PHYS 1441 – Section 501
Lecture #10
Monday, July 5, 2004
Dr. Jaehoon Yu
•
•
•
•
•
•
Energy Diagrams
Power
Linear Momentum & its conservation
Impulse & Collisions
Center of Mass
CM of a group of particles
Remember the second term exam, Monday, July 19!!
Monday, July 5, 2004
PHYS 1441-501, Summer 2004
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 would this diagram be?
1
U  kx2
2
1.
Minimum
 Stable
equilibrium
Maximum
unstable
equilibrium
1 2
kx
2
A Parabola
What does this energy diagram tell you?
Us
-xm
Us 
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 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.
Monday, July 5, 2004
PHYS 1441-501, Summer 2004
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 irreversible 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 place to another.
Total energy of universe is constant.
In any physical or chemical process, mass is neither created nor destroyed.
Principle of
Conservation of Mass Mass before a process is identical to the mass after the process.
Einstein’s MassEnergy equality.
Monday, July 5, 2004
ER  mc
2
How many joules does your body correspond to?
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
3
Power
• Rate at which work is done
– What is the difference for the same car with two different engines (4
cylinder and 8 cylinder) climbing the same hill?  8 cylinder car climbs
up faster
NO
The rate at which the same amount of work
performed is higher for 8 cylinder than 4.
Is the total amount of work done by the engines different?
Then what is different?
Average power
W
P
t
Instantaneous power
r
u
r
W
s
F
cos  
 lim
P  lim

t 0
t  0  t
t
ur r
 F v cos
Unit? J / s  Watts 1HP  746Watts
What do power companies sell? 1kWH  1000Watts  3600s  3.6  106 J
Monday, July 5, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
Energy
4
Energy Loss in Automobile
Automobile uses only at 13% of its fuel to propel the vehicle.
67% in the engine:
1. Incomplete burning
2. Heat
3. Sound
Why?
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
fa 
mcar  1450kg Weight  mg  14200 N
m n  m mg  227 N
1
1
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
Monday, July 5, 2004
Total Resistance
ft  f r  f a
P  ft v   691N   26.8  18.5kW
Pr  f r v  227   26.8  6.08kW
Pa 2004
 fav
PHYS 1441-501, Summer
Dr. Jaehoon Yu
 464.7 26.8  12.5kW5
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, July 5, 2004
1.
2.
3.
4.
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
ur
r
r
r
ur
r
m
v

v
0
 p mv  mv 0
v


ma   F
 m
t
t
t
t
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
6
Linear Momentum and Forces
ur
ur  p
 F  t
•
•
•
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, July 5, 2004
The relationship can be used to study
the case where the mass changes as a
function of time.
Motion of a meteorite
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Dr. Jaehoon Yu
Motion of a rocket
7
Linear Momentum Conservation
p1i  p 2i  m1 v1  m2 v 2
'
1 1
p1 f  p 2 f  m v  m2 v
Monday, July 5, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
8
'
2
Conservation of Linear Momentum in a Two
Particle System
Consider a system with two particles that does 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 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
And since net force
of this system is 0
ur
ur  p
F 21  1
t
F
and
 F 12  F 21
ur
ur  p
F 12  2
t
ur
ur
 p 2  p1   upr  upr


2
1

t
t t


0
Therefore p 2  p1  const The total linear momentum of the system is conserved!!!
Monday, July 5, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
9
More on Conservation of Linear Momentum in
a Two Particle 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.
 p p
ur
ur
ur
ur
p 2i  p1i  p 2 f  p1 f
Mathematically this statement can be written as
xi
system

P
xf
system
P
yi
system
This can be generalized into
conservation of linear momentum in
many particle systems.
Monday, July 5, 2004
 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 interaction
What does this mean?
P
2

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 1441-501, Summer 2004
Dr. Jaehoon Yu
10
Example for Linear Momentum Conservation
Estimate an astronaut’s resulting velocity after he throws his book to a
direction in the space to move to a direction.
vA
vB
From momentum conservation, we can write
p i  0  p f  mA v A  mB v B
Assuming the astronaut’s mass if 70kg, and the book’s
mass is 1kg and using linear momentum conservation
vA
1
mB v B
vB
 

70
mA
Now if the book gained a velocity
of 20 m/s in +x-direction, the
Astronaut’s velocity is
Monday, July 5, 2004
 
1
v A   70 20i   0.3 i m / s 
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
11
Impulse and Linear Momentum
ur
Net force causes change of momentum 
Newton’s second law
ur  p
ur ur
F
 p  Ft
t
By summing the above
equation in a time interval ti to
tf, one can obtain impulse I.
ur
 F t
So what do you
think an impulse is?
ur ur ur
 p  p f  pi
r ur
ur
I  Ft   p
Impulse of the force F acting on a particle over the time
interval t=tf-ti is equal to the change of the momentum of
the particle caused by that force. Impulse is the degree of
which an external force changes 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?
Monday, July 5, 2004
Defining a time-averaged force
ur 1
ur
F   F i t
t i
Impulse can be rewritten
If force is constant
I  Ft
I  F t
It is generally assumed that the impulse force acts on a
PHYS 1441-501, Summer 2004
short time
much
Dr. but
Jaehoon
Yu greater than any other forces present.
12
Example 7-5
(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
I  Ft  p  p f  pi  0  mv 
 70kg  7.7m / s  540 N  s
Monday, July 5, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
13
Example 7-5 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
0  vi
7.7

 3.8m / s
v
2
2
0.01m
d
 2.6  103 s
t  
v 3.8m / s
I  Ft  540N  s
I
540
5
The average force on the feet during


2.1

10
N
F
3
this landing is
t 2.6 10
2
2
How large is this average force? Weight  70kg  9.8m / s  6.9 10 N
Since the magnitude of impulse is
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.
Monday, July 5, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
14
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
p i  mvi  1500   15.0i  22500i kg  m / s
p f  mv f  1500  2.60i  3900i kg  m / s
Therefore the impulse on the
I   p  p  pi  3900  22500 i kg  m / s
automobile due to the collision is
 26400i kg  m / s  2.64 104 i kg  m / s
f
The average force exerted on the
automobile during the collision is
Monday, July 5, 2004
4
 p 2.64 10
F  t 
0.150
 1.76 105 i kg  m / s 2  1.76 105 i N
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
15
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 involves a
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
ur ur
 p1  F 21t
Likewise for particle 2 by particle 1
ur ur
 p 2  F 12 t
law we obtain
ur
 p 2  F 12 t
So the momentum change of the system in the
collision is 0 and the momentum is conserved
Monday, July 5, 2004
ur
  F 21t   p1
 p   p1   p 2  0
p system
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
 p1  p 2
 constant
16
Elastic and Inelastic Collisions
Momentum is conserved in any collisions as long as external forces negligible.
Collisions are classified as elastic or inelastic by the conservation of kinetic
energy before and after the collisions.
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 at a certain velocity together.
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.
Monday, July 5, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
17
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
m1 v1i  m2 v 2i  (m1  m2 )v f
vf 
m1 v1i  m2 v 2i
(m1  m2 )
m1 v1i  m2 v 2i  m1 v1 f  m2 v 2 f
How about elastic collisions?
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 
Monday, July 5, 2004
 2m1 
 m  m2 
v1i   1
v2i
v2 f  
 m1  m2 
 m1  m2 
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
18
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
p i  m1 v1i  m2 v 2i  0  m2 v 2i
m2
20.0m/s
m1
p f  m1 v1 f  m2 v 2 f  m1  m2 v f
After collision
Since momentum of the system must be conserved
m2
vf
m1
vf
What can we learn from these equations
on the direction and magnitude of the
velocity before and after the collision?
Monday, July 5, 2004
m1  m2 v f
pi  p f

 m2 v 2i
m2 v 2i
900  20.0i

 6.67i m / s
m1  m2  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 1441-501, Summer 2004
Dr. Jaehoon Yu
19
Two dimensional Collisions
In two dimension, one can use components of momentum to apply
momentum conservation to solve physical problems.
m1
m1 v 1i  m2 v 2 i  m1 v1 f  m2 v 2 f
v1i
m2

f
x-comp.
m1v1ix  m2 v2ix  m1v1 fx  m2 v2 fx
y-comp.
m1v1iy  m2 v2iy  m1v1 fy  m2 v2 fy
Consider a system of two particle collisions and scatters in
two dimension as shown in the picture. (This is the case at
fixed target accelerator experiments.) The momentum
conservation tells us:
m1 v 1i  m2 v 2i  m1 v1i
m1v1ix  m1v1 fx  m2 v2 fx  m1v1 f cos   m2 v2 f cos f
m1v1iy  0  m1v1 fy  m2 v2 fy  m1v1 f sin   m2 v2 f sin f
And for the elastic conservation,
the kinetic energy is conserved:
Monday, July 5, 2004
1
1
1
m1v 12i  m1v12f  m2 v22 f
2
2
2
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
What do you think
we can learn from
these relationships?
20
Example of Two Dimensional Collisions
Proton #1 with a speed 3.50x105 m/s collides elastically with proton #2 initially at
rest. After the collision, proton #1 moves at an angle of 37o to the horizontal axis and
proton #2 deflects at an angle f to the same axis. Find the final speeds of the two
protons and the scattering angle of proton #2, f.
m1
v1i
m2

Since both the particles are protons m1=m2=mp.
Using momentum conservation, one obtains
x-comp. m p v1i  m p v1 f cos   m p v2 f cos f
y-comp.
f
m p v1 f sin   m p v2 f sin f  0
Canceling mp and put in all known quantities, one obtains
v1 f cos 37  v2 f cos f  3.50 105 (1)
From kinetic energy
conservation:
3.50 10 
5 2
 v v
2
1f
Monday, July 5, 2004
2
2f
v1 f sin 37  v2 f sin f
Solving Eqs. 1-3
(3) equations, one gets
(2)
v1 f  2.80 105 m / s
v2 f  2.1110 m / s
5
f  53.0
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
Do this at
home
21
Center of Mass
We’ve been solving physical problems treating objects as sizeless
points with masses, but in realistic situation objects have shapes
with masses distributed throughout the body.
Center of mass of a system is the average position of the system’s mass and
represents the motion of the system as if all the mass is on the point.
What does above
statement tell you
concerning forces being
exerted on the system?
m2
m1
x1
x2
xCM
Monday, July 5, 2004
The total external force exerted on the system of
total mass M causes the center of mass to move at
an acceleration given by a   F / M as if all
the mass of the system is concentrated on the
center of mass.
Consider a massless rod with two balls attached at either end.
The position of the center of mass of this system is
the mass averaged position of the system
m x  m2 x2
CM is closer to the
xCM  1 1
m1  m2
heavier object
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
22
Center of Mass of a Rigid Object
The formula for CM can be expanded to Rigid Object or a
system of many particles
xCM
m x  m x      mn xn
 11 2 2
m1  m2      mn
m x

m
i i
i
i
yCM
i
mi
ri
rCM
i
zCM
i
i
i

m x
i
i
i   mi yi j   mi zi k
i
i
 mi r i
m
i
i
i
i
M
A rigid body – an object with shape
and size with mass spread throughout
the body, ordinary objects – can be
considered as a group of particles with
mass mi densely spread throughout
the given shape of the object
Monday, July 5, 2004
i i
i
r CM  xCM i  yCM j  zCM k
r CM 
m z

m
i
i
i
The position vector of the
center of mass of a many
particle system is
m y

m
xCM 
 m x
i
M
xCM  lim
m 0
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
i
i
i
r CM 
 m x
i
i
M
i

1
rdm

M
1
M
 xdm
23
Example 7-11
Thee people of roughly equivalent mass M on a lightweight (air-filled)
banana boat sit along the x axis at positions x1=1.0m, x2=5.0m, and
x3=6.0m. Find the position of CM.
Using the formula
for CM
m x

m
i
xCM
i
i
i
i
M 1.0 M  5.0 M  6.0  12.0 M
 4.0(m)

3M
M M M
Monday, July 5, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
24
Example for Center of Mass in 2-D
A system consists of three particles as shown in the figure. Find the
position of the center of mass of this system.
Using the formula for CM for each
position vector component
y=2 m (0,2)
1
m x

m
i
(0.75,4)
xCM
rCM
2 2
1
i
2
3 3
3

m2  2m3
m1  m2  m3
i
i
i
i
One obtains r CM  x i  y j 
CM
CM
i i
1 1
yCM
i
m x m x m x m x
xCM  m  m  m  m

i
i
m y

m
i
i
(2,0)
m3
x=2
(1,0)
m2
x=1
i
m2  2m3  i  2m1 j
m1  m2  m3
If m1  2kg; m2  m3  1kg
i
m y
yCM  m

i
i
i
i
i
Monday, July 5, 2004

m1 y1  m2 y2  m3 y3
2m1

m1  m2  m3
m1  m2  m3
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
r CM 
3i  4 j
 0.75i  j
4
25
Motion of a Diver and the Center of Mass
Diver performs a simple dive.
The motion of the center of mass
follows a parabola since it is a
projectile motion.
Diver performs a complicated dive.
The motion of the center of mass
still follows the same parabola since
it still is a projectile motion.
Monday, July 5, 2004
The motion of the center of mass
of the diver is always the same.
PHYS 1441-501, Summer 2004
26
Dr. Jaehoon Yu
Center of Mass and Center of Gravity
The center of mass of any symmetric object lies on an
axis of symmetry and on any plane of symmetry, if
object’s mass is evenly distributed throughout the body.
How do you think you can
determine the CM of
objects that are not
symmetric?
Center of Gravity
mi
Axis of
One can use gravity to locate CM.
symmetry
1. Hang the object by one point and draw a vertical line
following a plum-bob.
2. Hang the object by another point and do the same.
3. The point where the two lines meet is the CM.
Since a rigid object can be considered as collection
of small masses, one can see the total gravitational
force exerted on the object as
F g   F i   mi g  M g
i
mig
Monday, July 5, 2004
CM
What does this
equation tell you?
i
The net effect of these small gravitational
forces is equivalent to a single force acting on
a point (Center of Gravity) with mass M.
PHYS 1441-501, Summer 2004
27
The CoG is the point in an object
as
if
all
the
gravitational
force
is
acting
on!
Dr. Jaehoon Yu
Motion of a Group of Particles
We’ve learned that the CM of a system can represent the motion of a system.
Therefore, for an isolated system of many particles in which the total mass
M is preserved, the velocity, total momentum, acceleration of the system are
Velocity of the system
Total Momentum
of the system
Acceleration of
the system
External force exerting
on the system
If net external force is 0
Monday, July 5, 2004
v CM
r
r CM    1



t
M
t
r 
 mi r i   1
M
r
r i  mi v i
m
 i t  M
mv

 m v   p
p CM  M vCM  M
M
i i
a CM
r
vCM    1


t  M
t
 F ext  M aCM

r 
 mi vi   1
M
ur
 ptot
  mi a i 
t
ur
F ext  0   ptot
t
i i
p tot
i
 ptot
r
vi  mi a i
 mi t  M
What about the
internal forces?
System’s momentum
 const is conserved.
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
28