3 ms·

Okay 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

by dools 2y ago

Okay 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 ago
In 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.

by dools 2y ago

kazinator 2y ago

In 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.