Momentum Unit1. Momentum
Mass vs VelocitySpace Debris
2. ImpulseIncreasing momentumDecreasing momentum over a long timeDecreasing momentum over a short timeRebounding (Bouncing)
3. Collisions/ExplosionsConservation of pElasticInelasticExplosions
What is momentum?
The concept of momentum is closely related to Newton’s Laws of Motion
1. MOMENTUMThe 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 objectp = mv
• For a system of multiple objectsp = 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 = 10 g, v = 5 m/sM = 1000 kg, v = 1 m/s
Which has the most momentum?
M = 10 g, v = 3 m/sM = 1000 kg, v = 1 m/s
Truck: Butterfly:
p = mv p = mv
p = (1000 kg) (1 m/s) p = (.01 kg) (3 m/s)
Which has the most momentum?
M = 10 g, v = 3 m/sM = 1000 kg, v = 1 m/s
Truck: Butterfly:
P = mv 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?
M = 10 g, v = 3 m/sM = 1000 kg, v = 1 m/s
Truck: Butterfly:
p = mv 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 momentumAs the truck?
Which has the most momentum?
Truck: Butterfly:
p = 1000 kg m/s
How can the butterfly have the same momentumAs the truck?
M = 1000 kg, v = 1 m/s M = 10 g, v =
The mass of an object is usually constant, however, Velocity is easy
to change.
Which has the most momentum?
Truck: Butterfly:
p = 1000 kg m/s
M = 1000 kg, v = 1 m/s M = 10 g, v =
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: Butterfly:
p = 1000 kg m/s
M = 1000 kg, v = 1 m/s M = 10 g, v =
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 hasA large mass, even if it has a small velocity
Momentum Depends on mass AND velocity
An object can have a large momentum if it hasA large mass, even if it has a small velocity
However, a small object can also have a largemomentum 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 ShuttleAlthough 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 ShuttleWhy were the windows reduced in size?
To protect against micro meteor impacts. One holethrough the window can depressurize the entire cabin
This micro meteor traveled ¾ way through the glass before it stopped
The Space ShuttleThe Space Shuttle reaches speeds of around
18,000 MPH. At these speeds, even foam insulation (with very little mass) also posed problems.
The Space ShuttleThe Space Shuttle reaches speeds of around
18,000 MPH. At these speeds, even foam insulation (with very little mass) also posed problems.
The Space ShuttleThe 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
Often it is better to express momentumin relation to time.
It is useful to know how much time ittook to change or transfer momentum.
2. IMPULSE
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 finalmomentum of the object
Change in momentum.
mvf = mvi + Ft
Mvf is the finalmomentum of the object
Mvi was the initialmomentum of the object
Change in momentum.
mvf = mvi + Ft
Mvf is the finalmomentum of the object
Mvi was the initialmomentum of the object
F t is the change in momentum of the object.
Change in momentum.
mvf = mvi + Ft
Mvf is the finalmomentum of the object
Mvi was the initialmomentum of the object
F t is the change in momentum of the object.
We can rewrite this equation:
Ft = mvf – mvi
Impulse (J): The change in momentum of an object.
Mvf is the finalmomentum of the object
Mvi was the initialmomentum of the object
F t is the change in momentum of the object.
Ft = mvf – mvi
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.Increase Momentum2.Decrease Momentum over a long time3.Decrease Momentum over a short time4.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 tomaintain 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 maintaincontact 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 timeExample: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 timeExample: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 timeWhen 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 tWhen you land, do you want a large
impact force or a small one?
• Which would it be more safe to hit in a car ?
mv
mv
Ft
Ft
2. Decrease Momentum over a long time
Knowing the physics helps us understand why hitting a soft object is better than hitting a hard one.
• Which would it be more safe to hit in a car ?
Ft
Ft
2. Decrease Momentum over a long time
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 Hanging Cloth
∆p = Ft ∆p = Ft∆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 materialsEgg cartonsGym matsHelmets (interior padding)Nylon seat beltsNylon ropes used for rock climbingShoe padding and solesFootball padding / Hockey paddingSoft plastic/Rubber steering wheelsCar frames are designed to crumpleBullet Proof VestsCar airbagsEtcEtcEtc…
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 hammerinto 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
∆p = Ft
3. Decrease Momentum over a short time ∆p = Ft
∆p = Ft
4. Rebounding (Bouncing)
Vi = 2 m/s Vi = 2 m/sMb = Mr = 1 kg
Observe two objects, one bounces, the other doesn’t. Calculate their change in momentum:
4. Rebounding (Bouncing)
Vi = 2 m/s Vi = 2 m/s
Mb = Mr = 1 kg
Vf = 0 m/sVf = -2 m/s
Observe two objects, one bounces, the other doesn’t. Calculate their change in momentum:
4. Rebounding (Bouncing)
Vi = 2 m/s Vi = 2 m/s
Mb = Mr = 1 kg
Vf = 0 m/sVf = -2 m/s
Observe two objects, one bounces, the other doesn’t. Calculate their change in momentum:
∆p = m (vf – vi) = 1 kg (0 m/s – 2 m/s) = -2 kg m/s
∆p = m (vf – vi) = 1 kg (-2 m/s – 2 m/s) = -4 kg m/s
4. Rebounding (Bouncing)
Vi = 2 m/s Vi = 2 m/s
Mb = Mr = 1 kg
Vf = 0 m/sVf = -2 m/s
Observe two objects, one bounces, the other doesn’t. Calculate their change in momentum:
∆p = -2 kg m/s ∆p = = -4 kg 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
Pelton’s original patent for his 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 Wheel
The Pelton Wheel is still used today in many Turbines to generate electricity.
Impulse (J) on a graph
F(N)
t (ms)0 1 2 3 40
1000
2000
3000
area under curve
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/sCalculate 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 = p = 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 = Ft3.0 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.
SolutionFind the impulse from the area under the curve.A = ½ base * height = ½ (.1 s)(2500 N) = 125 NsJ = 125 N sSince 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.
p = 125 kg m/s p = pf - po = mvf - mvo
mvf = mvo + p
= (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