How to calculate dissipated energy of a ball bouncing back?
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There's seems to be some argument above about whether friction can be neglected (I note that the question does NOT say to neglect friction) and whether the gravitational potential energy can be depleted without touching the horizontal component of momentum.
Realistically, the ball probably loses a bit of speed and height with every bounce, and the process is impossible to calculate without more information about the ground-ball interaction. But we can still get a constraint on the time.
I see two limiting cases here:
1) The horizontal momentum is unaffected by bounces (maybe the ball is frictionless but "sticky"?). The ball bounces to a slightly lower height after each bounce, and finishes sliding horizontally along the ground with the same horizontal velocity it had initially until the end times. udiboy solved this in his answer, so I'll shamelessly steal his result, and call it the minimum time until bouncing stops:
Tmin=hig−−−√+∑1n2gEi−nmg−−−−−−−−√Tmin=hig+∑1n2gEi−nmg
2) The ball loses horizontal momentum with every bounce, but bounces to the same height every time, until it is out of horizontal momentum. Then it loses height with each bounce, until it is out of height, and finishes at rest. This isn't very realistic, but it is an upper bound on the time:
Tmax=Tmin+hi2g−−−√mv2iTmax=Tmin+hi2gmvi2
Taking gg to be 10ms−210ms−2 gives (provided I haven't botched computing the sum):
Tmin=472sTmin=472s
Tmax=1886s
I hope it helps you
Realistically, the ball probably loses a bit of speed and height with every bounce, and the process is impossible to calculate without more information about the ground-ball interaction. But we can still get a constraint on the time.
I see two limiting cases here:
1) The horizontal momentum is unaffected by bounces (maybe the ball is frictionless but "sticky"?). The ball bounces to a slightly lower height after each bounce, and finishes sliding horizontally along the ground with the same horizontal velocity it had initially until the end times. udiboy solved this in his answer, so I'll shamelessly steal his result, and call it the minimum time until bouncing stops:
Tmin=hig−−−√+∑1n2gEi−nmg−−−−−−−−√Tmin=hig+∑1n2gEi−nmg
2) The ball loses horizontal momentum with every bounce, but bounces to the same height every time, until it is out of horizontal momentum. Then it loses height with each bounce, until it is out of height, and finishes at rest. This isn't very realistic, but it is an upper bound on the time:
Tmax=Tmin+hi2g−−−√mv2iTmax=Tmin+hi2gmvi2
Taking gg to be 10ms−210ms−2 gives (provided I haven't botched computing the sum):
Tmin=472sTmin=472s
Tmax=1886s
I hope it helps you
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