Download Work and Energy

Work and Energy
An Introduction
Monday,October 27, 2008
Work Energy Theorem
Kinetic Energy Definition
Announcements
Test repair begins during lab sections
on Tuesday. Attendance is
MANDATORY unless you have a
perfect score!
Work
Work tells us how much a force or
combination of forces changes the energy
of a system.
Work is the bridge between force (a vector)
and energy (a scalar).
W = F ∆r cos θ
F: force (N)
∆r : displacement (m)
θ: angle between force and displacement
Units of Work
SI System: Joule (N m)
British System: foot-pound
(not used in Physics B)
cgs System: erg (dyne-cm)
1 Joule of work is done when 1 N acts on a
body moving it a distance of 1 meter
(not used in Physics B)
Atomic Level: electron-Volt (eV)
Force and direction of motion
both matter in defining work!
There is no work done by a force if it
causes no displacement.
Forces can do positive, negative, or zero
work. When an box is pushed on a flat floor,
for example…
The normal force and gravity do no work,
since they are perpendicular to the direction
of motion.
The person pushing the box does positive
work, since she is pushing in the direction of
motion.
Friction does negative work, since it
points opposite the direction of motion.
Conceptual Checkpoint
Question: If a man holds a 50 kg box at arms
length for 2 hours as he stands still, how much
work does he do on the box?
Conceptual Checkpoint
Question: If a man holds a 50 kg box at arms
length for 2 hours as he walks 1 km forward, how
much work does he do on the box?
Conceptual Checkpoint
Question: If a man lifts a 50 kg box 2.0 meters,
how much work does he do on the box?
Work and Energy
Work changes mechanical energy!
If an applied force does positive work
on a system, it tries to increase
mechanical energy.
If an applied force does negative
work, it tries to decrease mechanical
energy.
The two forms of mechanical energy are
called potential and kinetic energy.
Sample problem
Jane uses a vine wrapped around a pulley to lift a 70-kg Tarzan to
a tree house 9.0 meters above the ground.
a)How much work does Jane do when she lifts Tarzan?
b)How much work does gravity do when Jane lifts Tarzan?
Sample problem
Joe pushes a 10-kg box and slides it across the floor at constant velocity of
3.0 m/s. The coefficient of kinetic friction between the box and floor is
0.50.
a) How much work does Joe do if he pushes the box for 15 meters?
b)
How much work does friction do as Joe pushes the box?
Sample problem
A father pulls his child in a little red wagon with constant speed. If
the father pulls with a force of 16 N for 10.0 m, and the handle of
the wagon is inclined at an angle of 60o above the horizontal, how
much work does the father do on the wagon?
Kinetic Energy
Energy due to motion
K = ½ m v2
K: Kinetic Energy
m: mass in kg
v: speed in m/s
Unit: Joules
Sample problem
A 10.0 g bullet has a speed of 1.2 km/s.
a) What is the kinetic energy of the bullet?
b) What is the bullet’s kinetic energy if the speed is halved?
c) What is the bullet’s kinetic energy if the speed is doubled?
The Work-Energy Theorem
The
net work due to all forces equals
the change in the kinetic energy of a
system.
Wnet = ∆K
Wnet:
work due to all forces acting on an
object
∆K: change in kinetic energy (Kf – Ki)
Sample problem
A 15-g acorn falls from a tree and lands on the ground 10.0 m below
with a speed of 11.0 m/s.
a) What would the speed of the acorn have been if there had been
no air resistance?
b) Did air resistance do positive, negative or zero work on the
acorn? Why?
Sample problem
A 15-g acorn falls from a tree and lands on the ground 10.0 m below
with a speed of 11.0 m/s.
c) How much work was done by air resistance?
d)
What was the average force of air resistance?
Tuesday,
October 29, 2008
Work done by variable forces
Announcements
Test Repair begins TODAY!!!
4th period in Mr. Perkins’ room
5th period in Dr. Bertrand’s room
Constant force and work
The force shown is a
constant force.
W = F∆r can be used
to calculate the work
done by this force
when it moves an
object from xa to xb.
The area under the
curve from xa to xb can
also be used to
calculate the work
done by the force
when it moves an
object from xa to xb
F(x)
xa
xb
x
Variable force and work
The force shown is a
variable force.
W = F∆r CANNOT be
used to calculate the
work done by this
force!
The area under the
curve from xa to xb can
STILL be used to
calculate the work
done by the force
when it moves an
object from xa to xb
F(x)
xa
xb
x
Springs
When a spring is stretched or compressed from its equilibrium
position, it does negative work, since the spring pulls opposite the
direction of motion.
Ws = - ½ k x2
Ws: work done by spring (J)
k: force constant of spring (N/m)
x: displacement from equilibrium (m)
The force doing the stretching does positive work equal to the
magnitude of the work done by the spring.
Wapp = - Ws = ½ k x2
Springs: stretching
0
F(N)
200
m
100
00
m
x
Fs
-100
1
2
3
4
5
x (m)
Fs
-200
Ws = negative area
= - ½ kx2
Fs = -kx (Hooke’s Law)
Sample problem
It takes 180 J of work to compress a certain spring 0.10 m.
a) What is the force constant of the spring?
b) To compress the spring an additional 0.10 m, does it take 180 J,
more than 180 J, or less than 180 J? Verify your answer with a
calculation.
Sample problem
A vertical spring (ignore its mass) whose spring constant is
900 N/m, is attached to a table and is compressed 0.150 m.
a)
b)
What speed can it give to a 0.300 kg ball when released?
How high above its original position (spring compressed) will the ball fly?
Wednesday,
October 29, 2008
More work by variable forces
Announcements
Exam repair continues…
HW due today, pass to back of room
Sample Problem
How much work is done by the force shown when it
acts on an object and pushes it from x = 0.25 m to
x = 0.75 m?
Sample Problem
How much work is done by the force shown
when it acts on an object and pushes it from
x = 2.0 m to x = 4.0 m?
Power
Power is the rate of which work is
done.
P = W/∆t
W: work in Joules
∆t: elapsed time in seconds
When we run upstairs,
t is small so P is big.
When we walk upstairs,
t is large so P is small.
Unit of Power
SI unit for Power is the Watt.
1 Watt = 1 Joule/s
Named after the Scottish engineer
James Watt (1776-1819) who
perfected the steam engine.
British system
horsepower
1 hp = 746 W
How We Buy Energy…
The kilowatt-hour is a commonly used
unit by the electrical power company.
Power companies charge you by the
kilowatt-hour (kWh), but this not
power, it is really energy consumed.
1 kW = 1000 W
1 h = 3600 s
1 kWh = 1000J/s • 3600s = 3.6 x 106J
Sample problem
A record was set for stair climbing when a man ran up the 1600 steps of the
Empire State Building in 10 minutes and 59 seconds. If the height gain of
each step was 0.20 m, and the man’s mass was 70.0 kg, what was his
average power output during the climb? Give your answer in both watts and
horsepower.
Sample problem
Calculate the power output of a 1.0 g fly as it walks straight
up a window pane at 2.5 cm/s.
Thursday,
October 30, 2008
Force Types
Announcements
Force types
Forces acting on a system can be
divided into two types according to
how they affect potential energy.
Conservative forces can be related
to potential energy changes.
Non-conservative forces cannot be
related to potential energy changes.
So, how exactly do we distinguish
between these two types of forces?
Conservative forces
Work is path independent.
Work along a closed path is zero.
If the starting and ending points are the same, no work is
done by the force.
Work changes potential energy.
Examples:
Work can be calculated from the starting and ending
points only.
The actual path is ignored in calculations.
Gravity
Spring force
Conservation of mechanical energy holds!
Non-conservative forces
Work is path dependent.
Work along a closed path is NOT zero.
Work changes mechanical energy.
Examples:
Knowing the starting and ending points is not
sufficient to calculate the work.
Friction
Drag (air resistance)
Conservation of mechanical energy does
not hold!
Potential energy
Energy of position or configuration
“Stored” energy
For gravity: Ug = mgh
m: mass
g: acceleration due to gravity
h: height above the “zero” point
For springs: Us = ½ k x2
k: spring force constant
x: displacement from equilibrium position
Conservative forces and
Potential energy
Wc = -∆U
If a conservative force does positive work on a
system, potential energy is lost.
If a conservative force does negative work, potential
energy is gained.
For gravity
Wg = -∆Ug = -(mghf – mghi)
For springs
Ws = -∆Us = -(½ k xf2 – ½ k xi2)
More on paths and
conservative forces.
Q: Assume a conservative force
moves an object along the
various paths. Which two works
are equal?
A:
W2 = W3
(path independence)
Q: Which two works, when added
together, give a sum of zero?
A:
W1 + W2 = 0
or
W1 + W3 = 0
(work along a closed path is zero)
Friday,
October 31, 2008
Conservative Forces and Potential
Energy
Announcements
Sample problem
D
C
A
B
mg
A box is moved in the
closed path shown.
a) How much work is
done by gravity
when the box is
moved along the
∆r
path A->B->C?
b) How much work is
done by gravity
when the box is
moved along the
path A->B->C->D>A?
Sample problem
D
C
A
B
mg
A box is moved in the
closed path shown.
b) How much work is
done by gravity
when the box is
moved along the
∆r
path A->B->C->D>A?
Sample problem
A box is moved in the
closed path shown.
a) How much work is
done by gravity
when the box is
moved along the
path A->B->C?
b) How much work is
done by gravity
when the box is
moved along the
path A->B->C->D>A?
Solution
WG = 0 + F∆r
WG = 0 –mgh = -mgh
b)
WG = 0 -mgh + 0 + mgh
=0
The work in b) is zero because
work along a closed path is
zero for any conservative
force.
a)
Sample problem
A box is moved in the closed
path shown.
a)
How much work would
be done by friction if the
box were moved along
the path A->B->C?
b)
How much work is done
by friction when the box
is moved along the path
A->B->C->D->A?
Solution
Wf = -µkmgd - µkmgd
Wf = -2µkmgd
b)
Wf = -µkmgd - µkmgd µkmgd - µkmgd
= -4 µkmgd
Because friction is a
nonconservative force,
work along the closed
path in b) is not zero.
a)
Sample problem (#8.6)
As an Acapulco cliff diver drops to the water from a height
of 40.0 m, his gravitational potential energy decreases by
25,000 J. How much does the diver weigh?
Friday, October 31, 2008
Conservation of
Mechanical Energy
+ Pendulums
Announcements
• Turn in HW: Ch 7 (26, 27, 29)
• Lab next week: Pendulums and
Conservation of Energy
Sample problem
If 60.0 J of work are required to stretch a spring from a 2.00
cm elongation to a 5.00 cm elongation, how much
additional work is needed to stretch it from a 5.00 cm
elongation to a 8.00 cm elongation?
Law of Conservation of
Energy
In any isolated system, the total
energy remains constant.
Energy can neither be created nor
destroyed, but can only be
transformed from one type of energy
to another.
Law of Conservation of
Mechanical Energy
E = K + U = Constant
K: Kinetic Energy (1/2 mv2)
U: Potential Energy (gravity or spring)
∆E = ∆U + ∆K = 0
∆K: Change in kinetic energy
∆U: Change in gravitational or spring
potential energy
Conservation of Energy
Simulation
Energy Skate Park
http://phet.colorado.edu/simulations/sims.php?sim=Energy_Skate_Park
Sample problem (#8.15)
A 0.21 kg apple falls from a tree to the ground, 4.0 m below.
Ignoring air resistance, determine the apple’s gravitational potential
energy, U, kinetic energy, K, and total mechanical energy,
E, when its height above the ground is each of the following: 4.0 m,
2.0 m, and 0.0 m. Take ground level to be the point of zero
potential energy.
Pendulums and Energy
Conservation
Energy goes back and forth between
K and U.
At highest point, all energy is U.
As it drops, U goes to K.
At the bottom , energy is all K.
Pendulum Energy
½mvmax2 = mgh
For minimum and
maximum points of
swing
h
K1 + U1 = K2 + U2
For any points 1 and 2.
Sample problem
What is the speed of the pendulum bob at point B if it is
released from rest at point A?
40o
1.5 m
A
B
Springs and Energy
Conservation
Transforms energy back and forth
between K and U.
When fully stretched or extended, all
energy is U.
When passing through equilibrium, all
its energy is K.
At other points in its cycle, the energy
is a mixture of U and K.
Monday,
November 3, 2008
Energy conservation
Announcements
Conservation of Energy lab
in Dr. Bertrand’s room LC 210 this week
Turn in HW
Chapter 7 (33, 35, 37)
Spring Energy
0
All U
m
K1 + U1 = K2 + U2 = E
For any two points 1 and 2
-x
All K
m
All U
m
x
½kxmax2 = ½mvmax2
For maximum and minimum
displacements from
equilibrium
Spring Simulation
Spring Physics Simulation
Sample problem (#8.18)
A 1.60 kg block slides with a speed of 0.950 m/s on a
frictionless, horizontal surface until it encounters a spring with a
force constant of 902 N/m. The block comes to rest after
compressing the spring 4.00 cm. Find the spring potential
energy, U, the kinetic energy of the block, K, and the total
mechanical energy of the system, E, for the following
compressions: 0 cm, 2.00 cm, 4.00 cm.
Pendulum lab
Figure out how to demonstrate conservation of energy with
a pendulum using the equipment provided.
The photogates must be set up in “gate” mode this time.
The width of the pendulum bob is an important number. To
get it accurately, use the caliper.
Repeat for at least THREE different release heights.
Pendulum lab - Theory
h
Calculate the
velocity of the bob
at the bottom of the
swing using the
photogate.
Compare velocity
to the theoretical v
given by
conservation of
mechanical energy
for a set height of
release
Pendulum Lab write-up
In your handwritten report in your lab book, you
should have the following
1. A simple sketch of your apparatus.
2. A clearly labeled data table. This table should
include raw data plus calculated potential energy
and kinetic energy values for comparison
purposes. Measure multiple trials for at least
three different heights
3. A discussion regarding how well your results
support the Law of Conservation of Mechanical
Energy. If your results do not support this law
very well, what might be the explanation for the
discrepancy?
Sample Data Table
Trial
1
2
3
Xi Xf Ugi (J)
(m) (m)
Ugf (J)
∆U (J)
d (m)
t (s)
v
(m/s)
Kmax
(J)
Law of Conservation of
Energy
E = U + K + Eint= Constant
Eint is thermal energy.
∆U + ∆K + ∆ Eint = 0
Mechanical energy may be converted
to and from heat.
Work done by nonconservative forces
Wnet = Wc + Wnc
Net work is done by conservative and non-conservative
forces
Wc = -∆U
• Potential energy is related to conservative forces only!
Wnet = ∆K
• Kinetic energy is related to net force (work-energy theorem)
∆K = -∆U + Wnc
• From substitution
Wnc = ∆U + ∆K = ∆E
Nonconservative forces change mechanical energy. If
nonconservative work is negative, as it often is, the
mechanical energy of the system will drop.
Sample problem (#8.22)
Catching a wave, a 72-kg surfer
starts with a speed of 1.3 m/s,
drops through a height of 1.75 m,
and ends with a speed of 8.2 m/s.
How much non-conservative work
was done on the surfer?
Solution (#8.22)
Wnc = ∆U + ∆K
= Uf – Ui + Kf – Ki
= mghf – mghi + ½ mvf2 – ½ m vi2
= m[g(hf –hi) + ½ (vf2 –vi2)]
= 72[(9.8)(0 - 1.75) + ½ (8.22 – 1.32)]
= 1125 J
Tuesday,
November 4, 2008
Non-conservative forces and the
Law of Conservation of Energy
Announcements
Lab downstairs in Dr. Bertrand’s
room.
HW turned in at the back
Sample problem
If 60.0 J of work are required to stretch a spring from a 2.00 cm
elongation to a 5.00 cm elongation, how much additional work is
needed to stretch it from a 5.00 cm elongation to a 8.00 cm
elongation?
Sample problem (#8.29)
A 1.75-kg rock is released from rest at the
surface of a pond 1.00 m deep. As the
rock falls, a constant upward force of 4.10
N is exerted on it by water resistance.
Calculate the nonconservative work, Wnc,
done by the water resistance on the rock,
the gravitational potential energy of the
system, U, the kinetic energy of the rock,
K, and the total mechanical energy of the
system, E, for the following depths below
the water’s surface: d = 0.00 m, d = 0.500
m, d = 1.00 m. Let potential energy be zero
at the bottom of the pond.
Solution (#8.29) – for 0.00 m
Wnc = F∆r = 0
E=U+K
= mgh + 0 = mgh
= (1.75 kg)(9.8 m/s2)(1.00 m) = 17.15 J
Therefore
Wnc = 0 (the rock hasn’t moved yet)
E = 17.15 J(will be reduced by the drag force of water)
U = 17.15 J (maximum value)
K = 0 (minimum value)
Solution (#8.29) – for 0.50 m
Wnc = F∆r cos θ= (4.10 N)(0.50 m)cos(180°) = -2.05 J = ∆E
E = 17.15 J – ∆E = 17.15 J - 2.05 J = 15.1 J
E = U + K = mgh + K
15.1 J = (1.75 kg)(9.8 m/s2)(0.50 m) + K
15.1 J = 8.6 J + K
K = 15.1 – 8.6 J = 6.5 J
Therefore
Wnc = -2.05 J
E = 15.1 J(reduced by the drag force of water)
U = 8.6 J (determined by height)
K = 6.5 J (reduced by the drag force of water)
Solution (#8.29) – for 1.00 m
Wnc = F∆r = (4.10 N)(1.00 m) cosθ= -4.10 J = ∆E
E = 17.15 J – ∆E = 17.15 J - 4.10 J = 13.05 J
E =U+K =0+K=K
13.05 J = K
Therefore
Wnc = -4.10 J
E = 13.05 J (reduced by the drag force of water)
U = 0 (lowest point in problem)
K = 13.05 J (maximum value)
Wednesday,
November 5, 2008
Energy Review
Announcements
Thursday we will be UPSTAIRS, not in
the lab classrooms.
Turn in HW
Practice problem (not in packet)
A skeleton runner in the Winter Olympics drops 104 m
in elevation from top to bottom of the run.
a)
In the absence of nonconservative forces, what
would the speed of a rider be at the end of the
track. Assume initial velocity of zero.
b)
In reality, the best riders reach the bottom at a
speed of 35.8 m/s (80 mph). How much work is
done on an 86.0 kg rider and skeleton by
nonconservative forces?
Shown below is a graph of velocity versus time for an object that moves
along a straight, horizontal line under the, perhaps intermittent, action of
a single force exerted by an external agent.
Rank the intervals shown on the graph, from greatest to least, on the basis
of the work done on the object by the external agent.
B
A
C
D
G
-10 -8
-6
-4
-2
0
2
4
6
8
10
Velocity (m/s)
5
10
15
20
25
E
30
F
35
Time (s)
A
B
C
3m
5m
4m
0.600 kg
0.750 kg
0.400 kg
D
E
F
4m
3.5 m
3m
0.750 kg
0.400 kg
0.500 kg
Sample Problem (not in packet)
Two masses are suspended by a string over each side
of a pulley (Atwood’s machine). They are initially at rest
at the same height, after they are released, the large
mass m2 falls through a height h and hits the floor, and
the small mass m1 rises through a height h.
Find the speed of the masses just before m2 lands,
giving your answer in terms of m1, m2, g, and h. Assume
the ropes and pulley have negligible mass and that
friction can be ignored.
Thursday,
November 6, 2008
Special Speaker
Nick Antonas
Announcements
Friday, November 7, 2008
Linear Momentum
Announcements
Tuesday and Wednesday lab groups
have lab write-ups due TODAY
Thursday lab group has write-up due
MONDAY
Next week’s lab:
conservation of momentum
Turn in HW: Ch. 8 (23, 25, 28)
Which do you think has more
momentum?
Momentum
Momentum is a measure of how hard it is to stop or turn
a moving object.
What characteristics of an object would make it hard
to stop or turn?
Let’s watch Mad Scientist Guy on Momentum and
Newton’s Third Law of Motion!
Calculating Momentum
For one particle
p = mv
Note that momentum is a vector with the same
direction as the velocity!
For a system of multiple particles
p = Σpi --- add up the vectors
The unit of momentum is…
kg m/s or Ns
Sample Problem
Calculate the momentum of a 65-kg sprinter
running east at 10 m/s.
Sample Problem
Calculate the momentum of a system composed of a 65-kg
sprinter running east at 10 m/s and a 75-kg sprinter running north
at 9.5 m/s.
Change in momentum
Like any change, change in
momentum is calculated by looking at
final and initial momentums.
∆p = pf – pi
∆p: change in momentum
pf: final momentum
pi: initial momentum
Momentum change
demonstration
Using only a meter stick, find the momentum
change of each ball when it strikes the desk from a
height of exactly one meter.
Which ball, Bouncy or Lazy, has the greatest
change in momentum?
Wording
dilemma
In which case is the
magnitude of the
momentum change
greatest?
In which case is the
change in the
magnitude of the
momentum
greatest?
Shown below is a graph of velocity versus time for an object that moves
along a straight, horizontal line under the, perhaps intermittent, action of
a single force exerted by an external agent.
Rank the intervals shown on the graph, from greatest to least, on the basis
of the work done on the object by the external agent.
B
A
C
D
G
-10 -8
-6
-4
-2
0
2
4
6
8
10
Velocity (m/s)
5
10
15
20
25
E
30
F
35
Time (s)
Monday,
November 10, 2008
Impulse
Announcements
Lab in Dr. Bertrand’s room (LC 207)
all week
Set Packet FR Problem #2 out for
stamp check.
Impulse (J)
Impulse is the product of an external force
and time, which results in a change in
momentum of a particle or system.
J=F t
and J = ∆P
Therefore Ft = ∆P
Units: N s or kg m/s (same as momentum)
Impulsive Forces
Usually high
magnitude, short
duration.
Suppose the ball hits
the bat at 90 mph and
leaves the bat at 90
mph, what is the
magnitude of the
momentum change?
What is the change in
the magnitude of the
momentum?
Impulse (J) on a graph
F(N)
3000
2000
area under curve
1000
0
0
1
2
3
4
t (ms)
Sample Problem
Suppose a 1.5-kg brick is dropped on a glass
table top from a height of 20 cm.
a) What is the magnitude and direction of the
impulse necessary to stop the brick?
b) If the table top doesn’t shatter, and stops the
brick in 0.01 s, what is the average force it
exerts on the brick?
c) What is the average force that the brick exerts
on the table top during this period?
Sample Problem
F(N)
2,000
1,000
0.20
0.40
0.60
0.80
t(s)
This force acts on a 1.2 kg object moving at 120.0 m/s. The
direction of the force is aligned with the velocity. What is the new
velocity of the object?
Tuesday, November 11, 2008
Law of Conservation of Momentum
Announcements
Lab: Conservation of Momentum
HW due: Ch. 9 (15,17,18)
Impulsive Forces in Driving
Law of Conservation of
Momentum
If the resultant external force on a
system is zero, then the vector sum of
the momentums of the objects will
remain constant.
ΣPbefore = ΣPafter
Sample problem
A 75-kg man sits in the back of a 120-kg canoe that is at rest
in a still pond. If the man begins to move forward in the
canoe at 0.50 m/s relative to the shore, what happens to the
canoe?
External versus internal
forces
External forces: forces coming from
outside the system of particles whose
momentum is being considered.
External forces change the momentum of the
system.
Internal forces: forces arising from
interaction of particles within a system.
Internal forces cannot change momentum of
the system.
An external force in golf
The club head
exerts an external
impulsive force on
the ball and changes
its momentum.
The acceleration of
the ball is greater
because its mass is
smaller.
The System
An internal force in pool
The forces the balls
exert on each other
are internal and do
not change the
momentum of the
system.
Since the balls have equal
masses, the magnitude of
their accelerations is equal.
The System
Explosions
When an object separates suddenly, as in
an explosion, all forces are internal.
Momentum is therefore conserved in an
explosion.
There is also an increase in kinetic energy
in an explosion. This comes from a potential
energy decrease due to chemical
combustion.
Recoil
Guns and cannons “recoil” when fired.
This means the gun or cannon must move
backward as it propels the projectile forward.
The recoil is the result of action-reaction force
pairs, and is entirely due to internal forces. As the
gases from the gunpowder explosion expand,
they push the projectile forwards and the gun or
cannon backwards.
Wednesday,
November 12, 2008
Inelastic Collisions
Announcements
Labs today
Collisions
When two moving objects make contact
with each other, they undergo a collision.
Conservation of momentum is used to
analyze all collisions.
Newton’s Third Law is also useful. It tells us
that the force exerted by body A on body B
in a collision is equal and opposite to the
force exerted on body B by body A.
Collisions
During a collision,
external forces are
ignored.
The time frame of the
collision is very short.
The forces are
impulsive forces
(high force, short
duration).
Collision Types
Elastic collisions
Inelastic collisions
Also called “hard” collisions
No deformation occurs, no kinetic energy
lost
Deformation occurs, kinetic energy is lost
Perfectly Inelastic (stick together)
Objects stick together and become one
object
Deformation occurs, kinetic energy is lost
(Perfectly) Inelastic
Collisions
Simplest type of collisions.
After the collision, there is only one
velocity, since there is only one
object.
Kinetic energy is lost.
Explosions are the reverse of perfectly
inelastic collisions in which kinetic
energy is gained!
Sample Problem
An 80-kg roller skating
grandma collides
inelastically with a 40-kg
kid. What is their
velocity after the
collision?
How much kinetic
energy is lost?
Sample Problem
A fish moving at 2 m/s
swallows a stationary fish
which is 1/3 its mass.
What is the velocity of the
big fish after dinner?
Sample problem
A car with a mass of 950 kg and a speed of 16 m/s to the east
approaches an intersection. A 1300-kg minivan traveling north at
21 m/s approaches the same intersection. The vehicles collide and
stick together. What is the resulting velocity of the vehicles after the
collision?
Conservation of Momentum
Thursday,
November 13, 2008
Elastic Collisions
Announcements
Sample problem
Suppose a 5.0-kg projectile launcher shoots a
209 gram projectile at 350 m/s. What is the recoil
velocity of the projectile launcher?
Sample Problem
An exploding object breaks into three fragments. A 2.0 kg fragment
travels north at 200 m/s. A 4.0 kg fragment travels east at 100 m/s.
The third fragment has mass 3.0 kg. What is the magnitude and
direction of its velocity?
Elastic Collision
In elastic collisions, there is no deformation
of colliding objects, and no change in kinetic
energy of the system. Therefore, two basic
equations must hold for all elastic collisions
Σpb = Σpa (momentum conservation)
ΣKb = ΣKa (kinetic energy conservation)
Sample Problem
A 500-g cart moving at 2.0 m/s on an air track elastically strikes a
1,000-g cart at rest. What are the resulting velocities of the two
carts?
Sample Problem
Suppose three equally strong, equally
massive astronauts decide to play a game
as follows: The first astronaut throws the
second astronaut towards the third
astronaut and the game begins. Describe
the motion of the astronauts as the game
proceeds. Assume each toss results from
the same-sized "push." How long will the
game last?
2D-Collisions
Momentum in the x-direction is conserved.
Momentum in the y-direction is conserved.
ΣPy (before) = ΣPy (after)
Treat x and y coordinates independently.
ΣPx (before) = ΣPx (after)
Ignore x when calculating y
Ignore y when calculating x
Let’s look at a simulation:
http://surendranath.tripod.com/Applets.html
Sample problem
Calculate velocity of 8-kg ball after the collision.
2 m/s
y
2 kg
y
3 m/s
50o
x
2 kg
x
8 kg
0 m/s
8 kg
v
Before
After
Friday, November 14, 2008
Review of Momentum
Monday,
November 17, 2008
Energy and Momentum Review
Tuesday,
November 18, 2008
Energy and Momentum
EXAM