five, take comfort in the fact that your observation was probably obvious to most readers from the get-go.
<sigh>. One more try…
It amazes me how many people post to this thread saying, “If you do the math…” without actually doing the math. To everyone out there: read a book; specifically, a book on physics. If you can’t find a book, find a physicist, and I don’t mean some frosh straight out of introductory mechanics.
Honestly, I originally believed that two cars moving at 35 mph in opposite directions colliding, and one car moving at 70 mph hitting a wall and stopping were the same. I read a book (“Physics for Scientists & Engineers”, third edition, Raymond A. Serway). I did the math (see previous post). I was wrong, they weren’t the same, and I posted the results. Few listened, apparently.
Someone said that ‘no speedometer has a speed rating for “-35 mph.”’ Duh. Read a book. Velocity is a vector, therefore direction is significant. A car “moving at -35 mph” is moving at 35 mph in the direction opposite to the admittedly arbitrary frame of reference. For any two cars moving directly towards each other, each going 35 mph, to do the math, you pick one to be going in the positive direction, and the other will by definition be moving at -35 mph.
The Newtonian theory of relativity (as distinguished from Einstein’s special and general theories of relativity) is that, in any two inertial frames of reference, the laws of physics are the same. Notice I said inertial. Accelerating frames of reference have different rules, hence the appearence of the centrifugal force for observers in a rotating frame of reference. To any other observer in an inertial frame, there is no centrifugal force acting on the principles; rather, there is centripetal acceleration acting on them.
People who set up the gedankin experiment need to be very careful to pick two inertial frames of reference when comparing the (-35, 35) collision to the (0, 70) collision. I’ll attempt to demonstrate.
In the first experiment, one car is moving at -35 mph, the other at 35 mph. They collide, and stop moving. Boom. For the second experiment, we want to see one car moving at 0 mph, and the other at 70 mph towards it. Essentially, we’re following the car going -35 mph in the first experiment. So, we’re going -35 mph relative to the first experiment.
The apparent 70 mph car approaches the 0 mph car and they collide, and stop. But, wait! They stop relative to the first experiment’s frame of reference. We’re going -35 mph relative to the first frame of reference. The cars stop in the first frame of reference, but to us, moving at -35 mph relative to that frame, the cars appear to be moving at 35 mph.
For the cars to stop moving in our frame of reference, either we’re describing two separate collisions, i.e., not (necessarily) equal, or our frame of reference must accelerate by 35 mph to be moving at 0 mph relative to the first frame of reference. Therefore, it’s not an inertial frame of reference, therefore, again we are describing two different collisions.
Proof via math:
35 mph = 15.65 m/s
Cars masses are unstated, but assumed to be equal. Call them each half a ton. So:
1000 lbs = 453.6 kg (at 1 g, i.e., standard Earth gravity).
In the SI system (meters, kilograms, seconds; the standard scientific measurement system, sometimes called MKS), energy is in units of joules (J), where 1 J = 1 kg * m^2 / s^2.
Regardless of the measurement system, kinetic energy is 1/2 * mass * velocity^2. So, in the first experiment, car #1’s, kinetic energy is 1/2 * 453.6 kg * 244.9 m^2/s^2, or 55543.3 kg * m^2/s^2, which conveniently is equal to 55543.3 J. The #2 car’s kinetic energy is equal to the first car’s kinetic energy, as (15.65)^2 is equal to (-15.65)^2. Therefore, in the initial system, there is a total energy of approximately 111.1 kJ, to 4 significant figures.
In the second experiment, car #1’s kinetic energy is 1/2 * 1000 lbs * (70 mph)^2 = 1/2 * 453.6 kg * 979.7 m^2 / s^2 = 222.2 kJ, while car #2, since it’s at rest, has a kinetic energy of 0 kJ, so the total energy is 222.2 kJ.
Now, for the original comparison, where, at the end, both cars are not moving, the final kinetic energy in each system is 0 kJ. Therefore, the amount of energy dissipated (via heat, light, sound, damage to the cars and/or occupants, etc.) in the first experiment is 111.1 kJ, while in the second, is 222.2 kJ. Notice that these numbers are not the same.
Notice also that in the first experiment, the initial momentum is 453.6 kg * 15.65 m/s + 453.6 kg * -15.65 m/s = 0 kg * m/s (sorry, there’s no SI base unit for momentum). Since everything is at rest at the end, the ending momentum is also 0. However, in the second experiment, the initial momentum is 453.6 kg * 31.3 m/s = 1420 kg * m/s. Note that, if everything stops at the end, the final momentum will be 0 kg * m/s.
Notice that 1420 is not equal to 0.
By the standard laws of physics, momentum in any closed system is conserved. Therefore, either there is an outside body acting on the principles (e.g., the non-moving car is attached very firmly to the ground), or the frame of reference is non-inertial. In either case, the two experiments are not equivalent.
I hope by now I have shown why the two different experiments are just that – different. If you believe you have found a flaw in my reasoning, by all means, post why. But do the math first, and show your work in your post. Just saying “it’s obvious” without doing the math is equivalent to those math books saying, “it is easy to see that this equation can be derived from the earlier one” when it’s not easy, because it can’t.
So, please, if you have anything more to contribute, either perform the experiments in real life a statistically significant amount of times and show your experimental data. Or do the math and show your work. The most important facet of a real scientist is their willingness to accept when they’re wrong. You can’t be taken seriously any other way.
Here is a quote from the info page on the NHTSA site:
I think the people who do the crash tests know what they are testing. It also says elsewhere on the site that it is the equivilant of a car traveling 70 MPH colliding with an identical parked car.
The link is http://www.nhtsa.dot.gov/ncap/infopage.html#crash
CKDextHavn
Five
You both are pulling conclusions out of thin air. Nowhere has anyone said that hitting a wall at 1 mph is the same as hitting a car at 35 mph, or that hitting a wall that’s at rest is the same as hitting a wall that’s coming towards you at 35 mph. Neither of these statements are logical consequences of anything that anyone has said, either.
CKDextHavn:
A car going 1 mph would exert 1/35 the force on a wall as a car going 35 mph, and therefore would have 1/35 the force returned to it. Hitting a wall at 1 mph is the same as two cars hitting each other, each going 1 mph. What part of this do you not understand?
Five:
The question of what a baseball does is completely different from what a car does. Baseballs bounce much more than cars. If you were using a non-elastic ball (NEB) imnstead of a baseball, then: a NEB with a velocity v hitting a stationary bat (assuming that the batter were able to hold the ball still, and have it not bounce back from the impact at all) would be the same as a NEB hitting a NEB with a velocity of -v. A NEB going v hitting another NEB going -v would clearly not be the same as a NEB going v hitting a bat going -v; a bat is much more massive than a ball, and therefore would exert more force than a ball.
MetallicAsh:
according to the OP (emphasis mine):
It is quite clear to me that “knucklehead” is saying that two cars at 35= one car at 70 + wall at zero.
Revedge
[quote]
Unless I missed one of the posts. You all are on the right track butt are forgeting the formula. E= 1/2m times v squared. A car mass 1 at speed of one hits a wall and releases enrgy of 1/2. Another car going in the oppisite direction adds its energy to the equation. Total release of evergy of 1. Two times the energy of the brick wall. How ever, if you hit the wall at double the speed (2 squared) you have a release of energy that is four times the initial conditions. So both arguments are right in a way. Two cars hitting head on at 35 mph is not as bad as hitting a brick wall at 70mph. However, it is twice as bad as hitting a brick wall at 35 mph.
[/quoteh]
No, you’re the on that’s forgetting something: if you hit another car, there is twice as much energy, but also twice as many cars. So the energy per car is the same.
Guys, I have equations, pictures, etc. in a Word document proving that Bigtrout is correct. Only two people (including Bigtrout) have asked me to email this to them. I put enough time and effort into it to where if more of you don’t email me a request for this document, I’m going to start sending out to y’all at random.
I don’t follow the math well enough, so I may be wrong about this, but it seems to me that Oldman is right in terms of point of reference.
Car A @ 35 MPH hits wall @ 0. Car a sustains 35 MPH worth of impact/damage/inertia/whatever.
Car B @ 35 MPH hits wall @ 0. Car a sustains 35 MPH worth of impact/damage/inertia/whatever.
Car A @ 35 MPH hits Car B @ 35 MPH. Each car sustains 35 MPH worth of impact/damage/inertia/whatever, but the total amount of impact/damage/inertia/whatever = 70 MPH worth (A+B).
If you are measuring from the point of view or either A or B, then the impacts are the same. If you are an outside observer, the total combined impact to both cars = a 70 MPH crash into a brick wall.
I think that both Bigtrout and Knucklehead are both correct, it just depends on who is measuring.
I think that your post illustrates another problem that people have with respect to this problem: the idea that mph is a valid unit for measuring the severity of a crash. Not all crashes at x mph are the same. 70 mph into a stopped car is very different from 70 mph into a brick wall. You can’t just say “oh, they’re both at 70 mph, therefore they’ll cause the same amount of damage.” Well, you can, but you shouldn’t.
I think that another problem is the idea that you can just switch around frame of references. Once you choose a frame of refernce, you’re committed to that frame of reference for all the calculations. If you work out the problem with the ground being at rest, you see that the car goes from 35 to 0 mph, for a change of 35 mph. If you look at it from the point of view of the other car, the original car does indeed start out at 70 mph but (and this is what many people seem to be missing) it ends up at 35 mph. Remember, if you consider the car to be traveling at 70 mph, then the road must be traveling at 35 mph. So if the car ends up traveling at the same velocity as the road, it must be traveling 35 mph as well. So it goes from 70 mph to 35 mph, and the net change is still 35 mph not 70 mph. In the case of the wall, however, the car goes from 70 mph to 0 mph, and change of 70 mph, not 35 mph as in the original case.
Darnit Wee Man, you were almost right until you said this:
"If you are measuring from the point of view or either A or B, then the impacts are the same. If you are an outside observer, the total combined impact to both cars = a 70 MPH crash into a brick wall. "
WRONG WRONG WRONG WRONG.
Cmon people!!! The total combined impact, Wee Man, is equal to a 70mph crash into a stationary vehicle, NOT A BRICK WALL!!! It doesnt matter if you are watching that crash from the ground, from another car, from a blimp traveling backwards at 35mph, from a train or from a car in the crash.
You guys are putting your emphasis on the spead the cars are going. The object they crash into matters!!! So it is like a 70mph crash into a stationary car because 70 + 0 equals 70. But when you say it is like crashing into a wall at 70, then you are wrong because the wall will hit that car as hard as another car going 70mph. 70 + 70 is 140mph. Hitting a wall at 70 is like hitting a parked car at 140mph.
::: OK, everyone who thinks Bigtrout is right, Say, “Bigtrout is right”. Everyone who thinks his friend is right, say. “Bigtrout is a knucklehead”:::
BIGTROUT IS RIGHT
Bigtrout is right
Thanks Bear-Nunno, I stand corrected about the brick wall. All I meant was that from each car’s perspective, 35 MPH worth of damage is done, so the cumulative result is a 70 MPH crash. In that respect, Knucklehead could be said to be correct.
But I think that, given the spirit of the question, and the context of the debate,
Bigtrout is Correct
Bigtrout is correct*
Strainger, you can e-mail me your write-up. I’d hate for someone to start criticizing it, and me not have a copy.
I’ll send a copy to you tomorrow, ZenBeam. I have it saved on my drive at work.
What kills me is that Telemark has provided a very concise anwer from a very reputable source, and he’s getting blown off left and right.
Strainger, thanks. I spent 30 minutes searching the web sites of the people who run the crash tests for the right quote, I’m glad someone noticed it.
Look folks, read the rationale and test results of the actual crash tests. The people who conduct the tests know exactly what they are testing. That’s why there is all the handwaving about don’t compare tests for cars of different weight classes! Hitting a wall simulates hitting an identical car head on at the exact same speed.
Now if you’re talking about the 40% offset frontal tests into a deformable barrier, the physics gets really interesting…
Strainger, if you email it to me, I can convert it to html and post it on the web for you, if you’d like. No offense, but I’m not really terribly interested in reading it, because I already know that Bigtrout is correct.
There is some difference between a crash at 35 mph into a wall and a crash at 35 mph into an oncoming car travelling at 35 mph. In the crash between a car and a wall, the wall does not dissipate any energy, and will absorb some of the energy dissipated in the crash, due to being heated up. It does not deform at all, and therefore does not absorb a substantial protion of the energy. The difference is negligible.
OK, now to correct some of the opposing side’s assertions (I have a feeling that this will be the next Monty Hall problem):
You are looking at the energy that affects the entire system here, when the problem deals with energy that affects your car. Each car absorbs 0.5m(35 mph)[sup]2[/sup] worth of energy. This is the same as the amount of energy absorbed by the wall and the car in the case of the car hitting the wall at 35 mph. Since the wall does not deform, and the car does, the energy is almost entirely absorbed by the car. Thus, it is the same as the energy absorbed by your car in the case of oncoming cars hitting.
Read luckie’s post. Somehow you seem unable to grasp the difference between hitting a car and hitting a wall. Watch a football (American style) game. Watch someone who is not moving much get tackled. Now imagine the tackler trying the same thing with the White Cliffs of Dover. They wouldn’t go down quite so easily, would they? Maybe you should try studying the physics behind the problem, rather than dismissing Newtonian mechanics out of hand.
NO!!! In the case of hitting an approaching car, the analagous situation would be the baseball hitting another baseball, travelling at the same speed. The swinging bat is like a wall that is approaching your car at 35 mph. This is not the same as a car approaching you at 35 mph. What would you rather hit in an equal-speed, head-on collision, a car identical to your own, or a train?
And I will again stress the point, hopefully for the last time, HITTING A CAR IS NOT THE SAME AS HITTING A WALL!!!
I was trying to find something from the NTSB as well, but Telemark seems to have had more luck. Hopefully, with the professionals and the laws of physics supporting our claims, we can convince the knuckleheads of their wrongness. I kinda doubt it, though.
Except that this actually has a right answer.
Hitting a stationary car at 70mph is not the same as hitting a wall at 35mph because the stationary car is fixed to the ground by the grip of its tyres.
The tyres will transfer ,eventually, all of the impact energy to the ground but over a longer period than the wall would.
In a friction free world then maybe the first statement would be true.
So the impact would be worse at 70mph but not double the impact with wall at 35mph.
This is also a crucial differance to consider when trying to compare with a head on with two 35mph cars.
First time poster to SDMB…I couldn’t resist this one. I’m going to have to agree with the people that say two cars hitting each other, both traveling at 35 mph is the equivalent to one car hitting a wall at 70 mph. (Granted, it’s not exactly the same, but lighten up; it’s close enough)
Let’s forget the complex equations for a minute, and instead, try a little science experiment. All you have to do is clap your hands together twice. The first time, hold one hand still and clap. The second time, clap both hands, and have them move approx. the same speed as each other, but each hand moving one half the speed as your hand on the first clap.
Now, if the force you felt on your hands was (or very close to) the same, then you have your answer. The same thing that just happened to your hands would happen to two cars, just on a much larger scale. If it didn’t feel the same, I suggest you try it again. =)
You are completely wrong, which you would have realized had you bothered to do the math necessary to determine what the forces and energies involved in the collisions were.
In both of these examples, we are assuming a perfectly inelastic collision, which basically means that when it occurs, all objects involved stick together, without bouncing apart.
We can make any assumption about the weight of the cars, as long as we make both cars in the two-car collision weigh the same. Let’s assume the cars’ weight is half a ton apiece, 1000 lbs.
35 mph is approximately 15.65 m/s, and 70 mph is 31.3 m/s. 1000 lbs, at standard Earth gravity, is 453.6 kg.
In the first example, one car is moving at 70 mph towards the other car. Its momentum is therefore 31.3 m/s * 453.6 kg, ~14200 (missed a 0 in my previous post) kg * m/s. The momentum after the collision will therefore be the same. Dividing 14200 kg * m/s by 907.2 kg (the mass of both cars together, since this is an inelastic collision) we get 15.65 m/s.
The initial kinetic energy of the system (again, 1/2 * mass * velocity^2) is 1/2 * 453.6 kg * (31.3 m/s)^2 + 1/2 * 453.6 kg * (0 m/s)^2, which is about 222.2 kJ. The final kinetic energy is 1/2 * 907.2 kg * (15.65 m/s)^2, about 111.1 kJ. The amount of energy dissipated is, then 111.1 kJ. A reasonable assumption is that both cars, being identical, deform just as much as each other, so each car is deformed by about 55.55 kJ.
In the second example, one car moving at 35 mph towards a static wall, has kinetic energy of 1/2 * 453.6 * (15.65 m/s)^2, which is about 55.55 kJ. Upon hitting the wall, it stops, (Mostly true. The wall is attached to the ground. Divide 14200 kg * m/s by the mass of the Earth, and you get a negligible velocity.) resulting in a final kinetic energy of 0 kJ, meaning that 55.55 kJ of energy is dissipated on the car, since we’ve said the wall doesn’t deform.
Hey, look at that, the amount of energy dissipated on the car is 55.55 kJ for both. What a lovely coincidence. That means they’re equivalent.
This has nothing to do with the energy involved in the collision. An object inelastically colliding with a initially-stationary identical object at a given velocity has the same amount of force applied to it as the object at half that velocity inelastically colliding with an immovable, undeformable wall.
The crucial thing is to double-check your sodding answer to make sure you don’t look like a fool in front of the whole world. The crucial thing is to give references or to provide the math necessary to show your arrogant statements have some basis in reality. Put up or shut up.
Look, the math says that a car moving at 35 mph hitting a fixed barrier takes as much damage as if it had hit an oncoming identical car moving at it at 35 mph. The NTSB, an agency that performs these kinds of experiments, says the same thing. What sort of evidence do you need to believe these things?
P.S.: switch to the unopened door.
Whith the slight sidenote that if these are racing cars, the wall is this nice soft foam rubber and tires, and will absorb a lot of energy…but then again, any racecar driver who loses control at 35 mph…
Besides, all this physics could be avoided by not hitting the bloddy things; saving “knucklehead” the pain of having it proven.