• HaraldvonBlauzahn@feddit.orgOP
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    3 days ago

    The thing is: You are using velocities v1, v2 which are relative to Earth. But none of the two vehicles collide with Earth - they collide with each other, thus the thing that matters is their relative speed, thus the difference of their velocities relative to Earth.

    (That’s also why the speed at which both Earth, the car, and the motorized bike move around the sun does not matter - relative speed is all what matters).

    The other thing is that a human colliding with an object of several tons weight with a speed of, say, 36 km/h is not “elastic”. 36 km/h is 10 meter per second, which is equal to about one second of free fall (accelerating with a= 9.81 meter per square second to the ground), which is equivalent to a fall height of h = a/2 * s ^2 or 5 meters.

    Somebody falling from 5 meters hight on hard concrete ground will not bounce up but will likely have some broken bones, or a broken skull. What happens is that all parts of thier body is decelerated to a speed of zero within a distance of one or two centimeters, which involves massive forces that easily break bones.

    And a speed of 14 m/s, or 54 km/h corresponds to a fall of ten meters depth - almost certainly lethal if hitting a two-ton concrete block.

    • Successful_Try543@discuss.tchncs.de
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      3 days ago

      The thing is: You are using velocities v1, v2 which are relative to Earth.

      The formula includes the relative speed (v_2 - v_1) of both bodies. Derivation, see Wikipedia or a book on engineering mechanics.

      Somebody falling from 5 meters hight on hard concrete ground will not bounce up but will likely have some broken bones, or a broken skull. What happens is that all parts of thier body is decelerated to a speed of zero within a distance of one or two centimeters, which involves massive forces that easily break bones.

      case 1, k=0. Fortunately, a car is not solid rock. I don’t know about a typical of k for collissions of humans with a car, but if you say it’s 0, that’s actually good for the biker, as the forces then acting on their tissues is smaller than if that would not the case.

      So concluding. If the collision of the biker and the car is completely inelastic, it doesn’t matter if the biker crashes into a resting car or the car crashes into a standing biker. The only thing that matters is the relative velocity of the two objects.