Transcript PS Unit 2 Motion

Motion
Chapter 11
Standards
• Students will:
• SPS8. Determine the relationship between
force, mass and motion
• SPS8a Calculate velocity and acceleration
• SPS8c Relate falling objects to gravitational
force
• SPS8d Explain difference in mass and weight
Observing Motion
• Motion- change in position in relation to a
reference point.
Measuring Motion: Distance
• Distance- how far an object moves on a path
• Displacement- how far between starting and
ending points on a path
Measuring Motion: Speed
• Speed
– rate of motion
– distance traveled
per unit time
speed = distance
time
Measuring Motion: Speed (cont’d)
• Instantaneous
Speed
– speed at a given
instant
• Average Speed= total distance
total time
Measuring Motion: Velocity
• Velocity
– speed in a given
direction
– can change even
when the speed is
constant!
Calculating Velocity:
Distance/Speed/Time Triangle
Graphing Speed/Velocity
• X axis- usually independent variable (time)
• Y axis- usually dependent variable (distance)
• Slope of straight line= vertical change
horizontal change
Graphing (cont’d)
• slope = velocity
• steeper slope = faster
velocity
• straight line =
constant velocity
• flat line = 0 velocity
(no motion)
Calculating Slope
Distance (m)
1. Choose two points on
graph to calculate
slope.
2. Calculate the vertical
and horizontal
change.
3. Divide the vertical
change by the
horizontal change.
Distance Vs Time
16
14
.
12
10
8
.
6
4
2
0
1
2
3
Time (s)
4
5
Calculating Slope(cont’d)
16
14
.
12
Distance (m)
1. point 1: d= 6m
t= 1s
point 2: d= 12m
t= 4s
2. vert ∆- 12m-6m= 6m
horiz ∆- 4s-1s= 3s
Distance Vs Time
10
Vertical
change
8
.
6
Horizontal
Change
4
3. slope= 6m = 2m/s
3s
2
0
1
2
3
Time (s)
4
5
Practice Problem
A
B
• Who started out faster?
– A (steeper slope)
• Who had a constant speed?
–A
• Describe B from 10-20 min.
– B stopped moving
• Find their average speeds.
– A = (2400m) ÷ (30min)
A = 80 m/min
– B = (1200m) ÷ (30min)
B = 40 m/min
Measuring Motion: Acceleration
• Acceleration
– the rate of change of velocity
– change in speed or direction
• Centripetal acceleration
– circular motion (even if speed is constant,
direction is always changing)
ex. Moon accelerates around Earth
Measuring Motion: Acceleration
• Positive acceleration
-“speeding up”
•
Negative acceleration
-“slowing down”
Calculating Acceleration
• a= vf – vi
t
• a= ∆ v
t
•
•
•
•
a- acceleration
vf- final velocity
vi- initial velocity
t- time
Calculating Acceleration (cont’d)
1. List given, then
unknown values.
2. Write equation for
acceleration.
3. Insert known
values into
equation and solve.
t
Vf-Vi
a
t
Practice Problem 1
A flowerpot falls off a second-story windowsill.
The flowerpot starts from rest and hits Mr.
Mertz 1.5s later with a velocity of 14.7m/s.
Find the average acceleration of the flowerpot.
Given:
Remember:
Solve:
t= 1.5s
a= vf – vi
a= 14.7m/s-0m/s
Vi= 0m/s
t
1.5s
Vf= 14.7m/s
a= 14.7m/s
a= ?
1.5s
a= 9.8m/s2
Practice Problem 2
Joseph’s car accelerates at an average rate of
2.6m/s2. How long will it take his car to speed
up from 24.6m/s to 26.8m/s2?
Given:
Remember:
Solve:
a= 2.6m/s2
t= (vf-vi) ÷ a
vf= 26.8m/s2
vf-vi
t= 26.8m/s2-24.6m/s2
Vi= 24.6m/s2
a t
2.6m/s2
t= ?
t= 2.2m/s
2.6m/s2
t= 0.85s
Practice Problem 3
A cyclist travels at a constant velocity of 4.5m/s
westward and then speeds up with a steady
acceleration of 2.3m/s2. Calculate the cyclist’s
speed after accelerating for 5.0s.
Given:
Remember: Solve:
vi= 4.5m/s
vf= vi + a x t
vf= ?
vf-vi
vf= 4.5m/s + (2.3m/s2 x 5.0s)
a= 2.3m/s2
a
t
vf= 4.5m/s + 11.5m/s
t= 5.0s
vf= 16m/s
Graphing Acceleration
Distance/Time
400
On Distance-Time
graph:
• Acceleration is shown as
a curved line
Distance (m)
300
200
100
0
0
5
10
Time (s)
15
20
Graphing Acceleration
Speed/Time
3
Speed (m/s)
2
1
0
0
2
4
6
Time (s)
8
10
On a Speed-Time graph:
• Slope of straight line=
acceleration
• Positive slopespeeding up
• Negative slope- slowing
down
• Flat line- constant
velocity
(no acceleration)
Newton’s Laws of Motion
• 2nd Law of Motion:
The acceleration of an object is directly
proportional to the net force acting on it and
inversely proportional to its mass.
Force = mass x acceleration or
or
F= ma
Force
Force- push or pull that one body exerts on
another
Fundamental Forces
• 4 types:
1. gravity
2. electromagnetic
3. weak nuclear
4. strong nuclear
• Vary in strength
• Act through contact
or at a distance
Forces
• Force Pairs: forces
moving in opposite
directions
• Balanced forces: do not
move; push equally on
each other
• Unbalanced forces:
acceleration (movement)
in the direction of larger
force
Force Pair Examples
Friction
Friction: force that opposes
motion between 2
surfaces
• Static Friction: nonmoving
surfaces
• Kinetic Friction: moving
surfaces- sliding or rolling
(sliding friction is greater
than rolling friction)
Friction Facts
• Necessary for all
motion
• Rougher surfaces
create greater
friction
• Greater mass creates
greater friction
• Lubricants reduce
friction
Newton’s First and Second Laws
Chapter 12
Newton’s Laws of Motion
• 1st Law of Motion: Law of Inertia
• An object at rest will remain at rest;
• An object in motion will continue moving at
a constant velocity unless acted upon by a net
force.
Inertia
• Objects move only when
net force is applied.
• Objects maintain state of
motion.
• Inertia is related to mass.
(small mass can be
accelerated by small force
large mass can be
accelerated by large force)
Newton’s Laws of Motion
• 2nd Law of Motion:
The acceleration of an object is directly
proportional to the net force acting on it and
inversely proportional to its mass.
Force = mass x acceleration or
or
F= ma
Newton’s Second Law
• For equal forces, large
masses accelerate less
• Force is measured in
newtons (N)
• 1N= 1kg x 1m/s2
F
m
a
Practice Problem 1
Zoo keepers lift a stretcher that holds a sedate
lion. The lion’s mass is 175 kg and the upward
acceleration of the lion and stretcher is 0.657m/s2.
What force is needed to produce acceleration of the
lion and stretcher?
Given:
Remember:
Solve:
m= 175 kg
F=mxa
F
a= 0.657m/s2
F = 175 kg x 0.657m/s2
F
F= ?
m a
F = 11.49 N
Practice Problem 2
A baseball accelerates downward at 9.8 m/s2. If
gravity is the only force acting on the baseball and is
1.4N, what is the baseball’s mass?
Given:
Remember:
F= 1.4 kg/m/s2
F
a= 9.8 m/s2
m= ?
m a
Solve:
m=F
a
m = 1.4 N
9.8 m/s2
m = .14 kg
Weight and Mass
• Weight- measure of gravity on an object
• Not equal to mass (constant everywhere)
• Measured in Newtons
weight = mass x free-fall acceleration
(9.8m/s2)
w
m
g
free-fall
acceleration
Gravity
• Force of attraction
between 2 objects in
the universe
• Increases as:
- mass increases
- distance decreases
• Affects all matter
Gravity Quiz 1
Who experiences more gravity - the astronaut or
the politician?
More distance
Less distance
Gravity Quiz 2
Which exerts more gravity, your hand or your
pencil ?
Gravity Quiz 3
• Would you weigh more on Earth or Jupiter?
(Hint: Which planet has the greater mass?)
Free Fall Acceleration
• Occurs when Earth’s
gravity is only force
acting on an object
• In absence of air
resistance, all objects
accelerate at same
rate
• g = 9.8 m/s2
• g = 9.8 m/s2 g = 9.8
m/s2 g = 9.8 m/s2 g =
9.8 m/s2 g = 9.8 m/s2
Air Resistance
• Force of air on a
moving object which
opposes its motion
• Aka fluid friction or
drag
• Depends on objects:
- speed
- surface area
- shape
- density
Air Resistance (cont’d)
• Terminal velocity=
maximum velocity
reached by a falling
object
• Reached when…
F gravity = F air resistance
(no net force)
Projectile Motion
Projectile• Any object thrown in
air
• Only acted on by
gravity
• Follows parabolic
path- trajectory
• Has horizontal and
vertical velocities
V oy
V ox
Projectile Motion (cont’d)
• Horizontal Velocity
– depends on inertia
– remains constant
• Vertical Velocity
– depends on gravity
– accelerates
downward at
9.8 m/s2
Projectile Motion Quiz
• A moving truck launches a ball vertically
(relative to the truck). If the truck maintains a
constant horizontal velocity after the launch,
where will the ball land (ignore air resistance)?
A) In front of the truck
B) Behind the truck
C) In the truck
Answer: C because horizontal and vertical velocities are
independent of each other
Newton’s Laws of Motion
• 3rd Law of Motion:
When one object exerts a force on a second
object, the second object exerts an equal but
opposite force on the first.
Newton’s Third Law
• Forces always occur
in pairs
• Forces in a pair do
not act on the same
object
• Equal forces don’t
always have equal
effects
Newton’s Third Law
Problem: How can a horse pull a cart if the cart
is pulling with equal force back on the horse?
Newton’s Third Law
Answer:
1. Forces are equal and opposite but are acting
on different objects
2. Forces are not balanced
3. The movement of the horse depend on the
forces working on the horse.
Momentum
Momentum• quantity of motion
• = mass x velocity
• units: kg x m/s
p
m
v
Practice Problem 1
Find the momentum of a bumper car if it has a
total mass of 280 kg and a velocity of 3.2 m/s.
Given
m = 280 kg
v = 3.2 m/s
P=?
Remember
p
m
vv v
Solve
p = mv
p =(280kg)(3.2m/s)
P = 896 kg  m/s
Practice Problem 2
The momentum of a second bumper car is 675
kg·m/s. What is its velocity if its total mass is
300 kg?
Given
Remember
p = 675 kg·m/s
p
m = 300kg
m v
v=?
Solve
v=p÷m
v =(675kg·m/s)÷(300kg)
v = 2.25m/s
Conservation of Momentum
Law of Conservation of Momentum:
• Total momentum of two or more objects
before a collision is the same as it was before
the collision (momentum is conserved).
Jet Car Challenge
CHALLENGE:
Construct a car that will travel a least 3 meters
using only the following materials:
• scissors
• 2 skewers
• tape
• 1 balloon
• 4 plastic lids
• 1 cardboard base
• 2 plastic straws
Use your knowledge of Newton’s Laws to make this
thing go!