Tale of two cars (Physics question)

Factual Questions
141

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 E[sub]k[/sub] 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 ΔE[sub]k[/sub], because without Δ there will be no work done and E[sub]k[/sub] will most likely not be converted into another form of energy.

The fact that there is any remaining E[sub]k[/sub] 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, E[sub]k[/sub] is zero, and therfore ΔE[sub]k[/sub] 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, ΔE[sub]k[/sub] is < 100%.

142

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.

143

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.

144

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.

145

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 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.

146

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:

Post#32:

Post#41:

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.

147

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.

148

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.

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.

149

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.

150

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

151

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?

152

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.

153

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.

154

Ah, yes, applying the brakes works backwards from the wheels pushing to drive. I guess I missed that he meant braking, vs driving.

155

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!)

156

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.

157

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:

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.

158

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.

159

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.

160

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 :slight_smile: