Conservation of momentum when bouncing a ballby yoelhalb Tags: ball, bouncing, conservation, momentum 

#1
Apr2712, 05:59 PM

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Hi all.
I am learning physics and I have the following problem. Suppose there is a ball of mass m traveling with velocity v towards a wall at rest. The ball bounces back and it's now at velocity v. Since the wall is always at rest, it follows that the total momentum before the collision was mv, but after the collision it is mv, so how is the momentum conserved? It is clear that I am doing something wrong or missing something simple, but I can't spot what. Any help 



#2
Apr2712, 06:08 PM

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hi yoelhalb!
in this case, isn't there is a force keeping the wall at rest? 



#3
Apr2712, 06:32 PM

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Consider the proof of the conservation of momentum, it is based on that the force is the derivative of momentum, and that there equal and opposite forces, thus canceling out. But the wall force is there from the beginning and as such it should have no effect on the momentum? And in other words since the wall is at rest the net force must be zero, thus it has no effect here 



#4
Apr2712, 07:59 PM

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Conservation of momentum when bouncing a ball 



#5
Apr2712, 08:07 PM

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#6
Apr2812, 10:31 PM

P: 69

After all you are probably right, and although before and after the collision the net force on the wall is zero, still at the time of the collision we have to add the static friction force to keep it at rest.
As such the force that the ball exerts on the wall is canceled by the static friction force, and we are left with force that the wall exerts (the opposite reaction), thus the total forces do not cancel out, hence momentum is not conserved. 



#7
Apr2812, 10:45 PM

P: 69

But what triggers me is the fact that the change in the balls momentum is 2mv, and according to the second law this is the amount of force applied.
On the other hand the source that caused the force is the momentum of the ball (that resulted in an equal force from the wall on the ball). As such it turns out that a momentum of mv caused a force of 2mv, is this possible? Actually what is the push force that an object with a given momentum exerts? 



#8
Apr2912, 06:39 AM

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#9
Apr2912, 11:19 AM

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My problem is that this was caused by a momentum of mv! Suppose I have mass v and running with velocity v and momentum mv towards another object, is there a way to predicate the push force that I will exert on the object when I will push it? The reason that I asked it here is because I have seen that the momentum mv of the ball resulted in a force of 2mv. 



#10
Apr2912, 12:15 PM

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#11
Apr2912, 12:51 PM

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DaleSpam is correct  The ball moves towards the wall with momentum mv (kilogrammeters per second). It strikes the wall, and is in contact with the wall for, lets say, one tenth of a second. The average force it exerts on the wall is mv/10 (kilogram meters per second squared). By Newton's law, the wall exerts this same force on the ball. If the wall is rigidly connected to the earth, then the earth+wall system will gain momentum in the amount of 2mv, and the ball will lose momentum in the amount of 2mv, giving it a momentum of mv. Total momentum before was mv, total momentum after is 2mvmv=mv, so momentum is conserved. Adding a momentum of 2mv to the earthwall system is like adding a drop of water to the ocean  the mass of the earth is so large, that you cannot detect any change in the velocity of the earthwall system. But the ball itself will bounce back, its velocity is radically changed, since its mass is so much less than that of the earth.




#12
Apr2912, 01:41 PM

P: 69

And in other words the momentum caused an impulse twice as it's own. Consider for example that the ball pushed another ball with twice the mass as the first, and changed the velocity from rest to v, does this makes sense? But this is in effect an impulse of 2mv caused by a momentum of mv. And just as it doesn't make sense by the two balls, it doesn't make sense to me by the ball and the wall. Yet I understand that that's the fact, my question is if there is a logical reason for it. And as part of my question is if there is a logical reason to the fact that we cannot determine the force just by knowing the pushing objects momentum. 



#13
Apr2912, 02:20 PM

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#14
Apr2912, 03:06 PM

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I think you are still confusing force and momentum. If you push on an object with a gentle force of 1 newton for 10 seconds, it will have a momentum of 10 kgm/sec. If you push on that same object with a strong force of 100 newtons, but only for a tenth of a second, it will have the same momentum  10 kgm/sec.
Just because you know the momentum of something, doesn't mean you know the force that produced that momentum. It could be a gentle force over a long time, or a strong force for a short time. Its the same with an object that has momentum and transfers that momentum to another object. It transfers that momentum by exerting a force on that other object. But you don't know whether it did that by exerting a gentle force for a long time or a strong force for a short time. In other words, you just don't know what the force was when the momentum was transferred, even though you know what the final momentum is. 



#15
Apr2912, 07:44 PM

P: 69

Well I will have to settle my mind on that, after all this is probably the reality, although I still find it odd that a momentum of mv (and a collision just lasting a fraction of a second) will result in a change in momentum of 2mv (which is like saying that a momentum of mv caused an impulse of 2mv on another object in just an instant of time, something that all of you agree is impossible, and I don't really see the difference between another object and the same object such as a bouncing ball, unless the conversation of energy is used as a explanation), either way this is it.




#16
Apr2912, 08:04 PM

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Yoelhalb: consider the more realistic view that the ball is soft and springy. This avoids any infinite forces and allows us to analyse the stages of the process.
At first contact, there is no force (since no compression yet). As the ball continues against the wall, the force increases to a maximum, while the ball slows to a stop. Having stopped the ball, the momentum transferred is +mv to the wall, mv to the ball. Now the ball springs back from the wall. In practice, some energy will be lost, so the ball is only accelerated to speed w < v. Momentum transferred is mw, in same direction as before, so total transferred to the wall is +m(v+w). Now bring the thrower/catcher into the picture. The throw imparted momentum +mv to the ball, so mv to the thrower and thence to the wall/Earth/thrower system. On catching the returning ball, a further mw is so transferred. Net momentum change is 0. 



#17
Apr2912, 08:17 PM

P: 449

Hi.
consider a ball in a massive box in free space. When ball bounces box move pushed and has a too tiny velocity to observe. momentum conserves here. Another example when you pitch a ball you kick earth so earth moves with a tiny speed. Momentum conserves here. Regards. 



#18
Apr2912, 09:48 PM

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