chapter 9: 4-7 center of mass conservation of linear momentum
TRANSCRIPT
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
VmP
tFIP
Momentum is a vector quantityInitial
Final
mV
-mV
Change in momentum:
xmV
xmVxmV
PPP INITFINAL
2
ˆ)(ˆ)(
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 difference4. Sorry, I don’t believe in guns.
Relationship of Momentum and Kinetic Energy
m
PK
mVK
VmP
2
2
1
2
2
222
22
2
1
2
22
mVm
Vm
m
mV
m
PK
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!
M M
VBefore
M M
V/2After INELASTIC collision
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 ObjectAn 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?
M1 M2 M1 M2
V1V2
INITIAL: FINAL:
0
0
i
i
K
P
12
1
21
2
22
1122
22
KK
m
P
m
PK
xVmVmP
f
f
Conservation of Momentum: 1122 VmVm
In this problem, K2 = 2K1
Solution, Exploding object.
111222 PVmVmP
12 2KK
What is the ratio of the masses? Which piece has the larger mass?
2
1
121
222
1
2
2/
2/2
m
m
mP
mP
K
K
So
22
1 m
m
12
1
21
2
22
1122
22
0
KK
m
P
m
PK
xVmVmP
f
f
Use P1 = P2
Larger mass has smaller K.
M1 M2
V1V2
Conceptual Checkpoint.
1. Work, Impulse, Power, Force
2. Work, Power, Impulse, Force
3. Work, Power, Impulse, Energy
4. Work, Energy, Power, Impulse
What are the following quantities, in order?
TP
VF
TF
LF
/
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
Heavy atoms in substrate target
Light scattering atom.
m
M
Vi Vf
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?
UCF Heavy Ion Backscattering Spectrometer (HIBS)
mM
V1
mM
V2
V
Initial:
Final:
After some algebra….
21
1
2
1vmK
vmP
init
init
222
2
2
1
2
1MVvmK
vmMVP
final
final
Apply conservation of momentum and energy.
21
21
vv
vvmM
Is this reasonable?If v2=0, M=m.If V2=V1, M goes to infinity.
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.)
m
v
A rocket engine emits a certain mass of fuel per unit time. This results in a force, which is called Thrust.
thrustFt
P
t
vm
This is an important case where the change in momentum comes about because of the change in mass.
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:
...
...
21
2211
mm
RmRmRCM
For example, for the mobile,
21
2211
mm
XmXmXCM
Numerator is the “moment”
Total mass
Find the Center of Mass: 2nd try
1. R = .25 m2. R = .50 m3. R = .75 m4. R = 1.0 m
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?
1 3
1m
...
...
21
2211
mm
RmRmRCM
The choice of origin does not affect result.
Forces and Center of Mass
21
2211
mm
XmXmXCM
Suppose
221121 )( amamamm cm
21 FFFcm
F1
Fcm
F2