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MOMENTUM!
Momentum
Impulse
Conservation of Momentum in 1 Dimension
Conservation of Momentum in 2 Dimensions
Angular Momentum
Torque
Moment of Inertia
Momentum Facts
• p = mv
• Momentum is a vector quantity!
• Velocity and momentum vectors point in the same direction.
• SI unit for momentum: kg·m/s (no special name).
• Momentum is a conserved quantity (this will be proven later).
• A net force is required to change a body’s momentum.
• Momentum describes the tendency of a mass to keep going in
the same direction with the same speed.
• Something big and slow could have the same momentum as
something small and fast.
Momentum and Inertia
• Inertia is another property of mass that
resists changes in velocity; however,
inertia depends only on mass.
• Inertia is a scalar quantity.
• Momentum is a property of moving
mass that resists changes in a moving
object’s velocity.
• Momentum is a vector quantity.
Vocabulary
angular momentum
collision
law of conservation of momentum
elastic collision
gyroscope
impulse
inelastic collision
linear momentum
momentum
Momentum
• The momentum of a ball depends on its
mass and velocity.
• Ball B has more momentum than ball A.
Momentum
•Ball A is 1 kg moving 1m/sec,
•Ball B is 1kg at 3 m/sec.
•If a 1 N force is applied to deflect each ball's motion.
•What happens?
•Does the force deflect both balls equally?
 Ball B deflects much less
than ball A when the
same force is applied
because ball B had a
greater initial
momentum.
Calculating Momentum
The momentum of a moving object
is its mass multiplied by its
velocity.
That means momentum increases
with both mass and velocity.
Momentum
(kg m/sec)
Mass (kg)
p=mv
Velocity (m/sec)
Comparing momentum
A car is traveling at a velocity of 13.5 m/sec
(30 mph) north on a straight road. The
mass of the car is 1,300 kg.
A motorcycle passes the car at a speed of
30 m/sec (67 mph). The motorcycle (with
rider) has a mass of 350 kg.
Calculate and compare the momentum of
the car and motorcycle.
1.
You are asked for momentum.
2.
You are given masses and velocities.
3.
Use: p = m v
4.
Solve for car: p = (1,300 kg) (13.5 m/s) = 17,550 kg m/s
5.
Solve for cycle: p = (350 kg) (30 m/s) = 10,500 kg m/s
1.
The car has more momentum even though it is going much slower.
Momenta
Car: m = 1800 kg; v = 80 m /s
Bus: m = 9000 kg; v = 16 m /s
p = 144, 000 kg · m /s
p = 144 ,000 kg · m /s
Train: m = 3.6 ·104 kg; v = 4 m /s
p = 144,000 kg · m /s
Do
Momentum
Problems
Angular Momentum
Momentum resulting from
an object moving in linear
motion is called
linear momentum.
Momentum resulting from
the rotation (or spin) of an
object is called
angular momentum.
Conservation of Angular Momentum
Angular momentum is
important because it
obeys a conservation
law, as does linear
momentum.
The total angular
momentum of a closed
system stays the same.
Calculating Angular Momentum
Angular momentum is calculated in a
similar way to linear momentum, except
the mass and velocity are replaced by the
moment of inertia and angular velocity.
Angular
momentum
(kg m/sec2)
L=Iw
Moment of
inertia
(kg m2)
Angular
velocity
(rad/sec)
Calculating Angular Momentum
The moment of inertia of an
object is the average of
mass times radius squared
for the whole object.
Since the radius is measured
from the axis of rotation,
the moment of inertia
depends on the axis of
rotation.
Gyroscopes Angular Momentum
A gyroscope is a device that contains a spinning
object with a lot of angular momentum.
Gyroscopes can do amazing tricks because they
conserve angular momentum.
For example, a spinning gyroscope can easily
balance on a pencil point.
Gyroscopes
A gyroscope on the space shuttle is mounted at
the center of mass, allowing a computer to
measure rotation of the spacecraft in three
dimensions.
An on-board computer is able to accurately
measure the rotation of the shuttle and maintain
its orientation in space.
Comparison: Linear & Angular Momentum
Linear Momentum, p
Angular Momentum, L
• Tendency for a mass to continue • Tendency for a mass to continue
moving in a straight line.
rotating.
• Parallel to v.
• Perpendicular to both v and r.
• A conserved, vector quantity.
• A conserved, vector quantity.
• Magnitude is inertia (mass)
times speed.
• Magnitude is rotational inertia
times angular speed.
• Net force required to change it.
• Net torque required to change it.
• The greater the mass, the greater • The greater the moment of
the force needed to change
inertia, the greater the torque
momentum.
needed to change angular
momentum.
Impulse
Impulse Defined
F = ma
a = ∆v/t
F = m∆v/t
Ft = m∆v
(kg m/s2)s = kg m/s
.
Stopping Time
Ft = Ft
Impulse
Impulse
F = ∆mv
t
Impulse
F = ∆mv
t
Impulse
F = ∆mv
t
Impulse
F
t = ∆mv
Impulse
Ft= ∆mv
F t=
∆mv
Do
Impulse Problems
Conservation of Momentum
The law of conservation of momentum states
when a system of interacting objects is not
influenced by outside forces (like friction), the
total momentum of the system cannot change.
If you throw a rock forward from a
skateboard, you will move
backward in response.
Collisions in One Dimension
A collision occurs when two or more objects hit
each other.
During a collision, momentum is transferred from
one object to another.
Collisions can be elastic or inelastic.
Collisions
Conservation of Momentum in 1-D
Whenever two objects collide (or when they exert forces on each
other without colliding, such as gravity) momentum of the system
(both objects together) is conserved. This mean the total momentum
of the objects is the same before and after the collision.
(Choosing right as the +
before: p = m1 v1 - m2 v2
v2
v1
m1
direction, m2 has - momentum.)
m2
m1 v1 - m2 v2 = - m1 va + m2 vb
after: p = - m1 va + m2 vb
va
m1
m2
vb
Elastic collisions
Two 0.165 kg billiard balls roll
toward each other and collide
head-on.
Initially, the 5-ball has a velocity of
0.5 m/s.
The 10-ball has an initial velocity
of -0.7 m/s.
The collision is elastic and the 10ball rebounds with a velocity of 0.4
m/s, reversing its direction.
What is the velocity of the 5-ball
after the collision?
Elastic Collisions
You are asked for 10-ball’s velocity after collision.
You are given mass, initial velocities, 5-ball’s final velocity.
Diagram the motion, use m1v1 + m2v2 = m1v3 + m2v4
Solve for V3 :
(0.165 kg)(0.5 m/s) + (0.165 kg) (-0.7 kg)=(0.165 kg) v3 + (0.165 kg) (0.4 m/s)
V3 = -0.6 m/s
Directions after a collision
On the last slide the boxes were drawn going in the opposite direction
after colliding. This isn’t always the case. For example, when a bat hits
a ball, the ball changes direction, but the bat doesn’t. It doesn’t really
matter, though, which way we draw the velocity vectors in “after”
picture. If we solved the conservation of momentum equation (red box)
for vb and got a negative answer, it would mean that m2 was still moving
to the left after the collision. As long as we interpret our answers
correctly, it matters not how the velocity vectors are drawn.
v2
v1
m1
m2
m1 v1 - m2 v2 = - m1 va + m2 vb
va
m1
m2
vb
Sample Problem
A crate of raspberry donut filling collides with a tub of lime Kool Aid
on a frictionless surface. Which way on how fast does the Kool Aid
rebound? answer: Let’s draw v to the right in the after picture.
3 (10) - 6 (15) = -3 (4.5) + 15 v
v = -3.1 m/s
Since v came out negative, we guessed wrong in drawing v to the
right, but that’s OK as long as we interpret our answer correctly.
After the collision the lime Kool Aid is moving 3.1 m/s to the left.
before
3 kg
10 m/s
6 m/s
15 kg
after
4.5 m/s
3 kg
15 kg
v
Sample Problem 1
35 g
7 kg
700 m/s
v=0
A rifle fires a bullet into a giant slab of butter on a frictionless surface.
The bullet penetrates the butter, but while passing through it, the bullet
pushes the butter to the left, and the butter pushes the bullet just as hard
to the right, slowing the bullet down. If the butter skids off at 4 cm/s
after the bullet passes through it, what is the final speed of the bullet?
(The mass of the rifle matters not.)
35 g
v=?
4 cm/s
7 kg
continued on next slide
Sample Problem 1
(cont.)
Let’s choose left to be the + direction & use conservation of
momentum, converting all units to meters and kilograms.
35 g
p before = 7 (0) + (0.035) (700)
7 kg
= 24.5 kg · m /s
v=0
35 g
4 cm/s
v=?
p before = p after
7 kg
700 m/s
p after = 7 (0.04) + 0.035 v
= 0.28 + 0.035 v
24.5 = 0.28 + 0.035 v
v = 692 m/s
v came out positive. This means we chose the correct
direction of the bullet in the “after” picture.
Inelastic Collisions
A train car moving to the right at
10 m/s collides with a parked train
car.
They stick together and roll along
the track.
If the moving car has a mass of
8,000 kg and the parked car has a
mass of 2,000 kg, what is their
combined velocity after the
collision?
You are asked for the final velocity.
You are given masses, and initial velocity of moving
train car.
Inelastic Collisions
Diagram the problem
Use m1v1 + m2v2 = (m1v1 +m2v2) v3
Solve for v3 = m1v1 + m2v2
(m1v1 +m2v2)
v3= (8,000 kg)(10 m/s) + (2,000 kg)(0 m/s)
(8,000 + 2,000 kg)
v3= 8 m/s
The train cars moving together to right at 8 m/s.
Sample Problem 2
35 g
7 kg
700 m/s
v=0
Same as the last problem except this time it’s a block of wood rather than
butter, and the bullet does not pass all the way through it. How fast do
they move together after impact?
v
7. 035 kg
(0.035) (700) = 7.035 v
v = 3.48 m/s
Note: Once again we’re assuming a frictionless surface, otherwise there
would be a frictional force on the wood in addition to that of the bullet,
and the “system” would have to include the table as well.
Bouncing
Alfred went on a date with Mabel. When Alfred
dropped off Mabel after the date, he was anxious to
play Angry Birds, so he forgot to kiss her on the
cheek good night. She went up to her room, opened
the window and threw a flower pot at Alfred. On of
three things could happen.
1. The flower pot – head collision is elastic
2. The flower pot – head collision is inelastic
3. The flower pot bounces off his head
Which will hurt more?????
Elastic Collision
Before
After
Elastic Collision
Alfred + Flower pot = Alfred + Flower pot
m1v1 + m2v2 = m1v3 + m2v4
100kg(0m/s) + 10kg (15 m/s) = 100kg (v3) + 10kg (0m/s)
150kgm/s = 100kg (v3)
150kgm/s = 100kg (v3)
100kg 100kg
1.5 m/s = v3 (elastic)
Inelastic Collision
Before
After
Inelastic Collision
Alfred + Flower pot = (Alfred + Flower pot)
m1v1 + m2v2 = (m1 + m2)v3
100kg(0m/s) + 10kg(15 m/s) = (100kg + 10kg) (v3)
150kgm/s = 110kg(v3)
150kgm/s = 110kg(v3)
110kg 110kg
1.36 m/s = v3 (inelastic)
1.5 m/s = v3 (elastic)
Bouncing
Elastic Collision
Alfred + Flower pot = Alfred + Flower pot
m1v1 + m2v2 = m1v3 + m2v4
100kg(0m/s) + 10kg(15 m/s) = 100kg(v3) + 10kg(-5m/s)
150kgm/s = 100kg(v3) – 50kgm/s
200kgm/s = 100kg(v3)
100kg 100kg
2.0 m/s = v3 (bouncing)
1.5 m/s = v3 (elastic)
1.36 m/s = v3 (inelastic)
Do
Collision
Problems