Car Collisions and Gallilean Relativity ```Name: Dan H. Status: educator Age: 50s Location: N/A Country: N/A Date: 10/27/2003 ``` Question: I have never read a clear explanation of this apparent contradiction, and hope you can shed some light. Simple (Galilean) relativity tells us that 2 cars approaching at 50 mph are equivalent, in either's frames of reference, to one car at rest and the other approaching at 100 mph. But the total energy available, should they collide, is not equivalent. This is seen by comparing 2(0.5mv^2) when each of the cars is seen at 50 mph, to 0.5mv^2 with either at 100 mph. There seems to be more energy in the latter case by a factor of 2. So what is the answer to the question: Which is more dangerous? Two cars colliding at 50 mph, or one car, moving at 100 mph, colliding with another car at rest. Both cars, of course, have the same mass. Replies: The two situations are equivalent, at least in principle. Notice that after a collision with one car initially at rest, the two cars, locked together, are moving at a speed of v/2, by conservation of momentum: mv = 2mv/2 where m is the mass of either car and v is the initial speed of the moving car. They, of course, slow down rapidly due to friction with the ground. Note that the final kinetic energy of the two cars, locked together, before friction with the ground slows them down, is 0.5 (2m)(v/2)^2 = mv^2/4, just half the kinetic energy of the initial car. This means that the energy dissipated in the collision internally to the cars is also mv^2/4. If the two cars are each moving at speed v/2, their total kinetic energy, which is then all dissipated internally is 2*0.5*m(v/2)^2 = mv^2/4, exactly the same! This is utilized in all modern high energy particle accelerators. If a high energy proton collides with a stationary proton, most of the energy goes into the kinetic energy of the particles moving mostly in the direction of the initial proton to conserve momentum. Their kinetic energy can be much more than half the initial kinetic energy in a relativistic collision. Therefore we now use colliders to collide two protons (or other particles) head on so all the energy goes into probing the inner structure of the particles and is not "wasted" in the common downstream motion of the particles. Best, Dick Plano... Dan, Which is more dangerous actually depends more on friction than amount of energy. If the collision occurs on ice, there is essentially no difference. In the 50-50 collision, there is no motion after the collision. Kinetic energy is zero. In the 0-100 collision, the combined cars end up moving together at about 50mph. The "extra" energy is in the final kinetic energy. The drivers feel the same things in either case. In the 50-50 collision, all kinetic energy is used. In the 0-100 collision, only half is used. If the collision occurs on concrete, the 0-100 collision is more dangerous. The extra resistive force due to friction causes both cars to experience greater force. Consider hitting a punching bag free to move or a punching bag braced against a wall.In both collisions, all kinetic energy is used. The friction force depends on how the ground is moving. Since the ground moves differently in different reference frames, you get different outcomes in different frames. Dr. Ken Mellendorf Physics Professor Illinois Central College Click here to return to the Physics Archives

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