Losing energy during a collision of bodies vs conservation of momentum

Factual Questions
1

I’ve been asking my engineer father and the engineer math teacher across the hall about this, but I still don’t truly see why I’m wrong. I admit I’m essentially arguing against the conservation of momentum, but I don’t see how I’m mistaken. This is my first year teaching Physics, and I haven’t encountered all the weird questions from kids like I have for chemistry.

A 1.5 kg rock traveling 4.0 km/hr hits a 3.5 kg ball of clay.

A. What is the new velocity of the combined system (they stuck together)?

B. What is the lost energy from this action?

A. The book wants to use conservation of momentum to figure a new velocity. Easily done: 1.5 km/hr.

B. This v is then used in the KE formula to find the new KE for the system: 4.5 J. The old KE of the rock was 12 J. Subtracting these leaves a gap of 7.5 J that was lost.
My “argument” is that this is unknowable. The rock must have deformed the clay, creating work and lost heat/energy to an unknown amount. This could have only come from the KE of the system, right?? If it did, then the only value on the right side of the KE equation that could have decreased is the v, to an unknown extent. This means that I don’t know what the v really is.

I realize that this means I’m arguing that momentum is not conserved even if the two colliding objects are moving in a frictionless environment, but the more I think about it the more I convince myself. I had assumed that the textbooks were just cheating and assuming absolutely perfect conditions, where there is no friction, or energy lost, but then this problem goes and admits that energy is lost, but figures out the amount lost from a seemingly perfect collision during which no energy is lost

I’ll not be surprised to be shown that this has a sizable hole in its reasoning, but I can’t find it, and the people I’m asking are not really addressing my point directly. Can some of the physics/teachers on the Dope help?

2

Where are you getting the notion that no kinetic should be lost? The “lost” kinetic energy is converted to heat. Only in a perfectly elastic collision is kinetic energy preserved, but then the two objects do NOT end up traveling together in that case. The whole point of momentum is that it IS conserved in all colisions, but kenetic energy is not.

Aside: The OP is stunningly close to the pseudo science presented in a Linden Larouche publication in an attempt to discredit Newton.

3

First off the speed should be m/s not km/hr to get those Joule numbers. Second, the new velocity is 1.2 m/s, not 1.5. Making the final kinetic energy 3.6 J.

The energy is not lost but instead turned into sound and heat.

4

Sorry about the km/hr stuff. Of course, you’re right.

RATS. I meant it was a 2.5 kg piece of clay. Typo.
I know that heat is created by the loss of kinetic energy. That’s my entire point.

I guess I’m trying to ask how momentum is conserved but kinetic energy is not. If energy is lost to heat, we don’t know how much, and it seems to me that the v is unknowable, then, meaning that momentum is unknowable except in “cheater” problems, where facts like this are ignored.

5

I taught physics a few years back, (after several years of teaching chemistry, too, like the OP).

Momentum is always conserved. It’s a scientific law. No exceptions to this have ever been observed. There is no “how” here. As you should know from teaching chemistry, scientific laws are not explanations; they are a summary of observed behavior of the universe.

You can predict what the final velocity(ies) will be after a collision using conservation of momentum, and this matches the observed behavior if you do an experiment. Why do you think this final velocity is “unknowable”? You can measure it, for crying out loud! We’re not talking about quantum mechanics here!

Kinetic energy is generally not conserved (in macroscopic, observable collisions); it is only conserved in perfectly elastic collisions.

In your example, you have an inelastic collision. By definition, kinetic energy is not conserved. If you know the initial and final velocities, you can determine exactly how much kinetic energy is lost. That energy is converted into sound and heat as the clay deforms.

If you doubt all of this, for some reason, do an experiment!

What are you not getting here?

6

Momentum is conserved. That’s a scientific law.

The energy is also conserved: it’s just that some of it has been converted to heat energy. There’s no law of conservation of kinetic energy.

An alternative way of thinking of it is to only consider the kinetic energy, but then you have to recognize that you don’t have a closed system, so the law of conservation of energy does not apply.

7

We have two equations here:

Momentum:
M[sub]A1[/sub]*V[sub]A1[/sub]+M[sub]B1[/sub]*V[sub]B1[/sub]=M[sub]A2[/sub]*V[sub]A2[/sub]+M[sub]B2[/sub]*V[sub]B2[/sub]

Energy:

1/2*M[sub]A1[/sub]V[sub]A1[/sub][sup]2[/sup]+1/2M[sub]B1[/sub]V[sub]B1[/sub][sup]2[/sup]=1/2M[sub]A2[/sub]V[sub]A2[/sub][sup]2[/sup]+1/2M[sub]B2[/sub]*V[sub]B2[/sub][sup]2[/sup]+E
In these equations we have three unknowns: V[sub]A2[/sub], V[sub]B2[/sub], and E. E in this case represents the energy in the system that is not accounted for by the kinetic energy of the two masses. Two equations and three unknowns means that the system is impossible to solve. There are three ways to rectify the situation. (1) Assume that the masses stick together. Then V[sub]A2[/sub]=V[sub]B2[/sub]=V[sub]2[/sub], reducing the unknowns to two. (2) Assume that all of the energy remains in the two masses. This means E=0, and we are down to two unknowns with two equations. (3) Develop other equations based on the properties of the masses. (3) is very hard to do, so all the problems you see in basic physics use either (1) or (2) as an assumption.

In the case in the OP (1) is assumed, and therefore the equations are solvable without any voodoo.

8

If you would like to delve a little deeper into why momentum is conserved Google Emmy Noether symmetry / conservation law. Einstein praised her theorem as “penetrating mathematical thinking”

9

I think that what’s confusing the OP is the fact that heat is messy. It would, in fact, be very difficult to predict precisely how the lump of clay deforms, or exactly how much energy is deposited in what parts of the clay. But that doesn’t mean that we can’t tell anything about the energy lost to heat. What we can say, and about the only thing we can easily say, is the total amount of energy converted to heat.

10

In the real world, of course, things are messier. The problem, as originally stated, assumes that the two masses stick together – that’s a Perfectly Inelastic Collision. Momemntum is conserved, and after the collision the two bodies move off together with the same speed. All the parameters are known, so you can figure out what the velocities of the bodies are after the collision, if you know their velocities and masses beforehand.

At the other extreme is the Perfectly Elastic Collision, in which energy and momentum are conserved, and again you can solve the equations to figure out the velocities after the collision.
Almost all real collisions fall somewhere between these two ideal and extreme cases, with some energy lost to the collision, although you don’t know how much, in general, beforehand. I think this is the kind of uncertainty the OP (and his subsequent post) were talking about. But even in this case you can quantify things. You can define a coefficient of restitution that characterizes the collision. One useful measure is the velocity difference between the bodies after the collision. If you know that, you can mathematically figure out the energy lost, and the velocities of the bodies afterwards. Because, of course, momentum is still conserved.

Conservation of Momentum, by the way, isn’t just a fortuitous result of Cosmic Bookkeeping. It’s a natural consequence of the symmetry of space – conservation of mementum is a direct result of no explicit position term appearing in the Lagragian equation describing the condition of the system. If it doesn’t matter if the collision happens HERE or over THERE, then momentum is going to be conserved.

11

I would argue that real world [perfectly] inelastic collisions are not rare, uncommon thought experiments. Any real-world collision in which the two objects stick together is a [perfectly] inelastic collision. Heck in, the case of two similar objects with opposing velocities (that stick together), all of the kinetic energy is lost.

The only example of real-world perfectly elastic collisions that I can think of are gas molecules colliding. Any real-world macroscopic collision is going to involve some loss of energy due to sound and heat.

12

And I’d re-iterate that they’re not common. The standard physics-lab example of “two clay balls” that stick together isn’t even usual for clay balls, which tend to bounce apart. Colliding automobiles bounce off each other, after crumpling. Even if they “lock”, it’s after a bounce.

13

Would a baseball catcher receiving a pitch work as an example?

14

That’s not a case of any kind of simple collision – the baseball player is himself rooted to the ground, and he has all kinds of degrees of freedom with his limbs moving in different directions. “simple” is like two superballs running into each other, or two lumps of clay, or whatever. The bodies are small and not connected to anything, you assume they smack dead-on along the line of therir centers of mass (so you don’t have to worry about rotation). THAT’s a simple collision.

This was a live issue for me when I was doing my bachelor’s thesis – I was studying the effects of a Karate strike. If your hand hits a board of wood, is it elastic, inelastic, or somewhere in between? I would’ve guessed the last of those – the hand deforms, but it certainly doesn’t stick, and both bodies are free to move indepently. If you’ve got the wooden board supported at the ends (as the usually do for a demonstration), then it’s connected to something, but you can reason that over a sufficiently short time scale, the board acts like a simple mass on a spring (true, and easily modeled). Or you can study the case of a board that has been thrown into the air and is struck, if that bothers you.
So what is it? What kind of collision is it?

[spoiler]BZZZZZZ!
It’s d.) None of the Above

Using high-speed films and strobe photos, I found that the board broke while the hand was still in contact with the board, and the hand was still deforming and the board still accelerating. One of the unspoken assumptions of the classification of collisions is that the collision mechanism is over and done with before you look at the system again and measure their relative velocities, or whatever. But the hand and the board were in fact still interacting when the board broke and the pieces came away from the hand. The whole process took place in a couple of milliseconds. [/spoiler]

15

Another example is gravitational interactions between objects in space, such as gravitational slingshot maneuvers. One might not think of them as “collisions”, but all of the math is the same. In fact, they’re probably better examples than gas molecules, since most gas molecules (anything but noble gasses, and even those at high enough temperatures) have some internal degrees of freedom which could hold energy (the molecule can rotate, or the bonds between the atoms can vibrate in various ways, or (at high temperatures), the electrons can change orbitals).

And while perfectly inelastic collisions might not occur very often spontanously, they’re at least relatively easy to contrive. Much easier, certainly, than perfectly elastic ones.

16

OK, I talked to my physics major brother (Cal Poly San Luis Obispo, '03), and he said he’d never truly thought through the conservation of momentum before, but here is his take on it, as I understand it.

The rock hits the clay, and penetrates it until they are both going the same speed. This part is obvious.

His guess is that the work to penetrate a soft clay deeper or a hard clay shallower requires the same amount of work, hence losing the same amount of energy. If that’s true, since the masses have not changed, the only thing that could have changed to give up the energy is the velocity.

Is this why momentum in conserved? It still sounds kind of magical to me, and he wasn’t sure about it himself, but is the lost energy the same in all circumstances (it must be), and is this why? If not, then why is it always the same?

What I object to is being told to shut up and just accept that momentum is conserved, which is what some here have done. There absolutely IS a why for this kind of thing. Why plus attracts minus is currently a “because” situation, but I’ll bet there’s an explainable reason for that, too.

Chronos mentions that the energy lost to heat is what we can calculate, which means it must be the same in all possible rock-hits-clay type situations. I have the feeling there’s something of F=ma in there, about the materials’ qualities automatically resisting the deformation to differing extents, but always eventually using the same force to accelerate the new mass to the same velocity, thereby using up the same amount of energy.

17

It’s still not clear what you’re looking for here.
If I recall correctly (Chronos will probably correct me), with some mathematics you can show that any force with an inverse square law will result in momentum being conserved. So your question is either ‘explain the math’ or ‘why do gravity and electromagnetism vary as the inverse square of distance?’. I’m afraid I’m too rusty to do the first and don’t know if you have the math do follow even if I was fresh; I think we both understand that the only answer physics has to the second is ‘Because.’ If you can figure out how to go beyond that ‘Because’, well, then share it, and there’s a prize in Sweden waiting for you.

I think you may also be confused by thinking that all collisions are inelastic. That’s not true: if you throw a bunch of rocks at a bunch of pieces of clay, some of the time the rock and clay will bounce off of each other, sometimes the clay will break into pieces with part of it sticking to the rock and some bouncing off, sometimes the rock will tear right through the clay and keep going, and sometimes the rock and clay will stick together. Only the last case is inelastic. And in each case, the amount of energy lost to heat will be different.
The other thing that may be confusing you is calculating the energy lost to heat.
You’re right in that one way to calculate energy transferred to heat is by examining every detail of the collision, looking at the tiny deformations of the clay, calculating the work done, etc. and summing it all up. But the beautiful thing about conservation of energy is that we don’t have to do that if we just want a total. Energy in = energy out, so we know that the change in heat energy is just the difference in kinetic energy. If we know the beginning masses and velocities and ending masses and velocities, then we can bingo calculate how much energy was put into heat. Conservation of momentum is only used in this problem because we don’t know the final velocities, and it gives us a way to calculate them.

Does any of that help?

18

The conservation of momentum is a law because countless experiments and observations have shown it to be true. It’s not just a “because” statement, it’s backed up by tons of evidence. If you can manage to get two clay blobs to collide in a way that doesn’t conserve momentum, well, there might be a Nobel Prize in it for you.

Is there something you didn’t understand about my explanation?

19

From which first principles would you like to start? If Newton’s laws are good for you, then the conservation of momentum follows directly from Newton’s third law.

Assuming no external forces, and object A and B are interacting in some way. A has a mass m[sub]A[/sub] and initial velocity v[sub]0A[/sub], and B has a mass m[sub]B[/sub] and initial velocity v[sub]0B[/sub]. The two bodies interact in some way, so that there’s a force of A on B, and a force of B on A. At any given moment, due to Newton’s third law,
F[sub]AB[/sub] = -F[sub]BA[/sub].

The velocity of A at any given time is

v[sub]A/sub = v[sub]0A[/sub] + delta v[sub]A/sub, or V[sub]0A[/sub] + int(a[sub]A/sub, t’=0…t) .

Or, since F = ma,

v[sub]A/sub =V[sub]0A[/sub] + 1/m[sub]A[/sub] * int(F[sub]BA/sub, t’=0…t)

and the momentum of A is

P[sub]A/sub =P[sub]0A[/sub] + int(F[sub]BA/sub, t’=0…t)

By the same arguments, the momentum of B is

P[sub]B/sub =P[sub]0B[/sub] + int(F[sub]AB/sub, t’=0…t)

or, using F[sub]AB[/sub] = -F[sub]BA[/sub],

P[sub]B/sub =P[sub]0B[/sub] - int(F[sub]BA/sub, t’=0…t)

So, finally, at any moment of time, P[sub]A/sub + P[sub]B/sub = P[sub]0A[/sub] + P[sub]0B[/sub], so the total momentum is always the same as the initial total momentum.

Alternately, you can take conservation of momentum as your first principle, and derive Newton’s third law from that. Or you can start with the least-action principle, or Hamilton’s principle, or probably some other formulations of sets of first principles. But you have to start with something.

20

You said it; every time I try to make a substance that is perfectly elastic, I end up in the middle of a bunch of hijinks involving the military, a wealthy industrialist, a flying car, and a college basketball team, and I end up not missing my wedding to Nancy Olsen. (I don’t care what they say about her nose, it looks fine to me.) So stay away from materials with a coefficient of restitution of 1 or more; Maxwell doesn’t like them.

As for the o.p.'s issue with motus and the conservation thereof, this is a fundamental axiom of Newtonian mechanics. (It also holds in Special and General Relativity, and in its own stochastic way, in quantum mechanics as well.) I don’t know that I can add anything to the explanations already provided; if you want to challenge it, you’ll need to take on the whole system. Good luck with that, because it has and continues to work quite well, not only with discrete systems like a ball and a lump of clay, but also with multi-particle and fluid systems. The fact that you can’t directly account for where the energy goes i.e., it deforms the material resulting in heat and sound (and if you hit it hard enough, light) doesn’t mean that it is lost to the system as long as your system encompasses those emissions. Energy and momentum are both conserved, but energy can change “form” (i.e. from kinetic to potential, or chemical, or heat) while momentum remains as mass in motion.

I’m a little surprised that the o.p.'s “physics major brother…never truly thought through the conservation of momentum before”; although I didn’t go all the way through the physics sequence, there should have been at least a couple of higher level mechanics courses (classical and statistical) along with the entry-level general physics class which delved into the topic of momentum in detail. Momentum is one of the primary concepts in classical mechanics.

Stranger