04-14-04, 09:10 AM #1PolreanGuest
two cars crashing
I have a simple physics question thats been bothering me for a couple weeks. Assume a car is going 50mph and hits an unmoveable wall stopping the car dead in it's tracks, doing the car a certain amount of damage. Is the damage done the same as if the same car moving at the same speed crashes into another car head on also the same size travelling at the same speed? I know every action has an equal and opposite reaction. If the second car was moving any slower then the amount of damage done to the first wouldn't be compareable to it hitting a tree. Am I correct?
04-14-04, 10:49 AM #2
If it hit the second car head on (going at the same speed) the amount of damage would be almost twice that of hitting the tree. It won't exactly be twice because some of the doubled kinetic energy of the system will go into damaging the other car and also knocking it backwards
04-14-04, 07:14 PM #3
I do believe it is not double but is the product. The reason is that the stopping time is reduced.
That is if the damage had a mathematical value of 1 hitting the wall, hitting another car of the same weight and speed head on would not be 2 but would be 4.
Energy and momentum are conserved but the force of the crash is "Impulse". Impulse is change in momentum/time. By having two cars you double the momentum but you also cut the time in half or 2/0.5 = 4.
You can play with this simulator. It shows a car and a tree but you can see if you are going 30mph and get one result that two cars head on at 30 has a closing (stopping) rate equivelent to 60 mph. If you change the input by going from 30 to 60 you will see the force of impact is 4 times as great not twice as great.
Last edited by MacM; 04-14-04 at 07:49 PM.
04-14-04, 08:58 PM #4
Originally Posted by MacM
I believe the force computation (as in the link supplied by you) is based on the distance the car travels immediately after the collision until it stops. This stopping distance depends more on the construct of the car than the car speed. Impact force computed as kinetic energy devided by stopping distance is just an approximate result. In actual case, the force increases from zero to maximum and drop to zero again.
As kinetic energy is a function of v<SUP>2</SUP>, certainly doubling the car speed would increase the kinetic energy by 4 times. Combine kinetic energy of two colliding cars, each has speed of 30mph is not the same as the kinetic energy of a car having speed of 60mph.
04-15-04, 12:15 AM #5
I agree that kenetic energy is proportional to v^2. If you look at the rate of closure (relative velocity of collision) to the wall at 30 mph it is 30mph but if the wall (in this case the other car) is coming at you at 30 mph the closure rate (relative collision velocity) is 60 mph. The impact is proportional to the closure rate which is equivelent to 60 mph or 4 times the energy.
While it is absolutely true that cars with crushable or collapsable front ends will result in less collision force by extending the decelleration rate, it doesn't alter the quadrupling of the collision force in the case of two cars. You are simply quadrupling a smaller force.
Keeping the terms in line with your question you can see that if the collision force is for the amount of time to crush the car 12 inches that it only takes 1/2 the time since while car 1 moves forward 6 inches the other car (wall) also advances 6 inches, which results in the same 12 inch compression in only half the time. That is the equivelent to the car having twice the speed or 4 times the kenetic energy.
PS:That is the only Relativity I put full faith in.
Last edited by MacM; 04-15-04 at 12:32 AM.
04-15-04, 01:12 AM #6
Impulse is change in momentum/time.
You can double the impulse either by doubling the force applied while keeping the interaction time the same, or by doubling the interaction time while keeping the force the same.
If you look at one car hitting a wall, there is an impulse of mv, where m is the mass of the car, and v is its speed before the collision.
If you look at the same car colliding head-on with an identical car going at the same initial speed, the total impulse is 2mv.
If you want to look at the energy in each collision, then the energy released in the 2-car case is twice the energy in the 1 car case.
If you look at force, things get complicated, since the interaction times in the two cases may well be different.
04-15-04, 01:31 AM #7
As I had said earlier both momentum and energy are conserved. That is you don't have more energy. The energy is still two times one car at 30.
My posts may have been misleading in that respect.
My point was (as stated) the collision force of two cars going 30 in a head on collision is the same as one car going 60 and hitting a wall because the Delta v/t is not the same.
If a car going 30 comes to a stop in 12 inches of frontal compression it only takes 1/2 the time to collapse 12 inches if the collision impact (wall) is also moving toward you at the same speed.
((I think I see the flaw. The 2nd car also compresses hence the final stopping distance, hence time would be the same. But then that tends to say the head on collision and colliding with a stationary object are the same but the collision contains twice the energy? Its after midnight I'm going to bed ))
Last edited by MacM; 04-15-04 at 02:22 AM.
04-15-04, 06:30 AM #8
Actually Paul T is right. The collision contains not twice the energy but 4 times the energy. This would suggest the damage would be a little under 4 times that of hitting the tree
04-15-04, 07:48 PM #9
Originally Posted by MacM
04-15-04, 11:08 PM #10
Yes, there is a frog in the punch bowl I think. I know that two cars running side by side at 30 collectively equals the energy of one car at 30 mph times two.
Yet the issue of two cars going 30 mph in a head on collision suggests the equivelent of 60 mph collision which doesn't conserve energy. So I am not saying the energy is that of 60 mph but I do believe there is an impact "Force" that does varying in that regard.
After all the collision velocity is indeed 60 mph in the head on condition. To not think so would mean that there is no difference between hitting a wall at 30 mph and the head on collision of two cars each going 30 mph. That would not make sense.
04-16-04, 01:06 AM #11Originally Posted by Polrean
In the second scenario, the second car cops the same damage as the first.
Since "damage" effectively equates to "energy released", the second scenario involves twice as much total energy release (half to each car).
Originally Posted by John Connellan
Originally Posted by MacM
- Imagine an unbreakable, feather-light wall, sitting on frictionless rollers. When you drive into the wall, you push it ahead of your car. No damage is done.
- Now the same wall is immovably fixed to the ground. When you drive into the wall, you come to a damaging.
- Now, the same wall is back on frictionless rollers. This time when you drive into the wall, an identical car hits the wall from the other side at the same speed and the same time...
What damage occurs to your car in situation 3?
Same as 2? More than 2? Less than 2?
This problem is very similar to the tractor-rope problem, which everyone has probably heard before but I'll post in a new thread anyway.
Last edited by Pete; 04-16-04 at 01:56 AM.
04-16-04, 01:47 AM #12
Originally Posted by MacM
Similar situation applies for two cars colliding head-on. The speed of 60 mph is relative to the observer in one of the cars. While it is a valid view, it is unfortunately not very useful for addressing the problem discussed here. For example, when both cars eventually stop, the observer that originally observe the other car having speed of 60 mph is now stop too. You can't use the information taken by one of the moving observer to solve the collision problem. It complicates the issue. If you want to be consistent, you should measure the car kinetic energy from the moving car (relative to road) even though the cars already stop. This means you need to consider the reference point after collision moving away from the collision point at the speed of 30mph. Doesn't make sense, does it?
Choosing the right reference frame is important. Here, the right (or easy) one is the "at-rest reference frame", as from this reference frame, we can easily keep track the object speed and kinetic energy.
04-16-04, 06:04 AM #13Originally Posted by Pete
04-16-04, 09:20 AM #14
For two cars colliding head on, the total kinetic energy loss is twice (not four times) that of a car hitting a wall.
Half the energy is dissipated in each car, so the damage to the first car is the same as if it hit an unbreakable wall.
04-16-04, 01:37 PM #15
I think the easiest way to think of this problem is to imagine that the collision of two cars going 30 km/h is equal to one car hitting a stationary car at 60 km/h. Thus the KE of the wall collision is 1/4 that of the moving car collision split between where damage(deceleration) is split equally between each car. Thus, the KE absorbed by each car is twice that of the the wall collision.
KE(wall)=.5m(30)^2 thus KE car-wall = 450*m
KE(cars)=.5m(60)^2=2*KE absorbed by each car
Thus, KE each car=900*m
Now, the way the problem is phrased one would assume that this would imply that a wall collision is approximately half as damaging as a head on collision at the same speed.
In the real world, however, an actual collision is much more complex, due to the level of elasticity of the collision, the point(s) of impact etc...[added in edit] Not to mention the fact that the wall would absorb some of the impact in that collision as well.
Last edited by contrarian; 04-16-04 at 01:46 PM.
04-16-04, 11:36 PM #16I think the easiest way to think of this problem is to imagine that the collision of two cars going 30 km/h is equal to one car hitting a stationary car at 60 km/h.
One car of mass m and speed v has kinetic energy (1/2)mv<sup>2</sup>.
Two cars of mass m, each with speed v have a total kinetic energy of mv<sup>2</sup>.
One car of mass m and speed 2v has kinetic energy (1/2)m(2v)<sup>2</sup> = 2mv<sup>2</sup>.
Hence, the energy release in a head-on collision between 2 cars is half that involved in a collision of one car with twice the speed and a wall.
04-17-04, 12:11 AM #17
Originally Posted by Pete
04-17-04, 12:34 AM #18
04-17-04, 12:44 AM #19
That two cars having speed 30mph colliding head-on do not suffer more damage than just one car hitting the stationary wall may appear strange. This can be proven quite easily based on momentum and energy conservation principles.
Let's derive a relatively general solution for two cars (having mass respectively m and M) moving at velocity v and -v.
Momentum conservation: mv - Mv = (m+M) v<SUB>f</SUB> (equation 1)
Energy conservation: 0.5mv<SUP>2</SUP> + 0.5Mv<SUP>2</SUP> = E + 0.5(m+M) v<SUB>f</SUB><SUP>2</SUP> (equation 2)
E represents non-kinetic energy or energy involved in the collision.
Solution for the above equations 1 and 2 is:
E = 2mMv<SUP>2</SUP>/(m+M)
Condition 1: m=M. we'll get E = mv<SUP>2</SUP>, that is equal to total kinetic energy of both cars or 2 times kinetic energy of single car.
Condition 2: m small compare to M, we'll get E = 2mv<SUP>2</SUP> or 4 times kinetic energy of the car that has mass m. This condition is applicable, for example, for collision between a car and a train, which -- because the train mass is very much larger -- after collision, the car and train continue to move at velocity about -v.
For condition 1, the damage of car is really not worse than if the car collides with wall instead. Condition 2 would be as what MacM has in mind, analysing the problem based on velocity measured from one of the cars.
Last edited by Paul T; 04-17-04 at 06:52 AM.
04-19-04, 05:03 PM #20
Oops, mea culpa, mea culpa, I goofed! Ignore my last post!
Contrary to my initial post, the collision is not identical to one with a 60km/h car and a stationary one.
Will now delete my last post in shame.