Senior High School
General Physics 1
Quarter 1 - Module 8
Momentum and Collisions
Department of Education ● Republic of the Philippines
General Physics1 -Grade 12
Alternative Delivery Mode Self-Learning Module
Quarter 1 - Module 8: Momentum and Collisions
First Edition, 2020
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Senior High
Senior
HighSchool
School
General Physics 1
Quarter 1 - Module 8
Momentum and Collisions
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Table of Contents
What This Module is About
i
What I Need to Know
i
How to Learn from this Module
i
Icons of this Module
ii
What I Know
iii
Lesson 1: Momentum and Impulse
What I Need to Know
What’s New: Let’s Play I-Mom!
What Is It: Learning Concepts
What’s New: Collect Me!
What’s More: This is the Moment!
What’s New: Reflection
What I Have Learned
What I Can Do: Performance Task and Enrichment Activity
1
1
3
6
6
7
8
8
Lesson 2: Conservation of Momentum
What’s In
What I Need to Know
What’s New: Rocket Balloon
What Is It: Learning Concepts
What’s More: Hit it!
What I Have Learned
What I Can Do: Performance Task
Enrichment Activity
10
10
11
12
15
16
16
17
Lesson 3: Collisions
What’s In
What I Need to Know
What’s New: Count Me In
What Is It: Elastic and Inelastic Collisions
What’s More: Now I Know
What I Have Learned: My Insights
What I Can Do: Photos of the Day
Summary
Assessment: (Post-Test)
Key to Answers
References
18
18
18
19
22
22
23
23
24
26
28
What This Module is About
Momentum is a word that we use every day in a variety of situations. It is often used
by sports enthusiasts and announcers to mean changes in the flow of the game. In soccer,
players must consider an extreme amount of information the moment they set the ball or
themselves into motion. Once a player knows where the ball should go, he/she has to decide
how to get it there. It is the same in basketball, a team that has the “momentum” is the one
about to pull away from its opponent or has to come back from a big score deficit. So, what
exactly is momentum?
In this module, you will learn many things on momentum, impulse, and collisions. You
shall also make use of Newton’s second and third laws to obtain the law of conservation of
linear momentum. The laws of conservation provide essential explanations underlying distinct
physical phenomena. It holds true even when colliding objects become distorted and generate
heat during collisions and when they are moving at angles relative to one another. This module
will help you understand how these concepts relating to momentum and collisions can be
applied in real-life situations.
This module includes these lessons:
●
●
●
Lesson 1 – Momentum and Impulse
Lesson 2 – Conservation of Momentum
Lesson 3 – Collisions
What I Need to Know
At the end of this module, you should be able to:
1. Differentiate center of mass and geometric center (STEM_GP12WE-1h-i-56);
2. Relate the motion of center of mass of a system to the momentum and net external force
acting on the system (STEM_GP12MM1C-lh-57);
3. Relate the momentum, impulse, force, and time of contact in a system (STEM_GP12MM1Clh-58);
4. Compare and contrast elastic and inelastic collisions (STEM_GP12MM1C-li-60);
5. Apply the concept of restitution coefficient in collisions (STEM_GP12MM1C-li-61); and
6. Solve problems involving center of mass, impulse and momentum in context such as but
not limited to, rocket motions, vehicle collisions and ping-pong (STEM_GP12MM1C-li-63)
How to Learn from This Module
Below, are guide steps for you to attain the learning competencies in going about the module.
1. Read the lessons and follow the instructions carefully.
2. Take the pretest to determine how much you know about the content. A multiple-choice test
was provided for you. Be honest.
3. Perform all the activities diligently to help you understand the topic.
4. Take the assessment test (post-test) at the end of the module.
i
Icons of this Module
Here are the Icons used as your guide in every part of the lesson:
What I Need to
Know
This part contains learning objectives that
are set for you to learn as you go along the
module.
What I know
What’s In
This is an assessment as to your level of
knowledge to the subject matter at hand,
meant specifically to gauge prior related
knowledge
This part connects previous lesson with that
of the current one.
What’s New
An introduction of the new lesson through
various activities, before it will be presented
to you
What is It
These are discussions of the activities as a
way to deepen your discovery and understanding of the concept.
What’s More
These are follow-up activities that are intended for you to practice further in order to
master the competencies.
What I Have
Learned
Activities designed to process what you
have learned from the lesson
What I can do
These are tasks that are designed to showcase your skills and knowledge gained, and
applied into real-life concerns and situations.
ii
What I Know
Multiple Choice. Carefully read each item. Choose the letter that matches the best answer.
Write your best answer on a separate sheet of paper.
1. It is a point that represents the average location for the total mass of a system.
A. Vertex
B. Center of mass
C. Geometric center
D. Origin
2. The center of geometry is at the geometric center while the center of mass is at a lower
position. Because ___________
A. the density of the object is not homogenous; the greater mass is located at the lower
portion of the object.
B. the density of the object is homogenous; the greater mass is located at the lower portion
of the object.
C. the density of the object is not homogenous; the lesser mass is located at the lower
portion of the object.
D. the density of the object is homogenous; the lesser mass is located at the lower portion
of the object.
3. During a collision the center of mass remains constant. The velocity of the center of mass
in an isolated system, the total linear momentum does not change. Therefore, the velocity
of the center of mass ___________.
A. changes
B. does not change C. doubles
D. cannot be determined
4. The momentum of a system depends upon the system’s ___________ and ___________.
A. size; shape
B. mass; speed
C. mass; velocity
D. mass; energy
5. The coefficient of restitution (COR) describes the ratio of the difference in velocities before
and after the collision. Which one is correct?
A. A perfectly elastic collision, ek = 0
B. A perfectly inelastic collision, ek = 1
C. A perfectly inelastic collision, ek = 0
D. A perfectly elastic collision, ek = 1
6. The impulse-momentum theorem states that ___________.
A. the impulse on an object is equal to the change in momentum it causes
B. the impulse on an object is less than the change in momentum it causes
C. the impulse on an object is greater than the change in momentum it causes
D. the force on a moving object is equal to the magnitude of the impulse
7.
A. The longer the time the longer the impact force
B. The lesser the time the longer the impact force
C. The lesser the time the lesser the impact force
D. The longer the time the lesser the impact force
8. The Law of Conservation of Momentum states:
A. The total momentum before a collision is less than the total momentum after a collision.
B. The total momentum before a collision is greater than the total momentum after a
collision.
C. The total momentum before a collision is equal to the total momentum after a collision.
D. The total momentum before a collision is not related to the total momentum after a
collision.
iii
9. F1 is the force that m2 exerts on m1 during collision and F2 is the force that m1 exerts on m2
during collision. Additionally, the two forces from m1 and m2 act over the same time interval.
A. the impulse on m1 is equal to and opposite the impulse on m2.
B. the impulse on m1 is greater than equal to and the same direction the impulse on m2.
C. the impulse on m2 is not equal to and opposite the impulse on m1.
D. the impulse on m2 is greater than equal to and the same direction the impulse on m1.
10. The table shows the velocity and momentum of each ball both before and after the
collision. What does this table means in terms of momentum in a collision?
A. The momentum of each ball does not change once they collide and the total
momentum of the two balls together is decreased.
B. The momentum of each ball changes due to the collision but the total momentum of
the two balls together remains constant.
C. The momentum of the balls differs as to their total momentum.
D. The total momentum of both balls is greater compare to the momentum of each ball.
11. Which type of collision is described when two objects stick together and move with
common velocity after colliding?
A. elastic
B. inelastic
C. nearly elastic
D. perfectly inelastic
12. In an inelastic collision between two objects with unequal masses, which of the following
is true?
A. The kinetic energy of one object will increase by the amount that the kinetic energy of
the other object decreases.
B. The momentum of one object will increase by the amount that the momentum of the
other object decreases.
C. The total momentum of the system will increase.
D. The total momentum of the system will decrease.
13. A billiard ball collides with a stationary identical billiard ball in an elastic head-on collision.
After the collision, which of the following is true of the first ball?
A. It comes to rest.
B. It has one-half its initial velocity.
C. It maintains its initial velocity.
D. It moves in the opposite direction.
14. Two objects with different masses collide and bounce back after an elastic collision. Before
the collision, the two objects were moving at velocities equal in magnitude but opposite in
direction. Which will likely happen after the collision?
A. Both objects had the same momentum.
B. Both objects lost momentum
C. The less massive object had gained momentum.
D. The more massive object had gained momentum.
15. A big fish of mass 1.2 kg moving at speed 0.30 m/s to the right swallowed a small fish of
mass 0.15 kg moving at speed 0.90 m/s moving to the left. What will be the new speed of
the big fish (in m/s) after swallowing the small one?
A. 0.15
B. 0.17
C. 0.25
D. 0.60
iv
Lesson
1
Impulse and Momentum
What I Need to Know
Which is more difficult to stop – a baseball hit firmly by a base bat or one that is thrown
gently? How about -the truck or a butterfly moving with the same velocity?
If your answers are the baseball hit by the bat and the truck, then you do recognize
that two quantities affect an object’s tendency to continue in motion at a constant velocity. It’s
the object’s mass (𝑚) and velocity (𝑣⃑).
In this lesson, you are to define and calculate impulse and linear momentum, describe
changes in momentum in terms of force and time, compare the center of mass from the
geometric center, and find the center of mass in an irregular object.
Figure 7.1 (a) A baseball has a large momentum due to its high velocity (v).
(b) A truck has a large momentum due to its large mass (m)
What’s New
Activity 1.1 Let’s Play I-Mom!
Objectives:
1. Find the relationship between impulse and change in momentum.
2. Determine how impact force works with the time of impact if the change in momentum
is constant.
Materials:
2 balls of different masses, 1 raw egg, 1 piece of used fishing net (or saggy
sheet), inclined plane, protractor
1
Procedure:
A. Rolling ball
1. Let your friend roll the ball on an inclined plane placed about 30 o from the
horizontal. Using the palm of your hand, stop the ball at the lower end of the
inclined plane.
2. Repeat procedure 1 using the other ball released from the same position. Be
sure that its velocity is the same as that of the first ball just before it is stopped.
B. Catch the ball
1. Request someone to throw the ball twice going to you.
2. In your first catch, make sure you don’t move your hands backward. For the
second throw move your hands backwards. (Note: Be sure that you use the
same ball and that the force exerted by your classmate in throwing the ball is
the same.)
C. Ball hits the wall
1. Throw a ball horizontally against the concrete wall.
2. Mark the point where the ball first strikes the ground after hitting the wall.
3. From the same position, repeat procedure 1 with greater force.
4. Repeat procedure 2.
D. Egg Drop
1. Using a fishing net, try to catch a raw egg positioned from a height of about 3
meters.
2. Observe what happens.
Guide Questions. Write your answer on a separate sheet of paper.
A. 1. Which ball is harder to stop? Why?
2. What can you do to make the less massive ball harder to stop than the other ball?
3. What factors affect the ease or difficulty in stopping objects in motion?
4. Answer Question # 1 in terms of momentum.
B. 1. Which one has the greater force in stopping the ball, first or second throw?
2. Compare the amount of the first with the second impact force.
3. Which catch has the greater period of time in stopping the ball, first or second?
4. What relationship exists between impact force and duration of time?
C. 1. Which throw did the ball land farther from the wall, first or second?
2. What does longer distance travelled from the wall indicate about the velocity of
the
ball after impact with the wall?
3. Compare the momentum of the 1st and 2nd throw of the ball after it leaves the wall?
4. In which instance is the impact force greater?
D. What factors affect the ease or difficulty in catching the egg?
2
What Is It
Momentum describes an object’s motion
In physics, linear momentum of an object of mass (𝑚) moving with a velocity (𝑣⃑) is
⃑⃑.
defined as the product of the mass and velocity. Momentum is denoted by the symbol 𝒑
Momentum is the tendency of a moving particle/system to continue moving and the difficulty
it encounters in slowing down to rest. A common unit of this vector quantity is kilogram meter
per second (𝑘𝑔. 𝑚/𝑠).
𝑝⃑ = 𝑚𝑣⃑
𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑎𝑠𝑠 𝑥 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
Linear momentum is a vector quantity whose direction is the same as that of velocity
(𝑣⃑). It is sometimes referred to as inertia of a body in motion.
A change in momentum takes Force and Time
Now, consider a system acted upon by a force 𝐹⃑ . Momentum is closely related to force,
in fact, when Newton expressed mathematically his second law of motion, the force will
produce an acceleration (𝑎⃑), he wrote it not as 𝐹⃑ = 𝑚𝑎⃑, but in this form:
∆𝑝⃑
𝐹⃑ =
∆𝑡
𝐹𝑜𝑟𝑐𝑒 =
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
Rearranging the equation, we could find the change in momentum concerning the net
external force and the time interval required to produce this change.
Impulse-Momentum Theorem
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 ( 𝐽⃑) = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚
𝐽⃑ = 𝐹⃑ ∆𝑡 = ∆𝑝⃑
𝑤ℎ𝑒𝑟𝑒
∆𝑝⃑ = 𝑚𝑣
⃑⃑⃑⃑⃑𝑓 − 𝑚𝑣
⃑⃑⃑⃑𝑖
𝑎𝑛𝑑
𝐽⃑ 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑦𝑚𝑏𝑜𝑙 𝑓𝑜𝑟 𝐼𝑚𝑝𝑢𝑙𝑠𝑒
This equation above is called the Impulse-Momentum Theorem. It states that the net
external force, 𝐹⃑ , applied to an object for a certain time interval ∆𝑡, will cause a change in the
object’s momentum equal to the product of the force and the time interval. To simplify, small
force acting for a long time can produce the same change in momentum as large force acting
for a short time.
The expression 𝐹⃑ ∆𝑡 is called Impulse, denoted by 𝐽⃑, of the Force 𝐹⃑ for the time interval
∆𝑡 . We can see that Impulse equals the change in momentum. Impulse is an impelling force
that acts suddenly and produces motion. A common unit is Newton second (Ns).
Increasing Momentum
The equation 𝐹⃑ ∆𝑡 = ∆𝑝⃑ explains why follow-through is important in so many sports
like karate, golf, baseball, billiards, and boxing. Follow through increases the momentum of a
system. In increasing the momentum of an object, increasing the force is a requirement. The
change in momentum is greater so with the impulse when the contact time is increased. Follow
3
through is also important in everyday activities such as pushing the grocery cart or even
moving furniture.
Decreasing Momentum over a Long Time and Increasing Momentum over a Short Time
If you were asked to catch a raw egg with your bare
hands, how would you catch the egg without breaking it? When
particle starts to contact with our hands, generally we draw our
hands backward. Others who move their hands forward ended
up of breaking or crushing the egg. For these instances,
momentum is decreased by the same impulse. The only
difference is how long the egg touches the hand. The longer the
time of contact, the lesser the force applied; and the shorter the
time of contact, the greater force is applied.
This is why air bags and seatbelts are used in a vehicle
as safety devices. These safety devices make the time of
contact (you and the dashboard) longer which lessens the
Fig. 7.2. The air bag
impact force.
reduces the Force of
Impact during accidents
Center of Mass
Center of mass exhibits standard
location of the total mass of the object. For
example, a plumb line (a cord with a weight
attached, used to produce a vertical/reference
line) is said to be in stable equilibrium because
if you push the bob to the side, it will move to
return to its original position once it is released.
It is because when plumb line is pushed to one
side, its center of mass rises. Then it seeks a
most stable position, so gravity tries to pull it
back to its most stable position which is its
original position.
Another representation is when you
Fig. 7.3. The plumb line helps to locate the
carefully balance a ruler vertically on your finger.
center of gravity by (A) hanging the model
That instance has unstable equilibrium
from the first nail position and (B) hanging
condition, so a little force on it made its center of
the model from the second nail position.
mass fall. Therefore, the ruler fell. The center of
Image by Byron Inouye
gravity or mass remains the same when the
object is in neutral equilibrium.
Most of the objects we use and encounter everyday are on stable conditions. Is there
a difference between the center of geometry and center of mass? Yes, there is. Center of
geometry is different from a center of mass. They may be the same if the object has a
homogenous density; otherwise they are different. Figure 7.4 illustrates the center of geometry
does not coincide with the center of mass.
Figure 7.4
mass versus
Center of
center of
geometry
4
The center of geometry is at the geometric center while the center of mass is at a lower
position. The reason is that the density of the object is not homogenous; the greater mass is
positioned at the lower part of the object.
Figure 7.5 Plumb line method used to find the center of mass of an irregular object.
In finding the center of mass of an irregular object, plumb line method (Figure 7.5) is
very useful for systems which can be suspended freely about a point of rotation. The
cardboard shown in figure 7.5 pivots freely around the pin under gravity and reaches a stable
point. The plumb line (hung from the pin) was used to mark a line on the system. After marking
a line, the pin is moved to another location in the cardboard and the procedure is repeated. At
this time, the center of mass is just beneath the intersection point of the two lines.
On the other hand, during collision, the center of mass will remain constant. The
velocity of the center of mass in an isolated system and the total linear momentum does not
change. As a result, the velocity of the location of the total mass of the object does not change.
Sample Problems:
1. A bullet traveling at 500m/s is brought to rest by an impulse of 50Ns. What is the mass of
the bullet?
Given:
Unknown:
𝑣𝑖 = 500𝑚/𝑠
𝑣𝑓 = 0 (brought to rest)
⃑𝐽 = − 50𝑁𝑠 (in the negative direction with the bullet’s)
𝑚 =?
Solution:
𝐽⃑ = 𝐹⃑ ∆𝑡 = ∆𝑝⃑ = 𝑚𝑣
⃑⃑⃑⃑⃑𝑓 − 𝑚𝑣
⃑⃑⃑⃑𝑖
→
𝐽⃑ = 𝑚(𝑣
⃑⃑⃑⃑⃑𝑓 − ⃑⃑⃑⃑)
𝑣𝑖
Rearranging the equation to derive the formula of 𝑚:
𝑚=
𝐽⃑
⃑⃑⃑⃑⃑−𝑣
(𝑣
𝑓 ⃑⃑⃑⃑)
𝑖
2. A 1.3kg ball is coming straight at a 75kg soccer
player at 13m/s who kicks it in the opposite direction
at 22 m/s with an average Force of 1200N. How long
are his foot and the ball in contact?
𝑣𝑖 = 13𝑚/𝑠
𝑣𝑓 = −22𝑚/𝑠
⃑
𝐹 = −1200𝑁
𝑚 = 1.3𝑘𝑔
Note: The negative sign in 𝑣𝑓 and 𝐹⃑ indicates the
westward direction which is negative.
Given:
Unknown:
𝑡 =?
Solution:
𝐹⃑ ∆𝑡 = 𝑚(𝑣
⃑⃑⃑⃑⃑𝑓 − ⃑⃑⃑⃑)
𝑣𝑖
5
=
−50𝑁𝑠
(0−500𝑚/𝑠)
= 0.1𝑘𝑔
∆𝑡 =
𝑚(𝑣
⃑⃑⃑⃑⃑𝑓 − ⃑⃑⃑⃑)
𝑣𝑖
1.3𝑘𝑔[(−22𝑚/𝑠) − (13𝑚/𝑠)]
=
= 0.0379𝑠
−1200𝑁
𝐹⃑
𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑗𝑢𝑠𝑡 𝑢𝑛𝑑𝑒𝑟 40𝑚𝑠
What’s New
Collect Me! Place the 10 pictures on a clear paper and present the output to your teacher. (1
point each correct picture)
Task: Take/capture five (5) different pictures of stable and unstable objects found at home.
What’s More
Activity 1.2 This is the moment!
Word Problems: Write your answers on a separate sheet of paper.
1. A truck has a mass of 2050 kg travelling west with a velocity of 20.5 m/s. What is the
momentum of the truck?
2. A 1250 kg car moving eastward with a velocity of 12 m/s collides with a pole and is brought
to rest in 0.20 s. Find the force of the car during the collision.
3. A player passes a ball of mass 0.38 kg with a velocity of 22.0 m/s due north. If the player is
in contact with the ball for 0.045 s, what is the magnitude of the average force he exerts?
Rubrics
Criteria/Description
Given is complete which means that it has variable/s and magnitude with units
Unknown/required variable/s is/are identified
Equation/s or derivation is correct
Correct substitution of values
Correct final answer
Total
6
Points
2
1
1
2
1
7
What’s New
Reflection: Make a reflection of the question below. The reflection must consist of at least
300 words and would not exceed to 500 words. Write it on a clear sheet of paper.
Should car safety devices be required? Why?
What I Have Learned
From the concepts that you have learned, answer the following questions below.
Please use a clear paper as your answer sheet.
1. Which has more momentum, a huge and heavy truck that is at rest or a small toy car that
is in motion?
2. A moving racing car has momentum. If the racing car moves twice as fast, its momentum
would be __________ as much.
3. Two vehicles, one twice as heavy as the other, moves down a hill at the same time. The
heavier car would have a ________momentum.
What I Can Do
Performance Task:
Goal
Your objective is to construct a container or contraption that can keep an egg unbroken
when dropped from a three-storey building or taller to a concrete pavement.
Role
Now you have understood concepts on momentum and impulse. This time, you will try
to construct a container or contraption that can keep an egg unbroken when dropped from a
three-storey building or taller to a concrete pavement.
Audience
Your target individuals are the local folks, students and stakeholders (if available)
Situation
Upon gathering information and based on what you have understood in the lesson, our
country experiences a number of man-made and natural phenomena. The challenge now is
for you to construct a container or contraption that can keep an egg unbroken when dropped
from a three-storey building or taller to a concrete pavement. This is somehow significant in
times of delivering reliefs to the affected area where roads are not available but only air ways.
7
Product
Construct a container or contraption that can keep an egg unbroken when dropped
from a three-storey building or taller to a concrete pavement. You may use any inanimate and
harmless materials.
Standards
You will be given three (3) days to conceptualize and execute your ideas.
Your output will be assessed in accordance with the following criteria.
Performance Task:
Container or Contraption Construction
CRITERIA
PERCENTAGE
Presentation, Creativity and
40 %
Originality
Durability/Permanence: resist
60 %
stress or force
TOTAL
100 %
Enrichment Activity:
Try dropping a tennis ball and then a basketball from the same height onto a hard
surface. Assess the similarities and differences on how high each ball bounces. Explain your
observations.
If tennis ball and basketball are not available, you may use other balls with different
masses.
8
Lesson
2
Conservation of Momentum
What’s In
. You have learned in lesson 1 the relation of motion of a system to the momentum
and net external force acting on the system. You did activities and gain insights on momentum
that describes an object’s motion and impulse, a change in momentum where it takes force
and time of contact in a system. You have also appreciated how important it is to study
momentum and impulse; and understood the concepts which can be applied in real-life
situations.
So far in this module, we have considered the momentum of only one system at a time.
In this lesson, we will consider the momentum of two or more particles interacting with each
other.
What I Need to Know
Figure 7.6 shows a stationary ball set into motion by a collision with a moving ball. Let
us assume that both balls are on a smooth floor and that neither ball rotates before or after
the collision. Before the collision, the momentum of the ball B is equal to zero because the ball
is at rest. During collision, ball B gains momentum while ball A loses momentum. As it turns
out, the momentum that the ball A loses is exactly equal to the momentum that ball B gains.
Figure 7.6 (a) Before the collision, ball A has momentum pA and ball B has no momentum. (b)
After the collision, ball B gains momentum pB.
In this lesson, you will learn how two systems act upon one another in relation to the
change in momentum of each object, compare the total momentum of two particles before
and after they interact, state and explain the law of conservation of momentum, predict the
final velocities of the systems after collisions, given the initial velocities, and describe how
impact of collision between objects undergoes a period of deformation (change in shape) and
a period of restitution (return to its original shape).
9
What’s New
Activity 2.1 Rocket Balloon
Objective:
Explain how a rocket balloon works and how conservation of momentum explains
rocket motion.
Materials:
balloon (long shape); string (nylon, if available); tape; paper clip
Procedure:
1. Stretch the string over two posts. You can use chairs or iron stands as posts. Make sure
that the string is tight.
2. Inflate the balloon. Twine the open end of the balloon and secure it temporarily with a paper
clip.
3. Tape the straw to the balloon such that it is aligned with the balloon’s opening (see Figure
7.7).
Figure 7.7 Rocket Balloon set up
4. Inside the box, illustrate through diagram that shows momentum vectors of the rocket
balloon and the air.
Guide Questions
Answer the following questions and illustrate your diagram on a separate sheet of paper.
Q1. How do these momentums compare?
Q2. How does the velocity of the air that is pushed out of the rocket compare to the velocity
of the balloon rocket?
10
What Is It
To understand better, let us have this data inside the table. Data were based
on the illustration in figure 7.6. Table 7.1 shows the velocity and momentum of each ball both
before and after the collision. Because of the collision, ball’s momentum changes but the total
momentum of both balls remains constant.
Collision
Period
Mass (kg)
Before
After
0.35
0.35
Table 7.1 Momentum in a collision
Ball A
Velocity
Momentum
Mass (kg)
(m/s)
(kg.m/s)
0.72
0.252
0.35
0
0
0.35
Ball B
Velocity
(m/s)
0
0.72
Momentum
(kg.m/s)
0
0.252
In other words, the momentum of ball A (⃑⃑⃑⃑⃑⃑⃑
𝑝𝐴𝑖 ) plus the momentum of ball B (⃑⃑⃑⃑⃑⃑⃑
𝑝𝐵𝑖 )
before collision is equal to the momentum of ball A (⃑⃑⃑⃑⃑⃑⃑⃑
𝑝𝐴𝑓 ) plus the momentum of ball B (⃑⃑⃑⃑⃑⃑⃑⃑
𝑝𝐵𝑓 )
after collision.
⃑⃑⃑⃑⃑⃑⃑
𝑝𝐴𝑖 + ⃑⃑⃑⃑⃑⃑⃑
𝑝𝐵𝑖 = ⃑⃑⃑⃑⃑⃑⃑⃑
𝑝𝐴𝑓 + ⃑⃑⃑⃑⃑⃑⃑⃑
𝑝𝐵𝑓
This relationship is true for all interactions between isolated systems and is known as
the Law of Conservation of Momentum.
Conservation of Momentum
𝑝𝑠𝑦𝑠 = 𝑝′𝑠𝑦𝑠
𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = 𝑚1 𝑣1𝑓 + 𝑚2 𝑣2𝑓
𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑇𝑜𝑡𝑎𝑙 𝑓𝑖𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
Where:
𝑝𝑠𝑦𝑠 − 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
𝑝′𝑠𝑦𝑠 − 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
In its general form, the law of conservation of momentum can be stated as follows:
“The total momentum of all systems interacting with one another remains constant
regardless of the nature of the forces between the systems.”
Momentum is conserved in collisions
In the example shown in figure 7.6, we found out that both the momentum of ball A
and ball B changes, however, the total momentum of the system before is equal to the total
momentum after collision. This is true for a system with no other external forces involved.
Now, if a third ball exerted a force on either ball A or ball B during collision, the total momentum
of ball A, ball B, and the third ball would remain constant.
For 𝑛 number of colliding objects, the Conservation of Momentum is expressed as:
⃑⃑⃑⃑⃑⃑⃑
𝑝𝐴𝑖 + ⃑⃑⃑⃑⃑⃑⃑
𝑝𝐵𝑖 + ⃑⃑⃑⃑⃑⃑⃑
𝑝𝐶𝑖 + ⋯ + ⃑⃑⃑⃑⃑⃑⃑
𝑝𝑛𝑖 = ⃑⃑⃑⃑⃑⃑⃑⃑
𝑝𝐴𝑓 + ⃑⃑⃑⃑⃑⃑⃑⃑
𝑝𝐵𝑓 + ⃑⃑⃑⃑⃑⃑⃑⃑
𝑝𝐶𝑓 + ⋯ + ⃑⃑⃑⃑⃑⃑⃑⃑
𝑝𝑛𝑓
11
Sample Problem: Conservation of Momentum
A 70 kg traveler, initially at rest in a stationary
40 kg boat, steps out of the boat and landed onto the
dock. If the traveler moves out from the boat with a
velocity of 2.0 m/s going right, what is the final
velocity of the boat?
v1f = 2.0m/s
Diagram
Solution:
The system here is consists of the traveler (object 1) and the boat (object 2).
Given:
𝑚1 = 70𝑘𝑔
𝑣1𝑖 = 0
𝑣2𝑖 = 0
𝑚2 = 40𝑘𝑔
𝑣1𝑓 = +2.0𝑚/𝑠
𝑣2𝑓 =?
Note:
The initial velocity of the traveler and the boat is zero since they are both initially at rest.
The positive sign (+) in 𝑣1𝑓 denotes the positive direction of the traveler which is to the right
Using the Conservation of Momentum equation:
𝑝𝑠𝑦𝑠 = 𝑝′𝑠𝑦𝑠
𝑝1𝑖 + 𝑝2𝑖 = 𝑝1𝑓 + 𝑝2𝑓
𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = 𝑚1 𝑣1𝑓 + 𝑚2 𝑣2𝑓
Solving for 𝑣2𝑓 :
𝑣2𝑓 =
=
𝑚1 𝑣1𝑖 +𝑚2 𝑣2𝑖 −𝑚1 𝑣1𝑓
𝑚2
(70𝑘𝑔)(0) + (40𝑘𝑔)(0) − (70𝑘𝑔)(2.0𝑚/𝑠) 0 + 0 − 140𝑘𝑔. 𝑚/𝑠 −140𝑘𝑔. 𝑚/𝑠
=
=
40𝑘𝑔
40𝑘𝑔
40𝑘𝑔
𝑣2𝑓 = −3.5𝑚/𝑠
The negative sign here indicates that the boat is moving to the left.
Newton’s third law leads to conservation of momentum
Consider two isolated bumper cars, m1 and m2, colliding. Before collision, their
velocities are 𝑣1𝑖 and 𝑣2𝑖 . After collision, bumper cars’ velocities are 𝑣1𝑓 and 𝑣2𝑓 , respectively.
Figure 7.8 Due to collision, the force exerted on each bumper car causes a change in momentum for
each car. The total momentum is the same before and after the collision.
From the Impulse-Momentum Theorem, 𝐹⃑ ∆𝑡 = ∆𝑝⃑, the momentum of each car changes as
follows:
For 𝑚1 :
𝐹1 ∆𝑡 = 𝑚1 𝑣1𝑓 − 𝑚1 𝑣1𝑖 and
For 𝑚2 : 𝐹2 ∆𝑡 = 𝑚2 𝑣2𝑓 − 𝑚2 𝑣2𝑖
F1 is the force that m2 exerts on m1 during collision, and F2 is the force that m1 exerts
on m2 during collision (see figure 7.8). In the collision, both bumper cars exert force on each
12
other. Newton’s third law tells us that the force on m1 is equal to and opposite the force on m2,
𝐹1 = −𝐹2
Moreover, the two forces act over the same time interval, ∆𝑡. Thus, the force m2 exerts
on m1 multiplied by the time interval is equal to the force m1 exerts on m2 multiplied by the time
interval, 𝐹1 ∆𝑡 = −𝐹2 ∆𝑡.
Additionally, in every interaction between two isolated systems, the impulse of the first
system is equal to and opposite of the impulse of the second system. It can be expressed by
this equation.
𝐹1 ∆𝑡 = −𝐹2 ∆𝑡
𝑚1 𝑣1𝑓 − 𝑚1 𝑣1𝑖 = −(𝑚2 𝑣2𝑓 − 𝑚2 𝑣2𝑖 )
The equation means if the momentum of one system increases after collision, then the
momentum of the other system in this case must decrease by an equal amount. By
rearranging this equation would give us the equation for the conservation of momentum.
𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = 𝑚1 𝑣1𝑓 + 𝑚2 𝑣2𝑓
Forces in real collisions are not constant
As you learned in lesson 1, the forces involved in a collision are treated as constants.
In real collisions though, the forces may vary in time in a complex way. Figure 7.9 shows the
forces acting on two bumper cars during collision. In all instances during collision, the forces
are equal and opposite in direction. The magnitudes of forces change throughout the collision;
it could be increasing, reaching the maximum, and then decreasing.
Figure 7.9 The graph shows the force on each bumper car during collision. Though both forces vary
with time, F1 and F2 are always equal in magnitude and opposite in direction.
When solving impulse problems, you should use the average force during collision as
the value for force. In your previous lessons on Motion in One Dimension, you have learned
that the average velocity of an object having a constant acceleration is equal to the constant
velocity required for the object to travel the same displacement in the same time interval. In
the same manner, the average force during collision is equal to the constant force required to
cause the same change in momentum as the real, changing force.
Impact and Coefficients of Restitution
In many sports activities we use objects that collide with one another like a baseball
bat and a pitched ball and a bowling ball and pins. The impact of collision occurs over a very
short period of time. This involves contact forces that would result in rapid changes in
momentum of one or both colliding objects and would undergo a period of deformation
(change in shape) and a period of restitution (return to its original shape).
For perfectly elastic impact, the relative velocities of the two systems after impact
(separation velocities) are the same as their relative velocities before impact (approach
velocities) where the total energy of motion is not changed.
On the other hand, perfectly inelastic impact, the relative velocities of the colliding
systems after impact are less than those before impact, where some of the total energy of
motion is lost (transformed into heat and restitution process).
13
Coefficient of restitution (denoted by e) describes the relationship between the relative
velocities of two bodies before and after an impact. The term restitution describes the act of
restoring something that was lost or stolen, while the term coefficient means a single number
that describes how much.
The Coefficient of Restitution (e) is a variable number with no units, with values ranges
from 0 to 1. An e equal to 1 reflects a perfectly elastic collision, whereas e equal to 0 reflects
a perfectly inelastic collision where the bodies stick together after collision.
(𝑣 ′ 2 − 𝑣 ′1 )
(𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑓𝑡𝑒𝑟 𝑖𝑚𝑝𝑎𝑐𝑡) (𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛)
𝑒=|
|=
=
(𝑣2 − 𝑣1 )
(𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑣𝑒𝑙𝑖𝑐𝑡𝑦 𝑏𝑒𝑓𝑜𝑟𝑒 𝑖𝑚𝑝𝑎𝑐𝑡)
(𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ)
When an object drops on the ground, the coefficient of restitution is
𝑒=
𝑣′1
𝑣1
Since the velocity of the floor before and after impact is zero (𝑣′2 = 𝑣2 = 0).
Other expression of coefficient of restitution (using projectile motion equations)
ℎ𝑏
𝑒=√
ℎ𝑑
where
ℎ𝑑 is the height of drop
ℎ𝑏 is the height of bounce
and
What’s More
Activity 2.2 Hit it!
Objective:
Make qualitative observations of collisions between two systems
Materials:
Two identical coins
One object which is much heavier than the coins
Procedure:
1. Find an area that is best for the activity.
2. Observe what happens to the objects based on the given case.
3. Write your observations and draw diagrams to illustrate. Use a separate sheet to do it.
Case 1: Collision of the balls with same mass (m1 = m2), where one ball is stationary (v1 = 0)
Case 2: Collision of the balls with same mass (m1 = m2), where both have initial velocities of
same magnitude but opposite direction (v1 = -v2)
Case 3: Collision of ball with the heavier object m3, where m3 is stationary (v3 = 0)
14
What I Have Learned
From the concepts that you have learned, answer the following items below. Please use a
clear paper as your answer sheet.
1. An insect collided to a windshield of a fast-moving car. Write TRUE if the statement is
correct and FALSE if otherwise.
______ a. The impact force on the insect and on the car is the same.
______ b. The impulses on the insect and on the car are the same.
______ c. The changes in speed of the insect and of the car are the same.
______ d. The changes in momentum of the insect and of the car are the same.
2. When you ride a bicycle at full speed, which has the greater momentum – you or the
bike? Explain your side.
3. You cannot throw an egg to a wall without breaking it, but you can throw the egg with the
same speed into a saggy sheet without breaking it. Why?
What I Can Do
Performance Task:
Goal
Choose as many sports activities/events that apply and demonstrate conservation of
momentum. In every sport activity/event, explain how momentum is transferred from one
system to another yet the total momentum is conserved. Write your explanations on a long
size bond. Draw your diagrams on the white illustration board.
Standards
You will be given three (3) days to conceptualize and execute your ideas.
Your output will be assessed in accordance with the following criteria.
Performance Task:
Physics Sports – Conservation of momentum
CRITERIA
PERCENTAGE
Presentation and Creativity
20 %
Originality
30 %
Coherency of concepts/insights
50 %
TOTAL
100 %
Enrichment Activity:
About 500 years before Isaac Newton, the Chinese had already known about “recoil.”
They were making rockets! In fact, fireworks were the early precursors of rockets.
Research about how the Chinese developed fireworks and for what purpose these
were used. Research also about present-day fireworks technology.
15
Lesson
3
Collisions
What’s In
From the previous lesson, we have learned in the Law of Conservation of Momentum,
that in the absence of an external force, the total momentum of the system before and after a
collision remains unchanged. What are the types of collisions and how can we distinguish one
from the others?
What I Need to Know
In this lesson, you will be able to :
1. Compare and contrast elastic and inelastic collisions.
2. Solve problems involving momentum in context such as vehicle collisions and ping-pong.
What’s New
Activity 7.3.1 Count Me In
Mark the box of the word/phrases as an application of elastic collisions. Mark  the
inelastic collisions. 1 point each
1. When a dart pierces the dartboard and remains stuck in there.
2. A cue ball bouncing off another ball.
3. A fish eating a smaller one in feeding frenzy computer game.
4. A player hitting a baseball with a bat.
5. A soccer player kicking a soccer ball.
16
What Is It
What is collision? In your daily life you witness many collisions without actually thinking
about them. Collision happens when there is a strong interaction between two objects in a
relatively short time. When two objects collide from each other and no net force exists, the
total momentum of both carts is conserved. In some collisions, two objects collide and stick
together so that they travel together after the impact. In other cases, two objects collide and
bounce so that they move with two different velocities.
Elastic Collisions
In an elastic collision, the total kinetic energy of the system (all the objects that collide)
is conserved. An example of such collision might involve a super-bouncy ball; if you were to
drop it, it would bounce back up to the original height from which it was dropped. Another
elastic collision example may be observed in a game of pool. Watch a moving cue ball hit a
resting pool ball. At impact, the cue ball stops, but transfers all of its momenta to the other
ball, resulting in the hit ball rolling with the initial speed of the cue ball.
Figure 7.10a
Momentum is conserved
𝑝𝑠𝑦𝑠 = 𝑝′𝑠𝑦𝑠
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣′1 + 𝑚2 𝑣′2
Kinetic Energy is conserved
𝐾𝑠𝑦𝑠 = 𝐾′𝑠𝑦𝑠
𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑇𝑜𝑡𝑎𝑙 𝑓𝑖𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
For one-dimensional collision,
For two-dimensional collision,
x-component:
y-component
Elastic Collision
𝐾1 + 𝐾2 = 𝐾′1 + 𝐾′2
𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 = 𝑇𝑜𝑡𝑎𝑙 𝑓𝑖𝑛𝑎𝑙 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦
𝑚1 𝑣⃑1 + 𝑚2 𝑣⃑2 +. . . = 𝑚1 𝑣⃑′1 + 𝑚2 𝑣⃑′2 +. ..
𝑚1 𝑣⃑1𝑥 + 𝑚2 𝑣⃑2𝑥 +. . . = 𝑚1 𝑣⃑′1𝑥 + 𝑚2 𝑣⃑′2𝑥 +. ..
𝑚1 𝑣⃑1𝑦 + 𝑚2 𝑣⃑2𝑦 +. . . = 𝑚1 𝑣⃑′1𝑦 + 𝑚2 𝑣⃑′2𝑦 +. ..
Head-on Elastic Collision – one-dimensional collision where the initial and final relative
velocities have the same magnitude.
Figure 7.10b
Elastic Head-on Collision
The vector sum of the momenta just before the event equals the vector sum of the
momenta just after the event. Thus, the system’s total momentum is conserved.
𝑣1 − 𝑣2 = − (𝑣1 ′ − 𝑣2 ′)
17
Inelastic Collisions
In an inelastic collision, the total momentum of the system is conserved, but the total
kinetic energy of the system is not conserved. Instead, the kinetic energy is transferred to
another kind of energy such as heat or internal energy. A dropped ball of clay demonstrates
an extremely inelastic collision. It does not bounce at all and loses its kinetic energy. Instead,
all the energy goes into deforming the ball into a flat blob.
𝑝⃑𝑠𝑦𝑠 = 𝑝⃑′𝑠𝑦𝑠
and
𝐾𝑠𝑦𝑠 ≠ 𝐾′𝑠𝑦𝑠
Perfectly Inelastic Collision - A special case of inelastic collision, called perfectly inelastic
collision, happens when the colliding particles stick together and
move as one system after the collision. Thus, the objects move
with the same final velocity 𝑣⃑′.
𝑚1 𝑣⃑1 + 𝑚1 𝑣⃑1 +. . . = (𝑚1 + 𝑚1 +. . . )𝑣⃑′
The figures below illustrate perfectly inelastic collisions:
Figure 7.10c
Inelastic Non-Head-on Collision
Figure 7.10d
Inelastic Non-Head-on Collision
In the real world, there are no purely elastic or inelastic collisions. Rubber balls, pool
balls (hitting each other), and ping-pong balls may be assumed extremely elastic, but there is
still some bit of inelasticity in their collisions. If there were not, rubber balls would bounce
forever. The degree to which something is elastic or inelastic is dependent on the material of
the object.
Sample Problems:
1. A 1400 kg car moving westward along CM Recto Avenue
at 35.0 kph collides with a 2800 kg truck that is going
northward across the avenue at 50.0 kph. If the two
vehicles become coupled on collision, what are the
magnitude and direction of their velocity after colliding?
Neglect frictional force between the vehicles’ tires and the
road.
18
Solution:
The given problem is an example of 2-Dimensional Inelastic Collision since the two
vehicles coupled after collision, thus, the calculation should be done by component.
x-component:
𝑝⃑𝑥,𝑠𝑦𝑠 = 𝑝⃑ ′𝑥,𝑠𝑦𝑠
𝑚𝐶 𝑣𝑥𝐶 + 𝑚 𝑇 𝑣𝑥𝑇 = (𝑚𝐶 + 𝑚 𝑇 )𝑣𝑥 ′
(1400 𝑘𝑔)(−35.0 𝑘𝑚/ℎ) + (2800𝑘𝑔) (0) = (1400 𝑘𝑔 + 2800 𝑘𝑔) 𝑣𝑥 ’
𝑣𝑥 ’ = − 11.67 𝑘𝑚/ℎ
Note: The negative sign (-) in −35.0 𝑘𝑚/ℎ denotes that the car is moving westward, in the negative direction.
The value of 𝑣𝑥𝑇 is zero since the truck initially is moving along y. Thus, its x-component is zero.
𝑝⃑𝑦,𝑠𝑦𝑠 = 𝑝⃑ '𝑦,𝑠𝑦𝑠
y-component:
𝑚𝐶 𝑣𝑦𝐶 + 𝑚 𝑇 𝑣𝑦𝑇 = (𝑚𝐶 + 𝑚 𝑇 )𝑣𝑦 ′
(1400𝑘𝑔)(0) + (2800𝑘𝑔)(50.0 𝑘𝑚/ℎ) = (1400 𝑘𝑔 + 2800 𝑘𝑔) 𝑣𝑦 ’
𝑣 ′ 𝑦 = 33.33 𝑘𝑚/ℎ
Note: The value of 𝑣𝑦𝐶 is zero since the car initially is moving along x. Thus, its y-component is zero.
By Pythagorean Theorem:
Magnitude: 𝑣 ' = √(𝑣 ′𝑥 )2 + (𝑣 '𝑦 )2 = √(− 11.67 𝑘𝑚/ℎ)2 + (33.33 𝑘𝑚/ℎ)2 = 35.3 𝑘𝑚/ℎ
Direction:
𝜃 = 𝑡𝑎𝑛−1
𝑣′𝑦
= 𝑡𝑎𝑛−1
𝑣𝑥 ’
Final Answer: 𝑣⃑’ = 35.3
𝑘𝑚
, 70.7𝑜
ℎ
33.33
−11.67
= −70.7𝑜
𝑁 𝑜𝑓 𝑊
2. Proton 1 collides elastically with
proton 2 that is initially at rest.
Proton 1 has an initial speed of
3.50 x 105 m/s and makes a
glancing collision with proton 2,
as shown in the figure. After the
collision, proton 1 moves at an
angle of 37.0˚to the horizontal
axis, and proton 2 deflects at an
angle ∅ to the same axis. Find
the final speeds of the two protons and the angle ∅.
Solution:
The given problem is an example of 2-Dimensional Elastic Collision; thus, the calculation
should be done by component.
From the Conservation of Momentum:
x-component:
y-component:
′
′
𝑝⃑𝑠𝑦𝑠 𝑥 = 𝑝⃑ 𝑠𝑦𝑠 𝑥
𝑚1 𝑣1𝑥 + 𝑚2 𝑣2𝑥 = 𝑚1 𝑣1𝑥 ′ + 𝑚2 𝑣2𝑥 ′
𝑣1𝑥 = 𝑣1 ′ 𝑐𝑜𝑠37° + 𝑣2 ′ 𝑐𝑜𝑠∅
𝑝⃑𝑠𝑦𝑠 𝑦 = 𝑝⃑ 𝑠𝑦𝑠 𝑦
𝑚1 𝑣1𝑦 + 𝑚2 𝑣2𝑦 = 𝑚1 𝑣1𝑦 ′ + 𝑚2 𝑣2𝑦 ′
0 = 𝑣1 ′ 𝑠𝑖𝑛37° − 𝑣2 ′ 𝑠𝑖𝑛∅
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From the Conservation of Kinetic Energy:
1
1
1
1
𝑚1 𝑣1 2 + 𝑚2 𝑣2 2 = 𝑚1 𝑣1 ′2 + 𝑚2 𝑣2 ′2
2
2
2
2
𝑣1 2 = 𝑣1 ′2 + 𝑣2 ′2
Solving the three equations with three unknowns simultaneously gives
𝑣1 ′ = 2.80 𝑥 105
𝑚
𝑠
;
𝑣2 ′ = 2.11 𝑥 105
𝑚
𝑠
;
∅ = 53.0°
What’s More
Activity 7.3.2 Now I Know
A. Show how elastic and inelastic collisions are similar and different using a Venn
diagram. 1 point for each correct answer.
B. A 1500-kg blue car is travelling south, and a 2000-kg red sports car is travelling west.
If the momentum of the system consisting of the two cars is 8000kg∙m/s directed at 6
west of south, what is the speed of each car?
What I Have Learned
Activity 7.1.3 My Insights (Criteria: Critical Thinking-5, Communication-5)
Write an essay about your own experience, reaction and application regarding elastic
and inelastic collisions in three paragraphs.
What I Can Do
Activity 7.3.4 Photos of the Day (Criteria: Critical Thinking-5, Communication 5, ICT-5)
1. Take pictures around you featuring collisions. Choose the best three photos.
2. Find a friend on Facebook Messenger or somebody at home. Share with him for 2
minutes what you have learned about elastic and inelastic collisions.
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3. Encourage the person to ask 2-3 questions about what you have shared. If he has no
questions, you may ask him these:
a. Where can you use the concept and skills I have shared with you today in your
daily life?
b. Why is it important to know Elastic and Inelastic Collisions??
c. How can you develop desirable values and traits in life (i.e. respect, ability to know
right from wrong, scientific reasoning, critical thinking, etc.) with the topic that I
have shared?
Record questions and answers.
4. Show him the pictures that you have taken one at a time and ask him which pictures
describe elastic and inelastic collisions.
5. Send screenshots or submit a related output to your teacher.
Summary
•
Momentum is a vector quantity, which is defined as the product of an object’s mass and
velocity, 𝑝⃑ = 𝑚𝑣⃑
● A net external force applied constantly to a system for a certain time interval will cause a
change in the system’s momentum equal to the product of the force and the time interval,
𝐹⃑ ∆𝑡 = ∆𝑝⃑
● Impulse is the product of the force and the time during which the force acts, 𝐹⃑ ∆𝑡
● In all interactions between isolated objects, momentum is conserved.
● In every interaction between two isolated objects, the change in momentum of the first object
is equal to and opposite the change in momentum of the second object.
● In perfectly inelastic collisions, two objects stick together and move as one after a collision.
Momentum is conserved but kinetic energy is not.
● In inelastic collisions, kinetic energy is converted to internal elastic potential energy when
the objects deformed. Some kinetic energy is also converted to sound energy and internal
energy
● In an elastic collision, two objects return to their original shapes and move away from the
collision separately. Both momentum and kinetic energy are conserved.
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Assessment: (Post-Test)
Directions: Carefully read each item. Choose the letter that matches to the best answer.
Write your best answer on a separate sheet of paper.
1. The impulse-momentum theorem states that ___________.
A. the impulse on an object is equal to the change in momentum it causes
B. the impulse on an object is less than the change in momentum it causes
C. the impulse on an object is greater than the change in momentum it causes
D. the force on a moving object is equal to the magnitude of the impulse
2. Catching the ball without moving the hands and catching it with hands moving backward.
What relationship exists between impact force and duration of time?
A. The longer the time the longer the impact force
B. The lesser the time the longer the impact force
C. The lesser the time the lesser the impact force
D. The longer the time the lesser the impact force
3. The Law of Conservation of Momentum states:
A. The total momentum before a collision is less than the total momentum after a collision.
B. The total momentum before a collision is greater than the total momentum after a
collision.
C. The total momentum before a collision is equal to the total momentum after a collision.
D. The total momentum before a collision is not related to the total momentum after a
collision.
4. F1 is the force that m2 exerts on m1 during collision, and F2 is the force that m1 exerts on m2
during collision. Additionally, the two forces m1 and m2 act over the same time interval, Δt
A. The impulse on m1 is equal to and opposite the impulse on m2.
B. The impulse on m1 is greater than equal to and the same direction the impulse on m2.
C. The impulse on m2 is not equal to and opposite the impulse on m1.
D. The impulse on m2 is greater than equal to and the same direction the impulse on m1.
5. The table shows the velocity and momentum of each ball both before and after the collision.
What does this table mean in terms of momentum in a collision?
Collision
Period
Mass (kg)
Before
After
0.35
0.35
Ball A
Velocity
(m/s)
0.72
0.03
Momentum
(kg.m/s)
0.25
0.01
Mass (kg)
0.35
0.35
Ball B
Velocity
(m/s)
0
0.69
Momentum
(kg.m/s)
0
0.24
A. The momentum of each ball does not change once they collide, and the total
momentum of the two balls together is decreased.
B. The momentum of each ball changes due to the collision, but the total momentum of
the two balls together remains constant.
C. The momentum of the balls differs as to their total momentum.
D. The total momentum of both balls is greater compare to the momentum of each ball.
6. It is a point that represents the average location for the total mass of a system.
A. Vertex
B. Center of mass
C. Geometric center
D. Origin
7. The center of geometry is at the geometric center while the center of mass is at a lower
position. Because the density of the object is___________
A. not homogenous; the greater mass is located at the lower portion of the object.
B. homogenous; the greater mass is located at the lower portion of the object.
C. not homogenous; the lesser mass is located at the lower portion of the object.
D. homogenous; the lesser mass is located at the lower portion of the object.
8. During a collision the center of mass remains constant. The velocity of the center of mass
in an isolated system, the total linear momentum does not change. Therefore, the velocity
of the center of mass ___________.
A. changes
B. does not change C. doubles
D. cannot be determined
9. The momentum of a system depends upon the system’s ___________ and ___________.
A. size; shape
B. mass; speed
C. mass; velocity
D. mass; energy
10. The coefficient of restitution (COR) describes the ratio of the difference in velocities before
and after the collision. Which one is correct?
A. A perfectly elastic collision, ek = 0
B. A perfectly inelastic collision, ek = 1
C. A perfectly inelastic collision, ek = 0
D. A perfectly elastic collision, ek = 1
11. Which type of collision is described when two objects move separately after colliding, and
both the total momentum and total kinetic energy remain constant?
A. elastic
B. nearly elastic
C. inelastic
D. perfectly inelastic
12. What will happen to the colliding bodies in a completely inelastic collision?
A. stick together and move as one body after collision and the initial kinetic energy is greater
than the final kinetic energy
B. bounce away from each other after collision and the initial kinetic energy is equal to the
final kinetic energy
C. both stopped at the collision point and both momentum and total kinetic energy are
conserved
D. maintains its initial velocity after colliding and neither momentum nor KE is conserved
13. In the figure, determine the character of
the collision. The masses of the blocks
and the velocities before and after are
being given. The collision is
A. elastic
B. characterized by an increased in KE
C. inelastic
D. not possible because P is not conserved
14. A 1500-kg car is traveling south, and a 2000-kg SUV is traveling west. The total
momentum of the system consisting of the two cars is 8000 kg m /s directed 60.00 west of
south. The speed of the car in m/s is
A. 2.19
B. 2.67
C. 3.46
D. 4.89
15. A 0.150-kg glider A is moving to the right on a horizontal, frictionless air track with a speed
of 0.80 m/s. It makes a head-on collision with a 0.300-kg glider B which is moving to the
left with a speed of 2.20 m/s. Assume the collision is elastic. What is the magnitude of the
final velocity (in m/s) of glider A?
A. 0.48
B. 1.48
C. 2.48
D. 3.48
Key to Answers
References
Bernido, C. C. & Bernido, MV. C. 2008. Learning Physics as One Nation. Physics Essentials
Portfolio. Clavano Printers.
Caintic, Helen E. 2017. General Physics 1 For Senior High School. C & E Publishing,
Inc.
Ferrer, R. A. & Ungson, S. L. 2010. Physics. Science, Environment, Technology and Society.
Serway, R. A & Faughn, J. S. 2002. PHYSICS.
Young, Hugh. D. (1996). University physics (9th Edition). NY : Addison-Wesley Pub.
Co.
Department of Education. Grade 9 Science Learner’s Material.
Department of Education Central Office. Most Essential Learning Competencies (2020)
Department of Education. Project EASE Physics Module 10: Force and Motion (Learning
Resource Management Development Systems, 2003)
Suarez, V., et. al General Physics 1 Workbook, 8th ed, (Mindanao University of Science and
Technology, Cagayan de Oro City, 2015)
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