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Chapter 9: 4-7
Center of Mass
Conservation of Linear Momentum
Main Concepts
• Linear momentum is Mass times Velocity
• Momentum is a vector quantity
• Momentum is conserved in the absence of
an outside force
• Momentum is changed by an Impulse
• Impulse is Force times Time


P  mV
  
P  I  Ft
Momentum is a vector quantity
Initial
mV
Final
-mV
Change in momentum:
 

P  PFINAL  PINIT
 (mV ) xˆ  (mV ) xˆ

 2mVx
P
A block of wood is struck by a bullet. Is the block
more likely to be knocked over if the bullet is (A)
metal and embeds itself in the wood, or if the bullet
is (B) rubber and bounces off the wood? The mass
of the two bullets is the same.
1. (A), metal, because it would
transfer more energy to the
block.
2. (B), rubber, because it would
transfer more momentum to
the block (higher impulse).
3. Makes no difference
4. Sorry, I don’t believe in guns.
Relationship of Momentum and
Kinetic Energy


P  mV
1
K  mV 2
2
P2
K
2m
P 2 mV 
K

2m
2m
m 2V 2 1

 mV 2
2m
2
2
Consequences of Momentum Conservation in
Elastic and Inelastic Collisions
•Linear momentum is conserved (unchanged) in a collision
•Kinetic energy is only conserved in an elastic collision
•Linear momentum can stay the same, even though K changes!
Before
V
M
M
After INELASTIC collision
V/2
M
M
Momentum: P=mV (conserved, doesn’t change before and after)
Kinetic Energy:
Initial: K = ½ mV2
Final: K = ½ (2m) (V/2)2 = ¼ mV2
Problem: Exploding Object
An object initially at rest breaks into two pieces as the result of an
explosion. One piece has twice the kinetic energy of the other piece. What
is the ratio of the masses of the two pieces? Which piece has the larger
mass?
V2
V1
M1
M2
INITIAL:
Pi  0
Ki  0
Conservation of Momentum:
In this problem, K2 = 2K1
M1
FINAL:
M2


Pf  m2V2  m1V1 x
P22
P12
Kf 

2m2 2m1
 K 2  K1
m2V2  m1V1
Solution, Exploding object.


Pf  m2V2  m1V1 x  0
V2
V1
P22
P12
Kf 

2m2 2m1
M1
M2
P2  m2V2  m1V1  P1
 K 2  K1
K2  2K1
What is the ratio of the masses? Which piece has the larger mass?
K2
P22 / 2m2
2 2
K1
P1 / 2m1
So
m1
2
m2
m1

m2
Larger mass has smaller K.
Use P1 = P2
Conceptual Checkpoint.
What are the following quantities, in order?
F  L
F  T
F V
 P / T
1. Work, Impulse, Power,
Force
2. Work, Power, Impulse,
Force
3. Work, Power, Impulse,
Energy
4. Work, Energy, Power,
Impulse
Putting Momentum and Energy to Work
The masses m and M are known, and the incident velocity. Can we
determine how high the pendulum moves after the inelastic collision?
Work backwards from desired result…
If you knew the kinetic energy of the block and bullet after the
collision, you could use conservation of energy to convert the
kinetic energy into potential energy, and find the height H.
If you new the velocity of the block and bullet after collision, you
could find the kinetic energy.
You can use conservation of momentum to find the final velocity!
Work this one out!
A real-world example: Ion
scattering
Light scattering atom.
Vi
m
Vf
M
Heavy atoms in substrate target
VT
You select the scattering atom mass “m”, and the incident
velocity Vi. If you can measure the scattered velocity Vf, can you
tell what is the mass of the target, M?
Ion scattering: can it be done?
m
M
Initial:
K init 
V1
m
Pinit  m v1
Pfinal  MV  m v2
M
V
V2
1
m v12
2
Final:
K final 
1
1
m v22  MV 2
2
2
Apply conservation of momentum and energy.
After some algebra….
Is this reasonable?
If v2=0, M=m.
If V2=V1, M goes to infinity.
v1  v2
M m
v1  v2
UCF Heavy Ion
Backscattering
Spectrometer (HIBS)
The Rocket Problem
“Professor Goddard does not know the relation between action and reaction and
the need to have something better than a vacuum against which to react. He
seems to lack the basic knowledge ladled out daily in high schools."
(1921 New York Times editorial about Robert Goddard's revolutionary rocket
work.)
A rocket engine emits a certain mass
of fuel per unit time. This results in a
force, which is called Thrust.
m  v P

 Fthrust
t
t
m
This is an important case where the
change in momentum comes about
because of the change in mass.
v
Center of Mass: What it is, and why
it matters
“The center of mass is the point at which the external
forces acting on an object appear to act.”
Let’s look at some examples.
Center of Mass of a Mobile
The center of
mass can be
found by the
following
process:
RCM
For example, for the mobile,
X CM 
m R  m R  ...


 m  m  ...
1 1
2
1
2
Numerator is the “moment”
m1 X 1  m2 X 2
m1  m2
Total mass
2
Find the Center of Mass: 2nd try
A mass of 1 kg is located at the origin of a meter stick. A mass of 3
kg is at the other end of the meter stick. Where is the center-of-mass
located?
3
1
1m
RCM
1.
2.
3.
4.
m R  m R  ...


 m  m  ...
1 1
2
1
2
R = .25 m
R = .50 m
R = .75 m
R = 1.0 m
2
The choice of
origin does not
affect result.
Forces and Center of Mass
X CM 
Suppose
m1 X 1  m2 X 2
m1  m2
(m1  m2 )acm  m1a1  m2 a2

 
Fcm  F1  F2
Fcm
F1
F2