Does Newtons of an Egg drop vary accordingly to height?

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SUMMARY

The discussion centers on the physics of an egg drop, specifically analyzing the forces involved when an egg falls from varying heights. It is established that while the force acting on the egg during free fall is constant (F=mg), the impact force upon hitting the ground differs based on the height of the drop due to changes in momentum and kinetic energy. A drop from 2 meters results in greater velocity, momentum, and kinetic energy compared to a drop from 20 centimeters, leading to a higher likelihood of breakage. The discussion emphasizes that the time over which momentum changes significantly affects the force experienced by the egg upon impact.

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derfee
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According to my knowledge, I am able to apply Newton's second law on an egg for an egg drop; the resulted force from that can would be the force which occurs when it hits the ground. That force would also be reflected when it hits the ground according to Newton's third law.

What is confusing me is that the same force occurs to a 2m drop and a 2cm drop; one breaks and one doesn't yet the forces are still the same?

I do not understand.
 
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When the egg is falling to the Earth there is constant force acting on the egg and the Earth F=mg=const. But as they close together they pick the same momentum each P=mv in different directions so the whole momentum of the system stays 0. So the speed of the egg is increasing. The movement of the Earth is negligible.
If you drop the egg form height of 2 m it will have higher velocity than from 20 cm so bigger momentum and bigger kinetic energy.
When the egg hits the Earth the force with which the egg will act on the Earth will depend on the change of the momentum or in other words on how fast the speed will decrease over time ( F=dP/dt=m dv/dt ).
If the egg fall on something soft the time of change of the momentum will be bigger so the force will be smaller and maybe the egg won't break even from height of 2 m.
 
vlado_skopsko said:
when the egg is falling to the Earth there is constant force acting on the egg and the Earth f=mg=const. But as they close together they pick the same momentum each p=mv in different directions so the whole momentum of the system stays 0. So the speed of the egg is increasing. The movement of the Earth is negligible.
If you drop the egg form height of 2 m it will have higher velocity than from 20 cm so bigger momentum and bigger kinetic energy.
When the egg hits the Earth the force with which the egg will act on the Earth will depend on the change of the momentum or in other words on how fast the speed will decrease over time ( f=dp/dt=m dv/dt ).
If the egg fall on something soft the time of change of the momentum will be bigger so the force will be smaller and maybe the egg won't break even from height of 2 m.

thanks!
 
No problem, I hope I helped a bit :)
 

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