Post by dg on Jun 24, 2010 11:31:00 GMT -5
Recently on the tv, I watched a program where they tested the claim that two cars crashing head on with each going 50 mph in opposite directions would receive no more damage than one car crashing into a rigid wall at 50 mph -- contrary to the common viewpoint that it should be like hitting the wall at the total relative speed of 100 mph. Indeed, they showed the claim valid by actual crash tests.
Actually, any college physics student could have saved the show a good deal of time and money. By using simple collision physics, one can show why this is true, and why the common viewpoint is incorrect. Consider three different events: 1) two cars head on at 50 mph each, 2) one car stationary with the other crashing into it at 100 mph and 3) a car crashing into a massive rigid wall at 50 mph. For all three cases, it will be assumed that the collisions are completely inelastic (i.e. they will remain stuck together after the collision).
1) For the head on collision, each car initially has kinetic energy E = .5 mV^2. Thus the precollision kinetic energy is 2E. Because momentum of the 2 cars is preserved, the velocity of the 2 car mass after collision is zero. Thus the after collision kinetic energy is also zero. This means that 2E of kinetic energy was converted into damage work on 2 cars, or 1E per car.
2) For the situation where one car is stationary and the other is moving at 2V, the initial kinetic energy (all in the moving car) is .5m(2V)^2 , which is 4E. Conservation of momentum has the after collision 2 car mass moving at half the impact speed which is V. Thus the after collision kinetic energy is .5(2m)V^2 which is 2E. Thus, as before, 2E of kinetic energy was converted into damage work of 2 cars, or 1E per car.
3) Hitting the massive rigid wall at V provides the initial kinetic energy of .5mV^2 or E. Conservation of momentum for an inelastic collision with an infinite mass makes final velocity zero. Thus final kinetic energy is again zero. This means that 1E of kinetic energy was converted into damage work on 1 car.
Since for all three cases, the damage work per vehicle was the same, the damage experienced would be the same. Thus one learns that hitting an equal mass car head on at a relative velocity of 2V is equivalent to hitting a massive rigid wall at merely 1V, in terms of the damage realized by the car. So any drivers manual that tells you different is clearly wrong.
Actually, any college physics student could have saved the show a good deal of time and money. By using simple collision physics, one can show why this is true, and why the common viewpoint is incorrect. Consider three different events: 1) two cars head on at 50 mph each, 2) one car stationary with the other crashing into it at 100 mph and 3) a car crashing into a massive rigid wall at 50 mph. For all three cases, it will be assumed that the collisions are completely inelastic (i.e. they will remain stuck together after the collision).
1) For the head on collision, each car initially has kinetic energy E = .5 mV^2. Thus the precollision kinetic energy is 2E. Because momentum of the 2 cars is preserved, the velocity of the 2 car mass after collision is zero. Thus the after collision kinetic energy is also zero. This means that 2E of kinetic energy was converted into damage work on 2 cars, or 1E per car.
2) For the situation where one car is stationary and the other is moving at 2V, the initial kinetic energy (all in the moving car) is .5m(2V)^2 , which is 4E. Conservation of momentum has the after collision 2 car mass moving at half the impact speed which is V. Thus the after collision kinetic energy is .5(2m)V^2 which is 2E. Thus, as before, 2E of kinetic energy was converted into damage work of 2 cars, or 1E per car.
3) Hitting the massive rigid wall at V provides the initial kinetic energy of .5mV^2 or E. Conservation of momentum for an inelastic collision with an infinite mass makes final velocity zero. Thus final kinetic energy is again zero. This means that 1E of kinetic energy was converted into damage work on 1 car.
Since for all three cases, the damage work per vehicle was the same, the damage experienced would be the same. Thus one learns that hitting an equal mass car head on at a relative velocity of 2V is equivalent to hitting a massive rigid wall at merely 1V, in terms of the damage realized by the car. So any drivers manual that tells you different is clearly wrong.
