Chapter 7- Linear Momentum

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been posted on the web.

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Tutorial session will start at 5pm tomorrow

A very interesting seminar.

I encourage you to attend.

Chapter 7

• Momentum and Its Relation to Force

• Conservation of Momentum

• Collisions and Impulse• Collisions and Impulse

• Conservation of Energy and Momentum in Collisions

• Elastic Collisions in One Dimension

• Inelastic Collisions

• Collisions in Two or Three Dimensions

Recalling Recalling LastLast LectureLectureRecalling Recalling LastLast LectureLecture

When the work done by a force does NOT depend on the path taken, this FORCE is

said to be CONSERVATIVE.

When the work done by a force DOES depend on the path taken, this FORCE is

said to be NONCONSERVATIVE.

Nonconservative forces do NOT have a potential energy associated to them.

Conservative and Conservative and NonconservativeNonconservative ForcesForces

Thus, potential energy can be defined only for conservative forces.

Eq. 6-14 tells us that the work done by nonconservative forces is equal to the

total change in kinetic and potential energies.

(6-14)

“ If only conservative forces are acting, the TOTAL MECHANICAL ENERGY of a

system neither increases nor decreases in any process. It stays constant �� it

is CONSERVED”

POWER:

Power is the rate at which work is done, or the rate at which energy is

Mechanical Energy and its ConservationMechanical Energy and its Conservation

Power is the rate at which work is done, or the rate at which energy is

transformed.

In the SI system the unit of power is: 1 hp = 746 W

In general, if an object moves with average velocity , the power can be written as

(6-18)

(6-19)

TodayTodayTodayToday

Momentum is a vector symbolized by the symbol p, and is defined as

The momentum of an object tells how hard (or easy) is to change its state of

motion.

Example: It is easier to stop a car when it is moving at 10 km/h than when it is

Momentum and Its Relation to ForceMomentum and Its Relation to Force

(6-20)

Example: It is easier to stop a car when it is moving at 10 km/h than when it is

moving at 100 Km/h.

� But, note that it will also depend on the mass of the car: a heavy truck moving at

10 Km/h is more difficult to be brought to rest than a small Mercedes SMART

(which much lighter than a truck) moving at same speed.

Recalling that acceleration is the rate with which velocity changes, we can then write:

But,

Defining: and , we can write


The right side of the equation 6.21 is the net force acting on the object. Then:

(6-21)

(6-22)

Eq. 6-22 is another way of expressing Newton’s second law. However, it is a more

general definition because it introduces the situation where the mass may change.

Example: A rocket burning fuel when lifting off or maneuvering in space:

� As it burns fuel, the rocket becomes lighter


(6-22)

� So, its total mass changes and so does its momentum (even if you

manage to keep its velocity constant)

Collisions are common event in everyday life.

During a collision, objects are deformed due to the large forces involved.

These forces are generally very strong, much stronger than other interactions

between the colliding objects and their surrounding environment, and act for a very

short period of time ∆t.

We can use eq. 6-22 and define the impulse on an object as:

Collision and ImpulseCollision and Impulse

In general, the forces involved in the process is not constant

during ∆t, but we can approximate the resulting force by the

average force acting during this interval of time:

(6-23)

(6-24)

Let’s consider a collision between two billiard balls as shown in the figure.

Conservation of MomentumConservation of Momentum

Assume that the two balls form a system isolated

from the external world:

� in other words � there is NO net force acting

on the billiard balls other than the interaction

between them.

The momentum of each ball before and after the

collision is:

Object A: ,

Object B: ,

Recalling previous slide: The momentum of each ball before and after the collision is:


Object A: ,

Object B: ,

Now, ball A exerts a force on ball B.

According to the general form of Newton’s

second law:

But,

On the other hand, according to Newton’s third law, B exerts a force

We have then:


Combining the two equations, we find:

(6-25)

Equation 6-25 tells that the total momentum of the system (the sum of the

momentum of the two balls) before the collision is equal to the total momentum of

the system after the collision IF the net force acting on the system is zero ��

isolated system. This is known as Conservation of Total Momentum.

The above equation can be extended to include any number of objects such that the

only forces are the interaction between the objects in the system.


Momentum conservation works for a rocket as long as we consider the rocket and its

fuel to be one system, and account for the mass loss of the rocket.

Here you can consider a frame at rest relative to the rocket before it lifts off

� Its initial momentum is then zero. As it takes off, the fuel burns expelling gas in one

direction

� this gives momentum to the gas

� in order to obey momentum conservation, the rocket has to move in opposite

direction such that the total momentum (rocket + gas) remains zero.

Conservation of Energy and Momentum in CollisionsConservation of Energy and Momentum in Collisions

In general, we can identify two different types of collisions:

1. Elastic collision

2. Inelastic collision

In elastic collisions the total kinetic energy of a system is conserved.

� There is no energy dissipate in form of heat or other form of energy.

An example is the collision between the two billiard

balls discussed in the previous slides:balls discussed in the previous slides:

In inelastic collision, there is NO conservation of

kinetic energy.

� Some of the total initial kinetic energy is

transformed into some other form of energy.

(6-26)


Example of inelastic collision:

With inelastic collisions, some of the initial kinetic

energy is lost to thermal or potential energy. It may

also be gained during explosions, as there is the

addition of chemical or nuclear energy.

A completely inelastic collision is one where the

objects stick together afterwards, so there is only one

final velocity.final velocity.


Note:

Regardless whether we have inelastic or elastic collisions, the total momentum is

always conserved if the system is isolated.

For instance, the collision between two trains as depicted below is inelastic. Part of

the total initial kinetic energy might have be transformed into thermal or other form of

energy.

However, the total momentum of the closed (isolated) system (two trains) should be However, the total momentum of the closed (isolated) system (two trains) should be

conserved.

(The textbook

has a good

example of this

problem).


Note:

Also…

The total energy (the sum of all energies) in a closed (isolated system)

is ALWAYS conserved.is ALWAYS conserved.


Elastic collision in one dimension:

Here we have two objects colliding elastically. We know the masses and the initial

speeds.

Since both momentum and kinetic energy are

conserved, we can write two equations. This

allows us to solve for the two unknown final

speeds.

From conservation of momentum:

We can rearrange this equation and write:



Since it is an elastic collision, the total kinetic energy is conserved:

using that , then

We then have two equations from total momentum and kinetic energy conservation:

Replacing the last equation into the first one:



Eq. 6-27 tell us that in ONE DIMENSION elastic head-on

collision, the relative velocity between the objects have the same magnitude but in

opposite direction before and after the collision.

(6-27)


Collision in two or more dimensions

That is just a generalization of what

we have discussed in the previous

slides.

We have to make use of the concept of vector

components.

In the figure, considering the system isolated, the total momentum has to be

conserved in both x and y directions.

In this particular case, the reference system is taken such B is initially (before the

collision) at rest and A moves in the x direction before the collision.


Collision in two or more dimensions

We then have:

(i)

(ii)

It follows then, using the above expressions in (i) and (ii), that:

⇒ This gives a system of two equations

and three variables. if you can measure any

of these variables, the other two can be

calculated from system of equations.

Linear MomentumLinear Momentum

Problem 7-1 (textbook) A constant friction force of 25 N acts on a 65-kg skier for

20 s. What is the skier’s change in velocity?


Problem 7-1 :

From Newton’s second law,

For a constant mass object,

t∆ = ∆p Frr

m∆ = ∆p vr r

Equate the two expressions for

If the skier moves to the right, then the speed will decrease, because the friction force

is to the left.

The skier loses 7.7 m/s of speed.

∆pr

t

t mm

∆∆ = ∆ → ∆ =

FF v v

rr r r

( ) ( )25 N 20 s7 .7 m s

65 kg

F tv

m

∆∆ = − = − = −


Problem 7-4 (textbook) A child in a boat throws a 6.40-kg package out horizontally

with a speed of 10.0 m/s Fig. 7–31. Calculate the velocity of the boat immediately

after, assuming it was initially at rest. The mass of the child is 26.0 kg, and that of the

boat is 45.0 kg. Ignore water resistance.


Problem 7-4 :

The throwing of the package is a momentum-conserving

action, if the water resistance is ignored.

Let “A” represent the boat and child together, and let “B” represent the package.

Choose the direction that the package is thrown as the positive direction. Apply

conservation of momentum, with the initial velocity of both objects being 0.

The boat and child move in the opposite direction as the thrown package.

( )

( )( )

( )

i n i t i a l f i n a l

6 .4 0 k g 1 0 .0 m s0 .9 0 1 m s

2 6 .0 k g 4 5 .0 k g

A B A A B B

B B

A

A

p p m m v m v m v

m vv

m

′ ′= → + = + →

′′ = − = − = −

+


Problem 7-34 (textbook) An internal explosion breaks an object, initially at rest, into

two pieces, one of which has 1.5 times the mass of the other. If 7500 J were released

in the explosion, how much kinetic energy did each piece acquire?

.


Problem 7-34

Use conservation of momentum in one dimension, since the particles will separate

and travel in opposite directions. Call the direction of the heavier particle’s motion the

positive direction. Let A represent the heavier particle, and B represent the lighter

particle. We have

A B1.5m m=

A B0v v= =

B B 2 0 m v

p p m v m v v v′

′ ′ ′ ′= → = + → = − = −

The negative sign indicates direction.

Since there was no mechanical energy before the explosion, the kinetic energy of

the particles after the explosion must equal the energy added.

Thus:

2

initial final A A B B A B3

A

0 p p m v m v v vm

′ ′ ′ ′= → = + → = − = −

( ) ( ) ( )( )

22 2 2 25 51 1 1 2 1 1

added 2 2 2 3 2 3 2 3

3 3

added added5 5

1.5

7500 J 4500 J 7500 J 4500 J 3000 J

A B A A B B B B B B B B B

B A B

E KE KE m v m v m v m v m v KE

KE E KE E KE

′ ′ ′ ′ ′ ′ ′ ′= + = + = + = =

′ ′ ′= = = = − = − =

3 33.0 10 J 4.5 10 J

A BKE KE′ ′= × = ×

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