3 ms·
He has a reason to continue to disagree. In fact, if we assume a perfectly mirror-like collision between two identical cars going 50 mph in opposite directions,
by kazinator 2y ago
He has a reason to continue to disagree. In fact, if we assume a perfectly mirror-like collision between two identical cars going 50 mph in opposite directions, then each one experiences a brick wall collision at 50 mph. This is four times less severe than a 100 mph strike against an ideal wall (indestructible, infinitely rigid barrier with a vast inertial mass).
Animation frames in ASCII:
_____ . _____
/ \___ ____/ \
/ \ . / \
_____ . _____
/ \___ ____/ \
/ \./ \
_____ . _____
/ \___ ____/ \
/ | \
_____ . _____
/ \_/ \__/ \
/ | \
The dots at the center of the collision demarcate a virtual brick wall that either car experiences due to the exact and opposite momentum and deceleration of the other.
Each car crumples against the dotted line, not passing through it at all.
(In our idealized example with a perfectly mirror-like collision in which both cars crumple identically, it behaves like an ideal brick wall: the worst kind.)
Hitting an ideal brick wall at 50 is far less severe than doing so at 100.
If the collision is perfectly inelastic, then in the two car case, the energy dissipated is 2mv² where v is 50 mph. In in the one car case where it is going double (2v) it the energy is m(2v)² which is 4mv². We assume the brick wall has no kinetic energy to contribute. We also assume the brick wall is not damaged so the 4mv² is absorbed by one car. Whereas in the two car case 2mv² energy is absorbed by two cars, so mv² each.
Thus, from the point of view of one car, the 100 mph brick wall collision is 4 times more energetic than the 50 mph head-on collision. From that alone you can intuit that head-on is same as brick wall, since the damage scales 4 times with a doubling in velocity.
(Now yes, hitting an identical parked car at 100 mph would me more similar to the head-on collision. A parked car is not a brick wall. It will move in the direction of the collision in addition to crumpling. Books for drivers should probably not talk about either brick walls or parked cars, but just convey that kinetic energy follows a square law, quadrupling when speed doubles.)
- dools 2y agoOkay yep I get it now. The 4mv2 did it for me mathematically but I still couldn't get it intuitively until I imagined a car going 50mph rear ending a car going 40mph. If they wind up going the same speed as a result of the collision then the deceleration felt by the car in the back would only be 10mph if the car in front didn't change speed because it was like, an ocean liner. If the cars are the same size and the collision is perfect and all that then the car in front would gain 5mph and the car behind would lose 5mph. Since in a head on, they are both changing speed then the deceleration must be half the combined relative deceleration. The way I was imagining it was as if the car collides with another car going 50mph in a head on collision and the other car doesn't slow down due to the collision. So it appears that it is I who am confidently wrong about physics (although I still hold that Milton Friedman is confidently wrong about economics!).
- kazinator 2y agoIn a collision in space (no friction against the ground causing the pieces to stop), the combined center of mass of all the objects involved keeps moving at the same velocity at all times. Two approximately round space rocks of exactly equal size and density have a center of mass that is halfway between them. If one is moving 40 mph (according to the observer's frame of reference) and the other is catching up to it in exactly the same direction at 50 mph, then what it means is that the center of mass between the two rocks is moving at 45 mph. If the rocks collide in such a way that they clump together (inelastic collision) and no pieces fly off, the combined double rock will thereafter be moving at 45 mph. Vehicle collisions are more complicated because of cars skidding and rolling. In the milliseconds after the main impact though, those effects don't matter that much, I think. E.g. if you hit a parked car that is in neutral with no parking brake, versus one that has its transmission and brakes locked down, and this happens at 100 mph, I suspect those factors hardly make a difference.