I must have missed something, because the only response I can think of right now is "What?"

OK,but what issue/question are you addressing?

Re: the front end collision question at the start ofth show weekend of 23may08

Full analysis, without math

The points in question are: change in momentum and the time over which the change occurs.

Ignoring the time issue (at these speeds, it won't be real relevant), the change in momentum for a given vehicle will be half as much for the 60-60 head on than the 120 into a SOLID wall, making this the preferable (preferable? how about neither of these....).

If the wall is NOT solid, then it depends how solid it is. Assume it is a stopped, identical car, rather than the proposed SOLID wall. After the hit at 120, the formerly stopped car will now be moving, and the fast car will be slowed. This is the same as the 60-60 collision. As the target gets heavier/more solid, the change in momentum of the fast car increases to the maximum at the case of the unmoving target, making for a harder hit.

Viewing the two-car collision from a frame of reference with one car at rest is a nice way of making the point.

Also, discussions about energy are somewhat interesting, but the significant effect on occupants of each car will be determined by the acceleration experienced by each part of each person's body during the impact. A rough indication of the relative effect of different impact speeds on the resulting acceleration can be calculated from the loss of momentum of the car and its initial mass. Thus, in an impact with an immovable wall, the occupants of a car at 120 mph will undergo roughly twice the acceleration as they would in an impact with an identical car, each going 60 mph. The acceleration will be the determining factor for physical damage to the occupants.

Absolutely the Geo, from a physics standpoint.

Yes, the total velocity difference is important, but mass is also a key factor here. A train hitting a car at a crossing hardly notices the event due to the mass difference. Although the difference between the pickup and car is not nearly as great, you probably have at least a 2:1 advantage here, negating any velocity difference.

You also have to take into account "moveability" of the object. The Geo is only attached to the earth by four (or less depending how you hit it) small patches of rubber. The wall is attached to the earth along its entire base. Hitting the Geo wins here.

In addition, the Geo's metal body should absorb more of your pickup's energy than a brick/mortar wall. Hitting the Geo wins here.

So yes, hit the car - it's a much better choice (at least for the physics reasons). From the spiritual and judicial liability of probably killing the Geo driver to save yourself by doing this... I guess that's for your ethics class. :-)

1. Your grade does not impact the basic physics in consideration.

2. The entire chat and your reply deal with veocites as you point out. It is true that VIEWED FROM THE FRAME OF REFERENCE of one car at rest, the other car is driving 120 mph. [HUGE ASIDE: This is exactly why engineering experiments can simulate a moving aircraft in a wind tunnel. From a physics point of view, an aircraft moving through air is an unsteady problem while the air moving past the aircraft is a steady problem.] You could consider the thought experiment that Einstein discusses in Relativity. If you have a train moving and you toss bags out of the door, these bags are also moving at the speed of the train. So they are moving at (say) 60 MPG in the same direction of the train relative to the solid ground (track). If a train is coming opposite at 60 MPH, it will impact the ejected bag. Relative to the fixed ground, it appears that the second train hit the bag going 120 mph.

3. The issue that the caller had in his head deals with the energy equation, not the vector sum of velocities. You allude to this as well but no one has followed through.

Both cars are travelling with a kinetic energy of 1/2 * mass * velocity * velocity. Assuming an ideal elastic collision (head on and both come to a dead stop), this energy is absorbed. (An inelastic collision is like pool where the balls do not absorb the energy and bounce off each other) For the cars to come to a dead stop at the point of collision, as you point out, they must have the asme mass. If one has a greater kinetic energy, the system of two cars will move in the direction of motion of the heavier car. Assume equal mass.

Each car must absorb some of the combined energy. Cars have crumple zones to do this without harming the passenger. If both cars absorb the energy of equal amounts, then each car absorbs (dissipates) 1/2 mass * velocity * velocity [joules] of energy. If the wall absorbs 1/2 * mass * velocity * velocity joules of energy then the collision is the same as the cars. If the wall does not absorb the same quantity of energy, then the car hitting the wall at 120 MPH must absorb more (or all of the energy). Thus a car hitting a wall can absorb more of the energy then two cars colliding and this energy gets dissipated by deformation of the metal. The problem that the caller was having was separating velocity from energy.

Conclusions: (Winners)

As to who is right or wrong is up to the caller and his father and wife. As long as people learn, everyone wins. If a judgment were to be rendered, the wording of the actual dispute must be considered carefully. From the impact (vector) point of view, the father is correct, the sum of the relative velocities makes the apparent velocity double. From the consequence point of view (energy considerations, potential damage,...), the caller is correct, you experience double the energy if you hit a wall at the equivalent (double) velocity. It is a matter of thinking about what question you are really asking. Everyone is right.

In any case Tom and Ray were wrong. :)

Assuming an ideal elastic collision (head on and both come to a dead stop), this energy is absorbed. (An inelastic collision is like pool where the balls do not absorb the energy and bounce off each other)

You've got it backwards. Pool balls colliding would be a nearly elastic collision -- all energy that goes into deformation is returned and the balls end up none the worse for the wear. Two cars colliding would be inelastic -- there is much deformation and it's permanent.

Agreed, yes don't be hasty and don't write replys while trying to listen to the rest of the show. :)

Speed Bump,
Above, you wrote "kinetic energy of 1/2 * mass * velocity * velocity", then you wrote "you experience double the energy if you hit a wall at the equivalent (double) velocity. This is not quite correct. You experience the same impact at 60mph hitting a wall or an equal car head on. We agree on that. The total energy absorbed (or converted to heat) in the head-on is double that of one car hitting the wall.
However, one car hitting the wall at 120mph has 4 times the energey of one car hitting a wall at 60mph and still double the total energy of both cars together in the 60mph head-on. When the speed doubles, the enrgy squares. The impact to your body is 4 times greater at 120 than at 60 in either case. Staggering!

Thanks. Ditto on the hasty.

Tom and Ray did flunk mechanics, yes? I side with the caller on this question, and think they got it wrong.

Two cars of equal mass hitting each other head-on produces exactly the same result as one car hitting a stationary wall at 60 mph. In both cases, from the initial point of contact to the rest position, there is no net displacement. That is equivalen to one car hitting an immovable stone wall only if that car is going 60 mph.

If that single car was going 120 mph, it would have 4 times the energy as if it was going 60 mph and would not be equivalent to the two-vehicle 60 mph case.

My physicist son explains it as a matter of using equivalent frames of reference to determine the energy involved. Two vehicles each going 60 mph have twice as much energy as a single vehicle of the same mass going 60 mph, but that energy is dissipated in two cars. A
single vehicle going 120 mph has four times the energy, all of which is dissipated in that one vehicle. Definitely not equivalent.

In any event, if I had the choice as to which situation I'd rather be in, the answer is easy. I'd rather hit the stone wall because fewer people would die!

Speed bump got it right. If the wall is capable of absorbing energy, then the single car could be going faster than the cars in the head on collsion, but since the argument assumes an immovable object, we assume that all kinetic energy is absorbed by the car. The statement that a car traveling twice as fast actually carries 4 times the energy is also correct, which means that a single car going 120 mph must dissipate twice the sum of the energy in a 60 mph head on collision. But in the head-on collision the cars split the sum of their kinetic energy, while a single vehicle must dissipate the entire energy of the 120 mph single car collision, which is 4 times what either of the head on cars must dissipate (assuming equal mass of all cars). Only in the example of a single car collision with an immovable rock wall where the vehicle is traveling 60 mph does the vehicle dissipate the same energy of either of the two vehicle in the head-on collision.

On an interesting note: the energy in an impact of a single car crashing into an immovable wall at roughly 85 mph is the same as the energy of a two car head on collision where each car is traveling at 60 mph (85 = 60 times the square root of 2 - the result of the kinetic equivalence equation). In this case, the single car collision still has to dissipate twice the energy of what either car in a head on collision must dissipate.

Energy is measured in joules, not in "Newtons of energy" or newtons. Work requires that force be applied over a distance; hence, one joule = one newton-meter (one joule of energy is equivalent to one newton of force acting over a distance of one meter).

It looks like Speed Bump updated his old response with your fix within a day of your catching it. Good for both of you.

The vectors for a head on collision point in opposite directions: 60 - 60 = 0. This makes sense as we should expect two cars of equal mass colliding with each other, at equal velocities, to come to a stop.

Similarly, assuming an elastic collision, we expect that a vehicle colliding with an identical stationary vehicle would come to a stop and transfer its kinetic energy to the stationary vehicle, causing it to travel in the opposite direction at the same speed as the original moving vehicle. In the case of an elastic collision with a brick wall, we would expect the car to bounce off in the opposite direction with the same velocity that it hit it (think of a superball bouncing back up to the same height from which you dropped it). In other words a vehicle colliding with an immovable wall is equivalent to a stationary vehicle being hit by a moving vehicle.

Vehicle collisions are, of course, inelastic so what we actually observe, in the case of a brick wall, is that the vehicle "absorbs" the transferred kinetic energy by crumpling in heap of metal in front of the wall (which "absorbs" very little of the impact energy -- in each case they are actually transforming kinetic energy into heat). Since two identical colliding vehicles cannot transfer any more energy than they already have, the cars will each "absorb" the same amount of damage as if they had both run into immovable brick walls. Ergo the caller was correct.

BTW, I'm sure I did not get an A in college physics...

Your A has withstood the test of time.

As I understood it, either 1 car hits an >identical< car head-on, each going 60 for a combined speed of 120 - or one car going 60 hits an immovable wall. Keeping it to the simple, basic question that was posed, without adding crumple zones and comparing different possible wall hardnesses, the answer is very simple.

Whether it hit a wall or the other car, the caller's car goes from 60 to 0 in the same fraction of a second. The consideration of crumple zones was outside the scope of the question and only confuses the question but it wouldn't matter. The caller's car would benefit from it's own crumpling when it hit the wall. If two cars hit they would each beneift from their own crumpling. The combined benefit of two crumple zones cannot be greater than the sum of their individual benefit.

"- or one car going 60 hits an immovable wall." This is not equivalent to two cars hitting head on.

If "X" marks the point of impact - either the face of the wall or where the two bumpers touch, point X will be the same before, during, and after the impact. Unless something causes point X to be pushed one way or the other during the event, the energy from the impact will be absorbed at the same rate in either circumstance. It would be different if a 1-ton car hit a 2-ton car, (in which case the 2-ton car would cream the 1-ton car). Now, point X itself would be a different matter. If it had some kind of impact meter, it would register twice the impact from the two cars as it would from the car vs wall because both cars would impart their force to it.

From another view, either one car goes from 60 to 0 or two cars both go from 60 to 0 (losing a total of 120 mph). Each loses its own 60 mph worth of energy in the same amount of time. If that were not the case, do you think that each car would lose 120 mph worth of energy for a total of 240 mph worth? Whether a car goes from 60 to 0 in a split second because it hit a wall or another car is irrelevant, as long as it stops on the same dime.

"Now, point X itself would be a different matter. If it had some kind of impact meter, it would register twice the impact from the two cars as it would from the car vs wall because both cars would impart their force to it."

Q.E.D. the cases are not the same.