# How to explain independence of momentum and energy conservation in elementary terms?

I'm trying to explain to someone learning elementary physics (16 year old) that linear momentum and energy are conserved independently. I'm not a professional physicist and haven't tried to explain this stuff for years, and I can't think of any convincing elementary argument to show that this is the case. Does anyone know of an elementary approach to this? (i.e. one that does not contain the expressions "Lagrangian" and "Noether's Theorem".)

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If you want to give a convincing argument that this is true, the best one is that experiments show it to be true! If that's not a satisfactory argument, I'm not sure what will be. – Ted Bunn Apr 4 '11 at 14:21
At elementary (and beyond!) the proof of the physics pudding is in experiment. – Nic Apr 4 '11 at 14:27
By "independent," do you mean that you're trying to explain the fact that there are situations where energy is conserved but momentum is not? – David Z Apr 4 '11 at 15:44
What do you mean by 'independently'? Both are conserved quantities. Two numbers that dont change. Are you looking for an explanation of momentum conservation not implying energy conservation and other way around ? – user1708 Apr 4 '11 at 15:49
My original trail of thought started with the fact that to solve a 2 body collision, we need both conservation of energy and momentum, and that eqn we get from conservation of energy provides extra info (and the second eqn) that we require if we are to solve for the two velocities. I guess mathematically this comes about because $v$ and $v^2$ are linearly independent polynomials, but I'm trying to explain the physics not the maths. – user2925 Apr 4 '11 at 22:05

Give him an example of inelastic collisions and explain why momentum is conserved but kinetic energy is not. If you explain the reasoning, (all forces are internal hence momentum is conserved) and that there are losses so KE of the system is lost in heat/sound/other forms of energy.. he should get the idea that the two are different beasts. At such a level it is best to illustrate with counterexamples.

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I'm not sure that the inelastic collision approach is that useful, since the apparently non-conserved KE is actually conserved (as KE, as well). – user2925 Apr 4 '11 at 21:46
@englishphysics The conservation of energy is subtle enough that it took two centuries to get to it; as the Wikipedia page on Newton's laws makes plain, "Conservation of energy was discovered nearly two centuries after Newton's lifetime, the long delay occurring because of the difficulty in understanding the role of microscopic and invisible forms of energy such as heat and infra-red light." I take it from this that the inelastic case is worth understanding at some level. – Peter Morgan Apr 5 '11 at 13:07
@englishphysics : There is a slight ambiguity in this question as to whether you mean "Kinetic Energy" or "Energy" - and its conservation and use in physics. The top question says "Energy" and the remark here refers to total energy too. You will need to separate the two in the education to make progress. But clearly elastic/inelastic are important types of mechanical examples. – Roy Simpson Apr 5 '11 at 16:49
@englishphysics KE is conserved in a way but not in the system that the problem focuses on. If you look at just the system of the two balls, it is not conserved. However the whole balls colliding thing is an open system which can exchange energy with its environment so the KE of the balls will be transferred as KE in the faster jiggling of nearby air molecules,as heat.. if you think that is too obtuse or confusing for a 16 year old then you're really underestimating them. – yayu Apr 5 '11 at 17:06

For a particle in a 1D external time-dependent field there are no energy and momentum conservation laws, yet there are two independent conserved quantities. It is because the differential equation is of the second order and it is accompanied with two independent initial data - the initial position and initial velocity. See an example here.

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Suppose someone suggests that following a perfectly elastic collision, two billiard balls are each traveling twice as fast as they were before (and opposite to their original directions). You can't prove him wrong using conservation of momentum, but you can prove him wrong using conservation of energy. Therefore conservation of energy has implications that conservation of momentum does not.

Suppose we have two identical billiard balls, one traveling east and one traveling west, with the same velocities and they collide. Suppose someone claims that both balls will now travel west, at the same velocity. You can't prove him wrong using conservation of energy, but you can prove him wrong using conservation of momentum. Therefore conservation of momentum has implications that conservation of energy does not.

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All the following explanation is for elementary students. First ensure that he/she understands that momentum is a vector quantity and energy is scalar. Also s/he might be knowing Newton's 2nd law of motion. Momentum of a system is unique in given direction. For a given system if net external force (on the system) is zero then net momentum of the system does not change. He can argue that so what energy will also not change. Yes, but we can have situation where we apply force on a body and its momentum is changed but energy is not. (A force applied perpendicular to the direction of motion all times). I hope this sets the sail for discussion.

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This looks like a step in the right direction. But how would you respond if the student observed that the norm of the momentum is also a scalar and that it also doesn't change when one applies forces in perpendicular direction? – Marek Apr 4 '11 at 15:55
Right. So you're suggesting, say, the example of uniform circular motion where the constant force => constant rate of change of momentum, but KE is constant at all times since $v \cdot v$ is constant ? That sounds useful. – user2925 Apr 4 '11 at 21:51

Consider for simplicity a non-relativistic collision between two point particles of same masses in their center-of-mass frame. From total momentum conservation, we know that the center-of-mass (COM) frame is an inertial frame. Moreover, if particle $1$ at some instance $t$ has position ${\bf r}_1$ and velocity ${\bf v}_1$ (relative to the COM frame), then particle $2$ is completely dictated to have opposite position ${\bf r}_2=-{\bf r}_1$ and opposite velocity ${\bf v}_2=-{\bf v}_1$. So from the COM perspective, the two-particle system is completely determined by knowing the state of particle $1$ alone.

Up until now, we have only used momentum conservation, and it doesn't matter whether the collision is elastic, partially elastic, or inelastic. The above is true regardless.

Now let us investigate a collision at initial and final instances $t_i$ and $t_f$ well before and well after the collision takes place. Note that we have already completely extracted all the information in the momentum conservation law to conclude that whatever the particle $1$ does, the particle $2$ would do the opposite. There are no more information available. In particular, momentum conservation gives us no clue about how initial and final velocity of particle $1$ are related.

Finally, let us restrict to an elastic collision. The kinetic energy conservation is in this context the independent statement that the initial and final speed $v_{1i}=v_{1f}$ of particle $1$ are equal (still measured relative to the COM frame).

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Momentum and energy are both different depending on what I compare the motion of an object with. If I'm in a train, I have no momentum or kinetic energy relative to the train. Relative to the fields outside the train, however, I have lots of momentum and energy. If I jump off the train, I will come to a stop relative to the fields, so the momentum and energy relative to the fields have to go somewhere. On the other hand, I will then be moving relative to the train, so the momentum and energy to make that happen also have to come from somewhere.

The difference between the momentum and the energy comes from the fact that the forces that change my speed have to act both for a certain amount of time and over a certain distance.

Suppose the force that slows me down is constant, and that the force doesn't make me spin or break me up into pieces. This is the sort of wild idealization that gets Physics a bad name with 16 year olds, but it's a first approximation from which we can go on to a second approximation that's better, and no-one has yet thought up a better first approximation. If the train travels twice as fast, the force has to act for twice as much time to bring me to stop, that's the change of momentum, but the force has to act for four times the distance to bring me to a stop, that's the change of energy.

This gets very tricky, because someone in an airplane that's moving really fast sees the force acting for the same amount of time as someone standing in the field sees the force acting for, but the person in the airplane sees the force acting for much more distance, because when the force started I was right next to the airplane, say, but when the force ended I was a long way behind. So, the change of energy from the point of view of the person in the airplane was much bigger than someone in the fields thinks it was, even though everyone agrees that the change of momentum was the same.

To switch to a different analogy, the energy is important because it determines how far it takes me to stop a car using the brakes, so it determines whether I hit the brick wall that I suddenly see in front of me. The momentum determines how much time it takes to come to a stop, but I can't immediately think of a really graphic situation when that's important.

One can construct different situations endlessly. It can be done in equations, of course, but you'll have to decide whether that's appropriate. I'll be interested if there's anything about this Answer that you think could be made clearer. It certainly isn't complete. Welcome from an Englishman in the USA.

EDIT: Of course overnight I realize that I mention conservation not once. From the point of view of the above it's enough to note that both can be understood to be because of Newton's third law, which is, from Wikipedia, "The mutual forces of action and reaction between two bodies are equal, opposite and collinear". As a result, we can say that the energy added to an object is taken away from the other object, and the same for momentum. The independence of the two conservation laws is essentially just because the two quantities are independent.

I've decided to add a few simplified equations, $$Force = Mass \times Acceleration,$$ $$Kinetic\ Energy = \mathsf{The\ Sum\ Of}\ The\ Forces\ Applied \times The\ Distance\ each\ Force\ is\ Applied\ For,$$ $$Momentum = \mathsf{The\ Sum\ Of}\ The\ Forces\ Applied \times The\ Length\ of\ Time\ each\ Force\ is\ Applied\ For,$$ or, as vector equations, almost certainly beyond what you need, $$\underline{F}=m\underline{a}, \qquad E=\int \underline{F}(t,\underline{s}(t))\cdot\frac{d\underline{s}(t)}{dt}dt, \qquad \underline{P}=\underline{F}(t,\underline{s}(t))dt.$$

Really, this stuff should be left to specialist educators, the best of whom will take time not only to create new ways to explain ideas, but also to study how well different strategies of explanation work for different kinds of student, but I've always been interested occasionally to put myself in this mindset. It's always humbling to discover how much creativity is needed to do it well.

As often, trawling around in Wikipedia, starting from the page on Newton's laws that I mention above, will render up some gems amongst the too-much-information for the purposes of your Question. I particularly like the tail-end comment that "Conservation of energy was discovered nearly two centuries after Newton's lifetime, the long delay occurring because of the difficulty in understanding the role of microscopic and invisible forms of energy such as heat and infra-red light."

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I wonder if there is room here for the power derivation based on momentum. Starting from (power)=(force)*(velocity) you scale with mass (finite, or infinitesimal to have (power)=(force/mass)*(mass*velocity)=(acceleration)*(momentum). So as energy is the integral of force over all speeds. Similarly energy is the integral of acceleration of all momenta. – ja72 Apr 5 '11 at 13:00
Hi @ja72. Perhaps another Question. My EDIT went over the top already. I think Power is notoriously difficult to keep distinct from Momentum and Energy. – Peter Morgan Apr 5 '11 at 13:14

Consider elastic collision of a small mass with a large mass (say, fast bullet with (initially) non-moving heavy metall ball). I guess you are considering Newtonian mechanics, so there are following options here:

a) bullet stops after collision, and heavy ball starts moving. If you will require that energy is conserved than you will see that momentum is not conserved, and vise versa. That means that in elastic collision the bullet cannot stop and transfer all its energy (or all its momentum) to the heavy ball because in this case both energy and momentum cannot be conserved.

b) Heavy ball remains not moving after collision, while the bullet is moving with the same velocity in arbitrary direction. Obviously, the energy is conserved, but momentum is not.

Hence, we can imagine the number of outcomes where only one quantity is conserved. But nature leaves us only one choice out of many because both energy and momentum need to be conserved. There are many scenarios where only momentum (or only energy) is conserved, but if we require that both are conserved, only one scenario remains possible.

It is also interesting to consider the situation where the momentum is seemingly not conserved. The simplest case is when a man (or woman) stays on the floor and at some moment of time start moving (walking). The initial momentum of a man (women) is zero, and the final momentum is not. What happens to conservation of momentum? Here it is important to consider a concept of closed system, because only in closed system the momentum is conserved. In this case the closed system includes the Earth. So when we start walking we move the Earth! :-)

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edited

Conservation of energy is almost self-evident. Conservation of linear momentum comes from the fact the in an isolated system of particles there is no preferred location. To try to make a point that the two are related seems that would invoke some really convoluted arguments. One deals with invariance of energy and the other with invariance of location.

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Convoluted? If there are no forces present then $E = {p^2 \over 2m}$. To state that they are not related is what's convoluted. And how is conservation of energy more self-evident than conservation of momentum? Both need nontrivial assumptions and nontrivial theorems (at least to layman) to prove. Very strange answer... – Marek Apr 4 '11 at 15:46
And I just noticed that last sentence is also complete crap, so -1. Both have to do with invariance w.r.t. to translations. Orientation plays no role here. – Marek Apr 4 '11 at 15:48
I forgot how to read! I was thinking that linear momentum has a line and an orientation, so wrote down the wrong thought! Thanks for point it out. – ja72 Apr 4 '11 at 16:51
Hmm. Linear momentum has a "line" and an orientation ? So it's just like a vector, then ? :-) – user2925 Apr 4 '11 at 22:07
No it is more than a vector. It is a line (direction+location), just like a force. To describe the linear momentum of a rigid body you have to specify its components + the location of the center of gravity. To specify it away from the COG, one must include the cross-moment components $\vec{r}\times\vec{L}$. The conservation law includes the location as well as the components (orientation) of momentum. – ja72 Apr 5 '11 at 12:33

Check out Noether's Theorem.

Conservation laws are best explained as consequences of space and time symmetries: Because space is the same in every direction we see momentum as beng conserved, because time is uniform we see energy as being conserved. http://en.wikipedia.org/wiki/Noether's_theorem

Note that energy is not conserved in relativistic situations but momentum is - this is related to the nature of relativity.

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-1: Energy may or may not be conserved in relativistic systems just like nonrelativistic systems. – DJBunk Feb 7 '13 at 22:56

The 16 year old is on to something but no one realizes it because physicists and those who teach physics assume that kinetic energy is a valid scientific principle. This opening sentence will naturally solicite laughter and other reactions but take a moment anyway to consider the following analogy. It looks at what happens when you define energy or force with respect to distance. Energy (kinetic energy) is related to force acting through a distance. Imagine a passenger sitting in the rear seat of a car and he hands his cell phone to the driver. If the car is stationary, that phone travels about 3 feet with respect to the road. If the car is moving, the phone might travel 10, 20, 30 or more feet. The time it takes to hand the phone does not change; it is a constant. Now think about force, it causes things to accelerate. When a body accelerates, it changes its velocity/speed. As this occurs, that body will travel a certain distance and take TIME to do so. First, take wind resistance and other unrelated things out of the mix. Accelerate a body from 0 mph to 10 mph; this takes time and the force will act through a distance. Accelerate the same body from 10 mph to 20 mph using the same amount of force. The time to change that body's speed by 10 mph does not change; the distance it travels during that act will. In short, there is something wrong with the kinetic energy formula. Do not do what everyone does and assume that because something has been around for a long time that it is correct.

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Would you please express your doubts in formulas people can understand and explain precisely why do you think there is something wrong? – Bzazz Feb 7 '13 at 22:53
I think he's saying that kinetic energy is frame dependent, which is true and not at all a problem. More generally the total energy is frame dependent, but conserved. – Michael Brown Feb 8 '13 at 0:35