PHYS 1441 – Section 001
Lecture # 11
Wednesday, June 18, 2008
Dr. Jaehoon Yu
•
•
•
•
•
•
Wednesday, June 18,
2008
Linear Momentum
Linear Momentum and Impulse
Linear Momentum Conservation
Collisions
Center of Mass
Fundamentals of Rotational Motion
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
1
Announcements
• Quiz next Monday, June 23
– Beginning of the class
– Covers CH 6.7 – what we cover tomorrow
• Mid-term grade discussions
– Next Monday, June 23, bottom half of the class
– Do not miss this discussion!!
• Final exam
– 8- 10am, Monday, June 30, in SH103
– Comprehensive exam: Covers CH 1 – What we
complete next Thursday, June 25
Wednesday, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
2
Extra-Credit Special Project
• Derive the formula for the final velocity of two objects
which underwent an elastic collision as a function of
known quantities m1, m2, v01 and v02 in page 19 of
this lecture note. Must be done in far greater detail
than what is covered in the lecture note.
– 20 points extra credit
• Describe in detail what happens to the final velocities
if m1=m2.
– 5 point extra credit
• Due: Start of the class next Wednesday, June 25
Wednesday, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
3
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 you see from the
definition? Do you see force?
Wednesday, June 18,
2008
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 1441-001, Summer 2008
Dr. Jaehoon Yu
4
Impulse
There are many situations when the force on an
object is not constant.
Wednesday, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
5
Impulse and Linear Momentum
Net force causes change of momentum 
Newton’s second law
The quantity impulse is defined as
the change of momentum
So what do you
think the impulse is?
r
r p
F
t
r
r
p  F t
r
r
r
r
r r
J  p  p f  pi  mv f  mv0
The effect of a 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
Wednesday, June 18,
2008
Impulse can be rewritten
ur ur
J  Ft
r 1
r
F   Fi t
t i
Impulse
is a vector
quantity!!
PHYS 1441-001,
Summer
2008
Dr. Jaehoon Yu
If force is constant
ur ur
J  Ft
6
Ball Hit by a Bat
r r
r vf  vo
a
t
r
r
 F  ma
r mvr f  mvr o
 F  t
Multiply either side by t
 
r
r
r
r
 F t  mvf  mvo  J
Wednesday, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
7
Ex. 1 A Well-Hit Ball
A baseball (m=0.14kg) has an initial velocity of v0=-38m/s as it approaches a bat. We
have chosen the direction of approach as the negative direction. The bat applies an
average force F that is much larger than the weight of the ball, and the ball departs
from the bat with a final velocity of vf=+58m/s. (a) determine the impulse applied to the
ball by the bat. (b) Assuming that the time of contact is t=1.6x10-3s, find the average
force exerted on the ball by the bat.
What are the forces involved in this motion? The force by the bat and the force by the gravity.
Since the force by the bat is much greater than the weight, we ignore the ball’s weight.
(a) Using the impulsemomentum theorem
ur
r
r
r
J  p  mv f  mv0
 0.14  58  0.14   38  13.4kg  m s
(b)Since the impulse is known and the time during which the contact occurs are
know, we can compute the average force
r exerted on the ball during the contact
r
J
13.4

 8400 N
F
3
t 1.6 10
r
r
8400 r
How large is r
W  mg  0.14  9.8  1.37N
F 
W  6131 W
this force?
1.37
r
r
J  F t
Wednesday, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
8
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, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
9
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, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
10
Linear Momentum and Forces
r
r 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?
Wednesday, June 18,
2008
The relationship can be used to study
the case where the mass changes as a
function of time.
r
r
  mv  m r
r pr
v
 F  t  t  t v m t
Motion
of aSummer
meteorite
PHYS 1441-001,
2008
Dr. Jaehoon Yu
Motion of a rocket
11
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 thus the net force in the SYSTEM of the two particles is still 0.
Let’s say that the particle #1 has momentum
Now what would the momenta
p1 and #2 has p2 at some point of time.
of these particles look like?
Using momentumforce relationship
r
r
p1
F21 
t
and
r
r
p2
F12 
t
r
r
ur ur
ur
p2 p1  r r
 F  F 12  F 21  t  t  t  p2  p1 
And since net force
0
of this system is 0
ur
ur
Therefore p 2  p1  const The total linear momentum of the system is conserved!!!
Wednesday, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
12
Linear Momentum Conservation
Initial
r
r
r
r
p1i  p2i  m1v1  m2v2
Final
r
r
r
r
p1 f  p2 f  m1v1  m2v2
Wednesday, June 18,
2008
r
r
r
r
p1 f  p2 f
p

p

1
i
2
i
PHYS 1441-001, Summer 2008
13
Dr. Jaehoon 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, June 18,
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 1441-001, Summer 2008
Dr. Jaehoon Yu
14
Ex. 6 Ice Skaters
Starting from rest, two skaters push off
against each other on ice where friction is
negligible. One is a 54-kg woman and one
is a 88-kg man. The woman moves away
with a speed of +2.5 m/s. Find the recoil
velocity of the man.
No net external force  momentum conserved
Pf  Po
m1v f 1  m2 v f 2  0
Solve for Vf2
vf 2
vf 2  
m1v f 1
m2
54 kg  2.5m s 


 1.5m s
88 kg
Wednesday, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
15
How do we apply momentum conservation?
1. Decide which objects are included in the system.
2. Identify the internal and external forces relative to
the system.
3. Verify that the system is isolated.
4. Set the final momentum of the system equal to its
initial momentum. Remember that momentum is
a vector.
Wednesday, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
16
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
p1  F21t
Likewise for particle 2 by particle 1
r r
p2  F12t
law we obtain
ur
ur
r
r
 p 2  F12 t   F21t   p1
So the momentum change of the system in the
collision is 0, and the momentum is conserved
Wednesday, June 18,
2008
ur
ur
ur
 p   p1   p 2  0
ur
ur ur
p system  p1  p 2
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
 constant
17
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, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
18
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  v22 f  v22i 
are conserved. Therefore, the
final speeds in an elastic collision
m1 v1i  v1 f v1i  v1 f  m2  v2 f  v2i  v2i  v2 f 
can be obtained in terms of initial From momentum
m1  v1i  v1 f   m2  v2 f  v2i 
speeds as
conservation above
 m  m2 
 2m2 
v1i  
v2i
v1 f   1
 m1  m2 
 m1  m2 
Wednesday, June 18,
What
2008
 2m1 
 m  m2 
v1i   1
v2i
v2 f  
 m1  m2 
 m1  m2 
PHYS 1441-001, Summer 2008
happens when
the
two masses are the same?
Dr. Jaehoon Yu
19
Ex. 8 A Ballistic Pendulum
The mass of the block of wood is 2.50-kg and the
mass of the bullet is 0.0100-kg. The block swings
to a maximum height of 0.650 m above the initial
position. Find the initial speed of the bullet.
What kind of collision? Perfectly inelastic collision
No net external force  momentum conserved
m1v f 1  m2 v f 2  m1vo1  m2vo 2
 m1  m2 
Solve for V01
v f  m1vo1
vo1 
 m1  m2  v f
m1
What do we not know? The final speed!!
How can we get it? Using the mechanical
energy conservation!
Wednesday, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
20
Ex. 8 A Ballistic Pendulum, cnt’d
Now using the mechanical energy conservation
1
2
mv 2  mgh
 m1  m2  ghf

gh f 
1
2
 m1  m2  v2f
1
2
v 2f

Solve for Vf

v f  2 gh f  2 9.80 m s2  0.650 m 
Using the solution obtained previously, we obtain
vo1
m1  m2  v f


m1

 m1  m2 
2 gh f
m1
 0.0100 kg  2.50 kg 
2

 2  9.80 m s   0.650 m 
0.0100 kg


 896m s
Wednesday, June 18,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
21
Two dimensional Collisions
In two dimension, one needs to use components of momentum and
apply momentum conservation to solve physical problems.
m1
r
r
r
r
m1 v1i  m2 v 2i  m1 v1 f  m2 v 2 f
v1i
m2
q
f
x-comp.
m1v1ix  m2v2ix  m1v1 fx  m2v2 fx
y-comp.
m1v1iy  m2v2iy  m1v1 fy  m2v2 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:
r
r
r
m1 v1i  m2 v 2i  m1 v1i
m1v1ix  m1v1 fx  m2 v2 fx  m1v1 f cos q  m2 v2 f cos f
m1v1iy  0  m1v1 fy  m2 v2 fy  m1v1 f sin q  m2 v2 f sin f
And for the elastic collisions, the
kinetic energy is conserved:
Wednesday, June 18,
2008
1
1
1
m1v 12i  m1v12f  m2 v22 f
2
2
2
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
What do you think
we can learn from
these relationships?
22
Example for 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
q
Since both the particles are protons m1=m2=mp.
Using momentum conservation, one obtains
x-comp. m p v1i  m p v1 f cos q  m p v2 f cos f
y-comp.
f
m p v1 f sin q  m p v2 f sin f  0
Canceling mp and putting 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
2
2f
Wednesday, June 18,
2008
v1 f sin 37  v2 f sin f
(2)
v1 f  2.80 105 m / s
Solving Eqs. 1-3
5
(3) equations, one gets v2 f  2.1110 m / s
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
f  53.0
Do this at
home
23