PHYS 1443 – Section 001
Lecture #12
Monday, June 27, 2011
Dr. Jaehoon Yu
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•
•
•
Linear Momentum and Forces
Linear Momentum Conservation
Impulse and Linear Momentum
Collisions
Today’s homework is homework #7, due 10pm, Thursday, June 30!!
Monday, June 27, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
1
Announcements
• Quiz #3 tomorrow, Wednesday, June 29
– Beginning of the class
– Covers CH8.1 through what we learn tomorrow, Tuesday,
June 28
• Mid-term grade discussions
– Second half of the class
– I strongly urge you all to come and discuss your grades
• Bring your special projects to the grade discussion if
you haven’t already submitted
Monday, June 27, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
2
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 of mass m moving
at the velocity 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, June 27, 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
3
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, June 27, 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 aSpring
meteorite
PHYS
1443-001,
2011 Dr.
Jaehoon Yu
Motion of a rocket
4
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 a reaction force that
the particle #2 exerts on #1. Both the forces are internal forces, and thus
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, June 27, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
5
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, June 27, 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
6
Linear Momentum Conservation
Initial
r
r
r
r
p1i  p2i  m1v1  m2v2
Final
r
r
r
r
p1 f  p2 f  m1v1  m2v2
Monday, June 27, 2011
r
r
r
r
p1 f  p2 f
p

p

1
i
2
i
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
7
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, June 27, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu

8
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, June 27, 2011
 
r
r
r
1
20i  0.3i m / s
vA  
70
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu

9

Impulse
There are many situations where the force on an
object is not constant and in fact quite complicated
during the motion!!
Monday, June 27, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
10
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, June 27, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
11
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, June 27, 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 Dr.
short timeJaehoon
but much
Yu greater than any other forces present.
12
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, June 27, 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
5
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
2
5
13
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, June 27, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
14
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, June 27, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
15