Calling All Physics Majors - Energy transfer Question

Factual Questions
1

I have a question that I hope some of our Phyiscs people can answer.

My wife was involved in an accident yesterday. She was stopped at a red light and someone rear-ended a car and knocked that car into the car in front of them and then knocked that car into my wife.

So, which car had the most force (or whatever) acted upon it. Obviously damage can vary due to type of vehicle, weight, age, etc. But I will pose the assumptions and somebody help me with the proof. I have my thoughts on the matter, but can someone who can actually apply the formulas help me with the real answer.

Let’s number the cars 1 to 4 with 1 being the car that caused the accident and 4 being my wife’s car. Let’s assume that since the speed limit on the road was 45 that the car had slowed down to 30 before impact. I do not know the speed and maybe someone has a better estimate, but I think 30 is reasonable. Let’s also say everybody is driving standard mid-size cars. The first car actually was some big old car like a Crown Victoria or something. But I believe the rest were mid-size sedans. Let’s also assume that the distance between each car is 2 feet and that everyone has their foot on the brakes.

My thoughts are even in a system like the steel balls that bounce off one another by transferring energy that there is energy lost in the transfer, so the end ball does not get hit as hard as the balls in the middle. When you add in the factors of some small distance between cars and that the brakes were applied (all were stopped except car 1) and the fact that most cars today have crumple zones to absorb the impact, then it would seem to me that either car 1 or car 2 had the most force exerted upon it, but is this true? Is my logic flawed?

This is bugging me. Help me out you Physics gurus.

Jeffery

2

With both the steel “Newton” balls and the cars in the situation you describe, we’re talking about elastic collisions. With an elastic collision, the initial impact causes a deformation of the structure of the bodies, transferring the kinetic energy of motion to the energy of the connection between the molecules of the solid. Because the molecules “wish” to return to their lower energy state, they snap back.

Elasticity, especially in steel, is very efficient. In steel, the energy of the structure of the molecules is very near a local minimum. Therefore, any energy pumped in by a collision will be almost completely returned as the molecules return to their original state. When two very elastic objects of equal mass collide with the vector of motion passing through the centers of both objects,the elasticity of the objects transfer the momentum completely from one object to another. This process is extremely efficient; it generates very little heat. With the Newton balls, almost all the energy is lost to air resistance.

The steel frames of of cars are also very elastic. Even the “crumple zones”, which are designed to transform kinetic energy to a permanent deformation of the material, are very elastic at low forces. If this were not so, your car would permanently deform just due to air resistance when you’re driving and people leaning on your car. Without some opposing force (usually your own car’s momentum in a head-on collision), the car’s body “bounces” in the usual elastic manner. Without this opposing force most of the energy will be transmitted through the elasticity of the frame even at relatively high speeds; very little will be converted by the crumple zones.

Brakes are very good at using friction to convert the kinetic energy of the motion of the wheels to heat; this transfer is what makes your brakes stop. However, when the brake drum or disk is not moving relative to the brake pad, the brake itself becomes elastic; it stores and restores energy rather than converting it to heat. Tires are not so good at converting motion to heat. Thus you can stop faster by keeping the wheels moving and using the brakes than you can by just locking up the wheels.

All things considered, quite a lot of the kinetic energy of the first car will be tranferred to the first car. The brakes will not remove any energy since they’re usually just locked. Mostly the sliding of the tires (the spaces between the cars give the tires a little time to transform some of the kinetic energy to heat) and any crumpling done will remove the energy. Still, the impact on the last car, as you have seen, can still be quite severe. It’s really not possible to calculate the actual impact velocities of your specific situation without reference material, mostly the coefficient of friction of a generic tire on a generic road and the energy absorbtion behavior of typical crumple zones.

3

Now, I remember why I majored in Computer Science instead of Physics. I understand most of the general concepts, but dang it if I understood what you just explained.

Could you make some general guesses and arrive at a semi accuate number. I think the number would make your answer make more sense. Or could you make up some numbers and demonstrate what you are describing?

Thanks,

Jeffery

4

Not being an automotive/materials engineer, I can’t give you any exact answers. Generally, though, the system can lose (transform) measurable amounts kinetic energy only through crumpling of the sheet metal shell (minimal) and tire friction over a short distance. I would WAG that only about 10-20% of the energy would be lost by these means (most of that due to tire friction) and thus car #4 would receive between 73% (0.90^3) and 53% of the original car’s kinetic energy; essentially the equivalent of a 22-16 mph.

5

I’m afraid Feynman has made a few assumptions that are not that reasonable, which might explain your confusion. First or all, auto collisions are most certainly highly inelastic. Any collision in which one or the other of the bodies undergoes permanent deformation is inelastic by definition. Crumple zones are designed, in part, to reduce elastic bounce, and here’s why…

While energy may or may not be conserved in a collision of free bodies, momentum is always conserved. Momentum is the product of mass and velocity. Without going into a dissertation on collision dynamics, the ideal situation in an auto collision is if the two cars stick together. In your case, with one car stopped and the other moving (let’s neglect the brakes for a moment), and both of equal mass, if they stick together on collision, they end up moving with half the speed that the original moving car had. That means that the change in momentum (impulse) for each car is one-half the total initial momentum. In the completely elastic case, the moving car ends up stopped, and the stopped car ends up moving with the speed of the moving car. That means that both cars received an impulse equal to the entire original momentum. Which situation would you rather be in if you were in one of the cars-the elastic or inelastic case?

Now, as to your original situation. Let’s start by assuming that no one has their brakes on, and the crumple zones all work to force the cars to stick together at each impact, and that the initial momentum is P:

When car 1 hits car 2:
car 1 receives an impulse of P/2
car 2 receives an impulse of P/2

When the car 1/car 2 combination hits car 3:
car 3 receives an impulse of P/3
car 1 receives an impulse of (P/3)/2 = P/6
car 2 receives an impulse of (P/3)/2 = P/6

When the car 1/2/3 combination hits car 4:
car 4 receives an impulse of P/4
cars 1, 2, & 3 receive impulses of (P/4)/3 = P/12

Obviously, cumulatively, car 1 receives the most total magnitude of impulse (and this is what is key in assessing human and auto damage), and car 4 receives the least.

Now, obviously, the collisions are not completely inelastic, but I bet they’re a lot closer to that than to being elastic. Also, momentum is not completely conserved by the cars, since the tires transfer some of it to the ground, but that only further reduces the impact on the last car.

Was that any more understandable?

Rick

6

This statement is incorrect. Both energy and momentum are completely conserved. However, when considering the motions of macroscopic bodies, some momentum and energy are transfered to the motions of individual molecules as heat or to the higher-energy configurations of matter (crumpling). FTR: Momentum and velocity are vector quantities; you must take into account changes of directions as well as changes of scalar speeds.

This is true. But cars aren’t sticky, they’re bouncy. The crumple zones are not designed to make the car sticky, they’re designed to absorb energy by transforming kinetic energy into the deformation of their materials. Once they’ve been deformed, though, they transmit the force to the rigid (and thus elastic) frame and safety cage of the car (which must, for obvious reasons, retain its shape even under very high forces). If you carefully watch a crash test, you will see cars bouncing off each other (or bouncing off walls) in a crash.

Even the crumple zones are designed to be elastic under relatively low forces; you don’t want the body of your car permanently deformed by the forces of wind resistance or someone leaning against it. When the forces are not life-threatening, the preference is to retain the shape of the car and avoid expensive repairs.

It is difficult to specifically ascertain the contributions to crumple zones without detailed engineering specifications. Regardless, it seems more accurate to characterize auto collisions as being relatively elastic, especially at low speeds. However, if you want definitive engineering answers, you would be better to consult an automotive safety enginner; they probably could calculate your answer to the dyne and newton. In the meantime, it is probably not too ironic to note that your mileage may vary. :wink:

7

With regard to energy conservation, I noticed after I posted that what I should have said was that the center of mass kinetic energy of the colliding objects is not necessarily conserved. Of course, overall energy is conserved. But we’re talking about the kinetic energy of the center of mass of each car. Quite a bit of that can be lost in plastic deformation (crushing).

Yes, cars are designed to be elastic at very low speeds (bumpers are supposed to absorb a 5 mph crash, IIRC). But at 30 mph, we’re way into the inelastic zone.

I’ll concede that the collisions are partially elastic (or partially inelastic–is that glass half-full or half-empty?), but I’m guessing the losses are quite a bit higher than you are estimating. I could be wrong, of course :).

I’ll see if I can dig up anything specific over the weekend.

Rick

8

Quoth Feynman:

Exactly, and as RickG pointed out, the best possible way to transform as much of the energy as possible is to make it, effectively, “sticky”. Of course it doesn’t permanently deform from air resistance et al. That’s orders of magnitude less force than you get in a 30 mph collision with other cars.
I note, too, that we’re all agreed that car #1 had the greatest impulse, and #4 the least-- All that we’re arguing about is how much greater it is.

9

We’ve probably gone about as far as we can go on general principles. A more precise answer to the question would require either specialized engineering knowlege or actual experiments. Since I know none of the former, and the latter would be somewhat expensive and dangerous, we may never know for sure.

10

It is know that there is a infinte number of worlds, Simple because there is a infinte amount of space for them to be in. how ever, Not every one of them is inhabited. There for, There most be a finite number of inhabited nuber of inhabited nuber of worlds. Any finite number divided by infintey is as near to nothing as makes no odds, so the average population of all the plantes in the univers can be said to be zero. From this it follows that the population of the hole univers is also zero, And that any people you many meet from time to time are the producks of a deranged imagination. :wink:

There are test every day you fail you get slow then you die
Billy The Kid

11

Um, Billy, perhaps you meant to post that in the thread Should Life Exist? thread? And if I’m not mistaken, that’s from one of the Hitchhiker books, right?

12

Yes it is from One of the hitchhiker books I thought it would be funny.

There are test every day you fail you get slow then you die.
Billy The Kid

13

Thanks for all the help guys. So, the logical thought that #1 would get the greatest and #4 would get the least does hold.

How about, if #4 stopped suddenly and then #3 slammed into it and then #2 slammed into #3 and then #1 slammed into #2, then I would guess that #4 did get the most because then #4 absorbed the impact from #3 plus a portion of #2 and #1, while the others only got impact from the ones behind them.

Jeffery

14

Let’s work it out, using the original assumptions I made for totally inelastic collisions (essentially, treat all the cars as equal mass, freely moving bodies). Also, assume that the initial condition is that car #4 is stopped, and the other cars are moving with identical velocities (with momentum P), and they don’t apply the brakes before slamming into the cars in front of them. Also note that, once again, I’m only going to give magnitudes of impulse, not directions.

Car 3 hits car 4 (same as 1 hitting 2 in the original problem):

Car 4 gets an impulse of P/2
Car 3 gets an impulse of P/2

Now, car 2 slams into a mass equal to twice it’s own, but with total momentum P, so the final momentum of the 3 cars after this collision is 2P, but divided over 3 cars, so the momentum of each afterwards is 2P/3. Therefore, Cars 3 and 4 must have received impulses of P/6, and car 2 received impulse P/3.

For car 1, the initial momentum is P, and the final momentum is 3P/4, so it received an impulse of P/4, while 2, 3, and 4 received impulses of P/12 each.

Thus, the sum of the magnitudes of the impulses for each car is:

Car 1: P/4
Car 2: P/3 + P/12 = 5P/12
Car 3: P/2 + P/6 + P/12 = 3P/4
Car 4: P/2 + P/6 + P/12 = 3P/4

So the cumulative impulse magnitude is the same for the two cars in front, although for car 4, all the impulses are in one direction, while for 3, the largest component is backward, followed by two smaller jerks forward. Not sure which one is more damaging–probably the latter.

Of course, brakes and the partial elasticity of the collisions make some difference here. I’m guessing that car number 3 might get the worst of it in a real collision.

Rick