Download Tuesday, June 21, 2005

PHYS 1443 – Section 001
Lecture #12
Tuesday June 21, 2005
Dr. Andrew Brandt
•
•
•
Momentum
Impulse
Collisions
Tuesday June 21, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
1
Announcements
• Homework:
– HW7 on ch 8 due Tuesday 6/21 at 8pm
Tuesday June 21, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
2
Linear Momentum
The principle of energy conservation can be used to solve problems
that are harder to solve 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 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?
Tuesday June 21, 2005
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
 
dp d
dv
 ma  F

mv  m
dt
dt dt
PHYS 1443-001, Summer I 2005
3
Dr. Andrew Brandt
Linear Momentum and Forces
 
dp
d
F

mv
dt
dt
•
•
•
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
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.
What else can this
relationship be used for?
Can you think of a
few cases like this?
Tuesday June 21, 2005
The relationship can be used to study
the case where the mass changes as a
function of time.
Motion of a meteorite
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
Motion of a rocket
4
Linear Momentum Conservation
r
r
ur
ur
p1i  p 2i  m1 v1  m2 v 2
ur
ur
r'
r'
p1 f  p 2 f  m1 v1  m2 v 2
Tuesday June 21, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
5
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 a force on particle #2, there must be another force that
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’s say that the particle #1 has momentum
First define the momenta
p1 and #2 has p2 at a given time.
of the two particles
We use the momentum-force
relationship to define the force F21
(exerted on object 1 by object 2)
Since net force is 0
F
F 21 
 F 12  F 21
d p1
dt
this implies

F 12 

d p2
dt
d p 2 d p1  d p  p
0


2
1
dt
dt
dt
Therefore p 2  p1  const The total linear momentum of the system is conserved!!!
Tuesday June 21, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
6
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 acting 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.
Tuesday June 21, 2005
 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 1443-001, Summer I 2005
Dr. Andrew Brandt
7
Example for Linear Momentum Conservation
Estimate an astronaut’s resulting velocity after he throws a book.
vA
vB
From momentum conservation, we can write
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
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
Tuesday June 21, 2005
 
1
v A   70 20i   0.3 i m / s 
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
8
Example for Linear Momentum Conservation
A type of particle, a neutral kaon (K0), decays (breaks up) into a pair of particles
called pions (p+ and p-) that are oppositely charged but have equal mass. Assuming
the K0 is initially produced at rest, prove that the two pions must have momenta that
are equal in magnitude and opposite in direction.
This reaction can be written as
K0
pp
p
p
K p p
0
pp

Since this system consists of a K0 in the initial state
which results in two pions in the final state, the
momentum must be conserved. So we can write
Since the K0 is produced at
rest its momentum is 0.

p K 0  pp   pp 
p K 0  pp   pp   0
pp    pp 
Therefore, the two pions from this kaon decay have
momenta with same magnitude but opposite in direction.
Tuesday June 21, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
9
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
F
dp
dt
tf
d p  p f  p i  D p   F dt
t
i
d p  Fdt
tf
I   F dt  D p
ti
Impulse of the force F acting on a particle over the time
interval Dt=tf-ti is equal to the change of the momentum of
the particle caused by that force. Impulse is the degree to
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 units
of Impulse?
Defining a time-averaged force
1 tf
F
F dt

t
i
Dt
Tuesday June 21, 2005
Impulse can be rewritten
I  FDt
If the force is constant
I  F Dt
It is generally assumed that the impulse force acts for a short
PHYS 1443-001, Summer I 2005
10
time but
is
much
greater
than
any
other
forces
present.
Dr. Andrew Brandt
Example 9-6
(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  DPE
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  FDt  Dp  p f  pi  0  mv 
 70kg  7.7m / s  540 N  s
Tuesday June 21, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
11
Example 9 – 6 cont’d
In coming to rest, the body decelerates from 7.7m/s to 0m/s in a distance d=1.0cm=0.01m.
0  vi
7.7

 3.8m / s
v
2
2
0.01m
d
3
The time duration of the landing is

2.6

10
s

Dt 
v 3.8m / s
I  FDt  540N  s
Since the magnitude of impulse is
I
540
5
The average force on the feet during


2.1

10
N
F
3
this landing is
Dt 2.6 10
2
2
How large is this average force? Weight  70kg  9.8m / s  6.9 10 N
The average speed during this period is
F  2.1105 N  304  6.9 102 N  304  Weight
If landing is stiff-legged, the feet must sustain 300 times the body weight. The person will
likely break his leg(s).
d 0.50m

 0.13s
Dt 
3.8
m
/
s
For bent-legged landing:
v
540
F
 4.1103 N  5.9Weight
0.13
Tuesday June 21, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
12
Example for Impulse
A golf ball of mass 50g is struck by a club. The force exerted on the ball by the club
varies from 0, at the instant before contact, up to some maximum value at which the
ball is deformed and then back to 0 when the ball leaves the club. Assuming the ball
travels 200m, estimate the magnitude of the impulse caused by the collision.
vB
qB
m
The range R of a projectile is
vB2 sin 2q B
 200m
R
g
Let’s assume that launch angle qi=45o.
vB  200  g  1960  44m / s
Then the speed becomes:
(immediately before the collision)
Considering the time interval for the vi  0
collision, ti and tf , the initial and
v f  44m / s (immediately after the collision)
the final speed are
Therefore the magnitude of the impulse I  D p  mvBf  mvBi
on the ball due to the force of the club is
 0.05  44  2.2kg  m / s
Tuesday June
21,can
2005solve impulsePHYS
1443-001,through
Summer Iknowing
2005
Remember,
you
problems
Dr. Andrew Brandt
the force or knowing the change in momentum
13
Another Example for Impluse
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  D 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
Tuesday June 21, 2005
4
D p 2.64 10
F  Dt 
0.150
 1.76 105 i kg  m / s 2  1.76 105 i N
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
14
Collisions
Generalized collisions must cover not only the case of physical contact but also
collisions without physical contact such as particles on a microscopic scale.
The collisions of these ions never involves
Consider the case of a collision
physical contact because the electromagnetic
between a proton and a helium ion.
repulsive force between these two becomes great
as they get closer causing a collision.
F
F12
t
F21
Using Newton’s
3rd
Assuming no external forces, the force
exerted on particle 1 by particle 2, F21,
changes the momentum of particle 1 by
ur ur
d p 2  F 12 dt
Likewise for particle 2 on particle 1
law we obtain
ur
ur
d p 2  F 12 dt
So the momentum change of the system in the
collision is 0 and momentum is conserved
Tuesday June 21, 2005
ur ur
d p1  F 21dt
ur
ur
  F 21dt  d p1
ur
ur
D p  d p1  d p 2  0
p system
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
 p1  p 2
 constant
15
Elastic and Inelastic Collisions
Momentum is conserved in any collision as long as external forces are negligible.
A collisions is classified as elastic or inelastic depending on whether kinetic
energy is conserved during 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 the 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 constant only in elastic ones.
Tuesday June 21, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
16
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?
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 v2 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
dividing gives:
 m1  m2 
 2m2 
general solution:
v
v
  v
1i
1f
2i
Tuesday June 21, 2005
 v2 f

v1i  
v2i
v1 f  
 m1  m2 
 m1  m2 
 2
m m 
m  Summer
PHYS
1443-001,
v1i   1I 20052 v2i
v2 f   Dr. 1Andrew
Brandt
 m1  m2 
 m1  m2 
Special cases
17