experiment on law of conservation of momentum

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10 Conservation of Momentum

10–1 Newton’s Third Law

On the basis of Newton’s second law of motion, which gives the relation between the acceleration of any body and the force acting on it, any problem in mechanics can be solved in principle. For example, to determine the motion of a few particles, one can use the numerical method developed in the preceding chapter. But there are good reasons to make a further study of Newton’s laws. First, there are quite simple cases of motion which can be analyzed not only by numerical methods, but also by direct mathematical analysis. For example, although we know that the acceleration of a falling body is $32$ ft/sec², and from this fact could calculate the motion by numerical methods, it is much easier and more satisfactory to analyze the motion and find the general solution, $s=s_0+v_0t+16t^2$. In the same way, although we can work out the positions of a harmonic oscillator by numerical methods, it is also possible to show analytically that the general solution is a simple cosine function of $t$, and so it is unnecessary to go to all that arithmetical trouble when there is a simple and more accurate way to get the result. In the same manner, although the motion of one body around the sun, determined by gravitation, can be calculated point by point by the numerical methods of Chapter  9 , which show the general shape of the orbit, it is nice also to get the exact shape, which analysis reveals as a perfect ellipse.

Unfortunately, there are really very few problems which can be solved exactly by analysis. In the case of the harmonic oscillator, for example, if the spring force is not proportional to the displacement, but is something more complicated, one must fall back on the numerical method. Or if there are two bodies going around the sun, so that the total number of bodies is three, then analysis cannot produce a simple formula for the motion, and in practice the problem must be done numerically. That is the famous three-body problem, which so long challenged human powers of analysis; it is very interesting how long it took people to appreciate the fact that perhaps the powers of mathematical analysis were limited and it might be necessary to use the numerical methods. Today an enormous number of problems that cannot be done analytically are solved by numerical methods, and the old three-body problem, which was supposed to be so difficult, is solved as a matter of routine in exactly the same manner that was described in the preceding chapter, namely, by doing enough arithmetic. However, there are also situations where both methods fail: the simple problems we can do by analysis, and the moderately difficult problems by numerical, arithmetical methods, but the very complicated problems we cannot do by either method. A complicated problem is, for example, the collision of two automobiles, or even the motion of the molecules of a gas. There are countless particles in a cubic millimeter of gas, and it would be ridiculous to try to make calculations with so many variables (about $10^{17}$—a hundred million billion). Anything like the motion of the molecules or atoms of a gas or a block of iron, or the motion of the stars in a globular cluster, instead of just two or three planets going around the sun—such problems we cannot do directly, so we have to seek other means.

In the situations in which we cannot follow details, we need to know some general properties, that is, general theorems or principles which are consequences of Newton’s laws. One of these is the principle of conservation of energy, which was discussed in Chapter  4 . Another is the principle of conservation of momentum, the subject of this chapter. Another reason for studying mechanics further is that there are certain patterns of motion that are repeated in many different circumstances, so it is good to study these patterns in one particular circumstance. For example, we shall study collisions; different kinds of collisions have much in common. In the flow of fluids, it does not make much difference what the fluid is, the laws of the flow are similar. Other problems that we shall study are vibrations and oscillations and, in particular, the peculiar phenomena of mechanical waves—sound, vibrations of rods, and so on.

In our discussion of Newton’s laws it was explained that these laws are a kind of program that says “Pay attention to the forces,” and that Newton told us only two things about the nature of forces. In the case of gravitation, he gave us the complete law of the force. In the case of the very complicated forces between atoms, he was not aware of the right laws for the forces; however, he discovered one rule, one general property of forces, which is expressed in his Third Law, and that is the total knowledge that Newton had about the nature of forces—the law of gravitation and this principle, but no other details.

This principle is that action equals reaction .

What is meant is something of this kind: Suppose we have two small bodies, say particles, and suppose that the first one exerts a force on the second one, pushing it with a certain force. Then, simultaneously, according to Newton’s Third Law, the second particle will push on the first with an equal force, in the opposite direction; furthermore, these forces effectively act in the same line. This is the hypothesis, or law, that Newton proposed, and it seems to be quite accurate, though not exact (we shall discuss the errors later). For the moment we shall take it to be true that action equals reaction. Of course, if there is a third particle, not on the same line as the other two, the law does not mean that the total force on the first one is equal to the total force on the second, since the third particle, for instance, exerts its own push on each of the other two. The result is that the total effect on the first two is in some other direction, and the forces on the first two particles are, in general, neither equal nor opposite. However, the forces on each particle can be resolved into parts, there being one contribution or part due to each other interacting particle. Then each pair of particles has corresponding components of mutual interaction that are equal in magnitude and opposite in direction.

10–2 Conservation of momentum

Now what are the interesting consequences of the above relationship? Suppose, for simplicity, that we have just two interacting particles, possibly of different mass, and numbered $1$ and $2$. The forces between them are equal and opposite; what are the consequences? According to Newton’s Second Law, force is the time rate of change of the momentum, so we conclude that the rate of change of momentum $p_1$ of particle $1$ is equal to minus the rate of change of momentum $p_2$ of particle $2$, or \begin{equation} \label{Eq:I:10:1} dp_1/dt=-dp_2/dt. \end{equation} Now if the rate of change is always equal and opposite, it follows that the total change in the momentum of particle $1$ is equal and opposite to the total change in the momentum of particle $2$; this means that if we add the momentum of particle $1$ to the momentum of particle $2$, the rate of change of the sum of these, due to the mutual forces (called internal forces) between particles, is zero; that is \begin{equation} \label{Eq:I:10:2} d(p_1+p_2)/dt=0. \end{equation} There is assumed to be no other force in the problem. If the rate of change of this sum is always zero, that is just another way of saying that the quantity $(p_1+p_2)$ does not change. (This quantity is also written $m_1v_1+m_2v_2$, and is called the total momentum of the two particles.) We have now obtained the result that the total momentum of the two particles does not change because of any mutual interactions between them. This statement expresses the law of conservation of momentum in that particular example. We conclude that if there is any kind of force, no matter how complicated, between two particles, and we measure or calculate $m_1v_1+m_2v_2$, that is, the sum of the two momenta, both before and after the forces act, the results should be equal, i.e., the total momentum is a constant.

If we extend the argument to three or more interacting particles in more complicated circumstances, it is evident that so far as internal forces are concerned, the total momentum of all the particles stays constant, since an increase in momentum of one, due to another, is exactly compensated by the decrease of the second, due to the first. That is, all the internal forces will balance out, and therefore cannot change the total momentum of the particles. Then if there are no forces from the outside (external forces), there are no forces that can change the total momentum; hence the total momentum is a constant.

It is worth describing what happens if there are forces that do not come from the mutual actions of the particles in question: suppose we isolate the interacting particles. If there are only mutual forces, then, as before, the total momentum of the particles does not change, no matter how complicated the forces. On the other hand, suppose there are also forces coming from the particles outside the isolated group. Any force exerted by outside bodies on inside bodies, we call an external force. We shall later demonstrate that the sum of all external forces equals the rate of change of the total momentum of all the particles inside, a very useful theorem.

The conservation of the total momentum of a number of interacting particles can be expressed as \begin{equation} \label{Eq:I:10:3} m_1v_1+m_2v_2+m_3v_3+\dotsb=\text{a constant}, \end{equation} if there are no net external forces. Here the masses and corresponding velocities of the particles are numbered $1$, $2$, $3$, $4$, … The general statement of Newton’s Second Law for each particle, \begin{equation} \label{Eq:I:10:4} F=\ddt{}{t}(mv), \end{equation} is true specifically for the components of force and momentum in any given direction; thus the $x$-component of the force on a particle is equal to the $x$-component of the rate of change of momentum of that particle, or \begin{equation} \label{Eq:I:10:5} F_x=\ddt{}{t}(mv_x), \end{equation} and similarly for the $y$- and $z$-directions. Therefore Eq. ( 10.3 ) is really three equations, one for each direction.

In addition to the law of conservation of momentum, there is another interesting consequence of Newton’s Second Law, to be proved later, but merely stated now. This principle is that the laws of physics will look the same whether we are standing still or moving with a uniform speed in a straight line. For example, a child bouncing a ball in an airplane finds that the ball bounces the same as though he were bouncing it on the ground. Even though the airplane is moving with a very high velocity, unless it changes its velocity, the laws look the same to the child as they do when the airplane is standing still. This is the so-called relativity principle . As we use it here we shall call it “Galilean relativity” to distinguish it from the more careful analysis made by Einstein, which we shall study later.

We have just derived the law of conservation of momentum from Newton’s laws, and we could go on from here to find the special laws that describe impacts and collisions. But for the sake of variety, and also as an illustration of a kind of reasoning that can be used in physics in other circumstances where, for example, one might not know Newton’s laws and might take a different approach, we shall discuss the laws of impacts and collisions from a completely different point of view. We shall base our discussion on the principle of Galilean relativity, stated above, and shall end up with the law of conservation of momentum.

We shall start by assuming that nature would look the same if we run along at a certain speed and watch it as it would if we were standing still. Before discussing collisions in which two bodies collide and stick together, or come together and bounce apart, we shall first consider two bodies that are held together by a spring or something else, and are then suddenly released and pushed by the spring or perhaps by a little explosion. Further, we shall consider motion in only one direction. First, let us suppose that the two objects are exactly the same, are nice symmetrical objects, and then we have a little explosion between them. After the explosion, one of the bodies will be moving, let us say toward the right, with a velocity $v$. Then it appears reasonable that the other body is moving toward the left with a velocity $v$, because if the objects are alike there is no reason for right or left to be preferred and so the bodies would do something that is symmetrical. This is an illustration of a kind of thinking that is very useful in many problems but would not be brought out if we just started with the formulas.

The first result from our experiment is that equal objects will have equal speed, but now suppose that we have two objects made of different materials, say copper and aluminum, and we make the two masses equal. We shall now suppose that if we do the experiment with two masses that are equal, even though the objects are not identical, the velocities will be equal. Someone might object: “But you know, you could do it backwards, you did not have to suppose that. You could define equal masses to mean two masses that acquire equal velocities in this experiment.” We follow that suggestion and make a little explosion between the copper and a very large piece of aluminum, so heavy that the copper flies out and the aluminum hardly budges. That is too much aluminum, so we reduce the amount until there is just a very tiny piece, then when we make the explosion the aluminum goes flying away, and the copper hardly budges. That is not enough aluminum. Evidently there is some right amount in between; so we keep adjusting the amount until the velocities come out equal. Very well then—let us turn it around, and say that when the velocities are equal, the masses are equal. This appears to be just a definition, and it seems remarkable that we can transform physical laws into mere definitions. Nevertheless, there are some physical laws involved, and if we accept this definition of equal masses, we immediately find one of the laws, as follows.

Suppose we know from the foregoing experiment that two pieces of matter, $A$ and $B$ (of copper and aluminum), have equal masses, and we compare a third body, say a piece of gold, with the copper in the same manner as above, making sure that its mass is equal to the mass of the copper. If we now make the experiment between the aluminum and the gold, there is nothing in logic that says these masses must be equal; however, the experiment shows that they actually are. So now, by experiment, we have found a new law. A statement of this law might be: If two masses are each equal to a third mass (as determined by equal velocities in this experiment), then they are equal to each other. (This statement does not follow at all from a similar statement used as a postulate regarding mathematical quantities.) From this example we can see how quickly we start to infer things if we are careless. It is not just a definition to say the masses are equal when the velocities are equal, because to say the masses are equal is to imply the mathematical laws of equality, which in turn makes a prediction about an experiment.

As a second example, suppose that $A$ and $B$ are found to be equal by doing the experiment with one strength of explosion, which gives a certain velocity; if we then use a stronger explosion, will it be true or not true that the velocities now obtained are equal? Again, in logic there is nothing that can decide this question, but experiment shows that it is true. So, here is another law, which might be stated: If two bodies have equal masses, as measured by equal velocities at one velocity, they will have equal masses when measured at another velocity. From these examples we see that what appeared to be only a definition really involved some laws of physics.

In the development that follows we shall assume it is true that equal masses have equal and opposite velocities when an explosion occurs between them. We shall make another assumption in the inverse case: If two identical objects, moving in opposite directions with equal velocities, collide and stick together by some kind of glue, then which way will they be moving after the collision? This is again a symmetrical situation, with no preference between right and left, so we assume that they stand still. We shall also suppose that any two objects of equal mass, even if the objects are made of different materials, which collide and stick together, when moving with the same velocity in opposite directions will come to rest after the collision.

10–3 Momentum is conserved!

We can verify the above assumptions experimentally: first, that if two stationary objects of equal mass are separated by an explosion they will move apart with the same speed, and second, if two objects of equal mass, coming together with the same speed, collide and stick together they will stop. This we can do by means of a marvelous invention called an air trough, 1 which gets rid of friction, the thing which continually bothered Galileo (Fig.  10–1 ). He could not do experiments by sliding things because they do not slide freely, but, by adding a magic touch, we can today get rid of friction. Our objects will slide without difficulty, on and on at a constant velocity, as advertised by Galileo. This is done by supporting the objects on air. Because air has very low friction, an object glides along with practically constant velocity when there is no applied force. First, we use two glide blocks which have been made carefully to have the same weight, or mass (their weight was measured really, but we know that this weight is proportional to the mass), and we place a small explosive cap in a closed cylinder between the two blocks (Fig.  10–2 ). We shall start the blocks from rest at the center point of the track and force them apart by exploding the cap with an electric spark. What should happen? If the speeds are equal when they fly apart, they should arrive at the ends of the trough at the same time. On reaching the ends they will both bounce back with practically opposite velocity, and will come together and stop at the center where they started. It is a good test; when it is actually done the result is just as we have described (Fig.  10–3 ).

Now the next thing we would like to figure out is what happens in a less simple situation. Suppose we have two equal masses, one moving with velocity $v$ and the other standing still, and they collide and stick; what is going to happen? There is a mass $2m$ altogether when we are finished, drifting with an unknown velocity. What velocity? That is the problem. To find the answer, we make the assumption that if we ride along in a car, physics will look the same as if we are standing still. We start with the knowledge that two equal masses, moving in opposite directions with equal speeds $v$, will stop dead when they collide. Now suppose that while this happens, we are riding by in an automobile, at a velocity $-v$. Then what does it look like? Since we are riding along with one of the two masses which are coming together, that one appears to us to have zero velocity. The other mass, however, going the other way with velocity $v$, will appear to be coming toward us at a velocity $2v$ (Fig.  10–4 ). Finally, the combined masses after collision will seem to be passing by with velocity $v$. We therefore conclude that an object with velocity $2v$, hitting an equal one at rest, will end up with velocity $v$, or what is mathematically exactly the same, an object with velocity $v$ hitting and sticking to one at rest will produce an object moving with velocity $v/2$. Note that if we multiply the mass and the velocity beforehand and add them together, $mv+0$, we get the same answer as when we multiply the mass and the velocity of everything afterwards, $2m$ times $v/2$. So that tells us what happens when a mass of velocity $v$ hits one standing still.

In exactly the same manner we can deduce what happens when equal objects having any two velocities hit each other.

Suppose we have two equal bodies with velocities $v_1$ and $v_2$, respectively, which collide and stick together. What is their velocity $v$ after the collision? Again we ride by in an automobile, say at velocity $v_2$, so that one body appears to be at rest. The other then appears to have a velocity $v_1-v_2$, and we have the same case that we had before. When it is all finished they will be moving at $\tfrac{1}{2}(v_1-v_2)$ with respect to the car. What then is the actual speed on the ground? It is $v=\tfrac{1}{2}(v_1-v_2)+v_2$ or $\tfrac{1}{2}(v_1+v_2)$ (Fig.  10–5 ). Again we note that \begin{equation} \label{Eq:I:10:6} mv_1+mv_2=2m(v_1+v_2)/2. \end{equation}

Thus, using this principle, we can analyze any kind of collision in which two bodies of equal mass hit each other and stick. In fact, although we have worked only in one dimension, we can find out a great deal about much more complicated collisions by imagining that we are riding by in a car in some oblique direction. The principle is the same, but the details get somewhat complicated.

In order to test experimentally whether an object moving with velocity $v$, colliding with an equal one at rest, forms an object moving with velocity $v/2$, we may perform the following experiment with our air-trough apparatus. We place in the trough three equally massive objects, two of which are initially joined together with our explosive cylinder device, the third being very near to but slightly separated from these and provided with a sticky bumper so that it will stick to another object which hits it. Now, a moment after the explosion, we have two objects of mass $m$ moving with equal and opposite velocities $v$. A moment after that, one of these collides with the third object and makes an object of mass $2m$ moving, so we believe, with velocity $v/2$. How do we test whether it is really $v/2$? By arranging the initial positions of the masses on the trough so that the distances to the ends are not equal, but are in the ratio $2:1$. Thus our first mass, which continues to move with velocity $v$, should cover twice as much distance in a given time as the two which are stuck together (allowing for the small distance travelled by the second object before it collided with the third). The mass $m$ and the mass $2m$ should reach the ends at the same time, and when we try it, we find that they do (Fig.  10–6 ).

The next problem that we want to work out is what happens if we have two different masses. Let us take a mass $m$ and a mass $2m$ and apply our explosive interaction. What will happen then? If, as a result of the explosion, $m$ moves with velocity $v$, with what velocity does $2m$ move? The experiment we have just done may be repeated with zero separation between the second and third masses, and when we try it we get the same result, namely, the reacting masses $m$ and $2m$ attain velocities $-v$ and $v/2$. Thus the direct reaction between $m$ and $2m$ gives the same result as the symmetrical reaction between $m$ and $m$, followed by a collision between $m$ and a third mass $m$ in which they stick together. Furthermore, we find that the masses $m$ and $2m$ returning from the ends of the trough, with their velocities (nearly) exactly reversed, stop dead if they stick together.

Now the next question we may ask is this. What will happen if a mass $m$ with velocity $v$, say, hits and sticks to another mass $2m$ at rest? This is very easy to answer using our principle of Galilean relativity, for we simply watch the collision which we have just described from a car moving with velocity $-v/2$ (Fig.  10–7 ). From the car, the velocities are \begin{equation*} v_1'=v-v(\text{car})=v+v/2=3v/2 \end{equation*} and \begin{equation*} v_2'=-v/2-v(\text{car})=-v/2+v/2=0. \end{equation*} After the collision, the mass $3m$ appears to us to be moving with velocity $v/2$. Thus we have the answer, i.e., the ratio of velocities before and after collision is $3$ to $1$: if an object of mass $m$ collides with a stationary object of mass $2m$, then the whole thing moves off, stuck together, with a velocity $1/3$ as much. The general rule again is that the sum of the products of the masses and the velocities stays the same: $mv+0$ equals $3m$ times $v/3$, so we are gradually building up the theorem of the conservation of momentum, piece by piece.

Now we have one against two. Using the same arguments, we can predict the result of one against three, two against three, etc. The case of two against three, starting from rest, is shown in Fig.  10–8 .

In every case we find that the mass of the first object times its velocity, plus the mass of the second object times its velocity, is equal to the total mass of the final object times its velocity. These are all examples, then, of the conservation of momentum. Starting from simple, symmetrical cases, we have demonstrated the law for more complex cases. We could, in fact, do it for any rational mass ratio, and since every ratio is exceedingly close to a rational ratio, we can handle every ratio as precisely as we wish.

10–4 Momentum and energy

All the foregoing examples are simple cases where the bodies collide and stick together, or were initially stuck together and later separated by an explosion. However, there are situations in which the bodies do not cohere, as, for example, two bodies of equal mass which collide with equal speeds and then rebound. For a brief moment they are in contact and both are compressed. At the instant of maximum compression they both have zero velocity and energy is stored in the elastic bodies, as in a compressed spring. This energy is derived from the kinetic energy the bodies had before the collision, which becomes zero at the instant their velocity is zero. The loss of kinetic energy is only momentary, however. The compressed condition is analogous to the cap that releases energy in an explosion. The bodies are immediately decompressed in a kind of explosion, and fly apart again; but we already know that case—the bodies fly apart with equal speeds. However, this speed of rebound is less, in general, than the initial speed, because not all the energy is available for the explosion, depending on the material. If the material is putty no kinetic energy is recovered, but if it is something more rigid, some kinetic energy is usually regained. In the collision the rest of the kinetic energy is transformed into heat and vibrational energy—the bodies are hot and vibrating. The vibrational energy also is soon transformed into heat. It is possible to make the colliding bodies from highly elastic materials, such as steel, with carefully designed spring bumpers, so that the collision generates very little heat and vibration. In these circumstances the velocities of rebound are practically equal to the initial velocities; such a collision is called elastic .

That the speeds before and after an elastic collision are equal is not a matter of conservation of momentum, but a matter of conservation of kinetic energy . That the velocities of the bodies rebounding after a symmetrical collision are equal to and opposite each other , however, is a matter of conservation of momentum.

We might similarly analyze collisions between bodies of different masses, different initial velocities, and various degrees of elasticity, and determine the final velocities and the loss of kinetic energy, but we shall not go into the details of these processes.

Elastic collisions are especially interesting for systems that have no internal “gears, wheels, or parts.” Then when there is a collision there is nowhere for the energy to be impounded, because the objects that move apart are in the same condition as when they collided. Therefore, between very elementary objects, the collisions are always elastic or very nearly elastic. For instance, the collisions between atoms or molecules in a gas are said to be perfectly elastic. Although this is an excellent approximation, even such collisions are not perfectly elastic; otherwise one could not understand how energy in the form of light or heat radiation could come out of a gas. Once in a while, in a gas collision, a low-energy infrared ray is emitted, but this occurrence is very rare and the energy emitted is very small. So, for most purposes, collisions of molecules in gases are considered to be perfectly elastic.

As an interesting example, let us consider an elastic collision between two objects of equal mass . If they come together with the same speed, they would come apart at that same speed, by symmetry. But now look at this in another circumstance, in which one of them is moving with velocity $v$ and the other one is at rest. What happens? We have been through this before. We watch the symmetrical collision from a car moving along with one of the objects, and we find that if a stationary body is struck elastically by another body of exactly the same mass, the moving body stops, and the one that was standing still now moves away with the same speed that the other one had; the bodies simply exchange velocities. This behavior can easily be demonstrated with a suitable impact apparatus. More generally, if both bodies are moving, with different velocities, they simply exchange velocity at impact.

Another example of an almost elastic interaction is magnetism. If we arrange a pair of U-shaped magnets in our glide blocks, so that they repel each other, when one drifts quietly up to the other, it pushes it away and stands perfectly still, and now the other goes along, frictionlessly.

The principle of conservation of momentum is very useful, because it enables us to solve many problems without knowing the details. We did not know the details of the gas motions in the cap explosion, yet we could predict the velocities with which the bodies came apart, for example. Another interesting example is rocket propulsion. A rocket of large mass, $M$, ejects a small piece, of mass $m$, with a terrific velocity $V$ relative to the rocket. After this the rocket, if it were originally standing still, will be moving with a small velocity, $v$. Using the principle of conservation of momentum, we can calculate this velocity to be \begin{equation*} v=\frac{m}{M}\cdot V. \end{equation*} So long as material is being ejected, the rocket continues to pick up speed. Rocket propulsion is essentially the same as the recoil of a gun: there is no need for any air to push against.

10–5 Relativistic momentum

In modern times the law of conservation of momentum has undergone certain modifications. However, the law is still true today, the modifications being mainly in the definitions of things. In the theory of relativity it turns out that we do have conservation of momentum; the particles have mass and the momentum is still given by $mv$, the mass times the velocity, but the mass changes with the velocity , hence the momentum also changes. The mass varies with velocity according to the law \begin{equation} \label{Eq:I:10:7} m=\frac{m_0}{\sqrt{1-v^2/c^2}}, \end{equation} where $m_0$ is the mass of the body at rest and $c$ is the speed of light. It is easy to see from the formula that there is negligible difference between $m$ and $m_0$ unless $v$ is very large, and that for ordinary velocities the expression for momentum reduces to the old formula.

The components of momentum for a single particle are written as \begin{equation} \label{Eq:I:10:8} p_x=\frac{m_0v_x}{\sqrt{1-v^2/c^2}},\quad p_y=\frac{m_0v_y}{\sqrt{1-v^2/c^2}},\quad p_z=\frac{m_0v_z}{\sqrt{1-v^2/c^2}}, \end{equation} \begin{equation} \begin{aligned} p_x&=\frac{m_0v_x}{\sqrt{1-v^2/c^2}},\\[1ex] p_y&=\frac{m_0v_y}{\sqrt{1-v^2/c^2}},\\[1ex] p_z&=\frac{m_0v_z}{\sqrt{1-v^2/c^2}}, \end{aligned} \label{Eq:I:10:8} \end{equation} where $v^2=v_x^2+v_y^2+v_z^2$. If the $x$-components are summed over all the interacting particles, both before and after a collision, the sums are equal; that is, momentum is conserved in the $x$-direction. The same holds true in any direction.

In Chapter  4 we saw that the law of conservation of energy is not valid unless we recognize that energy appears in different forms, electrical energy, mechanical energy, radiant energy, heat energy, and so on. In some of these cases, heat energy for example, the energy might be said to be “hidden.” This example might suggest the question, “Are there also hidden forms of momentum—perhaps heat momentum?” The answer is that it is very hard to hide momentum for the following reasons.

The random motions of the atoms of a body furnish a measure of heat energy, if the squares of the velocities are summed. This sum will be a positive result, having no directional character. The heat is there, whether or not the body moves as a whole, and conservation of energy in the form of heat is not very obvious. On the other hand, if one sums the velocities , which have direction, and finds a result that is not zero, that means that there is a drift of the entire body in some particular direction, and such a gross momentum is readily observed. Thus there is no random internal lost momentum, because the body has net momentum only when it moves as a whole. Therefore momentum, as a mechanical quantity, is difficult to hide. Nevertheless, momentum can be hidden—in the electromagnetic field, for example. This case is another effect of relativity.

One of the propositions of Newton was that interactions at a distance are instantaneous. It turns out that such is not the case; in situations involving electrical forces, for instance, if an electrical charge at one location is suddenly moved, the effects on another charge, at another place, do not appear instantaneously—there is a little delay. In those circumstances, even if the forces are equal the momentum will not check out; there will be a short time during which there will be trouble, because for a while the first charge will feel a certain reaction force, say, and will pick up some momentum, but the second charge has felt nothing and has not yet changed its momentum. It takes time for the influence to cross the intervening distance, which it does at $186{,}000$ miles a second. In that tiny time the momentum of the particles is not conserved. Of course after the second charge has felt the effect of the first one and all is quieted down, the momentum equation will check out all right, but during that small interval momentum is not conserved. We represent this by saying that during this interval there is another kind of momentum besides that of the particle, $mv$, and that is momentum in the electromagnetic field. If we add the field momentum to the momentum of the particles, then momentum is conserved at any moment all the time. The fact that the electromagnetic field can possess momentum and energy makes that field very real, and so, for better understanding, the original idea that there are just the forces between particles has to be modified to the idea that a particle makes a field, and a field acts on another particle, and the field itself has such familiar properties as energy content and momentum, just as particles can have. To take another example: an electromagnetic field has waves, which we call light; it turns out that light also carries momentum with it, so when light impinges on an object it carries in a certain amount of momentum per second; this is equivalent to a force, because if the illuminated object is picking up a certain amount of momentum per second, its momentum is changing and the situation is exactly the same as if there were a force on it. Light can exert pressure by bombarding an object; this pressure is very small, but with sufficiently delicate apparatus it is measurable.

Now in quantum mechanics it turns out that momentum is a different thing—it is no longer $mv$. It is hard to define exactly what is meant by the velocity of a particle, but momentum still exists. In quantum mechanics the difference is that when the particles are represented as particles, the momentum is still $mv$, but when the particles are represented as waves, the momentum is measured by the number of waves per centimeter: the greater this number of waves, the greater the momentum. In spite of the differences, the law of conservation of momentum holds also in quantum mechanics. Even though the law $F=ma$ is false, and all the derivations of Newton were wrong for the conservation of momentum, in quantum mechanics, nevertheless, in the end, that particular law maintains itself!

• H. V. Neher and R. B. Leighton, Amer. Jour. of Phys. 31 , 255 (1963). ↩

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Conservation of linear momentum and energy.

A pendulum is allowed to "crash" into a bar, dramatically altering its motion, but energy is conserved as is evidenced by the return swing.

Requiring little or no skill and the aid of a carpenter's square, one shot sinks two balls into the pockets.

Currently we provide two different versions of this experiment. One version uses balls rolling down a long, inclined track (roughly 2 meters), and the other version uses balls that swing on a pendulum.

Same as previous except that mass ratio of balls is 1:3 (softball:basketball) leaving basketball dead and softball four times the height.

A tennis ball/basketball combination is dropped to the floor ... the tennis ball theoretically bounces to nine times the original drop height.

What it shows:

Demonstration of elastic collisions between metal balls to show conservation of momentum and energy.

How it works:

Newton's Cradle (less affectionately known as Newton's Balls) consists of six rigid balls hanging in a row with bifilar suspension. The balls hang so that they just barely touch their neighbor.

Various initial conditions can be employed. A single ball displaced will collide with the remaining four, sending the ball at the far end off. Same idea for two or three balls. Four balls, and only the first two will stop; the center two...

Using conservation of energy, calculate the height from which Barney must jump so that his head just barely kisses the floor at the bottom of his bungee cord jump. Then verify by experiment. Oops ... hate when that happens! It turns out that it's not so simple and there are important details that must be taken into account.

Barney (the friendly pink dinosaur) is "sandbagged" (with a 5 kg weight, duct-taped around his waist) and suspended from the sky-hook by a 3.1 meter-long (unstretched) spring. The spring constant has been measured...

Use conservation of energy to predict the height the arrow will reach.

When the string of a bow and arrow is pulled from equilibrium, the elastic potential energy in the bow is converted to kinetic energy of the arrow when...

A toy car rolling down a loop-the-loop track demonstrates the minimum height it must start at to successfully negotiate the loop.

For an object to move in a vertical circle, its velocity must exceed a critical value vc=(Rg) 1/2 , where R is the radius of the circle and g the acceleration due to gravity. This ensures that, at the top of the loop, the centripetal force balances the body's weight. This can be shown using a toy car on a looped track.

The car is released from the top of a ramp and runs down a slope towards...

Faith in the conservation of energy is tested by taking the demonstrator's nose to task.

The principle of conservation of energy ensures that a pendulum released at a particular amplitude will not exceed that amplitude on the return swing. A lecturer's faith in their subject is put to the test using a 50lb (22.7kg) iron ball.

Technique is very important here. The best method to employ is to stand with your back against the blackboard with your head also touching the board. This ensures that you don't lean forward after release....

Two cars have the same mass and same spring bumper. When given a push and allowed to collide with a wall, one car bounces off with only a small reduction in speed ("elastic" collison) whereas the other car comes nearly to a complere stop ("inelastic" collision).

There are two impulse cars made of identical materials and have the same mass. The car that models an elastic collision has all its lead sinkers securely attached to the frame so that they can't move. In contrast, the car that models an inelastic collision has the lead sinkers...

Radio controlled car moves one way while road moves the other.

We tell our students that, when a car drives down the road, the road and the Earth move in the opposite direction, albeit imperceptibly. This demonstration is a realization of that concept, made possible (and perceptible) by the fact that the road is not attached to the Earth.

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8.3 Conservation of Momentum

Learning objectives.

By the end of this section, you will be able to:

• Describe the law of conservation of linear momentum.
• Derive an expression for the conservation of momentum.
• Explain conservation of momentum with examples.
• Explain the law of conservation of momentum as it relates to atomic and subatomic particles.

The information presented in this section supports the following AP® learning objectives and science practices:

• 5.A.2.1 The student is able to define open and closed systems for everyday situations and apply conservation concepts for energy, charge, and linear momentum to those situations. (S.P. 6.4, 7.2)
• 5.D.1.4 The student is able to design an experimental test of an application of the principle of the conservation of linear momentum, predict an outcome of the experiment using the principle, analyze data generated by that experiment whose uncertainties are expressed numerically, and evaluate the match between the prediction and the outcome. (S.P. 4.2, 5.1, 5.3, 6.4)
• 5.D.2.1 The student is able to qualitatively predict, in terms of linear momentum and kinetic energy, how the outcome of a collision between two objects changes depending on whether the collision is elastic or inelastic. (S.P. 6.4, 7.2)
• 5.D.2.2 The student is able to plan data collection strategies to test the law of conservation of momentum in a two-object collision that is elastic or inelastic and analyze the resulting data graphically. (S.P.4.1, 4.2, 5.1)
• 5.D.3.1 The student is able to predict the velocity of the center of mass of a system when there is no interaction outside of the system but there is an interaction within the system (i.e., the student simply recognizes that interactions within a system do not affect the center of mass motion of the system and is able to determine that there is no external force). (S.P. 6.4)

Momentum is an important quantity because it is conserved. Yet it was not conserved in the examples in Impulse and Linear Momentum and Force , where large changes in momentum were produced by forces acting on the system of interest. Under what circumstances is momentum conserved?

The answer to this question entails considering a sufficiently large system. It is always possible to find a larger system in which total momentum is constant, even if momentum changes for components of the system. If a football player runs into the goalpost in the end zone, there will be a force on him that causes him to bounce backward. However, the Earth also recoils —conserving momentum—because of the force applied to it through the goalpost. Because Earth is many orders of magnitude more massive than the player, its recoil is immeasurably small and can be neglected in any practical sense, but it is real nevertheless.

Consider what happens if the masses of two colliding objects are more similar than the masses of a football player and Earth—for example, one car bumping into another, as shown in Figure 8.6 . Both cars are coasting in the same direction when the lead car (labeled m 2 ) m 2 ) size 12{m rSub { size 8{2} } \) } {} is bumped by the trailing car (labeled m 1 ) . m 1 ) . size 12{m rSub { size 8{1} } \) "." } {} The only unbalanced force on each car is the force of the collision. (Assume that the effects due to friction are negligible.) Car 1 slows down as a result of the collision, losing some momentum, while car 2 speeds up and gains some momentum. We shall now show that the total momentum of the two-car system remains constant.

Using the definition of impulse, the change in momentum of car 1 is given by

where F 1 F 1 size 12{F"" lSub { size 8{1} } } {} is the force on car 1 due to car 2, and Δ t Δ t size 12{Δt} {} is the time the force acts (the duration of the collision). Intuitively, it seems obvious that the collision time is the same for both cars, but it is only true for objects traveling at ordinary speeds. This assumption must be modified for objects travelling near the speed of light, without affecting the result that momentum is conserved.

Similarly, the change in momentum of car 2 is

where F 2 F 2 is the force on car 2 due to car 1, and we assume the duration of the collision Δ t Δ t size 12{?t} {} is the same for both cars. We know from Newton’s third law that F 2 = – F 1 F 2 = – F 1 size 12{F rSub { size 8{2} } = - F rSub { size 8{1} } } {} , and so

Thus, the changes in momentum are equal and opposite, and

Because the changes in momentum add to zero, the total momentum of the two-car system is constant. That is,

where p ′ 1 p ′ 1 size 12{ { {p}} sup { ' } rSub { size 8{1} } } {} and p ′ 2 p ′ 2 size 12{ { {p}} sup { ' } rSub { size 8{2} } } {} are the momenta of cars 1 and 2 after the collision. (We often use primes to denote the final state.)

This result—that momentum is conserved—has validity far beyond the preceding one-dimensional case. It can be similarly shown that total momentum is conserved for any isolated system, with any number of objects in it. In equation form, the conservation of momentum principle for an isolated system is written

where p tot p tot size 12{p rSub { size 8{"tot"} } } {} is the total momentum (the sum of the momenta of the individual objects in the system) and p ′ tot p ′ tot size 12{ ital "p'" rSub { size 8{"tot"} } } {} is the total momentum some time later. (The total momentum can be shown to be the momentum of the center of mass of the system.) An isolated system is defined to be one for which the net external force is zero F net = 0 . F net = 0 . size 12{ left (F rSub { size 8{ ital "net"} } =0 right ) "." } {}

Conservation of Momentum Principle

Isolated system.

An isolated system is defined to be one for which the net external force is zero F net = 0 . F net = 0 . size 12{ left (F rSub { size 8{ ital "net"} } =0 right ) "." } {}

Making Connections: Cart Collisions

Consider two air carts with equal mass (m) on a linear track. The first cart moves with a speed v towards the second cart, which is initially at rest. We will take the initial direction of motion of the first cart as the positive direction.

The momentum of the system will be conserved in the collision. If the collision is elastic, then the first cart will stop after the collision. Conservation of momentum therefore tells us that the second cart will have a final velocity v after the collision in the same direction as the initial velocity of the first cart.

The kinetic energy of the system will be conserved since the masses are equal and the final velocity of cart 2 is equal to the initial velocity of cart 1. What would a graph of total momentum vs. time look like in this case? What would a graph of total kinetic energy vs. time look like in this case?

Consider the center of mass of this system as the frame of reference. As cart 1 approaches cart 2, the center of mass remains exactly halfway between the two carts. The center of mass moves toward the stationary cart 2 at a speed v 2 v 2 . After the collision, the center of mass continues moving in the same direction, away from (now stationary) cart 1 at a speed v 2 v 2 . How would a graph of center-of-mass velocity vs. time compare to a graph of momentum vs. time?

Suppose instead that the two carts move with equal speeds v in opposite directions towards the center of mass. Again, they have an elastic collision, so after the collision, they exchange velocities (each cart moving in the opposite direction of its initial motion with the same speed). As the two carts approach, the center of mass is exactly between the two carts, at the point where they will collide. In this case, how would a graph of center-of-mass velocity vs. time compare to a graph of the momentum of the system vs. time?

Let us return to the example where the first cart is moving with a speed v toward the second cart, initially at rest. Suppose the second cart has some putty on one end so that, when the collision occurs, the two carts stick together in an inelastic collision. In this case, conservation of momentum tells us that the final velocity of the two-cart system will be half the initial velocity of the first cart, in the same direction as the first cart’s initial motion. Kinetic energy will not be conserved in this case, however. Compared to the moving cart before the collision, the overall moving mass after the collision is doubled, but the velocity is halved.

The initial kinetic energy of the system is:

The final kinetic energy of the two carts (2 m ) moving together (at speed v /2) is:

What would a graph of total momentum vs. time look like in this case? What would a graph of total kinetic energy vs. time look like in this case?

Consider the center of mass of this system. As cart 1 approaches cart 2, the center of mass remains exactly halfway between the two carts. The center of mass moves toward the stationary cart 2 at a speed v 2 v 2 . After the collision, the two carts move together at a speed v 2 v 2 . How would a graph of center-of-mass velocity vs. time compare to a graph of momentum vs. time?

Suppose instead that the two carts move with equal speeds v in opposite directions towards the center of mass. They have putty on the end of each cart so that they stick together after the collision. As the two carts approach, the center of mass is exactly between the two carts, at the point where they will collide. In this case, how would a graph of center-of-mass velocity vs. time compare to a graph of the momentum of the system vs. time?

Perhaps an easier way to see that momentum is conserved for an isolated system is to consider Newton’s second law in terms of momentum, F net = Δ p tot Δ t F net = Δ p tot Δ t . For an isolated system, F net = 0 F net = 0 ; thus, Δ p tot = 0 Δ p tot = 0 size 12{?p rSub { size 8{"tot"} } =0} {} , and p tot p tot is constant.

We have noted that the three length dimensions in nature— x x size 12{x} {} , y y size 12{y} {} , and z z size 12{z} {} —are independent, and it is interesting to note that momentum can be conserved in different ways along each dimension. For example, during projectile motion and where air resistance is negligible, momentum is conserved in the horizontal direction because horizontal forces are zero and momentum is unchanged. But along the vertical direction, the net vertical force is not zero and the momentum of the projectile is not conserved. (See Figure 8.7 .) However, if the momentum of the projectile-Earth system is considered in the vertical direction, we find that the total momentum is conserved.

The conservation of momentum principle can be applied to systems as different as a comet striking Earth and a gas containing huge numbers of atoms and molecules. Conservation of momentum is violated only when the net external force is not zero. But another larger system can always be considered in which momentum is conserved by simply including the source of the external force. For example, in the collision of two cars considered above, the two-car system conserves momentum while each one-car system does not.

Making Connections: Take-Home Investigation—Drop of Tennis Ball and a Basketball

Hold a tennis ball side by side and in contact with a basketball. Drop the balls together. (Be careful!) What happens? Explain your observations. Now hold the tennis ball above and in contact with the basketball. What happened? Explain your observations. What do you think will happen if the basketball ball is held above and in contact with the tennis ball?

Making Connections: Take-Home Investigation—Two Tennis Balls in a Ballistic Trajectory

Tie two tennis balls together with a string about a foot long. Hold one ball and let the other hang down and throw it in a ballistic trajectory. Explain your observations. Now mark the center of the string with bright ink or attach a brightly colored sticker to it and throw again. What happened? Explain your observations.

Some aquatic animals such as jellyfish move around based on the principles of conservation of momentum. A jellyfish fills its umbrella section with water and then pushes the water out resulting in motion in the opposite direction to that of the jet of water. Squids propel themselves in a similar manner but, in contrast with jellyfish, are able to control the direction in which they move by aiming their nozzle forward or backward. Typical squids can move at speeds of 8 to 12 km/h.

The ballistocardiograph (BCG) was a diagnostic tool used in the second half of the 20th century to study the strength of the heart. About once a second, your heart beats, forcing blood into the aorta. A force in the opposite direction is exerted on the rest of your body (recall Newton’s third law). A ballistocardiograph is a device that can measure this reaction force. This measurement is done by using a sensor (resting on the person) or by using a moving table suspended from the ceiling. This technique can gather information on the strength of the heart beat and the volume of blood passing from the heart. However, the electrocardiogram (ECG or EKG) and the echocardiogram (cardiac ECHO or ECHO; a technique that uses ultrasound to see an image of the heart) are more widely used in the practice of cardiology.

Applying Science Practices: Verifying the Conservation of Linear Momentum

Design an experiment to verify the conservation of linear momentum in a one-dimensional collision, both elastic and inelastic. For simplicity, try to ensure that friction is minimized so that it has a negligible effect on your experiment. As you consider your experiment, consider the following questions:

• Predict how the final momentum of the system will compare to the initial momentum of the system that you will measure. Justify your prediction.
• How will you measure the momentum of each object?
• Should you have two objects in motion or one object bouncing off a rigid surface?
• Should you verify the relationship mathematically or graphically?
• How will you estimate the uncertainty of your measurements? How will you express this uncertainty in your data?

When you have completed each experiment, compare the outcome to your prediction about the initial and final momentum of the system and evaluate your results.

Making Connections: Conservation of Momentum and Collision

Conservation of momentum is quite useful in describing collisions. Momentum is crucial to our understanding of atomic and subatomic particles because much of what we know about these particles comes from collision experiments.

Subatomic Collisions and Momentum

The conservation of momentum principle not only applies to the macroscopic objects, it is also essential to our explorations of atomic and subatomic particles. Giant machines hurl subatomic particles at one another, and researchers evaluate the results by assuming conservation of momentum (among other things).

On the small scale, we find that particles and their properties are invisible to the naked eye but can be measured with our instruments, and models of these subatomic particles can be constructed to describe the results. Momentum is found to be a property of all subatomic particles including massless particles such as photons that compose light. Momentum being a property of particles hints that momentum may have an identity beyond the description of an object’s mass multiplied by the object’s velocity. Indeed, momentum relates to wave properties and plays a fundamental role in what measurements are taken and how we take these measurements. Furthermore, we find that the conservation of momentum principle is valid when considering systems of particles. We use this principle to analyze the masses and other properties of previously undetected particles, such as the nucleus of an atom and the existence of quarks that make up particles of nuclei. Figure 8.8 below illustrates how a particle scattering backward from another implies that its target is massive and dense. Experiments seeking evidence that quarks make up protons (one type of particle that makes up nuclei) scattered high-energy electrons off of protons (nuclei of hydrogen atoms). Electrons occasionally scattered straight backward in a manner that implied a very small and very dense particle makes up the proton—this observation is considered nearly direct evidence of quarks. The analysis was based partly on the same conservation of momentum principle that works so well on the large scale.

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conservation of momentum

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• The Physics Classroom - Momentum Conservation Principle
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conservation of momentum , general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant. Momentum is equal to the mass of an object multiplied by its velocity and is equivalent to the force required to bring the object to a stop in a unit length of time. For any array of several objects, the total momentum is the sum of the individual momenta. There is a peculiarity, however, in that momentum is a vector , involving both the direction and the magnitude of motion, so that the momenta of objects going in opposite directions can cancel to yield an overall sum of zero.

Before launch, the total momentum of a rocket and its fuel is zero. During launch, the downward momentum of the expanding exhaust gases just equals in magnitude the upward momentum of the rising rocket, so that the total momentum of the system remains constant—in this case, at zero value. In a collision of two particles, the sum of the two momenta before collision is equal to their sum after collision. What momentum one particle loses, the other gains.

The law of conservation of momentum is abundantly confirmed by experiment and can even be mathematically deduced on the reasonable presumption that space is uniform—that is, that there is nothing in the laws of nature that singles out one position in space as peculiar compared with any other.

There is a similar conservation law for angular momentum , which describes rotational motion in essentially the same way that ordinary momentum describes linear motion . Although the precise mathematical expression of this law is somewhat more involved, examples of it are numerous. All helicopters , for instance, require at least two propellers (rotors) for stabilization. The body of a helicopter would rotate in the opposite direction to conserve angular momentum if there were only a single horizontal propeller on top. In accordance with conservation of angular momentum, ice skaters spin faster as they pull their arms toward their body and more slowly as they extend them.

Angular-momentum conservation has also been thoroughly established by experiment and can be shown to follow mathematically from the reasonable presumption that space is uniform with respect to orientation—that is, that there is nothing in the laws of nature that singles out one direction in space as being peculiar compared with any other.

Conservation of Momentum

Now you can perform the classic momentum lab with all the same calculations, but without the inconvenient physical air track and photogates. Investigate the basics of conservation of momentum, or take it further with elastic vs. inelastic collisions. We’ve even included partially elastic collisions so you can investigate the coefficient of restitution. All of the lab guides below may be freely reproduced for classroom use.

• The Conservation of Momentum
• Elastic vs. Inelastic Collisions
• The Coefficient of Restitution

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Conservation of Momentum: Explanation and Examples

• The Albert Team
• Last Updated On: April 5, 2023

You’ve learned about momentum for singular objects. The question now, is what happens if you have more than one object? How can momentum be used to understand what happens when two objects collide? The answer comes from a law known as the law of the conservation of momentum. In this post, we’ll cover the law of conservation of momentum, the difference between elastic and inelastic collisions, and examples of each.

What We Review

What Is Conservation Of Momentum?

Momentum is, effectively, a measure of how hard it will be to stop a moving object. This is a common value in physics as it applies to every moving object. We already learned how to calculate momentum so now we’ll learn how actions affect momentum.

Definition Of The Law Of Conservation Of Momentum

In summary, for all actions, momentum must be conserved. This is true when looking at the movement of a single object, or interactions between two or more objects. This is also true for linear momentum and angular momentum.

Momentum is the product of an object’s mass and speed. You may be able to see a link to kinetic energy with those values. Although momentum and energy are not equivalent, the same principles that apply to the conservation of energy also apply to the conservation of momentum. If you have some amount of something, that thing cannot vanish because of an action. It can be transferred from one object to another, but it cannot vanish. 

Applications Of The Conservation Of Momentum

The conservation of momentum appears in many areas of physics. It helps engineers better understand what happens during car crashes to help keep us safe. In sports, it explains what will happen when hitting a baseball or tackling another player. These would mostly be cases of linear momentum , but this law applies to angular momentum as well. We’ll be sticking with linear momentum for the rest of this post, but know that this same idea can help predict how objects move when in orbit or when changing direction.

Elastic And Inelastic Collisions

We often utilize the law of conservation of momentum when looking at collisions. This law applies to both elastic and inelastic collisions.

An elastic collision is one in which two objects collide and then bounce apart. This can be a basketball bouncing off the floor or one ball in a game of pool bouncing off another. An inelastic collision is the opposite of an elastic collision. In these, the two objects collide and then stick together. This can be a car crash or a cooked piece of pasta sticking to a wall.

Similarities And Differences Between Elastic And Inelastic Collisions

The main similarity in both of these collisions is that momentum is conserved. In reality, this usually has to account for energy lost to the production of sound or heat. For the sake of your physics courses, though, you genuinely won’t account for these things. The main difference between elastic and inelastic collisions is that elastic collisions bounce apart while inelastic collisions stick together.

If you have trouble figuring out how to know if a collision is elastic or inelastic, it’s important to try to picture what’s actually happening. With words like “bounce”, “moved apart after colliding”, or “separated”, you’ve got an elastic collision. On the other hand, if you have “stuck together”, “moved together after colliding”, or “becomes embedded”, then it’s an inelastic collision. 

The Conservation Of Momentum Equation

The conservation of momentum equation is a relatively straightforward one:

The reason we use the \Sigma symbol is to show that this is true for several momentum values. You could also substitute in the momentum equation and write the equation for the conservation of momentum using masses and velocities.

Here, we’ve used i to indicate initial values and f final values respectively. We often use numbers to represent these different values, but you’ll see why we used letters here in a moment.

How To Use The Conservation Of Momentum Equation

You’ve probably seen the conservation of momentum mentioned with collisions. This law can be applied to elastic or inelastic collisions by expanding our equation a bit to include two objects:

We can use numbers to note whether we are talking about object one or two and letters to note whether we are talking about the value before ( i for “initial”) or after ( f for “final”) an action has taken place. This can help us predict how an object will behave after a collision because we can see how momentum needs to be balanced.

If before the collision, object one had very little momentum while object two had a lot, and then after the collision, object one now has a lot of momentum, we can predict that object two should have less momentum after the collision. The more you work with the concept, the better you will get at making predictions around changes in mass or velocity after collisions.

Conservation Of Momentum In The Lab

There are many examples of the conservation of momentum in the real world. A famous one is Newton’s cradle, where momentum is transferred from one ball through several others to finally launch the ball on the opposite end. The Moon’s motion is an excellent example of the conservation of angular momentum if you’d like to learn more about it. For now, let’s look at an in-lab experiment where this can be proven.

A Classic Experiment

To run this experiment, you’ll need two carts, a frictionless air track, and a way to measure velocity. You can also utilize an online simulator . To start, we’ll model an elastic collision and then an inelastic collision.

Modeling an Elastic Collision

To do this, we’ll record the mass of our two carts before pushing them toward each other. You’ll need to measure the initial and final velocities each time you do this.

Now, we can use this data to check that the conservation of momentum equation works. To do this, we simply need to plug in our values and see if the equation is balanced. Keep in mind that your data likely has some degree of error so these values may not work out perfectly each time. So, we’ll set up a table to see how similar the two sides are. We’ll show our work for the first trial, but after that, we’ll just show the results.

As expected, not all of these values are a perfect match. They are close enough, however, that the differences could be due to human error or imprecise measurements. For this reason, we can conclude that our experiment would prove the law of the conservation of momentum holds for elastic collisions. Now let’s check if this is true for inelastic collisions.

Modeling an Inelastic Collision

Now, we’ll repeat the experiment for inelastic collisions. This time, we will make sure the two carts stick together after colliding. We will then record our data and check that the equation holds in the same way. One common way to make the two carts stick together is to add velcro to the ends of the carts.

Note: Because the carts stick together after, we will only have one mass and one velocity value on the right-hand side of our equation. This means our table and notation will look a little bit different.

Now that we have our data, we can set up the same table we’d had before to check the equation for the conservation of momentum. Again, we’ll show work for the first trial but none of the others.

Again, not all of our numbers match perfectly, but they are close enough to conclude the conservation of momentum applies to inelastic collisions.

Now we’ve learned about the conservation of momentum, the difference between elastic and inelastic collisions, and how to apply the conservation of momentum to different types of collisions. This is one of the fundamental laws that help shape the universe and you’ll continue to see it throughout your physics journey.

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For a collision occurring between object 1 and object 2 in an isolated system , the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

The Logic Behind Momentum Conservation

Consider a collision between two objects - object 1 and object 2. For such a collision, the forces acting between the two objects are equal in magnitude and opposite in direction ( Newton's third law ). This statement can be expressed in equation form as follows.

The forces act between the two objects for a given amount of time. In some cases, the time is long; in other cases the time is short. Regardless of how long the time is, it can be said that the time that the force acts upon object 1 is equal to the time that the force acts upon object 2. This is merely logical. Forces result from interactions (or contact) between two objects. If object 1 contacts object 2 for 0.050 seconds, then object 2 must be contacting object 1 for the same amount of time (0.050 seconds). As an equation, this can be stated as

Since the forces between the two objects are equal in magnitude and opposite in direction, and since the times for which these forces act are equal in magnitude, it follows that the impulses experienced by the two objects are also equal in magnitude and opposite in direction. As an equation, this can be stated as

But the impulse experienced by an object is equal to the change in momentum of that object ( the impulse-momentum change theorem ). Thus, since each object experiences equal and opposite impulses, it follows logically that they must also experience equal and opposite momentum changes. As an equation, this can be stated as

The Law of Momentum Conservation

The above equation is one statement of the law of momentum conservation. In a collision, the momentum change of object 1 is equal to and opposite of the momentum change of object 2. That is, the momentum lost by object 1 is equal to the momentum gained by object 2. In most collisions between two objects, one object slows down and loses momentum while the other object speeds up and gains momentum. If object 1 loses 75 units of momentum, then object 2 gains 75 units of momentum. Yet, the total momentum of the two objects (object 1 plus object 2) is the same before the collision as it is after the collision. The total momentum of the system (the collection of two objects) is conserved.

A useful analogy for understanding momentum conservation involves a money transaction between two people. Let's refer to the two people as Jack and Jill. Suppose that we were to check the pockets of Jack and Jill before and after the money transaction in order to determine the amount of money that each possesses. Prior to the transaction, Jack possesses $100 and Jill possesses $100. The total amount of money of the two people before the transaction is $200. During the transaction, Jack pays Jill $50 for the given item being bought. There is a transfer of $50 from Jack's pocket to Jill's pocket. Jack has lost $50 and Jill has gained $50. The money lost by Jack is equal to the money gained by Jill. After the transaction, Jack now has $50 in his pocket and Jill has $150 in her pocket. Yet, the total amount of money of the two people after the transaction is $200. The total amount of money (Jack's money plus Jill's money) before the transaction is equal to the total amount of money after the transaction. It could be said that the total amount of money of the system (the collection of two people) is conserved. It is the same before as it is after the transaction.

A useful means of depicting the transfer and the conservation of money between Jack and Jill is by means of a table.

The table shows the amount of money possessed by the two individuals before and after the interaction. It also shows the total amount of money before and after the interaction. Note that the total amount of money ($200) is the same before and after the interaction - it is conserved. Finally, the table shows the change in the amount of money possessed by the two individuals. Note that the change in Jack's money account (-$50) is equal to and opposite of the change in Jill's money account (+$50).

For any collision occurring in an isolated system , momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision. A common physics lab involves the dropping of a brick upon a cart in motion.

The dropped brick is at rest and begins with zero momentum. The loaded cart (a cart with a brick on it) is in motion with considerable momentum. The actual momentum of the loaded cart can be determined using the velocity (often determined by a ticker tape analysis) and the mass. The total amount of momentum is the sum of the dropped brick's momentum (0 units) and the loaded cart's momentum. After the collision, the momenta of the two separate objects (dropped brick and loaded cart) can be determined from their measured mass and their velocity (often found from a ticker tape analysis). If momentum is conserved during the collision, then the sum of the dropped brick's and loaded cart's momentum after the collision should be the same as before the collision. The momentum lost by the loaded cart should equal (or approximately equal) the momentum gained by the dropped brick. Momentum data for the interaction between the dropped brick and the loaded cart could be depicted in a table similar to the money table above.

Note that the loaded cart lost 14 units of momentum and the dropped brick gained 14 units of momentum. Note also that the total momentum of the system (45 units) was the same before the collision as it was after the collision.

Collisions commonly occur in contact sports (such as football) and racket and bat sports (such as baseball, golf, tennis, etc.). Consider a collision in football between a fullback and a linebacker during a goal-line stand . The fullback plunges across the goal line and collides in midair with the linebacker. The linebacker and fullback hold each other and travel together after the collision. The fullback possesses a momentum of 100 kg*m/s, East before the collision and the linebacker possesses a momentum of 120 kg*m/s, West before the collision. The total momentum of the system before the collision is 20 kg*m/s, West ( review the section on adding vectors if necessary). Therefore, the total momentum of the system after the collision must also be 20 kg*m/s, West. The fullback and the linebacker move together as a single unit after the collision with a combined momentum of 20 kg*m/s. Momentum is conserved in the collision. A vector diagram can be used to represent this principle of momentum conservation; such a diagram uses an arrow to represent the magnitude and direction of the momentum vector for the individual objects before the collision and the combined momentum after the collision.

Now suppose that a medicine ball is thrown to a clown who is at rest upon the ice; the clown catches the medicine ball and glides together with the ball across the ice. The momentum of the medicine ball is 80 kg*m/s before the collision. The momentum of the clown is 0 m/s before the collision. The total momentum of the system before the collision is 80 kg*m/s. Therefore, the total momentum of the system after the collision must also be 80 kg*m/s. The clown and the medicine ball move together as a single unit after the collision with a combined momentum of 80 kg*m/s. Momentum is conserved in the collision.

Momentum is conserved for any interaction between two objects occurring in an isolated system. This conservation of momentum can be observed by a total system momentum analysis or by a momentum change analysis. Useful means of representing such analyses include a momentum table and a vector diagram. Later in Lesson 2, we will use the momentum conservation principle to solve problems in which the after-collision velocity of objects is predicted.

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Check Your Understanding

Express your understanding of the concept and mathematics of momentum by answering the following questions. Click on the button to view the answers.

1. When fighting fires, a firefighter must use great caution to hold a hose that emits large amounts of water at high speeds. Why would such a task be difficult?

See Answer   The hose is pushing lots of water (large mass) forward at a high speed. This means the water has a large forward momentum. In turn, the hose must have an equally large backwards momentum, making it difficult for the firefighters to manage.  

2. A large truck and a Volkswagen have a head-on collision.

a. Which vehicle experiences the greatest force of impact? b. Which vehicle experiences the greatest impulse? c. Which vehicle experiences the greatest momentum change? d. Which vehicle experiences the greatest acceleration? See Answer   a, b, c: the same for each. Both the Volkswagon and the large truck encounter the same force, the same impulse, and the same momentum change (for reasons discussed in this lesson). d: Acceleration is greatest for the Volkswagon. While the two vehicles experience the same force, the acceleration is greatest for the Volkswagon due to its smaller mass. If you find this hard to believe, then be sure to read the next question and its accompanying explanation.  

3. Miles Tugo and Ben Travlun are riding in a bus at highway speed on a nice summer day when an unlucky bug splatters onto the windshield. Miles and Ben begin discussing the physics of the situation. Miles suggests that the momentum change of the bug is much greater than that of the bus. After all, argues Miles, there was no noticeable change in the speed of the bus compared to the obvious change in the speed of the bug. Ben disagrees entirely, arguing that that both bug and bus encounter the same force, momentum change, and impulse. Who do you agree with? Support your answer.

See Answer   Ben Travlun is correct. The bug and bus experience the same force, the same impulse, and the same momentum change (as discussed in this lesson). This is contrary to the popular (though false) belief which resembles Miles' statement. The bug has less mass and therefore more acceleration; occupants of the very massive bus do not feel the extremely small acceleration. Furthermore, the bug is composed of a less hardy material and thus splatters all over the windshield. Yet the greater "splatterability" of the bug and the greater acceleration do not mean the bug has a greater force, impulse, or momentum change.  

4. If a ball is projected upward from the ground with ten units of momentum, what is the momentum of recoil of the Earth? ____________ Do we feel this? Explain.

The earth recoils with 10 units of momentum . This is not felt by Earth's occupants. Since the mass of the Earth is extremely large, the recoil velocity of the Earth is extremely small and therefore not felt.

5. If a 5-kg bowling ball is projected upward with a velocity of 2.0 m/s, then what is the recoil velocity of the Earth (mass = 6.0 x 10 24 kg).

Since the ball has an upward momentum of 10 kg*m/s , the Earth must have a downward momentum of 10 kg*m/s. To find the velocity of the Earth, use the momentum equation, p = m*v. This equation rearranges to v=p/m. By substituting into this equation,

v = (10 kg*m/s)/(6*10 24 kg) v = 1.67*10 -24 m/s (downward)

Another way to write the velocity of the earth is to write it as

0.00000000000000000000000167 m/s

6. A 120 kg lineman moving west at 2 m/s tackles an 80 kg football fullback moving east at 8 m/s. After the collision, both players move east at 2 m/s. Draw a vector diagram in which the before- and after-collision momenta of each player is represented by a momentum vector. Label the magnitude of each momentum vector.

7. In an effort to exact the most severe capital punishment upon a rather unpopular prisoner, the execution team at the Dark Ages Penitentiary search for a bullet that is ten times as massive as the rifle itself. What type of individual would want to fire a rifle that holds a bullet that is ten times more massive than the rifle? Explain.

Someone who doesn't know much physics. In such a situation as this, the target would be a safer place to stand than the rifle. The rifle would have a recoil velocity that is ten times larger than the bullet's velocity. This would produce the effect of "the rifle actually being the bullet."

8. A baseball player holds a bat loosely and bunts a ball. Express your understanding of momentum conservation by filling in the tables below.

a: +40 (add the momentum of the bat and the ball)

c: +40 (the total momentum is the same after as it is before the collision)

b: 30 (the bat must have 30 units of momentum in order for the total to be +40)

9. A Tomahawk cruise missile is launched from the barrel of a mobile missile launcher. Neglect friction. Express your understanding of momentum conservation by filling in the tables below.

a: 0 (add the momentum of the missile and the launcher)

c: 0 (the total momentum is the same after as it is before the collision)

b: -5000 (the launcher must have -5000 units of momentum in order for the total to be +0)

Return to question #6.

• Momentum Conservation Times Two