at 100km/h has a velocity, relative to the wall, of 100km/h

a car travelling towards another car, both doing 50km/h, also has a velocity, relative to the other car, of 100km/h

think about it, if you're standing next to a 50km/h road, you see cars go past at a certain, apparent speed
if you're driving along a 50km/h road (at 50km/h, of course, cos you're a sensible driver), you see other cars going the other way at what seems to be a faster speed than when you watched them from a stationary position



btw, cruzsida, put it this way, if two cars collided, both travelling 50km/h, would it cause the same amount of collective damage as if one car hit a concrete wall at 100km/h?


edit: squid, do you want to have a say in this at all? i mean it is your thread

[Updated on: Thu, 24 March 2005 08:59]

skellator wrote on Fri, 25 March 2005 12:12

well, im not sure about the energy bit, but i have confirmation from two (out of two) fellow yr12 physics students, so far, that a car in a collision between two (identical) cars travelling 50km/h in opp directions will experience the same deceleration force as one car colliding with a wall at 100km/h note that the above is assuming that all energy is used to decelerate the cars, and that the wall is immovable and indestructable i can also get confirmation from my physics teacher on tuesday

Well they aren't going to pass Year 12 physics at that rate - sorry to say!

Okay let me break it down for you.

The "total" accelerative force is the zero vector. You just can't add up both forces like they are scalars because the are not. Force is a vector. This is why I'm saying that if you go from 50km/h to 0km/h it doesn't matter what you hit, it's only the RATE that you slow down (or the length of the crash) that counts in regards to FORCE that YOU inside the car experience - it's not about whether something else ALSO slows down, all that matters from INSIDE the car is that you went from 50km/h to 0km/h (and how quickly you did it too).

It doesn't matter whether you hit a truck going 20, a boeing 747 going 600km/h (although I imagine going from 50km/h to 0 is unlikely, more likely you will go from 50km/h to -590km/h) or a massive pile of feathers - all that matters is the TIME frame if you are worried about the FORCE of the collision.

I could go on for hours showing you equations but it's a waste of my time.

If you go from 100km/h to 0km/h in the same time frame then the force you experience will double.

Hitting a wall at 50km/h feels *almost* the same as hitting a car coming at you at 50km/h

END OF STORY.

And tell those physics students FORCE IS A DAMN VECTOR I HATE IT WHEN PEOPLE FUCK THAT UP FOR GOD'S SAKE.

[Updated on: Fri, 25 March 2005 08:22]