DanielDaniels
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A ball falls to the ground from height H and bounces to height h. Momentum is conserved in the ball-earth system if:

Can I say that momentum will be conserved even if h>H or whatever h is actually is?
But will I have to consider that there is no gravity force acting?
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uberteknik
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(Original post by Daniel Atieh)
A ball falls to the ground from height H and bounces to height h. Momentum is conserved in the ball-earth system if:

Can I say that momentum will be conserved even if h>H or whatever h is actually is?
But will I have to consider that there is no gravity force acting?
Look up the definitions of elastic and inelastic collisions.

http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html
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DanielDaniels
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(Original post by uberteknik)
Look up the definitions of elastic and inelastic collisions.

http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html
Yeah I had a look at it.

Now lets say the ball hits the earth with a momentum mv and then bounces with momentum mv1 and the earth also recoils with m1v2. But the thing is that till what height will this ball go after the bounce? It can keep going forever if gravity was not taken into account. I am really confused and I know that I am mixing things up :/
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uberteknik
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(Original post by Daniel Atieh)
Yeah I had a look at it.

Now lets say the ball hits the earth with a momentum mv and then bounces with momentum mv1 and the earth also recoils with m1v2. But the thing is that till what height will this ball go after the bounce? It can keep going forever if gravity was not taken into account. I am really confused and I know that I am mixing things up :/
The ball will accelerate when dropped under the force of gravity.

Because the earth is effectively infinite mass in comparison to the ball, virtually all of the kinetic energy at impact will compress the ball with some energy loss due to friction, heat, sound etc.

The ball then accelerates upwards as the stored compression energy is released.

But some of the original KE is has been lost during compression so the ball cannot bounce back to it's original height.
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langlitz
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(Original post by Daniel Atieh)
Yeah I had a look at it.

Now lets say the ball hits the earth with a momentum mv and then bounces with momentum mv1 and the earth also recoils with m1v2. But the thing is that till what height will this ball go after the bounce? It can keep going forever if gravity was not taken into account. I am really confused and I know that I am mixing things up :/
When you work out v1 then you can work out its maximum height using various methods: for example you could work out the kinetic energy of the ball using E=1/2mv^2 then equate this with the potential energy formula to find the height.
1/2mv1^2=mgh
h= v1^2/(2g)
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DanielDaniels
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(Original post by uberteknik)
The ball will accelerate when dropped under the force of gravity.

Because the earth is effectively infinite mass in comparison to the ball, virtually all of the kinetic energy at impact will compress the ball with some energy loss due to friction, heat, sound etc.

The ball then accelerates upwards as the stored compression energy is released.

But some of the original KE is has been lost during compression so the ball cannot bounce back to it's original height.
aha thank you.

Now if someone asks me how is momentum conserved in the ball-earth system?
If I replied in this way:
First of all total momentum is zero provided no external force acts on the system. Now initial momentum before impact is just the balls momentum and after impact, some momentum will be given to the earth but, as you said, due to its infinite mass, it's not noticeable and the rest is given to the ball. But the problem is that, shall I consider gravity as an external force? and is what I said the right physics?

I understood it in terms of energy.
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DanielDaniels
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(Original post by langlitz)
When you work out v1 then you can work out its maximum height using various methods: for example you could work out the kinetic energy of the ball using E=1/2mv^2 then equate this with the potential energy formula to find the height.
1/2mv1^2=mgh
h= v1^2/(2g)
Image
Yeah I got you.
See this question. The answer is that whatever height it goes doesn't matter.

But I want to know how can it be possible to be greater than initial height? This can be true if it's in space maybe?
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langlitz
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(Original post by Daniel Atieh)
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Yeah I got you.
See this question. The answer is that whatever height it goes doesn't matter.

But I want to know how can it be possible to be greater than initial height? This can be true if it's in space maybe?
Unless you impart an impulse on to it then it can never go above its initial height. Even in space. In this case it is dropped, therefore it will be the same height or less depending on how efficient the ball is at bouncing. Remember that momentum is ALWAYS conserved (at low speeds).
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DanielDaniels
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(Original post by langlitz)
Unless you impart an impulse on to it then it can never go above its initial height. Even in space. In this case it is dropped, therefore it will be the same height or less depending on how efficient the ball is at bouncing. Remember that momentum is ALWAYS conserved (at low speeds).
aha that's clear. Many thanks!

But how it can't go greater than it's initial height after bouncing if there was no gravity and no air resistance?
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langlitz
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(Original post by Daniel Atieh)
aha that's clear. Many thanks!

But how it can't go greater than it's initial height after bouncing if there was no gravity and no air resistance?
If there was no gravity and you dropped it then it would not fall to the earth, it would float. I see your logic though
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DanielDaniels
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(Original post by langlitz)
If there was no gravity and you dropped it then it would not fall to the earth, it would float. I see your logic though
aha. Thank you, man!
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