Transcript Momentum

Momentum
• Read Your Textbook: Introduction to Physical Science
– Chapter 3.5
– Chapter 4
• Practice Homework Exercises
Motion Review
Velocity = change in displacement = Dx
change in time
Dt
Speed
Acceleration = change in velocity = Dv
change in time
Dt
How fast are you
getting faster.
Force = mass x acceleration = m Dv
Dt
a = F/m
A look at the two definitions of acceleration….
Force and Acceleration
F = m Dv = m a
Dt
If a Force acts occurs over a short time,
a small acceleration results.
If a Force acts over a long time,
a large acceleration results.
Cannon Ball!
F = m Dv
Dt
F Dt = m Dv
For the same Force (amount of powder), why is the speed of
a cannon ball greater when fired from a longer cannon barrel?
Interaction Time
F Dt = m Dv
The longer cannon barrel gives the cannon ball a larger
impulse and therefore more momentum. The Force (F) is
allowed to act for a longer time Dt to build up velocity (Dv).
F Dt = m Dv
Impulse and Momentum
acceleration = acceleration
a = a
F = Dv
m
Dt
F Dt = m Dv
Impulse
Momentum
Impulse and Momentum
F = Dv
m
Dt
F Dt = m Dv
Impulse
Momentum
If a change in velocity (momentum) occurs over a short time,
a large force must act.
If the change in velocity (momentum) occurs over an extended
time, a small force is acting.
• Recall the Egg Toss Game
• A Boxer Bobs and Weaves His Head
• Bending Legs Upon a Parachute Landing
Conservation of Momentum
Momentum is a conserved quantity, that is, for any isolated
system, the total momentum remains unchanged.
Momentum = mass x velocity
P=mv
Conservation of Momentum
Momentum is a conserved quantity, that is, for any isolated
system, the total momentum remains unchanged.
Momentum = mass x velocity
P=mv
Consider the following collision:
Before
v
M
m
V
After
Conservation of Momentum
Momentum is a conserved quantity, that is, for any isolated
system, the total momentum remains unchanged.
Momentum = mass x velocity
P=mv
Consider the following collision:
Before
v
M
m
V
After
V’
M
m
v’
Conservation of Momentum
Momentum is a conserved quantity, that is, for any isolated
system, the total momentum remains unchanged.
Momentum = mass x velocity
P=mv
Consider the following collision:
Before
v
M
m
V
After
V’
Total Momentum
MV + mv = total momentum =
M
m
v’
MV’ + mv’
Ice Ball Toss
v
M
m
Total Momentum Before:
MV
+
mv
60 kg ( 0 km/hr) + 20 kg (10 km/hr) = 200
Ice Toss
Momentum After (must be identical to momentum before)
= 200
= (M+m) v’
200 = (M+m) v’
200 = (60+20) v’
v’ = 200/80 = 2.5 km/hr
Conservation of Momentum
What is the total momentum of the debris from a firecracker?
Before
After
m2
m1
M V = 0 = total momentum before
m3
Total Momentum After = m1v1 + m2v2 + m3v3 + …
m4
Conservation of Momentum
What is the total momentum of the debris from a firecracker?
Before
After
m2
m1
MV =
0 = total momentum before
m3
m4
Total Momentum After = m1v1 + m2v2 + m3v3 + …
=0
Rifle Shot
Let mbullet = 0.3 kg, Mrifle = 5 kg, and vbullet = 370 m/s
mbulletvbullet + MrifleVrifle = 0.3 kg (370 m/s) + 5kgVrifle
0 = 0.3 kg (370 m/s) + 5kgVrifle
Rifle Shot
If momentum is conserved, why doesn’t a rifle kill you upon
recoil after firing a bullet?
Before:
mbulletvbullet + MrifleVrifle = 0
Rifle Shot
If momentum is conserved, why doesn’t a rifle kill you upon
recoil after firing a bullet?
Before:
After:
mbulletvbullet + MrifleVrifle = 0
=0
Let mbullet = 0.3 kg, Mrifle = 5 kg, and vbullet = 370 m/s
Rifle Shot
Let mbullet = 0.3 kg, Mrifle = 5 kg, and vbullet = 370 m/s
mbulletvbullet + MrifleVrifle = 0.3 kg (370 m/s) + 5kgVrifle
0 = 0.3 kg (370 m/s) + 5kgVrifle
-0.3(370) = 5 kg Vrifle
Vrifle = - 0.3(370)/5 = - 2.2 m/s
Rifle Shot
Let mbullet = 0.3 kg, Mrifle = 5 kg, and vbullet = 370 m/s
mbulletvbullet + MrifleVrifle = 0.3 kg (370 m/s) + 5kgVrifle
0 = 0.3 kg (370 m/s) + 5kgVrifle
-0.3(370) = 5 kg Vrifle
Vrifle = - 0.3(370)/5 = - 2.2 m/s
Shoulder aches, BUT your alive!
M
riflevrecoil =
V
mbullet
bullet
Now you try:
• What is the velocity of a bullet (m = 0.15 kg) after being
fired from a 10 kg rifle (NOTE: rifle recoils with a
velocity of 3 m/s).
A. 30 m/s
B. 1.5 m/s
C. 200 m/s
D. 310 m/s
E. none of these
mbulletvbullet + MrifleVrifle = 0.15 kg (vbullet) + 10kg (3 m/s)
0 = 0.15 kg (vbullet) + 30 kg m/s
-30 kg m/s = 0.15 kg vbullet
vbullet = - 30 kg m/s /(0.15 kg) = - 200 m/s
Train Link
An train engine runs into a stationary box car
weighing 4x more than itself to link up. If the engine
was traveling 10 mph before link up, how fast does
the train move after?
Train Link
An train engine runs into a stationary box car
weighing 4x more than itself to link up. If the engine
was traveling 10 mph before link up, how fast does
the train move after?
MOMENTUM BEFORE
MVBC
+
0
=
MOMENTUM AFTER
=
(M + 4M) VAC
Train Link
An train engine runs into a stationary box car
weighing 4x more than itself to link up. If the engine
was traveling 10 mph before link up, how fast does
the train move after?
MOMENTUM BEFORE
MVBC
+
0
M(10)
=
MOMENTUM AFTER
=
(M + 4M) VAC
=
(5M) VAC
10 = 5 VAC
2 = VAC
Angular Momentum
L Angular Momentum: A combination of...
m Mass
v Speed of Rotation
r Mass Position (with respect to rotational axis)
L=mvr
• Conservation Examples:
–
–
–
–
Spins of Dancers or Ice Skaters
Those Funky Coin Vortexes in Stores
Tops and Gyroscopes
Riding a Bicycle
Precession and the Earth
• 1 complete cycle takes 26,000 years
Orbit Application
Angular Momentum L,
is the product of a planet's mass (m),
orbital velocity (v)
and distance from the Sun (R).
The formula is simple: L = m v R,
where R = a function of e the eccentricity.
Faster, Closer
Conservation of Angular Momentum:
L = L
mVr
mvR
Summary
• Impulse and Momentum
– F Dt = m Dv
• Conservation of Momentum
– Total Momentum (P = M V)
• Angular Momentum
–L=mvr