PHYS 1441 – Section 002
Lecture #17
Monday, Nov. 15, 2010
Dr. Jaehoon Yu
•
•
•
•
Linear Momentum Conservation
Collisions
Center of Mass
Fundamentals of Rotational Motion
Today’s homework is homework #10, due 10pm, Tuesday, Nov. 23!!
Monday, Nov. 15, 2010
PHYS 1441-002, Fall 2010
Dr. Jaehoon Yu
1
• Quiz #5
Announcements
– Beginning of the class this Wednesday, Nov. 17
– Covers: CH 6.5 – CH 7.8
• Two colloquia this week
– One at 4pm Wednesday, Nov. 17
– Another at 4pm Friday, Nov. 19
Monday, Nov. 15, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
2
Reminder: Special Project
1. A ball of mass M at rest is dropped from the height h above
the ground onto a spring on the ground, whose spring
constant is k. Neglecting air resistance and assuming that
the spring is in its equilibrium, express, in terms of the
quantities given in this problem and the gravitational
acceleration g, the distance x of which the spring is pressed
down when the ball completely loses its energy. (10 points)
2. Find the x above if the ball’s initial speed is vi. (10 points)
3. Due for the project is this Wednesday, Nov. 17.
4. You must show the detail of your OWN work in order to
obtain any credit.
Monday, Nov. 15, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
5
Reminder: Special Project II
P
L
θA
h{
B
m
T
m
mg
Monday, Nov. 15, 2010
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 with the vertical, and
the pivoting point P is frictionless.
A) Find the speed of the ball when it is at the lowest
point, B, in terms of the quantities given above.
B) Determine the tension T at point B in terms of the
quantities given above.
Each of these problem is 10 point. The due date is this
Wednesday, Nov. 17.
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
6
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 13 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 Monday, Nov. 29
Monday, Nov. 15, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
7
More on Conservation of Linear Momentum in
a Two Body System
ur
ur ur
 p  p2  p1 const
From the previous lecture 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, Nov. 15, 2010

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, Fall 2010 Dr. Jaehoon
Yu
8
How do we apply momentum conservation?
1. Define your system by deciding which objects
would be included in it.
2. Identify the internal and external forces with
respect to the system.
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.
Monday, Nov. 15, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
9
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
Monday, Nov. 15, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
10
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
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
ur
ur
r
r
Using Newton’s law we obtain
 p 2  F12 t   F21t   p1
ur
ur
ur
 p   p1   p 2  0
So the momentum change of the system in the
ur
ur ur
collision is 0, and the momentum is conserved
p system  p1  p2  constant
3rd
Monday, Nov. 15, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
11
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 momentum is the same before and
after the collision but not the total kinetic energy .
Two types of inelastic collisions:Perfectly inelastic and inelastic
Perfectly Inelastic: Two objects stick together after the collision,
moving together with the same 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.
Monday, Nov. 15, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
12
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 
Monday, Nov. 15, 2010
 2m1 
 m  m2 
v1i   1
v2i
v2 f  
 m1  m2 
 m1  m2 
PHYS 1441-002, Fall 2010 Dr. Jaehoon
What happens
when the two
masses are the same?
Yu
13
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  m2v f 2  m1v01  m2 v02
m1  m2 v f  m1v01

Solve for V01

v01 
 m1  m2  v f
m1
What do we not know? The final speed!!
How can we get it? Using the mechanical
energy conservation!
Monday, Nov. 15, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
14
Ex. A Ballistic Pendulum, cnt’d
Now using the mechanical energy conservation
1
2

mv 2  mgh

m1  m2 gh f 
gh f 
1
2


1
2
v 2f
Solve for Vf
m1  m2 v 2f


v f  2gh  2 9.80 m s2  0.650 m 
f
Using the solution obtained previously, we obtain
v01
m  m v
m m 




1
2
f
m1
1
2
m1
2gh f
 0.0100 kg  2.50 kg 
2

2
9.80m
s
0.650 m

0.0100
kg





 896m s
Monday, Nov. 15, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
15
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


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 v1 f  m2 v 2 f  m1 v1i
m1v1ix  m1v1 fx  m2 v2 fx  m1v1 f cos   m2 v2 f cos 
m1v1iy  0  m1v1 fy  m2 v2 fy  m1v1 f sin   m2 v2 f sin 
And for the elastic collisions, the
kinetic energy is conserved:
Monday, Nov. 15, 2010
1
1
1
m1v 12i  m1v12f  m2 v22 f
2
2
2
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
What quantities do you
think we can learn from
these relationships?
16
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  to the same axis. Find the final speeds of the two
protons and the scattering angle of proton #2,  .
m1
v1i
m2

Since both the particles are protons m1=m2=mp.
Using momentum conservation, one obtains
x-comp. m p v1i  m p v1 f cos   m p v2 f cos 
y-comp.

m p v1 f sin   m p v2 f sin   0
Canceling mp and putting in all known quantities, one obtains
v1 f cos 37  v2 f cos   3.50 105 (1)
v1 f sin 37  v2 f sin 
From kinetic energy
conservation:
3.50  10   v
5 2
2
1f
(2)
v1 f  2.80 105 m / s
v
2
2f
Monday, Nov. 15, 2010
Solving Eqs. 1-3
5
(3) equations, one gets v2 f  2.1110 m / s
  53.0
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
Do this at
home
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