A. one car at 100 MPH crashes with a wall (perfecly rigid).
B. two identical cars crashes front to front each one at 100 MPH.
The question is:
Are these problems equivalent ?. The damage for the cars are the same ??.
Somebody says that in B, the damage for the cars are due to 200 MPH, and in the A are due to 100 MPH. Other says that is the same because, as is a simetric problem and then the other car is the same that a wall, and for instance as consecuence of energy conservation. in A 1/2 m v^2 has to be converted in deformation energy, heat, cinetic energy of flying parts, etc. In B the total energy is 2 x (1/2 m v^2) (twice), but each half is converted in deformation energy heat, etc for each car, so the damage is the same.
A and B are identical due to symmetry. Consider Case B first. Two cars identical in mass, geometry, and speed that are traveling toward each other on a straight line collide head on. Since each has the same kinetic energy at the moment of impact, neither car can push the other backwards. They just crush in together.
Now view this scene with a screen obstructing your view of one side of the collision. You see one car rushing toward the center (which is also the edge of the screen) and crashing into something and crushing. Because the screen obstructs your view of what the car collided with, you can't tell if the car you can see collided with a perfectly rigid wall (Case A) or an identical car (Case B). Since you can't tell the difference, the two cases must be identical.
hello, I think that you are right, except that the case cannot happen. One of the cars will act more or less as arigid wall, moving at 100MPH. This is also impossible, or nearly (car agains truck? I mean a fast truck) But the truth is in the middle somewhere, and one of the cars will be stronger than the other. m is also different for the two cars. When I crashed into a truck, I don't remember at what speed (because I don't remember anything) The truck could easily carry on its way, I was taken to a hospital and my car destroyed.beyond repair. I advise against such experimentation. yours robert hirschel
"David Duerr" a écrit dans le message de news: snipped-for-privacy@mb-m07.aol.com...
If the two vehicles in scenario #2 have identical masses and stiffness then the two scenarios are theoretically identical. In each scenario the vehicle(s) are subjected to a delta V (speed change) of 100 mph. It's the magnitude of the speed change that determines the damage.
(13 years experience as an accident reconstructionist)
Is the deceleration in either space or time the same in both cases? I'm not sure this is the case. The overall energy change may be so, but I doubt the rate of change is the same.
"BretCahill" escribió en el mensaje news: snipped-for-privacy@mb-m17.aol.com...
So if there are two identical cars and one is stationary and the other is moving at for instance 160 MPH, then the crash is equivalent to a crash of one car at 80 MPH with a perfectly rigid wall, Isn't it ???
Newton. He said any inertial frame will work. If you try a couple of them, and you get different results, you're doing it wrong.
The part about momentum is true in inertial frames. (But notice that if you introduce idealizations such as 'perfectly rigid wall', and attach to it what you're going to use as an inertial frame, you've assumed away conservation of momentum, and any conclusions you draw from its application will be incorrect, misleading or both.)
The part about energy is not. Only for certain frames. The phrase (indeed the idea of) 'energy in this collision' is poorly defined. It'll get you into trouble. Objects have kinetic energy relative to inertial frames. Collisions don't.
Remember back in undergrad mechanics when the professor would turn to the chalk board and start banging his head into it, repeating over and over in a soft voice through the sobs, "1.) Define the problem, 2.) State your assumptions, 3.) Define the frame of reference, 4.) Define the system . . ."? This is what he was talking about, and threads like this are what he was trying to save us from.
OK, OK. The OP's homework is long past due, so this won't hurt. Turn on your Courier font.
The model: Two identical vehicles are approaching one another on a straight, level road that runs due east/west. We're sitting in a grandstand to the south of the line, looking north to the point of impact directly in front of us. The vehicle, L, on the left is traveling east at a speed of V. The vehicle, R, on the right is traveling west at a speed of V. (Note that because we have not yet mounted a frame, we can't call them velocities.)
The vehicles have identical mass, M.
Purpose: Find the changes in vehicle and system energy and momentum due to a collision. Cast the problem in several different Newtonian frames and compare the results.
Assumptions: 1.) We're going to assume that the surface of the earth is a Newtonian frame. Specifically, we're going to ignore effects due to the rotation of the earth about its axis or about the sun.
2.) We're going to assume that the vehicles are identical, have the same mass, distributed identically, and have the same perfectly plastic structure that stays intact during a collision.
Frames: We're going to look at three frames of reference. All are Newtonian (non-accelerating). All have a linear dimension, x, with its positive direction pointing east (to our right.) Frame 1 is mounted on the surface of the earth. Frame 2 is attached to the vehicle on our left. Frame
- Frame 1 Vehicle L Vehicle R System Velocity before collision V -V - Momentum before collision MV -MV 0 K. Energy before collision (1/2)MV^2 (1/2)MV^2 MV^2
Velocity after collision 0 0 - Momentum after collision 0 0 0 K. Enerergy after collision 0 0 0
Change in Momentum -MV +MV 0 Change in Kinetic energy -(1/2)MV^2 -(1/2)MV^2 -MV^2
-- Frame 2 Vehicle L Vehicle R System Velocity before collision 0 -2*V - Momentum before collision 0 -2*MV -2*MV K. Energy before collision 0 (1/2)M(2*V)^2
2*MV^2
Velocity after collision -V -V - Momentum after collision -MV -MV -2*MV K. Enerergy after collision (1/2)MV^2 (1/2)MV^2 MV^2
Change in Momentum -MV +MV 0 Change in Kinetic energy (1/2)MV^2 -(3/2)MV^2 -MV^2
The calculated change in total kinetic energy is independent of the frame selected. If you insist on talking about something called the 'energy of collision' I suppose this change would be it.
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