Transcript Wednesday, April 8, 2009

PHYS 1441 – Section 002
Lecture #17
Wednesday, Apr. 8, 2009
Dr. Jaehoon Yu
•
•
•
•
•
Wednesday, Apr. 8, 2009
Linear Momentum and Forces
Linear Momentum Conservation
Collisions
Center of Mass
Fundamentals of Rotational Motion
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
1
Announcements
• Quiz Results
– Class average: 2.5/6
• Previous quizzes:
– Top score: 5/6
• 2nd term exam
–
–
–
–
–
1 – 2:20pm, Wednesday, Apr. 22, in SH103
Non-comprehensive exam
Covers: Ch. 6.1 – what we complete next Wednesday, Apr. 15
A help session in class Monday, Apr. 20 by Humphrey
One better of the two term exams will be used for final grading
• Colloquium today @ 2:30pm in SH101
• And at 4:30pm in the planetarium
Wednesday, Apr. 8, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2
Wednesday, Apr. 8, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
3
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 14 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 next Wednesday, Apr. 15
Wednesday, Apr. 8, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
4
Example 7.6 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, Apr. 8, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
5
Example 7.6 cont’d
In coming to rest, the body decelerates from 7.7m/s to 0m/s in a distance d=1.0cm=0.01m.
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, Apr. 8, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
6
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 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?
Wednesday, Apr. 8, 2009
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 aSpring
meteorite
PHYS
1441-002,
2009 Dr.
Jaehoon Yu
Motion of a rocket
7
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.
Let say that the particle #1 has momentum
Now how 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, Apr. 8, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
8
Linear Momentum Conservation
Initial
r
r
r
r
p1i  p2i  m1v1  m2v2
Final
r
r
r
r
p1 f  p2 f  m1v1  m2v2
Wednesday, Apr. 8, 2009
r
r
r
r
p1 f  p2 f
p

p

1
i
2
i
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
9
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, Apr. 8, 2009

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 1441-002, Spring 2009 Dr.
Jaehoon Yu
10
Ex. 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
r
r
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, Apr. 8, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
11
How do we apply momentum conservation?
1. Decide which objects are included in the system.
2. Relative to the system, identify the internal and
external forces.
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, Apr. 8, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
12
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, Apr. 8, 2009
ur
ur
ur
 p   p1   p 2  0
ur
ur ur
p system  p1  p 2
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
 constant
13
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, Apr. 8, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
14
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 v22i  v22 f 
are conserved. Therefore, the
final speeds in an elastic collision
m1 v1i  v1 f v1i  v1 f   m2 v2i  v2 f v2i  v2 f 
can be obtained in terms of initial From momentum
m1 v1i  v1 f   m2 v2i  v2 f 
speeds as
conservation above
 m  m2 
 2m2 
v1i  
v2i
v1 f   1
 m1  m2 
 m1  m2 
Wednesday, Apr. 8, 2009
 2m1 
 m  m2 
v1i   1
v2i
v2 f  
 m1  m2 
 m1  m2 
Spring 2009 Dr.
What happens PHYS
when1441-002,
the
two
masses are the same?
Jaehoon Yu
15
Ex. 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, Apr. 8, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
16
Ex. 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, Apr. 8, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
17
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, Apr. 8, 2009
1
1
1
m1v 12i  m1v12f  m2 v22 f
2
2
2
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
What do you think
we can learn from
these relationships?
18
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, Apr. 8, 2009
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 1441-002, Spring 2009 Dr.
Jaehoon Yu
f  53.0
Do this at
home
19