Question

if two balls of the same mass drop on a surface and one sticks and the other bounces back up to it's original height.. which one had more force on impact and why is that? How do i relate this to momentum if two balls of the same mass drop on a surface and one sticks and the other bounces back up to it's original height.. which one had more force on impact and why is that? How do i relate this to momentum @Physics

Answer 1

the true formula for force is not F=ma, but is\[F={dp \over dt}\]This formula is more general than F=ma because it deals with objects of changing mass as well as constant.
Average force is then\[F_{avg}={\Delta p \over \Delta t}=m {\Delta v \over \Delta t}={m \over \Delta t}(v_f-v_i)\]for the bouncing (elastic) collision vf=-vi, so the average force is\[F_{avg}={2mv_i \over \Delta t}\](we, of course, ignore all negative signs here)
The force on the sticky (inelastic) collision is then\[ F_{avg}={mv_i \over \Delta t}\]So assuming the same time for impact, the force on the elastic collision is proportional to 2vi, which is twice that for the inelastic one.

Answer 2

I shouldn't say "true" formula. It's just more general than F=ma.