Time-reverse symmetry of the principle of relativity

[Chrisc]

It seems neither of you are not objecting to the principles I have raised, you are only objecting to the quantitative values of the mechanics in my example. You are right they do not represent "real" mechanics unless the conservation of kinetic energy is included in the calculation of the collisions. But including conservation of kinetic energy does not change the principles I am questioning. It changes the final velocities but not the relative dynamics of dissimilar masses in collision.
Given that these are point masses in constant linear motion, whether you include the "specific" velocities necessary to conserve kinetic energy or not, what remains is the proportion of motion to mass which is easier to see and address in the momentum, hence my omission of Ek.
I am questioning the following principles of Newton's laws:
(which will [right or wrong] address the relativity of the dynamics once the principles are clear)
The dynamical laws require a collision of dissimilar masses observed from a position of rest with respect to one and the other, result in the larger mass acquiring a velocity after collision that is "ALWAYS LESS" than the velocity of the smaller mass before the collision, when the observer is initially at rest with the larger, regardless the "specific" numerical values.
Likewise and by virtue of the same laws upholding the principle of the second law of thermodynamics, the smaller mass will acquire a velocity after collision that is "NEVER GREATER" than the velocity of the larger mass before collision, when the observer is initially at rest with the smaller.

Do you both agree with this or not?

 

[yuiop]

Likewise and by virtue of the same laws upholding the principle of the second law of thermodynamics, the smaller mass will acquire a velocity after collision that is "NEVER GREATER" than the velocity of the larger mass before collision, when the observer is initially at rest with the smaller.

Do you both agree with this or not?

In https://www.physicsforums.com/showpost.php?p=1799644&postcount=17" of this thread, I showed that that the final velocity of the smaller mass after collision (2.666 m/s) IS greater than the velocity of the larger mass before collision (2 m/s), when the observer is initially at rest with the smaller.

Did you mean "ALWAYS GREATER" rather than "NEVER GREATER"? Obviously "NEVER GREATER" is not a true statement as I have shown a counter example.

 

[JesseM]

I am questioning the following principles of Newton's laws:
(which will [right or wrong] address the relativity of the dynamics once the principles are clear)
The dynamical laws require a collision of dissimilar masses observed from a position of rest with respect to one and the other, result in the larger mass acquiring a velocity after collision that is "ALWAYS LESS" than the velocity of the smaller mass before the collision, when the observer is initially at rest with the larger, regardless the "specific" numerical values.

How come you never explicitly stated that this was what you were arguing before? Anyway, clearly this statement is true in the case of elastic collisions, since the total kinetic energy before the collision would just be the kinetic energy of the smaller mass, (1/2)*m*v^2, and if the larger mass had a greater velocity than that after the collision, then its kinetic energy would be larger. So the only way your principle could be violated would be in the case of an inelastic collision where an extremely unlikely statistical fluctuation caused heat to be converted into a "kick" given to the larger mass, increasing its kinetic energy. This is not a violation of the fundamental dynamical laws, but as I said it's very unlikely in a statistical sense (it would require an unusual set of initial conditions for all the molecules making up the masses, but it would be possible to find such a set of initial conditions such that this decrease in entropy would follow in a deterministic way from the initial conditions, assuming we are using deterministic Newtonian-style laws rather than quantum laws). But this doesn't suggest an asymmetry in the laws of nature either, because if you don't specify that the initial conditions of the universe are at low entropy and instead imagine some masses in a box that long ago has had time to go to equilibrium, it is equally unlikely to see the reverse of this, where two masses collide and a significant amount of their linear kinetic energy is converted to heat, because the only way this can happen is if the entropy of the system is significantly lower before the collision than after, which would require a previous statistical fluctuation in the system from its more probable maximum-entropy state.

Likewise and by virtue of the same laws upholding the principle of the second law of thermodynamics, the smaller mass will acquire a velocity after collision that is "NEVER GREATER" than the velocity of the larger mass before collision, when the observer is initially at rest with the smaller.

Again, how come you never stated that this was what you were arguing? This one is false even in an elastic collision. For example, suppose we have a 10 kg mass moving towards a smaller 1 kg mass at 10 m/s, with the smaller mass initially at rest. In this case, in an elastic collision the result will be that after the collision, the larger mass is moving at (450/55) m/s (about 8.18 m/s) while the smaller mass is moving at (1000/55) m/s (about 18.18 m/s) in the same direction. You can check for yourself that both before and after the collision the total momentum is 100 kg*m/s, and the total kinetic energy is 500 kg*m^2/s^2. And obviously if the smaller mass is moving at about 18.18 m/s after the collision, this is greater than the velocity of the larger mass before the collision, which was 10 m/s.

 

[DrGreg]

The dynamical laws require a collision of dissimilar masses observed from a position of rest with respect to one and the other, result in the larger mass acquiring a velocity after collision that is "ALWAYS LESS" than the velocity of the smaller mass before the collision, when the observer is initially at rest with the larger, regardless the "specific" numerical values.

I agree (provided there's no external energy being supplied).

Likewise and by virtue of the same laws upholding the principle of the second law of thermodynamics, the smaller mass will acquire a velocity after collision that is "NEVER GREATER" than the velocity of the larger mass before collision, when the observer is initially at rest with the smaller.

I disagree and I don't understand why you would think that. kev and JesseM have both given counter-examples and here's another that you can understand with intuition and no calculation. When considering large and small values, why not take an extreme example? Let's collide a human being with planet Earth. To avoid a completely inelastic collision, let's bounce you off a trampoline. Your rebound velocity upwards, relative to the Earth, is non-zero, let's pessimistically say v/2 where -v is your impact velocity. (That's not a very good trampoline, but the actual value doesn't matter; anything noticeably greater than 0 and less than v will serve my argument.) So, relative to someone else falling at the same speed -v as you but missing the trampoline and continuing to fall, your velocity after colliding with the massive Earth+trampoline system is 3v/2 which is greater than the speed v at which the Earth+trampoline hit you.

 

[JesseM]

Did you mean "ALWAYS GREATER" rather than "NEVER GREATER"? Obviously "NEVER GREATER" is not a true statement as I have shown a counter example.

That's a good point, "ALWAYS GREATER" would actually be correct here, and it illustrates the way elastic collisions (where no linear kinetic energy is converted to heat or vice versa) are totally time-symmetric.

In such a collision, then as Chrisc said the following must be true:

"a collision of dissimilar masses observed from a position of rest with respect to one and the other, result in the larger mass acquiring a velocity after collision that is "ALWAYS LESS" than the velocity of the smaller mass before the collision, when the observer is initially at rest with the larger"

If we assume that time-reversal symmetry applies, this tells us that whenever we have a collision where the larger mass comes to rest after the collision, it must be true that the velocity v1 of the larger mass before the collision is "ALWAYS LESS" than the velocity v2 of the smaller mass after the collision.

Without loss of generality, let's assume that the smaller mass was approaching the larger one from the left, and that the smaller mass was initially moving at speed v to the right, the larger mass was initially moving at v1 to the left (if the smaller mass was initially moving to the right and the larger mass came to rest after the collision, the only way this makes sense is if the larger mass was initially moving to the left), and after the collision the smaller mass was moving at v2 to the left (if the larger mass comes to rest after the collision, the smaller mass obviously can't continue to move to the right or it'd have to pass right through the larger one). Now let's take this same collision and look at it from the perspective of a second frame where the smaller mass was at rest before the collision, a frame where all the velocities must be boosted from the first frame by a velocity of v to the left. In the second frame, the larger mass was initially moving at (v1 + v) to the left, and after the collision the smaller mass was moving at (v2 + v) to the left. So regardless of the actual value of v, if we know v1 is "ALWAYS LESS" than v2, it must also be true that (v2 + v) is "ALWAYS GREATER" than (v1 + v). So, we get the conclusion that in a frame where the smaller mass was at rest before the collision, the smaller mass will acquire a velocity (v2 + v) after collision that is "ALWAYS GREATER" than the velocity (v1 + v) of the larger mass before collision. And we got this second conclusion just by assuming time-reversal symmetry would apply to the first conclusion I quoted Chrisc saying, and then applying a boost of v to the left to all velocities in order to switch to a frame where the smaller mass is initially at rest.

 

[Chrisc]

You all agree with my first statement and you all disagree with my second.
I understand and agree with the mathematical reasoning you've used to justify your objections, but there is something not quite right with the principles involved.

JesseM, I did not state the question this way from the beginning because this is not my question. I thought this part was clear and my question pertains to what follows from it, which I will get back to when this point is made clear.
Also, you're last post (#95) is beginning to follow the line of reasoning I am trying to make here. But you have "first" assumed the time symmetry.
If you follow what I am saying here (below) you will see that before you can make that assumption (even though it is mathematically sound), you must find the principles provide a consistent, valid path to it.

When the observer is initially at rest with the smaller mass, the smaller mass cannot (during collision) exert a force on the larger mass greater than its own mass times the incident velocity of the larger mass. According to Newton's third law, this is the force that is exerted on the smaller mass by the larger during collision.
i.e. the smaller mass does not, cannot, resist the force of collision to any extent greater than its own mass(being at rest with the observer) times the velocity of the collision. It must then acquire a velocity after collision that is "NEVER GREATER" than the velocity of the larger mass before collision or invalidate Newton's third law.

It is true that the larger mass has more than enough energy to accelerate the smaller beyond the original velocity of the larger, but it has no way "in principle" to apply such force unless the smaller "spontaneously" acquires inertial energy greater than the sum of its rest mass times the incident velocity of the larger mass.
I understand these are not real world examples, we are talking about the principles involved in POINT-MASS collisions.
In the real world example of this, the smaller "composite" mass can acquire a greater kinetic energy as the collision "acts over time" on the constituent parts to impart MORE than the (macro) inertial resistance of the smaller mass.
But I am questioning the principles involved here because when they are observed in this "theoretical" manner they lead an observation about the relative dynamics of POINT-MASSES that seems to indicate something very interesting about the principle of relativity and time symmetry. Again I will get back to this point once the point I am trying to make here with respect to the principles of the laws is clear.

 

[JesseM]

When the observer is initially at rest with the smaller mass, the smaller mass cannot (during collision) exert a force on the larger mass greater than its own mass times the incident velocity of the larger mass. According to Newton's third law, this is the force that is exerted on the smaller mass by the larger during collision.
i.e. the smaller mass does not, cannot, resist the force of collision to any extent greater than its own mass(being at rest with the observer) times the velocity of the collision. It must then acquire a velocity after collision that is "NEVER GREATER" than the velocity of the larger mass before collision or invalidate Newton's third law.

OK, now that you've finally (!) stated your argument explicitly it's obvious that the problem is just that you have totally wrong ideas about basic Newtonian mechanics. Where did you get the idea that the smaller mass cannot exert a force "greater than its own mass times the incident velocity of the larger mass"? That statement doesn't even make any sense in terms of units, since mass * velocity gives a quantity in units of kg * meters / second, whereas forces are always in units of kg * meters / second^2. Anyway, the force on object exerts is in no way limited by its own mass--consider the electromagnetic force from a charged object, which can be made arbitrarily large by making the charge arbitrarily strong, without any need to change the mass. Newton's third law only says that whatever the force exerted by the small mass on the large mass, it must be equal in magnitude and opposite in direction to the force exerted by the large mass on the small mass. But Newton's third law says nothing whatsoever about the strength of these "equal and opposite" forces.

It is true that the larger mass has more than enough energy to accelerate the smaller beyond the original velocity of the larger, but it has no way "in principle" to apply such force unless the smaller "spontaneously" acquires inertial energy greater than the sum of its rest mass times the incident velocity of the larger mass.

Again, this statement completely fails to make sense in terms of units. Energy has units of kg * meters^2 /second^2, while "rest mass times the incident velocity of the larger mass" would have units of kg * meters / second -- these are the units of momentum, not energy. You might as well say "the smaller mass cannot acquire a velocity greater than the density of the larger mass", or something equally nonsensical.

 

[Chrisc]

OK, now that you've finally (!) stated your argument explicitly it's obvious that the problem is just that you have totally wrong ideas about basic Newtonian mechanics. Where did you get the idea that the smaller mass cannot exert a force "greater than its own mass times the incident velocity of the larger mass"? That statement doesn't even make any sense in terms of units, since mass * velocity gives a quantity in units of kg * meters / second, whereas forces are always in units of kg * meters / second^2.

You're right, as I used the word force, I should have said, - the smaller mass cannot exert a force on the larger mass greater than its own mass times the square of the velocity of the larger, incident mass. This is the acceleration applied to the larger mass by the smaller, which is by Newton's third law, the acceleration applied to the smaller by the larger. As the smaller increases velocity, the larger decreases velocity. Unless you invoke action at a distance, the larger cannot impart motion to the smaller after losing contact which will occur as soon as the velocity of the smaller is greater than the decreased velocity of the larger. As this decreased velocity of the larger is less than its original velocity, it is impossible for the smaller to have acquired a velocity greater than the original velocity of the larger.

The principle and the mechanics remain the same: the velocity of the smaller mass after collision is "Never Greater" than the velocity of the larger mass before collision.

Anyway, the force on object exerts is in no way limited by its own mass--consider the electromagnetic force from a charged object, which can be made arbitrarily large by making the charge arbitrarily strong, without any need to change the mass.

I think you know we are talking only about the forces exerted by the energy of mass in relative motion.

Newton's third law only says that whatever the force exerted by the small mass on the large mass, it must be equal in magnitude and opposite in direction to the force exerted by the large mass on the small mass. But Newton's third law says nothing whatsoever about the strength of these "equal and opposite" forces.

I didn't say it did. I said that because they are equal and opposite, then the "maximum" force that can be applied by the smaller to the larger equals the "maximum" the larger can apply to the smaller.
If you agree the "maximum" applied by the smaller cannot exceed its mass times the square of the velocity of the incident mass, then you will agree the "maximum" applied by the larger to the smaller cannot exceed the same. This means the velocity of the smaller after collision "cannot exceed" the velocity of the larger before collision.

If you consider the mechanics of the collision from every pertinent (non-redundant) inertial frame and then consider all of them through time reversal, in every single frame except one, the time symmetry of the dynamics is upheld.
If you are not interested in the one frame, if you do not think this "exception" deserves consideration, then I am wasting your time and mine.

 

[yuiop]

..
Likewise and by virtue of the same laws upholding the principle of the second law of thermodynamics, the smaller mass will acquire a velocity after collision that is "NEVER GREATER" than the velocity of the larger mass before collision, when the observer is initially at rest with the smaller.

Do you both agree with this or not?

Here is a nice practical demonstration that the above is not true. Take a heavy "super-ball" and hold a lighter "super ball" about one centimeter above the heavier ball. Release both balls at the same time. The larger ball rebounds off the floor with about the same speed of the still falling smaller ball but in the opposite direction and then collides with the still falling smaller ball. After the collision the smaller ball rebounds to a height that is higher than its original release point. If that is not in agreement with the second law, then the second law can be violated any time you drop two balls onto a hard surface and so by your arguments the second law is not in agreement with (easily demonstrated) observations.

 

[JesseM]

You're right, as I used the word force, I should have said, - the smaller mass cannot exert a force on the larger mass greater than its own mass times the square of the velocity of the larger, incident mass. This is the acceleration applied to the larger mass by the smaller, which is by Newton's third law, the acceleration applied to the smaller by the larger.

This principle is still something that appears to have come only from your own imagination, having no logical connection whatsoever to Newton's third law, which only says that the forces must be equal and opposite, not that there is any limit to their size. What's more, your statement still doesn't even make sense in terms of units! Mass times the square of velocity has units of kilograms * meters^2 / second^2, which are the units of energy, whereas the units of force are kilograms * meters / second^2.

As the smaller increases velocity, the larger decreases velocity.

No, as the smaller gains velocity to the left, the larger gains velocity to the right. Since the smaller started out moving to the right and the larger started out moving to the left, if they're colliding for a finite time, each one's speed is decreasing until it hits zero, then increasing as it continues to acquire velocity in the direction opposite to the one it was moving before the collision.

Unless you invoke action at a distance, the larger cannot impart motion to the smaller after losing contact which will occur as soon as the velocity of the smaller is greater than the decreased velocity of the larger.

If the objects collide for a finite time, presumably it's because they're compressing as they come into contact (like rubber balls, or balls with springs affixed to each end), then decompressing as they bounce off in opposite directions. They will not lose contact until they have completely decompressed, this may not happen until after the velocity of the smaller has already exceeded the velocity of the larger. If you don't believe me we could do an actual analysis of the situation where one of the objects has a spring on the end that becomes compressed as they get closer than the rest length of the spring, using law[/url].

As this decreased velocity of the larger is less than its original velocity, it is impossible for the smaller to have acquired a velocity greater than the original velocity of the larger.

Nope, your argument is handwavey and not based on anything in Newtonian mechanics.

I think you know we are talking only about the forces exerted by the energy of mass in relative motion.

"Energy" is not a force. Physicists recognize only four forces in nature--the gravitational force, the electromagnetic force, the strong nuclear force, and the weak nuclear force. When physical objects like balls hit each other and rebound, this is understood as a consequence of electromagnetic forces between atoms on the surface of each during the period of contact. Of course, this is just what's true in fundamental physics, in Newtonian physics we do idealize other forces like the normal force and the spring force which depend on contact, even though at a microscopic level we know these forces are really due to the electromagnetic forces between atoms in the objects.

I didn't say it did. I said that because they are equal and opposite, then the "maximum" force that can be applied by the smaller to the larger equals the "maximum" the larger can apply to the smaller.

Sure, and if we idealize them as exerting constant opposite forces on each other for some finite time-interval (which would not actually true if they behave like they've got a compressing spring between them during the contact, but it makes calculations either), then for each object (change in momentum)/(time-interval) will be equal to this force, and of course each object's change in momentum is just its own change in velocity times its own mass.

If you agree the "maximum" applied by the smaller cannot exceed its mass times the square of the velocity of the incident mass

But I don't agree, that's a made-up principle which has no logical connection to Newton's third law, and doesn't even make sense in terms of units.

If you are not interested in the one frame, if you do not think this "exception" deserves consideration, then I am wasting your time and mine.

In Newtonian physics, Newton's laws hold in every inertial frame, there are no exceptions to any valid conclusion you get from the fundamental principles. The problem is just that your conclusion that the velocity of the smaller mass must be smaller than the larger after the collision if it's at rest before the collision is based on confused thinking, and does not follow in any logical way from Newton's laws of motion or any other Newtonian principles.

 

[MeJennifer]

If the objects collide for a finite time, presumably it's because they're compressing as they come into contact (like rubber balls, or balls with springs affixed to each end), then decompressing as they bounce off in opposite directions. They will not lose contact until they have completely decompressed, this may not happen until after the velocity of the smaller has already exceeded the velocity of the larger.

Elastic and inelastic collisions are two completely different things.

 

[JesseM]

Elastic and inelastic collisions are two completely different things.

Yes, we talked a lot about elastic vs. inelastic collisions earlier on the thread. Here I think Chrisc wanted to talk about elastic collisions, so I was imagining two idealized objects that collide without losing any linear kinetic energy to heat.

 

[Chrisc]

Here is a nice practical demonstration that the above is not true. Take a heavy "super-ball" and hold a lighter "super ball" about one centimeter above the heavier ball. Release both balls at the same time. The larger ball rebounds off the floor with about the same speed of the still falling smaller ball but in the opposite direction and then collides with the still falling smaller ball. After the collision the smaller ball rebounds to a height that is higher than its original release point. If that is not in agreement with the second law, then the second law can be violated any time you drop two balls onto a hard surface and so by your arguments the second law is not in agreement with (easily demonstrated) observations.

kev, you are talking about composite matter again. We are talking about the principles of the laws as they pertain
to the "theoretical" transfer of energy between point-masses in collision.
What you have demonstrated can be thought of as a "sling shot". The smaller mass is in contact with the larger
for a period of time while it acquires a greater velocity. It can remain in contact because of its composite nature.
Its constituent parts, flex, decelerate, change direction and accelerate, all the while it is still a ball in macro terms.
Change the super balls to point-masses and the principles of the laws dictate the smaller will not acquire a greater velocity than the larger had before collision.

 

[Chrisc]

This principle is still something that appears to have come only from your own imagination, having no logical connection whatsoever to Newton's third law, which only says that the forces must be equal and opposite, not that there is any limit to their size. What's more, your statement still doesn't even make sense in terms of units! Mass times the square of velocity has units of kilograms * meters^2 / second^2, which are the units of energy, whereas the units of force are kilograms * meters / second^2.

Again, you are right. I should have said the mass of the smaller times the meters per second per second of the larger.
All the mathematical mistakes I have made and stated in this thread are obviously due to my lack of knowledge and ability with mathematics.
Mistakes should not be left unchecked, and I am grateful for your corrections.
But I hope you will consider the principles more carefully.
Newton's third law does dictate the limit of the forces by virtue of the necessity that they be equal.
As the motion between the masses is relative, the only property that determines the energy, momentum of each is their mass. The smaller mass then sets the limit.

No, as the smaller gains velocity to the left, the larger gains velocity to the right. Since the smaller started out moving to the right and the larger started out moving to the left, if they're colliding for a finite time, each one's speed is decreasing until it hits zero, then increasing as it continues to acquire velocity in the direction opposite to the one it was moving before the collision.

You are talking about a different collision here. I stated the observer was at rest with the larger or smaller in each case. You are considering them both initially moving.

If the objects collide for a finite time, presumably it's because they're compressing as they come into contact (like rubber balls, or balls with springs affixed to each end), then decompressing as they bounce off in opposite directions. They will not lose contact until they have completely decompressed, this may not happen until after the velocity of the smaller has already exceeded the velocity of the larger. If you don't believe me we could do an actual analysis of the situation where one of the objects has a spring on the end that becomes compressed as they get closer than the rest length of the spring, using law[/url].

I used your term "constituent" part, then your term "point-like" constituents. I have continued to use the more recognized term "point-mass" as you had seemed to have recognized what we are talking about with this term.
Now you are right back to composite matter, electromagnetic fields and springs.
For clarification, a "point-mass" is not a "point-particle", it is an idealized, infinitely small object.

Nope, your argument is handwavey and not based on anything in Newtonian mechanics.

Consider the masses as Point-masses and you will see it is true and based on Newtonian mechanics.

"Energy" is not a force. Physicists recognize only four forces in nature--the gravitational force, the electromagnetic force, the strong nuclear force, and the weak nuclear force. When physical objects like balls hit each other and rebound, this is understood as a consequence of electromagnetic forces between atoms on the surface of each during the period of contact. Of course, this is just what's true in fundamental physics, in Newtonian physics we do idealize other forces like the normal force and the spring force which depend on contact, even though at a microscopic level we know these forces are really due to the electromagnetic forces between atoms in the objects.

I did not say it was. I said the energy of mass, and the forces they exert. When the small mass is at rest it has no kinetic energy and no momentum, yet it exerts a force on the larger. That would be the "force" exerted by the energy of mass. Call it resistance to motion or inertia if you prefer.

Sure, and if we idealize them as exerting constant opposite forces on each other for some finite time-interval (which would not actually true if they behave like they've got a compressing spring between them during the contact, but it makes calculations either), then for each object (change in momentum)/(time-interval) will be equal to this force, and of course each object's change in momentum is just its own change in velocity times its own mass.

Unless and until the Higgs particle is found and confirmed, all mechanics pertaining to the interaction of mass will be and must be idealized.

But I don't agree, that's a made-up principle which has no logical connection to Newton's third law, and doesn't even make sense in terms of units.

This statement is true when the correction you mentioned is applied. M*m/s/s.

In Newtonian physics, Newton's laws hold in every inertial frame, there are no exceptions to any valid conclusion you get from the fundamental principles. The problem is just that your conclusion that the velocity of the smaller mass must be smaller than the larger after the collision if it's at rest before the collision is based on confused thinking, and does not follow in any logical way from Newton's laws of motion or any other Newtonian principles.

You seem far more comfortable with the math than the principles.
If you are willing to consider the theory and principles behind the mechanics of point-masses in collision, you will understand my point.

 

[DrGreg]

kev, you are talking about composite matter again. We are talking about the principles of the laws as they pertain
to the "theoretical" transfer of energy between point-masses in collision.
What you have demonstrated can be thought of as a "sling shot". The smaller mass is in contact with the larger
for a period of time while it acquires a greater velocity. It can remain in contact because of its composite nature.
Its constituent parts, flex, decelerate, change direction and accelerate, all the while it is still a ball in macro terms.
Change the super balls to point-masses and the principles of the laws dictate the smaller will not acquire a greater velocity than the larger had before collision.

Rubbish.

If you are talking about point masses, then you are talking about elastic collisions, the equations of conservation of momentum and kinetic energy apply and has been pointed out many, many times, your conclusion is completely wrong. "Superballs" are a very good approximation to elastic collisions, that's what makes them "super".

 

[yuiop]

kev, you are talking about composite matter again. We are talking about the principles of the laws as they pertain
to the "theoretical" transfer of energy between point-masses in collision.
What you have demonstrated can be thought of as a "sling shot". The smaller mass is in contact with the larger
for a period of time while it acquires a greater velocity. It can remain in contact because of its composite nature.
Its constituent parts, flex, decelerate, change direction and accelerate, all the while it is still a ball in macro terms.
Change the super balls to point-masses and the principles of the laws dictate the smaller will not acquire a greater velocity than the larger had before collision.

It might be as well to note that there are probably no real point masses in nature, which have mass and zero volume. Physicists think black holes might be an example of genuine point masses (I disagree) and in the case of colliding black holes I would imagine the collision is very inelastic.

 

[JesseM]

Again, you are right. I should have said the mass of the smaller times the meters per second per second of the larger.

Meters per second per second are units of acceleration. Are you just saying that F=m*a, i.e. Newton's second law of motion? Is so why didn't you say so? And again, in no way can you use F=m*a to show that the velocity of the smaller mass can't be larger than the velocity of the larger one. For example, if both received the same constant force F during the time t of the collision, then F=m*a tells us that the smaller mass will experience a larger constant acceleration during this time interval, and since for constant acceleration (change in velocity) = (acceleration)*(time interval), the smaller mass will experience a larger change in velocity.

Newton's third law does dictate the limit of the forces by virtue of the necessity that they be equal.

The only limit is that one be the same size as the other, Newton's third law alone doesn't set any limit on the magnitude of the force that both experience. In fact for an idealized collision between point masses, we must assume that each experiences an infinite force because the collision only lasts for an instant!

As the motion between the masses is relative, the only property that determines the energy, momentum of each is their mass. The smaller mass then sets the limit.

Their energy and momentum are determined by both mass and velocity.

As the smaller increases velocity, the larger decreases velocity.

No, as the smaller gains velocity to the left, the larger gains velocity to the right. Since the smaller started out moving to the right and the larger started out moving to the left, if they're colliding for a finite time, each one's speed is decreasing until it hits zero, then increasing as it continues to acquire velocity in the direction opposite to the one it was moving before the collision.

You are talking about a different collision here. I stated the observer was at rest with the larger or smaller in each case. You are considering them both initially moving.

Sorry, I was just thinking about the collision from the perspective of a different reference frame. From the perspective of the center-of-mass rest frame, their momenta must be equal and opposite at every moment, including during the collision, so if the collision happens over an extended time because their shapes are compressing and uncompressing like perfectly elastic rubber balls, in the center-of-mass frame both their momenta are symmetrically going to zero as they compress, they're both at rest at the point of maximum compression, and then increasing symmetrically as they decompress...since momentum is mass*velocity, the velocity of the smaller mass must be dropping more rapidly during the compression, and increasing more rapidly during the decompression (i.e. the smaller mass experiences a larger change in velocity in both phases). We can then switch to the frame where the smaller mass was initially at rest, and in this frame the center of mass of the system will be moving to the left at some constant velocity v which is smaller than the velocity V of the larger mass to the left. So in this frame, during the compression the smaller mass' velocity must be increasing to v to the left, while the larger mass' velocity is decreasing to v (since at maximum compression they were both at rest in the center-of-mass frame, and in this frame the center of mass is moving at v, they both must be moving at v at the point of maximum compression in this frame). Then during the decompression the smaller mass' velocity must be increasing symmetrically, so if it went from 0 to v during the compression it must go from v to 2v during the decompression, while if the larger mass' velocity decreased by (V - v) during the compression (since V - (V - v) = v), it must decrease by a further (V - v) during the decompression, so since it was moving at v at the point of maximum compression it will now be moving at v - (V - v) = 2v - V.

So, we conclude from this that in the frame where the smaller mass is initially at rest, after the collision the smaller mass is moving at 2v to the left, whereas the larger mass is moving at 2v - V to the left. Provided that 2v - V is always positive, this is just another way of showing that the smaller mass will always have a larger speed after the collision (if 2v - V could be negative, it might turn out that its magnitude to the right was actually larger than v to the left). And in fact it is possible to show 2v - V will always be positive--remember that V was the velocity of the larger mass, and v was the velocity of the center of mass, in the frame where the small mass is initially at rest, and center of mass velocity = (total momentum)/(total mass). In this case the small mass had zero momentum so the total momentum is just the momentum of the larger mass, MV, and the total mass is (m + M). So if v = VM/(m + M), then 2v - V = [2VM/(m + M)] - V = [2VM/(m + M)] - [V*(m + M)/(m + M)] = (2VM - Vm - VM)/(m + M) = (VM - Vm)/(M + m). Since M is larger than m, VM - Vm must be positive, so that means the velocity to the left of the larger mass after the collision, 2v - V, must always be positive in this frame, and "ALWAYS SMALLER" than the velocity to the left of the smaller mass after the collision, 2v.

I used your term "constituent" part, then your term "point-like" constituents. I have continued to use the more recognized term "point-mass" as you had seemed to have recognized what we are talking about with this term.
Now you are right back to composite matter, electromagnetic fields and springs.
For clarification, a "point-mass" is not a "point-particle", it is an idealized, infinitely small object.

I know what a point-mass is, but I didn't catch that you wanted to talk exclusively about point masses in your last two posts to me, I thought you just wanted to avoid talking about situations where kinetic energy was lost to heat (i.e. you wanted to focus on elastic collisions), and we can imagine perfectly elastic non-point masses which is what I was doing. Of course your comments about the limits on the magnitude of forces makes even less sense in the context of point masses--in order for energy and momentum to be conserved throughout all finite time-intervals, in Newtonian physics the collisions of point masses must always be infinitely brief (if they stayed in contact for any finite time, then for momentum to be conserved they'd have to be at rest in their center-of-mass frame, but that would mean their kinetic energy would drop to zero during this time-interval), which naturally means the force needed to achieve a finite change in velocity during that instant must be treated as infinitely large (a dirac delta function).

Nope, your argument is handwavey and not based on anything in Newtonian mechanics.

Consider the masses as Point-masses and you will see it is true and based on Newtonian mechanics.

How? What part of Newtonian mechanics? Do you understand that in Newtonian mechanics, for point masses the collision must be infinitely brief, so the force and acceleration at that instant must be infinite? This is why it would be much better to talk about elastic collisions of non-point masses which behave like idealized springs during the collision, compressing when they meet and then decompressing until they begin to move apart.

I did not say it was. I said the energy of mass, and the forces they exert. When the small mass is at rest it has no kinetic energy and no momentum, yet it exerts a force on the larger. That would be the "force" exerted by the energy of mass. Call it resistance to motion or inertia if you prefer.

This is all totally confused, neither inertia nor energy exerts a force in Newtonian mechanics, every specific force has a name, whether it's the spring force or the normal force or whatever. In an elastic collision of a deformable non-point-mass like a perfect rubber ball, the force is basically a type of spring force, caused by the compression of the ball during the collision. The more rigid the spring is (i.e. the greater the spring constant), the more brief the time will be between when they first touch and when they depart (because it the force rises more quickly as they compress and thus they accelerate in opposite directions more quickly and begin moving apart more quickly), and the infinite force which happens at a single instant in the collision of two point masses (or two perfectly rigid non-point masses) is simply an idealized limiting case of this.

Now, for a collision which takes place over some finite time the size of the two masses does enter into considerations of how they long they remain in contact, but that's just because F=m*a, so the same force doesn't accelerate a larger mass as much. Suppose we have a collection of perfectly elastic rubber balls that all have the same radius, and also all have the same internal "spring constant", so that if the radius of each ball is 5 cm, if you smoosh two balls together so their centers are only 8 cm apart (meaning each one's radius in that direction is squashed to 4 cm), the force that each ball is exerting on the other will be the same regardless of which two balls you chose. It would nevertheless be true that if you collided two lower-mass balls, they would become less compressed before starting to move apart and would remain in contact for less total time, the reason just being that by F = m*a they will be accelerating faster at a given amount of compression than larger balls would be, so it will take less time until their velocities have become matched, which must be the point of maximum compression since afterwards their velocities begin to increase in opposite directions.

This statement is true when the correction you mentioned is applied. M*m/s/s.

OK, so the forces are equal and opposite by the third law, and by the second law the magnitude of each object's acceleration is given by F = m*a, where F is the force on it and m is its own mass. You still haven't given any coherent argument as to why you think this implies there is a limit on the size of the force, or why it implies that the velocity of the smaller mass will be smaller than the velocity of the larger one after the collision (when in fact any Newtonian analysis will show the opposite must be true in an elastic collision).

You seem far more comfortable with the math than the principles.
If you are willing to consider the theory and principles behind the mechanics of point-masses in collision, you will understand my point.

You seem unwilling to consider the possibility that your grasp of the principles is poor, and that your overall point doesn't make sense. I'm sure if you try to make your argument in the classical physics forum, all the experts there would agree you don't know what you're misunderstanding the principles of Newtonian mechanics.

 

[Chrisc]

You seem unwilling to consider the possibility that your grasp of the principles is poor, and that your overall point doesn't make sense. I'm sure if you try to make your argument in the classical physics forum, all the experts there would agree you don't know what you're misunderstanding the principles of Newtonian mechanics.

OK let's assume I'm wrong and you're right.
From post #95 you said:

"whenever we have a collision where the larger mass comes to rest after the collision, it must be true that the velocity v1 of the larger mass before the collision is "ALWAYS LESS" than the velocity v2 of the smaller mass after the collision."

From the first post on, I have been talking about one or the other masses being initially at rest with respect to the observer.
I have been talking about perfectly elastic collisions of perfectly rigid mass points.
In my example, the larger (2-kg) mass was moving from left to right toward the smaller (1-kg) mass that was at rest with respect to the observer.
Will this collision result in the larger coming to rest?
If not, can you give me an example where the larger mass comes to rest after the collision when the smaller is initially at rest before the collision?

 

[atyy]

Please look at the attached diagram and let me know
if there is a reason for the asymmetric dynamics due
to the relative position of rest or, if I have incorrectly
interpreted the mechanics.
Kev, I haven't forgotten.

Let's work within Newtonian physics, since your question makes sense there.

The collision conserves momentum.

However it does not conserve kinetic energy, and so it is not elastic.

The formula for kinetic energy is mass(velocity)²/2

The initial kinetic energy is m(2v)²/2=2mv².

The final kinetic energy is 2m(v)²/2=mv².

This means that energy is lost as heat, which consists of vibrations in random directions or photons flying random directions. So it will not be time reversible, unless you also time reverse all those random movements (which you did not include in your diagram).

It's interesting why momentum can be conserved even though energy is not. The formulas for momentum and energy both contain velocity, so how can you change velocity without changing momentum? The answer is that momentum is a number and a direction (vector), but energy is just a (positive) number. The random vibrations take away the energy (ordinary addition and subtraction), and also the momentum - but the vibrations are random, so the momentum in one direction is taken away as much as in the opposite direction, and there is no loss of momentum (vector addition and subtraction) .

To see what happens if the collision is elastic, I suggest you solve for the final velocities of both balls by assuming both momentum (mv) and kinetic energy conservation (mv²/2).

 

[DrGreg]

OK let's assume I'm wrong and you're right.
From post #95 you said:

"whenever we have a collision where the larger mass comes to rest after the collision, it must be true that the velocity v1 of the larger mass before the collision is "ALWAYS LESS" than the velocity v2 of the smaller mass after the collision."

From the first post on, I have been talking about one or the other masses being initially at rest with respect to the observer.
I have been talking about perfectly elastic collisions of perfectly rigid mass points.
In my example, the larger (2-kg) mass was moving from left to right toward the smaller (1-kg) mass that was at rest with respect to the observer.
Will this collision result in the larger coming to rest?
If not, can you give me an example where the larger mass comes to rest after the collision when the smaller is initially at rest before the collision?

The maths isn't that difficult (in this specific case) and it's the only rigorous way to solve this.

Consider mass m with velocities v₁ and v₂ before and after collision, and mass M with velocities V₁ and V₂ before and after collision. Assume there are no other forces acting, and we'll use Newtonian formulas rather than relativistic.

Conservation of momentum always holds, so

mv_1 + MV_1 = mv_2 + MV_2 ...(1)​

If you have finally decided you want to consider elastic collisions[/color] only then conservation of kinetic energy holds too, so

\frac{1}{2} m v_1^2 + \frac{1}{2} M V_1^2 = \frac{1}{2} m v_2^2 + \frac{1}{2} M V_2^2 ...(2)​

These are the only two equations you need. Once you have specified "elastic" then it doesn't matter whether these are "rigid point masses" or not.

Now you have asked if it is possible to have v₁ = V₂ = 0. The equations now simplify to

MV_1 = mv_2 ...(3)
\frac{1}{2} M V_1^2 = \frac{1}{2} m v_2^2 ...(4)​

Divide (4) by ½(3) to get V₁ = v₂, and substitute back into (3) to get M = m. That wasn't hard, was it?

So the situation you describe can happen only for equal masses (assuming elastic collisions). (Example: Newton's cradle[/color].)

 

[JesseM]

OK let's assume I'm wrong and you're right.
From post #95 you said:

"whenever we have a collision where the larger mass comes to rest after the collision, it must be true that the velocity v1 of the larger mass before the collision is "ALWAYS LESS" than the velocity v2 of the smaller mass after the collision."

From the first post on, I have been talking about one or the other masses being initially at rest with respect to the observer.
I have been talking about perfectly elastic collisions of perfectly rigid mass points.
In my example, the larger (2-kg) mass was moving from left to right toward the smaller (1-kg) mass that was at rest with respect to the observer.
Will this collision result in the larger coming to rest?
If not, can you give me an example where the larger mass comes to rest after the collision when the smaller is initially at rest before the collision?

As DrGreg said, if mass #1 is initially at rest and mass #2 is coming towards it, and they collide elastically, the only way that mass #2 can come to rest is if their masses are equal. You can also see this by looking at the formula I derived in my previous post--if the mass initially at rest has mass m, and the mass moving towards it to the left has mass M and velocity V, then the center of mass of the system will always be moving to the left at v = VM/(m + M) since this is the total momentum divided by the total mass. With the variables defined this way, after the collision the mass m initially at rest will always be moving at 2v to the left, and the mass M will always be moving at (2v - V) to the left. So, the only way to have 2v - V = 0 is if [2VM/(m + M)] - V = 0, which means [2VM/(m + M)] - [V(m + M)/(m + M)] = (2VM - Vm - VM)/(m + M) = V(M - m)/(m + M) = 0. Assuming V is not zero, the only way V(M - m)/(m + M) can be 0 is if (M - m) = 0, i.e. M = m.

 

[Chrisc]

Thank you atyy, DrGreg and JesseM.
That is very logical mathematical reasoning.
In plain English you're saying that the incident mass will only come to rest if the mass it collides with is of equal mass.
In the case of the incident mass being greater than the mass at rest, both will continue to move in the same direction after collision. In the example JesseM gave their velocities will be 2/3m/s and 8/3m/s respectively, when the incident mass is 2-kg with velocity 2m/s and the other is 1-kg at rest. This also conserves momentum and kinetic energy.
How do you conserver the momentum and kinetic energy when the incident mass is less than the mass at rest?
It would appear from the reasoning you've given, if the incident mass is smaller than the mass at rest, the incident mass would never come to rest either.

 

[atyy]

It would appear from the reasoning you've given, if the incident mass is smaller than the mass at rest, the incident mass would never come to rest either.

Yes. And since in this case, you are conserving energy and momentum, we will not be able to attribute lack of reversibility to random vibrations of particles that you did not explicitly include in your diagram and calculation. So if this new situation fails to be time reversible, we will have no way out, and we shall have to conclude that Newton's laws are not time reversible. (Let me hedge my bets, and say there is also a possibility I didn't think this reply through carefully - since I shall be quite upset if I find out Newton's laws are not time reversible)

Why don't you try it and see?

--------------------
BTW, there is a subtlety about Newton's laws for point particles. If you just ask "There are 2 point particles with masses and velocities m1,m2,v1,v2, what happens if we apply F=ma?" - then there is actually NO answer.

F=ma requires the explicit specification of what F is.
For example, if there is gravity, then F=Gm1m2/r²
Or if the particles have charges q1,q1, then F=Kq1q1/r².
The reason you may have been confused is you tried to solve the problem without explicit specification of F.

The reason we solve the problem using energy and momentum conservation is that it is equivalent to solving the problem with only "partial", non-explicit specification of F. I can try to explain more if you're interested, but first you should calculate the case where energy and momentum are conserved, and see if it is time reversible.
--------------------

 

[DrGreg]

Thank you atyy, DrGreg and JesseM.
That is very logical mathematical reasoning.
In plain English you're saying that the incident mass will only come to rest if the mass it collides with is of equal mass.
In the case of the incident mass being greater than the mass at rest, both will continue to move in the same direction after collision. In the example JesseM gave their velocities will be 2/3m/s and 8/3m/s respectively, when the incident mass is 2-kg with velocity 2m/s and the other is 1-kg at rest. This also conserves momentum and kinetic energy.
How do you conserver the momentum and kinetic energy when the incident mass is less than the mass at rest?
It would appear from the reasoning you've given, if the incident mass is smaller than the mass at rest, the incident mass would never come to rest either.

I won't go through the proof (exercise for the reader!) but, unless I've made a silly mistake, the solution to equations (1) and (2) in my post #110 is:

v_2 = \frac{m-M}{m+M} v_1 + \frac{2M}{m+M} V_1 ...(5)
V_2 = \frac{2m}{M+m} v_1 + \frac{M-m}{M+m} V_1 ...(6)​

If V_1 = 0 and m < M (small mass m hitting larger stationary mass M),

v_2 = \frac{m-M}{m+M} v_1 < 0 ...(7)
V_2 = \frac{2m}{M+m} v_1 < v_1 ...(8)​

The smaller incident mass m rebounds in the opposite direction.


On the other hand, if v_1 = 0 and m < M (large mass M hitting stationary small mass m),

v_2 = \frac{2M}{m+M} V_1 > V_1 ...(9)
V_2 = \frac{M-m}{M+m} V_1 > 0 ...(10)​

The smaller stationary mass m is shot with a velocity greater than the incident mass M originally had, the larger mass slows down but does not stop.

 

[yuiop]

Thank you atyy, DrGreg and JesseM.
That is very logical mathematical reasoning.
In plain English you're saying that the incident mass will only come to rest if the mass it collides with is of equal mass.
In the case of the incident mass being greater than the mass at rest, both will continue to move in the same direction after collision. In the example JesseM gave their velocities will be 2/3m/s and 8/3m/s respectively, when the incident mass is 2-kg with velocity 2m/s and the other is 1-kg at rest. This also conserves momentum and kinetic energy.
How do you conserver the momentum and kinetic energy when the incident mass is less than the mass at rest?
It would appear from the reasoning you've given, if the incident mass is smaller than the mass at rest, the incident mass would never come to rest either.

If you go back to https://www.physicsforums.com/showpost.php?p=1799644&postcount=17"you will see that I showed way back then that the final velocities would be "be 2/3m/s and 8/3m/s respectively" as shown in this quote:

...There seems to be an error in the calculations in your diagrams when the calculations are done using the Newtonian equations.

The equation for a head on elastic collision is given here: http://hyperphysics.phy-astr.gsu.edu/Hbase/elacol2.html#c1

Using the notation given in that link, you have initial conditions:

Ball B1: m_1=2m, v_1=2v
Ball B2: m_2=1m, v_2=0v

The final velocity of mass m1 is:

v_1' = v1 \frac{m1-m2}{m1+m2} = 2v\frac{2m-1m}{2m+1m} =2/3v

The final velocity of mass m2 is:

v_2' = v1 \frac{2m_1}{m_1+m_2} = v\frac{4m}{2m+1m} =8/3v

To find out the final velocities when the incident mass is less than the rest mass you use the exact same equations and just change the variables as required, so for the case where the incident mass (B1) is 1 Kg moving at 2 m/s and the mass at rest (B2) is 2 Kgs:

Ball B1: m_1 = 1 Kg, v_1 = 2 m/s
Ball B2: m_2 = 2 Kg, v_2 = 0 m/s

The final velocity of B1 is:

v_1 ' = v_1 \frac{m_1-m_2}{m_1+m_2} = 2 m/s \frac{1 Kg - 2 Kg}{1 Kg + 2 Kg} = -2/3 m/s

The final velocity of B2 is:

v_2 ' = v_1 \frac{2*m_1}{m_1+m_2} = 2 m/s \frac{2 Kg}{(1 Kg + 2 Kg)} = +4/3 m/s

The negative sign for the final velocity of the smaller mass B1 simply means the smaller mass rebounds in the opposite direction to its initial velocity.

The formulas automatically conserve energy and momentum because they are derived assuming the conservation of energy and momentum as shown in the hyperphysics link.

... So if this new situation fails to be time reversible, we will have no way out, and we shall have to conclude that Newton's laws are not time reversible. (Let me hedge my bets, and say there is also a possibility I didn't think this reply through carefully - since I shall be quite upset if I find out Newton's laws are not time reversible)

Why don't you try it and see?

-------------------

To run the movie backwards and find the time reverse of the above situation simply make the final velocities the initial velocities and reverse the signs. I am using the full equations here because the equations used above are the simplified version where it assumed the target ball B2 is stationary. See http://en.wikipedia.org/wiki/Elastic_collision

Ball B1: m_1 = 1 Kg, v_1 = +2/3 m/s
Ball B2: m_2 = 2 Kg, v_2 = -4/3 m/s

The final (time reversed) velocity of B1 is:

v_1 ' = \frac{v_1(m_1-m_2)+2*m_2*v_2}{m_1+m_2} = \frac{2/3 m/s (1 Kg -2 Kg)+2*2 Kg* -4/3 m/s}{1 Kg + 2 Kg} = -2 m/s

The final (time reversed) velocity of B2 is:

v_2 ' = v_1 \frac{v_2(m_2-m_1)+2*m_1*v_1}{m_1+m_2} = \frac{2/3 m/s (2 Kg-1 Kg) +2*1 Kg*-2/3 m/s}{1 Kg + 2 Kg} = 0 m/s

which is the exact time reversed symmetry.

 

[yuiop]

In my last post (#117) I showed how to do the time reversed calculation using the full hairy momentum calculation but there is a simpler, quicker way to do it if you are comfortable with switching reference frames (which you should be if you are interested in relativity )

The calculaton for the incident mass having less mass than the stationary mass was:

...
To find out the final velocities when the incident mass is less than the rest mass you use the exact same equations and just change the variables as required, so for the case where the incident mass (B1) is 1 Kg moving at 2 m/s and the mass at rest (B2) is 2 Kgs:

Ball B1: m_1 = 1 Kg, v_1 = 2 m/s
Ball B2: m_2 = 2 Kg, v_2 = 0 m/s

The final velocity of B1 is:

v_1 ' = v_1 \frac{m_1-m_2}{m_1+m_2} = 2 m/s \frac{1 Kg - 2 Kg}{1 Kg + 2 Kg} = -2/3 m/s

The final velocity of B2 is:

v_2 ' = v_1 \frac{2*m_1}{m_1+m_2} = 2 m/s \frac{2 Kg}{(1 Kg + 2 Kg)} = +4/3 m/s

Taking the sign reversed final velocities as the new initial velocities for the time reversed situation the initial conditions are now:

Ball B1: m_1 = 1 Kg, v_1 = +2/3 m/s
Ball B2: m_2 = 2 Kg, v_2 = -4/3 m/s

The equations are simpler when one of the masses is at rest so we switch to new reference frame (F2) where the observer is moving at -4/3 m/s relative to the original reference frame (F1) and we do that by adding 4/3 m/s to all the velocities to obtain the initial conditions in F2 which are :

Ball B1: m_1 = 1 Kg, v_1 = 2 m/s
Ball B2: m_2 = 2 Kg, v_2 = 0 m/s

We note that we already know the answer to this calculation because these are the same initial conditions described in the my quote above so the final velocities in frame F2 are:

v_1 ' = -2/3 m/s

v_2 ' = +4/3 m/s

Now we switch back to the original reference frame to obtain the final velocities by subtacting 4/3 m/s from all the velocities to obtain the final answer in F1:

v_1 ' = -2 m/s

v_2 ' = 0 m/s

which is the exact time reversed symmetry.

 

[JesseM]

Thank you atyy, DrGreg and JesseM.
That is very logical mathematical reasoning.
In plain English you're saying that the incident mass will only come to rest if the mass it collides with is of equal mass.
In the case of the incident mass being greater than the mass at rest, both will continue to move in the same direction after collision. In the example JesseM gave their velocities will be 2/3m/s and 8/3m/s respectively, when the incident mass is 2-kg with velocity 2m/s and the other is 1-kg at rest. This also conserves momentum and kinetic energy.
How do you conserver the momentum and kinetic energy when the incident mass is less than the mass at rest?

The formulas I gave for the velocities after the collision works regardless of whether M is larger than m or vice versa. My formulas said that after the collision you'd have a velocity of 2v to the left for the mass m that was initially at rest, and (2v - V) to the left for the mass M that was initially approaching the other one at V to the left, with v as the velocity of the center of mass of the system, i.e. VM/(M + m) to the left (total momentum/total mass). If you work things out, you see that (2v - V) works out to V(M - m)/(m + M). Obviously if M is smaller than m, then (M - m) will be negative, so this means the velocity of the mass M in the left direction must be negative, i.e. after the collision it will always be moving to the right.

 

[atyy]

To find out the final velocities when the incident mass is less than the rest mass you use the exact same equations and just change the variables as required, so for the case where the incident mass (B1) is 1 Kg moving at 2 m/s and the mass at rest (B2) is 2 Kgs:

In my last post (#117) I showed how to do the time reversed calculation using the full hairy momentum calculation but there is a simpler, quicker way to do it if you are comfortable with switching reference frames (which you should be if you are interested in relativity )

Beautiful! I suppose we should wait for Chrisc's calculation, since even Feynman and Schwinger got the same wrong result when they calculated the Lamb shift.

 

[Chrisc]

Thanks everyone.
The math works fine in the sense that it conserves both momentum and kinetic energy.
I checked the relativity of the math and mechanics by setting the observer co-moving with each mass.
But I don't imagine anyone lost sleep waiting for my answer.
As kev said, it conserves because the math is designed to conserve. If the mechanics are (theoretically) considered valid because they meet the criteria of conservation, then there is no point in questioning the math.
But my problem is that the mechanics do not seem (theoretically) to uphold the principles.

Everyone agrees the 2kg/m/s momentum of the smaller incident mass imparts a 8/3kg/m/s momentum on the larger mass.
Everyone agrees that Newton's third law requires the collision be an exchange of equal and opposite force.
Does everyone also agree with the following:
The force required to accelerate a 1kg mass to 2m/s from rest, is equal to the force required to bring a 1kg mass to rest from an initial velocity of 2m/s?
The force required to accelerate a 2kg mass to 1m/s from rest, is equal to the force required to accelerate a 1kg mass to 2m/s from rest?
If so, how does a 2kg mass at rest impart more force (8/3kg/m/s^2) on the 1kg incident mass than the 1kg incident mass imparts on the 2kg mass (6/3kg/m/s^2) and uphold Newton's third law?

 

[Mentz114]

If so, how does a 2kg mass at rest impart more force (8/3kg/m/s^2) on the 1kg incident mass than the 1kg incident mass imparts on the 2kg mass (6/3kg/m/s^2) and uphold Newton's third law?

The measured forces depend on the frame of reference you choose to measure them in. It's been explained to you several times that there's no contradiction here.

But my problem is that the mechanics do not seem (theoretically) to uphold the principles.

You must have a direct line to a higher power. How can you understand physics without the maths ? I suggest your 'principles' are just wrong but you won't accept it.