Conservation of Momentum
Article objectives
In many subfields of physics these days, it is possible to read an entire issue of a journal without ever encountering an equation involving force or a reference to Newton's laws of motion. In the last hundred and fifty years, an entirely different framework has been developed for physics, based on conservation laws.
The new approach is not just preferred because it is in fashion. It applies inside an atom or near a black hole, where Newton's laws do not. Even in everyday situations the new approach can be superior. We have already seen how perpetual motion machines could be designed that were too complex to be easily debunked by Newton's laws. The beauty of conservation laws is that they tell us something must remain the same, regardless of the complexity of the process.
So far we have discussed only two conservation laws, the laws of conservation of mass and energy. Is there any reason to believe that further conservation laws are needed in order to replace Newton's laws as a complete description of nature? Yes. Conservation of mass and energy do not relate in any way to the three dimensions of space, because both are scalars. Conservation of energy, for instance, does not prevent the planet earth from abruptly making a 90-degree turn and heading straight into the sun, because kinetic energy does not depend on direction.
Momentum
A conserved quantity of motion
Your first encounter with conservation of momentum may have come as a small child unjustly confined to a shopping cart. You spot something interesting to play with, like the display case of imported wine down at the end of the aisle, and decide to push the cart over there. But being imprisoned by Dad in the cart was not the only injustice that day. There was a far greater conspiracy to thwart your young id, one that originated in the laws of nature. Pushing forward did nudge the cart forward, but it pushed you backward. If the wheels of the cart were well lubricated, it wouldn't matter how you jerked, yanked, or kicked off from the back of the cart. You could not cause any overall forward motion of the entire system consisting of the cart with you inside.
In the Newtonian framework, we describe this as arising from Newton's third law. The cart made a force on you that was equal and opposite to your force on it. In the framework of conservation laws, we cannot attribute your frustration to conservation of energy. It would have been perfectly possible for you to transform some of the internal chemical energy stored in your body to kinetic energy of the cart and your body.
The following characteristics of the situation suggest that there may be a new conservation law involved:
A closed system is involved. All conservation laws deal with closed systems. You and the cart are a closed system, since the well-oiled wheels prevent the floor from making any forward force on you.
Something remains unchanged. The overall velocity of the system started out being zero, and you cannot change it. This vague reference to “overall velocity” can be made more precise: it is the velocity of the system's center of mass that cannot be changed.
Something can be transferred back and forth without changing the total amount. If we define forward as positive and backward as negative, then one part of the system can gain positive motion if another part acquires negative motion. If we don't want to worry about positive and negative signs, we can imagine that the whole cart was initially gliding forward on its well-oiled wheels. By kicking off from the back of the cart, you could increase your own velocity, but this inevitably causes the cart to slow down.
It thus appears that there is some numerical measure of an object's quantity of motion that is conserved when you add up all the objects within a system.
Momentum Although velocity has been referred to, it is not the total velocity of a closed system that remains constant. If it was, then firing a gun would cause the gun to recoil at the same velocity as the bullet! The gun does recoil, but at a much lower velocity than the bullet. Newton's third law tells us
$$F_{gun on bullet} = - F_{bullet on gun} $$
and assuming a constant force for simplicity, Newton's second law allows us to change this to
$$m_{bullet} ΔvbulletΔt=-mgunΔvgunΔt$$
Thus if the gun has 100 times more mass than the bullet, it will recoil at a velocity that is 100 times smaller and in the opposite direction, represented by the opposite sign. The quantity mv is therefore apparently a useful measure of motion, and we give it a name, momentum, and a symbol, p. (The letter “p” was just chosen at random, since “m” was already being used for mass.) The situations discussed so far have been one-dimensional, but in three-dimensional situations it is treated as a vector.
Definition of momentum for material objects
The momentum of a material object, i.e., a piece of matter, is defined as
$$p=mv$$
the product of the object's mass and its velocity vector.
The units of momentum are kg⋅m/s , and there is unfortunately no abbreviation for this clumsy combination of units.
The reasoning leading up to the definition of momentum was all based on the search for a conservation law, and the only reason why we bother to define such a quantity is that experiments show it is conserved:
Definition of momentum for material objects
In any closed system, the vector sum of all the momenta remains constant,
$$p_{1i}+p_{2i}+…=p_{1f}+p_{2f}+…$$ where i labels the initial and f the final momenta. (A closed system is one on which no external forces act.)
A subtle point about conservation laws is that they all refer to “closed systems,” but “closed” means different things in different cases. When discussing conservation of mass, “closed” means a system that doesn't have matter moving in or out of it. With energy, we mean that there is no work or heat transfer occurring across the boundary of the system. For momentum conservation, “closed” means there are no external forces reaching into the system.
Example 1: A cannon
A cannon of mass 1000 kg fires a 10-kg shell at a velocity of 200 m/s. At what speed does the cannon recoil?
◊ The law of conservation of momentum tells us that $$p_{cannon,i} + p_{shell,i} = p_{cannon,f} + p_{shell,f} $$
Choosing a coordinate system in which the cannon points in the positive direction, the given information is $$p_{cannon,i} = 0$$
$$p_{shell,i} = 0$$
$$p_{shell,f} = 2000 kg m/s$$
We must have \(p_{cannon,f}\)=-2000 kg m/s, so the recoil velocity of the cannon is -2 m/s.
Figure a: The ion drive engine of the NASA Deep Space 1 probe, shown under construction (left) and being tested in a vacuum chamber (right) prior to its October 1998 launch. Intended mainly as a test vehicle for new technologies, the craft nevertheless carried out a successful scientific program that included a flyby of a comet.
Example 2: Ion drive for propelling spacecraft
◊ The experimental solar-powered ion drive of the Deep Space 1 space probe expels its xenon gas exhaust at a speed of 30,000 m/s, ten times faster than the exhaust velocity for a typical chemical-fuel rocket engine. Roughly how many times greater is the maximum speed this spacecraft can reach, compared with a chemical-fueled probe with the same mass of fuel (“reaction mass”) available for pushing out the back as exhaust?
◊ Momentum equals mass multiplied by velocity. Both spacecraft are assumed to have the same amount of reaction mass, and the ion drive's exhaust has a velocity ten times greater, so the momentum of its exhaust is ten times greater. Before the engine starts firing, neither the probe nor the exhaust has any momentum, so the total momentum of the system is zero. By conservation of momentum, the total momentum must also be zero after all the exhaust has been expelled. If we define the positive direction as the direction the spacecraft is going, then the negative momentum of the exhaust is canceled by the positive momentum of the spacecraft. The ion drive allows a final speed that is ten times greater. (This simplified analysis ignores the fact that the reaction mass expelled later in the burn is not moving backward as fast, because of the forward speed of the already-moving spacecraft.)
Generalization of the momentum concept
As with all the conservation laws, the law of conservation of momentum has evolved over time. In the 1800's it was found that a beam of light striking an object would give it some momentum, even though light has no mass, and would therefore have no momentum according to the above definition. Rather than discarding the principle of conservation of momentum, the physicists of the time decided to see if the definition of momentum could be extended to include momentum carried by light. The process is analogous to the process outlined on page 281 for identifying new forms of energy. The first step was the discovery that light could impart momentum to matter, and the second step was to show that the momentum possessed by light could be related in a definite way to observable properties of the light. They found that conservation of momentum could be successfully generalized by attributing to a beam of light a momentum vector in the direction of the light's motion and having a magnitude proportional to the amount of energy the light possessed. The momentum of light is negligible under ordinary circumstances, e.g., a flashlight left on for an hour would only absorb about \(10^{-5}\) kg m/s of momentum as it recoiled.
Example 3: The tail of a comet
Figure b: Steam and other gases boiling off of the nucleus of Halley's comet. This close-up photo was taken by the European Giotto space probe, which passed within 596 km of the nucleus on March 13, 1986.
Figure c: Halley's comet, in a much less magnified view from a ground-based telescope.
Momentum is not always equal to mv. Like many comets, Halley's comet has a very elongated elliptical orbit. About once per century, its orbit brings it close to the sun. The comet's head, or nucleus, is composed of dirty ice, so the energy deposited by the intense sunlight boils off steam and dust, b. The sunlight does not just carry energy, however --- it also carries momentum. The momentum of the sunlight impacting on the smaller dust particles pushes them away from the sun, forming a tail, c. By analogy with matter, for which momentum equals mv, you would expect that massless light would have zero momentum, but the equation p=mv is not the correct one for light, and light does have momentum. (The gases typically form a second, distinct tail whose motion is controlled by the sun's magnetic field.)
The reason for bringing this up is not so that you can plug numbers into a formulas in these exotic situations. The point is that the conservation laws have proven so sturdy exactly because they can easily be amended to fit new circumstances. Newton's laws are no longer at the center of the stage of physics because they did not have the same adaptability. More generally, the moral of this story is the provisional nature of scientific truth.
It should also be noted that conservation of momentum is not a consequence of Newton's laws, as is often asserted in textbooks. Newton's laws do not apply to light, and therefore could not possibly be used to prove anything about a concept as general as the conservation of momentum in its modern form.
Momentum compared to kinetic energy
Momentum and kinetic energy are both measures of the quantity of motion, and a sideshow in the Newton-Leibnitz controversy over who invented calculus was an argument over whether mv (i.e., momentum) or \(mv^2\) (i.e., kinetic energy without the 1/2 in front) was the “true” measure of motion. The modern student can certainly be excused for wondering why we need both quantities, when their complementary nature was not evident to the greatest minds of the 1700's. The following table highlights their differences.
| Kinetic Energy... | Momentum... |
|---|---|
| is a scalar | is a vector |
| is not changed by a force perpendicular to the motion, which changes only the direction of the velocity vector. | is changed by any force, since a change in either the magnitude or the direction of the velocity vector will result in a change in the momentum vector. |
| is always positive, and cannot cancel out. | cancels with momentum in the opposite direction. |
| can be traded for other forms of energy that do not involve motion. KE is not a conserved quantity by itself. | is always conserved in a closed system. |
| is quadrupled if the velocity is doubled. | is doubled if the velocity is doubled. |
Example 4: A spinning top
A spinning top has zero total momentum, because for every moving point, there is another point on the opposite side that cancels its momentum. It does, however, have kinetic energy.
Example 5: Why a tuning fork has two prongs
A tuning fork is made with two prongs so that they can vibrate in opposite directions, canceling their momenta. In a hypothetical version with only one prong, the momentum would have to oscillate, and this momentum would have to come from somewhere, such as the hand holding the fork. The result would be that vibrations would be transmitted to the hand and rapidly die out. In a two-prong fork, the two momenta cancel, but the energies don't.
Example 6: Momentum and kinetic energy in firing a rifle
The rifle and bullet have zero momentum and zero kinetic energy to start with. When the trigger is pulled, the bullet gains some momentum in the forward direction, but this is canceled by the rifle's backward momentum, so the total momentum is still zero. The kinetic energies of the gun and bullet are both positive scalars, however, and do not cancel. The total kinetic energy is allowed to increase, because kinetic energy is being traded for other forms of energy. Initially there is chemical energy in the gunpowder. This chemical energy is converted into heat, sound, and kinetic energy. The gun's “backward” kinetic energy does not refrigerate the shooter's shoulder!
Example 7: The wobbly earth
As the moon completes half a circle around the earth, its motion reverses direction. This does not involve any change in kinetic energy, and the earth's gravitational force does not do any work on the moon. The reversed velocity vector does, however, imply a reversed momentum vector, so conservation of momentum in the closed earth-moon system tells us that the earth must also change its momentum. In fact, the earth wobbles in a little “orbit” about a point below its surface on the line connecting it and the moon. The two bodies' momentum vectors always point in opposite directions and cancel each other out.
Example 8: The earth and moon get a divorce
Why can't the moon suddenly decide to fly off one way and the earth the other way? It is not forbidden by conservation of momentum, because the moon's newly acquired momentum in one direction could be canceled out by the change in the momentum of the earth, supposing the earth headed the opposite direction at the appropriate, slower speed. The catastrophe is forbidden by conservation of energy, because both their energies would have to increase greatly.
Example 9: Momentum and kinetic energy of a glacier
A cubic-kilometer glacier would have a mass of about \(10^12\) kg. If it moves at a speed of \(10^{-5}\) m/s, then its momentum is \(10^{7}\) kg⋅m/s . This is the kind of heroic-scale result we expect, perhaps the equivalent of the space shuttle taking off, or all the cars in LA driving in the same direction at freeway speed. Its kinetic energy, however, is only 50 J, the equivalent of the calories contained in a poppy seed or the energy in a drop of gasoline too small to be seen without a microscope. The surprisingly small kinetic energy is because kinetic energy is proportional to the square of the velocity, and the square of a small number is an even smaller number.
Collisions in One Dimension
Figure e: This Hubble Space Telescope photo shows a small galaxy (yellow blob in the lower right) that has collided with a larger galaxy (spiral near the center), producing a wave of star formation (blue track) due to the shock waves passing through the galaxies' clouds of gas. This is considered a collision in the physics sense, even though it is statistically certain that no star in either galaxy ever struck a star in the other. (This is because the stars are very small compared to the distances between them.)
Physicists employ the term “collision” in a broader sense than ordinary usage, applying it to any situation where objects interact for a certain period of time. A bat hitting a baseball, a radioactively emitted particle damaging DNA, and a gun and a bullet going their separate ways are all examples of collisions in this sense. Physical contact is not even required. A comet swinging past the sun on a hyperbolic orbit is considered to undergo a collision, even though it never touches the sun. All that matters is that the comet and the sun exerted gravitational forces on each other.
The reason for broadening the term “collision” in this way is that all of these situations can be attacked mathematically using the same conservation laws in similar ways. In the first example, conservation of momentum is all that is required.
Example 10: Getting rear-ended
◊ Ms. Chang is rear-ended at a stop light by Mr. Nelson, and sues to make him pay her medical bills. He testifies that he was only going 35 miles per hour when he hit Ms. Chang. She thinks he was going much faster than that. The cars skidded together after the impact, and measurements of the length of the skid marks and the coefficient of friction show that their joint velocity immediately after the impact was 19 miles per hour. Mr. Nelson's Nissan weighs 3100 pounds, and Ms. Chang 's Cadillac weighs 5200 pounds. Is Mr. Nelson telling the truth?
◊ Since the cars skidded together, we can write down the equation for conservation of momentum using only two velocities, v for Mr. Nelson's velocity before the crash, and v' for their joint velocity afterward:
$$m_N v = m_N v' + m_C v' $$ Solving for the unknown, v, we find
$$v=(1+\frac{mC}{mN})v′.$$ Although we are given the weights in pounds, a unit of force, the ratio of the masses is the same as the ratio of the weights, and we find v=51 miles per hour. He is lying.
The above example was simple because both cars had the same velocity afterward. In many one-dimensional collisions, however, the two objects do not stick. If we wish to predict the result of such a collision, conservation of momentum does not suffice, because both velocities after the collision are unknown, so we have one equation in two unknowns.
Conservation of energy can provide a second equation, but its application is not as straightforward, because kinetic energy is only the particular form of energy that has to do with motion. In many collisions, part of the kinetic energy that was present before the collision is used to create heat or sound, or to break the objects or permanently bend them. Cars, in fact, are carefully designed to crumple in a collision. Crumpling the car uses up energy, and that's good because the goal is to get rid of all that kinetic energy in a relatively safe and controlled way. At the opposite extreme, a superball is “super” because it emerges from a collision with almost all its original kinetic energy, having only stored it briefly as potential energy while it was being squashed by the impact.
Collisions of the superball type, in which almost no kinetic energy is converted to other forms of energy, can thus be analyzed more thoroughly, because they have \(KE_f\)=\(KE_i\), as opposed to the less useful inequality \(KE_f\)<\(KE_i\) for a case like a tennis ball bouncing on grass.
Example 11: Pool balls colliding head-on
◊ Two pool balls collide head-on, so that the collision is restricted to one dimension. Pool balls are constructed so as to lose as little kinetic energy as possible in a collision, so under the assumption that no kinetic energy is converted to any other form of energy, what can we predict about the results of such a collision?
◊ Pool balls have identical masses, so we use the same symbol m for both. Conservation of momentum and no loss of kinetic energy give us the two equations
$$mv_{1i}+mv_{2i} =mv_{1f}+mv_{2f}$$ $$\frac{1}{2}mv_{1i} ^2+\frac{1}{2}mv_{2i} ^2 =\frac{1}{2}mv_{1f} ^2+\frac{1}{2}mv_{2f} ^2$$ The masses and the factors of 1/2 can be divided out, and we eliminate the cumbersome subscripts by replacing the symbols \(v_{1i}\),... with the symbols A,B,C, and D:
$$A+B =C+D$$ $$A^2+B^2 =C^2+D^2.$$ A little experimentation with numbers shows that given values of A and B, it is impossible to find C and D that satisfy these equations unless C and D equal A and B, or C and D are the same as A and B but swapped around. A formal proof of this fact is given in the sidebar. In the special case where ball 2 is initially at rest, this tells us that ball 1 is stopped dead by the collision, and ball 2 heads off at the velocity originally possessed by ball 1. This behavior will be familiar to players of pool.
Details of the Proof in Example 11
The equation A+B = C+D says that the change in one ball's velocity is equal and opposite to the change in the other's. We invent a symbol x=C-A for the change in ball 1's velocity. The second equation can then be rewritten as \(A^2\)+\(B^2\) = \((A+x)^2\)+\((B-x)^2\). Squaring out the quantities in parentheses and then simplifying, we get 0 = Ax-Bx + \(x^2\). The equation has the trivial solution x=0, i.e., neither ball's velocity is changed, but this is physically impossible because the balls can't travel through each other like ghosts. Assuming x≠ 0, we can divide by x and solve for x=B-A. This means that ball 1 has gained an amount of velocity exactly right to match ball 2's initial velocity, and vice-versa. The balls must have swapped velocities.
Often, as in the example above, the details of the algebra are the least interesting part of the problem, and considerable physical insight can be gained simply by counting the number of unknowns and comparing to the number of equations. Suppose a beginner at pool notices a case where her cue ball hits an initially stationary ball and stops dead. “Wow, what a good trick,” she thinks. “I bet I could never do that again in a million years.” But she tries again, and finds that she can't help doing it even if she doesn't want to. Luckily she has just learned about collisions in her physics course. Once she has written down the equations for conservation of energy and no loss of kinetic energy, she really doesn't have to complete the algebra. She knows that she has two equations in two unknowns, so there must be a well-defined solution. Once she has seen the result of one such collision, she knows that the same thing must happen every time. The same thing would happen with colliding marbles or croquet balls. It doesn't matter if the masses or velocities are different, because that just multiplies both equations by some constant factor.
The discovery of the neutron
This was the type of reasoning employed by James Chadwick in his 1932 discovery of the neutron. At the time, the atom was imagined to be made out of two types of fundamental particles, protons and electrons. The protons were far more massive, and clustered together in the atom's core, or nucleus. Attractive electrical forces caused the electrons to orbit the nucleus in circles, in much the same way that gravitational forces kept the planets from cruising out of the solar system. Experiments showed that the helium nucleus, for instance, exerted exactly twice as much electrical force on an electron as a nucleus of hydrogen, the smallest atom, and this was explained by saying that helium had two protons to hydrogen's one. The trouble was that according to this model, helium would have two electrons and two protons, giving it precisely twice the mass of a hydrogen atom with one of each. In fact, helium has about four times the mass of hydrogen.
Chadwick suspected that the helium nucleus possessed two additional particles of a new type, which did not participate in electrical forces at all, i.e., were electrically neutral. If these particles had very nearly the same mass as protons, then the four-to-one mass ratio of helium and hydrogen could be explained. In 1930, a new type of radiation was discovered that seemed to fit this description. It was electrically neutral, and seemed to be coming from the nuclei of light elements that had been exposed to other types of radiation. At this time, however, reports of new types of particles were a dime a dozen, and most of them turned out to be either clusters made of previously known particles or else previously known particles with higher energies. Many physicists believed that the “new” particle that had attracted Chadwick's interest was really a previously known particle called a gamma ray, which was electrically neutral. Since gamma rays have no mass, Chadwick decided to try to determine the new particle's mass and see if it was nonzero and approximately equal to the mass of a proton.
Unfortunately a subatomic particle is not something you can just put on a scale and weigh. Chadwick came up with an ingenious solution. The masses of the nuclei of the various chemical elements were already known, and techniques had already been developed for measuring the speed of a rapidly moving nucleus. He therefore set out to bombard samples of selected elements with the mysterious new particles. When a direct, head-on collision occurred between a mystery particle and the nucleus of one of the target atoms, the nucleus would be knocked out of the atom, and he would measure its velocity.
Figure f: Chadwick's subatomic pool table. A disk of the naturally occurring metal polonium provides a source of radiation capable of kicking neutrons out of the beryllium nuclei. The type of radiation emitted by the polonium is easily absorbed by a few mm of air, so the air has to be pumped out of the left-hand chamber. The neutrons, Chadwick's mystery particles, penetrate matter far more readily, and fly out through the wall and into the chamber on the right, which is filled with nitrogen or hydrogen gas. When a neutron collides with a nitrogen or hydrogen nucleus, it kicks it out of its atom at high speed, and this recoiling nucleus then rips apart thousands of other atoms of the gas. The result is an electrical pulse that can be detected in the wire on the right. Physicists had already calibrated this type of apparatus so that they could translate the strength of the electrical pulse into the velocity of the recoiling nucleus. The whole apparatus shown in the figure would fit in the palm of your hand, in dramatic contrast to today's giant particle accelerators.
Suppose, for instance, that we bombard a sample of hydrogen atoms with the mystery particles. Since the participants in the collision are fundamental particles, there is no way for kinetic energy to be converted into heat or any other form of energy, and Chadwick thus had two equations in three unknowns:
equation #1: conservation of momentum
equation #2: no loss of kinetic energy
unknown #1: mass of the mystery particle
unknown #2: initial velocity of the mystery particle
unknown #3: final velocity of the mystery particle
The number of unknowns is greater than the number of equations, so there is no unique solution. But by creating collisions with nuclei of another element, nitrogen, he gained two more equations at the expense of only one more unknown:
equation #3: conservation of momentum in the new collision
equation #4: no loss of kinetic energy in the new collision
unknown #4: final velocity of the mystery particle in the new collision
He was thus able to solve for all the unknowns, including the mass of the mystery particle, which was indeed within 1% of the mass of a proton. He named the new particle the neutron, since it is electrically neutral.
Relationship of momentum to the center of mass
Figure g: In this multiple-flash photograph, we see the wrench from above as it flies through the air, rotating as it goes. Its center of mass, marked with the black cross, travels along a straight line, unlike the other points on the wrench, which execute loops.
Using the concept of momentum we can now find a mathematical method for defining the center of mass, explain why the motion of an object's center of mass usually exhibits simpler motion than any other point, and gain a very simple and powerful way of understanding collisions.
The first step is to realize that the center of mass concept can be applied to systems containing more than one object. Even something like a wrench, which we think of as one object, is really made of many atoms. The center of mass is particularly easy to visualize in the case shown on the left, where two identical hockey pucks collide. It is clear on grounds of symmetry that their center of mass must be at the midpoint between them. After all, we previously defined the center of mass as the balance point, and if the two hockey pucks were joined with a very lightweight rod whose own mass was negligible, they would obviously balance at the midpoint. It doesn't matter that the hockey pucks are two separate objects. It is still true that the motion of their center of mass is exceptionally simple, just like that of the wrench's center of mass.
Figure h: Two hockey pucks collide. Their mutual center of mass traces the straight path shown by the dashed line.
Figure h: Moving your head so that you are always looking down from right above the center of mass, you observe the collision of the two hockey pucks in the center of mass frame.
The x coordinate of the hockey pucks' center of mass is thus given by \(x_cm\)=(\(x_1\)+\(x_2\))/2, i.e., the arithmetic average of their x coordinates. Why is its motion so simple? It has to do with conservation of momentum. Since the hockey pucks are not being acted on by any net external force, they constitute a closed system, and their total momentum is conserved. Their total momentum is
$$mv_1+mv_2 =m(v_1+v_2)$$ $$ =m(\frac{Δx_1}{ Δt}+\frac{Δx_2}{ Δt})$$ $$ =\frac{m}{Δt}Δ(x_1+x_2)$$ $$ =m\frac{2Δx_{cm}}{Δt} $$ $$=m_{total} v_{cm}$$
In other words, the total momentum of the system is the same as if all its mass was concentrated at the center of mass point. Since the total momentum is conserved, the x component of the center of mass's velocity vector cannot change. The same is also true for the other components, so the center of mass must move along a straight line at constant speed.
The above relationship between the total momentum and the motion of the center of mass applies to any system, even if it is not closed.
total momentum related to center of mass motion
The total momentum of any system is related to its total mass and the velocity of its center of mass by the equation
$$p_{total}=m_{total} v_{cm}.$$
What about a system containing objects with unequal masses, or containing more than two objects? The reasoning above can be generalized to a weighted average
$$xcm=\frac{m_1 x_1 +m_2 x_2+…}{m_1 +m_2 +…}$$
with similar equations for the y and z coordinates.
Momentum in different frames of reference
Absolute motion is supposed to be undetectable, i.e., the laws of physics are supposed to be equally valid in all inertial frames of reference. If we first calculate some momenta in one frame of reference and find that momentum is conserved, and then rework the whole problem in some other frame of reference that is moving with respect to the first, the numerical values of the momenta will all be different. Even so, momentum will still be conserved. All that matters is that we work a single problem in one consistent frame of reference.
One way of proving this is to apply the equation \(p_{total}\) = \(m_{total} v_{cm}\) . If the velocity of frame B relative to frame A is \(v_{BA}\) , then the only effect of changing frames of reference is to change \(v_{cm}\) from its original value to \(v_{cm}\)+\(v_{BA}\) . This adds a constant onto the momentum vector, which has no effect on conservation of momentum.
The center of mass frame of reference
A particularly useful frame of reference in many cases is the frame that moves along with the center of mass, called the center of mass (c.m.) frame. In this frame, the total momentum is zero. The following examples show how the center of mass frame can be a powerful tool for simplifying our understanding of collisions.
Example 12: A collision of pool balls viewed in the c.m. frame
If you move your head so that your eye is always above the point halfway in between the two pool balls, you are viewing things in the center of mass frame. In this frame, the balls come toward the center of mass at equal speeds. By symmetry, they must therefore recoil at equal speeds along the lines on which they entered. Since the balls have essentially swapped paths in the center of mass frame, the same must also be true in any other frame. This is the same result that required laborious algebra to prove previously without the concept of the center of mass frame.
Example 13: The slingshot effect
Figure j: The slingshot effect viewed in the sun's frame of reference. Jupiter is moving to the left, and the collision is head-on.
Figure k: The slingshot viewed in the frame of the center of mass of the Jupiter-spacecraft system.
It is a counterintuitive fact that a spacecraft can pick up speed by swinging around a planet, if it arrives in the opposite direction compared to the planet's motion. Although there is no physical contact, we treat the encounter as a one-dimensional collision, and analyze it in the center of mass frame. Figure j shows such a “collision,” with a space probe whipping around Jupiter. In the sun's frame of reference, Jupiter is moving.
What about the center of mass frame? Since Jupiter is so much more massive than the spacecraft, the center of mass is essentially fixed at Jupiter's center, and Jupiter has zero velocity in the center of mass frame, as shown in figure k. The c.m. frame is moving to the left compared to the sun-fixed frame used in j, so the spacecraft's initial velocity is greater in this frame.
Things are simpler in the center of mass frame, because it is more symmetric. In the complicated sun-fixed frame, the incoming leg of the encounter is rapid, because the two bodies are rushing toward each other, while their separation on the outbound leg is more gradual, because Jupiter is trying to catch up. In the c.m. frame, Jupiter is sitting still, and there is perfect symmetry between the incoming and outgoing legs, so by symmetry we have v1f=-v1i. Going back to the sun-fixed frame, the spacecraft's final velocity is increased by the frames' motion relative to each other. In the sun-fixed frame, the spacecraft's velocity has increased greatly.
The result can also be understood in terms of work and energy. In Jupiter's frame, Jupiter is not doing any work on the spacecraft as it rounds the back of the planet, because the motion is perpendicular to the force. But in the sun's frame, the spacecraft's velocity vector at the same moment has a large component to the left, so Jupiter is doing work on it.
Momentum transfer
The rate of change of momentum
As with conservation of energy, we need a way to measure and calculate the transfer of momentum into or out of a system when the system is not closed. In the case of energy, the answer was rather complicated, and entirely different techniques had to be used for measuring the transfer of mechanical energy (work) and the transfer of heat by conduction. For momentum, the situation is far simpler.
In the simplest case, the system consists of a single object acted on by a constant external force. Since it is only the object's velocity that can change, not its mass, the momentum transferred is $$Δp=mΔv,$$ which with the help of a=F/m and the constant-acceleration equation a=Δv/Δt becomes
$$Δp =maΔt$$ $$ =FΔt.$$ Thus the rate of transfer of momentum, i.e., the number of kg⋅m/s absorbed per second, is simply the external force,
$$F=\frac{Δp}{Δt} $$ $$ \text{[relationship between the force on an object and the rate of change of its momentum; valid only if the force is constant]}$$
This is just a restatement of Newton's second law, and in fact Newton originally stated it this way. As shown in figure l, the relationship between force and momentum is directly analogous to that between power and energy.
Figure l: Power and force are the rates at which energy and momentum are transferred.
The situation is not materially altered for a system composed of many objects. There may be forces between the objects, but the internal forces cannot change the system's momentum. (If they did, then removing the external forces would result in a closed system that could change its own momentum, like the mythical man who could pull himself up by his own bootstraps. That would violate conservation of momentum.) The equation above becomes
$$F_{total}=\frac{Δp_{total}}{Δt} \;$$ $$ \text{[relationship between the total external force on a system and the rate of change of its total momentum; valid only if the force is constant]}$$
Example 14: Walking into a lamppost
◊ Starting from rest, you begin walking, bringing your momentum up to 100 kg⋅m/s . You walk straight into a lamppost. Why is the momentum change of -100 kg⋅m/s caused by the lamppost so much more painful than the change of +100 kg⋅m/s when you started walking?
◊ The situation is one-dimensional, so we can dispense with the vector notation. It probably takes you about 1 s to speed up initially, so the ground's force on you is F=Δp/Δt≈100 N . Your impact with the lamppost, however, is over in the blink of an eye, say 1/10 s or less. Dividing by this much smaller Δ t gives a much larger force, perhaps thousands of newtons. (The negative sign simply indicates that the force is in the opposite direction.)
This is also the principle of airbags in cars. The time required for the airbag to decelerate your head is fairly long, the time required for your face to travel 20 or 30 cm. Without an airbag, your face would hit the dashboard, and the time interval would be the much shorter time taken by your skull to move a couple of centimeters while your face compressed. Note that either way, the same amount of mechanical work has to be done on your head: enough to eliminate all its kinetic energy.
Figure m: The airbag increases Δ t so as to reduce F=Δ p/Δ t.
Example 15: Ion drive for spacecraft
◊ The ion drive of the Deep Space 1 spacecraft, discussed in example 2, produces a thrust of 90 mN (millinewtons). It carries about 80 kg of reaction mass, which it ejects at a speed of 30,000 m/s. For how long can the engine continue supplying this amount of thrust before running out of reaction mass to shove out the back?
◊ Solving the equation F=Δ p/Δ t for the unknown Δ t, and treating force and momentum as scalars since the problem is one-dimensional, we find
$$Δt =\frac{Δp}{F}$$ $$ =\frac{m_{exhaust} Δv_{exhaust}}{F}$$ $$ =\frac{(80 kg)(30,000 m/s)}{0.090 N}$$ $$ =2.7×10^7 s $$ $$ =300 days$$
Example 16: A toppling box
If you place a box on a frictionless surface, it will fall over with a very complicated motion that is hard to predict in detail. We know, however, that its center of mass moves in the same direction as its momentum vector points. There are two forces, a normal force and a gravitational force, both of which are vertical. (The gravitational force is actually many gravitational forces acting on all the atoms in the box.) The total force must be vertical, so the momentum vector must be purely vertical too, and the center of mass travels vertically. This is true even if the box bounces and tumbles. [Based on an example by Kleppner and Kolenkow.]
The area under the force-time graph
Figure o: The F-t graph for a tennis racquet hitting a ball might look like this. The amount of momentum transferred equals the area under the curve.
Few real collisions involve a constant force. For example, when a tennis ball hits a racquet, the strings stretch and the ball flattens dramatically. They are both acting like springs that obey Hooke's law, which says that the force is proportional to the amount of stretching or flattening. The force is therefore small at first, ramps up to a maximum when the ball is about to reverse directions, and ramps back down again as the ball is on its way back out. The equation F=Δ p/Δ t, derived under the assumption of constant acceleration, does not apply here, and the force does not even have a single well-defined numerical value that could be plugged in to the equation.
As with similar-looking equations such as v=Δ p/Δ t, the equation F=Δ p/Δ t is correctly generalized by saying that the force is the slope of the p-t graph.
Conversely, if we wish to find Δ p from a graph such as the one in figure o, one approach would be to divide the force by the mass of the ball, rescaling the F axis to create a graph of acceleration versus time. The area under the acceleration-versus-time graph gives the change in velocity, which can then be multiplied by the mass to find the change in momentum. An unnecessary complication was introduced, however, because we began by dividing by the mass and ended by multiplying by it. It would have made just as much sense to find the area under the original F-t graph, which would have given us the momentum change directly.
Momentum in three dimensions
We discuss how the concepts applied previously to one-dimensional situations can be used as well in three dimensions. Often vector addition is all that is needed to solve a problem:
Example 17: An explosion
◊ Astronomers observe the planet Mars as the Martians fight a nuclear war. The Martian bombs are so powerful that they rip the planet into three separate pieces of liquified rock, all having the same mass. If one fragment flies off with velocity components
$$v_{1x} = 0$$ $$v_{1y} = 1.0 x 10^4 km/hr$$
and the second with $$v_{2x} = 1.0 x 10^4 km/hr$$ $$v_{2y} = 0$$
(all in the center of mass frame) what is the magnitude of the third one's velocity?
◊ In the center of mass frame, the planet initially had zero momentum. After the explosion, the vector sum of the momenta must still be zero. Vector addition can be done by adding components, so
$$mv_{1x}+mv_{2x}+mv_{3x} =0 \; \text{,and}$$ $$ mv_{1y}+mv_{2y}+mv_{3y} =0,$$ where we have used the same symbol m for all the terms, because the fragments all have the same mass. The masses can be eliminated by dividing each equation by m, and we find
$$v_{3x} =-1.0×10^4 km/hr$$ $$ v_{3y} =-1.0×10^4 km/hr$$
which gives a magnitude of
$$|v_3| =\sqrt{v_{3x} ^2+v_{3y} ^2}$$ $$ =1.4×10^4 km/hr$$
Center of mass
In three dimensions, we have the vector equations
$$F_{tota;} = \frac{\Delta p_total}{\Delta t}$$
The following is an example of their use.
Example 18: The bola
The bola, similar to the North American lasso, is used by South American gauchos to catch small animals by tangling up their legs in the three leather thongs. The motion of the whirling bola through the air is extremely complicated, and would be a challenge to analyze mathematically. The motion of its center of mass, however, is much simpler. The only forces on it are gravitational, so $$F_{total}=m_{total} g.$$ Using the equation \(F_total\)=Δ\(p_total\)/Δt , we find
$$Δp_total /Δt=m_total g,$$
and since the mass is constant, the equation \(p_{total}=m_{total} v_{cm}\) allows us to change this to \(m_{total} Δv_{cm}\)/Δt=\(m_{total} g\).
The mass cancels, and Δ\(v_cm\)/Δt is simply the acceleration of the center of mass, so $$a_cm =g.$$ In other words, the motion of the system is the same as if all its mass was concentrated at and moving with the center of mass. The bola has a constant downward acceleration equal to g, and flies along the same parabola as any other projectile thrown with the same initial center of mass velocity. Throwing a bola with the correct rotation is presumably a difficult skill, but making it hit its target is no harder than it is with a ball or a single rock. [Based on an example by Kleppner and Kolenkow.]
Counting equations and unknowns
Counting equations and unknowns is just as useful as in one dimension, but every object's momentum vector has three components, so an unknown momentum vector counts as three unknowns. Conservation of momentum is a single vector equation, but it says that all three components of the total momentum vector stay constant, so we count it as three equations. Of course if the motion happens to be confined to two dimensions, then we need only count vectors as having two components.
Example 19: A two-car crash with sticking
Suppose two cars collide, stick together, and skid off together. If we know the cars' initial momentum vectors, we can count equations and unknowns as follows:
unknown #1: x component of cars' final, total momentum
unknown #2: y component of cars' final, total momentum
equation #1: conservation of the total \(p_x\)
equation #2: conservation of the total \(p_y\)
Since the number of equations equals the number of unknowns, there must be one unique solution for their total momentum vector after the crash. In other words, the speed and direction at which their common center of mass moves off together is unaffected by factors such as whether the cars collide center-to-center or catch each other a little off-center.
Example 20: Shooting pool
Two pool balls collide, and as before we assume there is no decrease in the total kinetic energy, i.e., no energy converted from KE into other forms. As in the previous example, we assume we are given the initial velocities and want to find the final velocities. The equations and unknowns are:
unknown #1: x component of ball #1's final momentum
unknown #2: y component of ball #1's final momentum
unknown #3: x component of ball #2's final momentum
unknown #4: y component of ball #2's final momentum
equation #1: conservation of the total \(p_x\)
equation #2: conservation of the total \(p_y\)
equation #3: no decrease in total KE
Note that we do not count the balls' final kinetic energies as unknowns, because knowing the momentum vector, one can always find the velocity and thus the kinetic energy. The number of equations is less than the number of unknowns, so no unique result is guaranteed. This is what makes pool an interesting game. By aiming the cue ball to one side of the target ball you can have some control over the balls' speeds and directions of motion after the collision.
It is not possible, however, to choose any combination of final speeds and directions. For instance, a certain shot may give the correct direction of motion for the target ball, making it go into a pocket, but may also have the undesired side-effect of making the cue ball go in a pocket.
Calculations with the momentum vector
The following example illustrates how a force is required to change the direction of the momentum vector, just as one would be required to change its magnitude.
Example 21: A turbine
◊ In a hydroelectric plant, water flowing over a dam drives a turbine, which runs a generator to make electric power. The figure shows a simplified physical model of the water hitting the turbine, in which it is assumed that the stream of water comes in at a 45° angle with respect to the turbine blade, and bounces off at a 90° angle at nearly the same speed. The water flows at a rate R, in units of kg/s, and the speed of the water is v. What are the magnitude and direction of the water's force on the turbine?
◊ In a time interval Δt, the mass of water that strikes the blade is RΔt, and the magnitude of its initial momentum is mv=vRΔ t. The water's final momentum vector is of the same magnitude, but in the perpendicular direction. By Newton's third law, the water's force on the blade is equal and opposite to the blade's force on the water. Since the force is constant, we can use the equation
$$F_{blade on water}=\frac{Δp_{water}}{ Δt}$$.
Choosing the x axis to be to the right and the y axis to be up, this can be broken down into components as
$$F_{blade on water,x}=\frac{Δp_{water,x}}{Δt}$$ $$ =\frac{-vRΔt-0}{Δt}$$ $$ =-vR$$ and
$$F_{blade on water,y}=\frac{Δp_{water,y}}{Δt}$$ $$=\frac{0-(-vRΔt)}{Δt}$$ $$ =vR$$ The water's force on the blade thus has components
$$F_{water on blade,x} =vR$$ $$ F_{water on blade,y} =-vR.$$
In situations like this, it is always a good idea to check that the result makes sense physically. The x component of the water's force on the blade is positive, which is correct since we know the blade will be pushed to the right. The y component is negative, which also makes sense because the water must push the blade down. The magnitude of the water's force on the blade is
$$|F_{water on blade}|=\sqrt{2}vR$$
and its direction is at a 45-degree angle down and to the right.
Applications of calculus
The rate of change of momentum can be represented with a derivative,
$$F_{total}=\frac{dp_{total}}{dt}$$ And of course the business about the area under the F-t curve is really an integral, Δ\(p_{total}\)=∫\(F_{total} dt\) , which can be made into an integral of a vector in the more general three-dimensional case:
$$Δp_{total}=\int F_{total} dt$$
In the case of a material object that is neither losing nor picking up mass, these are just trivially rearranged versions of familiar equations, e.g., F=mdv/dt rewritten as F=d(mv)/dt. The following is a less trivial example, where F=ma alone would not have been very easy to work with.
Example 22: Rain falling into a moving cart
◊ If 1 kg/s of rain falls vertically into a 10-kg cart that is rolling without friction at an initial speed of 1.0 m/s, what is the effect on the speed of the cart when the rain first starts falling?
◊ The rain and the cart make horizontal forces on each other, but there is no external horizontal force on the rain-plus-cart system, so the horizontal motion obeys
$$F=\frac{d(mv)}{dt}=0$$ We use the product rule to find
$$0=\frac{dm}{dt}v+m\frac{dv}{dt}$$ We are trying to find how v changes, so we solve for dv/dt,
$$\frac{dv}{dt} =-\frac{v}{m}\frac{dm}{dt}$$ $$ =-(\frac{1 m/s}{10 kg})(1 kg/s)$$ $$ =-0.1 m/s^2$$ (This is only at the moment when the rain starts to fall.)
Finally we note that there are cases where F=ma is not just less convenient than F=dp/dt but in fact F=ma is wrong and F=dp/dt is right. A good example is the formation of a comet's tail by sunlight. We cannot use F=ma to describe this process, since we are dealing with a collision of light with matter, whereas Newton's laws only apply to matter. The equation F=dp/dt, on the other hand, allows us to find the force experienced by an atom of gas in the comet's tail if we know the rate at which the momentum vectors of light rays are being turned around by reflection from the atom.