momentum unit 1. momentum mass vs velocity space debris 2. impulse increasing momentum decreasing...

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



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





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?


It Depends!

Which has the most momentum?M = 10 g, v = 5 m/sM = 1000 kg, v = 1 m/s


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)


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


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?


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.


Truck: Butterfly:

p = 1000 kg m/s

M = 1000 kg, v = 1 m/s M = 10 g, v =


to change.

Butterfly:

p = mv

1000 kg m/s = (.01 kg) (Vb)


Truck: Butterfly:

p = 1000 kg m/s

M = 1000 kg, v = 1 m/s M = 10 g, v =


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

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

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, 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


• Pperson = (60kg) (130 m/s) = 7800 kg m/s

Sample Problem:• Pcycle = 26000 kg m/s


• 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?)


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?



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



v = at

All we have to do is multiply Force by time, and we get momentum…

F t = mat


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


Mvi was the initialmomentum of the object

Change in momentum.

mvf = mvi + Ft



F t is the change in momentum of the object.

Change in momentum.

mvf = mvi + Ft




We can rewrite this equation:

Ft = mvf – mvi

Impulse (J): The change in momentum of an 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 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:


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



How do we do this? Follow through with a swing

1.Increase Momentum



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


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:


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:


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


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


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.


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.


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.


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.


∆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:


Vi = 2 m/s Vi = 2 m/s

Mb = Mr = 1 kg

Vf = 0 m/sVf = -2 m/s



Vi = 2 m/s Vi = 2 m/s

Mb = Mr = 1 kg

Vf = 0 m/sVf = -2 m/s


∆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


Vi = 2 m/s Vi = 2 m/s

Mb = Mr = 1 kg

Vf = 0 m/sVf = -2 m/s


∆p = -2 kg m/s ∆p = = -4 kg m/s


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

momentum unit 1. momentum mass vs velocity space debris 2. impulse increasing momentum decreasing...

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