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Republic of the Philippines

Department of Education Regional Office IX, Zamboanga

Peninsula

Science Grade 9

Quarter 4 - Module 4 Week 4 Conservation of

Momentum

Name of Learner: _____________________________________ Grade & Section: _____________________________________ Name of School: ______________________________________

Zest for Progress

Zeal of Partnership

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What I Need to Know

This module was designed to help you master the concepts behind how momentum is conserved (S9FE-lVb-37). The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. The lessons are arranged to follow the standard sequence of the course. However, the order in which you read them can be changed to correspond with the textbook you are now using.

After going through this module, you are expected to :

1. Describe how momentum is conserved;

2. Differentiate Elastic Collision from Inelastic Collision;

3. Relate the effects of collisions in real-life situations.

What makes things move? Why do some objects move continuously while some stop suddenly? These might be some of the questions you had in mind but were not answered in last year’s science class.

You have learned in grade 8 that unbalanced forces caused stationary objects to move. According to Newton’s Second Law of Motion, when a net force acts on the body, the body will accelerate in the force direction. The acceleration is directly proportional to the force. The greater the force applied, the larger is its acceleration. So, an external force is required to change the momentum of the body. If there is no external force, then there will be no change in momentum.

You have gained insights in your previous module on the factors that affect momentum and what causes changes in momentum. In this module, you will observe the total momentum before collision is equal to the total momentum after collision and relate the effects of collision in real life situations.

What’s In

Activity 1. The momentum of an object

Learning Intention: To investigate the momentum of an object.

Module 4

CONSERVATION OF MOMENTUM

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Solve for the momentum of the given data below using the formula

p = mv.

Object Mass(kg) Velocity (m/s) Momentum (kg-m/s)

Bird 0.03 18

Basketball Player 100 5

Bullet 0.004 600

Baseball 0.14 30

Frog 0.9 12

Activity 2. Tell Me

Multiple Choices: Read each item carefully. Encircle the letter of the correct item in each question.

1. Momentum is operationally defined as A. Mass x velocity B. Mass x acceleration C. Force x distance D. Mass / velocity

For Nos. 2-4, the given data are: A large truck and a Wigo have a head-on collision. 2. Which vehicle experiences the most significant force of impact? A. A large truck

B. A Wigo C. Both a large truck and a Wigo D. none of the above

3. Which of the vehicle experiences the tremendous impulse? A. A large truck B. A Wigo C. Both a large truck and a Wigo D. none of the above

4. Which vehicle experiences the greatest acceleration.

A. A large truck

B. A Wigo C. Both a large truck and a Wigo D. none of the above

5. A heavy car moving at 30 km/h or a light car moving at 30 km/h, which has more momentum,

A. heavy car B. light car

C. Both have the same momentum D. Cannot be determined.

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What’s New

Activity 3 Balloon Rocket

Objectives: • Describe how a balloon rocket works and how conservation of

momentum explains rocket motion. Materials Needed:

• Balloon (long shape)

• String/nylon (at least 2cm)

• Tape

• Paper clip

• Straw

• Pinch of starch Procedure:

1. Insert the string into the straw before stretching it over two posts. You can use chairs or iron stands as posts. Make sure that the string is taut.

2. Put a pinch of starch inside the balloon before inflating it. Twist the opened end and temporarily secure it with a paper clip.

3. Tape the straw to the balloon such it is aligned with the balloon’s opening (see Figure 1)

Figure 1: Balloon Rocket

Source: Science Learner’s Module – Grade 9

4. Position the balloon at the middle of the string 5. Release the air from the balloon by removing the paper clip and observe

carefully

6. Draw a diagram showing the momentum vectors of your balloon rocket and the air

Answer the following questions:

1. What can you say about the system’s initial momentum before releasing the air from the balloon? _____________________________________________________________________

2. What did you observe after releasing the air from the balloon? _____________________________________________________________________

3. Observed what is the direction of the balloon compared to the direction of the air coming from the balloon you have observed? _____________________________________________________________________

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4. How do their momenta compare after releasing the air?

_____________________________________________________________________ 5. From your answer in Q4, how does the velocity of the air that is pushed out

of the rocket compare to the velocity that of the velocity of the air that is pushed out of the rocket ? _____________________________________________________________________

What Is It

Conservation of momentum According to physics general law, the quantity called momentum that

characterizes motion never changes in an isolated collection of objects; that is, the systems total momentum remains constant. Thus, for an isolated system, the total momentum before interaction equals the total momentum after interactions. This statement is called conservation of momentum.

Momentum is the mass and velocity product and is equivalent to the force required to bring the object to a stop. For an array of several things, the total momentum is the sum of the individual momenta. Momentum is a vector quantity, involving both the direction and the magnitude of motion. The rates of the bodies going in opposite directions can cancel to yield an overall sum of zero.

Source: Science Learner’s Module – Grade 9

Source: Science Learner’s Module – Grade 9

Refer to figure 3 as an example. The two children on skateboards are

initially at rest. As they push each other the boy moves eventually to the right while the girl moves in the opposite direction away from each other. Newton’s Third Law of Motion which states that the force exerted by the girl on the boy and the force that makes the girl move in the other direction are of equal magnitude but opposite direction. The boy and the girl make up a system – a collection of objects that affect one another (Figure 2). No net/unbalanced external force acts on the

Figure3. In this example, the total momentum of

the boy-girl system before pushing is zero. After

pushing, the total momentum of the boy-girl

system is still zero because the momentum of the

girl is of equal magnitude but opposite direction

to the momentum of the boy. Note that the

momentum of the alone is not the same before

and after pushing and the momentum of the girl

alone is not the same before and after pushing.

Figure2. A system is a group of objects that interact

and affect each other. Examples are (a) Bowling

ball pin and (b) two football players.

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boy-girl system, thus, the total momentum of the system does not change (Figure 4). Remember that momentum, like velocity and force, is a vector quantity. The momentum gained by the girl is of equal magnitude but opposite direction to the momentum gained by the boy. In this system, no momentum is gained or lost. We can say that momentum is conserved.

Explanation to Activity 3 Balloon Rocket

Our system which consists of the balloon and the air inside it is stationary so the system’s total momentum is zero. When we let the air inside the balloon out, we noticed that the balloon moved. The force that caused the balloon to move came from the air that was pushed out of it. There was no external force involved.

Thus, the total momentum of the system was conserved and remained zero. If the balloon had its momentum in one direction, the air must have an equal and opposite rate for the total momentum to remain zero. Change in momentum = 0 Total Initial Momentum= Total Final momentum 0 = P balloon + Pair -P balloon = P air

-(mv)balloon = (mv)air

It was observed that the balloon’s mass is larger than the mass of air, the

velocity of the air must be more extensive in magnitude than the speed of the balloon, and must be opposite in direction. Example 1a Two ice skaters stand together as shown in Figure 4. They boy pushes the girl with a velocity of 1.50 m/s. Similarly, the girl pushes the boy in opposite direction. If the boy weighs 784 N and the girl, 490 N, what is the girl’s velocity after they push off? (The ice is considered to be frictionless.)

Source: Science Learner’s Module – Grade 9

Solution: W = mg, thus, m = W/g (Use g = 9.8 m/s2) Mass Velocity Boy 80.00 kg 1.5 m/s Girl 50.0 kg ?

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There is no external force present; hence, the ice where they stand on is considered to be frictionless. The momentum of the boy-girl system is conserved. There is no change in the momentum of the system before and after the push- off. Total Initial Momentum = Total Final Momentum

0 = P boy + P girl -P boy = P girl

-(mv)boy = P girl -(80 kg x 1.5 m/s) =50 kg (vgirl)

-120.0 kg m/s = 50 kg (v girl) -120.0 kg m/s = Vgirl

50.0 kg -2.4 m/s = V girl

So the girl moves with a velocity of 2.4 m/s opposite to the direction of the boy. Example 1b Two ice skaters stand together. They boy pushes the girl with a speed of +0.50 m/s and the girl pushes the boy in opposite direction with a speed of -0.65 m/s. If the mass of the boy is 50 kg, what is the girl’s mass? (Consider the ice to be frictionless.)

Solution:

The momentum of the boy-girl system is conserved. There is no change in the momentum of the system before and after the push off. Total Initial Momentum = Total Final Momentum

0 = P boy + P girl -P boy = P girl -(mv)boy = (mv)girl -50 kg (+0.50 m/s) = (m girl)-0.65 m/s - 25.5 kg m/s = (m girl)- 0.65 m/s -25.5 kg m/s = m girl -0.65 kg m/s 39.23kg = m girl

Try solving this… Two carts are stationary at the beginning. If the cart having 10kg mass

starts to move to the east with a velocity of 5m/s, find the cart’s speed having mass 4kg with respect to the ground.

Elastic and Inelastic Collisions

A collision is an interaction between two objects resulting in the exchange

of impulse and momentum. The time of impact is usually small; the impulse provided by external forces like friction during this time is negligible. When two or

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more bodies collide, the momentum of the system is therefore approximately conserved. In an isolated system, the total momentum before the collision is equal to the total momentum of the system after the collision.

total momentum before collision = total momentum after collision There are two types of collisions and are categorized according to whether

the system’s total kinetic energy changes. It may or may not be conserved depending on the type of collision. It may lose during collisions when (1) it is converted to heat or other forms like binding energy, sound, light (if there is spark), etc. and (2) it is spent in producing deformation or damage, such as when two cars collide. The two types of collision are:

1. Elastic collision – states that the system’s total kinetic energy does not

change and colliding objects bounce off after collision. (no kinetic energy is loss, no damage, no heat) Examples:

- Motion of atoms and molecules

- Hitting billiard balls

- When a soccer player kicks a ball since the player’s foot and the ball do indeed remain completely separate after collision.

Equation: total momentum before collision = total momentum after collision (m1v1 + m2v2)before = (m1v1+ m2v2)after

Source: Science Learner’s Module – Grade 9

2. Inelastic collision – states that the system’s total kinetic energy changes (i.e., converted to some other form of energy). Collision is said to be perfectly inelastic if after collision the objects stick together and move as one mass with one velocity. Examples:

- Celestial bodies collide, like two asteroids, they fuse together to form a larger body.

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Equation: Total momentum before collision= total momentum after collision (m1v1 + m2v2) before = (m1+m2)v after

Source: Science Learner’s Module – Grade 9

Can you identify which type of collision is shown in each situation?

Source: Science Learner’s Module – Grade 9

Figure 7a shows a head on collision of a moving steel pendulum in another

steel ball. The collision is elastic, that is; the system’s total kinetic energy (2 steel balls) is the same before and after the collision. The system’s total momentum before the collision is equal to the product of the first ball’s mass and velocity. The total momentum of the system after the collision must be equal to the total momentum before the collision. The first ball comes to rest while the second ball moves away with a velocity equal to the velocity of the first ball. This is the case when the two balls have equal masses. The momentum of the first ball in (Figure 7a) is transferred to the second ball. The first ball loses its momentum while the second ball gains momentum equal to that of the first ball’s momentum. What do you think would happen if you pull two balls away and release them at the same time? Why is it so? Example 2 A 300 g bicycle moves on an air track at 1.2 m/s. It collides with and sticks to another bike of mass 500 g, which was stationary before collision. What is the velocity of the combined bikes after collision?

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Solution Convert the unit of mass in grams to kg. (1kg = 1000g)

Mass Velocity

(before collision) bicycle 1 0.30 kg 1.2 m/s

bicycle 2 0.50 kg 0 The total momentum of the system is conserved before and after the collision. Total Momentum (before collision)= Total Momentum(after collision) (Pbicycle1+ Pbicycle2)before = (Pbicycle1+ Pbicycle2)after

(mv)1, before+ 0 = (m1 + m2) Vafter 0.36 kg m/s = (0.80 kg.) Vafter

0.45m/s = Vafter

Since the two bicycles stuck together after collision, they have the same velocity after collision. The combined bikes move at 0.45 m/s after the collision

What’s More

Activity 4: Solve the Problems

Directions:

1. Solve the following problems.

2. Write the given facts and what is required in the problem.

3. Write the solutions on the space provided below the problem.

1. A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely opposite directions participate in a head on crash. If we consider this event to be a perfectly inelastic collision, what is the speed and direction of the cars after collision?

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2. A pool player is about to use the cue ball to hit the eight ball directly at rest. Each ball has a mass of 170.0 grams and the cue ball’s initial speed is 6.00 m/s. After the collision, the cue ball comes to a stop. If no momentum is lost in this collision, what is the total momentum of this system and how fast is the eight ball?

Remember that momentum is equal to mass x velocity and the unit is kg.m/s. To be consistent, convert the mass to kilograms.

What I Have Learned

Activity 5: Calculate Me

Problem Solving:

Direction: Fill out the blanks and calculate what is asked in the problem. Show computations on the space provided below the problem.

1. A 250 g grocery cart moves on air track at 1.5 m/s. It collides with and sticks to another grocery cart of mass 450 g, which was stationary before collision. What is the velocity of the combined carts after collision?

Mass Velocity Grocery Cart 1 _____ _______ Grocery Cart 2 _____ _______

Activity 6: Complete me!

Direction: Complete the statements below with the correct words.

1. Conservation of momentum states that the total momentum before collision is________ to the total momentum after collision.

2. In elastic collision, colliding objects______________ off after collision.

3. Bodies that stick together after collision is said to be _____________collision.

4. When momentum is neither gained nor lost, we say that momentum is _______________.

5. Equation of inelastic collision is ( m1v1 + m2v2) before = (___________)v after.

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What I Can Do

Activity 7: Fill Me Up

I. Directions: Express your understanding of momentum conservation by filling in the tables below. 1. A baseball player holds a bat loosely and bunts a ball.

2. A Tomahawk cruise missile is launched from the barrel of a mobile missile launcher. Neglect friction.

Assessment

TEST I. Multiple Choice Directions. Choose the letter of the correct answer. Write the letter on the space provided before each number. ________1.If the momentum lost by one object is gained by another object, then the total momentum is?

A.zero B.constant

C.change D.doubled

________2. For any collision occurring in an isolated system, what happens to the momentum?

A.zero B.lost

C.gained D.conserved

________3. Equation of Conservation of Momentum A. P = m x v B. Ft= ∆P

C. (P1 + P2) before = (P1 + P2)after D. P = m/v

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a c

b

a

b

c

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________4. It is one in which the total kinetic energy of the system does not change and colliding objects bounce off after collision

A. Impulse B. Momentum

C. Elastic Collision D. Inelastic Collision

________5. It is one in which the total kinetic energy of the system changes. Objects stick together after collision.

A. Impulse B. Momentum

C.Elastic Collision D.Inelastic Collision

________6. Which is a necessary condition for the total momentum of a system to be conserved?

A. Kinetic energy must not change.

B. No external force is present. C. An object must be at rest. D. Only the force of gravity acts on the system.

For numbers 7 and 8: Two 0.5 kg balls approach each other with the same speed of 1.0 m /s. _______7. What is the system’s total momentum of the before collision?

A. 0 kg m/s B. 0.50 kg m/s

C. 1.0 kg m/s D. -1.0 kg m/s

_______8. What is the total momentum of the system after collision If there is no external force acting on the system?

A.0 kg m/s B.0.50 kg m/

C. 1.0 kg m/s D. -1.0 kg m/s

_______9. Two billiard balls approach each other with the same speed and collide in a perfectly elastic collision, what would be their velocities after collision?

A. Zero B. Same in magnitude and direction. C. Same in magnitude but opposite in direction D. Different in magnitude and opposite in direction

______10. A 50-kg trainee for spaceflight ejects 100 g of gas from his propulsion pistol at a velocity of 50 m/s. What is his resulting velocity?

A. -0.10 m/s B. -0.50 m/s C. 0 m/s D. -100 m/s

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II. Answer the question (5 pts)

1. When fighting fires, a firefighter must use great caution to hold a hose

that emits large amounts of water at high speeds. Why would such a

task be difficult?

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Answer Key – Gr9Q4W4 Science

What’s In

Activity 1

1.0.54

2.500 3.2.40

4.4.20

5.10.80

Activity 2

1.A

2.A

3.A

4.B

5.A

What’s More

1.Given:

M1=1300kg v1=20m/s

M2=900 kg v2=15 m/s

Find: vafter

M1V1 + M2V2 = (m1 + m2)V after

1300kg(20m/s)+900kg(-15m/s) = (1300kg+900kg)V after

12,500 kgm/s = (2200 kg)Vafter

Vf=5.68 m/s in the direction of the 1st car’s motion

2.Given: M1=170g v1=6m/s before v1after

M2=170g v2=0 Find: Vafter

(M1V1+m2V2) = ( m1V1 + m2V2)after

0.170kg(6m/s) + 0.170kg(0) =0.170kg(0) +0.170kgV2

1.02kgm/s =0.170kgV2after 6m/s=V2after

What’s New

Students answer may vary based on the experiment

What I Have Learned

Activity 5

Cart 1 mass=0.250kg velocity=1.5m/s

Cart 2 mass=0.450kg velocity= ?

Total momentum before collision=Total momentum after collision

P1 + P2 = (P1 + P2)after

M1v1 + m2v2 = (m1+m2)V after

0.250kg(1.5m/s) + 0.450kg(0) =0.250kg+0.450kg)Vafter

0.375kgm/s = 0.70kgVafter

0.54m/s=Vafter

Activity 6: Complete Me

1.Equal 3. Inelastic 5. M1 + M2

2.Bounce 4. Conserved

What I Can Do

1.

A.+40 (add the momentum of the bat & the ball)

B.30 ( the bat must have 30 units of momentum in order for the total

to be +40)

C.+40(total momentum the same as before) 2.

A.0 (add the momentum of the missile and the launcher)

B.-5000 ( the launcher must have -5000 units of momentum in order

for the total to be + 0)

C.0 (the total momenyum is the same after as it is before collision)

Assessment

II. The hose is pushing lots of water (large mass)forward at a high

speed. This means the water has a large forward momentum. In turn, the

hose must have an equally large backwards momentum, making it difficult

for the fight fighters to manage.

Assessment

I.

1.B 2.D

3.C

4.C

5.D

6.B 7.A

8.A

9.C

10.A

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References

Books: Liza Alvares, Dave G. Angeles, Hernan L. Apurada, Ma. Pilar P.

Carmona, Oliver A. Lahorra, Judith F. Marcaida, Marivic S. Rosales, Ma. Teresa B. Delos Santos. Science Learning Guide Grade-9. First Edition Bookstore, Inc., Pasig City

Department of Education- Instructional Materials Council Secretariat (Deped IMCS). 2015

Electronic Resources:

https://www.forestville.com/cms/lib/NY19000591/Centricity/Domain/38

/6.2_Conservation_Momentum_Lecture_worksheet.pdf.January10/2021

https://www.physicstutorials.org/home/impulsemomentum/conservati

on-of-momentum.January.21/2021

https://www.physicsclassroom.com/class/momentum/Lesson2/Momen

tum-ConservationPrinciple.january .21/2021

https://www.physicsclassroom.com/Class/momentum/U4L2bc

fm#A6.january.25/2021

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