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Thitical Crinker
25th September 2007, 09:16 PM
I have a 12 pack riding on this: I contend that two identical vehicles traveling at 100 mph each, hitting directly head-on, is the equivalent of hitting a wall at 100 mph. My buddy says it is equal to hitting the wall at 200mph. Who is out the 12 pack?

Probably a silly question, but we are both convinced we are right.:o

wollery
25th September 2007, 09:29 PM
You are wrong, he is right.

Two cars each travelling 100mph in opposite directions.

The relative speeds of the cars is 200mph.

Imagine one of the cars is stationary, what speed does the other car need to be travelling for the impact to be the same?

Now replace the stationary car with a brick wall.

boooeee
25th September 2007, 09:32 PM
I think you're right. Do I get one of the beers?

I think it helps to define what exactly is "equivalent". A convenient measure in this case is impulse (http://en.wikipedia.org/wiki/Impulse).

Also, you need to define what type of collision this is. Is it elastic? Meaning that kinetic energy is conserved. An example of a (roughly) elastic collision is when two billiard balls collide. Or is the collision inelastic? In an inelastic collision, the colliding bodies stick together. Momentum is conserved, but some of the kinetic energy is lost to heat in the collision. Or is the collision somewhere in between?

A collision between two vehicles is probably going to be more inelastic than elastic, so let's go with inelastic. Also, you need to assume that the wall is an infinitely strong barrier.

So, with those assumptions in hand, the impulse delivered to a car traveling at 100 km/h that collides head on with an identical car travelling at 100 km/h is equivalent to colliding with a wall at 100 km/h.

PixyMisa
25th September 2007, 09:32 PM
In terms of momentum or kinetic energy?

In terms of momentum, your friend is right.

In terms of kinetic energy, you're both wrong. :) Two cars colliding head-on at 100mph would be equivalent to running into a brick wall at about 140mph.

boooeee
25th September 2007, 09:36 PM
Now replace the stationary car with a brick wall.

Now replace the stationary car with a kitten. Or with a feather. The "heft" of the object that you are colliding with matters.

Complexity
25th September 2007, 09:37 PM
Kinetic energy is proportional to the square of the velocity: E = ( mass * velocity * velocity ) / 2.

You're both wrong.

The kinetic energy of two identical cars when they hit each other head-on at 100 mph is equal to twice the kinetic energy of one car when it hits a wall at 100 mph.

The kinetic energy of one car when it hits a wall at 200 mph is equal to four times the kinetic energy of the car when it hits a wall at 100 mph.

Each of you should buy a 12-pack.

Complexity
25th September 2007, 09:39 PM
Controversy!

Cool.

wollery
25th September 2007, 09:47 PM
Doh!! :o

That'll teach me to try to reply quickly to a question in Newtonian mechanics without shifting gears out of relativistic mechanics!! :bwall

ETA, Yes, I know my answer is still wrong in relativistic mechanics, but I applied equivalence without applying transforms and, well, damn it!!

boooeee
25th September 2007, 09:51 PM
I still say "impulse" is the right way to look at it. Impulse is a quantitative measure of the "violence" of a collision.

Let's say you blindfold the passenger in the car. They won't be able to tell the difference between colliding head on with another car at 100 km/h and colliding head on with a perfectly rigid wall. In both cases they experience an (approximately) instantaneous change in their velocity of 100 km/h.

If it comes down to "Which collision would you rather be in?", the answer would definitely be colldiing head on with another car at a (relative speed) of 200 km/h.

boooeee
25th September 2007, 09:54 PM
Doh!! :o

That'll teach me to try to reply quickly to a question in Newtonian mechanics without shifting gears out of relativistic mechanics!! :bwall

ETA, Yes, I know my answer is still wrong in relativistic mechanics, but I applied equivalence without applying transforms and, well, damn it!!

Come on, this thread will never get to 15 pages with an attitude like that. Dig in! Concede no ground! Hold your position to the death!

Refer to the air marshal thread(s) for helpful instruction.

wollery
25th September 2007, 09:58 PM
Come on, this thread will never get to 15 pages with an attitude like that. Dig in! Concede no ground! Hold your position to the death!

Refer to the air marshal thread(s) for helpful instruction.:D

Thitical Crinker
25th September 2007, 10:04 PM
Wow, didn't expect so many replies so fast. While googling this question I came across this http://www.wiskit.com/marilyn/mistake.html

It seems someone asked Marilyn a similar question and she got it wrong.

Maybe I can get my buddy to agree on splitting the 12 pack!:D

Thitical Crinker
25th September 2007, 10:20 PM
I think you're right. Do I get one of the beers?

I think it helps to define what exactly is "equivalent". A convenient measure in this case is impulse (http://en.wikipedia.org/wiki/Impulse).

Also, you need to define what type of collision this is. Is it elastic? Meaning that kinetic energy is conserved. An example of a (roughly) elastic collision is when two billiard balls collide. Or is the collision inelastic? In an inelastic collision, the colliding bodies stick together. Momentum is conserved, but some of the kinetic energy is lost to heat in the collision. Or is the collision somewhere in between?

A collision between two vehicles is probably going to be more inelastic than elastic, so let's go with inelastic. Also, you need to assume that the wall is an infinitely strong barrier.

So, with those assumptions in hand, the impulse delivered to a car traveling at 100 km/h that collides head on with an identical car travelling at 100 km/h is equivalent to colliding with a wall at 100 km/h.

This is kinda what I was thinking except I didn't have all those scientific sounding words in my train of thought.

And I would be glad to share the beer!

Thanks for all the help so far. Learning is fun at 1 in the morning.:)

Night shift bites!!

Complexity
25th September 2007, 10:23 PM
Beer is always the answer.

That is why you should each buy a 12-pack.

Of something good (derail alert!)

It would be very important not to live through either accident. No matter how well you're strapped into the car, your brain and other organs would be under a lot of deceleration stress.

DevilsAdvocate
25th September 2007, 11:05 PM
I contend that two identical vehicles traveling at 100 mph each, hitting directly head-on, is the equivalent of hitting a wall at 100 mph.Was there a consensus on this? They are equivalent, right?

Assuming the wall is like a thousand feet thick and creates a dead stop. The unmovable wall and the counter momentum of an object of the same mass and momentum would result in the same sudden zero forward movement of the truck.

The force of an unmovable wall must be equal to the force against it. So any other object with equal force is equal to the force of the unmovable wall. Two objects colliding with equal force the the same as one object colliding with an unmovable wall with equal force. They both have equal force. In both cases the truck's velocity on collision is zero. From the truck's perspective, the truck simply stops cold in both cases. They are the same.

Thitical Crinker
25th September 2007, 11:43 PM
Was there a consensus on this? They are equivalent, right?

Assuming the wall is like a thousand feet thick and creates a dead stop. The unmovable wall and the counter momentum of an object of the same mass and momentum would result in the same sudden zero forward movement of the truck.

The force of an unmovable wall must be equal to the force against it. So any other object with equal force is equal to the force of the unmovable wall. Two objects colliding with equal force the the same as one object colliding with an unmovable wall with equal force. They both have equal force. In both cases the truck's velocity on collision is zero. From the truck's perspective, the truck simply stops cold in both cases. They are the same.

I agree with you here. If I can get my friend to agree, I win. Couldn't care less about the beer, just want to win the bet.

SomeGuy
26th September 2007, 12:11 AM
I agree with you here. If I can get my friend to agree, I win. Couldn't care less about the beer, just want to win the bet.

You don't care about the beer?

Is nothing sacred anymore!?

Thabiguy
26th September 2007, 12:17 AM
I have a 12 pack riding on this: I contend that two identical vehicles traveling at 100 mph each, hitting directly head-on, is the equivalent of hitting a wall at 100 mph. My buddy says it is equal to hitting the wall at 200mph. Who is out the 12 pack?

You are correct, your friend is wrong.

Two identical cars travelling at 100 mph each will each suffer damage equivalent to each car individually hitting a stationary wall at 100 mph (or both hitting the same wall from opposite sides). The kinetic energy liberated is the same in both cases (and, obviously, equal to twice the kinetic energy of only one car hitting a stationary wall at 100 mph).

But! - there is also a grain of truth in wollery's original answer: this is also equivalent to one car being stationary and the other hitting it at 200 mph.

Although the kinetic energy of a car moving at 200 mph is four times greater than the same vehicle travelling at 100 mph, and two times greater than two such vehicles travelling at 100 mph, one must not forget that the resulting metal wreck will not remain stationary: it will continue travelling at 100 mph, having the mass of two vehicles, and thus retaining half of the original kinetic energy. The kinetic energy liberated in the actual crash is therefore the same as each car travelling at 100 mph. (Although, realistically, the leftover energy of the travelling metal wreck won't be dissipated entirely peacefully and is likely to cause some additional damage.)

The moral is: if you are driving a car at 100 mph, and you can choose between hitting a stationary wall and a stationary (end empty!) vehicle of about the same size, choose the vehicle! It will cause far less damage and dramatically increase your chances of survival, as the crash itself will be equivalent to hitting the wall at 50 mph.

wollery
26th September 2007, 12:30 AM
Okay, my head is now hurting.

It appears that everyone is right, and everyone is wrong. :confused:

Roboramma
26th September 2007, 12:51 AM
Although the kinetic energy of a car moving at 200 mph is four times greater than the same vehicle travelling at 100 mph, and two times greater than two such vehicles travelling at 100 mph, one must not forget that the resulting metal wreck will not remain stationary: it will continue travelling at 100 mph, having the mass of two vehicles, and thus retaining half of the original kinetic energy.

The rest of your post made sense to me, but I don't follow this part. Why will the resulting metal wreck keep moving at 100mph, and in what dirrection?
It seems to me that since the momentum of the two cars sums to zero before the crash, it must sum to zero after - and so if they don't separate, they must be stationary... am I missing something?

dacium2007
26th September 2007, 01:13 AM
Some people here are very wrong.
You are correct and your buddy is wrong, they are equivalent.

Yes there is twice the amount of energy, but there are now 2 cars to dissipate the energy.

The only possible way it could be equivalent to hitting a wall at 200mph is if one car dissipated all the energy, and the other none, such that after collision your car is moving at 100mph BACKWARDS and the other car is moving 100mph forwards.

The reason the crashes are both of the same severity is in both cases your car goes from 100 to zero, and dissipates the same amount of energy.

wollery
26th September 2007, 01:16 AM
The rest of your post made sense to me, but I don't follow this part. Why will the resulting metal wreck keep moving at 100mph, and in what dirrection?
It seems to me that since the momentum of the two cars sums to zero before the crash, it must sum to zero after - and so if they don't separate, they must be stationary... am I missing something?Assuming the collision is perfectly inelastic (i.e. the cars stick together) the momentum before is equal to the momentum after, so the car arrives at 200mph, momentum is 1 car mass (Mc) times 200mph (200Mc*mph), afterwards the momentum is 200Mc*mph/2Mc which is 100mph.

Broes
26th September 2007, 04:15 AM
The physic's involved were rather easy but then you mentioned that beer was involved and that kinda screwed everything up :D

fuelair
26th September 2007, 05:15 AM
If the two vehicles are exactly the same (same mass particularly) and they hit perfectly head on, they are not going anywhere - they will perfectly entangle in that spot. Going anywhere requires that either mass or velocity or both be different - then movement will continue in the direction of the greater momentum (greater prior to contact of course).

Gib
26th September 2007, 05:30 AM
I think this question is a job for mythbusters !!!

I can just see poor old Buster the crash test dummy now..

Just thinking
26th September 2007, 06:06 AM
The problem here is that some are looking at the collisions (either one) as elastic --- they are NOT. A great deal of energy will go into deforming the vehicles (or vehicle if one car hits a wall).

If one car hits a wall at 100 mph, all of the energy (KE) will go into deforming that car. If it hits a stationary car, much of its KE will be split 50/50 into deforming both cars; what's left over will go into the motion of both cars now going in the original car's direction prior to the collision (as essentially a new mass double that of the first moving car). Since this mass is now double that of the original car (and much KE has been lost due to deformation) we can be certain it will not be 71% of the moving car's initial velocity --- it will be much less. (Conservation of momentum will put it at 50%.)

So what will cause the same deformation to the original car hitting an identical car head on at the same speed (the question that auto designers test for in collision ratings)? Answer: a stationary wall --- the proverbial brick wall.

Ziggurat
26th September 2007, 07:22 AM
Assuming the collision is perfectly inelastic (i.e. the cars stick together) the momentum before is equal to the momentum after,

The beauty of conservation of momentum for this problem is that it doesn't matter if the collision is elastic or inelastic. The center of mass motion will still be conserved throughout the collision. If the collision is inelastic, and the cars bounce apart (slight elasticity), their center of mass still ends up moving at 100 mph immediately following the collision. Furthermore, if the cars would bounce off each other, they'd also bounce off the wall, so it makes no difference for the comparison how inelastic the collision process is. The only assumption we need in order to conclude equivalency is that the wall doesn't crumple or compress upon impact.

Just thinking
26th September 2007, 08:29 AM
Another way to consider how the two scenarios are identical is to ask "How much of the car's KE goes into deformation of the car?" If the car hits the brick wall (which remains fully intact during and after the collision), then we can say all of it. If it hits an oncoming car with the initial car's speed head on and both cars become motionless from the impact, then we have twice as much KE applied to twice as many cars --- resulting in each car deforming from as much KE as the single car hitting the wall. Identical damage to each car.

Also, please note that the car striking the wall or an oncoming car at the same speed will come to rest in the same amount of time --- meaning the force(s) and acceleration(s) applied to the damaging of the vehicle(s) will be the same. This comes out of the impulse equation which was mentioned earlier.

uruk
26th September 2007, 08:38 AM
Yea, but what if one car is a Hummer and the other one is a Cooper Mini?

Just thinking
26th September 2007, 08:43 AM
Yea, but what if one car is a Hummer and the other one is a Cooper Mini?

Interesting.

What happens when the mass of one is much much greater than the other? Guess what ... if the collision is elastic enough, the smaller can actually fly away at a speed GREATER than that of the larger. Meaning if a Hummer (traveling at 30 mph) hits a parked Mini (velocity = 0), the Mini can actually leave the collision at a speed greater than 30 mph. But don't get too carried away --- the most it can ever be is twice that of the larger mass.

becomingagodo
26th September 2007, 08:51 AM
Every one is wrong, except me.
You need a conscious observer to witness the crash to collaspe the wave function, cars our not conscious observers and then you need a conscious observer to witness the conscious observer. The zeno paradox proves the cars will never crash into each other. And finally matter should pass through each other because it is 99.99999999% nothing, so the cars never crash.

You need to think outside the box.

Paul C. Anagnostopoulos
26th September 2007, 08:55 AM
So the upshot is that everyone owes me beer, right?

~~ Paul

becomingagodo
26th September 2007, 08:57 AM
Drinking beer is for the less evolved.

Just thinking
26th September 2007, 09:00 AM
... cars our not conscious observers ...

AHHHHH !!!!!

wollery
26th September 2007, 09:16 AM
Every one is wrong, except me.
You need a conscious observer to witness the crash to collaspe the wave function, cars our not conscious observers and then you need a conscious observer to witness the conscious observer. The zeno paradox proves the cars will never crash into each other. And finally matter should pass through each other because it is 99.99999999% nothing, so the cars never crash.

You need to think outside the box.

Drinking beer is for the less evolved.

:troll

Complexity
26th September 2007, 09:22 AM
Every one is wrong, except me.
You need a conscious observer to witness the crash to collaspe the wave function, cars our not conscious observers and then you need a conscious observer to witness the conscious observer. The zeno paradox proves the cars will never crash into each other. And finally matter should pass through each other because it is 99.99999999% nothing, so the cars never crash.

You need to think outside the box.

This is the funniest think you've said, and got me thinking that you've just been putting us on...

until you said this:

Drinking beer is for the less evolved.

Evolution lesson #1: There are no 'less evolved'.

You're going to miss out on a lot of life and a lot of fun if you don't get off of that pedestal.

Oh, yeah - you're quite wrong, I'm afraid.

jond
26th September 2007, 09:22 AM
Yea, but what if one car is a Hummer and the other one is a Cooper Mini?

The Mini would fit comfortably under the Hummer, thus negating any damage at all.

Crossbow
26th September 2007, 09:30 AM
Some people here are very wrong.
You are correct and your buddy is wrong, they are equivalent.

Yes there is twice the amount of energy, but there are now 2 cars to dissipate the energy.

The only possible way it could be equivalent to hitting a wall at 200mph is if one car dissipated all the energy, and the other none, such that after collision your car is moving at 100mph BACKWARDS and the other car is moving 100mph forwards.

The reason the crashes are both of the same severity is in both cases your car goes from 100 to zero, and dissipates the same amount of energy.

Welcome to JREF 'dacium2007'!

And just to add my bit, I think that the above is the best way to evaluate the scenario.

Good call 'dacium2007'!' :)

DRBUZZ0
26th September 2007, 09:52 AM
The question comes down to what you mean by "equivalent to" since you come to a stop from the same speed, you would be subjected to just as much deceleration force. However there is a greater amount of energy present and the dynamics are different, so even divided between the two cars the amount of damage would probably be higher.

Just thinking
26th September 2007, 10:16 AM
The question comes down to what you mean by "equivalent to" since you come to a stop from the same speed, you would be subjected to just as much deceleration force.

Yes ... and that would result in the same amount of damage.

However there is a greater amount of energy present and the dynamics are different, so even divided between the two cars the amount of damage would probably be higher.

No --- each car has inflicted upon it the same amount of energy (the two hitting head on and the one hitting the wall). If you agree that a car either hitting a wall or another car head on experiences the same forces and accelerations (decelerations if you like) as I said in post 28, then the dynamics are the same (for any given car).

IXP
26th September 2007, 12:15 PM
I believe the consensus opinion is wrong.

In the case of two identical cars colliding head on, all of the energy is available to be dissapated by the two vehicles. In the case of hitting a brick wall, it is not. Think about it. If all of the energy were dissapated by deformation of the car, the total momentum could not be conserved since nothing would be moving at the end. In the ideal case, (where the wall is not deformable etc...) the momentum has to be preserved by accelerating the entire earth. In the real case, the wall would deform as well. Either of these scenarios leaves less energy to be absorbed by the car.

IXP

Just thinking
26th September 2007, 12:21 PM
I believe the consensus opinion is wrong.

In the case of two identical cars colliding head on, all of the energy is available to be dissapated by the two vehicles. In the case of hitting a brick wall, it is not. Think about it. If all of the energy were dissapated by deformation of the car, the total momentum could not be conserved since nothing would be moving at the end. In the ideal case, (where the wall is not deformable etc...) the momentum has to be preserved by accelerating the entire earth. In the real case, the wall would deform as well. Either of these scenarios leaves less energy to be absorbed by the car.

IXP

Although you are technically correct, the differences are immeasurable. Meaning that if you ran a statistical analysis of cost damage done to the vehicles (say, 200 trials --- 100 into a wall vs. 100 into an oncoming car), you would be hard pressed to tell which were which afterwords.

Besides --- couldn't one then argue (on such points as you raised) that energy/momentum went into the Earth as the car accelerated toward the wall (assuming it used its tires to accelerate)? And this energy/momentum was then completely countered upon striking the wall?

IXP
26th September 2007, 12:24 PM
I didn't say the difference would be measurable, just that there was a difference!

IXP

Just thinking
26th September 2007, 12:29 PM
I didn't say the difference would be measurable, just that there was a difference!

IXP

Please note my addendum.

IXP
26th September 2007, 12:42 PM
That is a good point, and I had to think about it, but I don't think so. The affect during acceleration would be that it would take a little more energy to get to 100 mph, but once there, you are moving in the earth's frame at 100 mph, and some of the kinetic energy due to that speed would still be lost. In the case of the head on collision, none of the energy would be lost in accelerating the earth.

IXP

Just thinking
26th September 2007, 12:57 PM
In the head on collision, the Earth gains neither momentum nor energy as the two cars accelerate and then collide --- their tire forces cancel out. So the Earth experiences no net movement.

In accelerating toward the wall, the car will need to apply more energy than needed to get to 100 mph (or whatever) due to air resistance. This (I believe) is the only opposing force that will be added to the Earth's momentum that will not be countered. Added in that the car must continue to apply a force to the Earth (via its tires) just to maintain a constant velocity.

In a vacuum, however --- with electric cars --- ;)

genesplicer
26th September 2007, 02:54 PM
What if one car is traveling in reverse? How does that change things?

Thitical Crinker
26th September 2007, 04:04 PM
Thanks for all the replies. Not sure of the correct "scientific" answer, but I am going to argue that I am correct. In both cases, all three vehicles go from 100mph to 0mph in essentially the same amount of time. For the purpose of our bet, I feel that is the winning answer. None of the vehicles go from 200mph to 0mph which is basically his argument.

Now to convince my friend.

I am going to have to up the wages to a case of beer to share with all of you.:D

Thanks again.

casebro
27th September 2007, 07:35 AM
If the question is "how much damage to your car if...":

If we were to imagine an inertia-free 'barrier' between the two cars, then when each car hit the barrier, each would be resisted by the exactly appropriate 'mass' of the other car. Therefor, a head on would be exactly the same as hitting a solid object.
Logic, not physics.

If the question was "how much energy would be released...":

Doubling the speed would have 4x the energy, not 2x. Then one car would need to go 1.4x the speed to release the same amount of energy.
Physics, not logic.

Just thinking
27th September 2007, 10:22 AM
Yes, Casebro, but only if you keep your reference frames constant (the same for both trials). Otherwise, a seeming contradiction may arise.

Let's take the reference frame of that of an observer standing at the point of impact where two oncoming cars collide --- each having the same speed. The energy measured by him would work out as ...

mv2/2 + m(-v)2/2 = 2mv2/2 = mv2

But look what happens when we take the point of view of being in one of the cars, and assume it at rest with the other car coming at him twice as fast. (Relativity allows this.)

m(2v)2/2 = 4mv2/2 = 2mv2

They don't match. (Or is something being overlooked?)

Soapy Sam
27th September 2007, 10:43 AM
Given one car is a Reliant Robin and the other a Rolls Royce.
1. How did a Robin reach 100mph?
2. What was that bump the roller driver just felt?
3. Watch out for that wall...

boooeee
27th September 2007, 12:47 PM
Yes, Casebro, but only if you keep your reference frames constant (the same for both trials). Otherwise, a seeming contradiction may arise.

Let's take the reference frame of that of an observer standing at the point of impact where two oncoming cars collide --- each having the same speed. The energy measured by him would work out as ...

mv2/2 + m(-v)2/2 = 2mv2/2 = mv2

But look what happens when we take the point of view of being in one of the cars, and assume it at rest with the other car coming at him twice as fast. (Relativity allows this.)

m(2v)2/2 = 4mv2/2 = 2mv2

They don't match. (Or is something being overlooked?)

I think it's fine they don't match. The only thing that has to match is the change in energy. In your second reference frame, with one car being at rest, the final kinetic energy will not be zero. If the collision is inelastic, the final kinetic energy of the two cars is mv^2, leading to the same change in kinetic energy, -mv^2.

Just thinking
27th September 2007, 02:23 PM
I think it's fine they don't match. The only thing that has to match is the change in energy. In your second reference frame, with one car being at rest, the final kinetic energy will not be zero. If the collision is inelastic, the final kinetic energy of the two cars is mv^2, leading to the same change in kinetic energy, -mv^2.

Yes, that is correct. (Also, it doesn't matter be the collision elastic or inelastic.) The end result is that the energy dissipated into damaging the cars is the same.

Good call. :D

ponderingturtle
28th September 2007, 06:05 AM
Kinetic energy is proportional to the square of the velocity: E = ( mass * velocity * velocity ) / 2.

You're both wrong.

The kinetic energy of two identical cars when they hit each other head-on at 100 mph is equal to twice the kinetic energy of one car when it hits a wall at 100 mph.

The kinetic energy of one car when it hits a wall at 200 mph is equal to four times the kinetic energy of the car when it hits a wall at 100 mph.

Each of you should buy a 12-pack.

But the energy used to deform each car is the same at 100 mph if they hit an intentical car at that speed as if they hit a wall.

Now as the car has more give the energy will be spread out differently in the car vs wall colision than the car vs car collision.

ponderingturtle
28th September 2007, 06:09 AM
Okay, my head is now hurting.

It appears that everyone is right, and everyone is wrong. :confused:

The problem is that the initial dispute was not rigorously defined.

ponderingturtle
28th September 2007, 06:15 AM
What if one car is traveling in reverse? How does that change things?

Is its engine in the front or rear?

boooeee
28th September 2007, 07:15 AM
Is its engine in the front or rear?

Stick to the scenario.

Walrus32
28th September 2007, 07:24 AM
The problem seems to be that you're using an absolute frame of reference. (Einstein said you can't do that.) So, disregarding elasticity and assuming that acceleration (or deceleration, in this case) is instantaneous, and using Newton's second law, F=Ma, each car will see the speed of closure as 200 mph, with the consequent force at impact - the same as if one car were approaching a stationery object at 200 mph.

boooeee
28th September 2007, 08:41 AM
The problem seems to be that you're using an absolute frame of reference. (Einstein said you can't do that.) So, disregarding elasticity and assuming that acceleration (or deceleration, in this case) is instantaneous, and using Newton's second law, F=Ma, each car will see the speed of closure as 200 mph, with the consequent force at impact - the same as if one car were approaching a stationery object car at 200 mph.

Fixed above. It matters what the object is that you're hitting. Hitting a car is not the same as hitting a stationary wall.

wollery
28th September 2007, 09:31 AM
The problem seems to be that you're using an absolute frame of reference. (Einstein said you can't do that.) So, disregarding elasticity and assuming that acceleration (or deceleration, in this case) is instantaneous, and using Newton's second law, F=Ma, each car will see the speed of closure as 200 mph, with the consequent force at impact - the same as if one car were approaching a stationery object at 200 mph.This was precisely the mistake I made in the second post of this thread. You can't replace the second car with a wall and change the frame of reference to car 2 being stationary, because the wall approximates to an immovable object, which car 2 certainly isn't. When car 1 hits car 2 with both travelling at 100mph towards each other car 1 stops dead, as does car 2. From the frame of reference of car 1 it may as well have hit a wall at 100mph. Same with car 2.

Walrus32
28th September 2007, 10:09 AM
Fixed above. It matters what the object is that you're hitting. Hitting a car is not the same as hitting a stationary wall.

Exactly. I tried to speak of inelastic objects of equal mass. If you're going to talk about cars, you need an extensive analysis of the structure,"crumpleabilty", distance traveled to a dead stop (deceleration over time), and relative mass.

(I'm always amused by the sheer engineering ignorance of the people who test bumpers by backing into a fixed object at 5 mph. The luxury cars rate poorly, the little cars do well...disregarding the fact that 5000 pound Mercedes hits with twice the force of 2500 pound cheapo.)

ponderingturtle
28th September 2007, 10:15 AM
Exactly. I tried to speak of inelastic objects of equal mass. If you're going to talk about cars, you need an extensive analysis of the structure,"crumpleabilty", distance traveled to a dead stop (deceleration over time), and relative mass.

(I'm always amused by the sheer engineering ignorance of the people who test bumpers by backing into a fixed object at 5 mph. The luxury cars rate poorly, the little cars do well...disregarding the fact that 5000 pound Mercedes hits with twice the force of 2500 pound cheapo.)

Seems like a reasonable test. People back into poles all the time.

The question is not how much energy can the bumper take but how much dammage it gets if you back into a poll.

boooeee
28th September 2007, 12:11 PM
Exactly. I tried to speak of inelastic objects of equal mass. If you're going to talk about cars, you need an extensive analysis of the structure,"crumpleabilty", distance traveled to a dead stop (deceleration over time), and relative mass.

It may be a little different than hitting a wall at 100 mph, due to the way that the cars may crumple on impact. However, it will be nowhere near the same as running into a wall at 200 mph. Nitpicking aside, Thitical Crinker's answer is much closer to the pin than his friend's answer.

(I'm always amused by the sheer engineering ignorance of the people who test bumpers by backing into a fixed object at 5 mph. The luxury cars rate poorly, the little cars do well...disregarding the fact that 5000 pound Mercedes hits with twice the force of 2500 pound cheapo.)

Theres nothing ignorant about it. If you design a larger car, you should design a stronger bumper. The test seems perfectly appropriate.

Walrus32
28th September 2007, 12:41 PM
It may be a little different than hitting a wall at 100 mph, due to the way that the cars may crumple on impact. However, it will be nowhere near the same as running into a wall at 200 mph. Nitpicking aside, Thitical Crinker's answer is much closer to the pin than his friend's answer.

Theres nothing ignorant about it. If you design a larger car, you should design a stronger bumper. The test seems perfectly appropriate.

That's value judgment, and I tend to agree. However, to be completely accurate, the Mercedes should be tested, as is, at half the speed. (Not that I'm a big fan of Mercedes...I finally figured out what all those model numbers mean... 360, 500, etc. It's the minimum shop charge any time you bring it in for service.)

Now, to further muddy the waters, I read where the Large Hadron Collider will smash counter-rotating beams of protons into each other. For my own edification, isn't this to increase the available energy output? I'm not trying to be argumentative...I just find it an interesting concept (as a non-physicist.)

ponderingturtle
28th September 2007, 12:48 PM
That's value judgment, and I tend to agree. However, to be completely accurate, the Mercedes should be tested, as is, at half the speed.

Why? Do Mercedes go especialy slow in reverse and so will not hit the pole someone is backing into as quickly?

Also think about being bumped by a different vehical your innertia will play a role in how hard the bumper gets hit.

OnlyTellsTruths
28th September 2007, 01:31 PM
Walrus is right.......... if they were testing the bumpers themselves and not damage to something the bumper is supposed to protect, which is what they're doing............ so, Walrus is wrong.

Walrus32
28th September 2007, 01:56 PM
The bottom line is that anyone that backs up at 5 mph without knowing what's behind them is an idiot. :D

sthomson
28th September 2007, 02:00 PM
The bottom line is that anyone that backs up at 5 mph without knowing what's behind them is an idiot. :D

...or momentarily distracted :p

*sheepishly admits that she has backed into a fence at idling speed*

wollery
28th September 2007, 10:53 PM
That's value judgment, and I tend to agree. However, to be completely accurate, the Mercedes should be tested, as is, at half the speed.No, it should be tested at the speed you will use it in real life, otherwise the test is pointless.

Now, to further muddy the waters, I read where the Large Hadron Collider will smash counter-rotating beams of protons into each other. For my own edification, isn't this to increase the available energy output? I'm not trying to be argumentative...I just find it an interesting concept (as a non-physicist.)Yes it is to increase the total energy of the collision.

Walrus32
29th September 2007, 08:29 AM
"No, it should be tested at the speed you will use it in real life, otherwise the test is pointless."

The test is pointless. That's my point. They're not testing what the design should be, only as things are, using unequal forces.

wollery
29th September 2007, 12:40 PM
They're testing to see what happens if you drive a car into a pole or bollard at the sort of speed that you're likely to drive a car into a pole or bollard when parking or manoeuvring.

In what way is that a pointless test?

Unless you're suggesting that you should park small cars at higher speeds than you park big cars.......

Walrus32
29th September 2007, 01:01 PM
It's pointless test because it's waste of time, money, and automobiles to show what most sentient people would know already...a big car is going to hit harder than a small car. And it also probably has zero influence on which car people are going to buy anyway.

jimbob
29th September 2007, 02:02 PM
Beer is always the answer.

That is why you should each buy a 12-pack.

Of something good (derail alert!)

It would be very important not to live through either accident. No matter how well you're strapped into the car, your brain and other organs would be under a lot of deceleration stress.

In fact the only (infinetessimally small) chance you'd have is if you wern't strapped in and got thrown out onto a conveniently sloped and surface with appropriate friction and runout to gently decelerate you to a standstill.

maybe a lovely ice rink with an initial near-parapolic curve that gradually flattens out into a several mile runoff.

JetLeg
29th September 2007, 02:34 PM
deleted by user

OnlyTellsTruths
29th September 2007, 07:55 PM
In fact the only (infinetessimally small) chance you'd have is if you wern't strapped in and got thrown out onto a conveniently sloped and surface with appropriate friction and runout to gently decelerate you to a standstill.

maybe a lovely ice rink with an initial near-parapolic curve that gradually flattens out into a several mile runoff.

You forget body-deployed air bags..... it works with landing payload on Mars.

ponderingturtle
2nd October 2007, 05:01 AM
It's pointless test because it's waste of time, money, and automobiles to show what most sentient people would know already...a big car is going to hit harder than a small car. And it also probably has zero influence on which car people are going to buy anyway.

So the difference between a couple of thousand dollar bill and no repair bill is a pointless difference?

Personally if I had to pay for it I would like to know my car had the zero dollar bill.

Walrus32
2nd October 2007, 07:29 AM
Well, a lot of people have more money than brains. Besides, that's what insurance is for.

If you want zero repair bills, get no-deductible collision insurance.

dacium2007
4th October 2007, 03:08 AM
I can't believe peoeple are still going on with this...

it has nothing to do with how the object deforms.
it has nothing to do with elasticity
it has nothing to do with frames of reference

In both cases the car goes from 100 to zero. The same amount of energy is transfered away. The crashes are equal. If it were not so, there would not be conservation of energy - energy would have been created. If one car is deformed more in the 100 head on crash, so must be the other equally.

Cuddles
4th October 2007, 07:22 AM
Now, to further muddy the waters, I read where the Large Hadron Collider will smash counter-rotating beams of protons into each other. For my own edification, isn't this to increase the available energy output? I'm not trying to be argumentative...I just find it an interesting concept (as a non-physicist.)

To add to Wollery's answer, there are several reasons this is the case, and energy output is only one of them. However, in a relativisitc situation, energy doesn't work in quite the same way. When the cars collide, it is just as easy to accelerate both to 100mph as to accelerate one to 200mph. With relativistic effects, it is much easier to accelerate two particles to 500GeV than to accelerate one particle to 1TeV (1000GeV).

There are also other considerations that are at least as important. A charged particle loses energy when it changes direction that is proportional to a pwoer of it's current energy. This means that much less energy is lost using two beams than with using one and a target. With something like the LHC, that really is a lot of energy, and hence money and engineering.

There is also the issue of momentum, as with the cars. If you collide a particle with a stationary target, the momentum has to go somewhere. This means the products of the collision carry on going at high speed in the same direction as the beam. This causes two problems. The first is that it means a lot of energy goes into the kinetic energy of the products, and so not as much is available for actually creating the products. The other is that the detector generally needs to be much bigger because all the particles have so much more energy.

Thitical Crinker
4th October 2007, 01:16 PM
I won the bet! I got my friend to agree that in any of the 3 vehicles you would be decelerating from 100 mph to 0 mph. For some reason, he had a hard time comprehending, but then there was that moment when I could almost see the light bulb turn on and he "saw the light".

Squirrel season opens Saturday and he is going to meet me at the cabin with my beer. I hope to kill dinner and chase it down with a few cold ones.

Thanks again for all the help.

Michael Redman
4th October 2007, 01:42 PM
Good for you! Winning beer is always a commendable act.

As for testing bumpers, one reason is that the US government requires it: 49 CFR Part 581, "The bumper standard," prescribes performance requirements for passenger cars in low-speed front and rear collisions. It applies to front and rear bumpers on passenger cars to prevent the damage to the car body and safety related equipment at barrier impact speeds of 2½ mph across the full width and 1½ mph on the corners.

This is equivalent to a 5 mph crash into a parked vehicle of the same weight. The standard requires protection in the region 16 to 20 inches above the road surface, and the manufacturer can provide the protection by any means it wants. For example, some vehicles do not have a solid bumper across the vehicle, but meet the standard by strategically placed bumper guards and corner guards.

http://www.nhtsa.dot.gov/cars/problems/studies/Bumper/Index.html

Walrus32
5th October 2007, 07:43 AM
8) Why did NHTSA lower the bumper standard requirements from 5 mph to 2½ mph?

The agency concluded that reducing the impact speed from 5 mph to 2½ front and rear impact speed best satisfied the statutory criteria that the bumper standard "seek to obtain maximum feasible reduction in costs to the public and to the consumer." The agency also concluded that reducing the impact speed to 2½ mph and eliminating the Phase II damage criteria would not have an adverse effect on safety as measured by the number of crashes, deaths or injuries that occur annually.

The agency set the protection standard at 2½ mph after studying the comparable repair costs of a 5 mph bumper that has higher energy absorption capacity along with additional cost and weight.

After public hearings involving all parties, including consumers and manufacturers, NHTSA concluded that the public is assured of the largest net benefits under a standard that requires 2½ mph protection for both the front and rear bumpers.

(Op.cit)