# Learning Objectives

### Learning Objectives

By the end of this section, you will be able to do the following:

• Describe momentum, what can change momentum, impulse, and the impulse-momentum theorem
• Describe Newton’s second law in terms of momentum
• Solve problems using the impulse-momentum theorem
 change in momentum impulse impulse–momentum theorem linear momentum

# Momentum, Impulse, and the Impulse-Momentum Theorem

### Momentum, Impulse, and the Impulse-Momentum Theorem

Linear momentum is the product of a system’s mass and its velocity. In equation form, linear momentum p is

$p=mv . p=mv .$

You can see from the equation that momentum is directly proportional to the object’s mass (m) and velocity (v). Therefore, the greater an object’s mass or the greater its velocity, the greater its momentum. A large, fast-moving object has greater momentum than a smaller, slower object.

Momentum is a vector and has the same direction as velocity v. Since mass is a scalar, when velocity is in a negative direction (i.e., opposite the direction of motion), the momentum will also be in a negative direction; and when velocity is in a positive direction, momentum will likewise be in a positive direction. The SI unit for momentum is kg m/s.

Momentum is so important for understanding motion that it was called the quantity of motion by physicists such as Newton. Force influences momentum, and we can rearrange Newton’s second law of motion to show the relationship between force and momentum.

Recall our study of Newton’s second law of motion (Fnet = ma). Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. The change in momentum is the difference between the final and initial values of momentum.

In equation form, this law is

$F net = Δp Δt , F net = Δp Δt ,$

where Fnet is the net external force, $Δp Δp$ is the change in momentum, and $Δt Δt$ is the change in time.

We can solve for $Δp Δp$ by rearranging the equation

$F net = Δp Δt F net = Δp Δt$

to be

$Δp= F net Δt . Δp= F net Δt .$

$F net Δt F net Δt$ is known as impulse and this equation is known as the impulse-momentum theorem. From the equation, we see that the impulse equals the average net external force multiplied by the time this force acts. It is equal to the change in momentum. The effect of a force on an object depends on how long it acts, as well as the strength of the force. Impulse is a useful concept because it quantifies the effect of a force. A very large force acting for a short time can have a great effect on the momentum of an object, such as the force of a racket hitting a tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time.

# Newton’s Second Law in Terms of Momentum

### Newton’s Second Law in Terms of Momentum

When Newton’s second law is expressed in terms of momentum, it can be used for solving problems where mass varies, since $Δp=Δ(mv) Δp=Δ(mv)$. In the more traditional form of the law that you are used to working with, mass is assumed to be constant. In fact, this traditional form is a special case of the law, where mass is constant.$F net =ma F net =ma$ is actually derived from the equation:

$F net = Δp Δt F net = Δp Δt$

For the sake of understanding the relationship between Newton’s second law in its two forms, let’s recreate the derivation of $F net =ma F net =ma$ from

$F net = Δp Δt F net = Δp Δt$

by substituting the definitions of acceleration and momentum.

The change in momentum $Δp Δp$ is given by

$Δp=Δ(mv) . Δp=Δ(mv) .$

If the mass of the system is constant, then

$Δ(mv)=mΔv . Δ(mv)=mΔv .$

By substituting $mΔv mΔv$ for $Δp Δp$, Newton’s second law of motion becomes

$F net = Δp Δt = mΔv Δt F net = Δp Δt = mΔv Δt$

for a constant mass.

Because

$Δv Δt =a , Δv Δt =a ,$

we can substitute to get the familiar equation

$F net =ma F net =ma$

when the mass of the system is constant.

### Tips For Success

We just showed how $F net =ma F net =ma$ applies only when the mass of the system is constant. An example of when this formula would not apply would be a moving rocket that burns enough fuel to significantly change the mass of the rocket. In this case, you would need to use Newton’s second law expressed in terms of momentum to account for the changing mass.

### Snap Lab

#### Hand Movement and Impulse

In this activity you will experiment with different types of hand motions to gain an intuitive understanding of the relationship between force, time, and impulse.

Materials:
• one ball
• one tub filled with water
Procedure:
1. Try catching a ball while giving with the ball, pulling your hands toward your body.
2. Next, try catching a ball while keeping your hands still.
3. Hit water in a tub with your full palm. Your full palm represents a swimmer doing a belly flop.
4. After the water has settled, hit the water again by diving your hand with your fingers first into the water. Your diving hand represents a swimmer doing a dive.
5. Explain what happens in each case and why.
Grasp Check
What are some other examples of motions that impulse affects?
1. a football player colliding with another, or a car moving at a constant velocity
2. a car moving at a constant velocity, or an object moving in the projectile motion
3. a car moving at a constant velocity, or a racket hitting a ball
4. a football player colliding with another, or a racket hitting a ball

# Solving Problems Using the Impulse-Momentum Theorem

### Worked Example

#### Calculating Momentum: A Football Player and a Football

(a) Calculate the momentum of a 110 kg football player running at 8 m/s. (b) Compare the player’s momentum with the momentum of a 0.410 kg football thrown hard at a speed of 25 m/s.

### Strategy

No information is given about the direction of the football player or the football, so we can calculate only the magnitude of the momentum, p. (A symbol in italics represents magnitude.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum:

$p=mv p=mv$
Solution for (a)

To find the player’s momentum, substitute the known values for the player’s mass and speed into the equation.

Solution for (b)

To find the ball’s momentum, substitute the known values for the ball’s mass and speed into the equation.

The ratio of the player’s momentum to the ball’s momentum is

$p player p ball = 880 10.3 =85.9. p player p ball = 880 10.3 =85.9.$
Discussion

Although the ball has greater velocity, the player has a much greater mass. Therefore, the momentum of the player is about 86 times greater than the momentum of the football.

### Worked Example

#### Calculating Force: Venus Williams’ Racquet

During the 2007 French Open, Venus Williams (Figure 8.3) hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What was the average force exerted on the 0.057 kg tennis ball by Williams’ racquet? Assume that the ball’s speed just after impact was 58 m/s, the horizontal velocity before impact is negligible, and that the ball remained in contact with the racquet for 5 ms (milliseconds).

Figure 8.3 Venus Williams playing in the 2013 US Open (Edwin Martinez, Flickr)

### Strategy

Recall that Newton’s second law stated in terms of momentum is

$F net = Δp Δt . F net = Δp Δt .$

As noted above, when mass is constant, the change in momentum is given by

$Δp=mΔv=m( v f − v i ) , Δp=mΔv=m( v f − v i ) ,$

where vf is the final velocity and vi is the initial velocity. In this example, the velocity just after impact and the change in time are given, so after we solve for $Δp Δp$, we can use $F net = Δp Δt F net = Δp Δt$ to find the force.

Solution

To determine the change in momentum, substitute the values for mass and the initial and final velocities into the equation above.

8.1

Now we can find the magnitude of the net external force using $F net = Δp Δt F net = Δp Δt$

8.2
Discussion

This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact. This problem could also be solved by first finding the acceleration and then using Fnet = ma, but we would have had to do one more step. In this case, using momentum was a shortcut.

# Practice Problems

### Practice Problems

What is the momentum of a bowling ball with mass $5 kg$ and velocity $10 m/s$?
1. $0.5 kg ⋅ m/s$
2. $2 kg ⋅ m/s$
3. $15 kg ⋅ m/s$
4. $50 kg ⋅ m/s$
What will be the change in momentum caused by a net force of $120 N$ acting on an object for $2$ seconds?
1. $60 kg ⋅ m/s$
2. $118 kg ⋅ m/s$
3. $122 kg ⋅ m/s$
4. $240 kg ⋅ m/s$

# Check Your Understanding

### Check Your Understanding

Exercise 1

What is linear momentum?

1. the sum of a system’s mass and its velocity
2. the ratio of a system’s mass to its velocity
3. the product of a system’s mass and its velocity
4. the product of a system’s moment of inertia and its velocity
Exercise 2

If an object’s mass is constant, what is its momentum proportional to?

1. Its velocity
2. Its weight
3. Its displacement
4. Its moment of inertia
Exercise 3

What is the equation for Newton’s second law of motion, in terms of mass, velocity, and time, when the mass of the system is constant?

1. $F net = Δv ΔmΔt F net = Δv ΔmΔt$
2. $F net = mΔt Δv F net = mΔt Δv$
3. $F net = mΔv Δt F net = mΔv Δt$
4. $F net = ΔmΔv Δt F net = ΔmΔv Δt$
Exercise 4

Give an example of a system whose mass is not constant.

1. A spinning top
2. A baseball flying through the air
3. A rocket launched from Earth
4. A block sliding on a frictionless inclined plane