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santorum
03-02-2012, 07:50 AM
Pardon the ignorance. I really should know the answer to this, but it's been a long time.

Suppose we have two cars of equal mass:

1. A and B collide into eachother both going 50 km/h in opposite directions
2. A is going 100 km/h and collides with B is at rest

Which one is more detrimental?

Floater
03-02-2012, 08:05 AM
As far as I understand there should be no difference.

Sunspace
03-02-2012, 08:07 AM
From the viewpoints of the cars, both collisions would be the same. Only context would differ: road, surroundings, etc.

jjimm
03-02-2012, 08:24 AM
How would the resistance of B's parking brake (or B left in P) in scenario two compare to the residual driving momentum of B's transmission/engine in the first scenario?

DCnDC
03-02-2012, 08:50 AM
The answer is 2.

2 cars colliding head-on at 50 km/h ≠ a 100 km/h collision, it equals 2 50 km/h collisions, as in each car is experiencing a 50 km/h collision. It doesn't matter that they are colliding with an object that happens to be moving toward it at 50 km/h, it's still primarily the same as hitting a stationary object (there's some variation as a parked car is not a brick wall but the basic physics are the same), because the moving cars are applying the same amount of force to each other. Thus, the 100 km/h collision will be more severe.

Great Antibob
03-02-2012, 09:15 AM
The answer is 2.

2 cars colliding head-on at 50 km/h ≠ a 100 km/h collision, it equals 2 50 km/h collisions, as in each car is experiencing a 50 km/h collision. It doesn't matter that they are colliding with an object that happens to be moving toward it at 50 km/h, it's still primarily the same as hitting a stationary object (there's some variation as a parked car is not a brick wall but the basic physics are the same), because the moving cars are applying the same amount of force to each other. Thus, the 100 km/h collision will be more severe.

Actually, there shouldn't be as much difference as you think because of the crumple zones. And, most likely, the car at rest will roll back, absorbing even more energy. So, the answer may well be 1.

The biggest difference between two cars hitting each other at 50 km/h and one car hitting a solid (unbreakable) brick wall at 100 km/h is that in the first case BOTH cars crumple to absorb the energy. In the second case, the first car alone absorbs all that energy.

But if the 100km/h car is hitting another car, it will, depending on the brakes of the car at rest, push the other car back somewhat and crumple both cars. A crude approximation would be an object hitting a spring.

The energies involved would be different (depending on the frame of reference), but I'm not seeing how the second collision is necessarily worse than the first, unless the 2nd car has a crappier crumple zone, has awesome brakes, and is backed up against a solid steel wall.

Snnipe 70E
03-02-2012, 09:23 AM
In both cases the crunple zones of both cars are going to absorb a lot of energy. In case 2 one car can move back wards some.

So case 2 is the winner by a lot.

westom
03-02-2012, 09:36 AM
Which one is more detrimental? Previous answers were from the car's perspective. Nobody cares about cars in such collisions. Most concern is for what is inside each car.

Since the amount of damage varies significantly with where and at what angles each car collides, then the question becomes too hypothetical. Damage to the occupants is the only concern. Both cars are now scrap metal. Damage to each car is only relevant to energy is imparted inside.

leahcim
03-02-2012, 09:36 AM
You could argue that (2) is more severe because if you move the system in (1) into A's frame, then A is at rest, and B is moving at very slightly less than 100mph because of the relativistic correction.

robby
03-02-2012, 09:40 AM
The answer is 2.

2 cars colliding head-on at 50 km/h ≠ a 100 km/h collision, it equals 2 50 km/h collisions, as in each car is experiencing a 50 km/h collision. It doesn't matter that they are colliding with an object that happens to be moving toward it at 50 km/h, it's still primarily the same as hitting a stationary object (there's some variation as a parked car is not a brick wall but the basic physics are the same), because the moving cars are applying the same amount of force to each other. Thus, the 100 km/h collision will be more severe.This is incorrect. To determine the severity of a collision, you look at the change in momentum, which is equivalent to the impulse. The force experienced is then equal to the impulse divided by the time interval over which the force is applied. If we assume that the time interval is the same in both collisions, all we need to look at then is the change in momentum.

Note that a collision between cars is an inelastic collision, so kinetic energy is not conserved.

Anyway, let's make the following assumptions: the two cars involved have the same mass, and the collisions considered take the same amount of time.

In the first collision, the change in momentum for car A is m times the change in velocity, or m times 50 km/hr. (Note that the final velocity of both cars is zero, because of conservation of momentum and the fact that it is an inelastic collision.)

In the second case (a car traveling at 100 km/hr hitting a parked vehcicle), note that the velocity of the two cars immediately after the collision will be 50 km/hr (also due to conservation of momentum). So for car A, the change in momentum will be m times 50 km/hr. The two situations are identical.

As for car A running into an immovable object (like the proverbial brick wall), the equivalent situation would be hitting the brick wall at 50 km/hr. The final velocity for car A will be zero, so again the change in momentum for car A will be m times 50 km/hr.

Dog80
03-02-2012, 09:40 AM
Kinetic energy is 1/2 m u2

Total kinetic energy in the first case is 2*(1/2 m 502) = 2500m

In the second case the energy is 1/2 m 1002 = 5000M

Alka Seltzer
03-02-2012, 09:41 AM
I think we need a reply from an actual physicist. Kinetic energy is given by the formula 1/2 mv2, so the collision in scenario 2 is a lot more energetic. What always confuses me is how this relates to frames of reference, as the calculated energy is different depending on where it is observed from.

Looking at the details of the Euro NCAP Frontal Impact (http://www.euroncap.com/tests/frontimpact.aspx) test, a 64 kph collision into a barrier is used to simulate a car-to-car collision with both cars travelling at 55 kph. Presumably this is done because the energies involved are the same, which implies scenario 2 above would be a much more serious impact.

Corcaigh
03-02-2012, 09:43 AM
I'm pretty sure they did this on Mythbusters and #2 would cause more damage

Chessic Sense
03-02-2012, 09:44 AM
The answer is 2.

2 cars colliding head-on at 50 km/h ≠ a 100 km/h collision, it equals 2 50 km/h collisions, as in each car is experiencing a 50 km/h collision. It doesn't matter that they are colliding with an object that happens to be moving toward it at 50 km/h, it's still primarily the same as hitting a stationary object (there's some variation as a parked car is not a brick wall but the basic physics are the same), because the moving cars are applying the same amount of force to each other. Thus, the 100 km/h collision will be more severe.

WTF? How do you figure that? The only difference is how fast the scenery is moving relative to the cars. Two 50 km/h crashes certainly does sum to 100 km/h. Are you suggesting that in the 100/0 scenario, the stopped car is all hunky-dory because it experiences a "0 km/h" collision?

Imagine a car hitting you at 50 km/h. Now imagine if you decided to gun the accelerator before you were struck. Are you saying there's no difference?!

In both cases the crunple zones of both cars are going to absorb a lot of energy. In case 2 one car can move back wards some.

So case 2 is the winner by a lot.

How does moving "backwards"* change anything? It's still absorbing energy. That's why it's moving backwards. If you're moving at 50 km/h and all of a sudden, you're moving at 0 km/h, guess what...you've been moved "backwards".

*I put it in quotes because the idea that the scenery is a relevant reference point is silly.

robby
03-02-2012, 09:48 AM
P.S. Reading some of the other replies, there is some argument to be made that the brick wall collison may be somewhat more severe because there is only one car in which to crumple (instead of two). This may have the effect of shortening the time of collision (which changes one of my previous assumptions), thereby increasing the force experienced by the occupants of the vehicle.

This would not apply to the original two cases presented in the OP, however. They are still identical for the reasons stated in my previous post.

Baracus
03-02-2012, 09:49 AM
Kinetic energy is 1/2 m u2

Total kinetic energy in the first case is 2*(1/2 m 502) = 2500m

In the second case the energy is 1/2 m 1002 = 5000M
In the first case, all that kinetics energy is dissipated in the collision.

In the second case, immediate post crash when the two cars are traveling a 50 km/h, their kinetic energy will be (1/2)*2m*502 = 2500m. Thus there will again be 2500m of kinetic energy dissipated during the collision.

westom
03-02-2012, 09:50 AM
Total kinetic energy in the first case is 2*(1/2 m 502) = 2500m
In the second case the energy is 1/2 m 1002 = 5000M If detriment is determined by energy, then Dog80 has struck the correct answer.

A 100 Km/hr car striking a parked car is energy equivalent to two cars hitting each other at 70.7 Km/hr.

(Speeding kills.)

TripleFail
03-02-2012, 09:56 AM
I think if you look at the problem from a purely kinetic energy point of view case 2 would be worse off. We know E=1/2*m*v^2 assuming 2 vehicle have same mass we have total energy of E(1)=2(1/2*m*v^2)=m*v^2. For case 2 we get E(2)=1/2*m*(2*v)^2=2*m*v^2.

Though even system 2 have twice the kinetic energy it might have longer collision time so the occupant would experience less G force.

Chessic Sense
03-02-2012, 09:58 AM
If detriment is determined by energy, then Dog80 has struck the correct answer.

But it's not. Detriment is determined by change in energy over time, i.e. impulse. Sure, one starts out more energetic in a certain reference frame, but it also ends more energetic in that same reference frame. The change is equivalent.

Looking at the details of the Euro NCAP Frontal Impact (http://www.euroncap.com/tests/frontimpact.aspx) test, a 64 kph collision into a barrier is used to simulate a car-to-car collision with both cars travelling at 55 kph. Presumably this is done because the energies involved are the same, which implies scenario 2 above would be a much more serious impact.

That's because they're using a barrier instead of a car, not because of the vector addition.

Chessic Sense
03-02-2012, 10:01 AM
I think if you look at the problem from a purely kinetic energy point of view case 2 would be worse off. We know E=1/2*m*v^2 assuming 2 vehicle have same mass we have total energy of E(1)=2(1/2*m*v^2)=m*v^2. For case 2 we get E(2)=1/2*m*(2*v)^2=2*m*v^2.

Though even system 2 have twice the kinetic energy it might have longer collision time so the occupant would experience less G force.

Again, unless you're expecting the second car to come out immaculate, you have to average the energy. It's the same energy per car.

In scenario 1, the delta-V is +50 for car 1 and -50 for car 2.
In scenario 2, the delta-V is -50 for car 1 and -50 for car 2.

Each car changes velocity the same amount. You're either not averaging across the cars or you're not considering the end scenario. It's the same error either way, but you have to do one or the other.

Blaster Master
03-02-2012, 10:08 AM
I'm pretty sure they did this on Mythbusters and #2 would cause more damage

Yes, they did do this on Mythbusters. Jamie had previously incorrectly stated that two vehicles going 50 would be the same as one going 100 into a brick wall. They tested it by having one car crash into a wall at 50, an identical one into a wall at 100 and two more identical cars hitting head on both going 50. They took numbers, but the visual was enough to show that the head-on collision looked a lot like the first example and not at all like the latter. The math behind why was well explained upthread.

I'd find the episode name but, well, the only thing harder to find than a specific Mythbusters myth involving car crashes is one involving explosions.

westom
03-02-2012, 10:09 AM
But it's not. Detriment is determined by change in energy over time, i.e. impulse. The OP did not define 'detriment'. So every post is making assumptions. Recall my first post. Both cars are now scrap metal. Damage to each car is only relevant to energy [that] is imparted inside.

Dogma's assumption is detriment defined by the amount of energy absorbed entirely by each car. Acceleration and time has no relevance there.

Acceleration to a human head? That is completely different. But again, too many hypotheticals exist to make a reliable reply to the OP's question.

Great Antibob
03-02-2012, 10:11 AM
Yes, they did do this on Mythbusters.


Doesn't look like the same thing, actually. There's a difference between a brick wall and a car. The brick wall crumples differently, for one. And a car isn't made of bricks.

So, what happens when a car at 100mph crashes into a parked car?

robby
03-02-2012, 10:14 AM
I think we need a reply from an actual physicist. Kinetic energy is given by the formula 1/2 mv2, so the collision in scenario 2 is a lot more energetic.Well, for what it's worth, I used to teach physics. :)

Kinetic energy is 1/2 m u2

Total kinetic energy in the first case is 2*(1/2 m 502) = 2500m

In the second case the energy is 1/2 m 1002 = 5000M

If detriment is determined by energy, then Dog80 has struck the correct answer.

A 100 Km/hr car striking a parked car is energy equivalent to two cars hitting each other at 70.7 Km/hr.All of these posts are incorrect, because you have to look at the change in kinetic energy dissipated in the collision. In the first case, the initial total KE of the two vehicles is 2500m. The final KE is zero, so the KE dissipated in the actual collision is 2500m.

In the second case, the initial KE of car A is 5000m, but the KE of the two mangled vehicles immediately after the collision is 2500m. (The rest of the KE will then be dissipated as friction as the two mangled vehicles slide to a stop). So the KE dissipated in the collision will again be 2500m.

In the first case, all that kinetics energy is dissipated in the collision.

In the second case, immediate post crash when the two cars are traveling a 50 km/h, their kinetic energy will be (1/2)*2m*502 = 2500m. Thus there will again be 2500m of kinetic energy dissipated during the collision.This is correct.

DCnDC
03-02-2012, 10:18 AM
The OP did not define 'detriment'. So every post is making assumptions.

Yes. Are we talking force or energy?

The force of the collision in (2) is double that of (1).

The energy OTOH is the same in both.

robby
03-02-2012, 10:23 AM
But it's not. Detriment is determined by change in energy over time, i.e. impulse.Nitpick: Impulse is equal to the change in momentum of an object to which a force is applied.

And as I stated above, the detriment to the people in the vehicle would be due to the peak forces experienced by the occupants.

It's not actually relevant in an inelastic collision to consider energy (including kinetic energy), because kinetic energy in an inelastic collision is not conserved. Instead, it is dissipated as friction, noise, and the crumpling up of the vehicles in question.

Terr
03-02-2012, 10:24 AM
I'm pretty sure they did this on Mythbusters and #2 would cause more damage

No. On Mythbusters the second scenario was crashing into a wall. Very different.

Chessic Sense
03-02-2012, 10:25 AM
Well, for what it's worth, I used to teach physics. :)



All of these posts are incorrect, because you have to look at the change in kinetic energy dissipated in the collision. In the first case, the initial total KE of the two vehicles is 2500m. The final KE is zero, so the KE dissipated in the actual collision is 2500m.

In the second case, the initial KE of car A is 5000m, but the KE of the two mangled vehicles immediately after the collision is 2500m. (The rest of the KE will then be dissipated as friction as the two mangled vehicles slide to a stop). So the KE dissipated in the collision will again be 2500m.

This is correct.

robby, I understand and agree with your sentiment but I think your math is flawed. First, KE has a vector, does it not? I understand it doesn't have to, in an absolute sense, but in the context of summing energies, it has to. So in the first scenario, it's really 2500m for the first car and -2500m for the second car, summing to zero. After collision, it's still zero, but some negative KE and some positive KE has been converted to damage, friction, heat, etc.

In the second scenario, yes, we'd end with 2500m, but that's still "little m", i.e. a variable. But that little m is equal to 2 big-Ms, i.e. the fixed mass of an individual car. So we go from 5000M and 0M to 2500(2M), and energy is conserved. Of course, in the real world, we won't come out of the collision at that "conserved energy" speed, and the difference will be the damage to the two cars.

So my point is, the scenarios are equal as far as we can tell without knowing exactly how the cars are built and what safety features they have. In practice it may make a difference, but in theory, there is no difference. The situations are the same.

westom
03-02-2012, 10:25 AM
Well, for what it's worth, I used to teach physics.
The energy content of one car at 100 km/hr is two times larger than the energy content of two cars doing 50 km/hr.

A car at 100 km/hr has four times the energy of a car at 50 km/hr.

Again, if "detriment" is defined by energy dissipated, then a 100 km/hr car is a much more violent event.

To have same energy, two cars must be doing 70.7 km/hr; not 50 km/hr. As Dog80 posted.

I used to be a student.

Chessic Sense
03-02-2012, 10:28 AM
The force of the collision in (2) is double that of (1)

No, the force is equal, too. I mean, at least in a scalar sense. A car going from rest to 50 km/h is the same as a car going from -50 km/h to 0. If it takes the same time and the cars have the same mass between scenarios (and why wouldn't they?), and F=M*dV/t, then the force is the same, except maybe for the sign depending on how you draw the coordinate system.

robby
03-02-2012, 10:28 AM
...The force of the collision in (2) is double that of (1).No, it isn't. See my Post #10 above.

Chessic Sense
03-02-2012, 10:30 AM
The energy content of one car at 100 km/hr is two times larger than the energy content of two cars doing 50 km/hr.

A car at 100 km/hr has four times the energy of a car at 50 km/hr.

Again, if "detriment" is defined by energy dissipated, then a 100 km/hr car is a much more violent event.

To have same energy, two cars must be doing 70.7 km/hr; not 50 km/hr. As Dog80 posted.

I used to be a student.

Then you get half credit for stopping halfway through the problem. Yes, a car at 100 km/hr has four times the energy as a car at 50 km/hr. You don't get to just stop right there, put down the pencil, and say you have an answer. Now find the energy post-collision and find the "energy dissipated," which you admit is what is at issue here.

ETA: Also note that only one car is going at 100 km/hr. The other is at rest. If you intend to damage two cars, then you have to average the energy across both cars. If you consider only one car to be changing energy, then you can't claim the second car would be damaged. It'd have to be built of unobtanium.

DCnDC
03-02-2012, 10:33 AM
I see. My physics (and math) are rusty.

Baracus
03-02-2012, 10:35 AM
robby, I understand and agree with your sentiment but I think your math is flawed. First, KE has a vector, does it not? I understand it doesn't have to, in an absolute sense, but in the context of summing energies, it has to. So in the first scenario, it's really 2500m for the first car and -2500m for the second car, summing to zero. After collision, it's still zero, but some negative KE and some positive KE has been converted to damage, friction, heat, etc.
No, KE is not a vector. Consider a rubber ball fired at a brick wall. All of the KE is pointed at the wall prior to impact and most of it is pointed away from the wall afterwards.

robby
03-02-2012, 10:35 AM
robby, I understand and agree with your sentiment but I think your math is flawed. First, KE has a vector, does it not? I understand it doesn't have to, in an absolute sense, but in the context of summing energies, it has to.

...After collision, it's still zero, but some negative KE and some positive KE has been converted to damage, friction, heat, etc.Well, your starting point here is incorrect. Kinetic energy is a scalar quantity. Note that you are squaring the magnitude of the velocity.

There is no such thing as "negative" kinetic energy.

TripleFail
03-02-2012, 10:37 AM
Again, unless you're expecting the second car to come out immaculate, you have to average the energy. It's the same energy per car.

In scenario 1, the delta-V is +50 for car 1 and -50 for car 2.
In scenario 2, the delta-V is -50 for car 1 and -50 for car 2.

Each car changes velocity the same amount. You're either not averaging across the cars or you're not considering the end scenario. It's the same error either way, but you have to do one or the other.

humm Not sure what you trying to say. Are you saying in both scenario have the same amount of energy dissipated? I think it very clear that case 2 carries more energy. How the energy is dissipated is another question Of course I am don't think the 2nd car will come out immaculate in either case. The real questions are in case 2 how much of that extra energy would be converted back to kinetic energy then come to a rest due to friction with road, and would the materials crumple zone material act the same way in both scenarios. As collision time would be different the material might handle the different intensity differently.

JimboJamesJimmersonIII,Esq.
03-02-2012, 10:41 AM
Robby has been correct throughout the thread. The two cases are physically equivalent to each other.

westom
03-02-2012, 10:44 AM
Yes, a car at 100 km/hr has four times the energy as a car at 50 km/hr. ... Now find the energy post-collision and find the "energy dissipated," which you admit is what is at issue here. Energy dissipated is 1) energy of a car at 100 km/hr or 2) the energy of two cars at 50 km/hr. Conservation of energy. In both cases, all energy is absorbed equally both cars. But 1) means double the energy.

Again, I keep saying this. Otherwise junk science results. Define "detriment". a) Detriment is defined by the acceleration of a brain inside the skull? Or b) defined by the total amount of energy dissipated?

b) is total energy dissipated in each car. a) defines detriment in terms of *where* energy dissipates throughout each car.

Why are so many answering a question before first defining the question? A major mistake. Define your interpretation of detriment in every answer. Otherwise any answer is possible.

Meanwhile, the problem is framed by one constant. The amount of energy in the collision. One 100 km/hr car means double energy dissipates in both cars.

robby
03-02-2012, 10:50 AM
Westom, I have addressed all of your points in my earlier posts.

TonySinclair
03-02-2012, 10:51 AM
Wow. And here I always liked physics because, unlike politics, there was a definite right answer for elementary problems.

The guys who say the two collisions are equivalent are right, and you shouldn't even have to do any arithmetic to see it. You only have to remember that experiments done in inertial frames have identical outcomes (except for the infinitesimal relativistic effects that occur at such low speeds).

Cheesesteak
03-02-2012, 10:52 AM
To echo robby and Chessic Sense, the collisions are identical. What differs is post-collision. In case #1, post collision is two wrecked cars at rest. In case #2, you have two wrecked cars going 50km/h.

If the cars are on neat, weightless, fricitonless sleds, and are collided under controlled conditions, and are brought to a clean damage-free stop, you will see little difference. In the real world, two wrecked cars going 50km/h down a street will eventually stop, but will likely hit other things and become more damaged along the way.

From a "let's assume the cow is a sphere" standpoint, the cases are the same. From a "let's pry these poor bastards out of the car" standpoint, the first case will likely be somewhat less damaging than the second. However, even though the second case has twice the energy as the first, most of that extra energy will be dissipated by the cars scraping to a stop, and not via additional impact damage.

This is why the car going 100 into a brick wall is way worse than the 50-50 head on, all the extra energy is impact energy.

TonySinclair
03-02-2012, 10:56 AM
One 100 km/hr car means double energy dissipates in both cars.

No. It does if the car hits an immovable object (the proverbial brick wall), but not if it hits a stationary car of equal weight, which will be propelled backwards at close to 50 km/h, disregarding inelastic effects.

So -- a car going 100 hitting a stationary car is the same as two cars going 50 hitting head on, and also the same as a car going 50 hitting a brick wall.

Chessic Sense
03-02-2012, 10:58 AM
humm Not sure what you trying to say. Are you saying in both scenario have the same amount of energy dissipated?

Yes, that's what I'm saying.

In both cases, all energy is absorbed equally both cars.

Meanwhile, the problem is framed by one constant. The amount of energy in the collision. One 100 km/hr car means double energy dissipates in both cars.

Your first statement is where you err. The word "all" does not belong in that sentence. If we slam a 100 km/hr car into a stopped one, they don't end with zero energy. They still have energy at the end. That is what your "team" is failing to consider.

Second, it's not important how much energy is in the collision. What's important is what energy change there is. Here's a thought experiment for you:

1) There's a parked car and a car moving at .05 mph that wrecks into it. How much energy is in the collision? Colloquially, how "violent" would you describe it?
2) There are two cars in a NASCAR race. One is moving at 200 mph and another behind it at 200.05 mph. The faster, rear car catches up to the front one and bumps it perfectly squarely from behind. How much energy is there in that scenario? Forgetting for a moment how in the real world this might cause the front driver to lose control or spin out after impact , how violent is this collision?

Reviewing your answers, are you sure you want to stick with the assertion that initial energy is what matters?

robby
03-02-2012, 11:15 AM
OK, I actually switched from my iPad to my work computer (which I don't normally like to do), because extensive typing on the iPad is a pain in the neck. However, it is my lunch break, so here goes...

Energy dissipated is 1) energy of a car at 100 km/hr or 2) the energy of two cars at 50 km/hr. Conservation of energy. In both cases, all energy is absorbed equally both cars. But 1) means double the energy.OK, first off, realize that the starting energy in both cases is kinetic energy, and kinetic energy is not conserved in an inelastic collision (like this one).

Now, total energy is always conserved, but realize that in the first case, all of the kinetic energy is dissipated in the collision, because the vehicles end up at rest immediately after the collision (because momentum is conserved).

However, in the second case, some of the initial kinetic energy is dissipated in the collision, but realize that the two mangled vehicles then have some residual kinetic energy (because they slide away from the collision at a speed of 50 km/hr, again due to conservation of momentum). This residual kinetic energy is then dissipated by friction as the two cars slide to a stop after the collision. Note that this slide to a stop doesn't actually hurt the occupants because it will take place over a much longer time interval.

Again, I keep saying this. Otherwise junk science results. Define "detriment". a) Detriment is defined by the acceleration of a brain inside the skull? Or b) defined by the total amount of energy dissipated?As I have stated several times now, the detriment to the people in the vehicle would be due to the peak forces experienced by the occupants.

Chessic Sense
03-02-2012, 11:21 AM
OK, first off, realize that the starting energy in both cases is kinetic energy, and kinetic energy is not conserved in an inelastic collision (like this one).

Now, total energy is always conserved, but realize that in the first case, all of the kinetic energy is dissipated in the collision, because the vehicles end up at rest immediately after the collision (because momentum is conserved).

However, in the second case, some of the initial kinetic energy is dissipated in the collision, but realize that the two mangled vehicles then have some residual kinetic energy (because they slide away from the collision at a speed of 50 km/hr, again due to conservation of momentum). This residual kinetic energy is then dissipated by friction as the two cars slide to a stop after the collision. Note that this slide to a stop doesn't actually hurt the occupants because it will take place over a much longer time interval.

As I have stated several times now, the detriment to the people in the vehicle would be due to the peak forces experienced by the occupants.

Whoa, whoa, whoa. Hold up. First, how do you know that both cars come to rest in the first collision. Cars bounce, y'know. Second, how could the tangled mass in the second collision be going at 50 km/hr and also be damaged? They'd lose energy in the collision itself, not just from scraping.

I'm not sure where you're getting that momentum would be conserved. Unless the cars are in pristine, pre-crash condition, they're not coming away from this at 50 km/hr.

Cheesesteak
03-02-2012, 11:23 AM
Second, how could the tangled mass in the second collision be going at 50 km/hr and also be damaged? They'd lose energy in the collision itself, not just from scraping.As was noted, two cars moving at 50km/h have half the energy of one car moving 100km/h.

Just to add, momentum is always conserved.

robby
03-02-2012, 11:34 AM
...I'm not sure where you're getting that momentum would be conserved.:dubious: In this universe, momentum is always conserved. No exceptions.

westom
03-02-2012, 11:41 AM
:dubious: In this universe, momentum is always conserved. No exceptions.
And energy is also and always conserved - no exceptions. Does not matter if the collision is elastic or inelastic. Energy is always conserved. The term elastic only says which side of a collision gets more energy. Energy is always conserved. What comes out is always what goes in.

The 100 km/hr car puts in twice the energy. What results in both cars is twice the energy. Obviously conservation of energy cannot be denied.

westom
03-02-2012, 11:48 AM
Whoa, whoa, whoa. Hold up. First, how do you know that both cars come to rest in the first collision. Cars bounce, y'know.
So where do you post your definition of detrimental in each post? You don't. Detrimental. The energy imparted into both cars. Even if one car is still moving, the energy remains. Your 'whoas' come from this fact, You don't define 'detrimental' before making conclusions. And you are not the only one doing what is common in junk science.

Define detrimental.

robby
03-02-2012, 11:48 AM
Whoa, whoa, whoa. Hold up. First, how do you know that both cars come to rest in the first collision. Cars bounce, y'know.I was making the simplying assumption that in both cases, the collisions are perfectly inelastic (http://en.wikipedia.org/wiki/Inelastic_collision#Perfectly_inelastic_collision). (There is no question that the collisions are inelastic (http://en.wikipedia.org/wiki/Inelastic_collision).) However, considering an inelastic collision that is not perfectly inelastic adds a needless complication to the problem.

What the effect of making the collisions not perfectly inelastic is to make the collisions worse, because while momentum is still conserved, you end up with a larger change in momentum for each vehicle (because momentum, unlike kinetic energy, is a vector quantity).

...Second, how could the tangled mass in the second collision be going at 50 km/hr and also be damaged? They'd lose energy in the collision itself, not just from scraping.Easy, you'd have a big crash, followed by a mangled heap of two cars sliding away at 50 km/hr (and rapidly slowing to a stop).

...I'm not sure where you're getting that momentum would be conserved. Unless the cars are in pristine, pre-crash condition, they're not coming away from this at 50 km/hr.If the cars stick together, which is not a bad assumption, they must be moving away from the point of collison at 50 km/hr. In this universe, we follow the law of conservation of momentum. ;)

westom
03-02-2012, 11:49 AM
Reviewing your answers, are you sure you want to stick with the assertion that initial energy is what matters?
Energy of a 100 km/hr car completely dissipates in the scrap metal once called two cars. *All* kinetic energy dissipates in both cars - that are now at 0 km/hr.

Two 50 km/hr cars have one half that energy. *All* kinetic energy dissipates in both cars - that are now at 0 km/hr. The concept is well understood – conservation of energy.

If 'detriment' is defined by energy dissipation, then a 100 km/hr collision is doubly detrimental. Obviously. Notice – every post defines the term ‘detriment’. You do not.

All energy from moving cars must be conserved. All that energy dissipates in damage to both cars because all cars are now at 0 km/hr. Conservation of energy always exists. Always (despite someone earlier who denied it).

A collision of one car at 100 km/hr means double energy dissipates in both cars. No way around reality.

Energy change is obvious. One car at 100 km/hr is now 0 km/hr. Two cars at 50 km/hr are now at 0 km/hr. All energy once defined by mass and speed is now measured by damage.

BTW, your NASCAR example is bogus. Speed of both cars after the 0.05 mph collision is not defined. So no answer is possible. Obviously, without mass, another reason why no answer is possible. Please present your answer with numbers. Subjective reasoning is also called junk science.

Obviously initial energy matters. It is called conservation of energy. Why is that so difficult? All energy in once moving cars is now dissipated in what is called scrap metal.

robby
03-02-2012, 11:50 AM
And energy is also and always conserved - no exceptions. Does not matter if the collision is elastic or inelastic. Energy is always conserved. The term elastic only says which side of a collision gets more energy. Energy is always conserved. What comes out is always what goes in.

The 100 km/hr car puts in twice the energy. What results in both cars is twice the energy. Obviously conservation of energy cannot be denied.See Post #44 above.

Telemark
03-02-2012, 11:52 AM
The 100 km/hr car puts in twice the energy. What results in both cars is twice the energy. Obviously conservation of energy cannot be denied.
I think everyone else is saying that the extra energy is present in the system in the form of both cars moving at 50 km/hr after the collision. That energy isn't going to crumple the cars.

Cheesesteak
03-02-2012, 11:54 AM
All energy from moving cars must be conserved.True. All that energy dissipates in damage to both cars because all cars are now at 0 km/hr. Not true.

The energy that gets dissipated need not result in damage to the cars, it very much depends HOW it is dissipated. Example, the cars gently sideswipe each other, and the drivers use their brakes to stop both cars. Same energy dissipated, but much different levels of damage.

In these cases, the energy dissipated during the collision is the same, what differs is how much energy is left over post-collision to be dissipated in another way.

Chessic Sense
03-02-2012, 11:54 AM
As was noted, two cars moving at 50km/h have half the energy of one car moving 100km/h.

Just to add, momentum is always conserved.

How is momentum always conserved? When I push my couch across the carpet, it stops. The cars and the couch are not closed systems.

westom
03-02-2012, 11:58 AM
The energy that gets dissipated need not result in damage to the cars, it very much depends HOW it is dissipated. Example, the cars gently sideswipe each other, and the drivers use their brakes to stop both cars. Same energy dissipated, but much different levels of damage. So again you post without defining the word 'detrimental'. Energy dissipated in scrap metal or brakes is still the same energy conserved. At least I define detirimental. Energy in each vehicle. Burn brakes - conservatioin of energy - is still defined by detriment.

robby
03-02-2012, 11:59 AM
Energy of a 100 km/hr car completely dissipates in the scrap metal once called two cars. *All* kinetic energy dissipates in both cars - that are now at 0 km/hr.

Two 50 km/hr cars have one half that energy. *All* kinetic energy dissipates in both cars - that are now at 0 km/hr. The concept is well understood – conservation of energy.

If 'detriment' is defined by energy dissipation, then a 100 km/hr collision is doubly detrimental. Obviously. Notice – every post defines the term ‘detriment’. You do not.

All energy from moving cars must be conserved. All that energy dissipates in damage to both cars because all cars are now at 0 km/hr. Conservation of energy always exists. Always (despite someone earlier who denied it).

A collision of one car at 100 km/hr means double energy dissipates in both cars. No way around reality.

Energy change is obvious. One car at 100 km/hr is now 0 km/hr. Two cars at 50 km/hr are now at 0 km/hr. All energy once defined by mass and speed is now measured by damage.

BTW, your NASCAR example is bogus. Speed of both cars after the 0.05 mph collision is not defined. So no answer is possible. Obviously, without mass, another reason why no answer is possible. Please present your answer with numbers. Subjective reasoning is also called junk science.

Obviously initial energy matters. It is called conservation of energy. Why is that so difficult? All energy in once moving cars is now dissipated in what is called scrap metal.westom, you are thoroughly confusing yourself (and possibly others) by only considering energy.

Energy dissipated as noise or friction or the crumpling of metal does not necessarily hurt the occupants of the vehicles.

Instead, as I have stated repeatedly, the detriment to the occupants in the vehicles would be due to the peak forces experienced by the occupants. Period.

Also, nobody has denied that energy is not conserved. I did state earlier that kinetic energy is not conserved. Which it isn't in this type of inelastic collison, by definition (http://en.wikipedia.org/wiki/Inelastic_collision).

An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved.

TripleFail
03-02-2012, 12:08 PM
robby, I think you are forgetting the limits on the material used to construct car. I agree with you that in case 2 immediately contact that there will be crumpling to both car in some degree then both car would move. That's why I said "Though even system 2 have twice the kinetic energy it might have longer collision time so the occupant would experience less G force."
But I am just not sure both of them would them move at around 50 km/h immediately after contact then slide to a stop.

As car A slams into a stationary car B some of the 5000m of energy is going to be dissipated in the crumpling of both car but depending on the geometry and material used the whole car might just collapse and pancaking the passengers inside. It would take special engineering to assume half of the energy would be absorbed by materials and rest into friction with road. Different rigidity, materials would all give you different results in case 2.

robby
03-02-2012, 12:09 PM
How is momentum always conserved?Because it's the law (http://en.wikipedia.org/wiki/Conservation_of_momentum#Conservation_of_linear_momentum).

The law of conservation of linear momentum is a fundamental law of nature, and it states that if no external force acts on a closed system of objects, the momentum of the closed system remains constant.

When I push my couch across the carpet, it stops. The cars and the couch are not closed systems.In the case of your couch, there is an external force that acts on your couch to bring it to a stop.

During our hypothetical car collision(s), there is no external force acting on the cars during the collisions.

Baracus
03-02-2012, 12:10 PM
Wow. And here I always liked physics because, unlike politics, there was a definite right answer for elementary problems.

The guys who say the two collisions are equivalent are right, and you shouldn't even have to do any arithmetic to see it. You only have to remember that experiments done in inertial frames have identical outcomes (except for the infinitesimal relativistic effects that occur at such low speeds).
I think this is what the entire thread digests down to. If you change your frame of reference to put one of the cars in the center and keep it there, then both scenarios result in the other car smashing into it a 100 kph. Hence, they are the same, in the ideal case.

In the real world, there would be some additional forces in the stationary car scenario due to the frictional forces that must be overcome to get the car moving.

newme
03-02-2012, 12:10 PM
1. A and B collide into each other both going 50 km/h in opposite directions
2. A is going 100 km/h and collides with B is at rest

Which one is more detrimental?

I am going to add a few assumptions here so we don't get into another plane on treadmill mess.

The cars are identical models.
They collide squarely head-on.
One car does not go flying over the other.

One half second after the initial contact, the "detriment" to cars and occupants would be the same in both cases.

In case #1, I would assume both cars have come to a complete stop, and no further damage can occur.

In case#2, both cars are now moving at 50kmh. In a real life scenario, it is entirely possible that the cars could could flip and roll, causing additional "detriment." It is also possible that they skid harmlessly to a stop.

Machine Elf
03-02-2012, 12:13 PM
Energy of a 100 km/hr car completely dissipates in the scrap metal once called two cars. *All* kinetic energy dissipates in both cars - that are now at 0 km/hr.

This is not correct. After the collision in scenario #2, both cars are moving in the same direction at 50 km/hr. If you doubt this, go watch some 2-vehicle crash tests on YouTube.

After the collision in scenario #1 (both cars headed toward each other at 50 km/hr), both cars are at rest, so in that case yes, all of the initial kinetic energy is dissipated in the collision.

scenario 1:
Initial total kinetic energy of both vehicles: E1 = 2*(0.5*m*50^2) = 2500m (m = mass of one car)
Final total kinetic energy of both vehicles: E2 = 0
Energy dissipated in scenario 1 impact: E1 - E2 = 2500m


scenario 2:
Initial kinetic energy of both vehicles: E1 = .5*m*100^2 = 5000m
Final kinetic energy of both vehicles: E2 = 2*(.5*m*50^2) = 2500m
Energy dissipated in scenario 2 impact: E1-E2 = 2500m


Either way, the energy dissipated into the crumple zones is the same, so both scenarios present equally severe impacts for the vehicle occupants.

robby
03-02-2012, 12:14 PM
robby, I think you are forgetting the limits on the material used to construct car. I agree with you that in case 2 immediately contact that there will be crumpling to both car in some degree then both car would move. That's why I said "Though even system 2 have twice the kinetic energy it might have longer collision time so the occupant would experience less G force."
But I am just not sure both of them would them move at around 50 km/h immediately after contact then slide to a stop.

As car A slams into a stationary car B some of the 5000m of energy is going to be dissipated in the crumpling of both car but depending on the geometry and material used the whole car might just collapse and pancaking the passengers inside. It would take special engineering to assume half of the energy would be absorbed by materials and rest into friction with road. Different rigidity, materials would all give you different results in case 2.None of this really matters, because the fact that the cars have to be moving away from the collision at 50 km/hr comes from conservation of momentum (assuming they are stuck together, which is not a bad assumption in a case where the objects colliding tend to crumple up when they collide).

Also, assuming the cars hit head-on in both cases, there should not be any difference in the time of collision.

Chronos
03-02-2012, 12:23 PM
Both cases are equivalent. The single car going 100 km/hr has more energy, but not all of that energy will go into damaging the cars: Some of it will go into making the other car move. This is different from hitting a proverbial brick wall, because the brick wall can't move, so the energy has no place to go but into damage. Hitting a brick wall at 50 km/hr is equivalent to a head-on collision between two cars each going 50 km/hr.

robby
03-02-2012, 12:32 PM
Both cases are equivalent.Exactly. And for those arguing that the cases are not equivalent, realize that this is not some esoteric, obscure physics problem.

It's discussed in detail in every introductory physics class, and in every introductory physics textbook.

newme
03-02-2012, 12:42 PM
So again you post without defining the word 'detrimental'. Energy dissipated in scrap metal or brakes is still the same energy conserved. At least I define detirimental. Energy in each vehicle. Burn brakes - conservatioin of energy - is still defined by detriment.

Westom's definition of "detriment" makes my brain hurt. By this definition, a car slamming into a brick wall at 100kmh undergoes the same detriment as a similar car initially moving at 100kmh coasting to stop without hitting anything.

westom
03-02-2012, 01:10 PM
As I have stated several times now, the detriment to the people in the vehicle would be due to the peak forces experienced by the occupants. So you are talking about oranges while others discuss pears. If detriment is what you have defined, then every number you have posted is useless. We need mass, velocity, etc for the occupants. We need to know the elasticity of their collision. We need to know if the occupants are restrained. The 50 and 100 km/hr numbers tell us nothing useful for your definition of detriment. Worse, your posts are only subjective.

As first noted, what you are calling 'detrimental' cannot even be discussed due to too many hypotheticals. Due to missing numbers. Due to facts not found in the OP's original question. Irrelevant even to what the myth busters demonstrated.

Other answers are addressing a completely different "detrimental". Oranges and pears. Your detriment cannot be defined by anything in the OP's original post - a 100 km/hr car or two 50 km/hr cars. We can only say the car and occupant as a unit are confronted by twice the energy when one car does 100 km/hr. Nothing more.

To say more means you must define "where" energy dissipates in the car and in its occupants - with numbers and vectors. I posted "where" in at the start. Your question in post #57 was repeatedly answered by "where".

newme - it's simple. Is detriment a relationship between the occupant and the dashboard? Or detriment defined by the energy that turned metal into scrap? Even the now moving car dissipates energy 'detrimentally' as tire rubber is ripped off by the pavement. "Where" does energy dissipate? robby says he is concerned with physics inside the passenger compartment. But then subjectively argues about energy that converts metal to scrap. Why is he concerned about energy dissipated outside when his ‘detriment’ is should be concerned with how hard the head hits a dashboard?

Why do politicians so easily manipulate the naive? The naive forget to define a problem before answering the question - as we have here. Or have brain pain rather than learn how to do the work. Always define a problem long before trying to answer it. Always. Otherwise scammers and politicians seek you out.

For most, 'detrimental' is the energy expended destructively into both vehicles. That includes any brake wear used to stop a once parked vehicle. When is that energy highest – most detrimental? When one vehicle at 100 km/hr strikes a parked vehicle.

newme
03-02-2012, 01:25 PM
Perhaps the OP santorum can drop back in and clarify what he means by detriment.

Baracus
03-02-2012, 01:28 PM
It doesn't matter what the OP means by detriment because both collisions are the same for all intents and purposes and hence will have the same "detriment" however you want to define it.

Earl Snake-Hips Tucker
03-02-2012, 01:30 PM
I took it to mean:

If two 50 mph cars collide in a head-on, there would theoretically be somewhat equivalent damage to the two cars.

If a 100 mph car its a sitting duck, would the damage be somewhat equivalent, or would one of them be a lot more damaged (and why?).

ZenBeam
03-02-2012, 01:34 PM
For most, 'detrimental' is the energy expended destructively into both vehicles. That includes any brake wear used to stop a once parked vehicle. So you agree with newme's assessment of your position?
By this definition, a car slamming into a brick wall at 100kmh undergoes the same detriment as a similar car initially moving at 100kmh coasting to stop without hitting anythingI don't think most people would agree with that definition of "detriment ".

westom
03-02-2012, 01:45 PM
By this definition, a car slamming into a brick wall at 100kmh undergoes the same detriment as a similar car initially moving at 100kmh coasting to stop without hitting anything. You must be Captain Kirk. When you don't like the problem, then you change the conditions. It got him a promotion.

Meanwhile, the OP's problem defines a collision.

Chessic Sense
03-02-2012, 02:00 PM
In the case of your couch, there is an external force that acts on your couch to bring it to a stop.

During our hypothetical car collision(s), there is no external force acting on the cars during the collisions.

Sure there is. It's in the car components that break and snap and bend. Those are "external" in the sense that they're not motion-in-motion-out. I realize you might reject this notion at first by saying that the car parts are in the car and thus part of the system, but that's not true. If you ignore those breakable components of the car, you'll get your 50 km/hr output, but the cars won't be the total wrecks that we're talking about.

Exactly. And for those arguing that the cases are not equivalent, realize that this is not some esoteric, obscure physics problem.

It's discussed in detail in every introductory physics class, and in every introductory physics textbook.

In the intro textbooks, they assume that the cows are spherical, that cars don't break, that the ice is frictionless, that the rubber band is perfectly elastic, etc. And that's all well and good, but the topic of the thread requires that we talk about breakable cars and cow-shaped cows.

If the cars don't deform, then you get the typical "this skater grabs that skater at rest, how fast do they move?" problem. But these cars do deform, and that takes energy/momentum out of the system.

And lest you confuse me with other nay-sayers in this thread, remember that I'm agreeing that the situations are equivalent. I'm just saying they're not coming out at the typical ".5mv2 + .5mv2 = 2mv" result you'd get from the textbook.

robby
03-02-2012, 02:13 PM
In the case of your couch, there is an external force that acts on your couch to bring it to a stop.

During our hypothetical car collision(s), there is no external force acting on the cars during the collisions.

Sure there is. It's in the car components that break and snap and bend. Those are "external" in the sense that they're not motion-in-motion-out. I realize you might reject this notion at first by saying that the car parts are in the car and thus part of the system, but that's not true. If you ignore those breakable components of the car, you'll get your 50 km/hr output, but the cars won't be the total wrecks that we're talking about.Chessic Sense, I'm sorry, but you're just plain wrong here. None of the forces you are discussing would be considered to be "external" forces. They would be considered to be "internal" to our system under consideration.

And when you are discussing conservation of momentum, you don't need to worry about internal forces, even the ones that that bend car parts.

We can safely assume that in the situation under discussion here, there are no external forces acting on the cars during the collision.

Therefore the linear momentum of the two cars are conserved, and in the case where a car traveling at 100 km/hr strikes a parked car, the two cars (assuming they stick together) will move off at an initial speed after the collision of 50 km/hr.

Indeed, assuming the cars stick together, it doesn't matter whether or not the cars are unharmed or are a mangled wreck. Conservation of momentum holds in either case.

robby
03-02-2012, 02:17 PM
...But these cars do deform, and that takes energy/momentum out of the system.I missed this statement, which represents a fundamental misunderstanding.

Deformation does result in a dissipation of kinetic energy, which is why kinetic energy is not conserved in an inelastic collision.

However, deformation does NOT result in taking "momentum out of the system."

Again, if there is no external force acting on our cars, momentum is ALWAYS conserved.

Alka Seltzer
03-02-2012, 02:19 PM
That's because they're using a barrier instead of a car, not because of the vector addition.

I think you're right about that, as doing the KE calculation on 64kph and 55 kph the numbers don't add up. A car travelling at 64kph does not have twice the KE of one travelling at 55 kph. Looking at the pictures, the barrier has a thin deformable layer on the front and a solid structure behind it.

Chronos
03-02-2012, 02:21 PM
Chessic Sense, I can't make sense of that post. I see no conceivable way that the components of the cars breaking and snapping can be considered "external": They're pretty much the textbook example of internal forces. Remember, energy can "hide", being converted to hard-to-track forms like heat, but momentum can't hide. You can't take momentum out of the cars without putting it into something that isn't the cars, and that something will move (or at least, change its motion) as a result.

And the equation you have in your last line, ".5mv2 + .5mv2 = 2mv" isn't even dimensionally consistent, and thus can't make sense.

jeffgorc
03-02-2012, 02:26 PM
I am convinced that for the most part there is no difference between 1 car going 100 into a stationary car and two going 50 head on into each other. Just fix your frame of reference to the front bumper of one of the cars and the colision looks the same.

The only difference is the rotation of the wheels. In the case of the one car going 100, its wheels are spinning twice as fast as the wheels in each car in the other situation. Since the rotational kinetic energy is proportional to the square of the angular velocity, the total rotational kinetic energy of the two cars is twice as much in the case of one car going 100. This shouldn't matter a whole lot, but I believe it does introduce greater energy regardless of your frame of reference. The wheels will still want to continue to spin after the collision which will cause the car to get a little more push from the ground. I believe that this total push is double in the first case.

This change would be very minor, but as a thought experiment, think of lightweight cars with big, heavy tires and it makes a bigger difference.

Chessic Sense
03-02-2012, 02:27 PM
Chessic Sense, I'm sorry, but you're just plain wrong here.

Oh! Why didn't you just say so in the first place?

None of the forces you are discussing would be considered to be "external" forces. They would be considered to be "internal" to our system under consideration.

Would be. Are. I'm not sure why you think the friction between my couch and carpet is external but the stress on the metal isn't.

We can safely assume that in the situation under discussion here, there are no external forces acting on the cars during the collision.

If you could, then it wouldn't matter if the car was made of metal, cheese, gold, or unobtanium. But it does, because that's the crux of the problem.

Indeed, assuming the cars stick together, it doesn't matter whether or not the cars are unharmed or are a mangled wreck. Conservation of momentum holds in either case.

You've got to be kidding me. You honestly think it doesn't matter how the car bends and stresses? You think all that distortion comes for free? Have you ever bent a metal rod back and forth and felt it get hot? Why do you think that is? You don't get to bend metal for free.

Conservation of momentum holds when you have a closed system, i.e. where the only force is that Newton's-Third force that causes one object to slow down and the other to speed up.

robby
03-02-2012, 02:31 PM
So you are talking about oranges while others discuss pears. If detriment is what you have defined, then every number you have posted is useless. We need mass, velocity, etc for the occupants. We need to know the elasticity of their collision. We need to know if the occupants are restrained.You don't need to know any of these things if you assume that both collisions are perfectly inelastic and that the cars, occupants, restraints, etc. are identical in both cases.

...Why do politicians so easily manipulate the naive? The naive forget to define a problem before answering the question - as we have here. Or have brain pain rather than learn how to do the work. Always define a problem long before trying to answer it. Always. Otherwise scammers and politicians seek you out.:rolleyes: It's also possible to needlessly obfuscate a simple physics problem by trying to bring in irrelevant factors and considerations.

Chessic Sense
03-02-2012, 02:32 PM
And the equation you have in your last line, ".5mv2 + .5mv2 = 2mv" isn't even dimensionally consistent, and thus can't make sense.

You're right. That was supposed to be m1u1+m2u2=(m+m)v.

robby
03-02-2012, 02:41 PM
...I'm not sure why you think the friction between my couch and carpet is external but the stress on the metal isn't.Because in the case of your couch and carpet, the carpet (being connected to your house, which is in turn connected to the ground) is exerting a force on your couch.

However, there is no external force acting to bend the metal of our two colliding cars. All of the forces are internal.

...If you could, then it wouldn't matter if the car was made of metal, cheese, gold, or unobtanium.You're right. In considering what the velocity is of our two cars after colliding, it doesn't matter what they are made of (so long as they stick together, which makes our collision perfectly inelastic).

...You've got to be kidding me. You honestly think it doesn't matter how the car bends and stresses? You think all that distortion comes for free? Have you ever bent a metal rod back and forth and felt it get hot? Why do you think that is? You don't get to bend metal for free.It comes at the expense of energy, not momentum. (Specifically, it comes from the initial kinetic energy of the colliding cars, which is not conserved.) Again, momentum is always conserved.

...Conservation of momentum holds when you have a closed system...Indeed, that's pretty much the system we are describing during our car collisions. What external force do you think is acting on our two car system?

Baracus
03-02-2012, 02:45 PM
You've got to be kidding me. You honestly think it doesn't matter how the car bends and stresses? You think all that distortion comes for free? Have you ever bent a metal rod back and forth and felt it get hot? Why do you think that is? You don't get to bend metal for free.

Conservation of momentum holds when you have a closed system, i.e. where the only force is that Newton's-Third force that causes one object to slow down and the other to speed up.
robby is right. Kinetic energy is not conserved and is what is "lost" when the metal is bent. The momentum, however, is conserved.

Chessic Sense
03-02-2012, 02:47 PM
if you assume that both collisions are perfectly inelastic

Tell you what, let's just cut to the chase and you tell me why you get to make that assumption. We all agree what would happen if this were the case, but I'm unconvinced why this would be the case. I don't understand why the Coefficient of Restitution would be assumed to be zero.

Irishman
03-02-2012, 03:01 PM
The answer is 2.

2 cars colliding head-on at 50 km/h ≠ a 100 km/h collision, it equals 2 50 km/h collisions, as in each car is experiencing a 50 km/h collision. It doesn't matter that they are colliding with an object that happens to be moving toward it at 50 km/h, it's still primarily the same as hitting a stationary object (there's some variation as a parked car is not a brick wall but the basic physics are the same), because the moving cars are applying the same amount of force to each other. Thus, the 100 km/h collision will be more severe.

This is incorrect. From the theoretical textbook example, the cases are identical. The energy dissipated in the collision is the same. The momentum is the same. Ergo, identical.

The thing is, each car is experiencing a 50 km/h collision, but two cars are involved. In the 2 cars, one of which is at 100 km/h and one is at 0 km/hr, the result has two cars involved as well. Each car absorbs some of the damage from the first moving car.

For a non-textbook case, you have to consider whether the "parked" car is in neutral or with parking brake on or with drive train engaged, and how all that balances out resistance-wise.

I'm pretty sure they did this on Mythbusters and #2 would cause more damage

Yes, they did do this on Mythbusters. Jamie had previously incorrectly stated that two vehicles going 50 would be the same as one going 100 into a brick wall. They tested it by having one car crash into a wall at 50, an identical one into a wall at 100 and two more identical cars hitting head on both going 50. They took numbers, but the visual was enough to show that the head-on collision looked a lot like the first example and not at all like the latter. The math behind why was well explained upthread.

A solid wall is different than a parked car. The solid wall acted as a giant immovable object that was essentially undamaged afterwards. All the energy was dissipated through damage to the single car. Two cars impacting each other had two cars to absorb the same energy. The single car 50 mph into wall was very similar to 2 cars 50 mph each head to head, because the second car in the reverse direction was essentially the same as a brick wall. But if the car going 100 mph hit a car that is sitting at zero and can (a) collapse/get damaged, and (b) move backwards -- two things that the wall did not do -- then the situation for the single 100 mph car will look like an impact of a single car at 50 mph hitting a wall. Each car sees the impact as if it is going 50 mph and hitting an immovable object.

Energy of a 100 km/hr car completely dissipates in the scrap metal once called two cars. *All* kinetic energy dissipates in both cars - that are now at 0 km/hr.

This is wrong. You are assuming that the two cars are at rest immediately after the collision. This is false. The second car will start moving due to the impact from the first car. The textbook example shows that the momentum of state prior to crash is 1 car at 100 km/hr and one car at 0 km/h, the momentum of the state after crash is 2 cars at 50 km/hr. The still car does not have any way to resist moving, it is not bolted to the ground, tied down, concreted in place. It is sitting there, and can roll or slide. It's resistance to motion is either internal friction of engine/brakes, or external friction of tires to road.

Your case is the equivalent of hitting a car at rest that is parked against a giant wall. 2 cars absorb the energy, but all of the original KE is dissipated through the collision, rather than only half of the KE being dissipated.

BTW, your NASCAR example is bogus. Speed of both cars after the 0.05 mph collision is not defined. So no answer is possible. Obviously, without mass, another reason why no answer is possible. Please present your answer with numbers. Subjective reasoning is also called junk science.

No, it was a simple conceptual demo. Two cars lightly bumping are two cars lightly buming, whether one car is at rest or both cars are moving at near light speed. The delta V is the significant contributor if the masses are the same.

Obviously initial energy matters. It is called conservation of energy. Why is that so difficult? All energy in once moving cars is now dissipated in what is called scrap metal.

No, some of the original KE from one car is now KE in two cars moving away from the crash. The second car that was initially parked is now moving. That KE came from somewhere.

Nars Glinley
03-02-2012, 03:01 PM
In 1, why don't the energies cancel? Yes, I'm serious. I know that they don't, I just don't know why.

Irishman
03-02-2012, 03:06 PM
Tell you what, let's just cut to the chase and you tell me why you get to make that assumption. We all agree what would happen if this were the case, but I'm unconvinced why this would be the case. I don't understand why the Coefficient of Restitution would be assumed to be zero.

There will be some non-zero coefficient of restitution, in that the bent metal and plastic will have a slight give. But you will see that most of the damage to each car remains after the crash. You end up with two crumpled heaps, not two pristine cars. Ergo, the restitution is small, and the collision is effectively inelastic.

How much restitution? Is it 5% is it 0.5% Where is the significant cut off for the purposes of the conversation between assuming perfectly inelastic and saying that assumption isn't valid? From an engineer's perspective, it's close enough that the car is totaled. Who cares if the metal unflexed a half a percent? You still have to tow that mess away.

robby
03-02-2012, 03:07 PM
Tell you what, let's just cut to the chase and you tell me why you get to make that assumption. We all agree what would happen if this were the case, but I'm unconvinced why this would be the case. I don't understand why the Coefficient of Restitution would be assumed to be zero.First off, realize that momentum is always conserved, whether or not the collision is elastic or inelastic.

We know that the collision is not elastic, because that is an unrealistic ideal condition in which kinetic energy is conserved, like when perfectly rigid billiard balls collide.

Therefore the collision will either be inelastic or perfectly inelastic. A perfectly inelastic collision is one in which the two colliding objects stick togeter. It is also a simpler case to consider, and is not unrealistic. I'm sure you can picture two cars getting mangled together and sticking together after a collision.

Remember, the original point of the OP was to compare two cases. Considering an inelastic (but not perfectly inelastic) just makes the problem needlessly complex, like bringing air resistance into introductory parabolic trajectory problems.

Irishman
03-02-2012, 03:11 PM
Nars, in the first case, each car has it's own equivalent share of momentum and energy. Each car effectively dissipates it's own energy, and the momentum cancels. Momentum is a vector, direction matters, but kinetic energy is a scalar, direction is irrelevant.

In the second case, one car starts with all the KE and momentum, then both cars end with some KE and momentum. The momentum gained by the stationary car is momentum removed from the initially moving car. Thus the initially moving car sees exactly the same situation - a delta 50 km/h velocity change and and a momentum transfer of 50*m. The initially stationary car also sees a 50km/h velocity change and a momentum transfer of 50*m. It's just in the reverse direction.

robby
03-02-2012, 03:15 PM
In 1, why don't the energies cancel? Yes, I'm serious. I know that they don't, I just don't know why.Because energy (unlike momentum) is a scalar quantity. Therefore there is no such thing as negative energy (in the non-quantum macroscopic world), and therefore there is no such thing as energy "cancelling."

Also, in inelastic collisions, note that kinetic energy is NOT conserved.

Chessic Sense
03-02-2012, 03:18 PM
How much restitution? Is it 5% is it 0.5% Where is the significant cut off for the purposes of the conversation between assuming perfectly inelastic and saying that assumption isn't valid?
Remember, the original point of the OP was to compare two cases. Considering an inelastic (but not perfectly inelastic) just makes the problem needlessly complex, like bringing air resistance into introductory parabolic trajectory problems.

OK, well it's quittin' time at work, so let me just register my disagreement with "significant" and "needlessly". Seeing as how we agreed on page one about the OP's question and we agree on the physics anyhow, I see no reason to continue quibbling about whether we should un-assume the spherical cow.

OP, the collisions are equivalent. Peace out. ;)

Quercus
03-02-2012, 03:37 PM
How about this: There are two answer to the question. (I haven't taught physics, but I do have a degree in it, and actually do some physics for my work, so I think I'm a 'physicist').

In the big picture, robby is absolutely right. If the collision took place on a perfectly smooth and frictionless sheet of ice or something, the two collisions would be identical, except in the 100/0 case the two cars would be sliding along at 50mph, while in the 50/50 case, they'd be (more or less) at rest. In fact (again if the ice is physics-problem-land perfectly featureless) you couldn't tell whether you saw a 50/50 crash from a stationary position, or a 100/0 crash from a separate third vehicle coasting along at 50mph -- in each case you'd see two cars approaching one from each side at 50 mph relative to you, an earth-shattering kaboom, and then a mangled two-car mashup not moving relative to you.

Now, in the real-world, there's going to be some difference. Depending on how the stationary car is pointed, what kind of surface it's sitting on, whether it has brakes set, how soon the crash locks up the wheels or pushes parts of the car into the ground, there's going to be some kind of friction. So in the 100/0 crash the two cars will be sliding away at something less than 50mph, and they won't be sliding harmlessly forever either. Some of that frictional force could cause more damage, even if a lot of it will dissipate more or less harmlessly as heat. So, yes I think the 100/0 crash will probably -- though it all depends on the details -- do some amount more damage than the 50/50 crash. But, in the big picture, not much more. I'm sure a 75/0 crash will do less overall damage than a 50/50, for instance

(Of course we're assuming that the two cars sliding/rolling away in the aftermath of the 100/0 crash won't hit anything else, which could certainly cause more damage. But that is I think not the point of the OP).

RedSwinglineOne
03-02-2012, 03:43 PM
Suppose we have two cars of equal mass:

1. A and B collide into eachother both going 50 km/h in opposite directions
2. A is going 100 km/h and collides with B is at rest



It's been said before, but its worth repeating. Except for the frame of reference, these are the same. The two vehicles collide with a closing speed of 100km/h.
If anybody cares to disagree, I would be interested in why you think so.

BarryB
03-02-2012, 04:51 PM
It's been said before, but its worth repeating. Except for the frame of reference, these are the same. The two vehicles collide with a closing speed of 100km/h.
If anybody cares to disagree, I would be interested in why you think so.

The wheels. A stationary car's wheels aren't spinning, whereas a moving cars is.

RedSwinglineOne
03-02-2012, 05:02 PM
The wheels. A stationary car's wheels aren't spinning, whereas a moving cars is.

I hope we can agree that as the thought experiment this is, the effects of spinning wheels are negligible.

westom
03-02-2012, 07:13 PM
This is incorrect. From the theoretical textbook example, the cases are identical. The energy dissipated in the collision is the same. The momentum is the same. Ergo, identical. Scary that so many do not even know simple high school physics. Doubling velocity (speed) does not double energy. It quadruples energy. Everyone here should know that. Otherwise scammers and politicians make money and get elected because the public is uneducated and naive. These fundamental facts are taught to everyone in high school to make dishonesty and scams difficult.

Energy of one car at 100 km/hr is equal to the energy of two cars at 70.7 km/hr. Basic physics. Obviously. Posting again is what everyone should have known when the OP first asked his question:
Energy = one half mass times velocity squared.

Irishman - if you know that is wrong, then you have posted numbers and equations that says so. Subjective claim or denial suggests outright lying or lack of basic high school education. Honest answers always include the numbers. So, where are your numbers? The always required numbers when you insist someone is wrong.

Explain how two cars at 50 km/hr is energy equal to one car at 100 km/hr? Explain means posting the numbers.

Energy in a 100 Km/hr car is one half times its mass times 100 squared. Energy in each 50 km/hr car is one half times the same mass times 50 squared. Please show everyone how that energy is equal. You claimed it is so – but only subjectively. Prove it. A point I make often because scammers and politicians can only exists when numbers are replaced by subjective denials (also called hearsay or junk science).

Why must every adult know this? Because a car doing 100 km/hr (or 70 mph) requires twice as much braking to stop. As compared to a car only doing 70 .7 km/hr (50 mph). But then every high school graduate is required to know this. Otherwise jobs must go overseas to those who did learn what is critically important – the numbers. Show me your numbers or withdrawal subjective (empty) denials.

westom
03-02-2012, 07:29 PM
You don't need to know any of these things if you assume that both collisions are perfectly inelastic and that the cars, occupants, restraints, etc. are identical in both cases.
Your conservation of momentum exists only when a collision is elastic. Car crashes are not elastic. M1 * V1 = M2 * V2 only when kinetic energy before the collision is exactly same as energy after the collision. Obviously that does not exist for two crashing autos. Crash zones are about converting that kinetic energy into scrap metal. To not conserve momentum. To make the collision as inelastic as possible. So that momentum is not conserved. And so that less energy appears inside the passenger compartment.

Again, I make the point because it is why we need all citizens educated. To not know this stuff subverts the economy by empowering scammers and corrupt politicians.

Energy equivalent of one car at 100 km/hr is two cars at 70.7 km/hr. Again, the basic energy equation that must be in you every conclusion:
Energy = one half times mass times velocity squared.
Conservation of energy (not momentum) applies.

Energy is always conserved. Two cars at 50 km/hr or one car at 100 km/hr. When the collision is finished, that energy defined by velocity is somewhere else. Both cars are at 0 km/hr. Where did that energy go? Exactly why crumple zones are created.

No way around what is taught to everyone in high school physics. And fundamental to answering the OP's question IF that is what the OP meant by a vague term: "detrimental".

Chronos
03-02-2012, 07:34 PM
Quoth Chessic Sense:

Tell you what, let's just cut to the chase and you tell me why you get to make that assumption [a perfectly elastic collision].If you'd prefer, we could assume some other degree of elasticity for the collisions, or leave it as a free parameter. Assuming that it's perfectly inelastic is just a simplification to make the calculations easier, but we could use any other elasticity at all. And as long as the elasticity is the same for both cases (no reason it shouldn't be), the collisions will be equivalent.

Quoth westom:

Explain how two cars at 50 km/hr is energy equal to one car at 100 km/hr? Explain means posting the numbers.Let me make it a bit more general, even: A head-on collision of cars of mass M each going at speed v, vs. a car of mass M going at 2*v hitting a car at rest.

Case 1: Each car has an initial kinetic energy of 0.5*M*v^2. Both cars end up at rest, for 0 final kinetic energy. All of the kinetic energy is therefore absorbed and converted to heat by the crumple zones, and the crumple zone of each car has absorbed an energy of 0.5*M*v^2.

Case 2: The moving car has an initial kinetic energy of 0.5*M*(2*v)^2, or 2*M*v^2, while the initially-stationary car has no kinetic energy initially. After the the collision, from conservation of momentum, we find that both cars are moving at speed v, half the speed of the initially-moving car. The final kinetic energy of the two cars is therefore 0.5*(2*M)*v^2, or M*v^2. Thus, the kinetic energy has decreased by M*v^2, which means that much energy must have been absorbed by the crumple zones. This energy is divided between the two cars, which means that each car's crumple zone absorbed an energy of 0.5*M*v^2.

Note that in both cases, each car absorbs an energy of 0.5*M*v^2, so the damage is equivalent in both cases.

Telemark
03-02-2012, 07:34 PM
It quadruples energy. Everyone here should know that.
Since everyone does know that this isn't really helping. What you are missing is that after the collision both cars are moving at 50 km/hr. That's where the extra energy is that you keep ignoring. At the end of the collision with the stationary car all that extra energy is now moving the crumpled cars.

Chronos
03-02-2012, 07:35 PM
Your conservation of momentum exists only when a collision is elastic.This is not correct. Momentum is conserved always, in every form of interaction known to science, from subatomic particles to billion-solar-mass black holes. There is never any situation in which momentum is not conserved.

Terr
03-02-2012, 07:46 PM
Scary that so many do not even know simple high school physics. What I find scary is that you don't know simple high school physics and yet accuse others of it.

People explained to you, multiple times, some with words, some with equations, that what's important is not the initial energy, but the difference between initial energy (pre-collision) and final energy (after collision). And that difference is the same in both cases.
Why must every adult know this? Because a car doing 100 km/hr (or 70 mph) requires twice as much braking to stop. As compared to a car only doing 70 .7 km/hr (50 mph). True. But it has nothing to do with the question in the OP.

Snnipe 70E
03-02-2012, 07:58 PM
WTF? How do you figure that? The only difference is how fast the scenery is moving relative to the cars. Two 50 km/h crashes certainly does sum to 100 km/h. Are you suggesting that in the 100/0 scenario, the stopped car is all hunky-dory because it experiences a "0 km/h" collision?

Imagine a car hitting you at 50 km/h. Now imagine if you decided to gun the accelerator before you were struck. Are you saying there's no difference?!



How does moving "backwards"* change anything? It's still absorbing energy. That's why it's moving backwards. If you're moving at 50 km/h and all of a sudden, you're moving at 0 km/h, guess what...you've been moved "backwards".

*I put it in quotes because the idea that the scenery is a relevant reference point is silly.

It means that the crashing car speed is not 0. Energy is being disipated by the cars rolling and slowing. In the head to head crash all energy is disipated by the two cars chrunching.

westom
03-02-2012, 07:59 PM
What you are missing is that after the collision both cars are moving at 50 km/hr. . So all collisions are elastic. So every car crash is still moving down the road? Of course not. Each crash is two cars not moving. What happened to that energy? Crumple zones. No moving vehicles. And more energy dissipated as tire skid marks.

The quadrupled energy means four times more crumpled metal, tire wear, burned breakes, etc.

The numbers that so many others have also posted. You did read every post before replying - as expected.

A car at 100 km/hr is one hald mass times velocity squared. 100 sqaured times one half means energy is 5000 times mass.

A parked car is one half times mass times zero squared or zero energy.

A car at 50 km/hr is 50 square times one half or 1250 times mass.

The energy of a 100 km/hr car into a parked car is 5000 times mass plus 0 times mass. Or 5000 times mass.

Then eerngy of two 50 km/hr cars is 1250 time mass plus 1250 times mass. Or 2500 times mass. Again, this was posted by others multiple times.

Show me how those numbers are wrong. Your second equation of
0.5*(2*M)*v^2, or M*v^2. for v=50 is 2500 times mass.

Your first equation of 0.5*M*v^2 for v=100 is 5000 times mass.

So, if both cars absorb energy equally, then the 50 km/hr example means each car absorbs energy at 1250 time mass. A 100 km/hr car crashing into a park car means both cars absorb 2500 times mass. How does 1250 equal 2500.

BTW, this is the classic question used in SATs to separate those who learned their high school physics from those who foolishly assume the collisions are equal energy.

Even your own posted equations agree with me.

robby
03-02-2012, 08:38 PM
So all collisions are elastic. So every car crash is still moving down the road? Of course not. Each crash is two cars not moving. What happened to that energy? Crumple zones. No moving vehicles. And more energy dissipated as tire skid marks.

The quadrupled energy means four times more crumpled metal, tire wear, burned breakes, etc.First off, all collisions are not elastic.

However, what I think you are missing is that you are looking at the second case after the mangled wreck of two cars has slid to a stop. You need to look at the situation immediately after the collision, in which the car traveling at 100 km/hr has struck the parked car. By conservation of momentum, the two cars after the collision will have a resultant velocity of 50 km/hr immediately after the collision. They will then slide to a halt. During the slide, the KE of the two mangled cars will be dissipated as frictional heat, but this does not necessarily cause harm to the occupants.

If it helps, consider the ideal case where the collision takes place on a frictionless surface (idealized as a sheet of ice). There is no question here that the two cars after the collision will move off with a speed of 50 km/hr, which will then NOT dissipate (because of the frictionless surface). Yet there was still a horrific collision that caused damage when the vehicle traveling 100 km/hr struck the motionless vehicle.

A car at 100 km/hr is one hald mass times velocity squared. 100 sqaured times one half means energy is 5000 times mass.

A parked car is one half times mass times zero squared or zero energy.

A car at 50 km/hr is 50 square times one half or 1250 times mass.

The energy of a 100 km/hr car into a parked car is 5000 times mass plus 0 times mass. Or 5000 times mass.

Then eerngy of two 50 km/hr cars is 1250 time mass plus 1250 times mass. Or 2500 times mass. Again, this was posted by others multiple times.

Show me how those numbers are wrong. Your second equation of
0.5*(2*M)*v^2, or M*v^2. for v=50 is 2500 times mass.

Your first equation of 0.5*M*v^2 for v=100 is 5000 times mass.

So, if both cars absorb energy equally, then the 50 km/hr example means each car absorbs energy at 1250 time mass. A 100 km/hr car crashing into a park car means both cars absorb 2500 times mass. How does 1250 equal 2500.Because, as I stated back in Post #24, in the first case, the initial total KE of the two vehicles is 2500m. The final KE is zero, so the KE dissipated in the actual collision is 2500m.

In the second case, the initial KE of car A is 5000m, but the KE of the two mangled vehicles immediately after the collision is 2500m. (The rest of the KE will then be dissipated as friction as the two mangled vehicles slide to a stop). So the KE dissipated in the collision will again be 2500m.

robby
03-02-2012, 08:58 PM
Scary that so many do not even know simple high school physics.

...Otherwise scammers and politicians make money and get elected because the public is uneducated and naive. These fundamental facts are taught to everyone in high school to make dishonesty and scams difficult.

...Subjective claim or denial suggests outright lying or lack of basic high school education.

:rolleyes:

From the New York Times regarding Robert Goddard (http://en.wikipedia.org/wiki/Robert_H._Goddard) and the impossibility of a rocket functioning in a vacuum:

[A]fter the rocket quits our air and really starts on its longer journey it will neither be accelerated nor maintained by the explosion of the charges it then might have left. To claim that it would be is to deny a fundamental law of dynamics...

That Professor Goddard with his "chair" in Clark College and the countenancing of the Smithsonian Institution, does not know the relation of action and reaction, and of the need to have something better than a vacuum against which to react—to say that would be absurd. Of course he only seems to lack the knowledge ladled out daily in high schools.

Needless to say, like westom here, the Times was dead wrong, too. The Times, at least, published a retraction 49 years later following the launch of Apollo 11.

...scammers and politicians can only exists when numbers are replaced by subjective denials (also called hearsay or junk science).

...But then every high school graduate is required to know this. Otherwise jobs must go overseas to those who did learn what is critically important – the numbers. Show me your numbers or withdrawal subjective (empty) denials.I've repeatedly given both explanations as well as numbers in my posts.

westom
03-02-2012, 09:00 PM
People explained to you, multiple times, some with words, some with equations, that what's important is not the initial energy, but the difference between initial energy (pre-collision) and final energy (after collision).
So two cars at 50 Km/hr crash. And keep moving at 50 km/hr after the collision? Nonsense. Must be true for momentum to be conserved. If the collision were elastic (ie between pool table balls). But a collision between cars is inelastic. When done, both cars are at 0 km/hr. Momentum is not conserved. Energy is. When done, all cars are at 0 km/hr.

Where did energy go? Into crumple zones, burned tires, brakes, etc. Energy defined by 50 km/hr is now found in scrap metal. An inelastic crash.

OP's question defines inelastic collisions. When done, a 100 km/hr car, the parked car, and the 50 km/hr cars are all doing zero km/hr. No car is still moving at 50 km/hr.

Energy of one car at 100 km/hr is equal to two cars at 70.7 km/hr. Not two cars at 50 km/hr as so many try to claim subjectively. Subjective claims imply lying - as scammers and politicians love to do. Make bogus claims without numbers. Numbers - to be energy equal, both cars must be at 70.7; not at 50.

BTW, this has long been a typical question on SATs to separate those who think subjectively from those who learned their physics and do the numbers.

Show me one post (with equations and numbers) that dispute what I have posted. Please do not mock everyone by citing a post number. Quote the exact numbers and put them into the relevant equation. Otheriwise it is just anothger useless subjective post.

For example, Telemark’s equations and numbers agreed with me. Energy of 1250 times mass (for each car) is not same as 2500 times mass. Those numbers from his own equations. Or do you know something different? Good. Then your next post shows those equations with the appropriate numbers inserted.

If someone posted numbers that say otherwise., then you quote each number in your post. Honesty demands that. Do not just cite a post number. If you have knowledge, your post puts each number into the appropriate equation. The equation is this simple:
Energy = 0.5 times mass timed velocity squared.

Insert the numbers. Show me how total energy of 2500 times mass is the same as 5000 times mass. Telemark made that claim even though numbers in his own equations contradicted him.

Snnipe 70E - the crash is done when all cars are at 0 km/hr. But again, confusion created by the OP's vague question: define "detrimental". One car at 100 km/hr or two cars at 50 km/hr. The crash is done when all cars are at 0 km/hr.

robby
03-02-2012, 09:01 PM
...BTW, this is the classic question used in SATs to separate those who learned their high school physics from those who foolishly assume the collisions are equal energy.So you evidently missed that question. :rolleyes:

JimboJamesJimmersonIII,Esq.
03-02-2012, 09:02 PM
This thread went from frustrating to comical, and now it has entered the pathetic stage. It is certainly a testament to the motto of the board - "it's taking longer than we thought."

I applaud anyone who has the patience to continue. Hopefully those individuals who were confused have gained some insight from multiple knowledgeable posters.

To the OP, this is not a difficult problem. Please disregard the ignorance of the vocal minority. I hope it is apparent who that is.

Terr
03-02-2012, 09:08 PM
So two cars at 50 Km/hr crash. And keep moving at 50 km/hr after the collision? No. They *start out* moving at 50 km/hr after the collision. Then, due to friction, they slow down to 0.
But a collision between cars is inelastic. When done, both cars are at 0 km/hr. Momentum is not conserved. Energy is. When done, all cars are at 0 km/hr.Momentum is, in fact, conserved. After the collision and after the cars slowed down, the earth is spinning just a little bit faster.
Where did energy go? Into crumple zones, burned tires, brakes, etc. Energy defined by 50 km/hr is now found in scrap metal. An inelastic crash.Why burned tires? Let's say the tires are not locked - they will just keep rolling (unless crimpled metal obstructs them) Same with brakes.
OP's question defines inelastic collisions. When done, a 100 km/hr car, the parked car, and the 50 km/hr cars are all doing zero km/hr. No car is still moving at 50 km/hr.A car starts at 100 miles an hour. Then you put the car into neutral. It keeps rolling, and eventually stops. No crumple zones. No burned tires. No burned brakes. Is this equivalent to a collision with a brick wall at 100 mph? After all, according to your logic, the same energy is involved in both.
BTW, this has long been a typical question on SATs to separate those who think subjectively from those who learned their physics and do the numbers.Wrong.
Show me one post (with equations and numbers) that dispute what I have posted. Please do not mock everyone by citing a post number. Quote the exact numbers and put them into the relevant equation. Otheriwise it is just anothger useless subjective post.Chronos did a pretty good job.

robby
03-02-2012, 09:17 PM
Momentum is not conserved.Momentum is always conserved.

When done, a 100 km/hr car, the parked car, and the 50 km/hr cars are all doing zero km/hr. No car is still moving at 50 km/hr.Wrong. In the second case, immediately after the collision, the two cars must be moving together at 50 km/hr, by conservation of momentum (assuming they stick together). They then slide to a stop, but this takes place AFTER the collision.

Energy of one car at 100 km/hr is equal to two cars at 70.7 km/hr. True, but irrelevant here, because it assumes that both situations will end up with the cars at rest immediately after the collision. They don't.

...Show me one post (with equations and numbers) that dispute what I have posted. Please do not mock everyone by citing a post number. Quote the exact numbers and put them into the relevant equation. Otheriwise it is just anothger useless subjective post.I and others have posted (with equations and numbers) the correct answer repeatedly. I've only been referencing post numbers because it's getting exhausting repeating myself.

westom
03-02-2012, 09:48 PM
However, what I think you are missing is that you are looking at the second case after the mangled wreck of two cars has slid to a stop. . Situation after a collision is all cars stopped. You are now changing the definition of 'detrimental' to justify your conclusion.

What happens as each slides to a halt? "Detrimental" continues. "Detrimental" does not end until all cars have halted by scrapping pavement, tearing tires, burning brakes, rolling and killing passengers, etc. That is when the crash is done as defined by the word "detrimental".

If the cars continue rolling 3 miles down the road and into a bridge, that is still part of 'detrimental' - the event called a crash.

You also posted virtually no numbers. This equation you post with your numbers in it:.
Energy = .5 times mass times velocity squared.

Show me how one car at 100 km/hr is energy equivalent to two cars at 50 km/hr? Not a rhetorical question. One that demands your post include (if necessary) numbers you claim to have already posted.

Again the not so subtle hint. Two cars must be at 70.7 km/hr to be energy equivalent. To create a collision with energy at 5000 times mass.

It’s easy. 0.5 times 70.7 squared times mass is what? 2500 mass. One for each car. Total of 5000 times mass. Don’t take my word for it. Honesty also demands you confirm the arithmetic.

Easy. 0.5 times 100 squared times mass is 5000 times mass.

Both cars must be doing 70.7 km/hr to be energy equivalent to one car at 100 km/hr. Two cars at 50 km/hr is a much less violent crash. That will always confuse those who only think subjectively. Honest means you always do and post the math – so as to not be subjective like scammers and corrupt politicians.

Show me your numbers. Show me where those above numbers are wrong. Please post each numbers with specific corrections. You know those numbers are wrong? Then post the numbers. And then show which number is wrong.

robby
03-02-2012, 09:56 PM
...Show me one post (with equations and numbers) that dispute what I have posted. Please do not mock everyone by citing a post number. Quote the exact numbers and put them into the relevant equation. Otheriwise it is just anothger useless subjective post.In the interest of fighting ignorance, here goes:

Assume that both cars are identical and have a mass of 1000 kg. To simplify matters further, we will assume that the collision takes place on a frictionless surface. (If you find this too abstract, consider the case of train cars on a train track). We will assume that the cars stick together in both collisions. (In the case of train cars, assume that the cars lock together.)

Case 1 (Two cars hit head-on, each traveling at 50 km/hr):

The initial momentum of car A is mAvA = (1000 kg)(50 km/hr) = 50,000 kg km/hr. The initial momentum of car B is mBvB = (1000 kg)(-50 km/hr) = -50,000 kg km/hr. You will notice that the initial total momentum of the combined two-car system being considered is zero, when you add them together.

The initial KE of car A is 0.5mAvA2 = 0.5(1000 kg)(50 km/hr)2 = 1,250,000 kg km2/hr2. The initial KE of car B is 0.5mBvB2 = 0.5(1000 kg)(-50 km/hr)2 = 1,250,000 kg km2/hr2. The total initial KE is 2,500,000 kg km2/hr2.

After the collision, for momentum to be conserved, and because the initial total momentum was zero, both cars must have zero final velocity. In other words, both cars are at rest after the collision.

Because the cars are now at rest, the final kinetic energy of the two cars is also zero. Where did this energy go? It went into heat, noise, and/or the crumpling of metal. Note that kinetic energy was not conserved. Indeed, the change in KE is equal to 2,500,000 kg km2/hr2.

Case 2 (A car traveling at 100 km/hr strikes a car at rest):

The initial momentum of car A is mAvA = (1000 kg)(100 km/hr) = 100,000 kg km/hr. The initial momentum of car B is mBvB = (1000 kg)(0 km/hr) = zero. You will notice that the initial total momentum of the combined two-car system being considered is 100,000 kg km/hr, when you add them together.

The initial KE of car A is 0.5mAvA2 = 0.5(1000 kg)(100 km/hr)2 = 5,000,000 kg km2/hr2. The initial KE of car B is 0.5mBvB2 = 0.5(1000 kg)(0 km/hr)2 = zero. The total initial KE is 5,000,000 kg km2/hr2.

After the collision, for momentum to be conserved, the final momentum after the collision must be 100,000 kg km/hr. It cannot be zero, unless an outside force acts on our cars. With a frictionless surface, no outside force can act on our cars in the direction of motion. So the final momentum is equal to 100,000 kg km/hr = mABvAB = (1000 kg + 1000 kg)vAB. Solving for vAB, gives vAB = 50 km/hr. This is the final velocity of the two cars after the collision. On a frictionless surface, this velocity would remain unchanged. On a road, friction would quickly slow the cars down, but this takes place AFTER THE COLLISION.

So the final kinetic energy of the two cars is 0.5mABvAB2 = 0.5(1000 kg + 1000 kg)(50 km/hr)2 = 1,250,000 kg km2/hr2 = 2,500,000 kg km2/hr2. You will notice that the change in KE is equal to 5,000,000 kg km2/hr2 minus 2,500,000 kg km2/hr2 = 2,500,000 kg km2/hr2. You will also notice that the change in kinetic energy is equivalent to Case 1.

Q.E.D.

robby
03-02-2012, 10:00 PM
...If the cars continue rolling 3 miles down the road and into a bridge, that is still part of 'detrimental' - the event called a crash.No, that is another, subsequent collision. We are only considering the first collision here.

westom
03-02-2012, 10:12 PM
No, that is another, subsequent collision. We are only considering the first collision here. The first collision ends when all cars stop moving. You cannot change the problem to suit your needs.

westom
03-02-2012, 10:15 PM
Momentum is, in fact, conserved. If momentum is conserved, then no energy is lost in crumple zones. You cannot have it both ways.

Momentum is only conserved if the collision is elastic. If crumple zones do not crumple; do not absorb energy. Only then can MV before a collision equal MV after the collision.

We know 1250 times mass (for each 50 km/hr car) is reduced because energy dissipates in crumple zones. Or do you also deny that? For example, '1000 times mass' energy dissipates in the crumple zone. Leaving '250 times mass' to keep moving the car. The equation is simple.

E = .5 times mass times velocity squared.
A remaining 250 times mass = .5 times mass times velocity squared.
What is the new velocity as cars separate? About 11 km/hr.

So momentum is conserved - you say. MV=MV? Mass times 50 km/hr equals mass times 11 km/hr?

Why does the math not work? Because momentum is never conserved when a collision is inelastic. When crumple zones exist. But then I am only repeating - and again with numbers - high school physics.

Energy is always conserved. Or do you also deny that? Not a rhetorical question. You answer it – but you won’t.

Provided are equations. Rather than post subjectively, post your equations with your "corrected" numbers. Because honesty demands your every post do so. Honesty says you explain how a car moving less than 50 km/hr after a crash is conservation of momentum. Explain how conservation of momentum exists when energy dissipates in crumple zones.

BTW, good luck changing the laws of physics.

Tangent
03-02-2012, 10:36 PM
Momentum is only conserved if the collision is elastic. If crumple zones do not crumple; do not absorb energy. Only then can MV before a collision equal MV after the collision.

This is incorrect, as others have tried to explain. Momentum is always conserved, including during inelastic collisions.

Kinetic energy is not conserved during an inelastic collision (because of the crumpling), but MOMENTUM IS STILL CONSERVED.

If you don't believe me, look it up in ANY PHYSICS TEXTBOOK EVER PUBLISHED.

robby
03-02-2012, 10:41 PM
If momentum is conserved, then no energy is lost in crumple zones. You cannot have it both ways.Momentum is not the same thing as energy.

Momentum is always conserved. Energy is always conserved. Kinetic energy is only conserved in ideal, perfectly elastic collisions. In inelastic collisions, kinetic energy is not conserved.

Momentum is only conserved if the collision is elastic. If crumple zones do not crumple; do not absorb energy. Only then can MV before a collision equal MV after the collision.Wrong. Wrong. WRONG. :rolleyes:

Momentum is always conserved. It is conserved in elastic collisions as well as inelastic collisions.

Cite. (http://en.wikipedia.org/wiki/Momentum#Conservation_of_linear_momentum)

The law of conservation of linear momentum is a fundamental law of nature, and it states that if no external force acts on a closed system of objects, the momentum of the closed system remains constant.

We know 1250 times mass (for each 50 km/hr car) is reduced because energy dissipates in crumple zones. Or do you also deny that? For example, '1000 times mass' energy dissipates in the crumple zone. Leaving '250 times mass' to keep moving the car. The equation is simple.

E = .5 times mass times velocity squared.
A remaining 250 times mass = .5 times mass times velocity squared.
What is the new velocity as cars separate? About 11 km/hr.

So momentum is conserved - you say. MV=MV? Mass times 50 km/hr equals mass times 11 km/hr?What's wrong here is not the law of conservation of linear momentum, but your apparent confusion between energy and momentum, your apparent inability to comprehend velocity vectors, your idea that momentum is dissipated in crumple zones, and your erroneous conclusion that the final velocity of the cars is "11 km/hr." In other words, a bunch of gibberish resulted in an erroneous conclusion.

Why does the math not work? Because momentum is never conserved when a collision is inelastic. When crumple zones exist. But then I am only repeating - and again with numbers - high school physics.The math doesn't work because you set up the problem wrong. And if your high school physics teacher read your nonsensical posts, they would either weep or be rolling over in their grave.

Energy is always conserved. Or do you also deny that? Not a rhetorical question. You answer it – but you won’t.Sure I will. I agree that energy is always conserved. Note that kinetic energy is not always conserved, though.

Provided are equations. Rather than post subjectively, post your equations with your "corrected" numbers. Because honesty demands your every post do so. Honesty says you explain how a car moving less than 50 km/hr after a crash is conservation of momentum. Explain how conservation of momentum exists when energy dissipates in crumple zones.I have tried to explain, but I can't do much about willful ignorance.

BTW, good luck changing the laws of physics.Speak for yourself. :rolleyes:

Again, momentum is ALWAYS conserved.

Terr
03-02-2012, 10:43 PM
Situation after a collision is all cars stopped. You are now changing the definition of 'detrimental' to justify your conclusion.

Two cases:

1. A car is moving at 100 mph, in neutral. It slams into a brick wall.
2. A car is driving at 100 mph, in neutral. It slams into an empty cardboard box, then coasts until it stops.

Is (1) more or less "detrimental" than (2)? Note that EXACTLY the same kinetic energy is dissipated in both cases.

Feel free to ignore answering this if you feel uncomfortable doing so.

robby
03-02-2012, 11:12 PM
In the interest of fighting ignorance, here goes:

Assume that both cars are identical and have a mass of 1000 kg. To simplify matters further, we will assume that the collision takes place on a frictionless surface. (If you find this too abstract, consider the case of train cars on a train track). We will assume that the cars stick together in both collisions. (In the case of train cars, assume that the cars lock together.)

Case 1 (Two cars hit head-on, each traveling at 50 km/hr):

The initial momentum of car A is mAvA = (1000 kg)(50 km/hr) = 50,000 kg km/hr. The initial momentum of car B is mBvB = (1000 kg)(-50 km/hr) = -50,000 kg km/hr. You will notice that the initial total momentum of the combined two-car system being considered is zero, when you add them together.

The initial KE of car A is 0.5mAvA2 = 0.5(1000 kg)(50 km/hr)2 = 1,250,000 kg km2/hr2. The initial KE of car B is 0.5mBvB2 = 0.5(1000 kg)(-50 km/hr)2 = 1,250,000 kg km2/hr2. The total initial KE is 2,500,000 kg km2/hr2.

After the collision, for momentum to be conserved, and because the initial total momentum was zero, both cars must have zero final velocity. In other words, both cars are at rest after the collision.

Because the cars are now at rest, the final kinetic energy of the two cars is also zero. Where did this energy go? It went into heat, noise, and/or the crumpling of metal. Note that kinetic energy was not conserved. Indeed, the change in KE is equal to 2,500,000 kg km2/hr2.

Case 2 (A car traveling at 100 km/hr strikes a car at rest):

The initial momentum of car A is mAvA = (1000 kg)(100 km/hr) = 100,000 kg km/hr. The initial momentum of car B is mBvB = (1000 kg)(0 km/hr) = zero. You will notice that the initial total momentum of the combined two-car system being considered is 100,000 kg km/hr, when you add them together.

The initial KE of car A is 0.5mAvA2 = 0.5(1000 kg)(100 km/hr)2 = 5,000,000 kg km2/hr2. The initial KE of car B is 0.5mBvB2 = 0.5(1000 kg)(0 km/hr)2 = zero. The total initial KE is 5,000,000 kg km2/hr2.

After the collision, for momentum to be conserved, the final momentum after the collision must be 100,000 kg km/hr. It cannot be zero, unless an outside force acts on our cars. With a frictionless surface, no outside force can act on our cars in the direction of motion. So the final momentum is equal to 100,000 kg km/hr = mABvAB = (1000 kg + 1000 kg)vAB. Solving for vAB, gives vAB = 50 km/hr. This is the final velocity of the two cars after the collision. On a frictionless surface, this velocity would remain unchanged. On a road, friction would quickly slow the cars down, but this takes place AFTER THE COLLISION.

So the final kinetic energy of the two cars is 0.5mABvAB2 = 0.5(1000 kg + 1000 kg)(50 km/hr)2 = 1,250,000 kg km2/hr2 = 2,500,000 kg km2/hr2. You will notice that the change in KE is equal to 5,000,000 kg km2/hr2 minus 2,500,000 kg km2/hr2 = 2,500,000 kg km2/hr2. You will also notice that the change in kinetic energy is equivalent to Case 1.

Q.E.D.


It is also possible to show that the change in momentum for each vehicle is the same in both cases:


Case 1 (Two cars hit head-on, each traveling at 50 km/hr):

The initial momentum of car A is mAvA = (1000 kg)(50 km/hr) = 50,000 kg km/hr. The initial momentum of car B is mBvB = (1000 kg)(-50 km/hr) = -50,000 kg km/hr.

After the collision, for momentum to be conserved, and because the initial total momentum was zero, both cars must have zero final velocity. In other words, both cars are at rest after the collision.

For both vehicles, the change in momentum before and after the collision is 50,000 kg km/hr.


Case 2 (A car traveling at 100 km/hr strikes a car at rest):

The initial momentum of car A is mAvA = (1000 kg)(100 km/hr) = 100,000 kg km/hr. The initial momentum of car B is mBvB = (1000 kg)(0 km/hr) = zero.

After the collision, by the analysis presented in the quoted section above, the velocity of the two vehicles is 50 km/hr. The momentum of car A is mAvA = (1000 kg)(50 km/hr) = 50,000 kg km/hr. The momentum of car B is mBvB = (1000 kg)(50 km/hr) = 50,000 kg km/hr.

Note that the change in momentum for car A is 50,000 kg km/hr. The change in momentum for car B is also 50,000 kg km/hr. This is equivalent to Case 1.

Q.E.D.

This conclusion is significant because the true measure of damage to the vehicles is the force experienced by the vehicles over the course of the collision.

The force experienced by the vehicles is equal to the change in momentum (i.e. impulse) divided by the time interval over which the force is applied. If we assume that the time interval is the same in both collisions, all we need to look at then is the change in momentum.

And as I have demonstrated here, the change in momentum during the actual collision is the same in both cases.

westom
03-02-2012, 11:27 PM
Case 1 (Two cars hit head-on, each traveling at 50 km/hr):
The initial momentum of car A is mAvA = (1000 kg)(50 km/hr) = 50,000 kg km/hr.
...
After the collision, for momentum to be conserved, and because the initial total momentum was zero, both cars must have zero final velocity. In other words, both cars are at rest after the collision. You first said the initial momentum was 50,000 kg km/hr. For a 1000 kg car, that means initial velocity is 50 km/hr.

Later you change things to suit your needs. Now your initial total momentum changes to zero? When did sleight of hand become part of physics?

MV before the collision. 50,000 kg km/hr. That means 1,250,000 kg km2/hr2 of energy - your numbers. Cars collide. You said, crumple zones, et al absorb energy. Where did this energy go? It went into heat, noise, and/or the crumpling of metal. 1,250,000 must be smaller because you said energy is absorbed. Any number - the crumple zone absorbs 1,000,000. Now the car only has 250.000 kg km2/hr2 remaining.

E / .5 / mass = velocity squared.
250,000 / .5 / 1000 = 125 = velocity squared.

Due to energy absorbed in crumple zones, the new velocity is less. In this example, 11 km/hr. Where is your conservation of momentum?

MV for 50 km/hr is not equal to MV for 11 km/hr. Why? Because collisions are not elastic. Because crumple zones absorb energy. Since a crumple zone absorbed only 1,000,000 kg km2/hr2 of energy, the new velocity goes from 50 km/hr to 11 km/hr. Where is conservation of momentum?

The collision was not elastic. Cars are intentionally designed (for human safety) to be the least elastic. Conservation of momentum can only exist in elastic collisions. Your own numbers demonstrate that conservation of momentum does not exist.


Then you also invented an initial velocity of zero when your own numbers started with an initial velocity of 50.


I aggressively asked for numbers in every post from every poster. Now that you did, identified is where your numbers only make sense when conservation of momentum is disposed.

Using your own numbers, if a crash absorbs energy, then momentum is not conserved. I used your numbers. Crumple zones reduce velocity from 50 o any other numbers down to zero (ie 11 in the example). If velocity decreases, then conservation of momentum does not exist. We intentionally design cars to make inelastic collisions. To not conserve momentum.


Finally using your own numbers. First collision: each car at 50 km/hr has an energy content of 1,250,000 kg km2/hr2. Second collision: a car moving at 100 km/hr has an energy content of 5,000,000 kg km2/hr2. For a parked car: 0 kg km2/hr2

Total energy in the first collision: 2,500,000. Total energy in a 100 km/hr crash: 5,000,000. Energy numbers do not lie. A 100 km/hr car crash is twice as violent - your numbers.

And yet somehow you then spin those numbers to make 2,500,000 equal to 5,000,000. How? You ignored conservation of energy. You assumed a crash and crumpling zone absorbs no energy. Assume conservation of momentum exists. Let’s use your numbers.

Second collisioni: what happens when 5,000,000 kg km2/hr2 energy is applied to two 1000 kg cars? Do the numbers. Both cars are now moving at 70.7 km/hr.
E = .5 times mass of both cars times the velocity of both squared.
5,000,000 = .5 * 2000 Kg * velocity squared.= 5000

Velocity is 70.7 km/hr

What happened to your conservation of momentum? When did 1000 kg times 100 kg/hr become 2000 times 70.7 km/hr? It didn't. Why. Conservation of momentum here is bogus. Conservation of energy says so.

But again, using your numbers. Everything works as long as conservation of momentum is bogus. You cannot have it both ways. Either you believe in conservation of energy. Or you believe in conservation of momentum. Both cannot exist – as demonstrated by your numbers.

How do you get 2000 kg times 50 km/hr to equal 1000 kg times 100 km/hr? Somehow you must change initial energy from 5,000,000 kg km2/hr2 to 2,500,000. How do you do that? Change initial velocity from 100 km/hr to 70.7 km/hr.

Changing initial conditions is the only way to make conservation of momentum work. But the initial conditions are 100 km/hr – not 70.7. You cannot change facts to justify a myth - 'conservation of momentum'. Why? Because conservation of momentum does not apply to inelastic collisions. And because your own numbers said the incoming 100 km/hr car is 5,000,000 kg km2/hr2. Not 2,500,000 as your conservation of momentum claims.


Now that you have provided the numbers - finally - I can use your numbers to expose mistakes. Your own numbers says the energy of a 100 km/hr car is not equivalent to a collision of two 50 km/hr cars. And yet using a myth called conservation of momentum, you magically changed 5,000,000 kg km2/hr2 into 2,500,000 kg km2/hr2.

Again, either you have conservation of momentum OR conservation of energy. Your own numbers say both cannot exist. Your own numbers say the two collisions have vastly different energy. You even changed initial conditions from 50 km/hr to zero km/hr just to make a bogus ‘conservation of momentum’ work.

robby
03-02-2012, 11:54 PM
You first said the initial momentum was 50,000 kg km/hr. For a 1000 kg car, that means initial velocity is 50 km/hr.

Later you change things to suit your needs. Now your initial total momentum changes to zero? When did sleight of hand become part of physics?Read my post again, and realize that momentum is a vector.

For case 1, the initial momentum of car A is mAvA = (1000 kg)(50 km/hr) = 50,000 kg km/hr. The initial momentum of car B is mBvB = (1000 kg)(-50 km/hr) = -50,000 kg km/hr.

If you add these two momentum vectors together, you get a total momentum for the entire two-car system that is zero.

This does not change. After the collision, the total momentum is also zero.

robby
03-02-2012, 11:58 PM
...MV before the collision. 50,000 kg km/hr. That means 1,250,000 kg km2/hr2 of energy - your numbers. Cars collide. You said, crumple zones, et al absorb energy. 1,250,000 must be smaller because you said energy is absorbed. Any number - the crumple zone absorbs 1,000,000. Now the car only has 250.000 kg km2/hr2 remaining.

E / .5 / mass = velocity squared.
250,000 / .5 / 1000 = 125 = velocity squared.

Due to energy absorbed in crumple zones, the new velocity is less. In this example, 11 km/hr. Where is your conservation of momentum?

MV for 50 km/hr is not equal to MV for 11 km/hr. Why? Because collisions are not elastic. Because crumple zones absorb energy. Since a crumple zone absorbed only 1,000,000 kg km2/hr2 of energy, the new velocity goes from 50 km/hr to 11 km/hr. Where is conservation of momentum?You can't just look at one vehicle to apply conservation of momentum. This law only applies if no outside force is applied. If you consider only one vehicle at a time, then certainly a force is applied--that of the impact from the other vehicle.

Instead, you must consider the two-car system as a whole.

robby
03-03-2012, 12:11 AM
...The collision was not elastic. Cars are intentionally designed (for human safety) to be the least elastic. Conservation of momentum can only exist in elastic collisions. Your own numbers demonstrate that conservation of momentum does not exist.This is so wrong, and represents such a gross misunderstanding of the physics in question, that I don't even know where to begin.

...I aggressively asked for numbers in every post from every poster. Now that you did, identified is where your numbers only make sense when conservation of momentum is disposed.

Using your own numbers, if a crash absorbs energy, then momentum is not conserved. I used your numbers. Crumple zones reduce velocity from 50 o any other numbers down to zero (ie 11 in the example). If velocity decreases, then conservation of momentum does not exist. We intentionally design cars to make inelastic collisions. To not conserve momentum.See above comment.

...Finally using your own numbers. First collision: each car at 50 km/hr has an energy content of 1,250,000 kg km2/hr2. Second collision: a car moving at 100 km/hr has an energy content of 5,000,000 kg km2/hr2. For a parked car: 0 kg km2/hr2

Total energy in the first collision: 2,500,000. Total energy in a 100 km/hr crash: 5,000,000. Energy numbers do not lie. A 100 km/hr car crash is twice as violent - your numbers.

And yet somehow you then spin those numbers to make 2,500,000 equal to 5,000,000. How?Well, you need to compare the change in kinetic energy in each case.

...You ignored conservation of energy. You assumed a crash and crumpling zone absorbs no energy. Assume conservation of momentum exists. Let’s use your numbers.

Second collisioni: what happens when 5,000,000 kg km2/hr2 energy is applied to two 1000 kg cars? Do the numbers. Both cars are now moving at 70.7 km/hr.
E = .5 times mass of both cars times the velocity of both squared.
5,000,000 = .5 * 2000 Kg * velocity squared.= 5000

Velocity is 70.7 km/hrYou are apparently assuming that kinetic energy is conserved. It is not.

...What happened to your conservation of momentum? When did 1000 kg times 100 kg/hr become 2000 times 70.7 km/hr? It didn't. Why. Conservation of momentum here is bogus. Conservation of energy says so.In an inelastic collision, kinetic energy is not conserved, so your analysis is flawed. In addition, you have come to the erroneous conclusion that conservation of momentum is "bogus," which should be a clue that you have made a serious error.

...But again, using your numbers. Everything works as long as conservation of momentum is bogus. You cannot have it both ways. Either you believe in conservation of energy. Or you believe in conservation of momentum. Both cannot exist – as demonstrated by your numbers.Again, total energy and momentum are conserved. Kinetic energy is NOT conserved.

...How do you get 2000 kg times 50 km/hr to equal 1000 kg times 100 km/hr? Somehow you must change initial energy from 5,000,000 kg km2/hr2 to 2,500,000. How do you do that? Change initial velocity from 100 km/hr to 70.7 km/hr.Again, KINETIC ENERGY IS NOT CONSERVED.

...Changing initial conditions is the only way to make conservation of momentum work. But the initial conditions are 100 km/hr – not 70.7. You cannot change facts to justify a myth - 'conservation of momentum'. Why? Because conservation of momentum does not apply to inelastic collisions. And because your own numbers said the incoming 100 km/hr car is 5,000,000 kg km2/hr2. Not 2,500,000 as your conservation of momentum claims.This is a joke, right?

...Now that you have provided the numbers - finally - I can use your numbers to expose mistakes. Your own numbers says the energy of a 100 km/hr car is not equivalent to a collision of two 50 km/hr cars. And yet using a myth called conservation of momentum, you magically changed 5,000,000 kg km2/hr2 into 2,500,000 kg km2/hr2.No, seriously, where are the cameras?

...Again, either you have conservation of momentum OR conservation of energy. Your own numbers say both cannot exist. Your own numbers say the two collisions have vastly different energy. You even changed initial conditions from 50 km/hr to zero km/hr just to make a bogus ‘conservation of momentum’ work.:rolleyes:

robby
03-03-2012, 12:30 AM
Say, westom, why don't you come on down here (http://boards.straightdope.com/sdmb/showthread.php?t=643983) to the Pit?

westom
03-03-2012, 12:39 AM
You can't just look at one vehicle to apply conservation of momentum. . My mistake.

Initially posted was 250,000 /.5 / 1000

It should have read
250,000 / .5 / 2000 = 250 = velocity squared
of 15.8 km/hr

BTW, I am trying to get you to deal with energy numbers. Once you apply those numbers properly, then conservation of momentum applies. None of this was possible until your every post including the underlying numbers. Without numbers, only lies, propaganda, anger, miscommunication, myths, and even pagan religions occur.

When done, the single vehicle at 100 km/hr is twice as destructive (dissipates twice the energy) of two 50 km/hr vehicles.

The balance between momentum and energy can define how energy dissipates But the bottom line remains - the 100 km/hr vehicle dissipates twice as much energy destructively as two 50 km/hr vehicles. That answers to the OP's question - assuming energy defines what he has called detrimental.

Using numbers I have provided, you can better define how collision energy gets applied to two vehicles that must absorb or dissipate that energy detrimentally.

Remember, any and all vectors apply. The final answer must even assume collisions from all directions. Even that one vehicle is rolling after the initial contact. And that the crash occurs on vehicles with structually different crumple zones. For all we know, the 100 km/hr car snow plows throughthe parked car. The only definitive numbers are obtained from numbers the OP provided – 100 km/hr, 50 km/hr, and 0 km/hr. Limits all answers to ballpark conclusions. Only fact we can say with certainty is energy and speed in each car before the crash. And that, like all crashes (except hit and run), means vehicles end at 0 km/hr. Therefore all energy dissipates in the crash.

westom
03-03-2012, 12:57 AM
This is so wrong, and represents such a gross misunderstanding of the physics in question, that I don't even know where to begin.
:
Any decent and honest person post corrections with numbers and equations. Such subjective sentences exist only from extremists, liars, junk scientists, and the least educated. You know the answer has errors? Good. So do I as I was originally hinting at in an earliest post. Now finish what was necessary and not found in your first posts - the numbers and equations. Stop posting subjectively. Stop posting emotions. Post the corrected equations. Explain what happens with all 2,500,000 and 5,000,000 kg km2/hr2. Explain it with numbers; not subjective (junk science) attacks and denials.

Tangent
03-03-2012, 01:24 AM
westom, I teach physics. We do instructional labs where we create collisions like this on a smaller scale, using carts on flat tracks or gliders on very low-friction air tracks. Our data always confirms that momentum is conserved during both elastic and inelastic collisions. It also confirms that kinetic energy is NEVER conserved during an inelastic collision. (Kinetic energy is only conserved in perfectly elastic collisions).

Forget the fucking cars--I'm going to use smaller numbers to try to make this easier to write and easier to understand. But first, you MUST concede that MOMENTUM IS CONSERVED DURING INELASTIC COLLISIONS. Seriously, dude, do 20 seconds of research online and you will find that any reliable source will confirm this. If you can't get your head straight about that fact, nothing anybody says is going to help you and you're going to remain ignorant. [The quick justification is that during either type of collision, the forces the objects exert on each other are the same magnitude but in opposite directions. These "equal and opposite" forces also must act on the objects for exactly the same amount of time. This leads to equal and opposite impulses on the system during the collision, resulting in no net change in momentum of the system.]

Okay. Case 1

Two balls of soft clay, each with mass 1 kg are moving towards each other, each with a speed of 10 m/s. They undergo a head-on perfectly inelastic collision (they stick together). Friction is negligible for the small time interval just before to just after the collision.

Momentum
Before collison: net momentum = 1(10) + 1(-10) = 0
Immediately after the collision, the net momentum must be 0, because MOMENTUM IS ALWAYS CONSERVED, EVEN IN INELASTIC COLLISONS!
net momentum = (1+1)v = 0, so the final velocity of the stuck-together clay balls is 0.

Individually, each clay ball had a change in momentum of 10 kg m/s. But because they changed in opposite directions, the net change is 0.


Kinetic Energy
Before the collision: kinetic energy = .5(1)10^2 +.5(1)(-10)^2 = 100 Joules
After the collision: since final v = 0, the final kinetic energy = 0.

Individually, each clay ball "lost" 50 Joules of kinetic energy.
Where did the 100 Joules of kinetic energy go? Most of it becomes thermal energy during the collision. Energy is conserved! It's just not kinetic energy anymore.

**********************************************************

Case 2: Now lets say that one of the clay balls was sitting still and the other was moving towards it with an initial speed of 20 m/s.

Okay. Perfectly inelastic collision, same as before:

Momentum
Before collision: net momentum = 1(0) + 1(20) = 20 kg m/s
After collision: net momentum = same as before collision = 20 = (1+1)v; therefore v = 10 m/s (this is the speed of the stuck-together clay balls immediately after the collision)

Individually, each clay ball had a change in momentum of 10 kg m/s. But because they changed in opposite directions, the net change is 0.

Kinetic Energy
Before collision: total kinetic energy = .5(1)0^2 + .5(1)20^2 = 200 Joules
After collision: total kinetic energy = .5(1+1)10^2 = 100 Joules

Individually, each clay ball "lost" 50 Joules of kinetic energy.
Where did this 100 J of kinetic energy go? Most of it becomes thermal energy during the collision. Energy is conserved! And so is momentum!


The change in momentum and the change in kinetic energy for each object is the same in either case.

******************
Yes, the cars will eventually come to a stop after the "Case 2" collision. But that's most likely to be an insignificant event compared to the collision itself, unless something unpredictable and unfortunate happens--like the combined objects going over a cliff, or falling into a river, or being hit by a meteorite, or being stepped on by Godzilla.

Half Man Half Wit
03-03-2012, 03:28 AM
This is a bizarre thread...

Simple thought experiment: you sit on a motorcycle. First case, you are at rest, and a car (car 1) going 100mph slams into a stationary one (car 2). Second case, you move at 50mph in the same direction as car 1. From your point of view, this car now moves at 50mph, while car 2 also moves at 50mph towards you. Both cars crash into each other. The only thing that has changed is your state of motion. Does this affect the crash in any way? No, of course not. Both situations are the same.

In case that's not enough, let's look at the math for general collisions. First, we'll treat the elastic case -- which doesn't apply in this situation, but will help us in getting a general handle on the problem. In this case, momentum and (kinetic) energy are both conserved, so let car 1 have a mass of m1 and an initial velocity of u1, while car 2 has a mass of m2, and a velocity of u2. Conservation of momentum gives us:
m1u1 + m2u2 = m1v1 + m2v2,
where the vi are the velocities of the cars after the collision. Similarly, conservation of energy yields:
m1u12/2 + m2u22/2 = m1v12/2 + m2v22/2.
Thus, we have two equations with two unknowns, v1 and v2. So this can be solved to yield:
v1 = (m1u1 - m2u1 + 2m2u2)/(m1 + m2)
v2 = (m2u2 - m1u2 + 2m1u1)/(m1 + m2)

Now let's treat the other extreme, the perfectly inelastic collision. In this case, kinetic energy is not conserved -- a lot of it goes into the deformation of the cars, and into all the sound and fury associated with that. In the perfectly inelastic case, both cars will 'stick together' after the collision. Thus, conservation of momentum yields:
m1u1 + m2u2 = (m1 + m2)v,
because now, both cars form one object (i.e. a twisted mass of broken metal) of mass m1 + m2, going at some velocity v. This is elementary to solve:
v = (m1u1 + m2u2)/(m1 + m2).

In reality, no collision is ever perfectly elastic or inelastic; so we'll need a way to interpolate between both solutions, such that for perfect elasticity, we get the first one, and for perfect inelasticity, we get v1 = v2 = v. This is done by introducing the coefficient of restitution, c:
v1 = (c*m2(u2 - u1) + m2u2 + m1u1)/(m1 + m2)
v2 = (c*m1(u1 - u2) + m2u2 + m1u1)/(m1 + m2)

As you can see, for c = 0, this reduces to the inelastic case, and for c = 1, this covers the elastic one. Intermediate values describe realistic cases, which are partly elastic, and partly inelastic.

Now, how do we see that both collisions are equivalent? We look at the velocity change of the second car, |u2 - v2|. If car 2 is stationary before the collision, then v2 = (c*m1u1 + m1u1)/(m1 + m2) = (c*1000kg * 100mph + 1000kg * 100mph)/2000kg = c*50mph + 50mph. For car 2 moving at 50mph, we have v2 = (c*1000kg*(50mph - (-50mph)) - 1000kg*50mph + 1000kg*50mph)/2000kg = (c*1000kg*100mph)/2000kg = -c*50mph, where the minus signs come from the fact that the cars move in opposite directions.

In the first case, then, |u2 - v2| = |0 - (c*50mph + 50mph)| = c*50mph + 50mph; this is easily seen to be the right answer, because in the elastic case, all the energy gets transferred to the second car, thus causing it to move at 100mph, while in the inelastic case, the combined object made from both cars moves (immediately after the impact) at 50mph. In the second case, |u2 - v2| = |-50mph - c*50mph| = c*50mph + 50mph. Thus, both collisions are equivalent, regardless of how elastic or inelastic they are -- as of course must be the case.

FinnAgain
03-03-2012, 07:57 AM
Good job folks, especially Robby.
FWIW I also teach physics and, yep, both the math and the concepts stand up properly.

robby
03-03-2012, 09:19 AM
Any decent and honest person post corrections with numbers and equations. Such subjective sentences exist only from extremists, liars, junk scientists, and the least educated. You know the answer has errors? Good. So do I as I was originally hinting at in an earliest post. Now finish what was necessary and not found in your first posts - the numbers and equations. Stop posting subjectively. Stop posting emotions. Post the corrected equations. Explain what happens with all 2,500,000 and 5,000,000 kg km2/hr2. Explain it with numbers; not subjective (junk science) attacks and denials.Westom, I and others have posted numbers and equations repeatedly, but as the saying goes, there are "none so blind as those that will not see."

westom
03-03-2012, 11:43 AM
Kinetic Energy
Before collision: total kinetic energy = .5(1)0^2 + .5(1)20^2 = 200 Joules
After collision: total kinetic energy = .5(1+1)10^2 = 100 Joules
...
Yes, cars will eventually come to a stop after the "Case 2" collision. But that's most likely to be an insignificant event compared to the collision itself,
First, you have posted what everyone had to do to have a correct answer. Others assumed your same conclusion. But posted no numbers and other reasons why. So they were insulting an educated reader. In the real world, a correct answer also provides reasons why with numbers - as you have done and robby eventually did. A "scammer or politician" (a term used repeatedly to say this) could have posted the same answer - and is still 100% wrong. Because he did what makes him a "scammer or corrupt politician". He did not post the numbers, in equations, with underlying facts. That defines scammers and politicians. They tell everyone what to know rather than what is necessary to be informed.

Second, after collision, the total kinetic energy is 100 joules. So where are those 100 joules? The cars are now rolling in a jumbled heap or ‘out of control’ manner; maybe even killing pedestrians. A crash is not over until the second 100 joules dissipates. Parked cars are not clay balls. A crash is not done until everything stops moving.

In the ball world, car 2 keeps moving because of that 100 joules. In the real world, the car is either a rolling ball of scrap metal, is tearing tires down the road, is plowing through a nearby store, rolling with parts still flying off in many directions, etc. I even suggested car one might be snow plowing through car 2. And still that is only a second collision? Of course not. A car crash is over when everything stops moving. A crash is never done until all kinetic energy (the second 100 joules) dissipates. If car 2 is moving (with brakes applied), then that is not "detrimental"? Of course it is "detrimental". Detriment continues until a car finally rolls (as in overturned) to a stop.

Third, the OP did not say balls or sliding blocks. And he used a vague term "detrimental". If a stopped car 2 is moving due to an inelastic collision, then it is still crashing. Because a car (ie brakes applied) must still dissipate the second 100 joules. The second 100 joules may be an “insignificant event” or it may be the worst part of a crash. As repeatedly noted in earlier posts, too many hypotheticals.

Need for a definition of 'detrimental' was necessary.

Many assumed the entire collision was an instantaneous event when two cars (or clay balls) were in contact. Something 'detrimental' continues until everything stops moving - until all kinetic energies are zero. Until smashed cars stop moving as a threat to everything and one around them.

Thank you for doing what most before you refused to do and what robby only eventually did with prodding.

Momentum was conserved. But that was irrelevant. A car crash is not two clay balls. A parked car or one waiting at a stop light is not “detrimental” when moving or rolling at 50 km/hr due to a crash? Of course it is. If the intent was to demonstrate conservation of momentum, the original question was vague and misleading.

Colibri
03-03-2012, 12:24 PM
Any decent and honest person post corrections with numbers and equations. Such subjective sentences exist only from extremists, liars, junk scientists, and the least educated.

[Moderator Warning]

westom, insults, even oblique ones, are not permitted in General Questions. This is an official warning. Don't do this again.

Colibri
General Questions Moderator

FinnAgain
03-03-2012, 12:27 PM
So where are those 100 joules? The cars are now rolling in a jumbled heap or ‘out of control’ manner

Depending on the coefficient of friction due to the surface of the road, the deformation of their wheels, etc... it's being burned off as waste heat, sonic energy and probably some light if metal is sparking against the pavement at a certain rate. And yes, the collision is over once the object collide and either stick together or do not stick together. After that we are no longer dealing with the mathematics of collision, but the mathematics of a new object, Car1+Car2, and it is no longer a collision problem.

Pnet start = Pnet finish. Momentum would not be conserved if Pnet start = Pnet finish + or - X kg m/s. Adding an external force like friction no longer has it being a closed system. Momentum is always conserved in a closed system, it's one of the fundamental laws of reality. When adding friction over time it creates impulse. Impulse is, in turn, a vector quantity. As friction always works against the direction of motion, it will provide a vector which reduces and eventually cancels out P. P + -P = 0. Once we include μk in our system, Pnet start =P net finish due to vector addition, and momentum is still conserved.

The OP also did not give us full information in the beginning. As P is conserved, the after effects of the 50 mph crash will likewise depend on the surroundings if the cars are still moving after the collision is resolved, just as the 100 mph crash will. If you don't believe me, picture the results of a Mack truck traveling at 50 m/s having a head on collision with a carbon fiber bike traveling at -50 m/s.

Andy L
03-03-2012, 04:45 PM
Pnet start = Pnet finish. Momentum would not be conserved if Pnet start = Pnet finish + or - X kg m/s. Adding an external force like friction no longer has it being a closed system. Momentum is always conserved in a closed system, it's one of the fundamental laws of reality. When adding friction over time it creates impulse. Impulse is, in turn, a vector quantity. As friction always works against the direction of motion, it will provide a vector which reduces and eventually cancels out P. P + -P = 0. Once we include μk in our system, Pnet start =P net finish due to vector addition, and momentum is still conserved.

Is it fair to say that if we consider the Earth to be part of the system, that friction is transferring momentum from the cars to the Earth - but that due to the huge mass difference between the Earth and the cars, the additional velocity of the Earth after the relative motion has disappeared is essentially undetectable?

zwede
03-03-2012, 06:03 PM
I haven't the energy to read through all the posts, but look at it this way:

Case one: we watch the cars hit each other each doing 50 km/h. Our perspective is such that we look at the cars from the side.

Case two: let's do the EXACT SAME THING except that we are moving the background at 50 km/h towards the left. This makes it appear like one car is standing still and the other is going twice as fast.

The cars will still hit and crumple the same way regardless how we move the background, since their speed relative to each other is the same.

That's about as simple as I can make it.

RadicalPi
03-03-2012, 06:22 PM
BTW, this has long been a typical question on SATs to separate those who think subjectively from those who learned their physics and do the numbers.

Let me see if I'm getting the arguments straight.

It seems that everyone is in agreement that the collisions themselves between two cars going at 50 k/h on the one hand and one car going 100 k/h and another stopped on the other hand are essentially identical. Except for the very slight differences arising from relativity, there is a pair* of inertial reference frames where the one event is the same as the other in the sense that you could superimpose them on each other completely.

But it seems that one reference frame has the earth stopped and the other has the earth moving at 50 k/h. And the cars are ultimately going to come to rest in both reference frames. So, in one of the collisions, everything has to slow down another 50 k/h. And the question is whether or not that matters. There's one side that says that that is all post-collision stuff and that it is consequently mostly irrelevant. Then there's westom who thinks that that is part of the collision (which is reasonable)** but that somehow the law of conservation of momentum gets violated in the process, even though this has never been shown to ever happen (which is not reasonable).

Westom goes on and on about the need to define detrimental, but it seems that it would be helpful to define collision to avoid this.


* probably a whole bunch of pairs, no?

** but what gets lost here is that the collisions themselves are exactly the same in the sense that the collisions are not the same if two cars are going 20 k/h and to other cars are going 200 k/h. The key really seems to be the post-collision effects.


Also some random comments:

1. There is no physics section on the SAT.

2. While charlatans and corrupt politicians often hide behind flowery rhetoric, an intuitive understanding of various concepts and how they relate to each other is invaluable. The problem is that in physics as well as in life, one's first intuition may not be the proper one, and the calculations help home you in on the right one. When things can't be quantified, this gets a lot harder, but it remains all the less still advantageous.

3. Hi, Opal.

4. It's not always possible to define a proper form for the answers to various questions, especially if you don't know where the questions will lead. I am reminded of a phrase that I can't remember that encapsulates this apparently by Asimov: "The most exciting phrase to hear in science, the one that heralds new discoveries, is not 'Eureka!' but rather, 'Hmmmm, that's funny.'"

FinnAgain
03-03-2012, 06:25 PM
Is it fair to say that if we consider the Earth to be part of the system, that friction is transferring momentum from the cars to the Earth - but that due to the huge mass difference between the Earth and the cars, the additional velocity of the Earth after the relative motion has disappeared is essentially undetectable?

Not exactly.
The important thing to realize is that momentum is a vector quantity, not a scalar quantity. That is, depending on directionality, vectors can be negative. For example, speed is scalar. You cannot be going slower than 0 miles an hour. Velocity is a vector, if our coordinate system privileges right as the positive direction, then moving to the left at 30 m/s is -30 m/s. So when we calculate ΣP, we have to take the initial momentum of an object into account as well as the force of friction. Since the force of friction is a force (measured in Newtons) and it's applied for a certain amount of time, it produces impulse. Impulse is equal to FT but also ΔP. ΔP is still, in essence, a form of P, it just denotes which direction the vector's going.

So imagine we have a toy car with a mass of 1kg, moving at 5 m/s. Its P is 5 kg m/s. Acting against it is the force of friction. Assume 1N of force acting for 5 seconds. ΔP is -5 kg m/s.

ΣPstart: 5 kg m/s + -5 kg m/s = 0 kg m/s
ΣPfinish (the toy car comes to a stop): (1kg) (0 m/s) = 0 kg m/s
0 kg m/s = 0 kg m/s
Momentum is conserved.

Animastryfe
03-03-2012, 07:11 PM
I haven't the energy to read through all the posts, but look at it this way:

Case one: we watch the cars hit each other each doing 50 km/h. Our perspective is such that we look at the cars from the side.

Case two: let's do the EXACT SAME THING except that we are moving the background at 50 km/h towards the left. This makes it appear like one car is standing still and the other is going twice as fast.

The cars will still hit and crumple the same way regardless how we move the background, since their speed relative to each other is the same.

That's about as simple as I can make it.

Yes, this should be the correct summary.
Westom, I am a third year physics student. Although I doubt anything I can say or do will change your mind, there is still the possibility that something my professors can say will change your mind. Do you want me to ask them this question?

westom
03-03-2012, 07:53 PM
Do you want me to ask them this question? Actually, I agreed mostly with what was provided. Even suggested it in a first post. Conservation of momentum does exist. But momentum alone was insufficient. Where kinetic energy went was also relevant and ignored.

The original question was poor if its purpose was to explain physics that also explains clay ball collisions. Tangent and robby finally provided answers with numbers that were incomplete. A parked car suddenly at 50 km/hr sideways or into an intersection is not part of the crash? That energy also must be accounted for.

So many answers were devoid of numbers for both momentum and energy equations. An answer that only said (subjectively) that collisions were equivalent was misleading and incomplete. In this case, a parked or waiting car suddenly doing 50 km/hr is irrelevant to the crash? Energy numbers and the resulting velocity suggested otherwise.

Momentum says the 100 km/hr crash or two 50 km/hr vehicles crashing are similar at the initial impact. But undissipated energy after a 100 km/hr crash implies that higher energy crash has further consequences. That undissipated kinetic energy was being ignored. And explains why the two cases are not same.

Appreciate the problem with most answers. The always required numbers were missing until robby and Tangent provided some of them.

FinnAgain
03-03-2012, 09:55 PM
Momentum says the 100 km/hr crash or two 50 km/hr vehicles crashing are similar at the initial impact. But undissipated energy after a 100 km/hr crash implies that higher energy crash has further consequences. That undissipated kinetic energy was being ignored. And explains why the two cases are not same.

No, you still do not understand the basic math or concepts behind the problem.
I'll use simpler numbers to help you out here.


1) Two 2 kg toy cars are moving, one with a velocity of 5 m/s one with a velocity of -5 m/s.

ΣPstart and ΣPfinish both equal 0 kg m/s and momentum is conserved. ΣEk start = 50J. Upon collision, ΣEk finish = 0J as result, ΔEk = -50J
ΔEk = Work according to the Work-Energy Theorem. Therefore, -50J of work were done to the system, or 50J of negative work if you prefer. Conservation of energy tells us that these are most likely transformed to thermal and sonic energy.

2) One 2kg toy car is moving with a velocity of 10 m/s. A second 2 kg car is stationary.

Assuming a perfectly inelastic collision with negligible friction, Pnet start equals 20 kg m/s. Upon collision we know 100% of momentum must be conserved, so since the new object has a mass of 4 kg its V must be 5 m/s. , ΣP start = , ΣP finish
Prior to the collision, ΣEk start = 100J. Upon collision, ΣEk finish = 50J. Therefore
ΔEk = -50J. ΔEk = Work. Therefore, -50J of work were done to the system. Conservation of energy tells us that these are most likely transformed to thermal and sonic energy.

1)ΔEk = -50J
2)ΔEk = -50J

The only difference is that collision 2 will still have a velocity vector after its initial impact. However that is no longer part of the collision, but part of the mechanics for the new object, Car1 + Car2. Car1+2 may grind against the ground until it gently comes to rest, may smack into another car, may go off a cliff. But that is not part of the collision problem.

FinnAgain
03-03-2012, 11:16 PM
It should also be noted that the issue is not that "a higher energy crash" has further consequences. It is not the energy, but the momentum that is at issue. If both cars have positive or negative momentum, then the combined mass would continue moving as well, due to conservation of momentum. Even if they had much less Ek during a collision, conservation of momentum would still tell us that they'd continue moving. The only issue that matters in a "higher energy crash" is generally going to be ΔEk, because without Δ there will be no work done and Ek will most likely not be converted into another form of energy.

The fact that there is any remaining Ek at all after a collision is solely because of conservation of momentum and vector addition; a head on collision usually results in ΣP at or near zero. When V reaches zero, Ek is zero, and therfore ΔEk is 100%. With addition of vectors which have the same sign, or one with zero magnitude and one with non-zero magnitude, we'll see a ΣP that is greater than zero. As such, ΔEk is < 100%.

Chronos
03-04-2012, 12:15 AM
But it seems that one reference frame has the earth stopped and the other has the earth moving at 50 k/h. And the cars are ultimately going to come to rest in both reference frames. So, in one of the collisions, everything has to slow down another 50 k/h. And the question is whether or not that matters. If you're driving along at 50 km/hr, and slam on your brakes, you're not going to do any appreciable damage at all to the car (at worst, you'll wear off a bit of the tire tread). This is the same situation the cars are in after the collision: They've got energy which they are losing to friction with the road.

Reply
03-04-2012, 01:09 AM
I just have to say I've never been part of a more interesting physics discussion. That improperly stopped cars causes pagan religions just isn't something you learn in school.

williambaskerville
03-04-2012, 03:58 AM
I just have to say I've never been part of a more interesting physics discussion. That improperly stopped cars causes pagan religions just isn't something you learn in school.

I read this thread because I read the OP a while ago, assumed it would be solved in 2 posts, and was fascinated how it got to 3 pages.

I know the answer to this because I did physics when I was 15. I have to bring up something Carl Sagan wrote. "They laughed at Galileo :- they laughed at a lot of idiots too". Being in a minority doesn't make you right.

Cheesesteak
03-04-2012, 05:40 AM
westom, you're missing a key concept here, one that has nothing to do with equations or numbers.

Assume 2 cars of equal mass
1) Car A goes 50kph to the right, Car B goes 50kph to the left. They meet head on and come to rest immediately after the collision.

2) Car a goes 100kph to the right, Car B goes 0kph. They meet head on and are observed moving 50kph to the right immediately after the collision.

1 & 2 are identical. Not similar, not equivalent, not comparable, identical, right down to the tears of the drivers' mothers when they hear the news. Why?

....

There was only one crash. 1 & 2 are descriptions of the crash from two different reliable observers who happen to be moving 50kph relative to each other.

Neither of their frames of reference (http://en.wikipedia.org/wiki/Inertial_frame) are preferred, they are both correct, and the equations (whatever they are) will tie out exactly, or we've just proven Einstein wrong with a pathetically simple example.

This is why all the knowledgeable physics folks immediately start with "identical" until other assumptions start being made. Assumptions like:
The cars are on a street
The cars are driven by people
The speed is measured relative to the ground

These are assumptions that are unstated in the OP, but need to be stated very specifically when answering a physics question.

YamatoTwinkie
03-05-2012, 12:31 PM
Where kinetic energy went was also relevant and ignored.

Westom, I think if you go back and read the entire thread, you'll find that it wasn't ignored. Some examples (amongst many):

Post#24:
In the second case, the initial KE of car A is 5000m, but the KE of the two mangled vehicles immediately after the collision is 2500m. (The rest of the KE will then be dissipated as friction as the two mangled vehicles slide to a stop).

Post#32:
Then you get half credit for stopping halfway through the problem. Yes, a car at 100 km/hr has four times the energy as a car at 50 km/hr. You don't get to just stop right there, put down the pencil, and say you have an answer. Now find the energy post-collision and find the "energy dissipated," which you admit is what is at issue here.

Post#41:
What differs is post-collision. In case #1, post collision is two wrecked cars at rest. In case #2, you have two wrecked cars going 50km/h.

If the cars are on neat, weightless, fricitonless sleds, and are collided under controlled conditions, and are brought to a clean damage-free stop, you will see little difference. In the real world, two wrecked cars going 50km/h down a street will eventually stop, but will likely hit other things and become more damaged along the way.

Now, the reason why the leftover, post-collision KE wasn't treated with the same level of rigor as the actual collision is because it's impossible to know for certain. Maybe the leftover KE causes the vehicles to come to a rolling stop over a few hundred meters. Maybe the leftover KE causes the vehicles to flip over, hit a schoolbus, and careen off the side of the mountain into an active volcano. All that really matters is that it's addressed as a post-collision event that can be either totally harmless or *may* cause further damage, but there's not enough information to determine.

And it has been addressed as such in this thread already, multiple times over.

Andy L
03-05-2012, 07:24 PM
Not exactly.
The important thing to realize is that momentum is a vector quantity, not a scalar quantity. That is, depending on directionality, vectors can be negative. For example, speed is scalar. You cannot be going slower than 0 miles an hour. Velocity is a vector, if our coordinate system privileges right as the positive direction, then moving to the left at 30 m/s is -30 m/s. So when we calculate ΣP, we have to take the initial momentum of an object into account as well as the force of friction. Since the force of friction is a force (measured in Newtons) and it's applied for a certain amount of time, it produces impulse. Impulse is equal to FT but also ΔP. ΔP is still, in essence, a form of P, it just denotes which direction the vector's going.

So imagine we have a toy car with a mass of 1kg, moving at 5 m/s. Its P is 5 kg m/s. Acting against it is the force of friction. Assume 1N of force acting for 5 seconds. ΔP is -5 kg m/s.

ΣPstart: 5 kg m/s + -5 kg m/s = 0 kg m/s
ΣPfinish (the toy car comes to a stop): (1kg) (0 m/s) = 0 kg m/s
0 kg m/s = 0 kg m/s
Momentum is conserved.

I was thinking this way:

The 1kg car is initially moving 6 m/s in whichever direction we call positive, and the earth is not moving. After the car has stopped relative to the Earth (due to the friction), the car+Earth system is moving 6*10^(-24) m/s in the positive direction.

Irishman
03-06-2012, 01:47 PM
Scary that so many do not even know simple high school physics. Doubling velocity (speed) does not double energy. It quadruples energy. Everyone here should know that. Otherwise scammers and politicians make money and get elected because the public is uneducated and naive. These fundamental facts are taught to everyone in high school to make dishonesty and scams difficult.

What's scary is that you turn a high school physics problem into a screed on polititians and scammers, all the while mangling the physics.

I'm not going to point by point you. It has already been done in this thread several times, and the answers are clear and accurate, and they continue to show the same thing.

You are choosing to define the term "collision" in a layperson's casual usage equivalent to "car crash" rather than the technical physics term being used as in the classroom setting, the impact of the two objects. For the problem as stated in the OP, this appears to be about the basic physics question of "collision", i.e. the impact of two objects, not anything to do with what happens to cars in the real world.

For that classroom situation, the "collision" is over after the two objects are done impacting, regardless of what condition that leaves them. In the example in question, that means it can leave them moving at 50km/hr as a single object, ready for more "collisions" or other interactions.

In your casual use of "collision" to mean "car crash", yes, the car crash is not over until all objects are at rest. But that is a very different situation than the OP.

You do have one valid point, the statement "detrimental" is vague. Taken by the phrasing of the OP, it is assumed the OP is looking at the textbook example, in which case the only detriment relevant is that which occurs during the collision, i.e. the actual impact of the two objects. For this textbook case, as has been shown numerous times including actual numbers and equations (thanks robby), the two collisions have the same dissipation of kinetic energy, with one situation using all the kinetic energy, and one situation only using some of it.

By the way, I will define one other term that you apparently do not understand. "Dissipation" in this situation applies to the kinetic energy that is lost during the collision. It only applies during the actual collision, not to all the kinetic energy that may be lost due to subsequent interactions between the two cars and the pavement, or the two cars and other objects.

For the case where the equivalent energy is 70.7km/h or whatever, you have to park one of the cars against a barricade/wall/chained to the ground, so that it becomes in immovable object. Then when you slam the first moving car into the second still car, all the KE from the first car is dissipated in the collision (notcrash).

But you have been continually wrong with your assertion that kinetic energy is conserved and your assertion that momentum is not conserved. You have is completely backwards.

Total Energy is always conserved. Momentum is always conserved. Kinetic Energy may or may not be conserved. In an inelastic collision, by definition KE is not conserved. Just like dropping a ball will not conserve KE.

Here are some basic physics texts explaining Conservation of Momentum and Inelastic Collisions.

http://www.physicsclassroom.com/class/momentum/u4l2b.cfm

This one is really neat because it has a video of two carts colliding. One has mass m and is moving, the other has mass 3m and starts at rest. It is inelastic, i.e. the two carts stick together. After the collision, guess what? They both move together. (The velocity is lower because the masses are not equal, but the principle is exactly the same.)

http://www.batesville.k12.in.us/physics/apphynet/Dynamics/Collisions/inelastic_collisions.htm

This one shows that Conservation of Momentum is precisely the tool used to be able to solve an inelastic collision.

In short, you are wrong. You interpret the question oddly, and you are wrong in your application of the physics.

FixMyIgnorance
03-06-2012, 09:34 PM
Too lazy to read this thread (just read the first post) -- both 1 and 2 are identical. This is basic, basic high school physics, here.

Irishman
03-08-2012, 01:33 PM
Too lazy to read this thread ....

Then how are you going to fix your ignorance? :D

Trinopus
03-08-2012, 06:52 PM
Thank you, all y'all, for this thread! I've learned stuff from it!

I had always thought that momentum, like energy, could be "lost" -- i.e., converted to other forms. Now I *think* I've learned better!

If I brake my car, my kinetic energy is (in part) converted to heat energy.

But my momentum is simply transferred to the earth as a whole. Is that right? If I'm travelling eastward, I speed up the earth's rotation (by a tiny amount!) If I'm travelling westward, I slow it down. If I'm going north or south, I actually alter the earth's axial tilt... Yeah, so microscopically that no one will ever measure it, but enough to "balance the books" and make the equations all sum to zero.

Er...is that right?

Irishman
03-10-2012, 04:02 PM
Mostly right.

If traveling eastward, you are traveling into the Sun. The Earth rotates opposite of what you see the Sun do. Ergo, traveling eastward, you are actually slowing Earth's rotation, and traveling westward you are speeding it up.

ZenBeam
03-10-2012, 04:18 PM
No, I think Trinopus has it right, if I understand what he's saying correctly. If he's braking while traveling East, he's transferring momentum into Earth's rotation (where "Earth" doesn't include him or his car :)), speeding it up ever so slightly. Of course, to initially accelerate in the Eastward direction, he had to slow the Earth's rotation down first.

Irishman
03-12-2012, 10:56 AM
Ah, yes, applying the brakes works backwards from the wheels pushing to drive. I guess I missed that he meant braking, vs driving.

Trinopus
03-12-2012, 05:37 PM
Oops, yes, sorry, unclear: braking.

Again, for me, the eye-opener was that, unlike energy, momentum can't be converted into other forms. I can "throw away" lots of energy in the form of heat, but I can't do that with momentum.

(I guess I always thought it was just some kind of analogue of energy!)

The Niply Elder
03-14-2012, 07:05 AM
As an engineer, I have to kind of agree with westom. Somewhat.

Since the OP asked about colliding cars, it's not a simple thought experiment. Basic equations on conservation of momentum will only get you rough answers. If the OP meant to ask this as a thought experiment, then he should have used colliding billiard balls.

The key to the debacle here are secondary collisions. The only way to predict these is through simulation. YouTube LS-DYNA for some cool car crash simulations. The pros run thousands of simulations, testing many variables: different speeds, different masses, different car models, different collision angles, different bumper heights, different coefficients of friction, heck the list is long. All these results are aggregated in huge Monte Carlo simulations, and the specific shape and design of the structural components are driven to satisfy the greatest test cases. I read somewhere that Mercedes Benz test crashed 50000 cars or so. The amount of crash simulations must number in the hundreds of millions. And the real crashes are used to verify their simulation models.

Anyway, to summarize, from an engineering perspective, as posed by the OP, no: the two cases are not the same, and the 100kmh car will release more energy and be dissipated in a different way with a higher resulting structural damage.

Irishman
03-14-2012, 08:36 AM
Niply, it's all about assumptions. You're assuming "collision" = "real world car crash", which of course has to take in all those considerations of how each car is manufactured, and secondary results after the primary collision.

But the OP asked a basic physics 101 question that is answered with delta KE and conservation of momentum. Here is part of his post:

Pardon the ignorance. I really should know the answer to this, but it's been a long time.

That suggests he is trying to remember what he learned in school, not delve into a high fidelity crash study analysis to design a better automobile. So the assumption most of us have made is that is the answer he wants. I've posted links to those kinds of problems.

Cheesesteak
03-14-2012, 08:58 AM
The other issue is that the summarized version glosses over the fact that the Car vs. Car interactions are identical, and the differences occur in the Cars vs. Environment interaction that follows.

Cars vs. Environment is a totally undefined interaction, the cars can run into a curb or a tree, or go over a cliff, or get hit by a train, or avoid getting hit by a train. The possibilities are endless, we can make some guesses as to what would typically happen, but it's really a guess. Better to indicate what the situation is post Car vs Car interaction (where we know the exact answer) and state that we don't know what the additional damage, if any, will be.

The Niply Elder
03-14-2012, 11:23 PM
But the OP asked a basic physics 101 question that is answered with delta KE and conservation of momentum. Here is part of his post:


I'm not so sure. He proposed a scenario about colliding cars and finally asked which of two situations would "be more detrimental".

If the first test case was replaced with a red and green billiard balls, and the second case with a blue and yellow billiard ball, could you answer the question of "which one is more detrimental"? What would a postmortem forensic examination tell you? Nothing. There is nothing that you can conclude. Other than extreme simplifications of complex problems do not help you much.

santorum
03-15-2012, 08:51 AM
I'm not so sure. He proposed a scenario about colliding cars and finally asked which of two situations would "be more detrimental".

If the first test case was replaced with a red and green billiard balls, and the second case with a blue and yellow billiard ball, could you answer the question of "which one is more detrimental"? What would a postmortem forensic examination tell you? Nothing. There is nothing that you can conclude. Other than extreme simplifications of complex problems do not help you much.

I phrased the op rather poorly. About 20 years ago some kids in my rural hometown where playing chicken on the highway, and apparently they didn't set up ground rules and both 'chickened out' onto the same lane. I now accept that it's a frame of reference problem as most people here have mentioned (some with equations). I'll check out some of those crash test videos :)

Irishman
03-15-2012, 11:19 AM
If the first test case was replaced with a red and green billiard balls, and the second case with a blue and yellow billiard ball, could you answer the question of "which one is more detrimental"? What would a postmortem forensic examination tell you? Nothing. There is nothing that you can conclude. Other than extreme simplifications of complex problems do not help you much.

Billiard balls tend to have elastic collisions, so neither is very "detrimental". Unless you ramp the velocities up to 11, in which case the collisions stop being elastic. "Detriment" seems to require inelasticity.

It seems to me you're overparsing the word "detrimental". If you want to get into that level of detail, you're going to have to specify which make and model of car, and what angle they intersect, the orientation and position of each car, whether one car is empty and the other is loaded with exercise weights, etc, ad nauseum. There was nothing in the OP to suggest he wanted a run down of how much each car collapsed, or if the trunk was sturdier than the hood, or if the windshield cracked or not, etc. He asked a very generic question with limited constraints, suggesting that he was looking for the textbook situation.

The other issue is that the summarized version glosses over the fact that the Car vs. Car interactions are identical, and the differences occur in the Cars vs. Environment interaction that follows.

Exactly, thank you for coming up with this phrasing.