Momentum and Momentum and CollisionsCollisions

Linear Momentum

The linear momentum of a particle or an object that can be modeled as a particle of mass m moving with a velocity v is defined to be the product of the mass and velocity: p = m v

We will usually refer to this as “momentum”, omitting the “linear”.

Momentum is a VECTOR. It has components. Don’t forget that.

Linear Momentum

The dimensions of momentum are ML/T The SI units of momentum are kg · m / s Momentum can be expressed in component

form: px = m vx py = m vy pz = m vz

A 3.00-kg particle has a velocity of . (a) Find its x and y components of momentum. (b) Find the magnitude and direction of its momentum.

Newton and Momentum

Newton called the product mv the quantity of motion of the particle

Newton’s Second Law can be used to relate the momentum of a particle to the resultant force acting on it

with constant mass.

d md dm m

dt dt dt

vv pF a

Newton’s Second Law

The time rate of change of the linear momentum of a particle is equal to the net force acting on the particle This is the form in which Newton presented the Second

Law It is a more general form than the one we used previously This form also allows for mass changes

Applications to systems of particles are particularly powerful

Two Particles (1 dimension for now)

other. on the force

a producing particle one are that those

are system on this acting forcesonly the

thenforces external NO are there

21

IFdt

dp

dt

dp

dt

dptotal

Continuing

00 int

21

ernaltotal

external

total

Fdt

dpF

dt

dp

dt

dp

dt

dp

NO EXTERNAL FORCES: ptotal is constant Force on P1 is from particle 2 and is F21

Force on P2 is from particle 1 and is F12

F21=-F12

Summary

The momentum of a SYSTEM of particles that areisolated from external forces remains a constant ofthe motion.

Conservation of Linear MomentumTextbook Statement Whenever two or more particles in an

isolated system interact, the total momentum of the system remains constant The momentum of the system is conserved,

not necessarily the momentum of an individual particle

This also tells us that the total momentum of an isolated system equals its initial momentum

Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M? (b) Find the original elastic potential energy in the spring if M = 0.350 kg.

A particle of mass m moves with momentum p. Show that the kinetic energy of the particle is K = p2/2m. (b) Express the magnitude of the particle’s momentum in terms of its kinetic energy and mass.

An Easy One…

Consider a particle roaming around that is suddenly subjected to some kind of FORCE that looks something like the last slide’s graph.

change will then 0 If

0 ifconstant

:

pF

Fp

Favp

vp

mdt

dm

dt

d

m

NEWTON

Let’s drop the vector notation and stick to one dimension.

f

i

f

i

t

t

if

f

i

t

t

Fdtpp

Fdtdp

Fdtdp

Fdt

dp

Change of Momentum

Impulse

NEW LAW

IMPULSE = CHANGE IN MOMENTUM

NEW LAW

A 3.00-kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. P9.9). If the ball is in contact with the wall for 0.200 s, what is the average force exerted on the ball by the wall?

LETS TALK ABOUT

Consider two particles:

1 2

m1 m2

v1v2

V1 V2

1 2

1 2

What goes on during the collision? N3

Force on m2

=F(12)

Force on m1

=F(21)

The Forces

Equal and Opposite

More About Impulse: F-t The Graph Impulse is a vector quantity The magnitude of the

impulse is equal to the area under the force-time curve

Dimensions of impulse are M L / T

Impulse is not a property of the particle, but a measure of the change in momentum of the particle

Impulse

The impulse can also be found by using the time averaged force

I = t

This would give the same impulse as the time-varying force does

F

SKATEBOARD DEMO

Conservation of Momentum, Archer Example

The archer is standing on a frictionless surface (ice)

Approaches: Newton’s Second Law –

no, no information about F or a

Energy approach – no, no information about work or energy

Momentum – yes

Archer Example, 2

Let the system be the archer with bow (particle 1) and the arrow (particle 2)

There are no external forces in the x-direction, so it is isolated in terms of momentum in the x-direction

Total momentum before releasing the arrow is 0 The total momentum after releasing the arrow is p1f

+ p2f = 0

Archer Example, final

The archer will move in the opposite direction of the arrow after the release Agrees with Newton’s Third Law

Because the archer is much more massive than the arrow, his acceleration and velocity will be much smaller than those of the arrow

An estimated force-time curve for a baseball struck by a bat is shown in Figure P9.7. From this curve, determine (a) the impulse delivered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the ball.

Overview: Collisions – Characteristics We use the term collision to represent an event

during which two particles come close to each other and interact by means of forces

The time interval during which the velocity changes from its initial to final values is assumed to be short

The interaction force is assumed to be much greater than any external forces present This means the impulse approximation can be used

Collisions – Example 1

Collisions may be the result of direct contact

The impulsive forces may vary in time in complicated ways This force is internal to

the system Momentum is

conserved

Collisions – Example 2

The collision need not include physical contact between the objects

There are still forces between the particles

This type of collision can be analyzed in the same way as those that include physical contact

Types of Collisions

In an elastic collision, momentum and kinetic energy are conserved Perfectly elastic collisions occur on a microscopic level In macroscopic collisions, only approximately elastic

collisions actually occur In an inelastic collision, kinetic energy is not

conserved although momentum is still conserved If the objects stick together after the collision, it is a

perfectly inelastic collision

Collisions, cont

In an inelastic collision, some kinetic energy is lost, but the objects do not stick together

Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types

Momentum is conserved in all collisions

Perfectly Inelastic Collisions

Since the objects stick together, they share the same velocity after the collision

m1v1i + m2v2i =

(m1 + m2) vf

A 10.0-g bullet is fired into a stationary block of wood (m = 5.00 kg). The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet?

. Velcro couplers make the carts stick together after colliding. Find the final velocity of the train of three carts. (b) What If? Does your answer require that all the carts collide and stick together at the same time? What if they collide in a different order?

Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that experienced by the truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false. Newton’s third law tells us that both objects experience forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at 8.00 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 80.0 kg. Including the drivers, the total vehicle masses are 800 kg for the car and 4 000 kg for the truck. If the collision time is 0.120 s, what force does the seatbelt exert on each driver?

SPRING BREAK SPRING BREAK BREAKBREAK

March 20, 2006March 20, 2006

Welcome Home!!Welcome Home!!

• The pace now quickens!The pace now quickens!• 5 weeks (+ 1 session) remain in the 5 weeks (+ 1 session) remain in the

semestersemester• Next Exam March 31Next Exam March 31

– Details on next 2 slidesDetails on next 2 slides

• This week we return to momentumThis week we return to momentum• Watch for another WA soonWatch for another WA soon• QuizQuiz on Friday - momentum on Friday - momentum

APPROXIMATE APPROXIMATE SCHEDULESCHEDULE

20-Mar Ch 9: Momentum Ch 9: Momentum Ch 9: Momentum

27-Mar Ch 10: Rotation Ch 10: Rotation EXAM #3

3-Apr Ch 10: RotationCh 11: Angular Mom.

Ch 11: Angular Mom.

10-Apr

Ch 11: Angular Mom.

Ch 11: Angular Mom.+ Chapt 12. Ch 12: Statics

17-Apr Ch 12: Statics Ch 15: Oscillations Ch 15: Oscillations

24-Apr Ch 15: Oscillations No Class No Class

1-May FINAL EXAM No Class No Class

Next ExamNext Exam

•Will IncludeWill Include– Potential Energy ChapterPotential Energy Chapter– Conservation of MomentumConservation of Momentum– Rotation – Material covered Rotation – Material covered

prior to examprior to exam•Includes Wednesday before Includes Wednesday before exam! SORRY!exam! SORRY!

Now Where Were We??? Oh yeah …. Elastic Collisions Both momentum and

kinetic energy are conserved

1 1 2 2

1 1 2 2

2 21 1 2 2

2 21 1 2 2

1 1

2 21 1

2 2

i i

f f

i i

f f

m m

m m

m m

m m

v v

v v

v v

v v

Elastic Collisions, cont Typically, there are two unknowns to solve for and so you

need two equations The kinetic energy equation can be difficult to use The energy equation, along with conservation of

momentum, can be used to solve for the two unknowns It can only be used with a one-dimensional, elastic

collision between two objects The solution is shown on pages 262-3 in the

textbook. (Lots of algebra but nothing all that difficult. We will look at a special case.

One Dimension Energy is Conserved

Let’s look at the case of equal masses.

Second Particle initially at rest

Explains the demo!

Elastic Collisions, final

Example of some special cases m1 = m2 – the particles exchange velocities When a very heavy particle collides head-on with a very

light one initially at rest, the heavy particle continues in motion unaltered and the light particle rebounds with a speed of about twice the initial speed of the heavy particle

When a very light particle collides head-on with a very heavy particle initially at rest, the light particle has its velocity reversed and the heavy particle remains approximately at rest

Collision Example – Ballistic Pendulum Perfectly inelastic collision –

the bullet is embedded in the block of wood

Momentum equation will have two unknowns

Use conservation of energy from the pendulum to find the velocity just after the collision

Then you can find the speed of the bullet

As shown in the figure, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

Two Dimensional Collisions

Two-Dimensional Collisions

The momentum is conserved in all directions Use subscripts for

identifying the object indicating initial or final values the velocity components

If the collision is elastic, use conservation of kinetic energy as a second equation Remember, the simpler equation can only be used for one-

dimensional situations

Two-Dimensional Collision, example Particle 1 is moving at

velocity v1i and particle 2 is at rest

In the x-direction, the initial momentum is m1v1i

In the y-direction, the initial momentum is 0

Two-Dimensional Collision, example cont After 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

Problem-Solving Strategies – Two-Dimensional Collisions Set up a coordinate system and define your

velocities with respect to that system It is usually convenient to have the x-axis coincide

with one of the initial velocities In your sketch of the coordinate system, draw

and label all velocity vectors and include all the given information

Problem-Solving Strategies – Two-Dimensional Collisions, 2 Write expressions for the x- and y-components of

the momentum of each object before and after the collision Remember to include the appropriate signs for the

components of the velocity vectors Write expressions for the total momentum of the

system in the x-direction before and after the collision and equate the two. Repeat for the total momentum in the y-direction.

Problem-Solving Strategies – Two-Dimensional Collisions, 3 If the collision is inelastic, kinetic energy of

the system is not conserved, and additional information is probably needed

If the collision is perfectly inelastic, the final velocities of the two objects are equal. Solve the momentum equations for the unknowns.

Problem-Solving Strategies – Two-Dimensional Collisions, 4 If the collision is elastic, the kinetic energy of

the system is conserved Equate the total kinetic energy before the

collision to the total kinetic energy after the collision to obtain more information on the relationship between the velocities

Two-Dimensional Collision Example Before the collision, the

car has the total momentum in the x-direction and the van has the total momentum in the y-direction

After the collision, both have x- and y-components

Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity 13.0 m/s toward the east and the other is traveling north with speed v2i. Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of 55.0o north of east. The speed limit for both roads is 35 mi/h and the driver of the northward-moving vehicle claims he was within the speed limit when the collision occurred. Is he telling the truth?

New Topic: Center of Mass

Momentum of a SYSTEM of particles.

The Center of Mass

There is a special point in a system or object, called the center of mass, that moves as if all of the mass of the system is concentrated at that point

The system will move as if an external force were applied to a single particle of mass M located at the center of mass M is the total mass of the system

Consider a SYSTEM of N particles There are N point particles. Each point “I” has a mass mi

The TOTAL mass of the system is given by M where M=mi

Let the FORCE on particle “I” be given by Fi, a VECTOR Quantity.

Let’s see what Newton says about such a SYSTEM

The MATH …

Fsystem is the TOTAL force on the system of particles

asome… is some average acceleration that we need to deal with.

averageofkindsomesystem MaFNewton :

We use a vector approach …Consider a system of only two particles

2211

21

2on 11on 2

222on 12

111on 21

aaF

FFF

FF

aFF

aFF

mm

m

m

external

externalexternalexternal

external

external

cm

CM

external

external

M

andM

mm

mmM

whereM

mm

dt

dMmm

dt

d

mm

aF

rrR

R

aF

rrrrF

aaF

2211

21

22112

2

22112

2

2211

as vector mass ofcenter the

define weif m likelook wouldThis

)()(

We therefore define the center of mass as

ii

ii

CM

mM

mM

rR 1

1

Center of Mass, Coordinates

The coordinates of the center of mass are

where M is the total mass of the system

CM CM CM

i i i i i ii i i

m x m y m zx y z

M M M

Center of Mass, position

The center of mass can be located by its position vector, rCM

ri is the position of the i th particle, defined by

CM

i ii

m

M r

r

ˆ ˆ ˆi i i ix y z r i j k

Center of Mass, Example

Both masses are on the x-axis

The center of mass is on the x-axis

The center of mass is closer to the particle with the larger mass

Center of Mass, Extended Object

Think of the extended object as a system containing a large number of particles

The particle distribution is small, so the mass can be considered a continuous mass distribution

Center of Mass, Extended Object, Coordinates The coordinates of the center of mass of the

object are

CM CM

CM

1 1

1

x x dm y y dmM M

z z dmM

Center of Mass, Extended Object, Position The position of the center of mass can also

be found by:

The center of mass of any symmetrical object lies on an axis of symmetry and on any plane of symmetry

CM

1dm

M r r

Density

dvdm

Vm

mass

volume-unit

Usually, is a constant but NOT ALWAYS!

Center of Mass, Example

An extended object can be considered a distribution of small mass elements, m=dxdydz

The center of mass is located at position rCM

For a flat sheet, we define a mass per unit area …

dadm

mass

eunit volum

Linear Density

dxdm

Lm

Center of Mass, Rod

Find the center of mass of a rod of mass M and length L

The location is on the x-axis (or

yCM = zCM = 0)

xCM = L / 2

A golf club consists of a shaft connected to a club head. The golf club can be modeled as a uniform rod of length L and mass m1 extending radially from the surface of a sphere of radius R and mass m2. Find the location of the club’s center of mass, measured from the center of the club head.

m2m1

LR

Motion of a System of Particles

Assume the total mass, M, of the system remains constant

We can describe the motion of the system in terms of the velocity and acceleration of the center of mass of the system

We can also describe the momentum of the system and Newton’s Second Law for the system

Velocity and Momentum of a System of Particles The velocity of the center of mass of a system of

particles is

The momentum can be expressed as

The total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass

CMCM

i ii

md

dt M

vr

v

CM toti i ii i

M m v v p p

Acceleration of the Center of Mass The acceleration of the center of mass can

be found by differentiating the velocity with respect to time

CMCM

1i i

i

dm

dt M v

a a

Forces In a System of Particles The acceleration can be related to a force

If we sum over all the internal forces, they cancel in pairs and the net force on the system is caused only by the external forces

CM ii

M Fa

Newton’s Second Law for a System of Particles Since the only forces are external, the net external

force equals the total mass of the system multiplied by the acceleration of the center of mass:

Fext = M aCM

The center of mass of a system of particles of combined mass M moves like an equivalent particle of mass M would move under the influence of the net external force on the system

Momentum of a System of Particles The total linear momentum of a system of

particles is conserved if no net external force is acting on the system

MvCM = ptot = constant when Fext = 0

Motion of the Center of Mass, Example A projectile is fired into the

air and suddenly explodes With no explosion, the

projectile would follow the dotted line

After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion

Rocket Propulsion

The operation of a rocket depends upon the law of conservation of linear momentum as applied to a system of particles, where the system is the rocket plus its ejected fuel

Rocket Propulsion, 2

The initial mass of the rocket plus all its fuel is M + m at time ti and velocity v

The initial momentum of the system is pi = (M + m) v

Rocket Propulsion, 3

At some time t + t, the rocket’s mass has been reduced to M and an amount of fuel, m has been ejected

The rocket’s speed has increased by v

Rocket Propulsion, 4

Because the gases are given some momentum when they are ejected out of the engine, the rocket receives a compensating momentum in the opposite direction

Therefore, the rocket is accelerated as a result of the “push” from the exhaust gases

In free space, the center of mass of the system (rocket plus expelled gases) moves uniformly, independent of the propulsion process

Rocket Propulsion, 5

The basic equation for rocket propulsion is

The increase in rocket speed is proportional to the speed of the escape gases (ve) So, the exhaust speed should be very high

The increase in rocket speed is also proportional to the natural log of the ratio Mi/Mf

So, the ratio should be as high as possible, meaning the mass of the rocket should be as small as possible and it should carry as much fuel as possible

ln if i e

f

Mv v v

M

Thrust

The thrust on the rocket is the force exerted on it by the ejected exhaust gases

Thrust =

The thrust increases as the exhaust speed increases

The thrust increases as the rate of change of mass increases The rate of change of the mass is called the burn rate

e

dv dMM v

dt dt