PHYS 1441 – Section 501
Lecture #11
Wednesday, July 7, 2004
Dr. Jaehoon Yu
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Collisions
Center of Mass
CM of a group of particles
Fundamentals on Rotation
Rotational Kinematics
Relationships between linear and angular quantities
Today’s homework is HW#5, due 6pm next Wednesday!!
Remember the second term exam, Monday, July 19!!
Wednesday, July 7, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
1
Announcements
• Quiz results:
– Class average: 57.2
– Want to know how you did compared to Quiz #1?
• Average Quiz #1: 36.2
– Top score: 90
• I am impressed of your marked improvement
• Keep this trend up, you will all get 100% soon…
Wednesday, July 7, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
2
Elastic and Inelastic Collisions
Momentum is conserved in any collisions as long as external forces negligible.
Collisions are classified as elastic or inelastic by the conservation of kinetic
energy before and after the collisions.
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 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 only in elastic collisions.
Wednesday, July 7, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
3
Example for Collisions
A car of mass 1800kg stopped at a traffic light is rear-ended by a 900kg car, and the
two become entangled. If the lighter car was moving at 20.0m/s before the collision
what is the velocity of the entangled cars after the collision?
The momenta before and after the collision are
Before collision
p i  m1 v1i  m2 v 2i  0  m2 v 2i
m2
20.0m/s
m1
p f  m1 v1 f  m2 v 2 f  m1  m2 v f
After collision
Since momentum of the system must be conserved
m2
vf
m1
vf
What can we learn from these equations
on the direction and magnitude of the
velocity before and after the collision?
Wednesday, July 7, 2004
m1  m2 v f
pi  p f

 m2 v 2i
m2 v 2i
900  20.0i

 6.67i m / s
m1  m2  900  1800
The cars are moving in the same direction as the lighter
car’s original direction to conserve momentum.
The magnitude is inversely proportional to its own mass.
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
4
Two dimensional Collisions
In two dimension, one can use components of momentum to apply
momentum conservation to solve physical problems.
m1
m1 v 1i  m2 v 2 i  m1 v1 f  m2 v 2 f
v1i
m2
q
f
x-comp.
m1v1ix  m2 v2ix  m1v1 fx  m2 v2 fx
y-comp.
m1v1iy  m2 v2iy  m1v1 fy  m2 v2 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:
m1 v 1i  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 conservation,
the kinetic energy is conserved:
Wednesday, July 7, 2004
1
1
1
m1v 12i  m1v12f  m2 v22 f
2
2
2
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
What do you think
we can learn from
these relationships?
5
Example of 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 put 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, July 7, 2004
v1 f sin 37  v2 f sin f
Solving Eqs. 1-3
(3) equations, one gets
(2)
v1 f  2.80 105 m / s
v2 f  2.1110 m / s
5
f  53.0
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
Do this at
home
6
Center of Mass
We’ve been solving physical problems treating objects as sizeless
points with masses, but in realistic situation objects have shapes
with masses distributed throughout the body.
Center of mass of a system is the average position of the system’s mass and
represents the motion of the system as if all the mass is on the point.
What does above
statement tell you
concerning forces being
exerted on the system?
m2
m1
x1
x2
xCM
Wednesday, July 7, 2004
The total external force exerted on the system of
total mass M causes the center of mass to move at
an acceleration given by a   F / M as if all
the mass of the system is concentrated on the
center of mass.
Consider a massless rod with two balls attached at either end.
The position of the center of mass of this system is
the mass averaged position of the system
m x  m2 x2
CM is closer to the
xCM  1 1
m1  m2
heavier object
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
7
Center of Mass of a Rigid Object
The formula for CM can be expanded to Rigid Object or a
system of many particles
xCM
m x  m x      mn xn
 11 2 2
m1  m2      mn
m x

m
i i
i
i
yCM
i
mi
ri
rCM
i
zCM
i
i
i

 mi r i
m x
i
i
i   mi yi j   mi zi k
i
i
m
i
i
i
i
M
A rigid body – an object with shape
and size with mass spread throughout
the body, ordinary objects – can be
considered as a group of particles with
mass mi densely spread throughout
the given shape of the object
Wednesday, July 7, 2004
i i
i
r CM  xCM i  yCM j  zCM k
r CM 
m z

m
i
i
i
The position vector of the
center of mass of a many
particle system is
m y

m
xCM 
 m x
i
M
xCM  lim
m 0
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
i
i
i
r CM 
 m x
i
i
M
i

1
rdm

M
1
M
 xdm
8
Example 7-11
Thee people of roughly equivalent mass M on a lightweight (air-filled)
banana boat sit along the x axis at positions x1=1.0m, x2=5.0m, and
x3=6.0m. Find the position of CM.
Using the formula
for CM
m x

m
i i
xCM
i
i
i
M 1.0 M  5.0 M  6.0  12.0 M
 4.0(m)

3M
M M M
Wednesday, July 7, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
9
Example for Center of Mass in 2-D
A system consists of three particles as shown in the figure. Find the
position of the center of mass of this system.
Using the formula for CM for each
position vector component
y=2 m (0,2)
1
m x

m
i
(0.75,4)
xCM
rCM
One obtains r CM  x
CM
i i
1 1
2 2
1
i
yCM
i
m x m x m x m x
xCM  m  m  m  m

i
i
2
3 3
3

m2  2m3
m1  m2  m3
m y

m
i
i
(2,0)
m3
x=2
(1,0)
m2
x=1
i
i
i
i
i
r
r m2  2m3  i  2m1 j
i  yCM j 
m1  m2  m3
If m1  2kg; m2  m3  1kg
i
m y
yCM  m

i
i
i
i
i

m1 y1  m2 y2  m3 y3
2m1

m1  m2  m3
m1  m2  m3
Wednesday, July 7, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
r CM 
3i  4 j
 0.75i  j
4
10
Motion of a Diver and the Center of Mass
Diver performs a simple dive.
The motion of the center of mass
follows a parabola since it is a
projectile motion.
Diver performs a complicated dive.
The motion of the center of mass
still follows the same parabola since
it still is a projectile motion.
Wednesday, July 7, 2004
The motion of the center of mass
of the diver is always the same.
PHYS 1441-501, Summer 2004
11
Dr. Jaehoon Yu