Transcript Monday, April 11, 2011 - UTA HEP WWW Home Page

PHYS 1443 – Section 001
Lecture #16
Monday, April 11, 2011
Dr. Jaehoon Yu
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Collisions – Elastic and Inelastic Collisions
Collisions in two dimension
Center of Mass
Center of mass of a rigid body
Motion of a Group of Objects
Today’s homework is homework #9, due 10pm, Tuesday, Apr. 19!!
Announcements
• Second non-comprehensive term exam results
– Class average: 64/97
• Equivalent to 66/100
• Previous exams: 70/100 and 72/100
– Top score: 93/97
– Will take the better of the two non-comprehensive exam after
normalizing to the class average between the two exams
• Quiz next Wednesday, Apr. 20
– Beginning of the class
– Covers from CH9.5 to what we finish Monday, Apr. 18
• Colloquium Wednesday at 4pm in SH101
Monday, April 11, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
2
Monday, April 11, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
3
Reminder: 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 7 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 this Wednesday, Apr. 13
Monday, April 11, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
4
Extra Credit: 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, Φ. This must be done in much more detail than the
book or on page 13 of this lecture note.
•10 points
•Due beginning of the class Monday, Apr. 18.
Collisions
Generalized collisions must cover not only the physical contact but also the collisions
without physical contact such as that of electromagnetic ones on 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
dp1  F21dt
Likewise for particle 2 by particle 1
r r
dp2  F12dt
law we obtain
ur
ur
r
r
d p 2  F12dt  F21dt  d p1
So the momentum change of the system in a
collision is 0, and the momentum is conserved
Monday, April 11, 2011
ur
ur
ur
d p  d p1  d p 2  0
ur
ur ur
p system  p1  p 2  constant
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
6
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, April 11, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
7
Elastic and Perfectly Inelastic Collisions
r
r
r
In perfectly inelastic collisions, the objects stick
m1 v1i  m2 v 2i  (m1  m2 )v f
together after the collision, moving together.
r
r
r
m1 v1i  m2 v 2i
Momentum is conserved in this collision, so the
vf 
final velocity of the stuck system is
(m1  m2 )
r
r
r
r
How about elastic collisions?
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 
What
Monday, April 11, 2011
 2m1 
 m  m2 
v1i   1
v2i
v2 f  
 m1  m2 
 m1  m2 
happens when
the two masses are the same?
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
8
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
pi  m1v1i  m2v2i  0  m2v2i
m2
20.0m/s
m1
p f  m1v1 f  m2v2 f  m1  m2 v f
After collision
Since momentum of the system must be conserved
m2
vf
m  m v
pi  p f
m1
vf

1
m2 v2i
m
1
What can we learn from these equations
on the direction and magnitude of the
velocity before and after the collision?
Monday, April 11, 2011
 m2


2
f
 m2v2i
900  20.0
 6.67m / s
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 1443-001, Spring 2011
Dr. Jaehoon Yu
9
Ex.9 – 11: A Ballistic Pendulum
The mass of a 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  m2v02
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, April 11, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
10
Ex. A Ballistic Pendulum, cnt’d
Now using the mechanical energy conservation
mv 2  mgh
2
1
m

m
gh

m

m
v
 1 2 f 2 1 2 f
1
2
gh f 
1
2
Solve for Vf
v 2f
v f  2gh f  2  9.80 m s2   0.650 m 
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, April 11, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
11
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 v1i  m2 v 2i  m1 v1i
m1v1ix  m1v1 fx  m2v2 fx  m1v1 f cos  m2v2 f cosf
m1v1iy  0  m1v1 fy  m2v2 fy  m1v1 f sin  m2v2 f sin f
And for the elastic collisions, the
kinetic energy is conserved:
Monday, April 11, 2011
1
1
1
m1v 12i  m1v12f  m2 v22 f
2
2
2
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
What do you think
we can learn from
these relationships?
12
Ex. 9 – 13: 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, Φ.
m1
v1i
m2
Since both the particles are protons m1=m2=mp.
Using momentum conservation, one obtains
x-comp. m p v1i  mp v1 f cos  mp v2 f cosf

y-comp.

mp v1 f sin   mp v2 f sin f  0
Canceling mp and putting in all known quantities, one obtains
v1 f cos37  v2 f cosf  3.50105 (1)
v1 f sin 37  v2 f sin f (2)
From kinetic energy
conservation:
3.50  10   v
5 2
2
1f
v1 f  2.80105 m / s
v
2
2f
Monday, April 11, 2011
Solving Eqs. 1-3
5
(3) equations, one gets v2 f  2.1110 m / s
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
f  53.0
Do this at
home
13
Center of Mass
We’ve been solving physical problems treating objects as sizeless
points with masses, but in realistic situations 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 the
forces being exerted on the
system?
m2
m1
x1
x2
xCM
Monday, April 11, 2011
The total external force exerted on the system of
total mass M causes the center
ofumass
to move at
r
r
an acceleration given by a   F / Mas 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
m1x1  m2 x2
CM is closer to the
xCM 
heavier object
m1  m2
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
14
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.
Monday, April 11, 2011
The motion of the center of mass
of the diver is always the same.
PHYS 1443-001, Spring 2011
15
Dr. Jaehoon Yu
Example 9 – 14
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
xCM
i
i
i
i
12.0M
M 1.0  M  5.0  M  6.0

 4.0(m)

3M
M M M
Monday, April 11, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
16
Ex9 – 15: 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
(0.75,1)
xCM
rCM
i
r
One obtains r CM  x
CM
m x m x  m x  m x
xCM  m  m  m  m

i i
1 1
i
2 2
1
i
i
i
(2,0)
m3
x=2
(1,0)
m2
x=1
2
3 3
3

m2  2m3
m1  m2  m3
i
yCM 
 mi yi
i
m
i

m1 y1  m2 y2  m3 y3
2m1

m1  m2  m3
m1  m2  m3
Monday, April 11, 2011
i
i
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
m y

m
i
yCM
i
i
i
i
r
r
r
r  m2  2m3  i  2m1 j
i  yCM j 
m1  m2  m3
If m1  2kg; m2  m3  1kg
r r
r
r r
3i  4 j
r CM 
 0.75i  j
4
17
Center of Mass of a Rigid Object
The formula for CM can be extended to a system of many particles
or a Rigid Object
xCM 
m1x1  m2 x2   mn xn
m1  m2    mn
m x

m
m y

m
i
i i
i
yCM
Δmi
ri
rCM
zCM
i
i
i
r
r
r
r
r CM  xCM i  yCM j  zCM k

r
r
r
m
x
i

m
y
j

m
z
k
 ii  i i  ii
i
i
r
 mi r 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
Monday, April 11, 2011
i
i
i
r
r CM 
i i
i
i
The position vector of the
center of mass of a many
particle system is
m z

m
i
xCM 
 m x
i
M
xCM  lim
m 0
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
i
i
i
 m x
i
i
M
i

r
1 r
r CM 
rdm

M
1
M
 xdm
18
Ex 9 – 16: CM of a thin rod
Show that the center of mass of a rod of mass M and length L lies in midway
between its ends, assuming the rod has a uniform mass per unit length.
The formula for CM of a continuous object is
L
xCM 
x
dx
Therefore xCM
Δm=λdx
1

M

xL
x 0
1
M

xL
x 0
xdm
Since the density of the rod (λ) is constant;   M / L
The mass of a small segment dm  dx
xL
1 1 2 1 1  L
1 1 2 

xdx   x 
 L    ML  
M  2  x 0 M  2  M  2  2
Find the CM when the density of the rod non-uniform but varies linearly as a function of x, λ x
M

xL
x 0
dx  
xL
x 0
xL
xdx
1
1

  x 2 
 L2
2
 x 0 2
Monday, April 11, 2011
xCM
1

M

xL
x 0
1
xdx 
M

xL
x 0
1
1 3
 L  
 32011  M
PHYS 1443-001, Spring
xCM 
1
M
Dr. Jaehoon Yu
1
x dx 
M
2
xL
1 3 
 3 x 

 x 0
2
 2L
 ML  
3
3

19