Yes, they did do this on Mythbusters. Jamie had previously incorrectly stated that two vehicles going 50 would be the same as one going 100 into a brick wall. They tested it by having one car crash into a wall at 50, an identical one into a wall at 100 and two more identical cars hitting head on both going 50. They took numbers, but the visual was enough to show that the head-on collision looked a lot like the first example and not at all like the latter. The math behind why was well explained upthread.
I’d find the episode name but, well, the only thing harder to find than a specific Mythbusters myth involving car crashes is one involving explosions.
Doesn’t look like the same thing, actually. There’s a difference between a brick wall and a car. The brick wall crumples differently, for one. And a car isn’t made of bricks.
So, what happens when a car at 100mph crashes into a parked car?
Well, for what it’s worth, I used to teach physics.
All of these posts are incorrect, because you have to look at the change in kinetic energy dissipated in the collision. In the first case, the initial total KE of the two vehicles is 2500m. The final KE is zero, so the KE dissipated in the actual collision is 2500m.
In the second case, the initial KE of car A is 5000m, but the KE of the two mangled vehicles immediately after the collision is 2500m. (The rest of the KE will then be dissipated as friction as the two mangled vehicles slide to a stop). So the KE dissipated in the collision will again be 2500m.
Nitpick: Impulse is equal to the change in momentum of an object to which a force is applied.
And as I stated above, the detriment to the people in the vehicle would be due to the peak forces experienced by the occupants.
It’s not actually relevant in an inelastic collision to consider energy (including kinetic energy), because kinetic energy in an inelastic collision is not conserved. Instead, it is dissipated as friction, noise, and the crumpling up of the vehicles in question.
robby, I understand and agree with your sentiment but I think your math is flawed. First, KE has a vector, does it not? I understand it doesn’t have to, in an absolute sense, but in the context of summing energies, it has to. So in the first scenario, it’s really 2500m for the first car and -2500m for the second car, summing to zero. After collision, it’s still zero, but some negative KE and some positive KE has been converted to damage, friction, heat, etc.
In the second scenario, yes, we’d end with 2500m, but that’s still “little m”, i.e. a variable. But that little m is equal to 2 big-Ms, i.e. the fixed mass of an individual car. So we go from 5000M and 0M to 2500(2M), and energy is conserved. Of course, in the real world, we won’t come out of the collision at that “conserved energy” speed, and the difference will be the damage to the two cars.
So my point is, the scenarios are equal as far as we can tell without knowing exactly how the cars are built and what safety features they have. In practice it may make a difference, but in theory, there is no difference. The situations are the same.
No, the force is equal, too. I mean, at least in a scalar sense. A car going from rest to 50 km/h is the same as a car going from -50 km/h to 0. If it takes the same time and the cars have the same mass between scenarios (and why wouldn’t they?), and F=M*dV/t, then the force is the same, except maybe for the sign depending on how you draw the coordinate system.
Then you get half credit for stopping halfway through the problem. Yes, a car at 100 km/hr has four times the energy as a car at 50 km/hr. You don’t get to just stop right there, put down the pencil, and say you have an answer. Now find the energy post-collision and find the “energy dissipated,” which you admit is what is at issue here.
ETA: Also note that only one car is going at 100 km/hr. The other is at rest. If you intend to damage two cars, then you have to average the energy across both cars. If you consider only one car to be changing energy, then you can’t claim the second car would be damaged. It’d have to be built of unobtanium.
No, KE is not a vector. Consider a rubber ball fired at a brick wall. All of the KE is pointed at the wall prior to impact and most of it is pointed away from the wall afterwards.
humm Not sure what you trying to say. Are you saying in both scenario have the same amount of energy dissipated? I think it very clear that case 2 carries more energy. How the energy is dissipated is another question Of course I am don’t think the 2nd car will come out immaculate in either case. The real questions are in case 2 how much of that extra energy would be converted back to kinetic energy then come to a rest due to friction with road, and would the materials crumple zone material act the same way in both scenarios. As collision time would be different the material might handle the different intensity differently.
Energy dissipated is 1) energy of a car at 100 km/hr or 2) the energy of two cars at 50 km/hr. Conservation of energy. In both cases, all energy is absorbed equally both cars. But 1) means double the energy.
Again, I keep saying this. Otherwise junk science results. Define “detriment”. a) Detriment is defined by the acceleration of a brain inside the skull? Or b) defined by the total amount of energy dissipated?
b) is total energy dissipated in each car. a) defines detriment in terms of where energy dissipates throughout each car.
Why are so many answering a question before first defining the question? A major mistake. Define your interpretation of detriment in every answer. Otherwise any answer is possible.
Meanwhile, the problem is framed by one constant. The amount of energy in the collision. One 100 km/hr car means double energy dissipates in both cars.
Wow. And here I always liked physics because, unlike politics, there was a definite right answer for elementary problems.
The guys who say the two collisions are equivalent are right, and you shouldn’t even have to do any arithmetic to see it. You only have to remember that experiments done in inertial frames have identical outcomes (except for the infinitesimal relativistic effects that occur at such low speeds).
