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Motion. Unit 1: Motion Chapter 4: Conservation Laws. 4.1 Newton’s Third Law and Momentum 4.2 Energy and the Conservation of Energy 4.3 Collisions. Key Question : What happens when equal and opposite forces are exerted on a pair of  Energy Cars?. - PowerPoint PPT Presentation

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Page 1: Motion

Motion

Page 2: Motion

Unit 1: MotionChapter 4: Conservation Laws

4.1 Newton’s Third Law and Momentum

4.2 Energy and the Conservation of Energy

4.3 Collisions

Page 3: Motion

4.1 Investigation: Conservation of Energy

Key Question: What happens when equal and

opposite forces are exerted on a pair of Energy Cars?

Objectives: Explain the meaning of action–reaction forces. Apply knowledge of Newton’s first and second laws

to explain the resulting force when objects experience equal and opposite forces.

Use different numbers of marbles in each Energy Car to see how motion is affected.

Page 4: Motion

Newton’s Third Law Newton’s Third Law (action-

reaction) applies when a force is placed on any object, such as a book on a table.

The second law says the acceleration of an object is directly proportional to force and inversely proportional to the mass (a = F ÷ m).

Page 5: Motion

Newton’s Third Law Forces always come in pairs. Since the action and reaction forces are

equal in strength, Newton accelerates more because his mass is smaller.

Page 6: Motion

The Third Law: Action/Reaction Newton’s Third Law

states that every action force creates a reaction force that is equal in strength and opposite in direction.

There can never be a single force, alone, without its action-reaction partner.

Page 7: Motion

Action and reaction When sorting out

action and reaction forces it is helpful to examine or draw diagrams.

Here the action force is on the ________________, and the reaction force is on the _______________.

Page 8: Motion

Action and reaction forces don’t cancelThe forces do not cancel because they act on different objects.

Page 9: Motion
Page 10: Motion

A woman with a weight of 500 newtons is sitting on a chair. Describe one action-reaction pair of forces in this situation.

Action and reaction

1. Looking for: You are asked for a pair of action and reaction forces.

2. Given: You are given Fw = 500 N.3. Relationships: Action-reaction forces are equal

and opposite, and act on different objects.4. Solution: The downward contact force the

woman exerts on the chair is an action. This action force happens to equal her weight, which is 500 N. The reaction force is the chair acting on the woman with an upward force of 500 N. Fw = -500 N

Fc = 500 N

Page 11: Motion

MomentumMomentum is the mass of a object times its

velocity.The units for momentum are

kilogram•meters per second (kg·m/s).

Page 12: Motion

Momentum

Stopping a fast-moving object is harder than stopping a slow-moving one.

Page 13: Motion

Momentum Newton’s third law tells us that any time two

objects hit each other, they exert equal and opposite forces on each other.

The effect of the force is not always the same.

Page 14: Motion

Calculating momentumMomentum is

calculated with velocity instead of speed because the direction of momentum is always important.

Page 15: Motion

ImpulseThe change in momentum is often referred to

as impulse.We can use this relationship to solve equations

using changes in mass or velocity.

Page 16: Motion

ImpulseThe tennis ball’s

momentum changes after it is hit by the racquet with a force of 80 N for 0.1 seconds.

Both the change of momentum and the impulse are +8 N·s.

Page 17: Motion

A net force of 100 N is applied for 5 seconds to a 10-kg car that is initially at rest. What is the speed of the car at the end of the 5 s?

Force and momentum

1. Looking for: … the speed.2. Given: … the net force (100N), the time the force acts

(5s), and the mass of the car (10 kg).3. Relationships: Use: impulse (Δp) = force (F) × time (t)

momentum = mass (m) × velocity (v)4. Solution: Δp = 100 N × 5 s = 500 kg·m/s

Velocity is momentum divided by mass, orv = (500 kg·m/s) ÷ 10 kg = 50 m/s

Page 18: Motion

Momentum and the third law If we combine Newton’s third law with the

relationship between force and momentum, the result is a powerful new tool for understanding motion.

If you stand on a skateboard and throw a 5 kg ball, with a velocity of 4 m/s, you apply a force to the ball.

That force changes the momentum of the ball. If the ball gains +20 kg·m/s of forward momentum,

you must gain –20 kg·m/s of backward momentum assuming there is no friction.

Page 19: Motion

MomentumThe result of throwing

a 5-kg ball at a speed of 4 m/sec is that person and the skateboard with a total mass of 40 kg move backward at a speed of -0.5 m/sec (if you ignore friction).

We use positive and negative numbers to show opposite directions.

Page 20: Motion

Law of momentum conservationThe law says the

total momentum in a system of interacting objects cannot change as long as all forces act only between the objects in the system.

Page 21: Motion

An astronaut floating in space throws a 2-kilogram hammer to the left at 15 m/s. If the astronaut’s mass is 60 kilograms, how fast does the astronaut move to the right after throwing the hammer?

Using the momentum relationship

1. Looking for: …the velocity of the astronaut after throwing the hammer.

2. Given: … the mass of the hammer (2 kg) and the velocity of the hammer (15 m/s) and the mass of the astronaut (60 kg).

3. Relationships: The total momentum before the hammer is thrown must be the same as the total momentum after it is thrown.

momentum (p) = mass (m) × velocity (v)

A negative sign indicates the velocity is to the left.

Page 22: Motion

Using the momentum relationship4. Solution: Both the astronaut and hammer were initially at

rest, so the initial momentum was zero. Use subscripts a and h to distinguish between the astronaut and the hammer.

momentum after + momentum before = 0mava + mhvh = 0

— Substitute the known quantities:(60 kg)(va) + (2 kg)(–15 m/s) = 0

— Solve:(60 kg)(va) = +30 kg·m/s

va = +0.5 m/s— The astronaut moves to the right at a velocity of 0.5 m/s.

Page 23: Motion

Unit 1: MotionChapter 4: Conservation Laws

4.1 Newton’s Third Law and Momentum

4.2 Energy and the Conservation of Energy

4.3 Collisions

Page 24: Motion

4.2 Investigation: Conservation of Energy

Key Question: How can you predict the

maximum velocity of a pendulum?

Objectives: Describe the relationship between potential energy and kinetic

energy in a system. Apply the law of conservation of energy to predict the maximum

velocity of a pendulum when it is released from different heights. Use the time required for a swinging bob to break the photogate’s

beam to calculate the velocity of the pendulum at different heights. Compare the measured and predicted velocities.

Page 25: Motion

What is energy? Energy measures the ability for things to change

themselves or to cause change in other things. — A gust of wind has energy because it can move

objects in its path.— A piece of wood in a fireplace has energy

because it can produce heat and light.— You have energy because you can change the

motion of your own body.— Batteries have energy because they can be

used in a radio to make sound.— Gasoline has energy because it can be burned

in an engine to move a car.— A ball at the top of a hill has energy because it

can roll down the hill and move objects in its path.

Page 26: Motion

Units of energy

Pushing a 1-kilogram object with a force of one newton for a distance of one meter uses one joule of energy.

A joule (J) is the S.I. unit of measurement for energy.

Page 27: Motion

Joules One joule is a small amount of energy.

An ordinary 100 watt electric light bulb uses 100 joules of energy every second!

Page 28: Motion

Work In science, work is a

form of energy you either use or get when a force is applied over a distance.

You do 1 joule of work if you push with a force of 1 newton for a distance of 1 meter.

Page 29: Motion

Work When thinking about work, remember that

work is done by forces that cause movement.

If nothing moves (distance is zero), then no work is done.

Page 30: Motion

Work

W = F x dDistance (m)

Force (N)

Work (joules)

Page 31: Motion

Potential energy Systems or objects with

potential energy are able to exert forces (exchange energy) as they change.

Potential energy is energy due to position.

Page 32: Motion

Potential energyA stretched spring has potential energy. If released, the spring will use this

energy to move itself and anything attached to it back to its original length.

Page 33: Motion

Potential Energy

EP = mgh

height object raised (m)

gravity (9.8 m/sec2)

PE (joules)

mass of object (g)

Page 34: Motion

Kinetic energy Energy of motion is called

kinetic energy. A skateboard and rider

have kinetic energy because they can hit other objects and cause change.

The amount of kinetic energy is equal to the amount of work the moving board and rider do as they come to a stop.

Page 35: Motion

Kinetic Energy

EK = ½ mv2

mass of object (kg)

velocity (m/sec)

KE (joules)

Page 36: Motion

Kinetic EnergyKinetic energy increases as the square of the speed.A car going twice as fast has four times the kinetic

energy and needs four times the stopping distance.

Page 37: Motion

A 2-kg rock is at the edge of a cliff 20 m. above a lake. It becomes loose and falls toward the water below. Calculate its potential and kinetic energy when it is on the cliff edge and when it is halfway down. Its speed is 14 m/s halfway down.

Potential and kinetic energy

1. Looking for: … the potential energy (Ep) and kinetic energy (Ek) at two locations.

2. Given: … the mass (2 kg), the height at cliff edge (20 m), and the speed halfway down (14 m/s). You can assume the initial speed is 0 m/s because the rock starts from rest.

3. Relationships: Use: Ep = mgh and Ek = ½mv2

Page 38: Motion

Potential and kinetic energy4. Solution:

— Potential energy at the top: Ep= (2 kg)(9.8 N/kg)(20 m) = 392 J

— Potential energy halfway down: Ep= (2 kg)(9.8 N/kg)(10 m) = 196 J

— Kinetic energy at the top:Ek= (½)(2 kg)(0 m/s)2 = 0 J

— Kinetic energy halfway down: Ek= (½)(2 kg)(14 m/s)2 = 196 J

Page 39: Motion

Conservation of EnergyThe law of conservation of

energy states that energy can never be created or destroyed, just converted from one form into another.

If gravity is the only force acting on the ball, it returns to your hand with exactly the same speed and kinetic energy it started with, except that it returns to your hand from the opposite direction.

Page 40: Motion

Conservation of EnergyThe law of energy conservation says the total energy

before the change equals the total energy after it.

The law of energy conservation says the total energy before the change equals the total energy after it.

Page 41: Motion

A 2-kg car moving with a speed of 2 m/s starts up a hill. How high does the car roll before it stops?

Potential and kinetic energy

1. Looking for: … the height.2. Given: …the car’s mass (2 kg), and initial speed (2

m/s)

3. Relationships: The law of conservation of energy states that the sum of the kinetic and potential energy is constant. The ball keeps going uphill until all of its kinetic energy has been turned into potential energy.

Page 42: Motion

Potential and kinetic energy

4. Solution: — Find the kinetic energy at the start:

EK = (½)(2 kg)(2 m/s)2 = 4 J— Use the potential energy to find the height:

EP = 4 J = mgh— Therefore:

h = (4 J) ÷ (2 kg)(9.8 N/kg) = 0.2 m— The car rolls to a height of 0.2 m above where it

started.

Page 43: Motion

Using and conserving energyWhen you “use” energy by turning on a light,

you are really converting energy from one form (electricity) to other forms (light and heat).

In the “physics” sense, the energy is not “used up” but converted into other forms of energy.

The total amount of energy stays constant.

Page 44: Motion

Unit 1: MotionChapter 4: Conservation Laws

4.1 Newton’s Third Law and Momentum

4.2 Energy and the Conservation of Energy

4.3 Collisions

Page 45: Motion

4.2 Investigation: Collisions

Key Question: Why do things bounce back when they collide?

Objectives: Explore collisions and explain how they obey the

law of conservation of momentum. Describe how Newton’s laws explain collisions.

Page 46: Motion

CollisionsThere are two main

types of collisions, elastic and inelastic.

When an elastic collision occurs, objects bounce off each other with no loss in the total kinetic energy of the system.

Page 47: Motion

Collisions In an inelastic

collision, objects change shape or stick together, and the total kinetic energy of the system decreases. An egg hitting the floor is

one example of an inelastic collision.

Page 48: Motion

Collisions When two billiard balls collide, it looks like they bounce without a loss of kinetic energy.

But the sound of the collision tells you a small amount of kinetic energy is being changed into sound energy.

Page 49: Motion

MomentumAs long as there are no outside forces (such as

friction), momentum is conserved in both elastic and inelastic collisions.

Conservation of momentum makes it possible to determine the motion of objects before or after colliding.

Page 50: Motion

Momentum Using momentum to analyze collision problems takes

practice.

1. Draw a diagram.

2. Decide whether the collision is elastic or inelastic.

3. Let variables represent the masses and velocities of the objects before and after the collision.

4. Use momentum conservation to write an equation stating that the total momentum before the collision equals the total after. Then solve it.

Page 51: Motion

An 8,000-kg train car moves to the right at 10 m/s. It collides with a 2,000-kg parked train car. The cars get stuck together and roll along the track. How fast do they move after the collision?

Momentum and collisions

1. Looking for: … the velocity of the train cars after the collision.

2. Given: … both masses (m1= 8,000 kg; m2= 2000 kg) and the initial velocity of the moving car (10 m/s). You know the collision is inelastic because the cars are stuck together.

Page 52: Motion

Momentum and collisions3. Relationships: Apply the law of conservation of

momentum. Because the two cars are stuck together, consider them to be a single larger train car after the collision (m3). The final mass is the sum of the two individual masses: initial momentum of car1 + initial momentum of car2 = final momentum of combined cars.m1v1 + m2v2 = (m1+ m2)v3

4. Solution: — (8,000 kg)(10 m/s) + (2,000 kg)(0 m/s) = (8,000 kg + 2,000

kg)v3

— v3 = 8 m/s. — The train cars move together to the right at 8 m/s.

Page 53: Motion

Forces in collisionsCollisions create forces because the

colliding objects’ motion changes. Since most collisions take place

quickly, the forces change rapidly and are hard to measure directly.

The total change in momentum is equal to the force multiplied by the time during which the force acts.

Page 54: Motion

Solving impulse problems Impulse can be used to solve many practical

problems.

Page 55: Motion

A 1 kg clay ball hits the floor with a velocity of –5 m/s and comes to a stop in 0.1 second. What force did the floor exert on it?

Impulse

1. Looking for: …upward force exerted by floor.2. Given: …ball’s mass (1 kg) final velocity (-5 m/s) and

final time (0.1 s)3. Relationships:.Use: Ft = mv2 − mv1 ,then solve for F

4. Solution: F = (1 kg)(0 m/s) – (1 kg) (-5 m/s) = 50 N (0.1s)

Page 56: Motion

Car crash safety Air bags work together with seat

belts to make cars safer. An air bag inflates when the force

applied to the front of a car reaches a specific level.

Automakers use crash test dummies to study the effects of collisions on passengers.

Crash test dummies contain electronic sensors to measure the forces exerted at various places on the body.

Page 57: Motion

Forensic Engineering

We usually think of engineering as a science focused on designing and constructing things—like bridges, computers, automobiles, or sneakers.

However, there is one branch of engineering that focuses on how things fail, collapse, or crash.