Transcript Monday, Nov. 3, 2008

PHYS 1443 – Section 002
Lecture #15
Monday, Nov. 3, 2008
Dr. Jae Yu
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Linear Momentum
Conservation of Momentum
Impulse
Collisions – Elastic and Inelastic Collisions
Exam Problem Solving Session
Today’s homework is HW #9, due 9pm, Monday, Nov. 10!!
Monday, Nov. 3, 2008
PHYS 1443-002, Fall 2008
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 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, Nov. 3, 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 1443-002, Fall 2008 Dr. Jaehoon
Yu
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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.
Monday, Nov. 3, 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
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Example for Linear Momentum Conservation
What is ther astronaut’s
(M=70kg) resulting velocity after he throws his book
r
(m=1kg, v  20i  m s ) in the space to move to the opposite 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
Monday, Nov. 3, 2008
 
r
r
r
1
20i  0.3i  m / s 
vA  
70
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
4
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
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
Monday, Nov. 3, 2008
r 1
r
F   Fi t
t i
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
short time butYumuch greater than any other forces present.
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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
Monday, Nov. 3, 2008
PHYS 1443-002, Fall 2008 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 
3.8m / s
uvr
r
I  F t  540N  s
v 
540
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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
Monday, Nov. 3, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
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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?
assume that the force involved in the collision is a lot larger than any other forces in
the system during the collision. From thLet’s e 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
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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
Monday, Nov. 3, 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
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PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
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