Download Chapter 4: Conservation Laws

Document related concepts

Inertia wikipedia , lookup

Quantum vacuum thruster wikipedia , lookup

Classical mechanics wikipedia , lookup

Photon polarization wikipedia , lookup

Internal energy wikipedia , lookup

Hunting oscillation wikipedia , lookup

Eigenstate thermalization hypothesis wikipedia , lookup

Classical central-force problem wikipedia , lookup

Work (thermodynamics) wikipedia , lookup

Momentum wikipedia , lookup

Kinetic energy wikipedia , lookup

Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup

Newton's laws of motion wikipedia , lookup

Relativistic mechanics wikipedia , lookup

Transcript
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
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.

Newton’s Third Law

Newton’s Third Law (actionreaction) 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).
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.
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.
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 _______________.
Action and reaction forces don’t cancel
 The
forces do not cancel because they act on
different objects.
Action and reaction
A woman with a weight of 500 newtons is sitting on a chair.
Describe one action-reaction pair of forces in this situation.
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.
Fc = 500 N
Fw = -500 N
Momentum
 Momentum
is the mass of a object times its
velocity.
 The
units for momentum are kilogram•meters per
second (kg·m/s).
Momentum
 Stopping
a fast-moving
object is harder than
stopping a slow-moving one.
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.
Calculating momentum
 Momentum
is
calculated with velocity
instead of speed
because the direction
of momentum is
always important.
Impulse
 The
change in momentum is often referred to as
impulse.
 We
can use this relationship to solve equations using
changes in mass or velocity.
Impulse
 The
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.
Force and momentum
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?
1.
2.
3.
4.
Looking for: … the speed.
Given: … the net force (100N), the time the force acts (5s),
and the mass of the car (10 kg).
Relationships: Use: impulse (Δp) = force (F) × time (t)
momentum = mass (m) × velocity (v)
Solution: Δp = 100 N × 5 s = 500 kg·m/s
Velocity is momentum divided by mass, or
v = (500 kg·m/s) ÷ 10 kg = 50 m/s
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
 If
force changes the momentum of the ball.
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.
Momentum
 The
We use positive and negative
numbers to show opposite
directions.
result of throwing a 5kg 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).
Law of momentum conservation
 The
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.
Using the momentum relationship
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?
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 A negative sign
hammer is thrown must be the same as the total
momentum after it is thrown.
momentum (p) = mass (m) × velocity (v)
indicates the
velocity is to the
left.
Using the momentum relationship
4.
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 = 0
mava + 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.
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
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.
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.
Units of energy

A joule (J) is the S.I.
unit of measurement
for energy.

Pushing a 1-kilogram object with a force of
one newton for a distance of one meter uses
one joule of energy.
Joules

One joule is a small amount of energy.

An ordinary 100 watt
electric light bulb uses
100 joules of energy
every second!
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.
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.
Work
Work (joules)
Force (N)
W=Fxd
Distance (m)
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.
Potential energy
 A stretched
 If
spring has potential energy.
released, the spring will use this energy
to move itself and anything attached to it
back to its original length.
Potential Energy
mass of object (g)
PE (joules)
EP = mgh
height object raised (m)
gravity (9.8 m/sec2)
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.
Kinetic Energy
KE (joules)
mass of object (kg)
EK = ½ mv2
velocity (m/sec)
Kinetic Energy
 Kinetic
 A car
energy increases as the square of the speed.
going twice as fast has four times the kinetic
energy and needs four times the stopping distance.
Potential and kinetic energy
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.
2.
Looking for: … the potential energy (Ep) and kinetic
energy (Ek) at two locations.
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
1.
Potential and kinetic energy
4.
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
Conservation of Energy
 The
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.
Conservation of Energy
 The
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.
Potential and kinetic energy
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?
1.
2.
3.
Looking for: … the height.
Given: …the car’s mass (2 kg), and initial speed (2 m/s)
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.
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.
Using and conserving energy
 When
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.
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
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.
Collisions
 There
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.
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.
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.
Momentum
 As
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.
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.
Momentum and collisions
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?
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.
Momentum and collisions
3.
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.
Forces in collisions
 Collisions
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.
Solving impulse problems
 Impulse
can be used to solve many practical
problems.
Impulse
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?
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)
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.
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.