Unit 8: Momentum and Collisions

Lesson 8.2
Momentum
Contents
Introduction 1

Learning Objectives 2

Warm Up 2

Learn about It! 3

Momentum and the Second Law of Motion 4
Conservation of Momentum 5
Momentum and Center of Mass 9

Key Points 14

Key Formulas 15

Check Your Understanding 16

Challenge Yourself 18

Bibliography 19

Key to Try It! 19

Unit 8: Momentum and Collisions

Lesson 8.2

Momentum

Introduction
Scientific jargons rarely coincide with their counterparts in ordinary language in an
accurate sense. Thus, physical terms like impulse, tension, and force, among others, convey
different connotations from how they are used in everyday language. The word
momentum, however, is an exception. In layman’s terms, the expression “to gain
momentum” means to achieve more strength and speed in something, which is very closely
related to the scientific meaning that it carries. Take, for instance, the device shown above
called “Newton’s cradle.” Each ball contains an equal mass as with the rest of the balls; pull
away one ball and release it, and the entire system will “gain momentum.” Let’s dive into
the compelling physics behind it in the lesson that follows.

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Learning Objectives DepEd Competencies

● Relate the motion of the center of
In this lesson, you should be able to do the mass of a system to the
momentum and net external force
following:
acting on the system
● Explain the concept of momentum. (STEM_GP12MMIC-Ih-57).
● Solve problems involving
● Determine how momentum affects
momentum in contexts such as,
the motion of a system’s center of but not limited to, rocket motion,
mass. vehicle collisions, ping-pong,
whiplash, and the sliding, rolling,
● Explain how internal and external
and mechanical deformations in
forces affect the momentum in a given vehicle collisions
system. (STEM_GP12MMIC-Ii-63).

● Solve problems related to a system’s

center of mass.

Warm Up
Galilean Cannon 5 minutes

You have seen a “Newton’s cradle,” in which all the balls in the device are of equal masses.
What happens though when a system consists of unequal masses? A “Galilean cannon,” or
balls stacked on top of each other from the heaviest to the lightest, is a great way to
visualize it. In this activity, you will create a similar but simpler version of this device.

Materials
● water-filled balloon
● tennis ball

Procedure

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1. Prepare the materials. Hold the tennis ball on top of the balloon while both are
raised at a certain distance from the table, as shown in Fig. 8.2.1.

Fig. 8.2.1. A two-ball Galilean cannon set-up

2. Then, let go of both balloon and ball. Ensure that they are in contact and are stable
before you drop them. Observe what happens.
3. Answer the guide questions that follow.

Guide Questions
1. What happened to the tennis ball and the balloon when they were dropped?
2. What do you think is responsible for the motion of the tennis ball after it has been
dropped?
3. If momentum is a quantity that pertains to how difficult it will be to stop a moving
object, which between the ball and the balloon had a greater momentum? What do
you think are the factors that affect momentum?

Learn about It!

You have learned that the Second Law of Motion may be expressed
alternatively through the work-energy theorem, which was an extremely useful tool in
approaching energy-related problems. In this lesson, we will take Newton’s Second Law

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once more, and reformulate it according to a different set of principles: those related to
momentum.

What is linear momentum?

Momentum and the Second Law of Motion

Newton’s Second Law suggests that the net force exerted upon an object amounts to
the “time rate of change of the product of the object’s velocity and mass.” This product is
also referred to as the object’s linear momentum (or simply, momentum). It is given by:

Equation 8.2.1

where

is the momentum of an object;

is the object’s mass; and
is the object’s velocity.

Did You Know?

While the lowercase letter p seems like a far-fetched symbol for the
quantity momentum, its designation in fact carries an interesting
history. Because of the apparent confusion that it would have with
the quantity mass if the letter m is used, German physicists first used
the capital letter I since momentum locally translates to der Impuls.
This, however, also turned out to be confusing, since it conflicted
with the symbol for moment of inertia, which was denoted the
same way. The French and the Germans then agreed to use the letter
p, which was eventually adopted in American notation.

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Remember
Momentum (mv), which is a vector quantity, is proportional to both
mass (m) and speed (v). Its direction is also similar to the velocity’s
direction, as illustrated below:

An object’s momentum (measured in ) may also be expressed through its

components, thus, for the velocity components of an object vx, vy, and vz, its corresponding
momentum components are x-momentum (px), y-momentum (py), and z-momentum (pz).
Integrating these with equation 8.2.1 results in the components px = mvx, py = mvy, and pz =

mvz, respectively. Applying it to the equation for the Second Law of Motion
yield the following expression:

Equation 8.2.2

where

is the net force acting on a particle; and

is the rate of change of the object’s momentum.

Conservation of Momentum
As with the Third Law of Motion, momentum has practical applications in situations when
there is an interaction between two or more objects. Consider, for instance, two astronauts

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exerting forces (of equal magnitude and opposite directions) on each other in free space (a
zero-gravity environment), as shown in Fig. 8.2.2. Thus, the change in momentum acting on
the two astronauts are also of equal magnitude and opposite directions. Note that the
change in momentum is also called impulse Δp, which we will be discussing in more detail
in the proceeding lesson.

Fig. 8.2.2. Conservation of momentum between two astronauts in free space

Recall from your previous lessons that internal forces are the ones which objects or
particles exert on each other within a given system. On the contrary, external forces are
those emerging from outside the system’s scope. Fig. 8.2.2 illustrates an example of an
isolated system, where no external forces are involved. Thus, for the action-and-reaction
pair of the two astronauts, the internal forces at play are those exerted by astronaut A on B

and by astronaut B on A . When we apply this principle with the Third

Law of Motion, it can be inferred that and are of equal magnitudes and
opposite directions, thus:

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where the vector signifies the vector sum of all the objects’ momenta , i.e.,
it is the total momentum of the system. This phenomenon may also be observed in the
example in Fig. 8.2.3.

Fig. 8.2.3. Conservation of momentum in a system of two skaters

Remember
Always be mindful of the following considerations with regard to the
Conservation of Momentum:

1. The equation implies that the

total momentum in the system is zero, i.e., the time rate of

the change of is zero.

2. If there are external forces involved, they must join the
internal forces on the left side of the equation

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, thus making the total

momentum not constant.

Keeping in mind the equations and principles noted above, we can thus conclude that the
Conservation of Momentum signifies that the total momentum of a system is constant
when the summation of all the external forces acting upon that system amounts to
zero.

How is momentum conserved?

Did You Know?

The Conservation of Momentum can be observed in the launching
of rockets. Inside the rocket, fuel is burned, thus producing
chemical energy at high temperatures. When this is ejected from
the rocket’s nozzle, an equal and opposite momentum then
enables the rocket to speed right up.

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Momentum and Center of Mass

Recall the system’s center of mass, which you have learned about in the previous lesson, as
the mass-weighted average position of all the parts of a given system. Conservation of
Momentum may also be alternatively expressed using this concept, and the motion of the
center of mass, in turn, can be best understood in terms of the principles of momentum.
Take, for instance, the top view of a wrench in Fig. 8.2.4 as it spins on a horizontal surface
with negligible friction.

Fig. 8.2.4. The motion of the center of mass for an object moving with a constant velocity

The motion of the center of mass, signified by the dotted white lines, can be seen in a
straight line, despite the seemingly erratic movement of the wrench. The total momentum
of a system amounts to the product of the mass and the velocity of the center of mass. This
is given by:

Equation 8.2.3

where
is the total mass of the individual masses in the system;

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is the velocity of the center of mass;

is the individual momentum of a given object; and

is the total momentum of the system

How does momentum affect the motion of a

system’s center of mass?

Let’s Practice!

Example 1
Determine the momentum of a child riding a wagon with a velocity of 1.75 m/s if the
wagon-child system has a mass of 35.5 kg.

Solution
Step 1: Identify what is required in the problem.
You are asked to solve for the momentum of the system.

Step 2: Identify the given in the problem.

The system’s mass, 35.5 kg, and its velocity, 1.75 m/s are given.

Step 3: Write the working equation.

Step 4: Substitute the given values.

Step 5: Find the answer.

The momentum of the child-wagon system is 62.125 kg m/s.

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1 Try It!
Calculate the momentum of a car with a mass of 1000 kg and a velocity of 30.0 m/s.

Example 2
A 1350-kg vehicle’s parking brake has malfunctioned when it reached a momentum of 6 500
kg m/s. What is the vehicle’s velocity?

Solution
Step 1: Identify what is required in the problem.
You are asked to calculate the vehicle’s velocity.

Step 2: Identify the given in the problem.

The vehicle’s momentum, 6 500 kg m/s, along with its mass, 1 350 kg are given.

Step 3: Write the working equation.

Derive from the general equation , as follows:

Step 4: Substitute the given values.

Step 5: Find the answer.

The velocity of the vehicle is 4.81 m/s.

2 Try It!

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A 40-kg bowling ball has produced a momentum amounting to 317.5 kg m/s. What is
its velocity?

Example 3
A dart gun produces a 1.56-kg m/s momentum and an 11.41-mph velocity when it thrusts its
darts. Determine the mass of each dart.

Solution
Step 1: Identify what is required in the problem.
You are asked to identify the mass of the dart.

Step 2: Identify the given in the problem.

The momentum, 1.56 kg m/s, and the dart’s velocity, 11.41 mph are given.

Step 3: Write the working equation.

First, convert 11.41 mph to m/s using the following conversion factors: 1 mile = 1
609.34 m and 1 h = 3 600 s. Thus:

Then, derive from the general equation , as follows:

Step 4: Substitute the given values.

Step 5: Find the answer.

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The mass of each dart is 0.31 kg.

3 Try It!
Suppose that the momentum produced in each dart in Example 3 is now 2.33 kg m/s.
What will be its mass if its velocity is 14.60 mph?

Example 4
A small truck with mass 1 450 kg is moving with a velocity of 3.5 m/s when it hits another
truck at rest of mass 1 215 kg. Considering the trucks got locked together, determine the
final speed of the two trucks.

Solution
Step 1: Identify what is required in the problem.
You are asked to identify the final speed of the two trucks.

Step 2: Identify the given in the problem.

The mass of the first truck, 1 450 kg and its velocity, 3.5 m/s, and the mass of the
second truck, 1 215 kg are given. Its initial velocity is 0 since it is at rest prior to the
collision.

Step 3: Write the working equation.

Apply the conservation of momentum to infer that:

Since , then .

Determine m2 as the combined mass of both trucks since they have been locked
together, thus:

Derive from , as follows:

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Step 4: Substitute the given values.

Step 5: Find the answer.

The final speed of the two trucks is 1.90 m/s.

4 Try It!
Suppose another small truck with mass 1 025 kg is moving with a velocity of 5.7 m/s
when it collides head-on with another truck at rest of mass 1 300 kg. What will be
their final speed?

How do we determine the momentum of a given

system?

Key Points
___________________________________________________________________________________________

● The product of the object’s velocity and mass is also referred to as momentum.
● Momentum (mv), which is a vector quantity, is proportional to both mass (m) and
speed (v). Its direction is also similar to the velocity’s direction.

● An object’s momentum (measured in ) may also be expressed through its

components, thus, for the velocity components of an object vx, vy, and vz, its
corresponding momentum components are x-momentum (px), y-momentum (py),
and z-momentum (pz).

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● The Conservation of Momentum signifies that the total momentum of a system is

constant when the summation of all the external forces acting upon that system
amounts to zero.
● Always be mindful of the following considerations with regard the Conservation of
Momentum:

○ The equation implies that the total momentum

in the system is zero, i.e., the time rate of change of is zero.

○ If there are external forces involved, they must join the internal forces on the

left side of the equation , thus making the total

momentum not constant.
___________________________________________________________________________________________

Key Formulas
___________________________________________________________________________________________

Concept Formula Description

Momentum Use this formula to solve for

the momentum in a given
where: system.
● is the momentum of an
object;
● is the object’s mass;
and
● is the object’s velocity

Total Momentum Use this formula to solve for

the total momentum when
where: either the total mass M and
● is the total mass of the the velocity of the center of
individual masses in the mass or the individual
system; masses and their respective

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● is the velocity of the velocities are given.

center of mass;
● is the individual
momentum of a given
object; and
● is the total momentum
of the system

Check Your Understanding

A. Identify what is being referred to in each statement.

______________ 1. The ______________ is equal to the time rate of change of the

product of the object’s velocity and mass.

______________ 2. Momentum is represented by the symbol ______________.

______________ 3. ______________ are forces which objects or particles exert on

each other within a given system.

______________ 4. The ______________ of all the objects’ momenta is termed as

total momentum.

______________ 5. The direction of an object’s momentum is always

______________ to the velocity’s direction.

______________ 6. The product of the object’s velocity and mass is referred to

as ______________.

______________ 7. A system where only internal forces are at play is also

called an ______________.

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______________ 8. In an isolated system, the sum of the internal forces is

______________.

______________ 9. The total momentum of a system is ______________ when

the summation of all the external forces acting upon that
system amounts to zero.

______________ 10. The SI unit for momentum is ______________ .

B. Answer the questions that follow.

For Items 1-5, determine the momentum of:

1. a 0.45-caliber bullet with m = 0.133 kg hurled from a gun’s muzzle at 950 m/s.
2. a 145-kg hockey player hurrying across a horizontal path at 9.7 m/s.
3. an electron with m = 8.75✕10-31 kg moving at 3.1✕106 m/s.
4. a 7.94 ✕ 105-lb airplane taxiing down a runway with a speed of 1.80 m/s.
5. a school bus with a mass of 3.03✕104 lbs running at 40 km/h.

6. Consider a sedan with m = 2 000 kg. What is its momentum if it is moving with
a speed of 35 m/s?
7. How fast is an 8000-g bowling ball rolling in a horizontal surface if it has a
momentum of 16.0 kg m/s?
8. If a basketball bowls across the floor with a speed of 0.50 m/s and a
momentum of 0.25 kg m/s, what is its mass?
9. A 140-gram tennis ball rolls in a straight line with a speed of 67.11 mph.
Determine its momentum.
10. A 2.20-lb tortoise is crawling across a horizontal line with a velocity of 0.036
km/h. Solve for its momentum.

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Challenge Yourself

Answer the following questions in two or three sentences.

1. How is momentum conserved?

2. Consider a heavy and a light hammer being used to chop a piece of wood with a
wedge, as shown in Fig. 8.2.5. Which material will be more efficient and why?
Suppose all other factors are kept constant for both set-ups.

Fig. 8.2.5. Two hammers and a wedge

3. Consider two push carts with masses m and 2m, respectively, at rest on a horizontal
surface with negligible friction. If you push the first cart for 5 seconds, and the
second one for an equal amount of time and with an equal force, how will the two
total momenta compare? Explain your answer.
4. Infer what will happen to the momentum of raindrops as they fall upon the ground.
Justify your answer.
5. A Volkswagen has accelerated from rest. By which amount has its total momentum
changed relative to the total momentum of the earth?

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Bibliography
Faughn, Jerry S. and Raymond A. Serway. Serway’s College Physics (7th ed). Singapore:
Brooks/Cole, 2006.

Halliday, David, Robert Resnick and Kenneth Krane. Fundamentals of Physics (5th ed). USA:
Wiley, 2002.

Knight, Randall D. Physics for Scientists and Engineers: A Strategic Approach (4th ed). USA:
Pearson Education, 2017.

Serway, Raymond A. and John W. Jewett, Jr. Physics for Scientists and Engineers with Modern
Physics (9th ed). USA: Brooks/Cole, 2014.

Young, Hugh D., Roger A. Freedman, and A. Lewis Ford. Sears and Zemansky’s University
Physics with Modern Physics (13th ed). USA: Pearson Education, 2012.

Key to Try It!

1. 30 000 kg m/s
2. 7.94 m/s
3. 0.36 kg
4. 2.51 m/s

8.2. Momentum 19