3.1 Momentum In One Dimension
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3.1.1 Mechanical momentum
In the martial arts movie Crouching Tiger, Hidden Dragon, those who had received mystical enlightenment are able to violate the laws of physics. Some of the violations are obvious, such as their ability to fly, but others are a little more subtle. The rebellious young heroine/antiheroine Jen Yu gets into an argument while sitting at a table in a restaurant. A young tough, Iron Arm Lu, comes running toward her at full speed, and she puts up one arm and effortlessly makes him bounce back, without even getting out of her seat or bracing herself against anything. She does all this between bites.
Although kinetic energy doesn't depend on the direction of motion, we've already seen how conservation of energy combined with Galilean relativity allows us to make some predictions about the direction of motion. One of the examples was a demonstration that it isn't possible for a hockey puck to spontaneously reverse its direction of motion. In the scene from the movie, however, the woman's assailant isn't just gliding through space. He's interacting with her, so the previous argument doesn't apply here, and we need to generalize it to more than one object. We consider the case of a physical system composed of pointlike material particles, in which every particle interacts with every other particle through an energy \(U(r)\) that depends only on the distance \(r\) between them. This still allows for a fairly general mechanical system, by which I mean roughly a system made of matter, not light. The characters in the movie are made of protons, neutrons, and electrons, so they would constitute such a system if the interactions among all these particles were of the form \(U(r)\).^{1} We might even be able to get away with thinking of each person as one big particle, if it's a good approximation to say that every part of each person's whole body moves in the same direction at the same speed.
The basic insight can be extracted from the special case where there are only two particles interacting, and they only move in one dimension, as in the example shown in figure b. Conservation of energy says
\[\begin{equation*} K_{1i}+K_{2i}+U_i = K_{1f}+K_{2f}+U_f . \end{equation*}\]
For simplicity, let's assume that the interactions start after the time we're calling initial, and end before the instant we choose as final. This is true in figure b, for example. Then \(U_i=U_f\), and we can subtract the interaction energies from both sides, giving.
\[\begin{align*} K_{1i}+K_{2i} &= K_{1f}+K_{2f} \\ \frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2 &= \frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2 . \end{align*}\]
As in the one-particle argument on page 89, the trick is to require conservation of energy not just in one particular frame of reference, but in every frame of reference. In a frame of reference moving at velocity \(u\) relative to the first one, the velocities all have \(u\) added onto them:^{2}
\[\begin{multline*} \frac{1}{2}m_1(v_{1i}+u)^2+\frac{1}{2}m_2(v_{2i}+u)^2 = \frac{1}{2}m_1(v_{1f}+u)^2+\frac{1}{2}m_2(v_{2f}+u)^2 \end{multline*}\]
When we square a quantity like \((v_{1i}+u)^2\), we get the same \(v_{1i}^2\) that occurred in the original frame of reference, plus two \(u\)-dependent terms, \(2v_{1i}u+u^2\). Subtracting the original conservation of energy equation from the version in the new frame of reference, we have
\[\begin{align*} m_1v_{1i}u+m_2v_{2i}u = m_1v_{1f}u+m_2v_{2f}u , \\ \text{or, dividing by $u$,} m_1v_{1i}+m_2v_{2i} = m_1v_{1f}+m_2v_{2f} . \end{align*}\]
This is a statement that when you add up \(mv\) for the whole system, that total remains constant over time. In other words, this is a conservation law. The quantity \(mv\) is called momentum, notated \(p\) for obscure historical reasons. Its units are \(\text{kg}\cdot\text{m}/\text{s}\).
Unlike kinetic energy, momentum depends on the direction of motion, since the velocity is not squared. In one dimension, motion in the same direction as the positive \(x\) axis is represented with positive values of \(v\) and \(p\). Motion in the opposite direction has negative \(v\) and \(p\).
Example 1: Jen Yu meets Iron Arm Lu |
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\(\triangleright\) Initially, Jen Yu is at rest, and Iron Arm Lu is charging to the left, toward her, at 5 m/s. Jen Yu's mass is 50 kg, and Lu's is 100 kg. After the collision, the movie shows Jen Yu still at rest, and Lu rebounding at 5 m/s to the right. Is this consistent with the laws of physics, or would it be impossible in real life? \(\triangleright\) This is perfectly consistent with conservation of mass (50 kg+100 kg=50 kg+100 kg), and also with conservation of energy, since neither person's kinetic energy changes, and there is therefore no change in the total energy. (We don't have to worry about interaction energies, because the two points in time we're considering are ones at which the two people aren't interacting.) To analyze whether the scene violates conservation of momentum, we have to pick a coordinate system. Let's define positive as being to the right. The initial momentum is (50 kg)(0 m/s)+(100 kg)(\(-\)5 m/s)=\(-\)500 \(\text{kg}\cdot\text{m}/\text{s}\), and the final momentum is (50 kg)(0 m/s)+(100 kg)(5 m/s)=500 \(\text{kg}\cdot\text{m}/\text{s}\). This is a change of 1000 \(\text{kg}\cdot\text{m}/\text{s}\), which is impossible if the two people constitute a closed system. One could argue that they're not a closed system, since Lu might be exchanging momentum with the floor, and Jen Yu might be exchanging momentum with the seat of her chair. This is a reasonable objection, but in the following section we'll see that there are physical reasons why, in this situation, the force of friction would be relatively weak, and would not be able to transfer that much momentum in a fraction of a second. |
This example points to an intuitive interpretation of conservation of momentum, which is that interactions are always mutual. That is, Jen Yu can't change Lu's momentum without having her own momentum changed as well.
Example 2: A cannon |
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\(\triangleright\) 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? \(\triangleright\) The law of conservation of momentum tells us that \[\begin{equation*} p_{cannon,i} + p_{shell,i} = p_{cannon,f} + p_{shell,f} . \end{equation*}\] Choosing a coordinate system in which the cannon points in the positive direction, the given information is \[\begin{align*} p_{cannon,i} &= 0 \\ p_shell,i &= 0 \\ p_{shell,f} &= 2000\ \text{kg}\cdot\text{m/s} . \end{align*}\] We must have \(p_{cannon,f}=-\text{2000 kg}\cdot\text{m/s}\), so the recoil velocity of the cannon is 2 m/s. |
Example 3: Ion drive |
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\(\triangleright\) 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? \(\triangleright\) 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.) |
3.1.2 Nonmechanical momentum
So far, it sounds as though conservation of momentum can be proved mathematically, unlike conservation of mass and energy, which are entirely based on observations. The proof, however, was only for a mechanical system, with interactions of the form \(U(r)\). Conservation of momentum can be extended to other systems as well, but this generalization is based on experiments, not mathematical proof. Light is the most important example of momentum that doesn't equal \(mv\) --- light doesn't have mass at all, but it does have momentum. For example, a flashlight left on for an hour would absorb about \(10^{-5}\ \text{kg}\cdot\text{m}/\text{s}\) of momentum as it recoiled.
Example 4: Halley's comet |
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Momentum is not always equal to \(mv\). Halley's comet, shown in figure d, has a very elongated elliptical orbit, like those of many other comets. 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 gradually removes ice from the surface and turns it into water vapor. The bottom photo shows a view of the water coming off of the nucleus from the European Giotto space probe, which passed within 596 km of the comet's head on March 13, 1986. Figure d / Halley's comet. Top: A photograph made from earth. Bottom: A view of the nucleus from the Giotto space probe. (W. Liller and European Space Agency) The sunlight does not just carry energy, however. It also carries momentum. Once the steam comes off, the momentum of the sunlight impacting on it pushes it away from the sun, forming a tail as shown in in the top image. The tail always points away from the sun, so when the comet is receding from the sun, the tail is in front. 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. (Some comets also have a second tail, which is propelled by electrical forces rather than by the momentum of sunlight.) |
The reason for bringing this up is not so that you can plug numbers into 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. The momentum of light will be a natural consequence of the discussion of the theory of relativity in chapter 7.
3.1.3 Momentum compared to kinetic energy
Momentum and kinetic energy are both measures of the quantity of motion, and a sideshow in the Newton-Leibniz 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 1700s. The following table highlights their differences.
Kinetic energy... | Momentum... |
doesn't depend on direction. | depends on direction. |
is always positive, and cannot cancel out. | cancels with momentum in the opposite direction. |
can be traded for forms of energy that do not involve motion. Kinetic energy 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 5: A spinning top |
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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 6: Why a tuning fork has two prongs |
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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 7: Momentum and kinetic energy in firing a rifle |
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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 numbers, 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 8: The wobbly earth |
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As the moon completes half a circle around the earth, its motion reverses direction. This does not involve any change in kinetic energy. The reversed velocity does, however, imply a reversed momentum, so conservation of momentum in the closed earth-moon system tells us that the earth must also reverse 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' momenta always point in opposite directions and cancel each other out. |
Example 9: The earth and moon get a divorce |
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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 off in the opposite direction at the appropriate, slower speed. The catastrophe is forbidden by conservation of energy, because their energies would have to increase greatly. |
Example 10: Momentum and kinetic energy of a glacier |
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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 \(10^7\ \text{kg}\cdot\text{m}/\text{s}\). 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. |
Discussion Questions
◊ If all the air molecules in the room settled down in a thin film on the floor, would that violate conservation of momentum as well as conservation of energy?
◊ A refrigerator has coils in back that get hot, and heat is molecular motion. These moving molecules have both energy and momentum. Why doesn't the refrigerator need to be tied to the wall to keep it from recoiling from the momentum it loses out the back?
3.1.4 Collisions in one dimension
Physicists employ the term “collision” in a broader sense than in ordinary usage, applying it to any situation where objects interact for a certain period of time. A bat hitting a baseball, a cosmic ray 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 interacted gravitationally with 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 our first example, conservation of momentum is all that is required.
Example 11: Getting rear-ended |
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\(\triangleright\) 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 55 km 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 show that their joint velocity immediately after the impact was 30 km per hour. Mr.\ Nelson's Nissan has a mass of 1400 kg, and Ms.\ Chang 's Cadillac is 2400 kg. Is Mr.\ Nelson telling the truth? \(\triangleright\) 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: \[\begin{equation*} m_{N} v = m_{N} v' + m_{C} v' . \end{equation*}\] Solving for the unknown, \(v\), we find \[\begin{align*} v &= \left(1+\frac{ m_C}{ m_{N}}\right) v' \\ &= \text{80\ km/hr} . \end{align*}\] 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 interatomic electrical 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 \(K_f=K_i\), as opposed to the less useful inequality \(K_f\lt K_i\) for a case like a tennis ball bouncing on grass.
Example 12: Pool balls colliding head-on |
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\(\triangleright\) 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? \(\triangleright\) Pool balls have identical masses, so we use the same symbol \(m\) for both. Conservation of energy and no loss of kinetic energy give us the two equations \[\begin{align*} mv_{1i}+mv_{2i} &= mv_{1f}+mv_{2f} \\ \frac{1}{2}mv_{1i}^2+\frac{1}{2}mv_{2i}^2 &= \frac{1}{2}mv_{1f}^2+\frac{1}{2}mv_{2f}^2 \end{align*}\] 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 \(D\): \[\begin{align*} A+B &= C+D \\ A^2+B^2 &= C^2+D^2 . \end{align*}\] 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. |
Gory Details of the Proof in Example 12 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 \(x\ne 0\), we can divide by \(x\) and solve for \(x=B-A\). This means that ball 1 has gained an amount of velocity exactly sufficient to match ball 2's initial velocity, and vice-versa. The balls must have swapped velocities.}
Often, as in example 12, 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. Electrical attraction caused the electrons to orbit the nucleus in circles, in much the same way that gravity kept the planets from cruising out of the solar system. Experiments showed, for example, that twice as much energy was required to strip the last electron off of a helium atom as was needed to remove the single electron from a hydrogen 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 interactions 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.
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.
Discussion Questions
◊ Good pool players learn to make the cue ball spin, which can cause it not to stop dead in a head-on collision with a stationary ball. If this does not violate the laws of physics, what hidden assumption was there in the example in the text where it was proved that the cue ball must stop?
3.1.5 The center of mass
Figures i and k show two examples where a motion that appears complicated actually has a very simple feature. In both cases, there is a particular point, called the center of mass, whose motion is surprisingly simple. The highjumper flexes his body as he passes over the bar, so his motion is intrinsically very complicated, and yet his center of mass's motion is a simple parabola, just like the parabolic arc of a pointlike particle. The wrench's center of mass travels in a straight line as seen from above, which is what we'd expect for a pointlike particle flying through the air.
The highjumper and the wrench are both complicated systems, each consisting of zillions of subatomic particles. To understand what's going on, let's instead look at a nice simple system, two pool balls colliding. We assume the balls are a closed system (i.e., their interaction with the felt surface is not important) and that their rotation is unimportant, so that we'll be able to treat each one as a single particle. By symmetry, the only place their center of mass can be is half-way in between, at an \(x\) coordinate equal to the average of the two balls' positions, \(x_{cm}=(x_1+x_2)/2\).
Figure j makes it appear that the center of mass, marked with an \(\times\), moves with constant velocity to the right, regardless of the collision, and we can easily prove this using conservation of momentum:
Since momentum is conserved, the last expression is constant, which proves that \(v_{cm}\) is constant.
Rearranging this a little, we have \(p_{total}=m_{total}v_{cm}\). 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.
Sigma notation
When there is a large, potentially unknown number of particles, we can write sums like the ones occurring above using symbols like “\(+...\),” but that gets awkward. It's more convenient to use the Greek uppercase sigma, \(\Sigma\), to indicate addition. For example, the sum \(1^2+2^2+3^2+4^2=30\) could be written as
read “the sum from \(j=1\) to \(n\) of \(j^2\).” The variable \(j\) is a dummy variable, just like the \(d{}x\) in an integral that tells you you're integrating with respect to \(x\), but has no significance outside the integral. The \(j\) below the sigma tells you what variable is changing from one term of the sum to the next, but \(j\) has no significance outside the sum.
As an example, let's generalize the proof of \(p_{total}=m_{total}v_{cm}\) to the case of an arbitrary number \(n\) of identical particles moving in one dimension, rather than just two particles. The center of mass is at
where \(x_1\) is the mass of the first particle, and so on. The velocity of the center of mass is
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:
Example 13: The solar system's center of mass |
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In the discussion of the sun's gravitational field on page 99, I mentioned in a footnote that the sun doesn't really stay in one place while the planets orbit around it. Actually, motion is relative, so it's meaningless to ask whether the sun is absolutely at rest, but it is meaningful to ask whether it moves in a straight line at constant velocity. We can now see that since the solar system is a closed system, its total momentum must be constant, and \(p_{total}= m_{total}v_{cm}\) then tells us that it's the solar system's center of mass that has constant velocity, not the sun. The sun wobbles around this point irregularly due to its interactions with the planets, Jupiter in particular. |
In the discussion of the sun's gravitational field on page 99, I mentioned in a footnote that the sun doesn't really stay in one place while the planets orbit around it. Actually, motion is relative, so it's meaningless to ask whether the sun is absolutely at rest, but it is meaningful to ask whether it moves in a straight line at constant velocity. We can now see that since the solar system is a closed system, its total momentum must be constant, and \(p_{total}= m_{total}v_{cm}\) then tells us that it's the solar system's center of mass that has constant velocity, not the sun. The sun wobbles around this point irregularly due to its interactions with the planets, Jupiter in particular.
Example 14: The earth-moon system |
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The earth-moon system is much simpler than the solar system because it contains only two objects. Where is the center of mass of this system? Let \(x\)=0 be the earth's center, so that the moon lies at \(x=3.8\times10^5\ \text{km}\). Then \[\begin{align*} x_{cm} &= \frac{\sum_{ j=1}^{2}{ m_{j} x_{j}}} {\sum_{ j=1}^{2}{ m_j}} \\ &= \frac{ m_{1} x_{1}+ m_2 x_{2}} { m_{1}+ m_2} , \\ \text{and letting 1 be the earth and 2 the moon, we have} x_{cm} &= \frac{ m_{earth}\times0+ m_{moon} x_{moon}} { m_{earth}+ m_{moon}} \\ &= 4600\ \text{km} , \end{align*}\] or about three quarters of the way from the earth's center to its surface. |
Example 15: The center of mass as an average |
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\(\triangleright\) Explain how we know that the center of mass of each object is at the location shown in figure p. \(\triangleright\) The center of mass is a sort of average, so the height of the centers of mass in 1 and 2 has to be midway between the two squares, because that height is the average of the heights of the two squares. Example 3 is a combination of examples 1 and 2, so we can find its center of mass by averaging the horizontal positions of their centers of mass. In example 4, each square has been skewed a little, but just as much mass has been moved up as down, so the average vertical position of the mass hasn't changed. Example 5 is clearly not all that different from example 4, the main difference being a slight clockwise rotation, so just as in example 4, the center of mass must be hanging in empty space, where there isn't actually any mass. Horizontally, the center of mass must be between the heels and toes, or else it wouldn't be possible to stand without tipping over. |
Example 16: Momentum and Galilean relativity |
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The principle of Galilean relativity states that 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 one frame relative to the other is \(u\), then the only effect of changing frames of reference is to change \(v_{cm}\) from its original value to \(v_{cm}+u\). This adds a constant onto the momentum, which has no effect on conservation of momentum. |
self-check:
The figure shows a gymnast holding onto the inside of a big wheel. From inside the wheel, how could he make it roll one way or the other?
(answer in the back of the PDF version of the book)
3.1.6 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 17: A collision of pool balls viewed in the c.m. frame |
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If you move your head so that your eye is always above the point halfway in between the two pool balls, as in figure r, 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 18: The slingshot effect |
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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. 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 3.1.6. The c.m. frame is moving to the left compared to the sun-fixed frame used in figure 3.1.6, so the spacecraft's initial velocity is greater in this frame than in the sun's frame. Things are simpler in the center of mass frame, because it is more symmetric. In the 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 \(v_{1f}=- v_{1i}\). 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. |
Example 19: Einstein's motorcycle |
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We've assumed we were dealing with a system of material objects, for which the equation \( p=mv\) was true. What if our system contains only light rays, or a mixture of light and matter? As a college student, Einstein kept worrying about was what a beam of light would look like if you could ride alongside it on a motorcycle. In other words, he imagined putting himself in the light beam's center of mass frame. Chapter 7 discusses Einstein's resolution of this problem, but the basic point is that you can't ride the motorcycle alongside the light beam, because material objects can't go as fast as the speed of light. A beam of light has no center of mass frame of reference. |
Discussion Questions
◊ Make up a numerical example of two unequal masses moving in one dimension at constant velocity, and verify the equation \(p_{total}=m_{total}v_{cm}\) over a time interval of one second.
◊ A more massive tennis racquet or baseball bat makes the ball fly off faster. Explain why this is true, using the center of mass frame. For simplicity, assume that the racquet or bat is simply sitting still before the collision, and that the hitter's hands do not make any force large enough to have a significant effect over the short duration of the impact.
Contributors
Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CC-BY-SA license.