# Conservation of momentum when bouncing a ball

by yoelhalb
Tags: ball, bouncing, conservation, momentum
 P: 69 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
PF Patron
HW Helper
Thanks
P: 25,505
hi yoelhalb!
 Quote by yoelhalb 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?
conservation of momentum only applies if there are no external forces …

in this case, isn't there is a force keeping the wall at rest?
P: 69
 Quote by tiny-tim hi yoelhalb! conservation of momentum only applies if there are no external forces … in this case, isn't there is a force keeping the wall at rest?
Good idea, but I really doubt if this is considered an external force.
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

HW Helper
Thanks
P: 7,923

## Conservation of momentum when bouncing a ball

 Quote by yoelhalb Good idea, but I really doubt if this is considered an external force.
You can take it as an external force or not. If not then you must include what happens to the wall. The wall is attached to the ground - another external or internal force - until you get to include the whole Earth. And yes, the whole Earth will feel the impact of the ball and change momentum, but it would be very hard to detect. The change to the Earth's momentum will exactly compensate for other associated actions: your throwing the ball in the first place, and how the ball eventually comes to rest.

 But the wall force is there from the beginning and as such it should have no effect on the momentum?
Eh? Why would the wall be exerting a horizontal force before the ball strikes it?
Mentor
P: 15,601
 Quote by yoelhalb I really doubt if this is considered an external force.
What is considered internal or external depends on what you have defined to be your system, and that is a completely arbitrary choice. If your system is the ball and wall then the force of the earth on the wall is indeed an external force.
 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.
 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?
Mentor
P: 15,601
 Quote by yoelhalb As such it turns out that a momentum of mv caused a force of 2mv, is this possible?
2mv is an impulse, not a force. It has different units than force.

 Quote by yoelhalb Actually what is the push force that an object with a given momentum exerts?
It can be anything from just larger than 0 to just smaller than infinity, depending inversely on the amount of time that the force is applied and depending proportionally on the change in momentum.
P: 69
 Quote by DaleSpam 2mv is an impulse, not a force. It has different units than force.
If my understanding if the concepts is correct, then impulse is the change of momentum, while force is the change of momentum with respect to time, and since here the collision was just an instant and the impulse is 2mv then the force must be 2mv/s.
My problem is that this was caused by a momentum of mv!

 Quote by DaleSpam It can be anything from just larger than 0 to just smaller than infinity, depending inversely on the amount of time that the force is applied and depending proportionally on the change in momentum.
The momentum that you speak is the resulting momentum, however I am asking about the force caused by a given momentum.
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.
Mentor
P: 15,601
 Quote by yoelhalb and since here the collision was just an instant and the impulse is 2mv then the force must be 2mv/s.
This is incorrect reasoning. If the collision was 1s then the average force would be 2mv/(1 s), but as you said the collision was just an instant. So the average force would be 2mv/(1 instant). Presumably 1 instant < 1 s so the average force would be greater than what you wrote.

 Quote by yoelhalb My problem is that this was caused by a momentum of mv!
So what? The force and the momentum are different units, comparing them is like comparing apples and oranges.

 Quote by yoelhalb 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?
No, at least, not without additional information.

 Quote by yoelhalb 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.
Again, this is incorrect. The 2mv is in units of impulse, not force. No matter how many times you repeat it 2mv will never be a force, so you may as well learn it now and stop repeating this mistake you have in your mind. Impulse and force are not the same thing, they have different units.
 P: 786 DaleSpam is correct - The ball moves towards the wall with momentum mv (kilogram-meters 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 2mv-mv=mv, so momentum is conserved. Adding a momentum of 2mv to the earth-wall 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 earth-wall 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.
P: 69
 Quote by DaleSpam This is incorrect reasoning. If the collision was 1s then the average force would be 2mv/(1 s), but as you said the collision was just an instant. So the average force would be 2mv/(1 instant). Presumably 1 instant < 1 s so the average force would be greater than what you wrote. So what? The force and the momentum are different units, comparing them is like comparing apples and oranges. No, at least, not without additional information. Again, this is incorrect. The 2mv is in units of impulse, not force. No matter how many times you repeat it 2mv will never be a force, so you may as well learn it now and stop repeating this mistake you have in your mind. Impulse and force are not the same thing, they have different units.
sorry for my incorrect notation, but it is an impulse of 2mv and a force of minimum 2mv/s, and it was caused by a momentum of 2mv.
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.
Mentor
P: 15,601
 Quote by yoelhalb sorry for my incorrect notation, but it is an impulse of 2mv and a force of minimum 2mv/s, and it was caused by a momentum of 2mv.
No, it is not a minimum force of 2mv/(1 s). If you have a 10 s collision then the average force will be less than 2mv/(1 s) by a factor of 10.

 Quote by yoelhalb And in other words the momentum caused an impulse twice as it's own.
Correct, impulse is the change in momentum, not the momentum itself. In this case the change in momentum is twice the original momentum (and in the opposite direction).

 Quote by yoelhalb 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.
I am not certian of this new scenario, they seem rather non-analogous. In one case you have an elastic collision for a non-isolated system where momentum is not conserved due to the external force, and in this new scenario it seems that you are considering a non-elastic collision (energy is produced) for an isolated system where momentum is not conserved but there is no external force. If I understand your new scenario correctly then it violates Newton's laws.

 Quote by yoelhalb Yet I understand that that's the fact, my question is if there is a logical reason for it.
Newton's laws are the logical reason. Also, you need to be careful not to mix up different terms. In this thread I have seen you mix up momentum and change in momentum and mix up force and momentum. Keep your different concepts straight and apply Newton's laws correctly and you will get the right answer.

 Quote by yoelhalb 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.
First, it is the pushing object's change in momentum which is important, not its momentum. Second, the units are wrong, so you need some other quantity with the right units. In this case the units are off by a factor of time, so you need some characteristic time. Specifically, the duration of the push. With that information you can determine the average force.
 P: 786 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 kg-m/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 kg-m/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.
 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.
 HW Helper Sci Advisor Thanks P: 7,923 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.
 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.
Mentor
P: 15,601
 Quote by yoelhalb I don't really see the difference between another object and the same object such as a bouncing ball
If I correctly understand the different scenarios, then the difference is the presence or absence of the external force. I.e. You could replace the wall with a second bouncing ball firmly anchored to the ground and still get a wall-like result.

 Related Discussions Introductory Physics Homework 3 Introductory Physics Homework 4 Classical Physics 8 Introductory Physics Homework 0 Introductory Physics Homework 5