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Contents

Section 14.1 - Momentum

Section 14.2 - Collisions in one dimension

Section 14.3 - Relationship of momentum to the center of mass (optional)

Section 14.4 - Momentum transfer

Section 14.5 - Momentum in three dimensions

Section 14.6 - Applications of calculus (optional calculus-based section)

Section 14.7 - Summary

Section 14.1 - Momentum

Section 14.2 - Collisions in one dimension

Section 14.3 - Relationship of momentum to the center of mass (optional)

Section 14.4 - Momentum transfer

Section 14.5 - Momentum in three dimensions

Section 14.6 - Applications of calculus (optional calculus-based section)

Section 14.7 - Summary

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. In this chapter, we develop a new conserved quantity, called momentum, which is a vector.

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.

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

\[\begin{equation*}
F_{gun\ on\ bullet} = - F_{bullet\ on\ gun} ,
\end{equation*}\]

and assuming a constant force for simplicity, Newton's second law allows us to change this to

\[\begin{equation*}
m_{bullet}\frac{\Delta v_{bullet}}{\Delta t}
= -m_{gun}\frac{\Delta v_{gun}}{\Delta t} .
\end{equation*}\]

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, \(p\). (As far as I know, 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.

The momentum of a material object, i.e., a piece of matter, is defined as

\[\begin{equation*}
\mathbf{p} = m\mathbf{v} ,
\end{equation*}\]

the product of the object's mass and its velocity vector.

The units of momentum are \(\text{kg}\!\cdot\!\text{m}/\text{s}\), 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:

In any closed system, the vector sum of all the momenta remains constant,

\[\begin{equation*}
\mathbf{p}_{1i}+ \mathbf{p}_{2i}+ ... = \mathbf{p}_{1f} + \mathbf{p}_{2f} + ... ,
\end{equation*}\]

where \(i\) labels the initial and \(f\) the final momenta. (A closed system is one on which no external forces act.)

This chapter first addresses the one-dimensional case, in which the direction of the momentum can be taken into account by using plus and minus signs. We then pass to three dimensions, necessitating the use of vector addition.

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.

\(\triangleright\) 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

\[\begin{align*}
p_{cannon,i} &= 0 \\
p_{shell,i} &= 0 \\
p_{shell,f} &= 2000\ \text{kg}\!\cdot\!\text{m}/\text{s} .
\end{align*}\]

We must have \(p_{cannon,f}=-2000\ \text{kg}\!\cdot\!\text{m}/\text{s}\), so the recoil velocity of the cannon is \(-2\) m/s.

\(\triangleright\) 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.)

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 289 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 \(10^{-5}\ \text{kg}\!\cdot\!\text{m}/\text{s}\) of momentum as it recoiled.

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 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. |

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!

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.

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.

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.

◊

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?

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.

\(\triangleright\) 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 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?

\(\triangleright\) 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

\[\begin{equation*}
v = \left(1+\frac{m_C}{m_N}\right)v' .
\end{equation*}\]

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 \(KE_f\ltKE_i\) for a case like a tennis ball bouncing on grass.

\(\triangleright\) 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

\[\begin{align*}
mv_{1i}+mv_{2i} &= mv_{1f}+mv_{2f} \\
\frac{1}{2}mv_{1i}^2+\frac{1}{2}mv_{2i}^2 &= \frac{1}{2}mv_{1f}^2+\frac{1}{2}mv_{2f}^2
\end{align*}\]

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.

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.

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.

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.

◊

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 above?

We have already discussed the idea of the center of mass on p. 67, but 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.

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

\[\begin{align*}
mv_1+mv_2 &= m(v_1+v_2) \\
&= m \left(\frac{\Delta x_1}{\Delta t}+\frac{\Delta x_2}{\Delta t}\right)\\
&= \frac{m}{\Delta t}\Delta\left(x_1+x_2\right)\\
&= m\frac{2\Delta x_{cm}}{\Delta t}\\
&= m_{total}v_{cm}
\end{align*}\]

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.

The total momentum of any system is related to its total mass and the velocity of its center of mass by the equation

\[\begin{equation*}
\mathbf{p}_{total} = m_{total}\mathbf{v}_{cm} .
\end{equation*}\]

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

\[\begin{equation*}
x_{cm} = \frac{m_1x_1+m_2x_2+...}{m_1+m_2+...} ,
\end{equation*}\]

with similar equations for the \(y\) and \(z\) coordinates.

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 \(\mathbf{p}_{total}=\)

[4] \(m_{total}\mathbf{v}_{cm}\).
If the velocity of frame B relative to frame A is
\(\mathbf{v}_{BA}\), then the only effect of changing frames of
reference is to change \(\mathbf{v}_{cm}\) from its original value to
\(\mathbf{v}_{cm}+\mathbf{v}_{BA}\). This adds a constant onto the momentum
vector, which has no effect on conservation of momentum.

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.

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.

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 \(v_{1f}=-v_{1i}\). 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.

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.

◊

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.

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

\[\begin{equation*}
\Delta\mathbf{p} = m\Delta\mathbf{v} ,
\end{equation*}\]

which with the help of \(\mathbf{a}=\mathbf{F}/m\) and the constant-acceleration equation \(\mathbf{a}=\Delta\mathbf{v}/\Delta t\) becomes

\[\begin{align*}
\Delta\mathbf{p} &= m\mathbf{a}\Delta t \\
&= \mathbf{F}\Delta t .
\end{align*}\]

Thus the rate of transfer of momentum, i.e., the number of \(\text{kg}\!\cdot\!\text{m}/\text{s}\) absorbed per second, is simply the external force,

\[\begin{equation*}
\mathbf{F} = \frac{\Delta\mathbf{p}}{\Delta t} .
\end{equation*}\]

[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.

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

\[\begin{equation*}
\mathbf{F}_{total} = \frac{\Delta\mathbf{p}_{total}}{\Delta t} .
\end{equation*}\]

[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]

\(\triangleright\) Starting from rest, you begin walking, bringing your momentum up to \(100\ \text{kg}\!\cdot\!\text{m}/\text{s}\). You walk straight into a lamppost. Why is the momentum change of \(-100\ \text{kg}\!\cdot\!\text{m}/\text{s}\) caused by the lamppost so much more painful than the change of \(+100\ \text{kg}\!\cdot\!\text{m}/\text{s}\) when you started walking?

\(\triangleright\) 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 \(F=\Delta p/\Delta t\approx 100\ \text{N}\). 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 \(\Delta t\) 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.

\(\triangleright\) The ion drive of the Deep Space 1 spacecraft, pictured on page 351 and 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?

\(\triangleright\) Solving the equation \(F=\Delta p/\Delta t\) for the unknown \(\Delta t\), and treating force and momentum as scalars since the problem is one-dimensional, we find

\[\begin{align*}
\Delta t &= \frac{\Delta p}{F} \\
&= \frac{m_{exhaust}\Delta v_{exhaust}}{F} \\
&= \frac{(80\ \text{kg})(30,000\ \text{m}/\text{s})}{0.090\ \text{N}}\\
&= 2.7\times10^7\ \text{s} \\
&= 300\ \text{days}
\end{align*}\]

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 \(F=\Delta p/\Delta t\), 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 \(v=\Delta p/\Delta t\), the equation \(F=\Delta p/\Delta t\) is correctly generalized by saying that the force is the slope of the \(p-t\) graph.

Conversely, if we wish to find \(\Delta p\) 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.

◊

Many collisions, like the collision of a bat with a baseball, appear to be instantaneous. Most people also would not imagine the bat and ball as bending or being compressed during the collision. Consider the following possibilities:

- The collision is instantaneous.
- The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact.
- The collision takes a finite amount of time, during which the ball and bat are bending or being compressed.

How can two of these be ruled out based on energy or momentum considerations?

In this section 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:

\(\triangleright\) 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

\[\begin{align*}
v_{1x}&=0 \\
v_{1y}&=1.0\times10^4\ \text{km}/\text{hr} ,
\text{and the second with}
v_{2x}&=1.0\times10^4\ \text{km}/\text{hr}\\
v_{2y}&=0 ,
\end{align*}\]

(all in the center of mass frame) what is the magnitude of the third one's velocity?

\(\triangleright\) 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

\[\begin{align*}
mv_{1x}+ mv_{2x}+ mv_{3x} &= 0 , \text{and} \\
mv_{1y}+ mv_{2y}+ mv_{3y} &= 0 ,
\end{align*}\]

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

\[\begin{align*}
v_{3x} &= -1.0\times10^4\ \text{km}/\text{hr} \\
v_{3y} &= -1.0\times10^4\ \text{km}/\text{hr}
\end{align*}\]

which gives a magnitude of

\[\begin{align*}
|\mathbf{v}_3| &= \sqrt{v_{3x}^2+v_{3y}^2}\\
&= 1.4\times10^4\ \text{km}/\text{hr}
\end{align*}\]

In three dimensions, we have the vector equations

\[\begin{align*}
\mathbf{F}_{total} &= \frac{\Delta\mathbf{p}_{total}}{\Delta t} \\
\text{and}
\mathbf{p}_{total} &= m_{total}\mathbf{v}_{cm} .
\end{align*}\]

The following is an example of their use.

\[\begin{equation*}
\mathbf{F}_{total} = m_{total}\mathbf{g} .
\end{equation*}\]

Using the equation \(\mathbf{F}_{total} = \Delta\mathbf{p}_{total}/\Delta t\), we find
\[\begin{equation*}
\Delta \mathbf{p}_{total}/\Delta t = m_{total}\mathbf{g} ,
\end{equation*}\]

and since the mass is constant, the equation \(\mathbf{p}_{total}=m_{total}\mathbf{v}_{cm}\)
allows us to change this to
\[\begin{equation*}
m_{total}\Delta \mathbf{v}_{cm}/\Delta t = m_{total}\mathbf{g} .
\end{equation*}\]

The mass cancels, and \(\Delta \mathbf{v}_{cm}/\Delta t\) is simply
the acceleration of the center of mass, so
\[\begin{equation*}
\mathbf{a}_{cm} = \mathbf{g} .
\end{equation*}\]

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 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.

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.

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.

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.

\(\triangleright\) In a time interval \(\Delta \)t, the mass of water that strikes the blade is \(R\Delta \)t, and the magnitude of its initial momentum is \(mv=vR\Delta t\). 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

\[\begin{equation*}
F_{\text{blade on water}} = \frac{\Delta \mathbf{p}_{water}}{\Delta t} .
\end{equation*}\]

Choosing the \(x\) axis to be to the right and the \(y\) axis to be up, this can be broken down into components as

\[\begin{align*}
F_{\text{blade on water},x} &= \frac{\Delta p_{\text{water},x}}{\Delta t} \\
&= \frac{-vR\Delta t-0}{\Delta t} \\
&= -vR
\end{align*}\]

and

\[\begin{align*}
F_{\text{blade on water},y} &= \frac{\Delta p_{\text{water},y}}{\Delta t} \\
&= \frac{0-(-vR\Delta t)}{\Delta t} \\
&= vR .
\end{align*}\]

The water's force on the blade thus has components

\[\begin{align*}
F_{\text{water on blade},x} &= vR \\
F_{\text{water on blade},y} &= -vR .
\end{align*}\]

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

\[\begin{equation*}
|F_{\text{water on blade}}| = \sqrt{2}vR
\end{equation*}\]

and its direction is at a 45-degree angle down and to the right.

◊

The figures show a jet of water striking two different objects. How does the total downward force compare in the two cases? How could this fact be used to create a better waterwheel? (Such a waterwheel is known as a Pelton wheel.)

By now you will have learned to recognize the circumlocutions I use in the sections without calculus in order to introduce calculus-like concepts without using the notation, terminology, or techniques of calculus. It will therefore come as no surprise to you that the rate of change of momentum can be represented with a derivative,

\[\begin{equation*}
\mathbf{F}_{total} = \frac{d \mathbf{p}_{total}}{dt} .
\end{equation*}\]

And of course the business about the area under the \(F-t\) curve is really an integral, \(\Delta p_{total}=\int F_{total}dt\), which can be made into an integral of a vector in the more general three-dimensional case:

\[\begin{equation*}
\Delta \mathbf{p}_{total} = \int \mathbf{F}_{total}dt .
\end{equation*}\]

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.

\(\triangleright\) 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?

\(\triangleright\) 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

\[\begin{equation*}
F = \frac{d(mv)}{dt} = 0
\end{equation*}\]

We use the product rule to find

\[\begin{equation*}
0 = \frac{dm}{dt}v + m\frac{dv}{dt} .
\end{equation*}\]

We are trying to find how \(v\) changes, so we solve for \(dv/dt\),

\[\begin{align*}
\frac{dv}{dt}&= -\frac{v}{m}\frac{dm}{dt} \\
&= -\left(\frac{1\ \text{m}/\text{s}}{10\ \text{kg}}\right)\left(1\ \text{kg}/\text{s}\right)\\
&= -0.1\ \text{m}/\text{s}^2 .
\end{align*}\]

(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.

*momentum* — a measure of motion, equal to \(mv\) for material objects

*collision* — an interaction between moving objects that
lasts for a certain time

*center of mass* — the balance point or average position of
the mass in a system

\notationitem{\(\mathbf{p}\)}{the momentum vector}

\notationitem{cm}{center of mass, as in \(x_{cm}\), \(a_{cm}\), etc.}

impulse, \(I\), \(J\) — the amount of momentum transferred, \(\Delta p\)

elastic collision — one in which no KE is converted into other forms of energy

inelastic collision — one in which some KE is converted to other forms of energy

{}

If two objects interact via a force, Newton's third law guarantees that any change in one's velocity vector will be accompanied by a change in the other's which is in the opposite direction. Intuitively, this means that if the two objects are not acted on by any external force, they cannot cooperate to change their overall state of motion. This can be made quantitative by saying that the quantity \(m_1 \mathbf{v}_1+m_2 \mathbf{v}_2\) must remain constant as long as the only forces are the internal ones between the two objects. This is a conservation law, called the conservation of momentum, and like the conservation of energy, it has evolved over time to include more and more phenomena unknown at the time the concept was invented. The momentum of a material object is

\[\begin{equation*}
\mathbf{p} = m\mathbf{v} ,
\end{equation*}\]

but this is more like a standard for comparison of momenta rather than a definition. For instance, light has momentum, but has no mass, and the above equation is not the right equation for light. The law of conservation of momentum says that the total momentum of any closed system, i.e., the vector sum of the momentum vectors of all the things in the system, is a constant.

An important application of the momentum concept is to collisions, i.e., interactions between moving objects that last for a certain amount of time while the objects are in contact or near each other. Conservation of momentum tells us that certain outcomes of a collision are impossible, and in some cases may even be sufficient to predict the motion after the collision. In other cases, conservation of momentum does not provide enough equations to find all the unknowns. In some collisions, such as the collision of a superball with the floor, very little kinetic energy is converted into other forms of energy, and this provides one more equation, which may suffice to predict the outcome.

The total momentum of a system can be related to its total mass and the velocity of its center of mass by the equation

\[\begin{equation*}
\mathbf{p}_{total} = m_{total} \mathbf{v}_{cm} .
\end{equation*}\]

The center of mass, introduced on an intuitive basis in book 1 as the “balance point” of an object, can be generalized to any system containing any number of objects, and is defined mathematically as the weighted average of the positions of all the parts of all the objects,

\[\begin{equation*}
x_{cm} = \frac{m_1x_1+m_2x_2+...}{m_1+m_2+...} ,
\end{equation*}\]

with similar equations for the \(y\) and \(z\) coordinates.

The frame of reference moving with the center of mass of a closed system is always a valid inertial frame, and many problems can be greatly simplified by working them in the inertial frame. For example, any collision between two objects appears in the c.m. frame as a head-on one-dimensional collision.

When a system is not closed, the rate at which momentum is transferred in or out is simply the total force being exerted externally on the system. If the force is constant,

\[\begin{equation*}
\mathbf{F}_{total} = \frac{\Delta \mathbf{p}_{total}}{\Delta t} .
\end{equation*}\]

When the force is not constant, the force equals the slope of the tangent line on a graph of \(p\) versus \(t\), and the change in momentum equals the area under the \(F-t\) graph.

\begin{homeworkforcelabel}{keintermsofp}{1}{}{1}Derive a formula expressing the kinetic energy of an object in terms of its momentum and mass.(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{rowboat}{1}{}{2}Two people in a rowboat wish to move around without causing the boat to move. What should be true about their total momentum? Explain. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{planes}{1}{}{3} A learjet traveling due east at 300 mi/hr collides with a jumbo jet which was heading southwest at 150 mi/hr. The jumbo jet's mass is five times greater than that of the learjet. When they collide, the learjet sticks into the fuselage of the jumbo jet, and they fall to earth together. Their engines stop functioning immediately after the collision. On a map, what will be the direction from the location of the collision to the place where the wreckage hits the ground? (Give an angle.)(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{gun}{1}{}{4}A bullet leaves the barrel of a gun with a kinetic energy
of 90 J. The gun barrel is 50 cm long. The gun has a mass of
4 kg, the bullet 10 g.

(a) Find the bullet's final velocity. (answer check available at lightandmatter.com)

(b) Find the bullet's final momentum. (answer check available at lightandmatter.com)

(c) Find the momentum of the recoiling gun.

(d) Find the kinetic energy of the recoiling gun, and
explain why the recoiling gun does not kill the shooter. (answer check available at lightandmatter.com)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{rocket}{1}{}{5}(answer check available at lightandmatter.com) The graph shows the force, in meganewtons, exerted by a rocket engine on the rocket as a function of time. If the rocket's mass is 4000 kg, at what speed is the rocket moving when the engine stops firing? Assume it goes straight up, and neglect the force of gravity, which is much less than a meganewton. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{cosmicray}{1}{}{6}Cosmic rays are particles from outer space, mostly protons and atomic nuclei, that are continually bombarding the earth. Most of them, although they are moving extremely fast, have no discernible effect even if they hit your body, because their masses are so small. Their energies vary, however, and a very small minority of them have extremely large energies. In some cases the energy is as much as several Joules, which is comparable to the KE of a well thrown rock! If you are in a plane at a high altitude and are so incredibly unlucky as to be hit by one of these rare ultra-high-energy cosmic rays, what would you notice, the momentum imparted to your body, the energy dissipated in your body as heat, or both? Base your conclusions on numerical estimates, not just random speculation. (At these high speeds, one should really take into account the deviations from Newtonian physics described by Einstein's special theory of relativity. Don't worry about that, though.) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{cmaverage}{1}{}{7}Show that for a body made up of many *equal* masses,
the equation for the center of mass becomes a simple average
of all the positions of the masses.
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{hockey-pucks}{1}{}{8}(solution in the pdf version of the book) The figure shows a view from above of a collision about to happen between two air hockey pucks sliding without friction. They have the same speed, \(v_i\), before the collision, but the big puck is 2.3 times more massive than the small one. Their sides have sticky stuff on them, so when they collide, they will stick together. At what angle will they emerge from the collision? In addition to giving a numerical answer, please indicate by drawing on the figure how your angle is defined. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{rope-over-edge}{2}{1}{9}A flexible rope of mass \(m\) and length \(L\) slides
without friction over the edge of a table. Let \(x\) be the
length of the rope that is hanging over the edge at a
given moment in time.

(a) Show that \(x\) satisfies the equation of motion \(d^2x/dt^2=gx/L\).
[Hint: Use \(F=dp/dt\), which allows you to handle the two parts of the rope
separately even though mass is moving out of one part and into the other.]

(b) Give a physical explanation for the fact that a larger
value of \(x\) on the right-hand side of the equation leads to
a greater value of the acceleration on the left side.

(c) When we take the second derivative of the function
\(x(t)\) we are supposed to get essentially the same function
back again, except for a constant out in front. The function
\(e^x\) has the property that it is unchanged by differentiation,
so it is reasonable to look for solutions to this problem
that are of the form \(x=be^{ct}\), where \(b\) and \(c\) are
constants. Show that this does indeed provide a solution for
two specific values of \(c\) (and for any value of \(b)\).

(d) Show that the sum of any two solutions to the equation
of motion is also a solution.

(e) Find the solution for the case where the rope starts at
rest at \(t=0\) with some nonzero value of \(x\).
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{recoil-double-speed}{1}{}{10}A very massive object with velocity \(v\) collides head-on with an object at rest whose mass is very small. No kinetic energy is converted into other forms. Prove that the low-mass object recoils with velocity \(2v\). [Hint: Use the center-of-mass frame of reference.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{fridge-recoil}{1}{}{11}(solution in the pdf version of the book) When the contents of a refrigerator cool down, the
changed molecular speeds imply changes in both momentum and
energy. Why, then, does a fridge transfer *power*
through its radiator coils, but not *force*?
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{bowling-collision}{1}{}{12} A 10-kg bowling ball moving at 2.0 m/s hits a 1.0-kg bowling pin, which is initially at rest. The other pins are all gone already, and the collision is head-on, so that the motion is one-dimensional. Assume that negligible amounts of heat and sound are produced. Find the velocity of the pin immediately after the collision. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{exhaust}{2}{1}{13}A rocket ejects exhaust with an exhaust velocity \(u\). The rate at which the exhaust mass is used (mass per unit time) is \(b\). We assume that the rocket accelerates in a straight line starting from rest, and that no external forces act on it. Let the rocket's initial mass (fuel plus the body and payload) be \(m_i\), and \(m_f\) be its final mass, after all the fuel is used up. (a) Find the rocket's final velocity, \(v\), in terms of \(u\), \(m_i\), and \(m_f\). Neglect the effects of special relativity. (b) A typical exhaust velocity for chemical rocket engines is 4000 m/s. Estimate the initial mass of a rocket that could accelerate a one-ton payload to 10% of the speed of light, and show that this design won't work. (For the sake of the estimate, ignore the mass of the fuel tanks. The speed is fairly small compared to \(c\), so it's not an unreasonable approximation to ignore relativity.) (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{firework}{1}{}{14}(solution in the pdf version of the book) A firework shoots up into the air, and just before it explodes it has a certain momentum and kinetic energy. What can you say about the momenta and kinetic energies of the pieces immediately after the explosion? [Based on a problem from PSSC Physics.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{energy-frames}{2}{}{15}(solution in the pdf version of the book) Suppose a system consisting of pointlike particles
has a total kinetic energy \(K_{cm}\) measured in the
center-of-mass frame of reference. Since they are pointlike,
they cannot have any energy due to internal motion.

(a) Prove that in a different frame of reference, moving with
velocity \(\mathbf{u}\) relative to the center-of-mass frame, the total
kinetic energy equals \(K_{cm}+M|\mathbf{u}|^2/2\), where \(M\) is the
total mass. [Hint: You can save yourself a lot of writing if
you express the total kinetic energy using the dot product.]

(b) Use this to prove that if energy is conserved in one
frame of reference, then it is conserved in every frame of
reference. The total energy equals the total kinetic energy
plus the sum of the potential energies due to the particles'
interactions with each other, which we assume depends only
on the distance between particles. [For a simpler numerical
example, see problem 13 on p. 300.]
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{funkosity}{1}{}{16}(solution in the pdf version of the book) The big difference between the equations for momentum and kinetic energy is that one is proportional to \(v\) and one to \(v^2\). Both, however, are proportional to \(m\). Suppose someone tells you that there's a third quantity, funkosity, defined as \(f=m^2v\), and that funkosity is conserved. How do you know your leg is being pulled? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{max-heat-released-in-collision}{1}{}{17}A mass \(m\) moving at velocity \(v\) collides with a stationary target having the same mass \(m\). Find the maximum amount of energy that can be released as heat and sound. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{colliding-putty}{1}{}{18}Two blobs of putty collide head-on. The collision is completely symmetric: the blobs are of equal mass, and they collide at equal speeds. What becomes of the energy the blobs had before the collision? The momentum? \end{homeworkforcelabel}

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.