Is momentum really conserved in a perfectly inelastic collision?

[V0ODO0CH1LD]

Is momentum really conserved in a perfectly inelastic collision?

By definition there is no conservation of kinetic energy, but isn't conservation of momentum dependent on conservation of energy? Shouldn't a system whose model takes into account conversion of kinetic energy into heat and deformation, not take into account conservation of momentum? Or is it that the loss of momentum is so negligible compared to the loss in kinetic energy that we can assume it's not existent?

 

[rude man]

Momentum is always conserved, even in aperfectly inelastic collision.

The thing is, you have to be carefuil to include all the mass involved.

Example 1: a squishy pool ball hits the side of the table and sticks to it. In this case the momentum of th entire Earth must be included!

Example 2: perfectly inelastic collision between two squishy balls on ice with no friction. Then the only mass involved is the two balls.

 

[Nugatory]

Is momentum really conserved in a perfectly inelastic collision?

By definition there is no conservation of kinetic energy, but isn't conservation of momentum dependent on conservation of energy? Shouldn't a system whose model takes into account conversion of kinetic energy into heat and deformation, not take into account conservation of momentum? Or is it that the loss of momentum is so negligible compared to the loss in kinetic energy that we can assume it's not existent?

Conservation of momentum does not depend in any way on conservation of kinetic energy. Even in an inelastic collision in which kinetic energy is converted into heat (or used to crumple sheet metal, or shatter solid materials, or whatever) momentum is completely absolutely 100% conserved.

 

[V0ODO0CH1LD]

Conservation of momentum does not depend in any way on conservation of kinetic energy. Even in an inelastic collision in which kinetic energy is converted into heat (or used to crumple sheet metal, or shatter solid materials, or whatever) momentum is completely absolutely 100% conserved.

And why is that?

EDIT: So in any collision ranging from perfectly elastic to perfectly inelastic is momentum conserved?

 

[Nugatory]

EDIT: So in any collision ranging from perfectly elastic to perfectly inelastic is momentum conserved?

Yes.

For some collisions, it's easiest to visualize this if you first transform into a frame in which the total momentum is zero.

 

[jtbell]

And why is that?

The change in an object's momentum when a force acts on it is given by the impulse of the force: basically force times the length of time during which the force acts.

According to Newton's Third Law, the forces that the two colliding objects exert on each other are equal in magnitude and opposite in direction. These two forces obviously act for the same length of time.

Therefore the impulses delivered to the two objects are equal in magnitude and opposite in direction. The objects' momenta change by equal magnitudes in opposite directions, therefore the total momentum must be the same before and after. The two changes "cancel out" as far as the total momentum is concerned.

 

[imiuru]

The condition for conservation of momentum is "no net force acting on the system".

Take, for example, two different balls A and B which move towards each other and collide. During the duration of collision, A exerts a force on B and vice versa. The forces are equal in magnitude and different in direction. (Newton's 3rd law) And of course, the duration of the force acting on A and the force acting on B is the same.

Therefore we have

See, the conservation of momentum is derived from Newton's third law. Hence, it doesn't matter if there is energy loss.

 

[rude man]

And why is that?

EDIT: So in any collision ranging from perfectly elastic to perfectly inelastic is momentum conserved?

Yes. The center of mass does not move unless an external force is applied. This immediately leads to the conservation of momentum.

 

[V0ODO0CH1LD]

So does conservation of momentum not allow for a collision with complete loss of kinetic energy? Because if a particle A collides with a stationary particle B, and all the kinetic energy is lost during the collision, there will be no energy left for the displacement of either particles, meaning no conservation of momentum. Is it the case that this never happens?

Also, are you telling me that the vibrations of particles in the atmosphere around the collision that cause all sorts of things, like sound waves and elevations in temperature, do not carry momentum? I guess I could see conservation of momentum being applicable if you took all of those into account, or if you used an approximation in which the loss of momentum is so negligible in this collisions that you could just completely ignore it.

Are you guys sure that the latter is not the case?

 

[D H]

So does conservation of momentum not allow for a collision with complete loss of kinetic energy?

Sure. Pick a frame of reference in which the total linear momentum prior to the collision is zero. It will be zero after a perfectly inelastic collision. Since the system now comprises but one object, that one object isn't moving. That means there is no (macroscopic) kinetic energy after the collision -- in that frame.

Remember that both momentum and energy are frame dependent quantities.

Because if a particle A collides with a stationary particle B, and all the kinetic energy is lost during the collision, there will be no energy left for the displacement of either particles, meaning no conservation of momentum. Is it the case that this never happens?

Suppose your stationary particle B is *huge*. Imagine a one gram ball of glue colliding with a one megaton asteroid. In a frame in which the asteroid is initially stationary, it's going to remain essentially stationary post-collision. (Note: This initially stationary asteroid defines our frame of reference.) Even if the glueball is moving a million meters/second pre-collision, the glueball+asteroid is moving at a micrometer per second post-collision. How much energy does this have? Not much. In this frame, the pre-collision glueball had 500 megaJoules of kinetic energy while the post-collision glueball+asteroid has 500 microJoules of kinetic energy.

Momentum is always conserved. Kinetic energy is not.

 

[imiuru]

Also, are you telling me that the vibrations of particles in the atmosphere around the collision that cause all sorts of things, like sound waves and elevations in temperature, do not carry momentum?

Which sentences you read were telling you that particles do not carry momentum? I read everything in this thread and found no places you could pick up this idea.

The particles certainly carry momentum.

...used an approximation in which the loss of momentum is so negligible in this collisions that you could just completely ignore it.

The loss of momentum in the system is zero instead of "negligible".

 

[rude man]

So does conservation of momentum not allow for a collision with complete loss of kinetic energy? Because if a particle A collides with a stationary particle B, and all the kinetic energy is lost during the collision, there will be no energy left for the displacement of either particles, meaning no conservation of momentum. Is it the case that this never happens?

That is incorrect. Remember, momentum is a vector, i.e. with direction. Say your particles collide with zero remaining kinetic energy, i.e. v3 = 0. Assuming no external forces on the particles (frictionless plane, no air friction, etc.) then we have

momentum before the collision: m1 v1 + (-m2 v2)
momentum after the collision: (m1 + m2)v3 = 0

So this indicates that the only way your blob after the collision would come to rest is if their prior separate momenta equated to zero. If m1 v1 ≠ m2 v2 then the blob would have some remaining velocity v3 and k.e. = 1/2 (m1 + m2)(v3)^2.

Also, are you telling me that the vibrations of particles in the atmosphere around the collision that cause all sorts of things, like sound waves and elevations in temperature, do not carry momentum? I guess I could see conservation of momentum being applicable if you took all of those into account, or if you used an approximation in which the loss of momentum is so negligible in this collisions that you could just completely ignore it.

Are you guys sure that the latter is not the case?

If you introduce air friction into the collision, that constitutes an external force to the system defined by the two particles alone, and the momentum is changed: Δp = ∫F dt where p is momentum, F is the force of air friction on both masses, and t is time.

If on the other hand you include the air molecules as part of your system, then the total momentum of the system would remain unchanged.

So you have to be careul to define your system. As I pointed out before, the whole Earth must sometimes be considered part of the "system".

 

[V0ODO0CH1LD]

While you can assume that in a perfectly inelastic collision kinetic energy is lost and therefore that energy conservation fails, in a better model you would have to specify where that energy went, because in an isolated system there will AWAYLS be conservation of energy. So for simplistic purposes some models can completely ignore conservation of energy, like the ones that deal with friction, and still give accurate enough results even though we know that is NOT what is actually happening.

Obviously in an isolated system momentum is conserved, but then again so is energy, by definition of an isolated system.

My point is, if your model takes into account loss of energy to things like heat and sound, and those things carry momentum, two particles that collide and loose energy to that will also loose momentum. I am accepting the fact that in most cases you can assume that loss of momentum is zero, but it's just hard to ignore the fact that you use heat and sound to justify the loss of kinetic energy but completely forget about those when talking about momentum.

It just bothers me that the source of squander of energy is NOT a source of squander of momentum, even though the things that use up that energy clearly HAVE momentum.

Are you completely sure that conservation of momentum in an inelastic collision isn't just a useful approximation?

 

[A.T.]

It just bothers me that the source of squander of energy is NOT a source of squander of momentum, even though the things that use up that energy clearly HAVE momentum.

As rude man said, it's because energy is a scalar that has no direction information, while momentum is a vector with a direction.

Since KE doesn't care about the direction, you can convert macroscopic KE (bulk movement in one direction) into microscopic KE (chaotic movement in all directions).

But this doesn't work for momentum, because it has direction. So you cannot diffuse and hide momentum that way, because conservation of momentum implies the conservation of the direction of the original bulk movement (center of mass).

Also note that the macroscopic KE is converted into internal potential energy as well, which is not associated with movement and momentum. But there is no such other type of momentum.

 

[nasu]

My point is, if your model takes into account loss of energy to things like heat and sound, and those things carry momentum, two particles that collide and loose energy to that will also loose momentum. I am accepting the fact that in most cases you can assume that loss of momentum is zero, but it's just hard to ignore the fact that you use heat and sound to justify the loss of kinetic energy but completely forget about those when talking about momentum.

It just bothers me that the source of squander of energy is NOT a source of squander of momentum, even though the things that use up that energy clearly HAVE momentum.

Don't forget that momentum is conserved if the system has no interaction with external systems.
If it heats up the medium or produces sound, there is interaction, the system is not perfectly isolated.
So for an approximately isolated system, the momentum is approximately conserved, sure.
Conservation of momentum is not an unqualified property. It only works for perfectly isolated systems. If you want it to be "perfect".

But you are right. If a moving body hits a body at rest and they stick together, the final speed cannot be exactly zero so kinetic energy is not completely converted in other forms.

 

[D H]

Don't forget that momentum is conserved if the system has no interaction with external systems.

Something that hasn't been mentioned yet: Collisions occur quickly. What this means is that even if external forces are present, these external forces can oftentimes be ignored in a collision event because the rapidity of the event makes FΔt rather tiny.

But you are right. If a moving body hits a body at rest and they stick together, the final speed cannot be exactly zero so kinetic energy is not completely converted in other forms.

In that particular frame of reference, in which one of the bodies was at rest. You can always find a frame in which the pre-collision total momentum is zero, and in this frame, kinetic energy *is* completely converted into other forms.

 

[sophiecentaur]

If you want to understand a basic concept like the conservation of momentum then it is simple models that will give you that understanding. Merely introducing greater and greater complexity doesn't help and can tempt one to assume that the Momentum, somehow 'sneaks' out of the system. That is looking at things the wrong way round. If the principle works between two bodies then it will work everywhere.
All successful Science experiments attempt to keep things as simple as possible and that also applies to scientific thinking.

 

[nasu]

Something that hasn't been mentioned yet: Collisions occur quickly. What this means is that even if external forces are present, these external forces can oftentimes be ignored in a collision event because the rapidity of the event makes FΔt rather tiny.

Yes, but tiny is not "perfectly" zero. And he wants to be perfectly conserved.

In that particular frame of reference, in which one of the bodies was at rest. You can always find a frame in which the pre-collision total momentum is zero, and in this frame, kinetic energy *is* completely converted into other forms.

I expected you will say this.

 

[imiuru]

Are you completely sure that conservation of momentum in an inelastic collision isn't just a useful approximation?

Well, conservation of momentum is derived from Newton's third law of motion. So, it holds true whenever Newton's third law of motion is applicable. As far as I know, Newton's third law of motion is true when the frame of reference is inertial. (There is another condition for 3rd law to be true but it is not the point of this discussion.)

Momentum is conserved before and after collision. To say that it is approximately conserved is to say that action and reaction is approximately equal.

 

[nasu]

Well, conservation of momentum is derived from Newton's third law of motion. So, it holds true whenever Newton's third law of motion is applicable.

Not quite. It's derived by applying Newton's third law to an isolated system.
External forces can change a system's momentum without violating any law.

Collisions satisfy the "isolated system" condition pretty well, most of the time.
However a car's momentum is not usually conserved (if it were the car would be useless ) even though every internal interaction satisfies Newton's third law.

 

[imiuru]

Not quite. It's derived by applying Newton's third law to an isolated system.

Conservation of momentum is the natural result of Newton's third law (or 2nd law as some may say). When we apply Newton's third law to derive the conservation of momentum, we are essentially looking at two objects acting on each other.

Since we are studying only the two objects, they form a closed system. (Not necessarily isolated system)

Basically we are saying the same thing

 

[D H]

Not quite. It's derived by applying Newton's third law to an isolated system.

Since we are studying only the two objects, they form a closed system. (Not necessarily isolated system)

Basically we are saying the same thing

No, you're not saying the same thing.

Consider a collision between a 4 gram ball of glue moving at 100 m/s and a 1 gram thin sheet of steel. The behavior that results if the two objects are two isolated, free-floating objects in space is quite different from the behavior that results if the thin sheet of steel is firmly attached to a 10 tonne brick wall.

How one draws system boundaries is important.

 

[imiuru]

Yeah, it's my mistake. It should be isolated system, not closed system.

 

[sophiecentaur]

Something that hasn't been mentioned yet: Collisions occur quickly. What this means is that even if external forces are present, these external forces can oftentimes be ignored in a collision event because the rapidity of the event makes FΔt rather tiny.


In that particular frame of reference, in which one of the bodies was at rest. You can always find a frame in which the pre-collision total momentum is zero, and in this frame, kinetic energy *is* completely converted into other forms.

It's important to note, however, that, for a given momentum change, the Force will be inversely proportional to the time it's applied. FΔt is only small when the momentum change is small.
Momentum conservation also applies to very 'soggy' springs where the time constant can be a second or more. The same amount of momentum would be transferred to an 'external' object, however long the collision lasts. (Have I missed something here? I'm sure that point is obvious.)

If there are external forces acting at times outside the collision time, the momentum transfer will be that force times the long time for which it applies. But this added complication really needs to be avoided until the basic, simple, collision has been understood fully.

 

[V0ODO0CH1LD]

I think some of you are missing the point of my original question.

In a model where after two particles collide some kinetic energy disappears from the system, and it does just disappear because the system does NOT include whatever the two particles lost their energy to, it seems to me that momentum conservation should be only an approximation.

I know that by definition if my system is just the two particles their momentum should always be conserved, but if the system is just the two particles their energy should ALSO always be conserved.

I get that in a isolated collision both momentum and energy are conserved, what I don't get is how momentum can be conserved if the same things that get energized after the collision obtain momentum they didn't have before.

I'm sure that if the energy lost was all potential, then conservation of momentum could actually still be there, because the things the system lost energy to don't have momentum at all.

Also, doesn't conserving momentum and not KINETIC energy break down mathematically at some point??

 

[sophiecentaur]

I tried to deal with your question directly when I wrote that you have to deal with the very simplest system if you want to understand things like this.
Momentum is always conserved. That statement is always correct for any isolated system. If you want to introduce an outside influence then all bets are off - unless you include the new addition as part of the system.
Two equal masses collide, going in opposite directions (wrt an Earth frame of reference) and with the same intiial speeds. Total momentum was zero and ends up at zero. The KE, wrt the Earth frame , was two times half mvsquared and ends up as zero. The energy lost must go somewhere, of course, but net motion of the masses doesn't contribute.
You cannot introduce Potential Energy (except between the two masses) in your argument or the system is not isolated. Every interaction involves some Potential Energy (even the elastic deformation during collisions). If you hit an infinitely massive object then you cannot consider the Momentum rigorously because, somewhere along the line, you will end up dividing by infinity and be comparing one zero with another zero (very crude maths here, but I won't apologise)
Instead of trying to find loopholes, it might be better if you were to try to spot where your suggestions are flawed. They have to be!
Forget the "Mathematically" bit. Maths is only a tool for working things out and it tends to stick to the model you use (assuming it's the right model).

 

[atyy]

Also, doesn't conserving momentum and not KINETIC energy break down mathematically at some point??

In some collisions, kinetic energy and momentum are lost.

However, it is mathematically exactly consistent to have loss of kinetic energy without any loss of momentum. Physically, you can think of it that the missing energy is carried away by particles you do not see. Consider the simple case in which there is a pair of unobserved particles. The pair of particles carries away kinetic energy, but because each particle moves in opposite directions, they carry away no net momentum. (Energy is a scalar, momentum is a vector.)

 

[nasu]

In addition to what sophiecentaur already explained, it seems that the OP is not aware of a significant difference in the effects of internal forces.

For an ISOLATED (100% isolated) system, the internal forces cannot change the total momentum but the same internal forces can change the kinetic energy of the system.
We don't need any mysterious or unmeasured particles to explain this. It just follows from Newton's laws for system of particles.