chapter 5: conservation of linear momentum -...

Etkina/Gentile/Van Heuvelen Process Physics 1/e, Chapter 5 5-1

Chapter 5: Conservation of Linear momentum

! How does jet propulsion work?

! How can you estimate the risk of injury when falling out of a bathtub?

! Can a meteorite collision significantly change the Earth’s orbit?

Make sure you know how to:

1. Construct a force diagram for an object.

2. Use Newton’s second law in component form.

3. Use kinematics to describe an object’s motion.

[Chapter opening:]

Cars use an interaction between their tires and the road to change their velocity (to

accelerate). To move faster, the tires rotate more rapidly and push back harder on the road. In

turn, the road pushes forward on the tires, causing the car to speed up. To move slower, the tires

turn more slowly and push forward on the road. In turn, the road pushes backward on the tires,

causing the car to slow down. The road is critical for changing the car’s velocity. How does a

rocket far above Earth’s atmosphere change velocity, since there is no object for the rocket to

push against? Physicist Robert Goddard, a US rocket pioneer, in 1920 published an article about

rocketry and even suggested a rocket flight to the Moon. Goddard suddenly became a subject of

ridicule by the press. A New York Times editorial dismissed his idea saying: “… even a schoolboy

knows that rockets cannot fly in space because a vacuum is devoid of anything to push on.” We

know now that Goddard was correct—but why? What does the rocket push on?

[Lead:]

In Chapter 3, we learned how to use Newton’s second law (F

am

"#

!!

) to relate the

acceleration of a system object to the forces being exerted on it. However, to use this law

effectively we need to have quantitative information about those external forces. There are many

situations in which this information is not accessible. For example, if two cars collide, how do

you determine the force that one car exerts on the other during the collision? When fireworks

explode, how do you calculate the forces that are exerted on the pieces flying apart? One way to

analyze situations such as this is to use an approach that does not involve forces but involves a

very important and powerful new idea: conserved quantities. Investigating this idea is the goal of

this chapter.


5.1 Conservation of Mass

We begin our investigation of conserved quantities by exploring a simple example, the

quantity mass. Mass was an important quantity in Chapter 2; we found that the acceleration of an

object depended on its mass—the greater its mass, the less it accelerated due to an unbalanced

external force. We ignored the possibility that an object’s mass might change during some

process. Is the mass in a system always a constant value?

You have probably observed physical processes in which mass seems to change (Fig.

5.1). For example, the mass of a log in a campfire decreases as the log burns. In contrast, the mass

of a seed planted in soil increases as the plant grows. Does this mean that mass can disappear or

reappear for different processes? To help answer this question, scientists in the 1700s used the

idea of a carefully defined system. In Chapter 3, we defined a system as an object that we would

focus on as we analyzed a process. Interactions between objects outside the system and the

system object were called external interactions and could produce changes in the system object’s

motion. In this and future chapters, we will often choose systems that consist of more than one

object. If two objects inside a system interact with each other, it is called an internal interaction.

The choices we make about which objects to include in a system of interest have a profound

effect on how we analyze a process.

Figure 5.1

Let’s use the idea of a system to consider what happens to the mass of a burning log. If

we choose only the log as the system, the mass of the system decreases as it burns. However, you

probably know that air is needed for burning. What happens to the mass if we choose air and the

log as the system? To conduct such an experiment, we would need to use a closed container.

In the 1700s, French chemist Antoine Lavoisier was the first to use the idea of a system

to measure the masses of reactants and products of chemical reactions. Lavoisier placed a closed

flask containing a reactant on one side of a balance scale and a metal block of equal mass on the

other side; then he burned the reactant (Fig. 5.2). For example, in one experiment, he burned iron

in the closed flask (the flask also contained air). Lavoisier found that the total mass of the closed

container with the burned iron (iron oxide) was the same as before burning. Next, he burned the

iron in an open flask and observed that the mass of the flask increased. Lavoisier made a


hypothesis that the increase in mass was due to the mass of additional air absorbed by the iron as

it burned.

Figure 5.2 The mass of the reactants in the closed jar did not change when burned. Mass of contents in

closed flask did not change when burned.

Scientists believed Lavoisier’s idea made sense, but they were skeptical about the

accuracy of his experiments. In the late 1800s, Hans Landolt performed careful tests of

Lavoisier’s hypothesis by doing numerous experiments, making precision measurements of the

masses of systems involved in chemical reactions. He failed to disprove Lavoisier’s hypothesis.

He concluded:

“The final result of the investigation is that no change in total weight can be found in

chemical reaction. … The experimental test of the law of conservation of mass may be

considered complete. If there exist any deviations from it, they would have to be less than

a thousandth of a milligram.” [Physics, the Human Adventure, Holton and Brush

(2001)]”

The idea Landolt described is called the conservation of mass. Another way of saying this

is that mass is a conserved quantity. The idea that mass is conserved means that it can neither be

created nor destroyed in any process. However, it is important to note that just because mass is a

conserved quantity does not mean that the mass of a system is always constant. As we learned,

Lavoisier’s experiments on iron oxide produced different results depending on the choice of

system. In the closed flask, the total mass of the reactant (iron) and the enclosed air before and

after burning did not change – it was constant. In this case Lavoisier had chosen the system of

interest as the iron and the enclosed air. In the open flask, the mass after burning was greater than

the mass before burning; thus, mass was not constant. Lavoisier had initially chosen the system of

interest to be just the solid material inside the flask. During burning, air from outside the system

joined the system. The mass of the system was constant only if that system was isolated. But if

we could keep track of the mass entering or leaving the system, and the amount of mass


entering/leaving the system would account for the change in the mass of the system. This is what

makes mass a conserved quantity.

To understand this idea better, think of an analogy. The system of interest will be the

money in your pocket. When you go to a supermarket to buy groceries, the money in your pocket

decreases; thus it is not constant. Choosing the supermarket as the system, we find that the money

in its account is not constant – it increases by the same amount as your money decreased.

However there was no money created or destroyed in the process – it decreased in your pocket

and increased on the supermarket’s account by the same amount. If we chose the system of

interest to be your pocket and the supermarket’s account, then the money in the system would be

constant. In other words, money is conserved no matter what choice of system you choose, but

money is constant only if you choose the supermarket and you as the system – an isolated system.

Tip! The mass of a system is constant only if you include in the system all objects that are

interacting. If you do not include all of the interacting objects, the mass of the system might

change. However, even if the mass of the system is not constant, the mass is always conserved.

Why are we interested in whether or not a physical quantity is conserved? If we know that

a quantity is conserved, we can construct a powerful relationship involving that quantity. For

example: mass is a conserved quantity. Therefore we can write an equation representing that idea:

mass of system at mass entering/leaving system mass of system at

earlier clock reading between the two clock readings later clock reading

$ % $ % $ %& #' ( ' ( ' (

) * ) * ) *

Each time we encounter a new conserved quantity, we will be able to write a similar equation

representing it.

Just as with every idea in physics, even the idea of the conservation of mass has limits to

its applicability. We will discover later in this book that in situations involving microscopic

particles, mass stops being a conserved quantity; instead a new quantity that includes mass as a

component will end up being conserved. You will learn more about this in Chapters 28 and 29.

An example of such a process is the production of light by the Sun. The mass of the Sun

decreases by about 94.2 10 kg+ each second and it does not appear anywhere else as mass.

Where does it go? You will have to wait to get the answer to that question.

Review Question 5.1

Give an everyday example of a situation and a choice of system where the mass of the system is

constant but some other physical quantity is not.


5.2 Linear Momentum

In the previous section, we found that mass is a conserved quantity, at least for the

macroscopic processes observed in our everyday life. What about motion? Is there a

quantity related to motion that is conserved? When you kick a stationary ball, there seems to

be a transfer of some “motional stuff” from your foot to the ball. When you knock bowling

pins down with a bowling ball, a similar thing seems to be happening. Can we quantify this

“motional stuff” as a conserved quantity?

For many years, physicists investigated different combinations of mass and speed, and

mass and velocity until they found what seemed to work as a conserved quantity relating to

motion. The experiments in Observational Experiment Table 5.1 are similar to those they

conducted. Here we observe two carts of different mass on a smooth track. For these experiments,

the system of interest will include both carts, which means the forces that the carts exert on each

other during the collision are internal forces and therefore would not appear in a force diagram for

the system. There are also external vertical forces exerted on the carts—an upward normal force

exerted by the track, and a downward gravitational force exerted by Earth. The sum of these two

forces for each cart is zero; thus, they do not affect the motion of the system. Since the carts are

on a smooth track, we’ll assume that friction has no significant affect on the situation. When the

external forces exerted on a system add to zero, the system is considered isolated.

Isolated system: An isolated system is one whose mass is constant and whose

interactions with objects outside the system are balanced; that is, the forces due to these

external interactions add to zero.

Tip! The word ‘isolated’ means something different in physics than it means in everyday life.

‘Isolated’ doesn’t mean that nothing is interacting with the system. It means that the interactions

between the system and the environment are balanced—add to zero.

ALG

5.1.1-

5.1.2


Observational Experiment Table 5.1. Collisions in a system of two carts (all velocities are

recorded with respect to the track).

Patterns One quantity remains the same before and after the collision in each experiment—the sum of the products

of the mass and x-velocity component of the system objects.

We explored several different combinations of physical quantities describing moving

objects to find a quantity that is constant in the system. In the three experiments in Observational

Observational

experiments

Analysis

I. Cart A (0.2 kg)

moving right at 1.0

m/s collides with

cart B (0.2 kg),

which is stationary.

Cart A stops and

cart B moves right

at 1.0 m/s.

The direction of motion is indicated with a plus and a minus sign.

• Speed: The speed of the system objects is the same before and after the collision:

(1.0 m/s + 0 m/s = 0 m/s + 1.0 m/s ).

• Mass•speed: The product of mass and speed is the same before and after the

collision: 0.2 kg(1.0 m/s) + 0.2 kg(0 m/s) = 0.2 kg(0 m/s) + 0.2 kg(1.0 m/s) .

• Mass•velocity: The product of mass and the x-velocity component is the same

before and after the collision:

0.2 kg(+1.0 m/s) + 0.2 kg(0) = 0.2 kg(0) + 0.2 kg(+1.0 m/s).

II. Cart A (0.4 kg)

moving right at 1.0

m/s collides with

cart B (0.2 kg),

which is stationary.

After the collision,

both carts move

right, cart B at 1.2

m/s, and cart A at

0.4 m/s.

• Speed: The speed of the system objects is not the same before and after the

collision: (1.0 m/s + 0 m/s 0.4 m/s + 1.2 m/s), .

• Mass•speed: The product of mass and speed is the same before and after the

collision: 0.4 kg(1.0 m/s) + 0.2 kg(0 m/s) = 0.4 kg(0.4 m/s) + 0.2 kg(1.2 m/s) .



0.4 kg(+1.0 m/s) + 0.2 kg(0) = 0.4 kg(+0.4 m/s) + 0.2 kg(+1.2 m/s) .

III. Cart A (0.2 kg)

with a piece of clay

attached to the

front moves right at

1.0 m/s. Cart B (0.2

kg) moves left at

1.0 m/s. The carts

collide, stick

together, and stop.

• Speed: The speed of the objects in the system is the not the same before and after

the collision: (1.0 m/s + 1.0 m/s 0 m/s + 0 m/s), .

• Mass•speed: The product of mass and speed is not the same before and after the

collision: 0.2 kg(1.0 m/s) + 0.2 kg(1.0 m/s) 0.2 kg(0 m/s) + 0.2 kg(0 m/s), .



0.2 kg(+1.0 m/s) + 0.2 kg( – 1.0 m/s) = 0.2 kg(0) + 0.2 kg(0) .


Experiment Table 5.1, only one quantity—the sum of the products of mass and the x-component

of velocity x

mv" —remained the same before and after the carts collided. Note also that the sum

of the products of the mass and the y-component of velocity y

mv" also did not change—it

remained zero. Since the x and y Perhaps the quantity mv"!

might be the “motional stuff” that is

conserved. In the three experiments in Table 5.1, this quantity is constant for the system. But, will

this pattern continue to hold in other situations? Let’s test this idea by using it to predict the

outcome of the experiment in Testing Experiment Table 5.2.

Testing Experiment Table 5.2 Testing the idea that the sum of the products of mass and velocity

mv"!

in a system remains constant (all velocities are with respect to the track).

Testing experiment Prediction Outcome

Cart A (0.4 kg) has a

piece of modeling clay

attached to its front and

is moving right at 1.0

m/s. Cart B (0.2 kg) is

moving left at 1.0 m/s.

The carts collide and

stick together. Predict

the velocity of the carts

after the collision.

The system consists of the two carts. The direction of velocity is

noted with a plus or minus sign of the velocity component:

f (0.4 kg)(+1.0 m/s) + (0.2 kg)( –1.0 m/s) = (0.4 kg + 0.2 kg)x

v

or

f = (+0.2 kg • m/s)/(0.6 kg) = +0.33 m/sx

v .

After the collision, the two carts should move right at a speed of

about 0.33 m/s.

After the

collision, the

carts move

together toward

the right at the

speed close to

the predicted

speed

(0.32m/s).

Conclusion

Using the idea that the sum of the products of mass and velocity mv"!

of an isolated system is constant, we

predicted an outcome of the experiment; the prediction matched the outcome. This gives us increased

confidence that this new quantity, mv!

, might be a conserved quantity.

Several simple observations allowed us to devise a new quantity characterizing the behavior of

a single object or a system. This quantity is called linear momentum p!

.

Linear Momentum The linear momentum p!

of a single object is the product of its mass

m and velocity v!

:

p mv#! !

(5.1)

Linear momentum is a vector quantity that points in the same direction as the object’s

velocity v!

(Fig. 5. 3). The SI unit of momentum is (kg)(m/s). The total linear momentum

of a system containing multiple objects is the vector sum of the momentums of the

individual objects.

netp mv# "! !

ALG

5.2.1-

5.2.3


Figure 5.3

There are three important points to mention here.

1. Unlike mass, which is a scalar quantity, p mv#! !

is a vector quantity; therefore, it is

important to consider the direction the colliding objects are moving before and after the

collision. For example, as cart B was moving left along the x-axis, the x-component of its

velocity was negative before the collision.

2. As momentum depends on the velocity of the object, and the velocity depends on the

choice of the reference frame, different observers will measure different momenta for the

same object. If you sit a car – it’s momentum with respect to you is zero, while it is not

zero for an observer on the ground when the car is moving away from him.

3. We chose an isolated system (the two carts) for our investigation. The sum of the

products of mass and velocity mv"!

of all objects in the isolated system remained

constant even though the carts collided with each other. Thus we can hypothesize that the

product of mass and velocity of an object is a conserved quantity. However, if we choose

the system to be just one of the carts, we see that the linear momentum p mv#! !

of the

cart before the collision is different than it is after the collision. Thus to establish that

momentum p!

is a conserved quantity, we need to make sure that the momentum of a

system changes in a predictable way for systems that are not isolated.

The system in the experiments in Observational Experiment Table 5.1 was chosen so that

the sum of the external forces was zero, making it an isolated system. Based on these

observational experiments and the testing experiment in Testing Experiment Table 5.2, it appears

that the total momentum of an isolated system is constant.

Momentum constancy of an isolated system: The momentum of an isolated system is

constant. For an isolated two-object system:

1 1 2 2 1 1f 2 2fi im v m v m v m v& # &! ! ! !

(5.2)

The x- and y-component forms of Eq. (5.2) are:

1 1i 2 2i 1 1f 1 1f 2 2f x x x x xm v m v m v m v m v& # & & (5.3x)

1 1i 2 2i 1 1f 1 1f 2 2f y y y y ym v m v m v m v m v& # & & (5.3y)


Note that momentum is a vector quantity and Eq. (5.2) is a vector equation. That is why

we will work with its x and y-component forms - Eqs. (5.3x) and (5.3y). Note also that for a

system of more than two objects, the momentum constancy of an isolated system can be extended

by adding additional terms so that there is one term on each side of the each equation for each

object in the system. Let’s test the idea that the momentum of an isolated system is constant in

another situation.

Example 5.1 Two rollerbladers Jen (50 kg) stands on roller blades facing David (75 kg), also on

rollerblades. They push off each other abruptly. After the contact is over, each person coasts

backwards at approximately constant speed. During a certain time interval, Jen travels 3.0 m.

How far does David travel during that same time interval?

Sketch and Translate The process is sketched in Fig. 5.4a. We consider all motion with respect to

the floor. Initially the two rollerbladers are at rest. After pushing off each other, they move in

opposite directions, Jen in the positive direction and David in the negative direction. We choose

the two rollerbladers as the system of interest. We will relate the momentum of the system at the

initial state just before the push off to the momentum of the system at the final state when Jen has

traveled 3.0 m from her starting position. All motion is along the x-axis. We can write Jen’s

velocity component as Jf xv # +(3.0 m)/ t- , where t- is the time interval needed for Jen to travel

the 3.0 m after the pushing is over. We can use momentum constancy to calculate David’s

velocity component and use it to predict the distance he will travel during that same time interval.

Figure 5.4(a)

Simplify and Diagram We model each person as a point particle and assume that the friction

forces exerted on the rollerbladers doesn’t affect their motion significantly. Thus there are no

horizontal external forces exerted on the system. In addition, two forces are exerted on each

person (see Fig. 5.4b): an upward normal force F on P N!

that the floor exerts on each person and an

equal magnitude downward gravitational force E on PF!

that the Earth exerts on each person. These

forces cancel. So the system can be considered isolated since the net external force on the system


is zero. What about the forces that the rollerbladers exert on each other? These are internal forces;

thus they should not affect the momentum of the system.

Figure 5.4(b)

Represent Mathematically We label Jen as J and David as D. Momentum constancy applied to the

system is:

J Ji D Di J Jf D Df .x x x x

m v m v m v m v& # &

Here the initial state (labeled i) of the system is before they start pushing on each other and the

final state (f) is when Jen has traveled 3.0 m. We’ll choose the positive direction toward the right.

Because the initial velocity of each person is zero, the above equation becomes:

J Jf D Df 0 0 x xm v m v& # &

or

D Df J Jf –x x

m v m v#

Solve and Evaluate The x-component of Jen’s velocity after the push-off is Jf (3.0 m)/x

v t# & -

where t- is the time interval needed for her to travel 3.0 m. We solve the above equation for

David’s final x velocity component to determine how far David should travel during that same

time interval:

J Jf JDf Jf

D D

(50 kg) (3.0 m) (2.0 m)– – – – .

(75 kg)

x

x x

m v mv v

m m t t# # # #

- -

Using the idea of momentum constancy, we predict that David will travel 2.0 m in the negative

direction during that same time interval that Jen travels 3.0 m in the positive direction. When the

experiment is performed, the measured value is very close to the predicted value. This increases

or confidence in the idea of momentum constancy for isolated systems.

Try It Yourself: Suppose your mass is 60 kg. Estimate the magnitude of your momentum when

walking and when jogging.

Answer: When walking, you travel at a speed of about 1 to 2 m/s. So the magnitude of your

momentum will be . /. /60 kg 1.5 m/s 90 kg m sp mv# 0 0 1 . When jogging, your speed is

about 2 to 5 m/s or a momentum of magnitude . /. /60 kg 3.5 m/s 200 kg m sp mv# 0 0 1 .

Notice in the last example that we were able to determine the velocity of David (at least

an expression for how far he traveled in a certain time interval) by using the idea that the total


momentum of the objects in an isolated system is constant. We did not need to have any

information about the forces that the two people exerted on each other while pushing off. This is a

very powerful result, since in all likelihood the forces they exerted on each other were not

constant. Analyzing situations with non-constant forces previously presented problems for us

since the kinematics we developed required that the acceleration of the system be constant. Using

the idea of momentum constancy has allowed us to analyze a situation involving non-constant

forces.

So far, we have investigated situations involving isolated systems. In the next section, we

will investigate how to use momentum ideas to analyze situations where the system is not

isolated.

Review question 5.2

Two identical carts are traveling toward each other at the same speed. One of the carts has a piece

of modeling clay on its front. The carts collide, stick together, and stop; the momentum of each

cart is now zero. If the system includes both carts, did the momentum of the system disappear?

Explain your answer.

5.3 Impulse and Momentum

In the last section, we constructed a new physical quantity to describe the motion of

objects—linear momentum. We found that the momentum of a system is constant if that

system is isolated (the net force exerted on the system is zero). How do we account for

the change in momentum of a system when the net force exerted on it is not zero? In

other words, what is the relationship between the momentum of a system and the forces being

exerted on it? We can actually use Newton’s laws to derive an expression relating forces and

momentum change.

When you push a bowling ball, you exert a force on it causing the ball to accelerate. The

average acceleration a!

is defined as the change in velocity f i

v v2! !

divided by the time interval

t –f i

t t t- # during which that change occurs:

–

f i

f i

v va

t t

2#! !

!

Recall that in Chapters 2 and 3, Newton’s second law was used to determine an object’s

acceleration if we know its mass and the sum of the forces that other objects exert on it:

Fa

m# 3

!

!

We now have two expressions for an object’s acceleration. Setting these two expressions

for acceleration equal to each other, we get:

ALG

5.3.3


–

f i

f i

v v F

t t m

2# 3

!! !

Now multiply both sides by ( – )f i

m t t and get the following:

( – )f i f i f i

mv mv p p F t t2 # 2 #3!! ! ! !

(5.4)

We see that the left side of the above equation represents the change in momentum of the object.

This change depends on the product of the net external force and the time interval during which

the forces are exerted on the object (the right side of the equation). Therefore, it is not the force

itself but the product of the force and the time interval that is equal to the change in the

momentum of the object. There are two important points:

1. Equation (5.4) is just Newton’s second law written in a different form—one that involves the

physical quantity of momentum.

2. The time interval during which the forces are exerted on an object affects the change in its

momentum—the longer the time interval, the greater the momentum change. In addition, a

small force exerted for a long time interval can change the momentum of an object by the

same amount as a large force exerted for a short time interval.

Let’s explore this last point further. Equation (5.4) applied to a situation involving one

external force exerted on an object would be: ( – )f i f i

p p F t t2 #!! !

. The right side of this

equation is called the impulse of the force. When you kick a ball or hit a baseball with a bat, your

foot or the bat exerts an impulse on the ball. The forces in these situations are not constant but

rather, they vary in time. (Fig 5.5a) We can approximate a varying force as the average force

exerted during the time interval (Fig. 5.5b). The shaded area under the curve represents the

impulse of the force.

Figure 5.5


Impulse The impulse J!

of a force is the product of the average force avF!

exerted on an object

during a time interval (tf – ti) and that time interval:

av ( – )f i

J F t t#! !

. (5.5)

Impulse is a vector quantity that points in the direction of the force. For linear motion, the

impulse has a plus or minus sign depending on the orientation of the force relative to a coordinate

axis. The SI unit for impulse is N•s = (kg•m/s2)•s = kg•m/s, the same units as momentum.

It is often difficult to measure directly the impulse or the net average force during a time

interval. However, if we look at Eq. (5.4) again,

( – )f i f i

p p F t t2 # "!! !

we see that the net force on the right side of the equation can be determined indirectly by

measuring or calculating the momentum change on the left side of the equation. For this reason,

the combination of impulse and momentum change provides a powerful tool for analyzing

interactions between objects. We now write Eq. (5.4) as what is called the impulse–momentum

equation.

Impulse–momentum equation If there are several external objects exerting forces on the system

during a time interval ( – )f i

t t , the sum of their impulses J"!

causes a change in momentum of

the system:

f ip p J2 # "

!! !

In terms of the external forces and the time interval,

. /on System– –f i f i

p p F t t#3!! !

(5.6)

The x- and y-scalar component forms of the impulse-momentum equation are:

. /on System – –fx ix x f i

p p F t t#3 (5.7x)

. /on System y– –fy iy f i

p p F t t#3 (5.7y)

There are several points that are important to consider when you start applying the idea of

impulse. First, notice that Eq. (5.6) is a vector equation, as both the momentum and the impulse

are vector quantities. However, vector equations are not as easy to manipulate mathematically as

scalar equations. Therefore, we will usually use the scalar component forms of Eq. (5.6)––Eqs.

(5.7x) and (5.7y).

Second, the time intervals in Eqs. (5.6) and (5.7x and y) are very important. When object 2

exerts a force on object 1, the momentum of object 1 changes by an amount equal to:

. /1 1 1 2 on 1 2 on 1f i f ip p p F t t F t- # 2 # 2 # -

! !! ! !.


The longer that object 2 exerts the force on object 1, the more the momentum change of object 1.

This explains why a fast-moving object might have less of an effect on a stationary object during

a collision than a slow-moving object interacting with the stationary object over a longer time

interval. For example, a fast moving bullet passing from the inside of a room through a partially

closed wooden door might not open the door (it will just punch a hole in the door) whereas your

little finger, moving much slower than the bullet, could open the door. Although the bullet moves

at high speed and exerts a large force on the door, the time interval during which it interacts with

the door is very small (milliseconds). Hence, it exerts a relatively small impulse on the door—too

small to significantly change the door’s momentum. A photo of a bullet shot through an apple

illustrates the effect of a short impulse time. (Fig. 5.6) The impulse exerted by the bullet on the

apple was too small to knock the apple off its support.

Third, if the force changes during the time interval considered in the process, one needs to

use the average force.

Figure 5.6 The bullet’s impulse time is very short, causing small impulse

SOURCE: Harold and Ester Edgarton Foundation , 2006, Palm Press. www.photocritic.org/high-speed-

photography/

Tip! If the same amount of force is exerted for the same time interval on a large-mass object and

on a small-mass object, the objects will have an equal change in momentum (the same impulse

was exerted on them). However, the small-mass object would experience a greater change in

velocity than the large-mass object.

Example 5.2 Abrupt stop in a car A 60-kg person is traveling in a car that is moving at a

speed of 16 m/s with respect to the ground when the car hits a barrier. The person is stopped

by an air bag in a time interval of 0.20 s. Determine the average force that the air bag exerts

on the person while stopping him.

Sketch and Translate We choose the person as the system of interest since we are investigating a

force being exerted on him. An initial-final sketch of the process is shown in Fig. 5.7a.

Simplify and diagram A force diagram for the person (the system) is shown in Fig. 5.7b. Here

B on PF!

(bag on person) is the average force. The vertical normal force and gravitational forces

cancel. The air bag exerts a force in the negative x-direction on the person.

ALG

5.4.1


Figure 5.7

Represent Mathematically The x-component form of the impulse–momentum equation (including

the force exerted by the air bag on the person B on PF!

) is:

P P P P P B on P ( – ) ( – )f x i x f x i x x f i

p p m v v F t t2 # #

Solve and Evaluate Solve for the force exerted by the air bag on the person:

P P P

B on P

( – )

( – )

f x i x

x

f i

m v vF

t t# .

The person’s initial x-component of velocity vPi x = +16 m/s decreases to the final x-component of

velocity vPf x = 0 in a time interval ( – ) 0.20 sf i

t t # . Thus the average force exerted by the air

bag on the person in the x-direction is:

P P P

B on P

( – ) (60 kg)(0 –16 m/s)–4800 N

( – ) (0.20 s – 0)

fx ix

x

f i

m v vF

t t# # # .

The negative sign in –4800 N indicates that the average force exerted by the air bag points in the

negative x-direction. The magnitude of this force is about 1000 lb!

Try It Yourself: Suppose a crash-test dummy that is not wearing a seat belt in a car without air

bags flies forward and stops when it hits a hard surface (the steering wheel or dashboard or

windshield). The stopping time interval for the dummy is 0.02 s (a much shorter stopping time

interval than if the dummy were wearing a seat belt). What is the average magnitude of the

stopping force that this harder surface exerts on the dummy?

Answer: The average force that the hard surface exerts on the dummy would be about 50,000 N,

extremely unsafe for a human. Note that the momentum change of the person is the same.

However, since the change occurs during a shorter time interval, the force exerted on the person

is much greater. This is why it is so important for a person to wear a seat belt and shoulder strap

and to have an air bag.

Momentum conservation comes from Newton’s laws

In Section 5.2, we found experimentally that in an isolated system with two objects, the

momentum of the system is constant. We have just seen how to account for the change in


momentum of a single object by applying the concepts of impulse and momentum, which were

developed using Newton’s second law and kinematics for a single-object system [Eq. (5.4)]. Can

we use the impulse-momentum equation [Eq. (5.4)] to develop the momentum constancy idea for

a two-object isolated system?

We investigate the same type of situation described in Observational Experiment Table

5.1 but this time we will not conduct the experiments but instead will analyze their motion using

Newton’s laws, impulse, and momentum. The two carts travel toward each other at different

speeds, collide, and rebound backward (Fig. 5.8). First we will analyze each cart as a separate

system, and then analyze them together as a single system. Assume that the vertical forces

exerted on the carts are balanced and that the friction force exerted by the surface on the carts

does not significantly affect their motion.

Figure 5.8

Cart 1: Before the collision, cart 1 with mass m1 travels in the positive direction at

velocity 1 iv!

. After the collision, cart 1 moves with a different velocity 1 fv!

in the opposite

direction. To determine the effect that the impulse exerted by cart 2 on cart 1, we apply the

impulse–momentum equation to cart 1 only:

1 1 1 2 on 1( ) ( )f i f i

m v v F t t2 # 2!! !

Cart 2: We repeat this analysis with cart 2 as the system. Its velocity and momentum

changes because of the impulse exerted on it by cart 1:

2 2 2 1 on 2( ) ( )f i f i

m v v F t t2 # 2!! !

Newton’s third law provides a connection between our analyses of the two carts;

interacting objects at each instant exert equal magnitude but oppositely directed forces on each

other:

1 on 2 2on 1F F# 2! !

Substituting the expressions for the forces from above and simplifying we get:


1 on 2 2 on 1F F# 2! !

2 2 2 1 1 1

2 2 2 1 1 1

( ) ( )

( ) ( )

f i f i

f i f i

f i f i

m v v m v v

t t t t

m v v m v v

2 24 # 2

2 2

4 2 # 2 2

! ! ! !

! ! ! !

Moving the initial momentum for both objects to the left side and the final momentum for both

objects to the right side:

1 1 2 2 1 1 2 2 = i i f f

m v m v m v m v& &! ! ! !

This is the same equation we arrived at to describe the constancy of momentum for an

isolated system in Section 5.2, where we observed and analyzed cart collisions. We have now

arrived at the result using only our knowledge of Newton’s laws, momentum, and impulse. In

other words, we used a piece of physics that we already understood (Newton’s third law) to arrive

at the idea of the momentum constancy of an isolated system. We actually didn’t need to perform

any observational experiments. This doesn’t mean that we were wasting time going through those

earlier experiments. Oftentimes in physics it’s not clear that a new phenomenon (like colliding

objects) can actually be understood with previous ideas. Sometimes that revelation only comes

later, after trying to understand it using other ideas. This is part of the process of constructing

knowledge.

Review question 5.3

An apple is falling from a tree. Why does its momentum change? Specify the external force

responsible. Find a system in which the momentum is constant during this process.

5.4 Generalized Impulse–Momentum Principle and Bar Charts

We now have two principles that involve momentum: (1) the constancy of momentum for

isolated systems; and (2) the impulse–momentum equation for non-isolated systems. We can

summarize them together: the change in momentum of a system is equal to the net external

impulse exerted on it. If the net impulse is zero, then the momentum of the system is constant.

This idea, expressed mathematically as the generalized impulse–momentum principle, accounts

for situations when the system includes one or more objects and may or may not be isolated.

Momentum before

collisionMomentum after

collision


Generalized impulse-momentum principle For a one- or multi-object system, the initial

momentum of the system plus the sum of the impulses that external objects exert on the system

objects during the time interval (tf – ti) equals the final momentum of the system:

1 1 2 2 on Sys 1 1 2 2( ...) ( – ) ( ...)i i f i f f

m v m v F t t m v m v& & &" # & &!! ! ! !

(5.8)

The x- and y-component forms of the generalized impulse–momentum principle are:

1 1 2 2 on Sys 1 1 2 2( ...) ( – ) ( ...)ix ix x f i fx fx

m v m v F t t m v m v& & &" # & & (5.9x)

1 1 2 2 on Sys y 1 1 2 2( ...) ( – ) ( ...)iy iy f i fy fy

m v m v F t t m v m v& & &" # & & (5.9y)

Note: If the net impulse exerted in a particular direction is zero, then the component of the

momentum of the system in that direction is constant.

The above equation is useful in two ways. First, any time we choose to analyze a situation

using the ideas of impulse and momentum, we can start from a single principle, regardless of what

the situation is. Second, the equation reminds us that we need to consider all the interactions

between the environment and the system that might cause a change in the momentum of the

system.

Tip! Recall from Chapter 2 that Newton’s laws are applicable if the observer is in an inertial

reference frame. Since we just learned that the generalized impulse–momentum principle actually

follows from Newton’s laws, then it too is only applicable if the observer is in an inertial reference

frame.

Impulse–momentum bar charts

We can mathematically describe an impulse–momentum process using Eqs. (5.9x and y).

However, it is often difficult to translate directly from a word description of a process (such as

might be given in an end-of-chapter problem) to a mathematical solution to that problem. Often,

it is helpful to make intermediate representations that bridge the gap between words and the

equations. This bridging occurred using sketches and force diagrams as intermediate

representations when analyzing processes with Newton’s second law.

When we analyze processes with the generalized impulse-momentum principle, an

initial-final sketch of the process will continue to be important. In addition, we will use a new

intermediate representation called an impulse–momentum bar chart. The skill box below shows

the steps for constructing an impulse-momentum bar chart.

Net impulse exerted on the system Initial momentum of the system Final momentum of the system


The example in the above skill box represents a system that includes two carts of equal

mass traveling toward each other (we’re using this familiar situation so you can focus solely on

how to construct the corresponding bar chart). In this example, all motion is along the x-axis with

respect to the track. Cart 1, which is traveling at high speed to the right in the positive direction,

collides with cart 2, which is moving slower in the negative direction. Thus, the bar for the initial

state x component of momentum of cart 1 points up (positive component, since cart 1 is moving

in the positive direction), and the bar for cart 2 points down (negative component, since cart 2 is

moving in the negative direction). In the final state, the carts are stuck together and are moving in

the positive direction; thus the momentum bars for both carts point up. Because the carts are stuck

together, they have the same final velocity. Since they also have the same mass, they each have

the same final momentum.

The middle shaded column in the bar chart represents the net external impulse exerted on

the system objects during the time interval ( – )f i

t t . This time interval is the time that passes


between the initial state of the system and the final state of the system. The shading reminds us

that impulse is caused by external forces. For the example shown, we consider the system

isolated and not affected by external forces for two reasons. First, both carts are included in the

system; therefore, the forces that they exert on each other during the collision are internal.

Second, if we assume that the collision occurs during a short time interval (typically 0.1 s or less),

then the impulse due to surface friction on the system during that short time interval is very small

compared to the momentum of the system. Thus, the external impulse due to the force of friction

for this situation is regarded as zero. In addition, the upward normal force exerted by the surface

is balanced by the downward gravitational force exerted by Earth on each cart. Lastly, even

though objects may become attached to one another, as they often do in collisions, they still each

get their own final momentum bar.

Converting a bar chart to a mathematical equation

Once we have constructed the bar chart, we can convert it into a mathematical equation

using the generalized impulse–momentum equation. Notice that in the reasoning skill box, each

non-zero bar in the chart corresponds to a term in the equation. Each term has a sign depending

on the orientation of the bar. The upward or downward orientation of the bars (positive or

negative) is related to the direction of the momentum or impulse relative to the chosen coordinate

axis.

Using impulse–momentum to investigate internal forces

We’ve used impulse-momentum bar charts and the generalized impulse–momentum

principle to analyze collisions between two objects—for example to determine the velocity of two

carts locked together after a collision. We were able to do this without knowing about the forces

the two objects exerted on each other during the collision. Can we use the ideas of impulse and

momentum to learn something about those forces? Consider a collision between two cars (Fig.

5.9).

Figure 5.9

To analyze the force that each car exerts on the other, define the system to include only

one of the cars. Let’s choose car 1 and construct a bar chart for it. Car 2 exerts an impulse on car


1 during the collision that changes the momentum of car 1. If the initial momentum of car 1 is in

the positive direction, then the impulse exerted by car 2 on car 1 points in the negative direction.

Because of this, the impulse bar on the bar chart points downward (Fig. 5.9). The direction and

size of the impulse bar is adjusted so that the total size of the initial momentum bar on the left

side of the chart, plus the size of the impulse bar, adds up to the total size of the final momentum

bar on the right side. The bar chart we just constructed can now be used to write the component

form of the impulse–momentum equation

1 1 1 1 i x x f xm v J m v& #

The components of the initial and final momentum are positive. As the force is exerted in the

negative direction, the x component of the impulse is negative and equal to 2 on 1F t2 - . Thus

1 1 2 on 1 1 1( )i f

m v F t m v& & 2 - # &

We could now solve this equation for 2 on 1F and determine the magnitude of the average force

that car 2 exerted on car 1 during the collision. Newton’s third law tells us that the average force

that car 1 exerts on car 2 is equal in magnitude and opposite in direction to this.

Different system choices can be used to answer different questions. If we know the

masses and the initial velocities of the cars, and the final velocity for one of the cars, we could

find the final velocity of the other car using the two-object constancy of momentum approach.

Similarly, if we know the mass of one of the cars, and its initial and final velocities, then we can

determine the impulse delivered by the other car on the system car and possibly determine the

magnitude of the force that each car exerts on the other.

Tip! Always specify the reference frame when you draw a bar chart (the object of reference and

the coordinate system). The direction of the bars on the bar chart (up for positive and down for

negative) should match the direction of the momentum or impulse based on the chosen coordinate

system.

Figure 5.10

Example 5.3 Happy and sad balls You have two identical mass and size balls that behave very

differently. When you drop the so-called “sad” ball it thuds on the floor and does not bounce at

all (Fig. 5.10). When you drop the so-called “happy” ball from the same height, it bounces back

to almost the same height from which it was dropped. When you hang each ball from identical

length strings, and let them swing back and forth like pendulums, you see that their motions are


exactly the same, each moving at the same speed with respect to the ground at each moment.

Next, you place a wood board standing on its end in the path of each swinging ball. Which ball

has the best chance of knocking their respective board over—the sad ball or the happy ball?

Sketch and Translate An initial-final sketch of the process is shown in Fig. 5.11a. We choose the

initial moment to be just before a ball hits the board, when moving horizontally toward the left

(the same for each ball). The final moment will be just after the collision with the board has

occurred. We choose each ball separately as a system of interest and analyze each of them.

Quantities corresponding to the happy ball have a subscript ‘H’ and for the sad ball have subscript

‘S’. Finally, we use an x-axis that points left in the direction of the initial velocities of the balls.

Figure 5.11(a)

Simplify and Diagram Assume that the two balls move with the same initial horizontal velocity

iv!

with respect to the ground before hitting the boards. Also assume that the collision time

interval t is about the same for each ball and that the happy ball bounces back at about the same

speed as it had before the collision and that the sad ball does not bounce back at all. We’ll analyze

only the horizontal x-component of the process, the component that is relevant to whether or not

either of the boards is knocked over. The boards exert impulses on their respective balls (in the x-

direction) that cause the momentum of each ball to change. Therefore each ball, according to

Newton’s third law, exerts an impulse of the same magnitude on the board that it hits. A larger

force exerted on the board has more chances to tip the board. A bar chart for each ball-board

collision is shown in Fig. 5.11b.


Figure 5.11(b)

Represent Mathematically The x-component form of the impulse–momentum Eq. (5.5x) applied

to each ball is as follows ( Board on Ball xF is the x-component of the force exerted by the board on

the ball):

Sad ball: S B on S Six x fxmv F t mv& - # or B on S • 0

i xmv F t m& - #

Happy ball: H B on H Hix x fxmv F t mv& - # or B on H (– )

i x imv F t m v& - #

Note that the x- component of the final velocity of the sad ball is S 0fx

v # (it does not bounce),

and that the x-component of the final velocity of the happy ball is H –fx i

v v# .

Solve and Evaluate We can now get an expression for the force exerted by each board on each

ball:

Sad ball: B on S

(0 – )–i i

x

m v mvF

t t# #

- -.

Happy ball: B on H

[(– ) – ] 2–i i i

x

m v v mvF

t t# #

- -.

The board exerts twice as large a force on the happy ball as on the sad ball. This makes

sense because the board is causing the momentum of the happy ball to change by twice as much

as the sad ball’s momentum. According to Newton’s third law, this means the happy ball will

exert twice as large a force on the board than the sad ball will. Thus, the happy ball has a greater

chance of tipping the board.

Try It Yourself: A football player runs into an unpadded goal post. Is it safer for him to bounce

backwards off the goal post or to just hit the goal post and stop?

Answer: Better to hit the goal post and stop. If the football player bounces back off the goal post,

his momentum will have changed by a greater amount (like the happy ball in the last example).

This means the goal post exerts a greater force on him, which means there is a greater chance for

injury.


The pattern we found in the example above is true for all collisions – for an object to

bounce back after the collision there must be a larger magnitude force exerted on it compared

with an object that stops after the collision. For example, if you were designing a bulletproof vest

for law-enforcement agents, you would want the bullet to embed in the vest rather than bounce

off it.

Conceptual Exercise 5.4 Jumping off a skateboard A small child is traveling slowly on her

skateboard. The mass of the skateboard is 0.25 times the mass of the child. Suddenly she jumps

off the skateboard and moves backward at 0.25 times her original speed. Draw an impulse–

momentum bar chart for (a) a system that includes both the child and the skateboard, (b) a system

that includes only the child, and (c) a system that includes only the skateboard.

Sketch and Translate A sketch of the situation shows the child and skateboard initially traveling

in the positive x-direction; then the child changing direction and traveling in the negative x-

direction at 25 percent her original speed while the skateboard continues to move in the positive

direction. (Fig. 5.12a) The velocities are with respect to the ground.

Figure 5.12(a)

(a) Child and skateboard system bar chart (Fig. 5.12b): There are no external forces exerted on

the system in the horizontal direction. The system’s momentum in the initial state is in the

positive direction. The child and the skateboard have the same initial velocity with respect to the

ground; however, the mass of the skateboard is 0.25 times that of the child so the momentum bar

for the initial state of the skateboard is shorter. In the final state, the child is moving backward in

the negative direction at 25 percent of her initial speed; therefore, the momentum bar for her final

state points down and is one-fourth as long as the upward momentum bar for her initial state.

Since the momentum in the system is constant, the final momentum bar of the skateboard must

point upward and be quite long.

Figure 5.12(b)


(b) Child only system bar chart (Fig. 5.12c): In this case, the skateboard is in the environment and

exerts an impulse in the negative direction on the child, causing her momentum to change. In the

final state, she is moving in the negative direction and her final momentum bar points down; the

board must have exerted a negative force and negative impulse on her.

Figure 5.12(c)

(c) Skateboard only system bar chart (Fig. 5.12d): Now the child is in the environment and exerts

a positive impulse on the board, causing its momentum to change. The board continues to move

in the positive direction, thus, the final momentum bar points up.

Figure 5.12(d)

Notice in Fig. 5.12c that the impulse exerted by the board on the child has the same

magnitude but opposite sign of the impulse exerted by the child on the board shown in Fig. 5.12d.

This is consistent with Newton’s third law.

Try It Yourself: You drop a small metal ball from height h above the ground. The initial state is

just before you release the ball, and the final state is just before it touches the ground. Draw an

impulse–momentum bar chart for (1) the ball alone as the system and (2) the ball and the Earth

together as the system.

Answer: Figure 5.13 shows the momentum bar charts. Note that you can choose to have the

positive y-axis point down. In that case, the gravitational force exerted by the Earth on the ball

points in the positive direction as does the downward final velocity of the ball. In each case we’ve

made the simplifying assumption that air resistance does not have a significant effect on the

system.


Figure 5.13

Review Question 5.4

In part (a) of Conceptual Exercise 5.4, we drew a bar chart with no external impulse, and in parts

(b) and (c), we drew bar charts for the same process but with non-zero external impulses. Why

were the bar charts different?

5.5 Impulse–Momentum Problem Solving Skills

Initial–final sketches and bar charts are useful tools to help analyze situations using the

impulse-momentum principle. Let’s investigate further how these tools work together. A general

strategy for analyzing processes using the impulse-momentum principle is shown on the left side

of Table 5.3 and illustrated for the following example on the right side of the table.

Example 5.5 Bullet hits wood block A 0.020-kg bullet traveling horizontally at 250 m/s embeds

in a 1.0-kg block of wood resting on a table. Determine the speed of the bullet and wood block

together immediately after the bullet embeds in the block.


Table 5.3 Problem-solving strategy: Impulse–momentum problems

Sketch and Translate:

! Sketch the initial and final states and include

appropriate coordinate axes. Label the sketches with

the known information. Decide what the object of

reference is.

• Choose a system. If you know all of the masses

and velocities (except one), then the system usually

includes multiple objects and you can use the

constancy of momentum to determine the unknown

quantity. If you are interested in the force exerted on

an object, then the system usually includes that one

object and you use the impulse-momentum equation

to determine the unknown.

The left side of the sketch above shows the bullet

traveling in the positive x-direction with respect to the

ground; it then joins the wood. All motion is along the x-

axis, the object of reference is the Earth. The system

includes the bullet and wood; it is an isolated system since

the vertical forces balance.

Simplify and Diagram:

! Determine if there are any external impulses

exerted on the system. A force diagram could help

determine the external forces and their directions.

! Draw an impulse–momentum bar chart(s) for the

system for the chosen direction(s).

Assume that the friction force exerted by the tabletop on

the bottom of the wood does not change the momentum of

the system during the very short collision time interval.

The initial state is immediately before the collision; the

final state is immediately after. The bar chart represents

the process. The bar for the bullet is shorter than for the

block – their velocities are the same afte the collision but

the mass of the bullet is much smaller.

Represent Mathematically:

! Use the bar chart to construct an equation using

the generalized impulse–momentum equation along

the chosen axis. Each non-zero bar becomes a non-

zero term in the equation. The orientation of the bar

determines the sign in front of the corresponding

term in the equation.

W WB B B B-W B-W0 ( )

xix fx fxm v m J m v m v&& & #!

Since 0x

J # , B B

B WB-W ( + )fx

ixm vv

m m#

Solve and Evaluate

! Insert the known information to determine the

unknown quantity.

! Check if your answer is reasonable with respect

to sign, unit, and magnitude. Also make sure it

applies for limiting cases, such as objects of very

small or very large masses.

B-W

(0.020 kg)(250 m/s)

(0.020 kg + 1.00 kg)5.0 m/s= +

fxv #

The magnitude of the answer seems reasonable given how

fast the bullet was initially traveling. The plus sign

indicates the direction, which makes sense too. The units

are also correct (m/s). An example of a limiting case is a

zero mass or speed for the bullet, in which case the block

remains stationary after the collision.

Try It Yourself: A 0.020-kg bullet is fired horizontally into a 2.00-kg block of wood resting on a

table. Immediately after the bullet joins the block, the block and bullet move in the positive x-

direction at 4.0 m/s. What was the initial speed of the bullet?

Answer: 400 m/s.

pBix pWix Jx pBfx pWfx


Just after the bullet embedded into the wood block in Example 5.6, the block was

traveling at 5.0 m/s in the positive direction. Because the collision time interval was very short

(0.1 s or less) and because the friction force exerted by the surface was not very large compared

to the force of the bullet on the block, the impulse due to friction could be ignored during that

short time interval. Let’s analyze the situation after the bullet embeds in the block and they

slide together along the tabletop.

Example 5.6 Sliding to a stop Imagine that we are observing the motion of the 0.02-kg

bullet and the 1.00-kg block of wood after the bullet enters the wood in Example 5.6.

Immediately after the collision, the bullet and wood block were traveling to the right at 5.0 m/s

until the friction force exerted by the table’s surface causes them to stop. Determine the time

interval needed for the bullet–wood block to stop and how far the bullet-block slides during that

time interval. The coefficient of friction k

5 between the wood block and surface is 0.40.

Sketch and Translate A sketch of the process is shown in Fig. 5.14a. Since the motion is along the

x-direction we will only investigate that component of the process. The initial state is just after

the bullet has embedded in the block. The final state is at the instant the system stops. The system

is the bullet and the wood block (labeled with subscript WB). The time interval ( – )f i

t t that the

system travels is unknown. The object of reference is the Earth, thus all velocities are with respect

to the Earth.

Figure 5.14(a)

Simplify and diagram A force diagram shows that in the y-direction, the normal force

T on WBN!

exerted by the surface on the bullet–wood block system balances the gravitational force

E on WBF!

exerted by Earth on it. (Fig. 5.14b) The table exerts a horizontal friction force T on WBf!

on

the system in the negative x-direction, providing the negative impulse that causes the bullet–wood

ALG

5.4.7-

5.4.17


block system to stop. There are no other horizontal external forces. The situation is represented

with an impulse–momentum bar chart in Fig. 5.14c.

Figure 5.14(b)(c)

Represent mathematically We now use the bar chart to help apply the x-component form of the

generalized impulse–momentum principle for the process.

B W WB T on WB( ) ( )( – ) 0ix x f i

m m v f t t& & #

The friction force points in the negative direction; thus the component of the impulse that it exerts

on the system is negative:

B W WB T on WB( ) (– )( – ) 0i f i

m m v f t t& & #

From the force diagram, we see that the magnitude of the normal force is the same as the

magnitude of the force exerted by Earth on the system, or T on WB B W( )N m m g# & . Thus, the

magnitude of the kinetic friction force exerted on the system is

T on WB T on WB B W( )k k

f N m m g5 5# # & . We then substitute this friction force into the above

equation:

B W WB B W( ) [– ( ) ]( – ) 0i k f i

m m v m m g t t5& & & #

where ( )f i

t t2 is the time interval needed for the system to stop.

We can determine the stopping time interval from the above application of the impulse–

momentum equation. After dividing all parts of the equation by B W( )m m& , we get:

WB – ( )( – ) 0i k f i

v g t t5 #

Solve and Evaluate Next, solve the above equation for the time interval ( – )f i

t t :

WB

2

5.0 m/s= 1.3 s.

(0.40)(9.8 m/s )

i

f i

k

vt t

g52 # #

How far did the block with bullet travel while stopping? Its speed just after the bullet

entered was 5.0 m/s and its final speed was zero. As the force exerted on the system was constant,

the acceleration while it was stopping was constant. The average speed of the wood block with

bullet while stopping was:

average

(5.0 m/s 0)2.5 m/s

2 2

i fv v

v& &

# # # .

The block traveled this average speed for 1.3 s, or a distance of:


average– ( – ) (2.5 m/s)(1.3 s) = 3.2 mf i f i

x x v t t# # .

This stopping distance is reasonable for the block sliding on a surface with friction. Notice that

the stopping time interval and the stopping distance do not depend on the masses of the bullet and

the wood block, only on the initial speed of the system.

Try It Yourself: In Example 5.5, the bullet traveling at 250 m/s stopped when it entered the wood

block. Estimate the time interval for the bullet to stop.

Answer: Suppose the bullet traveled 2 cm (0.02 m) into the wood block. This is its stopping

distance. Using kinematics we can determine the stopping time interval:

. /. /

. / . /

1

2

2 2 0.02 m0.00016 s

0 250 m s

f i f i f i

f i

f i

f i

x x v v t t

x xt t

v v

2 # & 2

24 2 # # #

& &

This is further justification for ignoring the effect of friction in our analysis. This time interval is

so short that friction will exert an extremely small impulse on the system.

Ballistic measuring device

In the previous two examples, we used information about the initial speed of the bullet,

the masses of the bullet and block, and the coefficient of friction between the block and the

surface to determine how far the block would slide before it came to rest. If we had known the

distance the bullet-wood block system slid while stopping, we could have worked those problems

backwards to determine the initial speed of the bullet before hitting the block. This would be

useful since the bullet travels so fast, it is difficult to measure its speed. Variations of this method

are used to, for example, to decide whether or not golf balls conform to the necessary rules. The

balls are hit by the same mechanical launching impulse and allowed to embed in another object.

The ball speeds are determined by measuring the response of the object they embed in.

Summary of strategies used to analyze impulse-momentum processes

! Choose a system based on the quantity you are interested in; for example, a multi-object

isolated system to determine the velocity of an object, or a single-object non-isolated

system to determine an impulse or force.

! Identify the initial and final states of the system. Think with respect to what object you

are defining these states.

! Remember that momentum and impulse are vector quantities, so include the plus or

minus signs of the components based on the chosen coordinate system.

! Include a bar chart to better understand the situation, to help formulate a mathematical

representation of the process, and to evaluate your results.

Stopping time interval from stopping distance

The impulse exerted on an object in the impulse-momentum principle depends on the

time interval that an external force was exerted on the object and on the time interval during


which its momentum changed. During collisions, this time interval is usually very short, one-

tenth of a second or less. Examples include a car colliding with a tree or a wall, a person jumping

and landing on a solid surface, and even a meteorite colliding with the Earth. During this short

time interval, the object of interest travels what is called its stopping distance. If this distance can

be estimated, it is then possible to use it to estimate the time interval for the collision. Suppose

that a car runs into a large tree, and an examination of the car indicates that its body has been

crumpled about 0.5 m in the front. This 0.5 m, the distance that the center of the car traveled from

the beginning of the impact to the end, is the car’s stopping distance. Similarly, the depth of the

hole left by a meteorite provides a rough estimate of its stopping distance when it collided with

the Earth. However, to use the impulse-momentum principle, we need the time interval associated

with the collision, not the stopping distance. Here’s how we can use a known stopping distance to

estimate the stopping time interval.

• First, assume that the acceleration of the object while stopping is constant. In that case

the average velocity of the object while stopping is just the sum of the initial and final

velocities divided by 2: . /average / 2x fx ix

v v v# & .

• Thus, the stopping displacement ( )f i

x x2 and the stopping time interval ( )f i

t t2 are

related by the kinematics equation:

. / . / . /average2

fx ix

f i f i f i

v vx x v t t t t

&2 # 2 # 2

• Rearrange this equation to determine the stopping time interval:

2( – )–

f i

f i

fx ix

x xt t

v v#

&. (5.10)

Equation (5.10) provides a method to convert stopping distance –f i

x x into stopping time

interval –f i

t t . Equation (5.10) can be applied to horizontal or to vertical stopping.

Using the stopping distance to analyze an impulse-momentum process

We will apply the procedure just discussed in order to help estimate the force that a

cushion exerts on a stunt diver while the cushion stops his fall.

Example 5.7 Stopping the fall of a movie stunt diver The record for the highest fall

made by a movie stunt person without a parachute is 71 m (230 ft.), held by 80-kg A.J.

Bakunas. His fall was stopped as he sank about 4.0 m into a large air cushion. His speed

was about 36 m/s (80 mph) when he reached the top of the air cushion. Estimate the

average force that the cushion exerted on his body while stopping him.

Sketch and Translate We’ll focus only on the part of the fall when Bakunas is sinking into the

cushion. The situation is sketched in Fig. 5.15a. We choose Bakunas as the system, and the y-axis

pointing up (we can choose any coordinate system we like). The initial state is just as he touches

ALG

5.4.1


the cushion at position 4.0 mi

y # & , and the final state is when the cushion has stopped him, at

position 0f

y # . All motion is with respect to the Earth. The other information about the process

is given in the figure. Be sure to pay attention to the signs of the quantities (especially the initial

velocity).

Figure 5.15(a)

Simplify and Diagram We model Bakunas as a point-like particle. A force diagram for Bakunas

as he sinks into the cushion is shown in Fig. 5.15b. Since Bakunas’ downward speed decreases,

the cushion must be exerting an upward force on Bakunas of greater magnitude than the

downward force that the Earth exerts on him. Thus, the net force exerted on him points upward,

in the positive y-direction. A qualitative impulse-momentum bar chart for the process is shown in

Fig. 5.15c. Notice that the force diagram helps us identify external objects that exert impulses on

the diver while stopping.

Figure 5.15(b)(c)

Represent Mathematically Since all motion and all of the forces are in the vertical direction, we

use the bar chart to help construct the vertical y-component form of the impulse–momentum

equation [Eq. (5.6y)] to determine the force that the cushion exerts on Bakunas as he sinks into it:


B C on B E on B B( )( – )iy y y f i fy

m v N F t t m v& & # .

Using the force diagram, we see that the y-components of the forces are:

C on B C on B yN N# & and E on B E on B B– –

yF F m g# # where C on B N is the magnitude of the

average normal force that the cushion exerts on Bakunas, the force we are trying to estimate.

Noting that 0fy

v # and substituting the force components into the above equation, we get:

B C on B B B

B C on B B

[( ) (– )]( – ) 0

( – )( – ) 0

iy f i

iy f i

m v N m g t t m

m v N m g t t

& & & #

4 & #.

We can find the time interval that the cushion takes to stop Bakunas using Eq. (5.10) and

noting that 0fy

v # :

2( – )–

0

f i

f i

iy

y yt t

v#

&

Solve and evaluate The stopping time interval while Bakunas sinks 4.0 m into the cushion is:

2(0 – 4.0 m)– 0.22 s

0 (–36 m/s)f i

t t # #&

Solving for C on BN we get:

B

C on B B

(80 kg)(–36 m/s)(80 kg)(9.8 N/kg)

( – ) (0.22 s)

iy

f i

m vN m g

t t

2 2# & # &

13,000 N + 780 N 14,000 N# & !

Wow, that is a huge force!!! Notice that we’ve only included two significant digits since

that’s how many the data had. To reduce the risk of injury, stunt divers practice landing so that

the stopping force that a cushion exerts on them is distributed evenly over their entire body. The

cushions must be deep enough so that they provide a long stopping time interval, therefore a

smaller stopping force. The same strategy is applied to developing airbags and collapsible frames

for automobiles to make them safer for passengers during collisions.

Notice three important points. First, it’s very easy to make sign mistakes. A good way to

avoid these is to have a carefully drawn sketch that includes a coordinate system and labels

showing the values of known physical quantities, including their signs. Second, note that the

impulse due to the Earth’s gravitational force is small in magnitude compared to the impulse

exerted by the air cushion. Lastly, the force exerted by the air cushion would be even greater if

the stopping distance and consequently the stopping time interval were shorter.

Try it yourself: Suppose that the cushion in the last example stopped Bakunas in 1.0 m instead of

4.0 m. What would be the stopping time interval, and the magnitude of the average force of the

cushion on Bakunas?

Answer: The stopping time interval is 0.056 s, and the average stopping force is approximately

50,000 N.


Order-of-magnitude estimate—will bone break?

The strategy that we used in the previous example helps analyze injuries, that might lead

to concussions when the bone of a skull fractures. Laboratory experiments indicate that a bone

can fracture if the compressive force exerted on it per unit area is 1.7 x 108 N/m2. The surface

area of the skull is much smaller than 21 m so square centimeters is a more reasonable unit of

area to use. Since 1 m2 = 1 x 104 cm2, the compressive force per area that some object needs to

exert on a 1 cm2 area of a bone in order to break it is about:

28 2 4 2

4 2

1 m(1.7 x 10 N/m ) 1.7 x 10 N/cm

1 x 10 cm

$ %#' (

) *.

Consider the following collision and try to decide if the rollerblader’s skull will fracture.

.

Example 5.8 Bone fracture estimation A rollerblader is gliding along a smooth path looking to

the side when his forehead runs into and stops when he hits a metal pole across the pathway. Is

there a significant chance that his skull will fracture?

Sketch and Translate The process is sketched in Fig. 5.16a. The initial state is at the instant that

the head initially contacts the pole; the final state is when the head and body have stopped. We

can apply the impulse-momentum principle to estimate the force exerted by the pole on the skull

during the time interval that the person is stopping. The person is the system of interest. We have

been given little information, so we’ll have to make some reasonable estimates of various

quantities in order to make a decision about a possible skull fracture.

Figure 5.16(a)

Simplify and Diagram The bar chart in Fig. 5.16b illustrates the momentum change of the system

and the impulse exerted by the pole that caused the change. The person was initially moving in

the horizontal x-direction with respect to the Earth, and not moving at all after the collision was

over. The pole exerted an impulse in the negative x-direction on the rollerblader. We’ll need to

estimate the following quantities: the mass and speed of a typical rollerblader, the time interval

that is used to stop the person, and the area of contact. Let’s assume that this is a 70-kg

rollerblader moving at 3 m/s (we are not using 3.0 m/s since we’re making a rough estimate). The


person’s body keeps moving forward for a short distance after the bone makes contact with the

pole. The skin indents some during the collision. Because of these two factors, we will assume a

stopping distance of about 10 cm (this is a very uncertain number—perhaps off by one order of

magnitude in either direction). However, with this assumption, we can estimate a stopping time

interval. Finally, we will assume an area of contact of about 4 cm2. All of these numbers have

large uncertainties, which is why this is an estimate.

Figure 5.16(b)

Represent Mathematically We now apply the generalized impulse–momentum principle:

Person Person Pole on Person x Person Person

Person Person Pole on Person x

( )( – ) 0

–

i x f i f x

i x

f i

m v F t t m v

m vF

t t

& # #

24 #

We can use the strategy from the last example to estimate the stopping time interval –f i

t t from

the stopping distance –f i

x x :

2( – )–

f i

f i

fx ix

x xt t

v v#

&,

where ix

v is the initial velocity of the rollerblader and 0fx

v # is his final velocity.

Solve and Evaluate Substituting the estimated initial velocity and the stopping distance into the

above, we get an estimate for the stopping time interval:

2( – ) 2(0.1 m – 0)– 0.067 s.

0 3 m/s

f i

f i

fx ix

x xt t

v v# # #

& &

Since this stopping time interval is an intermediate calculated value, we don’t need to worry

about its number of significant digits. When we complete our estimate though, we will keep just

one significant digit.

We can now insert our estimated values of quantities in the expression for the force

exerted by the pole on the person:

Person Person Pole on Person x

(70 kg)(3 m/s)– – –30,000 N.

( – ) (0.067 s)

ix

f i

m vF

t t# # #

Our estimate of the force per area will be:


2

2

Force 30,000 N10,000 N/cm .

Area 4 cm# 0

Is the person likely to break his skull bone? As noted earlier, the force per area needed to break a

bone is about 1.7 x 104 N/cm2 = 17,000 N/cm2, on the order of magnitude of our estimated 10,000

N/cm2. Our estimate could have been off by at least a factor of ten. It seems there is a reasonable

chance that the skull will be fractured. Watch where you are going.

Try It Yourself: Would slipping in the bathtub and landing on the edge of the tub on your ribs be

more or less dangerous for fracturing a rib than the potential skull fracture in the last example?

Answer: When landing on the edge of the tub, the stopping distance is probably shorter—perhaps

0.01 m instead of 0.1 m—making the fall more dangerous. However, the area of contact is

probably greater, making the fall less dangerous. A more careful analysis would be needed to

determine which of these two competing effects is more significant.

Review question 5.5

As the bullet enters the block in Example 5.6, the block exerts a force on the bullet, causing the

bullet’s speed to decrease to almost zero. Why did we use the idea of momentum constancy to

analyze this situation?

5.6 Putting It All Together

In this section we will apply the impulse-momentum ideas that were developed and used

in this chapter to analyze a variety of interesting phenomena: meteorites colliding with the Earth,

radioactive decay of radon in the lungs, two-dimensional car collisions, launching rockets, and

space travel. We start by analyzing a real meteorite collision with the Earth that occurred about

50,000 years ago.

Choosing a system of interest based on your goals

In the next example we use two separate choices of systems to answer two different

questions about a meteorite collision with the Earth. Pay special attention to when it is best to

choose a two object system and use momentum constancy and when it is best to choose a one-

object system and use the impulse-momentum equation.

Example 5.9 Meteorite Impact Meteor Crater (also called Diablo Canyon Crater shown in Fig.

5.17) is approximately 43 miles east of Flagstaff in northern Arizona and was produced 50,000

years ago by the impact of a 3 x 108-kg meteorite traveling at 1.3 x 104 m/s (29,000 mph). The

crater is about 200 m deep. Estimate the change in the Earth’s velocity as a result of the impact

and also estimate the average force the meteorite exerted on the Earth during the collision.


Figure 5.17

Sketch and Translate An initial-final sketch of the process is

shown in Fig. 5.18a.

As the Earth is the object in whose motion we are interested, we cannot use it as the object of

reference in this problem. Instead we will use an imaginary object that is at the location of the

Earth’s surface just before the meteorite hits it; it is not anchored to the Earth but instead is at rest

with respect to very distant stars. The origin of the coordinate axis is at the point on this

imaginary object where the meteorite hits the Earth, and the axis points in the direction of the

meteorite’s motion with respect to that object. We will choose a different system of interest to

answer each question. The first question concerned the change in velocity of the Earth and was

not concerned about the force exerted by the meteorite on the Earth. To estimate the Earth’s

change in velocity with respect to our imaginary object of reference, we choose the Earth and the

meteorite as the system and use momentum constancy to answer the question. However,

momentum constancy does not provide information about the forces exerted by colliding objects

in a system and cannot be used to answer the second question. To estimate the average force that

the meteorite exerted on the Earth during the collision (and that the Earth exerted on the

meteorite), we will choose the meteorite alone as the system and use the impulse-momentum

equation to answer that question.

Figure 5.18(a)

Simplify and Diagram Assume that the meteorite hits perpendicular to the Earth’s surface

(meaning its velocity before impact is in the positive y-direction), and that the force exerted on it

by Earth was constant. An impulse-momentum bar chart in Fig. 5.18b represents the process for

the first question where we chose the system to be the Earth and the meteorite. We assume that

since the collision time interval is very short, that no other forces (such as the gravitational force


exerted by the Sun) will cause significant impulses on the system. This means the system is

isolated and its momentum will be constant.

Figure 5.18(b)

For the second question where we are interested in the average force exerted on the Earth

by the meteorite, the momentum of the system will not be constant since the Earth exerts a very

large impulse on the meteorite during the collision. We choose the meteorite as the system. The

bar chart in Fig. 5.18c represents this process.

Figure 5.18(c)

Represent mathematically The y-component of the meteorite’s initial velocity is

4

M 1.3 10 m/siy

v # & + . The Earth’s initial velocity is zero (with respect to the object of

reference). The y-component of the meteorite’s final velocity is equal to the Earth’s since the

meteorite embeds into the Earth. The meteorite’s mass is about 3 x 108 kg and the Earth’s mass is

6 x 1024 kg. We use momentum constancy to determine the speeds of the Earth and meteorite

after they join together:

E M M E M M

M M E M M

( • 0 ) [0( – )] ( )iy f i Efy fy

iy Efy fy

m m v t t m v m v

m v m v m v

& & # &

4 # &

Since the Earth and the meteorite have the same final velocities we can simplify this even further:

. /M M E Miy fym v m m v# &

Now for the second question: to estimate the force that the meteorite exerts on the Earth

during the collision. We use the y-component form of the impulse-momentum equation with the

meteorite alone as the system:

M M E on M M M( – ) .iy y f i fy

m v F t t m v& #


The Earth exerts an impulse on the meteorite. According to Newton’s third law this will be equal

in magnitude and opposite in direction to the force we are interested in—the force exerted by the

meteorite on the Earth. The time interval required for the collision (using Eq. (5.9) is:

M M

2( – )–

f i

f i

fy iy

y yt t

v v#

&

Solve and Evaluate To answer the first question, we solve for the final velocity of the Earth and

meteorite together:

. /8

4 13MM 24 8

E M

3 10 kg1.3 10 m s 7 10 m s

6 10 kg 3 10 kgfy iy

mv v

m m

2+# # + # +

& + & +.

This is so slow that it would take the Earth about 50,000 years to travel just 1 m. Since the Earth

is so much more massive than the meteorite, the meteorite impact has extremely little effect on

the Earth’s motion.

We can now use the impulse-momentum equation for the meteorite as the system to

determine the Earth’s force on the meteorite. First, the time interval for the impact is about:

4 –13

M M

2( – ) 2(200 m)– 0.015 s.

(1.3 x 10 m/s + 7 x 10 m/s)

f i

f i

fy iy

y yt t

v v# # #

&

This collision, like most impulsive collisions, was over quickly! Note that we’ve estimated the

displacement of the meteorite to be the depth of the crater. Rearranging the impulse-momentum

equation as applied to the collision, we find the average force exerted by the Earth on the

meteorite:

M M M

E on M

( – )

( – )

fy iy

y

f i

m v vF

t t#

8 –13 414(3 10 kg)(7 10 m/s –1.3 10 m/s)

–3 10 N(0.015 s)

+ + +# # + .

The component of the force exerted by the Earth on the meteorite is negative—it means that the

force points opposite to the meteorite’s initial velocity. According to Newton’s third law, the

force that the meteorite exerts on the Earth is positive and has the same magnitude:

14

M on E +3 10 Ny

F # +

This sounds like a very large force, but since the mass of the Earth is 6 x 1024 kg, this force will

cause an acceleration of about 10 210 m s2

, a very small number.

Try it yourself: Estimate the change in the Earth’s velocity and acceleration if it were hit by a

meteorite traveling at the same speed as in the last example, stopping in the same distance, but

having mass of 6 x 1019 kg instead of 3 x 108 kg.


Answer: About 0.1 m/s. If the change occurred in 0.001 s, the Earth’s acceleration would be about

10 m/s2!

Tip! Notice how the choice of system was motivated by the question being investigated. Always

be thinking about your goal when deciding what your system of interest will be.

An object breaks in parts (radioactive decay)

We will learn in Chapter 29 that the nuclei of some atoms are unstable and spontaneously

break apart into a daughter nucleus and a much smaller particle. One such process is called alpha

decay. The parent nucleus emits a so-called alpha particle (which is a helium nucleus) and turns

into a daughter nucleus that is slightly smaller and lighter than the parent nucleus (the alpha

particle is much smaller than both the parent and daughter nuclei). An example of alpha decay is

depicted in Fig. 5.19 in which a radon nucleus (the parent) decays into a polonium nucleus (the

daughter) and an alpha particle.

Figure 5.19

Radon is a gas that does not bind with other atoms to form molecules (it’s a noble gas). It is

produced in the Earth by a series of decay reactions starting with heavy elements in the soil, such

as uranium. When radon is formed, it diffuses out of the soil possibly entering a home through its

foundation. If inhaled, the radon can undergo radioactive decay inside a person’s lungs. If that

happens, the alpha particle streaks through the lungs causing ions to be formed, which can lead to

mutations that may cause cancer. The body has potent repair mechanisms to defend against this,

but even then it is best to avoid the risk. Home testing kits can detect radon concentrations above

recommended levels. If too high, contractors can take remedial action. In the next example, we

will analyze alpha decay by radon using the idea of momentum constancy.

Example 5.10 Radioactive decay of radon in lungs An inhaled radioactive radon nucleus

resides more or less at rest in a person’s lungs when it undergoes radioactive decay. The radon

nucleus is converted to a less massive daughter polonium nucleus plus a much smaller alpha

particle (see Fig. 5.19). With what speed does the alpha particle move if the daughter polonium

nucleus moves away at 4.0 x 106 m/s relative to the lung tissue? The mass of the polonium

nucleus is 54 times greater than the mass of the alpha particle.


Sketch and Translate An initial-final sketch of the situation is shown in Fig. 5.20a. The problem

statement asks for the velocity of the alpha particle and is not concerned about forces exerted by

the particles on each other. Thus, we choose the system of interest to be the radon nucleus in the

initial state, which converts to the polonium nucleus and the alpha particle in the final state. We

choose the lung as the object of reference with respect to which all motion occurs and coordinate

system with the positive x-axis pointing in the direction of motion of the alpha particle. The initial

velocity of the radon nucleus along the x-axis is 0 and the final velocity component of the

polonium daughter nucleus is 6

Po –4.0 x 10 m/sfx

v # . The final velocity component of the alpha

particle fx

v6 is unknown. If the mass of the alpha particle is m, then the mass of the polonium is

54 m.

Figure 5.20(a)

Simplify and Diagram Assume that the idea momentum constancy applies to isolated systems of

elementary particles. Assume the lungs to be associated with an inertial reference frame. Also

assume that there are no external forces exerted on the system, meaning the system is isolated and

thus its momentum is constant. The impulse-momentum bar chart in Fig. 5.20b represents the

process.

Figure 5.20(b)

Represent Mathematically We use the bar chart together with the impulse-momentum equation to

represent the situation mathematically:


. / . /. /Rn Po Po

Po Po

0 0

0

f i fx fx

fx fx

m t t m v m v

m v m v

6 6

6 6

& 2 # &

4 # &

Solve and Evaluate Rearranging, we get an expression for the final velocity of the alpha particle

in the x-direction:

Po Po– .

fx

fx

m vv

m6

6

#

The x component of the velocity of the alpha particle after radon decay is:

. / 6

Po Po 754 (–4.0 x 10 m/s)

– – 2.2 x 10 m/s.fx

fx

m v mv

m m

66

6 6

# # # &

The sign indicates that the alpha particle is traveling in the positive x-direction opposite the

direction of the polonium. The magnitude of this velocity is huge - about one-tenth of light speed!

As the alpha particle passes through lung tissue at this high speed, it collides with atoms and

molecules along the way, knocking electrons out of these atoms and molecules. The alpha particle

creates a path of ionized atoms and molecules as it streaks through the lung tissue. Supposedly,

about 15,000 cases of lung cancer occur each year due to radon alpha decay in the lungs. It is

important to monitor radon concentrations in homes.

Try it yourself: Francium nuclei undergo radioactive decay either by emitting an alpha particle or

by emitting a beta particle (an electron). The alpha particle is about 8000 times more massive

than a beta particle. If the particles are emitted with the same speed, in which case is the recoil

speed of the daughter nucleus greatest?

Answer: Since the mass of the alpha is much greater than the mass of the beta, and they are

traveling with the same speed, the momentum of the alpha is much greater than the momentum of

the alpha. Therefore, the daughter nucleus would have a greater recoil speed during alpha decay.

Collision in two dimensions

So far, the collisions we investigated occurred along one axis. Often a motor vehicle

accident involves two vehicles traveling along perpendicular paths. For these two-dimensional

collisions, we can still apply the ideas of impulse and momentum, but we will one impulse-

momentum equation for each coordinate axis.

Example 5.11 A 1600-kg Chevy traveling east at 20 m/s collides with a 1300-kg Honda traveling

north at 16 m/s. The cars remain tangled together after the collision. Determine the velocity

(magnitude and direction) of the combined wreck immediately after the collision.

Sketch and Translate A sketch of the situation shows the initial and final situations of the cars.

(Fig. 5.21a) The initial state is just before the collision; the final state is just after the cars collide

and are moving off together. We choose the two cars as the system of interest. The object of

reference is the Earth; positive x-axis points east and the positive y-axis points north.


Figure 5.21(a)

Simplify and Diagram Force diagrams representing the side view for each car just before the

collision are shown in Fig. 5.21b. We assume that the friction force exerted by the road is very

small compared to the force that each car exerts on the other. Thus, the impulse due to surface

friction during the short collision time interval of about 0.1 s can be ignored, and the system can

be considered isolated. Momentum constancy can be applied. Impulse–momentum bar charts for

the x-direction and for the y-direction are shown in Fig. 5.21c. The momentum of the system in

each direction is constant.

Figure 5.21(b)

Figure 5.21(c)

Represent Mathematically Now, convert each momentum bar in the x-component bar chart into a

term in the x-component form of the impulse–momentum equation [Eq. (5.7a)] and each bar in


the y-component bar chart into a term in the y-component form of the impulse–momentum

equation [Eq. (5.7b)]. (A C subscript is for the Chevy and an H subscript for the Honda.) Notice

that the x-component of the final velocity vector is C-H C-H cosf x f

v v 7# and the y-component is

C-H C-H sinf y f

v v 7# :

x-component equation: C C H H C H C-H( ) cos .

i x i x fm v m v m m v 7& # &

y-component equation: C C H H C H C-H( ) sin .

i y i y fm v m v m m v 7& # &

We have two equations and two unknowns (C-H and

fv 7 ). With some careful algebra we can

solve for both unknowns.

Solve and Evaluate

x-component equation:

. / C-H(1600 kg)(20 m/s) (1300 kg) 0 m/s (2900 kg) cos .f

v 7& #

y-component equation:

. / C-H(1600 kg) 0 m/s (1300 kg)(16 m/s) (2900 kg) sin .f

v 7& #

Dividing the left side of the second equation by the left side of the first equation and the right side

the second equation by the right of the first and canceling the (2900 kg) on the top and bottom of

the right side, we get:

(1300 kg)(16 m/s) sintan

(1600 kg)(20 m/s) cos

77

7# # .

Solving the above we find that:

tan 0.657 # .

A 330 angle has a 0.65 tangent. Thus, the cars move off at 330 above the +x axis (the north of the

east direction). We can now use this angle with either the x-component equation or the y-

component equation above to determine the speed of the two vehicles immediately after the

collision. Using the x-component equation, we get:

C-H 0

(1600 kg)(20 m/s)13 m/s.

(2900 kg)cos 33f

v # #

From the y-component equation, we have:

C-H 0

(1300 kg)(16 m/s)13 m/s.

(2900 kg)sin 33f

v # #

The two equations give the same result for the final speed, a good consistency check. For

collisions in which cars lock together like this, police investigators commonly use the lengths of

the skid marks along with the motion of the cars after the collision to determine the initial speeds

of the cars. This allows them to determine if either car was exceeding the speed limit before the

collision.


Try It Yourself: Use a limiting case analysis, and the x- and y-component forms of the impulse–

momentum equation to predict what would happen during the collision if the Chevy had infinite

mass. Is the answer reasonable?

Answer: If we place 8 in Eq. (5.11) in place of the 1600-kg mass of the Chevy, the left side of

the equation becomes zero. Then tan = 0. The Chevy would move straight ahead when hitting

the Honda. In other words, the collision with the Honda would not change the direction of travel

of the Chevy. The result seems reasonable if the mass of the Chevy were very large compared to

the mass of the Honda.

Jet propulsion

In the chapter opening, we said that the interaction between car tires and the road is

necessary for cars to change velocity. Other examples of motion on Earth operate by the same

principle. For example, a ship’s propellers push water backward; in turn, water pushes the ship

forward. Once the ship or car is moving, the external force due to the water or the road has to

balance the opposing friction force or the vehicle’s velocity will change.

What does a rocket push against in empty space to change its velocity? There is nothing

physical to play the role of the road or water. We all know that rockets carry fuel that they ignite

and then eject at high speed out of the exhaust nozzles (see the jet airplane in Fig. 5.22). Could

this burning fuel ejected from the rocket provide the push to change its velocity? This idea does

seem reasonable from the point of view of momentum conservation. Choose the system of

interest to be rocket and fuel together. If the rocket and fuel are at rest before the rocket fires its

engines, then its momentum is zero. If there are no external impulses, then even after the rocket

fires its engines, the momentum of the rocket-fuel system should still be zero. However the

burning fuel is ejected backward at high velocity from the exhaust nozzle and has a backward

momentum. The rocket must now have a non-zero foreward velocity and momentum. Maybe the

momentum of the system is still zero. Let us test the idea that momentum conservation can

quantitatively account for rocket propulsion (see Testing Experiment Table 5.4)

Figure 5.22 A rocket expelling fuel


Testing Experiment Table 5.4 Rocket propulsion

Testing

experiment

Prediction Outcome

You are

traveling

through space

and observe

another rocket

moving next to

you. All of a

sudden you

notice a burst of

burning fuel

that is ejected

from it. Predict

what happens to

that rocket’s

velocity.

Choose the other rocket and its fuel as the

system. Our rocket serves as the object of

reference. The other rocket has zero

velocity in the initial state with respect to

the object of reference. Its final state is

just after it expels fuel backward at high

speed; the rocket in turn gains equal

magnitude of momentum in the forward

direction. The initial-final sketch and

momentum bar chart for the rocket–fuel

system represents this process.

Assuming that fuel is ejected all at once at constant speed, the

speed of the rocket should be

rocket rocket fuel fuel

fuel fuel rocket

rocket

0x x

x

x

m v m v

m vv

m

# &

4 # 2

Here rocketm is the mass of the rocket without the fuel.

We can also choose the rocket alone as the system. The

rocket pushes back on the fuel expelling it backward at high

speed; the fuel in turn pushes forward against the rocket,

exerting an impulse that causes the rocket’s momentum to

increase.

The constancy

of momentum

and the

impulse-

momentum

change ideas

can both

account for the

change in

velocity of a

rocket ship in a

vacuous space.

Conclusion

Independently of the choice of the system, we can now see that when a rocket expels fuel in one

direction, it gains velocity and therefore momentum in the opposite direction. This mechanism of

accelerating a rocket or spaceship is called jet propulsion.


The force exerted by the fuel on a rocket is called thrust. Typical rocket thrusts measure

in mega-newtons (106 N), and exhaust speeds are more than 10 times the speed of sound. Thrust

provides the necessary impulse to change a rocket’s momentum. You can observe the principles

of jet propulsion using a long narrow balloon. Blow up the balloon; then open the valve and

release it. The balloon will shoot away rapidly in the opposite direction of the air streaming out of

the balloon’s valve.

In reality, a rocket burns its fuel gradually rather than in one short burst, thus its mass is

not a constant number but changes gradually. However, the same methods we used in Testing

Experiment Table 5.4, together with some calculus, can be applied to calculate the change in the

rocket’s velocity.

The main idea of the jet propulsion method is that when an object ejects some its mass in

one direction, it accelerates in the opposite direction. This means that the same method that is

used to speed up a rocket can also be used to slow it down. To do this the fuel needs to be ejected

in the same direction that the rocket is traveling. Also, you can become your own jet propulsion

machine by standing on rollerblades and throwing a medicine ball or a heavy book forward or

backward.

Review question 5.6

When a meteorite hits the Earth, the meteorite’s motion apparently disappears completely. How

can we claim that momentum is conserved?


Summary

Words Pictorial and physical representations Mathematical

representations

System: The object(s) of interest.

Interactions between objects within

the system are called internal

interactions. Interactions with

objects outside the system are called

external interactions. An isolated

system is one whose mass is

constant and for which the net

external force exerted on it is zero.

(Sections 5.1–5.2)

Mass conservation: Mass is a

conserved quantity in the universe

but may not be constant in non-

isolated systems. (Section 5.1)

Linear Momentum p!

: A vector

quantity that is the product of an

object’s mass m and velocity v!

.

Linear momentum is a conserved

quantity. (Section 5.2)

p mv#! !

Impulse J!

: Impulse is the product

of the average force AvF!

exerted on

an object during a time interval !t

times that time interval (Section 5.3)

Av– )(

f iJ F t t#! !

a

Generalized impulse–momentum

equation: If the system is isolated,

its momentum is constant. If it is not

isolated, the change in the system’s

momentum equals the sum of the

impulses exerted on the system

during the time interval

( – )f i

t t t- # . (Section 5.4)

1 1 2 2( ...)i i

m v m v& &! !

+ext

tF -"!

1 1 2 2( ...)f f

m v m v# & &! !

x- and y-component forms:

1 1 2 2 Ext ix ix xm v m v F t& & " -

1 1 2 2fx fxm v m v# &

1 1 2 2 Ext yiy iym v m v F t& & " -

1 1 2 2fy fym v m v# &

chapter 5: conservation of linear momentum -...

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