7.1

The Impulse–Momentum Theorem

There are many situations in which the force acting on an object is not constant, but varies with time. For instance, Figure 7-1a shows a baseball being hit, and part b of the figure illustrates approximately how the force applied to the ball by the bat changes during the time of contact. The magnitude of the force is zero at the instant t0 just before the bat touches the ball. During contact, the force rises to a maximum and then returns to zero at the time tf when the ball leaves the bat. The time interval during which the bat and ball are in contact is quite short, being only a few-thousandths of a second, although the maximum force can be very large, often exceeding thousands of newtons. For comparison, the graph also shows the magnitude of the average force exerted on the ball during the time of contact. Figure 7-2 depicts other situations in which a time-varying force is applied to a ball.

Figure 7-1 (a) The collision time between a bat and a ball is very short, often less than a millisecond, but the force can be quite large. (b) When the bat strikes the ball, the magnitude of the force exerted on the ball rises to a maximum value and then returns to zero when the ball leaves the bat. The time interval during which the force acts is , and the magnitude of the average force is ·(a, Chuck Savage/Corbis Images)

Figure 7-2 In each of these situations, the force applied to the ball varies with time. The time of contact is small, but the maximum force can be large. (left, Jonathan Daniel/MLS/Getty Images; right, Robert Laberge/Getty Images Sport Services)

CONCEPTS AT A GLANCE To describe how a time-varying force affects the motion of an object, we will introduce two new ideas: the impulse of a force and the linear momentum of an object. As the Concepts-at-a-Glance chart in Figure 7-3 illustrates, these ideas will be used with Newton’s second law of motion to produce an important result known as the impulse–momentum theorem. This theorem plays a central role in describing collisions, such as that between a ball and a bat. Later on, we will see also that the theorem leads in a natural way to one of the most fundamental laws in physics, the conservation of momentum.

Figure 7-3 CONCEPTS AT A GLANCE Two new concepts, impulse and momentum, will be combined with Newton’s second law of motion to produce the impulse–momentum theorem, a theorem useful for describing collisions. In the photograph, when the fighting red kangaroos collide, each imparts an impulse to the other. In response to the impulse, the momentum of each kangaroo changes, in accord with the impulse–momentum theorem. (Martin Harvey/Gallo Images/Corbis Image)

If a baseball is to be hit well, both the magnitude of the force and the time of contact are important. When a large average force acts on the ball for a long enough time, the ball is hit solidly. To describe such situations, we bring together the average force and the time of contact, calling the product of the two the impulse of the force.

| Definition of Impulse
| The impulse of a force is the product of the average force and the time interval during which the force acts:

| (7.1)

Impulse is a vector quantity and has the same direction as the average force.
SI Unit of Impulse: newton · second (N·s)

When a ball is hit, it responds to the value of the impulse. A large impulse produces a large response; that is, the ball departs from the bat with a large velocity. However, we know from experience that the more massive the ball, the less velocity it has after leaving the bat. Both mass and velocity play a role in how an object responds to a given impulse, and the effect of each of them is included in the concept of linear momentum, which is defined as follows:

| Definition of Linear Momentum
| The linear momentum of an object is the product of the object’s mass m and velocity :

| (7.2)

Linear momentum is a vector quantity that points in the same direction as the velocity.
SI Unit of Linear Momentum: kilogram · meter/second (kg·m/s)

Newton’s second law of motion can now be used to reveal a relationship between impulse and momentum. Figure 7-4 shows a ball with an initial velocity approaching a bat, being struck by the bat, and then departing with a final velocity . When the velocity of an object changes from to during a time interval , the average acceleration is given by Equation 2.4 as

According to Newton’s second law, , the average acceleration is produced by the net average force . Here represents the vector sum of all the average forces that act on the object. Thus,

| (7.3)

In this result, the numerator on the far right is the final momentum minus the initial momentum, which is the change in momentum. Thus, the net average force is given by the change in momentum per unit of time.* Multiplying both sides of Equation 7.3 by yields Equation 7.4, which is known as the impulse–momentum theorem.

Figure 7-4 When a bat hits a ball, an average force is applied to the ball by the bat. As a result, the ball’s velocity changes from an initial value of (top drawing) to a final value of (bottom drawing).

| Impulse–Momentum Theorem
| When a net force acts on an object, the impulse of this force is equal to the change in momentum of the object:

| (7.4)

During a collision, it is often dificult to measure the net average force , so it is not easy to determine the impulse directly. On the other hand, it is usually straightforward to measure the mass and velocity of an object, so that its momentum just after the collision and just before the collision can be found. Thus, the impulse–momentum theorem allows us to gain information about the impulse indirectly by measuring the change in momentum that the impulse causes. Then, armed with a knowledge of the contact time , we can evaluate the net average force. Examples 1 and 2 and Examples 1 and 2 illustrate how the theorem is used in this way.

| Example 1 | | | A Well-Hit Ball

A baseball () has an initial velocity of as it approaches a bat. We have chosen the direction of approach as the negative direction. The bat applies an average force hat is much larger than the weight of the ball, and the ball departs from the bat with a final velocity of = + 58 m/s. (a) Determine the impulse applied to the ball by the bat. (b) Assuming that the time of contact is , find the average force exerted on the ball by the bat.
Reasoning Two forces act on the ball during impact, and together they constitute the net average force: the average force exerted by the bat, and the weight of the ball. Since is much greater than the weight of the ball, we neglect the weight. Thus, the net average force is equal to , or . In hitting the ball, the bat imparts an impulse to it. We cannot use Equation 7.1 () to determine the impulse directly, since is not known. We can find the impulse indirectly, however, by turning to the impulse–momentum theorem, which states that the impulse is equal to the ball’s final momentum minus its initial momentum. With values for the impulse and the time of contact, Equation 7.1 can be used to determine the average force applied by the bat to the ball.
Solution

a. | According to the impulse–momentum theorem, the impulse applied to the ball is

b. | Now that the impulse is known, the contact time can be used in Equation 7.1 to find the average force exerted by the bat on the ball:

The force is positive, indicating that it points opposite to the velocity of the approaching ball. A force of 8400 N corresponds to 1900 lb, such a large value being necessary to change the ball’s momentum during the brief contact time.

This karate expert applies an impulse to the stack of concrete bricks. As a result, the momentum of the stack changes as the bricks break. (Photodisc Royalty Free Photograph/Fotosearch, LLC)

Problem solving insight Momentum is a vector and, as such, has a magnitude and a direction. For motion in one dimension, be sure to indicate the direction by assigning a plus or minus sign to it, as in this example.

| Example 2 | | | A Rainstorm

During a storm, rain comes straight down with a velocity of and hits the roof of a car perpendicularly (see Figure 7-5). The mass of rain per second that strikes the car roof is 0.060 kg/s. Assuming that the rain comes to rest upon striking the car (), find the average force exerted by the rain on the roof.

Figure 7-5 A raindrop falling on a car roof has an initial velocity of just before striking the roof. The final velocity of the raindrop is , because it comes to rest on the roof.

Reasoning This example differs from Example 1 in an important way. Example 1 gives information about the ball and asks for the force applied to the ball. In contrast, the present example gives information about the rain but asks for the force acting on the roof. However, the force exerted on the roof by the rain and the force exerted on the rain by the roof have equal magnitudes and opposite directions, according to Newton’s law of action and reaction (see Section 4.5). Thus, we will find the force exerted on the rain and then apply the law of action and reaction to obtain the force on the roof. Two forces act on the rain while it impacts with the roof: the average force exerted by the roof and the weight of the rain. These two forces constitute the net average force. By comparison, however, the force is much greater than the weight, so we may neglect the weight. Thus, the net average force becomes equal to , or . The value of can be obtained by applying the impulse–momentum theorem to the rain.
Solution The average force needed to reduce the rain’s velocity from to is given by Equation 7.4 as

The term is the mass of rain per second that strikes the roof, so that . Thus, the average force exerted on the rain by the roof is

This force is in the positive or upward direction, which is reasonable since the roof exerts an upward force on each falling drop in order to bring it to rest. According to the action–reaction law, the force exerted on the roof by the rain also has a magnitude of 0.90 N but points downward: Force on roof = .

As you reason through problems such as those in Examples 1 and 2 and Examples 1 and 2, take advantage of the impulse–momentum theorem. It is a powerful statement that can lead to significant insights. The following Conceptual Example further illustrates its use.

| Conceptual Example 3 | | | Hailstones Versus Raindrops

In Example 2 rain is falling on the roof of a car and exerts a force on it. Instead of rain, suppose hail is falling. The hail comes straight down at a mass rate of and an initial velocity of and strikes the roof perpendicularly, just as the rain does in Example 2. However, unlike rain, hail usually does not come to rest after striking a surface. Instead, the hailstones bounce off the roof of the car. If hail fell instead of rain, would the force on the roof be smaller than, equal to, or greater than that calculated in Example 2?
Reasoning and Solution The raindrops and the hailstones fall in exactly the same way. That is, they both fall with the same initial velocity and mass rate, and they both strike the roof perpendicularly. However, there is an important difference: the raindrops come to rest (see Figure 7-5), whereas the hailstones bounce upward after striking the roof (see Figure 7-6). According to the impulse–momentum theorem, the impulse that acts on an object is given by the change in the momentum of the object. This change is and is proportional to the change in velocity . For a hailstone, the change is from (downward) to (upward). This is a larger change than for the raindrop, for which the change is only from (downward) to zero. Therefore, the change in momentum is greater for the hailstone, and, correspondingly, a greater impulse must act on it. But an impulse is the product of the average force and the time interval . Since the same amount of mass falls in the same time interval in either case, is the same for hailstones as for raindrops. The greater impulse acting on the hailstones, then, means that the car roof must exert a greater force on the hailstones than on the raindrops. According to Newton’s action– reaction law then, the car roof experiences a greater force from the hailstones than from the raindrops.

Figure 7-6 Hailstones have a downward velocity of just before striking this car roof. They rebound with an upward velocity of .

Related Homework: Problems 8 and 48

Need more practice?

Interactive LearningWare7.1
A 1200-kg car has an initial velocity of 13 m/s, eastward. The driver applies the brakes and brings the car to rest in 6.0 s. Determine the net average force (magnitude and direction) exerted on the car by using (a) the impulse–momentum theorem and (b) kinematics and Newton’s second law.
Related Homework: Problem 2

Check Your Understanding 1

Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter than if you land on water. Using the impulse–momentum theorem as a guide, determine which one of the following statements is correct. (The answer is given at the end of the book.)

a. | In bringing you to a halt, the sand exerts a greater impulse on you than does the water.
b. | In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force.
c. | In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a smaller average force.

Background: The concept of impulse and its relation to the change in an object’s momentum are pertinent to this question. Think carefully about the definition of impulse.
For similar questions (including calculational counterparts), consult Self-Assessment Test 7.1. This test is described at the end of Section 7.2.

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