Momentum Chapter 6. Underlined words are WOD.. Momentum Momentum: mass in motion. Abbreviated with a rho which looks like a “p” Momentum is a vector!!

Momentum Chapter 6. Underlined words are WOD.

Momentum Momentum: mass in motion .

Abbreviated with a rho which looks like a “p” Momentum is a vector!!

Momentum = mass * velocity(momentum is in the same

direction as the velocity)

p = mv

Units : kilogram * meter / second

Momentum is not Inertia

In common language, “momentum” and “inertia” are often used interchangeably.

In physics, Inertia = Mass

Momentum = Mass * Velocity

Because velocities are often similar, often the most mass is the most momentum (football players); BUT NOT ALWAYS!!!!!! (football player vs. a bullet)

Quickly, work these practice problems:

.

WOD Definition: Net momentum before collision = net momentum after collisionQuestion: What will the momentum be after the shot?

WOD

CollisionsElastic collisions – total kinetic energy is conserved. No permanent change in shape of objects, they bounce off perfectly

2 billiard balls collide head on.

Time Line:

Before

During

After

If the collisions are 1.) elastic and 2.) masses are the same,Then the objects swap velocities.

These are BIG BIG BIG if’s and apply to the next 3 slides.

Collisions: Balls moving at same speed,opposite directions.

Net momentum before collision = net momentum after collision

2 billiard balls collide head on.Momentum is zero before and after.

Note their speeds before and after.

Before

During

After

Show on Newton’s Cradle on each slide

Collisions: 1 ball is stationary

Net momentum before collision = net momentum after collision

Before

During

After

1 billiard ball collides with a stationary one.Momentum is the same before and after.

Collisions: Balls moving at different speeds, in same direction.

2 billiard balls moving in the same direction collide. Momentum is the same before and after.

Note which is fast and slow, before and after.

Before

During

After

Net momentum before collision = net momentum after collision

Collisions: Summary SlideElastic collisions- no permanent change in shape of objects, they bounce off perfectly

2 billiard balls collide head onmomentum is zero before and after

1 billiard balls collide with a stationary onemomentum is the same before and after

2 billiard balls moving in the same direction collide momentum is the same before and after

1. Always momentum before collision = momentum after collision.2.Elastic means KE before = KE after

Collisions

Inelastic collisions – Kinetic Energy changes.Momentum is conserved, but KE is not. Examples: objects permanently change shape or stick together after impact. Some of the momentum might be used up in work, shape deformation, sound generation, friction, etc.Sound ipod

last longer with sound off. Upon collision, the cars stick together. The total mass moves slower, but the momentum of the 2 cars together is the same as the momentum of the system before the collision.

Net momentum before collision = net momentum after collision

CollisionsA toy train car, A, with a mass of 0.515 kg moves with a velocity of 1.10 m/s. It collides with and

sticks to another train car, B, which has a mass of 0.450 kg. Train car B is at rest. How fast do the

two train cars move immediately after the collision?

CollisionsA toy train car, A, with a mass of 0.515 kg moves with a velocity of 1.10 m/s. It collides with and sticks to another train car, B, which has a mass of 0.450 kg. Train car B is at rest. How fast do the two train cars move

immediately after the collision?

pinitial

= pfinal

pcar Ainitial

+ pcar Binitial

= pcar A final

+ p

carB final

mAv

Ai + m

Bv

Bi = m

Av

Af + m

bv

bf

.515kg(1.1m/s) + .450 kg (0) = .515kg*vAf

+ .450*vbf

Wait, since they stick together vAf =

vbf

.5665 kg*m/s = (.515kg + .450kg)* v

f

Vf = .578 m/s

CollisionsA truck with a mass of 3000 kg moves with a

velocity of 10 m/s. It collides with a car which has a mass of 1000 kg. The car was at rest. After the collision, the car is moving at 15 m/s. What is

the final velocity of the truck?

CollisionsA truck with a mass of 3000 kg moves with a velocity of 10 m/s. It collides a car which has a mass of 1000 kg. The car was at rest. After the collision, the car is moving at 15 m/s. What is the final velocity of

the truck?

pinitial

= pfinal

pcar initial

+ ptruck

initial

= pcar final

+ p

truck final

mcv

ci + m

tv

ti = m

cv

cf + m

tv

tf

1000kg(0) + 3000 kg (10m/s) = 1000kg*(15m/s) +

3000*vtf

30000kg*m/s = 15000kg*m/s + 3000* v

ft

Vf = 5 m/s

The ArcherAn archer stands at rest on frictionless ice and fires a 0.5-kg arrow horizontally at 50.0 m/s. The combined mass of the archer and bow is 60.0 kg. With what velocity does the archer move across the ice after firing the arrow?

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0.60

5.02

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21

Explosions

Explosion

What is an explosion, really?

Explosions

Explosions – Really just collisions, only backwards.

Explosions

Explosions – Really just collisions, only backwards.

Imagine a bomb, stationary in space with no gravity

What is the total momentum?

ExplosionsExplosions – Really just collisions, only backwards.

Imagine a bomb, stationary in space with no gravity.

Big arrow is NOT movement, but change in time.

What is the total momentum after the explosion?

Explosions

Notice: The “center of mass” didn’t move.

What is Center of Mass?? (see next slide.)

Center of mass

The “average location of the masses” of a system.

The “average position” of a system.

Your system is just a bunch of atoms. They have an “average” position.

Case 1: A large object is orbited by a small object.

Questions:

Does the earth orbit around the sun or does the sun orbit around the earth?

Do electrons orbit around the nucleus

or nucleus orbit the electrons?

Case 1: A large object is orbited by a small object.

No net force. The center of mass follows Newton’s 1st law. The center of mass cannot change since no force is acting on it. So the large object will make a small orbit around the center of mass and the small object will make a large orbit.

Red + is Center of Mass.Circles are orbits around it.

+

+

Select all. Ctrl A. Shift select this box.Grab the rotate tag.Rotate 360 circles to show rotation of objects. Undo when done to reset picture.

++ +

+

Case 1: A large object is orbited by a small object.

Questions:

Does the earth orbit around the sun or does the sun orbit around the earth?

Do electrons orbit around the nucleus

or the nucleus orbit around the electrons?

Answer: Neither. They both orbit around the system’s center of mass.

Case 2: External force is applied. The center of mass moves according that

force.

The hammer is a

System of particles.

What does the center of mass do?

Answer next slide.

What does the center of mass do?

Only the center of mass will follow the parabolic, free fall path from kinematics that we studied in chapters 2 and 3.

Fireworks from 4th of July. What part of the firework travels in a parabolic, freefall arc?

Answer: The center of mass.

(The blue line is path of the center of mass.)

Explosions

We have a cannon:

What is the total momentum of the cannon & tennis ball?

Explosion

The cannon goes off:

Cannon = 1.27 kg Ball = .0562 kg

The ball now has v = 63.2 m/s

What is the recoil velocity of the cannon?

Explosions

pcannon initial

+ p ball initial

= pcannon final

+ p

ball final

mcv

ci + m

bv

bi = m

cv

cf + m

bv

bf

mc*0

+ m

b*0

= m

cv

cf + m

bv

bf

0= 1.27 kg*vcf

+ 0562 kg* 63.2 m/s

Vcf

= 2.80m/s

Problem Solving #1A 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that

is at rest. Find the velocity of the fish immediately after “lunch”.

net momentum before = net momentum after

(net mv)before = (net mv)after

(6 kg)(1 m/sec) + (2 kg)(0 m/sec) = (6 kg + 2 kg)(vafter)

6 kg.m/sec = (8 kg)(vafter)

6

8 kg

vafter = ¾ m/sec

vafter = 6 kg.m/sec

Problem Solving #3 & #4Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg

fish that is swimming towards it at 3 m/sec.

Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 4 m/sec.

Problem Solving #3Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg

fish that is swimming towards it at 3 m/sec.

(net mv)before = (net mv)after

(6 kg)(1 m/sec) + (2 kg)(-3 m/sec) = (6 kg + 2 kg)(vafter)

6 kg.m/sec + -6 kg.m/sec = (8 kg)(vafter)

vafter = 0 m/sec

Problem Solving #4

Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 4 m/sec.

(net mv)before = (net mv)after

(6 kg)(1 m/sec) + (2 kg)(-4 m/sec) = (6 kg + 2 kg)(vafter)

6 kg.m/sec + -8 kg.m/sec = (8 kg)(vafter)

vafter = -1/4 m/sec

Before After

Skip SlideProblem Solving #3 & #4

Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 3 m/sec.

(net mv)before = (net mv)after

(6 kg)(1 m/sec) + (2 kg)(-3 m/sec) = (6 kg + 2 kg)(vafter)

6 kg.m/sec + -6 kg.m/sec = (8 kg)(vafter)

vafter = 0 m/sec

Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 4 m/sec.

(net mv)before = (net mv)after

(6 kg)(1 m/sec) + (2 kg)(-4 m/sec) = (6 kg + 2 kg)(vafter)

6 kg.m/sec + -8 kg.m/sec = (8 kg)(vafter)

vafter = -1/4 m/sec

Skip Slide Problem Solving #3 & #4

Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 3 m/sec.

(net mv)before = (net mv)after

(6 kg)(1 m/sec) + (2 kg)(-3 m/sec) = (6 kg + 2 kg)(vafter)

6 kg.m/sec + -6 kg.m/sec = (8 kg)(vafter)

vafter = 0 m/sec

Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 4 m/sec.

(net mv)before = (net mv)after

(6 kg)(1 m/sec) + (2 kg)(-4 m/sec) = (6 kg + 2 kg)(vafter)

6 kg.m/sec + -8 kg.m/sec = (8 kg)(vafter)

vafter = -1/4 m/sec

Rockets

How do rockets work?

Rockets

How do rockets work?

The engine pushes some kind of exhaust backwards, giving it negative momentum, so the rest of the rocket gains positive momentum to counter act that.

Momentum Momentum: mass in motion .

Momentum is a vector!! So it has 3 components.

We are only going to work with 2.

Conservation of Momentum applies in all 3 directions!!!!Net momentum before collision = net momentum after collisionWe are only going to work with 2 D.

Momentum

2 D rule of Conservation of Momentum

X: Net momentum before collision in X= net momentum after collision in X.

Y: Net momentum before collision in Y= net momentum after collision in Y.

i.e., Momentum is a VECTOR and is solved using VECTOR SUMS!!!px is the same before and after. py is also conserved.

Two-Dimensional CollisionsFor a general collision of two objects in two-

dimensional space, the conservation of momentum principle implies that the total momentum of the system in each direction is conserved

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March 24, 2009

Glancing Collisions

The “after” velocities have x and y components.Momentum is conserved in the x direction and in the y direction.Apply conservation of momentum separately to each direction.(Done on next slides.)

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March 24, 2009

2-D Collision, example

Particle 1 is moving at velocity and particle 2 is at rest

In the x-direction, the initial momentum is m1v1i

In the y-direction, the initial momentum is 0

1iv

March 24, 2009

2-D Collision, example contAfter the collision, the momentum in the x-direction is m1v1f cos m2v2f cos

After the collision, the momentum in the y-direction is m1v1f sin m2v2f sin

222111

22211111

sinsin00

coscos0

ff

ffi

vmvm

vmvmvm

Collision at an IntersectionA car with mass 1.5×103 kg traveling east at a speed of 25 m/s collides at an intersection with a 2.5×103 kg van traveling north at a speed of 20 m/s. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision and assuming that friction between the vehicles and the road can be neglected.

??,/20,/25

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vsmvsmv

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March 24, 2009

Collision at an Intersection

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cos)( fvcvfxvcfxcxf vmmvmvmp

cos)1000.4(/1075.3 34fvkgsmkg

smkgvmvmvmp viyvviyvciycyi /1000.5 4

sin)( fvcvfyvcfycyf vmmvmvmp

sin)1000.4(/1000.5 34fvkgsmkg

March 24, 2009

Collision at an Intersection

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sin)1000.4(/1000.5 34fvkgsmkg

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/1000.5tan

4

4

smkg

smkg

1.53)33.1(tan 1

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smkgv f /6.15

1.53sin)1000.4(

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