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PHYS 1443 – Section 001
Lecture #15
Monday, April 4, 2011
Dr. Jaehoon Yu
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Linear Momentum and Forces
Linear Momentum Conservation
Collisions and Impulse
Collisions – Elastic and Inelastic Collisions
Center of Mass
Announcements
• Second non-comprehensive term exam date
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Time: 1 – 2:20pm, Wednesday, Apr. 6
Location: SH103
Covers: CH6.4 – CH9.3
Please do NOT miss the exam!!
The worse of the two non-comprehensive term exams will be
dropped!
• Colloquium Wednesday at 4pm in SH101
Monday, April 4, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
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Monday, April 4, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
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Reminder: Special Project
L
θA
h{
B
m
T
m
mg
Monday, April 4, 2011
A ball of mass m is attached to a light cord of
length L, making up a pendulum. The ball is
released from rest when the cord makes an
angle θA which is greater than 90 degrees
with the vertical, and the pivoting point P is
frictionless. Find the speed of the ball when
it is at the lowest point, B. 10 points
Due: beginning of the class Monday, Apr. 11
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
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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 20 of this
lecture note in a far greater detail than the note.
– 20 points extra credit
• Show mathematically what happens to the final
velocities if m1=m2 and describe in words the resulting
motion.
– 5 point extra credit
• Due: Start of the class Wednesday, Apr. 13
Monday, April 4, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
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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 the 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, April 4, 2011
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 1443-001, Spring 2011
Dr. Jaehoon Yu
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Linear Momentum and Forces
r
r
r
p dp
 F  lim t  dt
t 0
•
•
•
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 the net force is 0, the particle’s linear momentum is a
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, April 4, 2011
The relationship can be used to study
the case where the mass changes as a
function of time.
 
r
r
r dpr
d mv
d
v
dm r

v m

F

 dt
dt
dt
dt
Motion
of a Spring
meteorite
PHYS 1443-001,
2011
Dr. Jaehoon Yu
Motion of a rocket
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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.
Now how would the momenta
of these particles look like?
Using momentumforce relationship
Let say that the particle #1 has momentum
p1 and #2 has p2 at some point of time.
r
r
r
r
dp
dp
F21  1 and F12  2
dt
dt
r
r
ur ur
ur
dp2 dp1 d r r
 F  F12  F 21  dt  dt  dt  p2  p1   0
And since net force
of this system is 0
ur
ur
Therefore p  p  const The total linear momentum of the system is conserved!!!
2
1
Monday, April 4, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
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More on Conservation of Linear Momentum in
a Two Body System
ur
ur ur
 p  p2  p1 const
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.
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.
Monday, April 4, 2011

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 1443-001, Spring 2011
Dr. Jaehoon Yu
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Linear Momentum Conservation
Initial
r
r
r
r
p1i  p2i  m1v1  m2v2
Final
r
r
r
r
p1 f  p2 f  m1v1  m2v2
Monday, April 4, 2011
r
r
r
r
p1 f  p2 f
p

p

1
i
2
i
PHYS 1443-001, Spring 2011
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Dr. Jaehoon Yu
Example 9.4: Rifle Recoil
Calculate the recoil velocity of a 5.0kg rifle that shoots a 0.020kg bullet at a
speed of 620m/s.
From
ur momentum
ur conservation, we can write
r
r
r r
p i  0  p f  PR  PB mR v R  mB v B
The x-comp
PR  PB  mR v R  mB v B  0
Solving the above for vR and using the rifle’s mass and the bullet’s mass, we
obtain
mB
0.020
vR 
vB 
 620  2.5m s
mR
5.0
r
r
v R  2.5i m s 
Monday, April 4, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
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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
ur
r
r
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
mB v B
1
 
vA  
vB
mA
70
Now if the book gained a velocity
of 20 m/s in +x-direction, the
Astronaut’s velocity is
Monday, April 4, 2011
 
r
r
r
1
20i  0.3i m / s
vA  
70
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu

12

Impulse
There are many situations when the force on an
object is not constant and in fact quite complicated!!
Monday, April 4, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
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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 Δt
 
ur
r
r
r
 F t  mvf  mvo  J
Monday, April 4, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
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Impulse and Linear Momentum
Net force causes change of momentum 
Newton’s second law
r
r dp
F
dt
r
r
dp  Fdt
By integrating the above
tf r
r
r
ur
r
tf r
equation in a time interval ti to
dp  p f  pi  p  F dt  J
ti
ti
tf, one can obtain impulse I.
Effect of the force F acting on an object over the time
So what do you
interval Δt=tf-ti is equal to the change of the momentum of
think an impulse is?
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
Monday, April 4, 2011
r
r
1
F   Fi t
t i
Impulse can be rewritten
If force is constant
ur ur
J  F t
ur ur
J  F t
It is generally assumed that the impulse force acts on a
PHYS 1443-001, Spring 2011
short time
much
Dr. but
Jaehoon
Yu greater than any other forces present.
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An 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
Therefore the impulse on the
automobile due to the collision is
The average force exerted on the
automobile during the collision is
Monday, April 4, 2011
ur
J
r
ur ur ur
  p  p f  pi 3900  22500i kg  m / s
r
r
4
 26400i u
kg
r  m / s  2.64  10 i kg  m / s
ur
2.64  10 4 r
p
i
F  t 
r 0.150
r
 1.76 10 i kg  m / s  1.76 10 i N
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
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5
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Another 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
ur ur
ur
ur ur
r
J  F t   p  p f  p i  0  mv 
r
r
 70kg  7.7m / s j  540 j N  s
Monday, April 4, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
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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 
ur
uvr 3.8m / s
J  F t  540N  s
v 
540
5
The average force on the feet during F  J

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
Monday, April 4, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
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