PHYS 1443 – Section 002
Lecture #14
Wednesday, Oct. 29, 2008
Dr. Mark Sosebee
(Disguised as Dr. Yu)
•
•
•
•
•
•
Energy Diagram
General Energy Conservation & Mass Equivalence
More on gravitational potential energy
• Escape speed
Power
Linear Momentum
Conservation of Momentum
Wednesday, Oct. 29, 2008
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
1
Energy Diagram and the Equilibrium of a System
One can draw potential energy as a function of position  Energy Diagram
Let’s consider potential energy of a spring-ball system
What shape is this diagram?
1
U  kx2
2
What does this energy diagram tell you?
Us
1.
Minimum
 Stable
equilibrium
Maximum
unstable
equilibrium
-xm
A Parabola
1 2
U s  kx
2
xm
x
2.
3.
Potential energy for this system is the same
independent of the sign of the position.
The force is 0 when the slope of the potential
energy curve is 0 at the position.
x=0 is the stable equilibrium position of this
system where the potential energy is minimum.
Position of a stable equilibrium corresponds to points where potential energy is at a minimum.
Position of an unstable equilibrium corresponds to points where potential energy is a maximum.
Wednesday, Oct. 29, 2008
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
2
General Energy Conservation and
Mass-Energy Equivalence
General Principle of
Energy Conservation
What about friction?
The total energy of an isolated system is conserved as
long as all forms of energy are taken into account.
Friction is a non-conservative force and causes mechanical
energy to change to other forms of energy.
However, if you add the new forms of energy altogether, the system as a
whole did not lose any energy, as long as it is self-contained or isolated.
In the grand scale of the universe, no energy can be destroyed or created but just
transformed or transferred from one to another. The total energy of universe is
constant as a function of time!! The total energy of the universe is conserved!
Principle of
Conservation of Mass
Einstein’s MassEnergy equality.
In any physical or chemical process, mass is neither created nor destroyed.
Mass before a process is identical to the mass after the process.
2
mc
ER 
Wednesday, Oct. 29, 2008
How many joules does your body correspond to?
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
3
The Gravitational Field
The gravitational force is a field force. The force exists everywhere in the universe.
If one were to place a test object of mass m at any point in the
space in the existence of another object of mass M, the test object
will feel the gravitational force exerted by M, Fg  mg .
Therefore the gravitational field g is defined as
g
Fg
m
In other words, the gravitational field at a point in the space is the gravitational force
experienced by a test particle placed at the point divided by the mass of the test particle.
So how does the Earth’s
gravitational field look like?
Far away from the
Earth’s surface
Wednesday, Oct. 29, 2008
g 
E
Fg
m

GM E
rˆ
RE2
Where r̂ is the unit vector pointing
outward from the center of the Earth
Close to the
Earth’s surface
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
4
The Gravitational Potential Energy
What is the potential energy of an object at the
height y from the surface of the Earth?
U  mgy
Do you think this would work in general cases?
No, it would not.
Why not?
Because this formula is only valid for the case where the gravitational force
is constant, near the surface of the Earth, and the generalized gravitational
force is inversely proportional to the square of the distance.
OK. Then how would we generalize the potential energy in the gravitational field?
m
Fg
rf
m
ri
RE
Wednesday, Oct. 29, 2008
Fg
Since the gravitational force is a central force, and a
central force is a conservative force, the work done by
the gravitational force is independent of the path.
The path can be considered as consisting of
many tangential and radial motions.
Tangential motions do not contribute to work!!!
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
5
More on The Gravitational Potential Energy
Since the gravitational force is a radial force, it performs work only when the path
has component in radial direction. Therefore, the work performed by the gravitational
force that depends on the position becomes:
dW  F  dr  F r dr
Potential energy is the negative change
of the work done through the path
Since the Earth’s gravitational force is
Thus the potential energy
function becomes
For the whole path
W  r F r dr
i
U  U f  U i   r F r dr
rf
i
F r   
U f Ui  
rf
ri
GM E m
r2
 1 1
GM E m
dr  GM E m   
2
r
 rf ri 
Since only the difference of potential energy matters, by taking the
infinite distance as the initial point of the potential energy, we obtain
For any two
particles?
Gm1m2
U
r
Wednesday, Oct. 29, 2008
rf
The energy needed
to take the particles
infinitely apart.
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
For many
particles?
U
GM E m
r
U  U i, j
i, j
6
Example of Gravitational Potential Energy
A particle of mass m is displaced through a small vertical distance y near the Earth’s
surface. Show that in this situation the general expression for the change in gravitational
potential energy is reduced to the U=-mgy.
Taking the general expression of
gravitational potential energy
Reorganizing the terms w/
the common denominator
Since the situation is close to
the surface of the Earth
Therefore, U becomes
 GM E
ri  RE

r
m
and
U  GM E m
GM E
Since on the surface of the
g
RE2
Earth the gravitational field is
Wednesday, Oct. 29, 2008
U
 1 1
 GM E m   
r

r
f
i


f
 ri 
rf ri
y
 GM E m
rf ri
rf  RE
y
RE2
The potential
energy becomes
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
U   mgy
7
Escape Speed
vf=0 at h=rmax
m
h
vi
RE
Consider an object of mass m is projected vertically from the surface of
the Earth with an initial speed vi and eventually comes to stop vf=0 at
the distance rmax.
Since the total mechanical
energy is conserved
ME
ME  K  U 
Solving the above equation
for vi, one obtains
vi
Therefore if the initial speed vi is known, one can use
this formula to compute the final height h of the object.
In order for an object to escape
Earth’s gravitational field completely, vesc 
the initial speed needs to be

1 2 GM E m
GM E m
mvi 

2
RE
rmax
 1
1
2GM E 

 RE rmax
hr
2GM E

RE
max



vi2 RE2
 RE 
2GM E  vi2 RE
2  6.67 10 11  5.98 10 24
6.37 106
 1.12 10 4 m / s  11.2km / s
This is called the escape speed. This formula is
valid for any planet or large mass objects.
Wednesday, Oct. 29, 2008
How does this depend
on the mass of the
escaping object?
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
Independent of
the mass of the
escaping8object
Power
• Rate at which the work is done or the energy is transferred
– What is the difference for the same car with two different engines (4
cylinder and 8 cylinder) climbing the same hill?
–  The time… 8 cylinder car climbs up the hill faster!
Is the total amount of work done by the engines different? NO
Then what is different?
The rate at which the same amount of work
performed is higher for 8 cylinders than 4.
Average power P  W
t
Instantaneous power
Unit? J / s  Watts
r sr
W dW
 F  t 

 lim
P  lim
t 0
t  0
t
dt


1HP  746Watts


r r
F v 
r r
 F v cos
What do power companies sell? 1kWH  1000Watts  3600s  3.6  106 J
Wednesday, Oct. 29, 2008
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
Energy
9
Energy Loss in Automobile
Automobile uses only 13% of its fuel to propel the vehicle.
Why?
67% in the engine:
•
•
•
Incomplete burning
Heat
Sound
16% in friction in mechanical parts
4% in operating other crucial parts
such as oil and fuel pumps, etc
13% used for balancing energy loss related to moving vehicle, like air
resistance and road friction to tire, etc
Two frictional forces involved in moving vehicles
Coefficient of Rolling Friction; m=0.016
Air Drag
mcar  1450kg Weight  mg  14200 N
m n  m mg  227 N
1
1
f a  D  Av 2   0.5 1.293  2v 2  0.647v 2
2
2
Total power to keep speed v=26.8m/s=60mi/h
Power to overcome each component of resistance
Wednesday, Oct. 29, 2008
Total Resistance
ft  f r  f a
P  ft v   691N   26.8  18.5kW
Pr  f r v  227   26.8  6.08kW
PHYS 1443-002, FallP2008
a  fav
Dr. Jaehoon Yu
 464.7 26.8  12.5kW
10
Linear Momentum
The principle of energy conservation can be used to solve problems
that are harder to solve just using Newton’s laws. It is used to
describe motion of an object or a system of objects.
A new concept of linear momentum can also be used to solve physical problems,
especially the problems involving collisions of objects.
Linear momentum of an object whose mass is m
and is moving at a velocity of v is defined as
What can you tell from this
definition about momentum?
What else can use see from the
definition? Do you see force?
Monday, Oct. 22, 2007
1.
2.
3.
4.
ur
r
p  mv
Momentum is a vector quantity.
The heavier the object the higher the momentum
The higher the velocity the higher the momentum
Its unit is kg.m/s
The change of momentum in a given time interval
r r
r
r
r
r
r
r
m  v  v0 
p
mv  mv0
v


ma   F
 m
t
t
t
t
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
11
Linear Momentum and Forces
r dpr
 F  dt
•
•
•
What can we learn from this Force-momentum
relationship?
The rate of the change of particle’s momentum is the same as
the net force exerted on it.
When net force is 0, the particle’s linear momentum is
constant as a function of time.
If a particle is isolated, the particle experiences no net force.
Therefore its momentum does not change and is conserved.
Something else we can do
with this relationship. What
do you think it is?
Can you think of a
few cases like this?
Monday, Oct. 22, 2007
The relationship can be used to study
the case where the mass changes as a
function of time.
r
r
d  mv  dm r
r dpr
dv
 F  dt  dt  dt v  m dt
Motion
of a meteorite
PHYS 1443-002,
Fall 2007
Dr. Jaehoon Yu
Motion of a rocket
12
Conservation of Linear Momentum in a Two
Particle System
Consider an isolated system with two particles that do not have any
external forces exerting on it. What is the impact of Newton’s 3rd Law?
If particle#1 exerts force on particle #2, there must be another force that
the particle #2 exerts on #1 as the reaction force. Both the forces are
internal forces, and the net force in the entire SYSTEM is still 0.
Let say that the particle #1 has momentum
Now how would the momenta
p1 and #2 has p2 at some point of time.
of these particles look like?
Using momentumforce relationship
r
r
dp1
F21 
dt
and
r
r
dp2
F12 
dt
r
r
ur ur
ur
And since net force
dp
dp
d r r
2
1
F
  F 12  F 21  dt  dt  dt  p2  p1   0
of this system is 0
ur
ur
Therefore p 2  p1  const The total linear momentum of the system is conserved!!!
Wednesday, Oct. 29, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
13
Linear Momentum Conservation
Initial
r
r
r
r
p1i  p2i  m1v1  m2v2
Final
r
r
r
r
p1 f  p2 f  m1v1  m2v2
Wednesday, Oct. 29, 2008
r
r
r
r
p1 f  p2 f
p

p

1
i
2
i
PHYS 1443-002, Fall 2008 Dr. Jaehoon
14
Yu
More on Conservation of Linear Momentum in
a Two Body System
From the previous slide we’ve learned that the total
momentum of the system is conserved if no external
forces are exerted on the system.
ur
ur ur
 p  p 2  p1 const
As in the case of energy conservation, this means
that the total vector sum of all momenta in the
system is the same before and after any interactions
What does this mean?
r
r
r
r
p2i  p1i  p2 f  p1 f
Mathematically this statement can be written as
P
xi
system
P

xf
system
P
yi
system
This can be generalized into
conservation of linear momentum in
many particle systems.
Wednesday, Oct. 29, 2008

P
yf
system
P
zi
system

P
zf
system
Whenever two or more particles in an
isolated system interact, the total
momentum of the system remains constant.
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
15
Example for Linear Momentum Conservation
Estimate an astronaut’s (M=70kg) resulting velocity after he throws his book
(m=1kg) to a direction in the space to move to another direction.
vA
vB
From momentum conservation, we can write
ur
r
r
ur
p i  0  p f  mA v A  mB v B
Assuming the astronaut’s mass is 70kg, and the book’s
mass is 1kg and using linear momentum conservation
r
r
1 r
mB v B
vB
 
vA  
70
mA
Now if the book gained a velocity
of 20 m/s in +x-direction, the
Astronaut’s velocity is
Wednesday, Oct. 29, 2008
 
r
r
r
1
20i  0.3i  m / s 
vA  
70
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
16
Impulse and Linear Momentum
Net force causes change of momentum 
Newton’s second law
By integrating the above
equation in a time interval ti to
tf, one can obtain impulse I.
So what do you
think an impulse is?

tf
ti
r dpr
F
dt
r
r r
r
dp  p f  pi  p 
r
r
dp  Fdt

tf
ti
r
r
Fdt  I
Effect of the force F acting on an object over the time
interval t=tf-ti is equal to the change of the momentum of
the object caused by that force. Impulse is the degree of
which an external force changes an object’s momentum.
The above statement is called the impulse-momentum theorem and is equivalent to Newton’s second law.
What are the
dimension and
unit of Impulse?
What is the
direction of an
impulse vector?
Defining a time-averaged force
r 1
r
F   Fi t
t i
Wednesday, Oct. 29, 2008
Impulse can be rewritten
If force is constant
r ur
I  Ft
r ur
I  F t
It is generally assumed that the impulse force acts on a
PHYS 1443-002, Fall 2008 Dr. Jaehoon
17
short time butYumuch greater than any other forces present.
Example for Impulse
(a) Calculate the impulse experienced when a 70 kg person lands on firm ground
after jumping from a height of 3.0 m. Then estimate the average force exerted on
the person’s feet by the ground, if the landing is (b) stiff-legged and (c) with bent
legs. In the former case, assume the body moves 1.0cm during the impact, and in
the second case, when the legs are bent, about 50 cm.
We don’t know the force. How do we do this?
Obtain velocity of the person before striking the ground.
KE PE
1 2
mv  mg  y  yi   mgyi
2
Solving the above for velocity v, we obtain
v  2 gyi  2  9.8  3  7.7m / s
Then as the person strikes the ground, the
momentum becomes 0 quickly giving the impulse
r ur
ur
ur ur
r
I  F t   p  p f  p i  0  mv 
r
r
 70kg  7.7m / s j  540 jN  s
Wednesday, Oct. 29, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
18
Example cont’d
In coming to rest, the body decelerates from 7.7m/s to 0m/s in a distance d=1.0cm=0.01m.
The average speed during this period is
The time period the collision lasts is
Since the magnitude of impulse is
0  vi
7.7

 3.8m / s
2
2
0.01m
d
3

2.6

10
s

t 
3.8m / s
uvr
r
I  F t  540N  s
v 
540
5
The average force on the feet during F  I 

2.1

10
N
3
this landing is
t 2.6 10
How large is this average force?
Weight  70kg  9.8m / s 2  6.9 102 N
F  2.1105 N  304  6.9 102 N  304  Weight
If landed in stiff legged, the feet must sustain 300 times the body weight. The person will
likely break his leg.
d 0.50m
 0.13s

t 
3.8
m
/
s
For bent legged landing:
v
540
F
 4.1 103 N  5.9Weight
0.13
Wednesday, Oct. 29, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
19
Another Example for Impulse
In a crash test, an automobile of mass 1500kg collides with a wall. The initial and
final velocities of the automobile are vi= -15.0i m/s and vf=2.60i m/s. If the collision
lasts for 0.150 seconds, what would be the impulse caused by the collision and the
average force exerted on the automobile?
Let’s assume that the force involved in the collision is a lot larger than any other
forces in the system during the collision. From the problem, the initial and final
momentum of the automobile before and after the collision is
r
r
r
r
pi  mvi  1500   15.0  i  22500i kg  m / s
r
r
r
r
p f  mv f 1500   2.60  i  3900i kg  m / s
ur ur ur
r
r
Therefore the impulse on the
I   p  p f  p i   3900  22500  i kg  m / s
r
r
4
automobile due to the collision is
 26400i kg  m / s  2.64 10 i kg  m / s
The average force exerted on the
automobile during the collision is
Wednesday, Oct. 29, 2008
ur
2.64 104 r
p
ur
F  t  0.150 i
r
r
 1.76 10 i kg  m / s  1.76 10 i N
5
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
2
5
20
Collisions
Generalized collisions must cover not only the physical contact but also the collisions
without physical contact such as that of electromagnetic ones in a microscopic scale.
Consider a case of a collision
between a proton on a helium ion.
F
F12
t
F21
Using Newton’s
3rd
The collisions of these ions never involve
physical contact because the electromagnetic
repulsive force between these two become great
as they get closer causing a collision.
Assuming no external forces, the force
exerted on particle 1 by particle 2, F21,
changes the momentum of particle 1 by
r r
dp1  F21dt
Likewise for particle 2 by particle 1
r r
dp2  F12dt
law we obtain
ur
ur
r
r
d p 2  F12dt   F21dt  d p1
So the momentum change of the system in the
collision is 0, and the momentum is conserved
Wednesday, Oct. 29, 2008
ur
ur
ur
d p  d p1  d p 2  0
ur
ur ur
p system  p1  p 2
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
 constant
21
Elastic and Inelastic Collisions
Momentum is conserved in any collisions as long as external forces are negligible.
Collisions are classified as elastic or inelastic based on whether the kinetic energy
is conserved, meaning whether it is the same before and after the collision.
Elastic
Collision
A collision in which the total kinetic energy and momentum
are the same before and after the collision.
Inelastic
Collision
A collision in which the total kinetic energy is not the same
before and after the collision, but momentum is.
Two types of inelastic collisions:Perfectly inelastic and inelastic
Perfectly Inelastic: Two objects stick together after the collision,
moving together at a certain velocity.
Inelastic: Colliding objects do not stick together after the collision but
some kinetic energy is lost.
Note: Momentum is constant in all collisions but kinetic energy is only in elastic collisions.
Wednesday, Oct. 29, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
22
Elastic and Perfectly Inelastic Collisions
In perfectly Inelastic collisions, the objects stick
together after the collision, moving together.
Momentum is conserved in this collision, so the
final velocity of the stuck system is
How about elastic collisions?
r
r
r
m1 v1i  m2 v 2i  (m1  m2 )v f
r
r
r
m v1i  m2 v 2i
vf  1
(m1  m2 )
r
r
r
r
m1 v1i  m2 v 2i  m1 v1 f  m2 v 2 f
1
1
1
1
In elastic collisions, both the
m1v12i  m2 v22i  m1v12f  m2 v22 f
2
2
2
2
momentum and the kinetic energy
m1 v12i  v12f   m2 v22i  v22 f 
are conserved. Therefore, the
final speeds in an elastic collision
m1 v1i  v1 f v1i  v1 f   m2 v2i  v2 f v2i  v2 f 
can be obtained in terms of initial From momentum
m1 v1i  v1 f   m2 v2i  v2 f 
speeds as
conservation above
 m  m2 
 2m2 
v1i  
v2i
v1 f   1
 m1  m2 
 m1  m2 
Wednesday, Oct. 29, 2008
 2m1 
 m  m2 
v1i   1
v2i
v2 f  
 m1  m2 
 m1  m2 
PHYS 1443-002, Fall 2008 Dr. Jaehoon
What happens
when the two
masses are the same?
Yu
23
Example for Collisions
A car of mass 1800kg stopped at a traffic light is rear-ended by a 900kg car, and the
two become entangled. If the lighter car was moving at 20.0m/s before the collision
what is the velocity of the entangled cars after the collision?
The momenta before and after the collision are
Before collision
ur
r
r
r
p i  m1 v1i  m2 v 2i  0  m2 v 2i
ur
r
r
r
p f  m1 v1 f  m2 v 2 f   m1  m2  v f
m2
20.0m/s
m1
After collision
Since momentum of the system must be conserved
m2
vf
m1
ur ur
pi  p f
r
r
m2 v 2i

vf
 m1  m2 
What can we learn from these equations
on the direction and magnitude of the
velocity before and after the collision?
Wednesday, Oct. 29, 2008
r
r
 m1  m2  v f  m2 v 2i
r
r
900  20.0i

 6.67i m / s
900  1800
The cars are moving in the same direction as the lighter
car’s original direction to conserve momentum.
The magnitude is inversely proportional to its own mass.
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
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