Download Momentum

Momentum Unit
1. Momentum
Mass vs Velocity
Space Debris
2. Impulse
Increasing momentum
Decreasing momentum over a long time
Decreasing momentum over a short time
Rebounding (Bouncing)
3. Collisions/Explosions
Conservation of p
Elastic
Inelastic
Explosions
What is momentum?
The concept of momentum is closely related to
Newton’s Laws of Motion
1. MOMENTUM
The concept of momentum is closely related to
Newton’s Laws of Motion
Momentum – Inertia of a moving object
What is momentum?
The concept of momentum is closely related to
Newton’s Laws of Motion
Momentum – Inertia of a moving object
Momentum is a measure of how hard it is to
Stop or turn a moving object
What is momentum?
The concept of momentum is closely related to
Newton’s Laws of Motion
Momentum – Inertia of a moving object
Momentum is a measure of how hard it is to
Stop or turn a moving object
It is related to both Mass and Velocity
Only moving objects have momentum
What is momentum?
The concept of momentum is closely related to
Newton’s Laws of Motion
Momentum – Inertia of a moving object
Momentum is a measure of how hard it is to
Stop or turn a moving object
It is related to both Mass and Velocity
Only moving objects have momentum
What is the equation for momentum?
Calculating Momentum
• For one object
p = mv
• For a system of multiple objects
p = pi = mivi
(total p is the sum of all individual p’s)
The unit for momentum
p = mv
The unit for momentum
p = mv
Therefore, the unit for p = kg m/s
Remember, momentum is a vector
Which has the most momentum?
Which has the most momentum?
It Depends!
Which has the most momentum?
M = 1000 kg, v = 1 m/s
M = 10 g, v = 5 m/s
Which has the most momentum?
Truck:
M = 1000 kg, v = 1 m/s
p = mv
p = (1000 kg) (1 m/s)
Butterfly:
M = 10 g, v = 3 m/s
p = mv
p = (.01 kg) (3 m/s)
Which has the most momentum?
Truck:
M = 1000 kg, v = 1 m/s
P = mv
Butterfly:
M = 10 g, v = 3 m/s
p = mv
P = (1000 kg) (1 m/s)
p = (.01 kg) (3 m/s)
P = 1000 kg m/s
p = .03 kg m/s
Which has the most momentum?
Truck:
M = 1000 kg, v = 1 m/s
p = mv
Butterfly:
M = 10 g, v = 3 m/s
p = mv
p = (1000 kg) (1 m/s)
p = (.01 kg) (3 m/s)
p = 1000 kg m/s
p = .03 kg m/s
How can the butterfly have the same momentum
As the truck?
Which has the most momentum?
Truck:
M = 1000 kg, v = 1 m/s
Butterfly:
M = 10 g, v =
p = 1000 kg m/s
How can the butterfly have the same momentum
As the truck?
The mass of an object is usually
constant, however, Velocity is easy to
change.
Which has the most momentum?
Truck:
M = 1000 kg, v = 1 m/s
Butterfly:
M = 10 g, v =
p = 1000 kg m/s
The mass of an object is usually
constant, however, Velocity is easy to
change.
Butterfly:
p = mv
1000 kg m/s = (.01 kg) (Vb)
Which has the most momentum?
Truck:
M = 1000 kg, v = 1 m/s
Butterfly:
M = 10 g, v =
p = 1000 kg m/s
The mass of an object is usually
constant, however, Velocity is easy to
change.
Butterfly:
p = mv
1000 kg m/s = (.01kg) (Vb)  Vb = (1000 kg m/s)/(.01 kg)
Vb = 100,000 m/s
Momentum Depends on mass AND velocity
An object can have a large momentum if it has
A large mass, even if it has a small velocity
Momentum Depends on mass AND velocity
An object can have a large momentum if it has
A large mass, even if it has a small velocity
However, a small object can also have a large
momentum if it has a large velocity
Grains of dust in space pose a big
problem for satellites and
spacecraft.
Grains of dust in space pose a big
problem for satellites and
spacecraft.
Pieces of debris come from
manmade objects, such as fuel
drops, ice crystals, parts of
spacecraft, or from natural things,
such as stars and asteroids.
Pieces of debris come from
manmade objects, such as fuel
drops, ice crystals, parts of
spacecraft, or from natural things,
such as stars and asteroids.
A dust particle can be .1 mm in size, and travel up
to 158,000 miles per second
Pieces of debris come from
manmade objects, such as fuel
drops, ice crystals, parts of
spacecraft, or from natural things,
such as stars and asteroids.
A dust particle can be .1 mm in size, and travel up
to 158,000 miles per second
Pieces of debris come from
manmade objects, such as fuel
drops, ice crystals, parts of
spacecraft, or from natural things,
such as stars and asteroids.
A dust particle can be .1 mm in size, and travel up
to 158,000 miles per second
Space Debris
Notice all the “junk” as they are looking for the MIR
ESA Space Debris
The Space Shuttle
Although the fleet is retired, it is a good example of
spacecraft window design
Look closely at the windows…
The Space Shuttle
The Space Shuttle
The Space Shuttle
Why were the windows reduced in size?
Damage on heat shielding due to micro meteor impacts
The Space Shuttle
Why were the windows reduced in size?
This micro meteor traveled ¾ way
through the glass before it stopped
To protect against micro meteor impacts. One hole
through the window can depressurize the entire cabin
The Space Shuttle
The Space Shuttle reaches speeds of around
18,000 MPH. At these speeds, even foam insulation
(with very little mass) also posed problems.
The Space Shuttle
The Space Shuttle reaches speeds of around
18,000 MPH. At these speeds, even foam insulation
(with very little mass) also posed problems.
The Space Shuttle
The Space Shuttle reaches speeds of around
18,000 MPH. At these speeds, foam insulation
(with very little mass) also posed problems.
This will continue to be a problem with ANY
vehicle that is designed to return to Earth.
The point is, momentum depends on
mass AND velocity
Momentum Review
(with Bill Nye)
:
Momentum and rockets:
Sample Problem:
• A 200 kilogram motorcycle is moving at a
speed of 130 m/s. What is the momentum of
the cycle?
Sample Problem:
• A 200 kilogram motorcycle is moving at a
speed of 130 m/s. What is the momentum of
the cycle?
• P = mv = (200 kg)(130 m/s) = 26000 kg m/s
Sample Problem:
• A 200 kilogram motorcycle is moving at a
speed of 130 m/s. What is the momentum of
the cycle?
• P = mv = (200 kg)(130 m/s) = 26000 kg m/s
• A 60 kg person is riding the motorcycle at a
speed of 130 m/s. What is the total
momentum of the system?
Sample Problem:
• Pcycle = 26000 kg m/s
• A 60 kg person is riding the motorcycle at a
speed of 130 m/s. What is the total
momentum of the system?
• Pperson = (60kg) (130 m/s) = 7800 kg m/s
Sample Problem:
• Pcycle = 26000 kg m/s
• A 60 kg person is riding the motorcycle at a
speed of 130 m/s. What is the total
momentum of the system?
• Pperson = (60kg) (130 m/s) = 7800 kg m/s
• ptotal = pcycle + pperson = 26000 kg m/s + 7800 kg m/s
= 33800 kg m/s
2. IMPULSE
Often it is better to express momentum
in relation to time.
It is useful to know how much time it
took to change or transfer momentum.
Lets play with some equations a little...
p = mv
and
(what equation looks similar?)
Lets play with some equations a little...
p = mv
and
F = ma
(Newton’s 2nd law)
These two equations look very similar. The
only difference is the “v” and the “a”.
What’s the difference between “v” and a?
Lets play with some equations a little...
p = mv
and
F = ma
(Newton’s 2nd law)
These two equations look very similar. The
only difference is the “v” and the “a”.
What’s the difference between “v” and a?
a = v/t  v = at
Lets play with some equations a little...
p = mv
and
F = ma
(Newton’s 2nd law)
v = at
All we have to do is multiply Force by time,
and we get momentum…
F t = mat
Lets play with some equations a little...
p = mv
and
F = ma
v = at
All we have to do is multiply Force by time,
and we get momentum…
F t = mv
So if we apply force over a period of time, an
object will accelerate (move). Any object
that moves has momentum.
F t = mv
So if we apply force over a period of time, an
object will accelerate (move). Any object
that moves has momentum.
F t = mv
Therefore, anytime a force is applied to an
object, it will change it’s velocity, and
momentum will change. Therefore, a more
convenient way to express this equation is by
looking at the Change in momentum.
mvf = mvi + Ft
Change in momentum.
mvf = mvi + Ft
Change in momentum.
mvf = mvi + Ft
Mvf is the final
momentum of the object
Change in momentum.
mvf = mvi + Ft
Mvf is the final
momentum of the object
Mvi was the initial
momentum of the object
Change in momentum.
mvf = mvi + Ft
Mvf is the final
momentum of the object
F t is the change in
momentum of the object.
Mvi was the initial
momentum of the object
Change in momentum.
mvf = mvi + Ft
Mvf is the final
momentum of the object
F t is the change in
momentum of the object.
Mvi was the initial
momentum of the object
We can rewrite this equation:
Ft = mvf – mvi
Impulse (J): The change in momentum of an object.
Ft = mvf – mvi
F t is the change in
momentum of the object.
Mvi was the initial
momentum of the object
Mvf is the final
momentum of the object
Anytime a force is applied to an object, the
momentum of the system changes.
A shorthand way to write this equation is :
Ft = ∆p
Impulse: The change in momentum.
Ft = mvf – mvi
A shorthand way to write the impulse equation is:
Ft = ∆p
(The ∆(Delta) means “Change in” pf – pi)
Impulse: The change in momentum.
Impulse is actually common sense, we are just
not use to thinking of it this way. For
example, there are several ordinary
circumstances we use impulse:
Impulse: The change in momentum.
Circumstances where we use impulse:
1.
2.
3.
4.
Increase Momentum
Decrease Momentum over a long time
Decrease Momentum over a short time
Rebounding (Bouncing)
1. Increase Momentum
Increasing momentum will speed something
up:
1. Increase Momentum
Increasing momentum will speed something
up:
Examples:
Hitting a Golf Ball / Baseball / Kicking a
Ball, etc…
1. Increase Momentum
Impulse ( ft = ∆p) is directly proportional to
force and time.
So to make something move as fast as possible
one needs to apply a maximum force for a
maximum period of time.
1. Increase Momentum
Impulse ( ft = ∆p) is directly proportional to
force and time.
So to make something move as fast as possible
one needs to apply a maximum force for a
maximum period of time.
How do we do this? Follow through with a swing
1. Increase Momentum
Impulse ( ft = ∆p) is directly proportional to
force and time.
So to make something move as fast as possible
one needs to apply a maximum force for a
maximum period of time.
How do we do this? Follow through with a swing
Following through with a swing or kick allows one to
maintain the contact force for the maximum period
of time.
1. Increase Momentum
Baseball Example:
To hit a ball as far as possible, do you want to bunt or
swing?
1. Increase Momentum
Golf Club example:
If one wants to hit a golf ball as far as possible, one
Follows through with the swing.
1. Increase Momentum
Soccer Example:
To hit the ball as far as possible, the foot must maintain
contact with the ball for as long as possible.
Soccer kick
Football Punt
Impulse: The change in momentum.
Circumstances where we use impulse:
2. Decrease Momentum over a long time
Imagine you are falling from an elevated height,
and you have a choice of falling on a concrete
floor or a cushioned bed.
The choice is obvious, and impulse explains why.
2. Decrease Momentum over a long time
Example:
When you fall on the concrete floor, the time of
impact is VERY small, therefore the force of impact
must be VERY large:
Concrete Floor: ∆p =
F
t
2. Decrease Momentum over a long time
Example:
When you fall on the concrete floor, the time of
impact is VERY small, therefore the force of impact
must be VERY large:
Concrete Floor: ∆p =
F
t
When you fall on the padded bed, the time of
impact is VERY large, therefore the force of
impact must be VERY small:
Padded Matress: ∆p = F
t
2. Decrease Momentum over a long time
When you fall on the concrete floor, the time of
impact is VERY small, therefore the force of impact
must be VERY large:
Concrete Floor: ∆p =
F
t
When you fall on the padded bed, the time of
impact is VERY large, therefore the force of
impact must be VERY small:
Concrete Floor: ∆p = F
t
When you land, do you want a large impact
force or a small one?
2. Decrease Momentum over a long time
• Which would it be more safe to hit in a car ?
mv
mv
Ft
Ft
Knowing the physics helps us understand why hitting
a soft object is better than hitting a hard one.
2. Decrease Momentum over a long time
• Which would it be more safe to hit in a car ?
Ft
Ft
Knowing the physics helps us understand why hitting
a soft object is better than hitting a hard one.
2. Decrease Momentum over a long time
Example: Cars are designed to crumple. This increases
the time of impact, and therefore reduces the force of
impact.
2. Decrease Momentum over a long time
Example: A NASCAR racecar crashes.
The plastic frame of the car is designed to crumple, to maximize the impact time.
The sidewall on the course is also designed to flex, to maximize the impact time.
2. Decrease Momentum over a long time
Example: A paintball is shot at two materials:
Concrete
∆p = Ft
Hanging Cloth
t
∆p = F
∆p is the same, Vi is muzzle velocity, Vf is 0 m/s
2. Decrease Momentum over a long time
Example: Bullet proof vest:
Notice, the bullet and vest acts exactly like the
paintball striking the cloth.
2. Decrease Momentum over a long time
Example: Airbags in cars and Nylon seatbelts.
Airbags deploy BEFORE your head hits the steering
column. Your head then hits the airbag, this increases
your stopping time, thus decreases the force of impact.
2. Decrease Momentum over a long time
Example: Airbags in cars and Nylon seatbelts.
The nylon seatbelts stretch, which increases the time it
takes for your body to stop, thereby decreasing the
force of impact.
2. Decrease Momentum over a long time
Applications of decreasing p over a long time period
includes anything with padding:
Packing materials
Egg cartons
Gym mats
Helmets (interior padding)
Nylon seat belts
Nylon ropes used for rock climbing
Shoe padding and soles
Football padding / Hockey padding
Soft plastic/Rubber steering wheels
Car frames are designed to crumple
Bullet Proof Vests
Car airbags
Etc
Etc
Etc…
3. Decrease Momentum over a short time
(This is reversed from situation 2)
If you catch a high–speed ball while your hands move
toward the ball instead of away upon contact; or
jumping from an elevated position onto the floor
instead of a cushioned bed; or while boxing, if you
move into a punch instead of away–you're really in
trouble. In these cases of short impact times, the
impact forces are large.
∆p = Ft
3. Decrease Momentum over a short time
∆p = Ft
For an object brought to rest, the impulse is the
same, no matter how it is stopped. But if the time is
short, the force will be large.
3. Decrease Momentum over a short time
∆p = Ft
Example: A hammer and a nail
You want to swiftly transfer the p of the hammer
into the nail. The smaller the impact time, the
Larger the force on the nail.
3. Decrease Momentum over a short time
∆p = Ft
Example: A jackhammer
A jackhammer pushes on a chisel a few times per second.
The small time of impact dramatically increases the force of impact.
3. Decrease Momentum over a short time
∆p = Ft
The idea of short time of contact explains how a
karate expert can break a stack of concrete slabs
with one blow.
Although the human arm muscles can only apply a maximum of
a few hundred pounds, when the arm strikes a target with a
fast blow, the force can be tremendously increased, a few
hundred pounds can turn into a few thousand pounds.
3. Decrease Momentum over a short time
∆p = Ft
t
∆p = F
3. Decrease Momentum over a short time
∆p = Ft
∆p = Ft
4. Rebounding (Bouncing)
Observe two objects, one bounces, the other doesn’t. Calculate their
change in momentum:
Mb = Mr = 1 kg
Vi = 2 m/s
Vi = 2 m/s
4. Rebounding (Bouncing)
Observe two objects, one bounces, the other doesn’t. Calculate their
change in momentum:
Mb = Mr = 1 kg
Vi = 2 m/s
Vi = 2 m/s
Vf = 0 m/s
Vf = -2 m/s
4. Rebounding (Bouncing)
Observe two objects, one bounces, the other doesn’t. Calculate their
change in momentum:
Mb = Mr = 1 kg
Vi = 2 m/s
Vi = 2 m/s
∆p = m (vf – vi)
= 1 kg (0 m/s – 2 m/s)
= -2 kg m/s
Vf = 0 m/s
∆p = m (vf – vi)
= 1 kg (-2 m/s – 2 m/s)
= -4 kg m/s
Vf = -2 m/s
4. Rebounding (Bouncing)
Observe two objects, one bounces, the other doesn’t. Calculate their
change in momentum:
Mb = Mr = 1 kg
Vi = 2 m/s
Vi = 2 m/s
∆p = -2 kg m/s
Vf = 0 m/s
∆p = = -4 kg m/s
Vf = -2 m/s
4. Rebounding (Bouncing)
The impulse required to bring an object to a
stop and then to throw it back upward again
is greater than the impulse required to
merely bring the object to a stop.
4. Rebounding (Bouncing)
Example: The Pelton Wheel
4. Rebounding (Bouncing)
Example: The Pelton Wheel
The Pelton wheel used scoop
shaped paddles instead of flat
paddles. This caused the water to
rebound up when water hit the
wheel.
Rebounding caused the impulse to
double, which caused the force to
double.
His wheels spun twice as fast as
competitors wheels
By the 1880’s his wheel was
almost exclusively used for mining
and farming.
Pelton’s original patent for
his wheel
Pelton Wheel
The Pelton Wheel is still used today in many
Turbines to generate electricity.
Impulse (J) on a graph
F(N)
3000
2000
area under curve
1000
0
0
1
2
3
4
t (ms)
Sample Problem
• Suppose a 1.5-kg brick is dropped on a
glass table top from a height of 20 cm.
a) What is the magnitude and direction of the
impulse necessary to stop the brick?
b) If the table top doesn’t shatter, and stops
the brick in 0.01 s, what is the average
force it exerts on the brick?
c) What is the average force that the brick
exerts on the table top during this period?
Solution a)
Find the velocity of the brick when it strikes the table
using conservation of energy.
mgh = ½ mv2
v = (2gh)1/2 = (2*9.8 m/s2*0.20 m) 1/2 = 2.0 m/s
Calculate the brick’s momentum when it strikes the
table.
p = mv = (1.5 kg)(-2.0 m/s) = -3.0 kg m/s (down)
The impulse necessary to stop the brick is the impulse
necessary to change to momentum to zero.
J = Dp = pf – po = 0 – (-3.0 kg m/s) = +3.0 kg m/s
or 3.0 kg m/s (up)
Solution b) and c)
b) Find the force using the other
equation for impulse.
J = Ft
3.0 N s = F (0.01 s)
F = 300 N (upward in the same direction
as impulse)
c) According the Newton’s 3rd law, the
brick exerts an average force of 300
N downward on the table.
Solution
Find the impulse from the area under the curve.
A = ½ base * height = ½ (.1 s)(2500 N) = 125 Ns
J = 125 N s
Since impulse is equal to change in momentum and it is
in the same direction as the existing momentum,
the momentum increases by 125 kg m/s.
Dp = 125 kg m/s
Dp = pf - po = mvf - mvo
mvf = mvo + Dp
= (1.2 kg)(120 m/s) + 125 kg m/s = 269 kg m/s
vf = (269 kg m /s) / (1.2 kg) = 224 m/s