chapter 9 linear momentum

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Chapter 9

Linear Momentum

Announcements

• Assignments due Saturday

• Midterm Exam II: October 17 (chapters 6-9)

• Formula sheet will be posted

• Practice problems

• Practice exam next week

Linear Momentum

Momentum is a vector; its direction is the same as the direction of the velocity.

Momentum and Newton’s Second Law

Newton’s second law, as we wrote it before:

is only valid for objects that have constant mass. Here is a more general form (also useful when the mass is changing):

Impulse

Impulse is a vector, in the same direction as the average force.

The same change in momentum may be produced by a large force acting for a short time, or by a smaller force acting for a longer time.

Impulse quantifies the overall change in momentum

Impulse

We can rewrite

as

So we see that

The impulse is equal to the change in momentum.

Why we don’t dive into concreteThe same change in momentum may be

produced by a large force acting for a short time, or by a smaller force acting for a longer time.

Going Bowling II

p

p

a) the bowling ball

b) same time for both

c) the Ping-Pong ball

d) impossible to say

A bowling ball and a Ping-Pong ball are rolling toward you with the same momentum. If you exert the same force to stop each one, which takes a longer time to bring to rest?

Going Bowling II

We know:

Here, F and p are the same for both balls!

It will take the same amount of time to stop them. p

p so p = Fav t

a) the bowling ball

b) same time for both



A bowling ball and a Ping-Pong ball are rolling toward you with the same momentum. If you exert the same force to stop each one, which takes a longer time to bring to rest?

av tp

F

Going Bowling III

p

p

A bowling ball and a Ping-Pong

ball are rolling toward you with

the same momentum. If you

exert the same force to stop each

one, for which is the stopping

distance greater?

a) the bowling ball

b) same distance for both



Going Bowling III

p

p

Use the work-energy theorem: W = KE. The ball with less mass has the greater speed, and thus the greater KE. In order to remove that KE, work must be done, where W = Fd. Because the force is the same in both cases, the distance needed to stop the less massive ball must be bigger.

A bowling ball and a Ping-Pong

ball are rolling toward you with

the same momentum. If you

exert the same force to stop each

one, for which is the stopping

distance greater?

a) the bowling ball

b) same distance for both



Conservation of Linear Momentum

The net force acting on an object is the rate of change of its momentum:

If the net force is zero, the momentum does not change!

•A vector equation•Works for each coordinate separately

With no net force:

Internal Versus External Forces

Internal forces act between objects within the system.

As with all forces, they occur in action-reaction pairs. As all pairs act between objects in the system, the internal forces always sum to zero:

Therefore, the net force acting on a system is the sum of the external forces acting on it.

Momentum of components of a systemInternal forces cannot change the momentum of a system.

However, the momenta of pieces of the system may change.

An example of internal forces moving components of a system:

With no net external force:

Kinetic Energy of a SystemAnother example of internal forces moving components of a system:

The initial momentum equals the final (total) momentum.

But the final Kinetic Energy is very large

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Birth of the neutrino

Beta decay fails momentum conservation?

Pauli “fixes” it with a new ghost-like, undetectable particle

Bohr scoffs

First detection 1956

Lecture 11Momentum, Energy, and

Collisions

Linear Momentum

Impulse


Nuclear Fission I

A uranium nucleus (at rest) A uranium nucleus (at rest)

undergoes fission and splits undergoes fission and splits

into two fragments, one into two fragments, one

heavy and the other light. heavy and the other light.

Which fragment has the Which fragment has the

greater momentum? greater momentum?

a) the heavy one a) the heavy one

b) the light oneb) the light one

c) both have the same c) both have the same

momentummomentum

d) impossible to sayd) impossible to say

1 2

1 2

Nuclear Fission I

A uranium nucleus (at rest)

undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater momentum?

a) the heavy one

b) the light one

c) both have the same

momentum


The initial momentum of the uranium

was zero, so the final total momentum

of the two fragments must also be

zero. Thus the individual momenta

are equal in magnitude and opposite

in direction.

Nuclear Fission II

a) the heavy one

b) the light one

c) both have the same speed


1 2


undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater speed?

Nuclear Fission II

We have already seen that the

individual momenta are equal and

opposite. In order to keep the

magnitude of momentum mv the

same, the heavy fragment has the

lower speed and the light fragment

has the greater speed.

a) the heavy one

b) the light one

c) both have the same speed


1 2


undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the

greater speed?

A plate drops onto a smooth floor and shatters into three pieces of equal mass. Two of the pieces go off with equal speeds v along the floor, but at right angles to one another. Find the speed and direction of the third piece.

We know that px=0, py = 0 in initial stateand no external forces act in the horizontal

looking from above: x

y

v1

v2

v3

An 85-kg lumberjack stands at one end of a 380-kg floating log, as shown in the figure. Both the log and the lumberjack are at rest initially. (a) If the lumberjack now trots toward the other end of the log with a speed of 2.7 m/s relative to the log, what is the lumberjack’s speed relative to the shore? Ignore friction between the log and the water. (b) If the mass of the log had been greater, would the lumberjack’s speed relative to the shore be greater than, less than, or the same as in part (a)? Explain.

Center of Mass

Center of MassThe center of mass of a system is the point where the

system can be balanced in a uniform gravitational field.

For two objects:

The center of mass is closer to the more massive object.

Think of it as the “average location of the mass”

Center of Mass

The center of mass need not be within the object

Momentum of components of a system

Internal forces cannot change the momentum of a system.

However, the momenta of pieces of the system may change.

An example of internal forces moving components of a system:


RECALL:

Motion about the Center of MassThe center of mass of a complex or composite object follows a trajectory as if it were a single particle - with mass equal to the complex object, and experiencing a force equal to the sum of all external forces on that complex object

Motion of the center of mass

Action/Reaction pairs inside the system cancel out

The total mass multiplied by the acceleration of the center of mass is equal to the net external force

The center of mass accelerates just as though it were a point particle of mass M acted on by

Momentum of a composite object

Recoil Speed

a) 0 m/s

b) 0.5 m/s to the right

c) 1 m/s to the right

d) 20 m/s to the right

e) 50 m/s to the right

A cannon sits on a stationary

railroad flatcar with a total

mass of 1000 kg. When a 10-

kg cannonball is fired to the left

at a speed of 50 m/s, what is

the recoil speed of the flatcar?

Recoil Speed

Because the initial momentum of the

system was zero, the final total

momentum must also be zero. Thus, the

final momenta of the cannonball and the

flatcar must be equal and opposite.

pcannonball = (10 kg)(50 m/s) = 500 kg-

m/s

pflatcar = 500 kg-m/s = (1000 kg)(0.5

m/s)

a) 0 m/s







mass of 1000 kg. When a 10-

kg cannonball is fired to the left

at a speed of 50 m/s, what is

the recoil speed of the flatcar?

Recoil Speed

a) 0 m/s







mass of 1000 kg. When a 10-kg

cannonball is fired to the left at

a speed of 50 m/s, what is the

speed of the center of mass?

Recoil Speed

a) 0 m/s






railroad flatcar with a total mass

of 1000 kg. When a 10-kg

cannonball is fired to the left at

a speed of 50 m/s, what is the

speed of the center of mass?

Internal forces cannot change the motion of the center of mass. The CM was originally motionless at zero, and remains so after the gun is fired.

Rolling in the Rain

a) speeds up

b) maintains constant speed

c) slows down

d) stops immediately

An open cart rolls along a

frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)

Because the rain falls in vertically,

it adds no momentum to the box,

thus the box’s momentum is

conserved. However, because the

mass of the box slowly increases

with the added rain, its velocity has

to decrease.

Rolling in the Rain

a) speeds up

b) maintains constant speed

c) slows down

d) stops immediately

An open cart rolls along a

frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)

Two objects collide... and stick

A completely inelastic collision: no “bounce back”

No external forces... so momentum of system is conservedinitial px = mv0

final px = (2m)vf

mv0 = (2m)vf

vf = v0 / 2

mass m mass m

Inelastic collision: What about energy?

mass m mass m vf = v0 / 2

Kinetic energy is lost! KEfinal = 1/2 KEinitial

initial

final

Collisions

This is an example of an “inelastic collision”

Collision: two objects striking one another

Elastic collision ⇔ “things bounce back”

⇔ energy is conserved

Inelastic collision: less than perfectly bouncy

⇔ Kinetic energy is lost

Time of collision is short enough that external forces may be ignored so momentum is conserved

Completely inelastic collision: objects stick together afterwards. Nothing “bounces back”. Maximal energy loss

Elastic vs. Inelastic

Inelastic collision: momentum is conserved but kinetic energy is not

Completely inelastic collision: colliding objects stick together, maximal loss of kinetic energy

Elastic collision: momentum and kinetic energy is conserved.

Completely Inelastic Collisions in One Dimension

Solving for the final momentum in terms of initial velocities and masses, for a 1-dimensional, completely inelastic collision between unequal masses:

Completely inelastic only(objects stick together, so have same final velocity)

KEfinal < KEinitial

Momentum Conservation:

Crash Cars I

a) I

b) II

c) I and II

d) II and III

e) all three

If all three collisions below

are totally inelastic, which

one(s) will bring the car on

the left to a complete halt?

Crash Cars I

In case I, the solid wall clearly stops the car.

In cases II and III, because

ptot = 0 before the

collision, then ptot must

also be zero after the collision, which means that the car comes to a halt in all three cases.

a) I

b) II

c) I and II

d) II and III

e) all three

If all three collisions below

are totally inelastic, which

one(s) will bring the car on

the left to a complete halt?

Crash Cars II

If all three collisions below are

totally inelastic, which one(s)

will cause the most damage (in

terms of lost energy)?

a) I

b) II

c) III

d) II and III

e) all three

Crash Cars II

a) I

b) II

c) III

d) II and III

e) all three

The car on the left loses the same KE in all three cases, but in case III, the car on the right loses the most KE because KE = mv2 and the car in case III has the largest velocity.

If all three collisions below are

totally inelastic, which one(s) will

cause the most damage (in

terms of lost energy)?

Ballistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block.

vf = m v0 / (m+M)

momentum conservation in inelastic collision

KE = 1/2 (mv0)2 / (m+M)

energy conservationafterwards

PE = (m+M) g h

hmax = (mv0)2 / [2 g (m+M)2]

Velocity of the ballistic pendulum

Pellet Mass (m): 2 gPendulum Mass (M): 3.81 kgWire length (L): 4.00 m

approximation

Inelastic Collisions in 2 Dimensions

Energy is not a vector equation: there is only 1 conservation of energy equation

Momentum is a vector equation: there is 1 conservation of momentum equation per dimension

For collisions in two dimensions, conservation of momentum is applied separately along each axis:

Elastic CollisionsIn elastic collisions, both kinetic

energy and momentum are conserved.

One-dimensional elastic collision:

Elastic Collisions in 1-dimension

solving for the final speeds:

Note: relative speed is conserved for head-on (1-D) elastic collision

We have two equations: conservation of momentum conservation of energy

and two unknowns (the final speeds).

For special case of v2i = 0

Limiting cases of elastic collisions

note: relative speed conserved

Limiting cases

note: relative speed conserved

Toy Pendulum

Could two balls recoil and conserve both momentum and energy?

Incompatible!

Elastic Collisions II

v

v

m

M

Carefully place a small rubber ball (mass m) on

top of a much bigger basketball (mass M) and

drop these from the same height h so they

arrive at the ground with the speed v. What is

the velocity of the smaller ball after the

basketball hits the ground, reverses direction,

and then collides with the small rubber ball?

a) zero

b) v

c) 2v

d) 3v

e) 4v

• Remember that relative velocity has to be equal before and after collision! Before the collision, the basketball bounces up with v and the rubber ball is coming down with v, so their relative velocity is –2v. After the collision, it therefore has to be +2v!!

Elastic Collisions II

v

v

v

v

3v

v

(a) (b) (c)

m

M

Carefully place a small rubber ball (mass m) on

top of a much bigger basketball (mass M) and

drop these from the same height h so they

arrive at the ground with the speed v. What is

the velocity of the smaller ball after the

basketball hits the ground, reverses direction,

and then collides with the small rubber ball?

a) zero

b) v

c) 2v

d) 3v

e) 4v

Elastic Collisions I

v 2v

1at rest

at rest

a) situation 1

b) situation 2

c) both the same

Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on.

In which case does the golf ball have the greater speed after the collision?

Remember that the magnitude of the relative velocity has to be equal before and after the collision!

Elastic Collisions I

v1

In case 1 the bowling ball will almost remain at rest, and the golf ball will bounce back with speed close to v.

v 22v

In case 2 the bowling ball will keep going with speed close to v, hence the golf ball will rebound with speed close to 2v.

a) situation 1

b) situation 2

c) both the same

Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on.

In which case does the golf ball have the greater speed after the collision?

chapter 9 linear momentum

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