7.2

The Principle of Conservation of Linear Momentum

It is worthwhile to compare the impulse–momentum theorem to the work–energy theorem discussed in Chapter 6. The impulse–momentum theorem states that the impulse produced by a net force is equal to the change in the object’s momentum, while the work–energy theorem states that the work done by a net force is equal to the change in the object’s kinetic energy. The work–energy theorem leads directly to the principle of conservation of mechanical energy (see Figure 6-15), and, as we will shortly see, the impulse–momentum theorem also leads to a conservation principle, known as the conservation of linear momentum.
We begin by applying the impulse–momentum theorem to a midair collision between two objects. The two objects (masses m1 and m2) are approaching each other with initial velocities and , as Figure 7-7a shows. The collection of objects being studied is referred to as the “system.” In this case, the system contains only the two objects. They interact during the collision in part b and then depart with the final velocities and shown in part c. Because of the collision, the initial and final velocities are not the same.

Figure 7-7 (a) The velocities of the two objects before the collision are and . (b) During the collision, each object exerts a force on the other. These forces are and . (c) The velocities after the collision are and .

Two types of forces act on the system:

1. | Internal forces—Forces that the objects within the system exert on each other.
2. | External forces—Forces exerted on the objects by agents external to the system.

During the collision in Figure 7-7b, is the force exerted on object 1 by object 2, while is the force exerted on object 2 by object 1. These forces are action–reaction forces that are equal in magnitude but opposite in direction, so . They are internal forces, since they are forces that the two objects within the system exert on each other. The force of gravity also acts on the objects, their weights being and . These weights, however, are external forces, because they are applied by the earth, which is outside the system. Friction and air resistance would also be considered external forces, although these forces are ignored here for the sake of simplicity. The impulse–momentum theorem, as applied to each object, gives the following results:

Adding these equations produces a single result for the system as a whole:

On the right side of this equation, the quantity is the vector sum of the final momenta for each object, or the total final momentum of the system. Likewise, the quantity is the total initial momentum . Therefore, the result above becomes

| (7.5)

The advantage of the internal/external force classification is that the internal forces always add together to give zero, as a consequence of Newton’s law of action–reaction; so that . Cancellation of the internal forces occurs no matter how many parts there are to the system and allows us to ignore the internal forces, as Equation 7.6 indicates:

| (7.6)

We developed this result with gravity as the only external force. But, in general, the sum of the external forces on the left includes all external forces.
With the aid of Equation 7.6, it is possible to see how the conservation of linear momentum arises. Suppose that the sum of the external forces is zero. A system for which this is true is called an isolated system. Then Equation 7.6 indicates that

| (7.7a)

In other words, the final total momentum of the isolated system after the objects in Figure 7-7a collide is the same as the initial total momentum.* Explicitly writing out the final and initial momenta for the two-body collision, we obtain for Equation 7.7a that

| (7.7b)

This result is an example of a general principle known as the principle of conservation of linear momentum.

| Principle of Conservation of Linear Momentum
| The total linear momentum of an isolated system remains constant (is conserved). An isolated system is one for which the vector sum of the average external forces acting on the system is zero.

CONCEPTS AT A GLANCE The Concepts-at-a-Glance chart in Figure 7-8 presents an overview of how concepts that we have studied lead to the conservation of linear momentum. The chart is an extension of that shown earlier in Figure 7-3. In the middle, it shows that the origin of the impulse–momentum theorem lies in Newton’s second law and the ideas of the impulse of a force and the linear momentum of an object. The theorem is stated at the right of the chart in the form (Sum of average external forces). If the sum of the average external forces is zero, then the final and initial total momenta are equal, or , which is the mathematical statement of the conservation of linear momentum.

Figure 7-8 CONCEPTS AT A GLANCE The impulse–momentum theorem leads to the principle of conservation of linear momentum when the sum of the average external forces acting on a system is zero. To the extent that this principle applies to the fireworks explosions in the photograph, it follows that the total linear momentum of a rocket before it explodes is equal to the total linear momentum of the fragments that exist after it explodes. (Norbert Schwerin/The Image Works)

The conservation-of-momentum principle applies to a system containing any number of objects, regardless of the internal forces, provided the system is isolated. Whether the system is isolated depends on whether the vector sum of the external forces is zero. Judging whether a force is internal or external depends on which objects are included in the system, as Conceptual Conceptual Example 4 illustrates.

| Conceptual Example 4 | | | Is the Total Momentum Conserved?

Imagine two balls colliding on a billiard table that is friction-free. Use the momentum-conservation principle in answering the following questions. (a) Is the total momentum of the two-ball system the same before and after the collision? (b) Answer part (a) for a system that contains only one of the two colliding balls.
Reasoning and Solution (a) Figure 7-9a shows the balls colliding, and the dashed outline emphasizes that both balls are part of the system. The collision forces are not shown, because they are internal forces and cannot cause the total momentum of the system to change. The external forces include the weights and of the balls. There are two additional external forces, however: the upward-pointing normal force that the table exerts on each ball. These normal forces are and . Since the balls do not accelerate in the vertical direction, the normal forces must balance the weights, so that the vector sum of the four external forces in Figure 7-9a is zero. Furthermore, the table is friction-free. Thus, there is no net external force to change the total momentum of the two-ball system, and it is the same before and after the collision. The total momentum of this two-ball system is conserved.

Figure 7-9 (a) Two billiard balls collide. They have weights and . The dashed box emphasizes that both balls are included in the system. and are the normal forces that the table applies to the balls. (b) Now only ball 1 is included in the system.

(b) In Figure 7-9b only one ball is included in the system, as the dashed outline indicates. The forces acting on this system are all external and include the weight of the ball and the normal force . As in part a of the drawing, these two forces balance. However, there is a third-external force to consider. Ball 2 is outside the system, so the force that it applies to the system during the collision is now an external force. As a result, the vector sum of the three external forces is not zero, and the net external force causes the total momentum of the one-ball system to be different after the collision than it was before the collision. The total momentum of this one-ball system is not conserved.

Next, we apply the principle of conservation of linear momentum to the problem of assembling a freight train.

| Example 5 | | | Assembling a Freight Train

A freight train is being assembled in a switching yard, and Figure 7-10 shows two boxcars. Car 1 has a mass of and moves at a velocity of . Car 2, with a mass of and a velocity of , overtakes car 1 and couples to it. Neglecting friction, find the common velocity vf of the cars after they become coupled.

Figure 7-10 (a) The boxcar on the left eventually catches up with the other boxcar and (b) couples to it. The coupled cars move together with a common velocity after the collision.

Reasoning The two boxcars constitute the system. The sum of the external forces acting on the system is zero, because the weight of each car is balanced by a corresponding normal force, and friction is being neglected. Thus, the system is isolated, and the principle of conservation of linear momentum applies. The coupling forces that each car exerts on the other are internal forces and do not affect the applicability of this principle.
Solution Momentum conservation indicates that

Problem solving insight The conservation of linear momentum is applicable only when the net external force acting on the system is zero. Therefore, the first step in applying momentum conservation is to be sure that the net external force is zero.

This equation can be solved for vf, the common velocity of the two cars after the collision:

In the previous example it can be seen that the velocity of car 1 increases, while the velocity of car 2 decreases as a result of the collision. The acceleration and deceleration arise at the moment the cars become coupled, because the cars exert internal forces on each other. The powerful feature of the momentum-conservation principle is that it allows us to determine the changes in velocity without knowing what the internal forces are. Example 6 further illustrates this feature.

| Example 6 | | | Ice Skaters

Starting from rest, two skaters “push off” against each other on smooth level ice, where friction is negligible. As Figure 7-11a shows, one is a woman (), and one is a man (). Part b of the drawing shows that the woman moves away with a velocity of . Find the “recoil ” velocity vf2 of the man.

Figure 7-11 (a) In the absence of friction, two skaters pushing on each other constitute an isolated system. (b) As the skaters move away, the total linear momentum of the system remains zero, which is what it was initially.

Reasoning For a system consisting of the two skaters on level ice, the sum of the external forces is zero. This is because the weight of each skater is balanced by a corresponding normal force and friction is negligible. The skaters, then, constitute an isolated system, and the principle of conservation of linear momentum applies. We expect the man to have a smaller recoil speed for the following reason: The internal forces that the man and woman exert on each other during pushoff have equal magnitudes but opposite directions, according to Newton’s action – reaction law. The man, having the larger mass, experiences a smaller acceleration according to Newton’s second law. Hence, he acquires a smaller recoil speed.
Solution The total momentum of the skaters before they push on each other is zero, since they are at rest. Momentum conservation requires that the total momentum remains zero after the skaters have separated, as in part b of the drawing:

Solving for the recoil velocity of the man gives

The minus sign indicates that the man moves to the left in the drawing. After the skaters separate, the total momentum of the system remains zero, because momentum is a vector quantity, and the momenta of the man and the woman have equal magnitudes but opposite directions.

It is important to realize that the total linear momentum may be conserved even when the kinetic energies of the individual parts of a system change. In Example 6, for instance, the initial kinetic energy is zero, since the skaters are stationary. But after they push off, the skaters are moving, so each has kinetic energy. The kinetic energy changes because work is done by the internal force that each skater exerts on the other. However, internal forces cannot change the total linear momentum of a system, since the total linear momentum of an isolated system is conserved in the presence of such forces.
When applying the principle of conservation of linear momentum, we have been following a definite reasoning strategy that is summarized as follows:

Problem solving insight

| Reasoning Strategy

Applying the Principle of Conservation of Linear Momentum
1. | Decide which objects are included in the system.
2. | Relative to the system that you have chosen, identify the internal forces and the external forces.
3. | Verify that the system is isolated. In other words, verify that the sum of the external forces applied to the system is zero. Only if this sum is zero can the conservation principle be applied. If the sum of the average external forces is not zero, consider a different system for analysis.
4. | Set the total final momentum of the isolated system equal to the total initial momentum. Remember that linear momentum is a vector. If necessary, apply the conservation principle separately to the various vector components.

Need more practice?

Interactive LearningWare7.2
An astronaut is motionless in outer space. On command, the propulsion unit strapped to his back ejects some gas with a velocity of +32 m/s, and the astronaut recoils with a velocity of −0.30 m/s. After the gas is ejected, the mass of the astronaut is 160 kg. What is the mass of the ejected gas?
Related Homework: Problem 15

Check Your Understanding 2

A canoe with two people aboard is coasting with an initial momentum of +110 kg·m/s. Then, one of the people (person 1) dives off the back of the canoe. During this time, the net average external force acting on the system (canoe and the two people) is zero. The table lists four possibilities for the final momentum of person 1 and the final momentum of person 2 plus the canoe, immediately after person 1 leaves the canoe. Only one possibility could be correct. Which is it? (The answer is given at the end of the book.)

Final Momenta
| Person 1 | Person 2 and Canoe
a. | −60 kg·m/s | +170 kg·m/s
b. | −30 kg·m/s | +110 kg·m/s
c. | −40 kg·m/s | −70 kg·m/s
d. | +80 kg·m/s | −30 kg·m/s

Background: Momentum and the principle of conservation of linear momentum play a role in this question. Remember that momentum is a vector quantity.
For similar questions (including conceptual counterparts), consult Self-Assessment Test 7.1. The test is described next.

Four important ideas are discussed in Sections 7.1 and 7.2:

· | The Impulse of a Force
· | The Linear Momentum of an Object
· | The Impulse—Momentum Theorem
· | The Conservation of Linear Momentum

Use Self-Assessment Test 7.1 to check your understanding of these concepts.

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.