Does a free falling ball drops on Earth violate this theory?

AI Thread Summary
The discussion centers on whether a free-falling ball's behavior on Earth violates momentum conservation principles. It asserts that in an isolated system, the total momentum of the ball and Earth remains unchanged during a collision. While the ball may bounce back with the same speed, the momentum transferred to Earth is significant, albeit resulting in a negligible change in its velocity due to its mass. The conversation highlights that momentum and energy conservation must consider all objects involved, including the Earth's response to the ball's impact. Ultimately, the ball's interaction with Earth illustrates the complexities of momentum conservation in real-world scenarios.
rajivrout
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Does a free falling ball drops on Earth violate this theory?

If we consider the ball and Earth in an isolated system then the ball should continue to drop over time.

If the impulse given is extremely negligible that it doesn't cause any significant change in the path of earth, then the whole momentum must be conserved by the ball and it should bounce back with the same velocity.

Which is the missing point?
 
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The total momentum of the 'ball + earth' system remains unchanged. Whether the ball bounces back with the same speed depends on the nature of the collision, not on momentum conservation. Regardless of the speed of the bouncing ball, the total momentum of the system remains unchanged.
 
The change of momentum of the Earth isn't exactly negligible. It is exactly the same change in momentum as for the ball. However, you are right that the change in momentum of the Earth corresponds to a negligible change in velocity of the Earth due to its large mass.
Usually a ball won't bounce back with the full velocity, since it will transfer momentum to particles in the Earth (which will scatter in all directions). This could create a tiny shock wave in the earth, but again due to the spread it is not detectable.
 
@pf mentor
My point is m(e)*v(e)+m(b)*v(b) must be unchanged after colission. Since m(e)*v(e) is equal to zero and there is no (negligible) change in m(b), the final velocity before colission must be equal to the initial velocity after colission. If the initial velocity while bouncing is same as the final velocity at the time of drop then there can not be any change in displacement. Yes if the material is more elastic then it will help the object to restore the kinectic energy and it will bounce back better, but let's talk abt only momentum.
@gerenuk
Do you think the momentum transferred to the particles should be considered? Why can't we just consider these simply as two objects? Let's forget Earth consider a big stone.
 
rajivrout said:
@gerenuk
Do you think the momentum transferred to the particles should be considered? Why can't we just consider these simply as two objects? Let's forget Earth consider a big stone.
I'm not sure what you want to point out. The Earth is a big object only?
Maybe I make another comment. Momentum and energy is always perfectly conserved if only you take all objects into account.

You are right if you mean that, provided the Earth does not absorb any energy and does not deform, then energy and momentum conservation will dictate that the ball will have the same magnitude of velocity when it reaches a point that it was at before.
 
rajivrout said:
@pf mentor
My point is m(e)*v(e)+m(b)*v(b) must be unchanged after colission.
That's true.
Since m(e)*v(e) is equal to zero and there is no (negligible) change in m(b), the final velocity before colission must be equal to the initial velocity after colission.
Not true. Don't confuse having the same speed with having the same velocity. If the ball bounces back with the same speed but in the opposite direction, its momentum is not the same.
If the initial velocity while bouncing is same as the final velocity at the time of drop then there can not be any change in displacement. Yes if the material is more elastic then it will help the object to restore the kinectic energy and it will bounce back better, but let's talk abt only momentum.
Let's work out the details:
Initial velocity of ball: - v(b) (it's moving towards the earth)
Initial momentum of ball: -m(b)v(b)
Initial velocity of earth: 0 (assume it's zero)
Initial momentum of earth: 0

Final velocity of ball: + v(b) (assume it bounces back with same speed, so now it's moving away from earth)
Final momentum of ball: +m(b)v(b)

Change in momentum of ball = final - initial = +m(b)v(b) - -m(b)v(b) = +2m(b)v(b)

Thus momentum conservation allows us to conclude that the final momentum of the Earth is now -2m(b)v(b). The Earth is now moving away from the ball. The speed of the Earth might be tiny, but you cannot neglect its momentum.
 
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