Is Momentum Conserved in Elastic and Inelastic Collisions with Walls?

[suchal]

Elastic:
A ball of mass much less than the wall collides with the wall. If the collision is elastic the kinetic energy of ball is unchanged. It means that kinetic energy of wall is zero before and after the collision. Therefore velocity of wall HAS NOT CHANGED. When Velocity is Zero momentum is zero.
The change of momentum of ball is 2mv while the change of momentum of wall is ZERO.
Momentum is not conserved.

Inelastic: ball collides with the wall and due to the glue applied on the surface of ball the ball sticks on the wall after collision. The change of momentum is mv. The change of momentum is zero as the wall doesn't move. The kinetic energy is converted into internal energy and sound energy. Momentum is although not conserved again.

As mass of wall is much bigger the change in velocity of ball has to be small but not zero. Here it is zero.
In a video Professor walter lewin said that in this case the momentum of wall is NOT zero but kinetic energy of wall is zero. He said that it can be mathematically proven it but i don't this it is possible.

 

[WannabeNewton]

The assumption that the wall doesn't move after the collision is only valid in the limit as $\frac{m}{M}$ goes to zero, where $m$ is the mass of the ball and $M$ is the mass of the wall. If you set everything up in the general case wherein you make no assumptions about $\frac{m}{M}$ and then take the limit as this dimensionless parameter goes to zero then you will see the proper behavior of the conservation laws.

 

[suchal]

The assumption that the wall doesn't move after the collision is only valid in the limit as $\frac{m}{M}$ goes to zero, where $m$ is the mass of the ball and $M$ is the mass of the wall. If you set everything up in the general case wherein you make no assumptions about $\frac{m}{M}$ and then take the limit as this dimensionless parameter goes to zero then you will see the proper behavior of the conservation laws.

There is a property of matter called inertia and it can not be ignored. Such a force can not make it move.

 

[dauto]

Professor walter lewin is correct. Let's do the elastic case. The ball momentum changes by -2mv. The momentum of the wall must change by 2mv due to momentum conservation. Assuming that the wall was initially at rest, the speed of the wall will be V_w = 2mv/M_w. It's kinetic energy will be K_w= (1/2) M_w V_w² = (1/2) M_w (2mv/M_w)² = 2(mv)²/M_w. Now if you take M_w → ∞, you get K_w → 0 while the momentum of the wall will be p_w = 2mv ≠ 0.

 

[arildno]

For a fixed wall, you can perfectly well have an elastic collision (meaning that kinetic ENERGY is conserved), even though momentum is not conserved.
If external reaction forces holding the wall fixed are comparable in magnitude to the collision forces, there is no reason why momentum should be conserved.
---
Another familiar, somewhat analogous example is a ball hitting a pendulum.
In this case, you will have conservation of angular momentum around the pivot point, but the impulse acting AT the pivot point is comparable to the collision impulse where the ball hits the pendulum. Linear momentum is certainly not conserved in this scenario, although the angular momentum is. Such a scenario can perfectly well be elastic, though.

 

[dauto]

For a fixed wall, you can perfectly well have an elastic collision (meaning that kinetic ENERGY is conserved), even though momentum is not conserved.
If external reaction forces holding the wall fixed are comparable in magnitude to the collision forces, there is no reason why momentum should be conserved.

I think it is implied in the question that there are no external forces (A wall floating in space) but the Original poster might want to clarify that point.

 

[arildno]

I think it is implied in the question that there are no external forces (A wall floating in space) but the Original poster might want to clarify that point.

I ought to have said that for a wall floating in space, momentum might certainly well be conserved, and the reason why we may regard the velocity of the wall to be zero, is that it is many orders of magnitude less than the velocity of the ball. (The resulting momenta, though, will be of the same order, indeed perfectly equal but oppositely directed)

 

[atyy]

Professor walter lewin is correct. Let's do the elastic case. The ball momentum changes by -2mv. The momentum of the wall must change by 2mv due to momentum conservation. Assuming that the wall was initially at rest, the speed of the wall will be V_w = 2mv/M_w. It's kinetic energy will be K_w= (1/2) M_w V_w² = (1/2) M_w (2mv/M_w)² = 2(mv)²/M_w. Now if you take M_w → ∞, you get K_w → 0 while the momentum of the wall will be p_w = 2mv ≠ 0.

This is very unintuitive. It also says the velocity of the wall V_w = 2mv/M_w goes to zero, then it would seem the momentum of the wall is finite although its velocity is zero?

 

[WannabeNewton]

The momentum is $p_w = M_w v_w$. We are taking the limit as $v_w$ goes to zero and $M_w$ goes to infinity so the momentum remains finite. This is why I said in post #2 to write things out generally and then take limits, which is what dauto later did anyways.

 

[atyy]

The momentum is $p_w = M_w v_w$. We are taking the limit as $v_w$ goes to zero and $M_w$ goes to infinity so the momentum remains finite. This is why I said in post #2 to write things out generally and then take limits, which is what dauto later did anyways.

I see. So for the kinetic energy, Mv² goes to zero as M goes to inifnity and v goes to zero, because v² goes to zero much faster than M gets large?

 

[WannabeNewton]

Yep.

 

[atyy]

And strictly speaking, do we have to say we take the double limit?

 

[dauto]

I see. So for the kinetic energy, Mv² goes to zero as M goes to inifnity and v goes to zero, because v² goes to zero much faster than M gets large?

That's right. The energy is proportional to the square of the speed, so it is an "infinitesimal" of higher order.

 

[A.T.]

This is very unintuitive.

Infinities are. When you keep M constant, and let m -> 0, the effect on m/M is the same, but you don't get the mismatch between momentum and energy limits.

 

[kaplan]

Elastic:
A ball of mass much less than the wall collides with the wall. If the collision is elastic the kinetic energy of ball is unchanged. It means that kinetic energy of wall is zero before and after the collision. Therefore velocity of wall HAS NOT CHANGED. When Velocity is Zero momentum is zero.
The change of momentum of ball is 2mv while the change of momentum of wall is ZERO.
Momentum is not conserved.

Inelastic: ball collides with the wall and due to the glue applied on the surface of ball the ball sticks on the wall after collision. The change of momentum is mv. The change of momentum is zero as the wall doesn't move. The kinetic energy is converted into internal energy and sound energy. Momentum is although not conserved again.

As mass of wall is much bigger the change in velocity of ball has to be small but not zero. Here it is zero.
In a video Professor walter lewin said that in this case the momentum of wall is NOT zero but kinetic energy of wall is zero. He said that it can be mathematically proven it but i don't this it is possible.

Momentum is conserved if the laws of physics are invariant under spatial translations (i.e. if they are the same everywhere in space). If you have an immovable wall somewhere, the laws of physics are NOT invariant under spatial translations, because there is a law that says that when you arrive at the wall something happens (you elastically rebound, or stick to it, or whatever).

So why would you expect momentum to be conserved when you treat the wall as immovable? It's not - you're right. (Of course in the real world there's no such thing as an immovable wall, but you're not asking about the real world.)

 

[UltrafastPED]

This ball-wall problem is worked in detail here:
http://www.lhup.edu/~dsimanek/ideas/bounce.htm

Momentum is conserved in the absence of external forces.

Both cases are discussed here:
http://www.colorado.edu/physics/phys1110/phys1110_sp01/Notes/Chapter11.htm

 

[suchal]

When talking about conservation of momentum it is already clear that NO EXTERNAL FORCES ACT ON OBJECTS so those who raised this point should rethink about the concept of conservations of momenta.
Secondly, we say momentum of ball is 2mv. 2mv=Mw.Vw, Mw->infinity, Vw->0 but their product still is 2mv.

Atyy hope u got the answer.
Thanks to everyone for helping me out.

 

[atyy]

Yes, I got the answer - thanks to everyone from me too. And suchal, thanks for asking the question - I'd never heard this crazy thing before!

 

[suchal]

The next problem is that when we talk about momentum, the concept of inertia doesn't make sense. a little force will have some change on the momentum of the body. I think inertia is related to momentum. An object with large mass will have larger momentum so it will require more force to change it's velocity.

All forces i talk about are internal forces and there are no external forces acting on the closed system.

 

[suchal]

It is good to see how helpful you people are. And please don't be shy to ask question no matter how much knowledge you have, you won't be insulted. These discussions clear the concepts and answer the question which otherwise disturb us.

Yes, I got the answer - thanks to everyone from me too. And suchal, thanks for asking the question - I'd never heard this crazy thing before!

 

[mfb]

The next problem is that when we talk about momentum, the concept of inertia doesn't make sense.

No.
If you don't understand something, your first, second, and third approach should be "I don't get the concept" instead of "it has to be wrong, it does not make sense".

a little force will have some change on the momentum of the body.

Sure

I think inertia is related to momentum.

There is a relation, indeed.

An object with large mass will have larger momentum so it will require more force to change it's velocity.

No, it will require more force (or more time) to change its velocity by the same amount. Tiny velocity changes require tiny forces and short timescales, and there is no "minimal force" required to do anything (unless you have friction).

 

[suchal]

No.
If you don't understand something, your first, second, and third approach should be "I don't get the concept" instead of "it has to be wrong, it does not make sense".

Sure
There is a relation, indeed.
No, it will require more force (or more time) to change its velocity by the same amount. Tiny velocity changes require tiny forces and short timescales, and there is no "minimal force" required to do anything (unless you have friction).

F=ma, a=F/m, if m increase a decrease if F is same.
So for the same force, if the object with larger mass will have smaller acceleration.
When i say "doesn't make sense" it means i am talking about myself that the concept doesn't make sense to my weak mind. I never said it is wrong.

 

[mfb]

F=ma, a=F/m, if m increase a decrease if F is same.
So for the same force, if the object with larger mass will have smaller acceleration.

Right.

When i say "doesn't make sense" it means i am talking about myself that the concept doesn't make sense to my weak mind. I never said it is wrong.

Okay.

 

[suchal]

Right.
Okay.

That is exactly what i said in simple words.
If the object has more mass a greater force will be required to accelerate it by the same amount.

 

[Vanadium 50]

When i say "doesn't make sense" it means i am talking about myself that the concept doesn't make sense to my weak mind. I never said it is wrong.

The very title of this thread, "Momentum is NOT conserved" is an example.

Posting wrong statements in the hope someone will correct them is a very inefficient way to learn. It also tends to annoy those who are trying to help you. That's why PF strongly discourages this. It's much better to ask questions.

 

[suchal]

The very title of this thread, "Momentum is NOT conserved" is an example.

Posting wrong statements in the hope someone will correct them is a very inefficient way to learn. It also tends to annoy those who are trying to help you. That's why PF strongly discourages this. It's much better to ask questions.

Sorry then. I meant by my calculation which I showed on first page the momentum was not conserved. I will avoid this kind of statements. :)