Pardon the ignorance. I really should know the answer to this, but it’s been a long time.
Suppose we have two cars of equal mass:
Which one is more detrimental?
As far as I understand there should be no difference.
From the viewpoints of the cars, both collisions would be the same. Only context would differ: road, surroundings, etc.
How would the resistance of B’s parking brake (or B left in P) in scenario two compare to the residual driving momentum of B’s transmission/engine in the first scenario?
The answer is 2.
2 cars colliding head-on at 50 km/h ≠ a 100 km/h collision, it equals 2 50 km/h collisions, as in each car is experiencing a 50 km/h collision. It doesn’t matter that they are colliding with an object that happens to be moving toward it at 50 km/h, it’s still primarily the same as hitting a stationary object (there’s some variation as a parked car is not a brick wall but the basic physics are the same), because the moving cars are applying the same amount of force to each other. Thus, the 100 km/h collision will be more severe.
Actually, there shouldn’t be as much difference as you think because of the crumple zones. And, most likely, the car at rest will roll back, absorbing even more energy. So, the answer may well be 1.
The biggest difference between two cars hitting each other at 50 km/h and one car hitting a solid (unbreakable) brick wall at 100 km/h is that in the first case BOTH cars crumple to absorb the energy. In the second case, the first car alone absorbs all that energy.
But if the 100km/h car is hitting another car, it will, depending on the brakes of the car at rest, push the other car back somewhat and crumple both cars. A crude approximation would be an object hitting a spring.
The energies involved would be different (depending on the frame of reference), but I’m not seeing how the second collision is necessarily worse than the first, unless the 2nd car has a crappier crumple zone, has awesome brakes, and is backed up against a solid steel wall.
In both cases the crunple zones of both cars are going to absorb a lot of energy. In case 2 one car can move back wards some.
So case 2 is the winner by a lot.
Previous answers were from the car’s perspective. Nobody cares about cars in such collisions. Most concern is for what is inside each car.
Since the amount of damage varies significantly with where and at what angles each car collides, then the question becomes too hypothetical. Damage to the occupants is the only concern. Both cars are now scrap metal. Damage to each car is only relevant to energy is imparted inside.
You could argue that (2) is more severe because if you move the system in (1) into A’s frame, then A is at rest, and B is moving at very slightly less than 100mph because of the relativistic correction.
This is incorrect. To determine the severity of a collision, you look at the change in momentum, which is equivalent to the impulse. The force experienced is then equal to the impulse divided by the time interval over which the force is applied. If we assume that the time interval is the same in both collisions, all we need to look at then is the change in momentum.
Note that a collision between cars is an inelastic collision, so kinetic energy is not conserved.
Anyway, let’s make the following assumptions: the two cars involved have the same mass, and the collisions considered take the same amount of time.
In the first collision, the change in momentum for car A is m times the change in velocity, or m times 50 km/hr. (Note that the final velocity of both cars is zero, because of conservation of momentum and the fact that it is an inelastic collision.)
In the second case (a car traveling at 100 km/hr hitting a parked vehcicle), note that the velocity of the two cars immediately after the collision will be 50 km/hr (also due to conservation of momentum). So for car A, the change in momentum will be m times 50 km/hr. The two situations are identical.
As for car A running into an immovable object (like the proverbial brick wall), the equivalent situation would be hitting the brick wall at 50 km/hr. The final velocity for car A will be zero, so again the change in momentum for car A will be m times 50 km/hr.
Kinetic energy is 1/2 m u[sup]2[/sup]
Total kinetic energy in the first case is 2*(1/2 m 50[sup]2[/sup]) = 2500m
In the second case the energy is 1/2 m 100[sup]2[/sup] = 5000M
I think we need a reply from an actual physicist. Kinetic energy is given by the formula 1/2 mv[sup]2[/sup], so the collision in scenario 2 is a lot more energetic. What always confuses me is how this relates to frames of reference, as the calculated energy is different depending on where it is observed from.
Looking at the details of the Euro NCAP Frontal Impact test, a 64 kph collision into a barrier is used to simulate a car-to-car collision with both cars travelling at 55 kph. Presumably this is done because the energies involved are the same, which implies scenario 2 above would be a much more serious impact.
I’m pretty sure they did this on Mythbusters and #2 would cause more damage
WTF? How do you figure that? The only difference is how fast the scenery is moving relative to the cars. Two 50 km/h crashes certainly does sum to 100 km/h. Are you suggesting that in the 100/0 scenario, the stopped car is all hunky-dory because it experiences a “0 km/h” collision?
Imagine a car hitting you at 50 km/h. Now imagine if you decided to gun the accelerator before you were struck. Are you saying there’s no difference?!
How does moving “backwards”* change anything? It’s still absorbing energy. That’s why it’s moving backwards. If you’re moving at 50 km/h and all of a sudden, you’re moving at 0 km/h, guess what…you’ve been moved “backwards”.
*I put it in quotes because the idea that the scenery is a relevant reference point is silly.
P.S. Reading some of the other replies, there is some argument to be made that the brick wall collison may be somewhat more severe because there is only one car in which to crumple (instead of two). This may have the effect of shortening the time of collision (which changes one of my previous assumptions), thereby increasing the force experienced by the occupants of the vehicle.
This would not apply to the original two cases presented in the OP, however. They are still identical for the reasons stated in my previous post.
In the first case, all that kinetics energy is dissipated in the collision.
In the second case, immediate post crash when the two cars are traveling a 50 km/h, their kinetic energy will be (1/2)2m50[sup]2[/sup] = 2500m. Thus there will again be 2500m of kinetic energy dissipated during the collision.
If detriment is determined by energy, then Dog80 has struck the correct answer.
A 100 Km/hr car striking a parked car is energy equivalent to two cars hitting each other at 70.7 Km/hr.
(Speeding kills.)
I think if you look at the problem from a purely kinetic energy point of view case 2 would be worse off. We know E=1/2mv^2 assuming 2 vehicle have same mass we have total energy of E(1)=2(1/2mv^2)=mv^2. For case 2 we get E(2)=1/2m*(2v)^2=2m*v^2.
Though even system 2 have twice the kinetic energy it might have longer collision time so the occupant would experience less G force.
But it’s not. Detriment is determined by change in energy over time, i.e. impulse. Sure, one starts out more energetic in a certain reference frame, but it also ends more energetic in that same reference frame. The change is equivalent.
That’s because they’re using a barrier instead of a car, not because of the vector addition.
Again, unless you’re expecting the second car to come out immaculate, you have to average the energy. It’s the same energy per car.
In scenario 1, the delta-V is +50 for car 1 and -50 for car 2.
In scenario 2, the delta-V is -50 for car 1 and -50 for car 2.
Each car changes velocity the same amount. You’re either not averaging across the cars or you’re not considering the end scenario. It’s the same error either way, but you have to do one or the other.