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Seventh Edition

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICSCHAPTER

.

E. Russell Johnston, Jr. Kinetics of Particles:

Energy and MomentumJ. Walt Oler

Texas Tech University

Methods

2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: DynamicsSeventh

Edition

ContentsIntroduction

Work of a Force

Princi le of Work & Ener

Sample Problem 13.6

Sample Problem 13.7

Sam le Problem 13.9

Applications of the Principle of Work& Energy

Power and Efficienc

Principle of Impulse and MomentumImpulsive Motion

Sample Problem 13.1

Sample Problem 13.2

.

Sample Problem 13.11


.

Sample Problem 13.4

Sample Problem 13.5

Direct Central Impact

Oblique Central Impact

o en a nergyConservative Forces

Conservation of Energy

Pro ems Invo v ng Energy an MomentumSample Problem 13.14


2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 2

Motion Under a Conservative Central

Force

Sample Problems 13.16

Sample Problem !3.17

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Introduction

Previously, problems dealing with the motion of particles were

solved through the fundamental equation of motion, .amF

= .

Method of work and energy: directly relates force, mass,

.

Method of impulse and momentum: directly relates force,mass, velocity, and time.


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Work of a Force

Differential vector is theparticle displacement.rd

or o e orce s

dsF

rdFdU

=

=

cos

dzFdyFdxF zyx ++=

, . .,

sign but not direction.

force.len th Dimensions of work are Units are( ) ( )( ) J1.356lb1ftm1N1J1 ==joule


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Work of a Force

Work of a force during a finite displacement,

2A =

22

1

21

ss

A

r

=

==

2

11

A

s

t

s

1A

Work is re resented b the area under the

curve of Ftplotted against s.


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Work of a Force

Work of a constant force in rectilinear motion,

( ) xFU = cos21

Work of the force of gravity,

dzFdyFdxFdU zyx ++=

dyWU

dyW

y

=

=

212

( ) yWyyW

y

== 12

1

Work of the weightis equal to product ofweight Wand vertical displacement y.


or o t e we g t s pos ve w en y < ,

i.e., when the weight moves down.

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Work of a Force

Work of a gravitational force (assume particleM

shown),

2 dr

MmGFdrdU ==

12221

2

r

MmG

r

MmGdr

r

MmGU

r

r

==


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Work of a Force

Forces which do notdo work (ds = 0 or cos = 0):

reaction at frictionless surface when body in contact

reaction at frictionless pin supporting rotating body,

reaction at a roller moving along its track, and

moves along surface,

weight of a body when its center of gravity moves

horizontally.


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Particle Kinetic Energy: Principle of Work & Energy

dt

dvmmaF tt ==

Consider a particle of mass m acted upon by force F

dvmvdsF

ds

dv

mvdt

ds

ds

dv

m

t ===

Integrating fromA1 toA2 ,

mvmvdvvmdsF

vs

t == 2121222122

energykineticmvTTTU

vs

===2

21

1221

11

e wor o t e orce s equa to t e c ange nkinetic energy of the particle.

Units of work and kinetic energy are the same:


JmNms

mkg

s

mkg

2

2

2

1 ==

=

== mvT

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Applications of the Principle of Work and Energy

Wish to determine velocity of pendulum bob

atA2. Consider work & kinetic energy.

Force acts normal to path and does no

work.

P

TUT 2211 =+

vg

WWl

2

10 22

=

=+

Velocity found without determining

expression for acceleration and integrating.

All quantities are scalars and can be added

directly.


Forces which do no work are eliminated from

the problem.

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Applications of the Principle of Work and Energy

Principle of work and energy cannot be

applied to directly determine the acceleration

of the endulum bob.

Calculating the tension in the cord requires

supplementing the method of work and energy

with an application of Newtons second law.

As the bob passes throughA2 ,

vWWP

amF nn

22=

=

Wl

gl

g

WWP 3

2=+=

glv 22=


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Power and Efficiency

rate at which work is done.

rdFdU

Power

==

=

vF

tt

=

mens ons o power are wor me or orce ve oc y.

Units for power are

W746lbft550h1ormN1J1wattW1 ====s

ss

efficiency=

outputpower

input workkoutput wor=


inputpower

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Sample Problem 13.1

SOLUTION:

Evaluate the change in kinetic energy.

Determine the distance required for thework to equal the kinetic energy change.

driven down a 5o incline at a speed of

60 mi/h when the brakes are applied

of 1500 lb.

Determine the distance traveled by the


au omo e as comes o a s op.

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Sample Problem 13.1

SOLUTION:

Evaluate the change in kinetic energy.

lbft481000882.324000

sft88s3600mi

t

h

m

60

2121

1

=====

mvT

v

00 22 == Tv

equal the kinetic energy change.

( ) ( )( )xxU 5sinlb4000lb150021 +=

2211 =+ TUT( )xlb1151=


= x

ft418=x

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Sample Problem 13.2

SOLUTION:

Apply the principle of work and

energy separately to blocks andB.

When the two relations are combined,

Two blocks are joined by an inextensible

.

Solve for the velocity.

ca e as s own. e sys em s re ease

from rest, determine the velocity of block

A after it has moved 2 m. Assume that the

and the plane is k= 0.25 and that the

pulley is weightless and frictionless.


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Sample Problem 13.2SOLUTION:

Apply the principle of work and energy separately

to blocksA andB.

( )( )

2

N490N196225.0N1962sm81.9kg200

WNF

W

AkAkA

A

======

( ) ( )

2

2

1

2211

m2m20

:

vmFF

TUT

AAC =+

=+

( ) ( )( ) ( )21 kg200m2N490m2 vFC =

2 ==

( ) ( ) 212211

m2m20

:

.

vmWF

TUT

BBc =+

=+


( ) ( )( ) ( ) 2

2

1 kg300m2N2940m2 vFc =+

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Apply the principle of work and energy

A spring is used to stop a 60 kg package

point at which the spring is fullycompressed and the velocity is zero.

The onl unknown in the relation is thewhich is sliding on a horizontal surface.

The spring has a constant k = 20 kN/m

and is held by cables so that it is initially

friction coefficient.

Apply the principle of work and energy

compresse mm. e pac age as a

velocity of 2.5 m/s in the position shown

and the maximum deflection of the spring

.

only unknown in the relation is the

velocity at the final position.

.

Determine (a) the coefficient of kinetic

friction between the package and surface


and ( ) the velocity of the package as it

passes again through the position shown.

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Apply principle of work and energy between initial

position and the point at which spring is fully compressed.

( )( ) 0J5.187sm5.2kg60 22

2

121

2

11 ==== TmvT

( ) kf xWU 21 =

( )( )( ) ( ) kk J377m640.0sm81.9kg60 2 ==

( )( ) N2400m120.0mkN200min ===kxP( ) ( )( )

( ) ( )

N3200m160.0mkN20

1

maxmin21

21

0max

==

+=

==+=

xPPU

xxkP

e

..2

( ) ( ) ( ) J112J377212121 =+= kef UUU

2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 20( ) 0J112J377-J5.187

:2211

=

=+

k

TUT

20.0=k

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Sample Problem 13.3

Apply the principle of work and energy for the rebound

of the package.

2121 ===22

( ) ( ) ( ) J112J377323232

=

+=+= kef UUU

( ) 2313322

kg60J5.360

:

v

TUT

=+

=+

sm103.13=v


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Sample Problem 13.4

SOLUTION:

Apply principle of work and energy to

eterm ne ve oc ty at po nt 2.

Apply Newtons second law to find

normal force by the track at point 2.A 2000 lb car starts from rest at point 1

and moves without friction down the

track shown.

Apply principle of work and energy to

determine velocity at point 3.

Determine:

a) the force exerted by the track on

pp y ewton s secon aw to n

minimum radius of curvature at point 3

such that a positive normal force is

,

b) the minimum safe value of the

radius of curvature at point 3.

.


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Apply principle of work and energy to determine

velocity at point 2.

( )ft40

2

1

0

21

2

2

2

22

1

21

+====

WU

vg

W

mvTT

( )

sft8.50ft2.32ft402ft402

2

1ft400:

22

222211

===

=+=+

vsv

vg

WWTUT

Apply Newtons second law to find normal force by

the track at point 2.

:nn amF =+ ( )gWvW

amNW nft40222 ===+


WN 5

2

= lb10000=N

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Sample Problem 13.4 Apply principle of work and energy to determine

velocity at point 3.

1 2W

( ) ( )( ) sft1.40sft2.32ft252ft252

2

323

33311

===

==

vgv

g

Apply Newtons second law to find minimum radius of

curvature at point 3 such that a positive normal force is

exerted by the track.

:nn amF =+

amW n=

( )33

23 ft252

g

g

Wv

g

W == ft503=


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Sample Problem 13.5

SOLUTION:

Force exerted by the motor

the dumbwaiter velocity.Power delivered by motor is

e ual to Fv , v = 8 ft/s.

The dumbwaiterD and its load have a In the first case, bodies are in uniformcombined weight of 600 lb, while the

counterweight C weighs 800 lb.

Determine the ower delivered b the

.

motor cable from conditions for static

equilibrium.

electric motorMwhen the dumbwaiter(a) is moving up at a constant speed of

8 ft/s and (b) has an instantaneous

n t e secon case, ot o es areaccelerating. Apply Newtons

second law to each body to


velocity of 8 ft/s and an acceleration of

2.5 ft/s2, both directed upwards.

force.

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Sample Problem 13.5

In the second case, both bodies are accelerating. Apply

Newtons second law to each body to determine the required

motor cable force.

===2

212

sft25.1sft5.2 DCD aaa

ree- o y :

:CCy amF =+ ( ) lb5.38425.12.32

8002800 == TT

Free-body D:

:DDy amF =+ ( )5.22.32

600600=+ TF

lb1.2626.466005.384 ==+ FF( )( ) slbft2097sft8lb1.262 === DFvPower


( ) hp81.3

slbft550

hp1slbft2097 =

=Power

V M h i f E i D iSE

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Potential Energy

2121 yWyWU =

Work of the force of gravity ,W

Work is independent of path followed; dependsonly on the initial and final values of Wy.

=

= yg

potential energy of the body with respect

toforce of gravity.

2121 gg VVU =

measured is arbitrary.


JmN === WyVg

V t M h i f E i D iSE

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Potential Energy

Previous expression for potential energy of a body

with respect to gravity is only valid when the

wei ht of the bod can be assumed constant.

For a space vehicle, the variation of the force of

ravit with distance from the center of the earth

should be considered.

Work of a ravitational force

1221

r

GMm

r

GMmU =

Potential energy Vg when the variation in the

force of gravity can not be neglected,

2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 29r

WR

r

GMmVg ==

V t M h i f E i D iSE

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Vector Mechanics for Engineers: Dynamicseventh

dition

Potential Energy

Work of the force exerted by a spring depends

only on the initial and final deflections of the

s rin

2

2212

121

21 kxkxU =

The potential energy of the body with respect

to the elastic force,

21=

( ) ( )2121

2

ee

e

VVU =

Note that the preceding expression for Ve isvalid only if the deflection of the spring is

measured from its undeformed position.


V t M h i f E i D iSe

Ed

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dition

Conservative Forces Concept of potential energy can be applied if the

work of the force is independent of the path

followed by its point of application.

( ) ( )22211121 ,,,, zyxVzyxVU =Such forces are described as conservative forces.

,

0= rdF

Elementary work corresponding to displacementbetween two neighboring points,

( ) ( )

zxdV

dzzdyydxxVzyxVdU

,,

,,,,

=

+++=

dzz

Vdy

y

Vdx

x

VdzFdyFdxF zyx

+

+

=++


Vz

V

y

V

x

VF grad=

+

+

=

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dition

Conservation of Energy

Work of a conservative force,

2121 VVU =

,

1221 TTU = Follows that

constant

2211

=+=

+=+

VTE

VTVT

When a article moves under the action ofWVT == 0 conservative forces, the total mechanical

energy is constant.WVT =+ 11

W2 1

WVT

gg

mv

=+

====

22

222

22

r c on orces are no conserva ve. o a

mechanical energy of a system involving

friction decreases.


Mechanical energy is dissipated by friction

into thermal energy. Total energy is constant.


Ed

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dition

Motion Under a Conservative Central Force When a particle moves under a conservative central

force, both the principle of conservation of angular

momentum

and the principle of conservation of energy

sinsin000

rmvmvr =

r

GMm

mvr

GMm

mv

VTVT

=

+=+

2

2

12

02

1

00

may be applied.

, .

At minimum and maximum r, = 90o. Given the

launch conditions the e uations ma be solved for


rmin, rmax, vmin, and vmax.


Ed

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Sample Problem 13.6

SOLUTION:

Apply the principle of conservation of

energy between positions 1 and 2.

The elastic and gravitational potential

the given information. The initial kinetic

energy is zero.

co ar s es w ou r c on

along a vertical rod as shown. The

spring attached to the collar has an

Solve for the kinetic energy and velocity

at 2.

.

constant of 3 lb/in.

If the collar is released from rest at


pos t on 1, eterm ne ts ve oc ty a ter

it has moved 6 in. to position 2.

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Vector Mechanics for Engineers: Dynamicsventh

ition


Apply the principle of conservation of energy between

positions 1 and 2.

Position 1: ( )( )

lbft20lbin.24

lbin.24in.4in.8in.lb3

1

2

2

12

12

1

=+=+=

===

VVV

kxV

ge

e

1=

Position 2: ( )( )

( )( )

2

212

221

lbin.120in.6lb20

lbin.54in.4in.01in.lb3

WyV

kxVe

===

===

22

22

222

12

2

311.0201

lbft5.5lbin.6612054

vvmvT

VVV ge

===

===+=

.

Conservation of Energy:

2211 +=+ VTVT


.. 2 =+ v

= sft91.42v

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tion


Setting the force exerted by the loop to zero, solve for the

minimum velocity atD.

:nn maF =+

( )( ) 222 sft4.64sft32.2ft2 ===

==

rgv

rvmmgmaW

D

Dn

Apply the principle of conservation of energy between

pointsA andD.

( ) 18ftlb360 221211 ==+=+= xxkxVVV e

01=T

( )( ) lbft2ft4lb5.002 ==+=+= ge WyVVV

( ) lbft5.0sft4.64sft2.32lb5.0

21 22

22

21

2 === DmvT


25.0180 22211

+=+

+=+

x in.47.4ft3727.0 ==x


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tion


For motion under a conservative central

energy and conservation of angular

momentum may be applied simultaneously.

A satellite is launched in a direction

parallel to the surface of the earth

minimum and maximum altitude to

determine the maximum altitude.

with a velocity of 36900 km/h from

an altitude of 500 km.

Determine a the maximum altitude

Apply the principles to the orbit insertion

point and the point of minimum altitude to

determine maximum allowable orbit

reached by the satellite, and (b) themaximum allowable error in the

direction of launching if the satellite

nser on ang e error.


is to come no closer than 200 km to

the surface of the earth


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Vector Mechanics for Engineers: Dynamicsenth

tion

Sample Problem 13.9 Apply the principles of conservation of energy and

conservation of angular momentum to the points of minimum

and maximum altitude to determine the maximum altitude.

Conservation of energy:

12121

02021 r

GMmmv

r

GMmmvVTVT AAAA =+=+

onserva on o angu ar momen um:

1

0011100

r

rvvmvrmvr ==

Combinin

2001

0

1

0

02

1

202

021 2111

vr

GM

r

r

r

r

r

GM

r

rv =+

=

23122622

60

0

sm10398m1037.6sm81.9

sm1025.10hkm36900

===

==

RGM

v


km60400m104.60 6

1

==r

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ion

Sample Problem 13.9 Apply the principles to the orbit insertion point and the point

of minimum altitude to determine maximum allowable orbit

insertion angle error.

Conservation of energy:

min

2max21

0

202100 r

GMmmvr

GMmmvVTVT AA =+=+

Conservation of angular momentum:

min

000maxmaxmin000 sinsin

r

rvvmvrmvr ==

Combining and solving for sin0,

=

=

5.1190

9801.0sin 0

= 5.11errorallowable



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on

Principle of Impulse and Momentum

From Newtons second law,

( ) == vmvmd

F

linear momentumt

( )2

vmddtFt

=

2t

12

1

vmvmdtF

t

=

Dimensions of the impulse of

2211

21 forcetheofimpulse

1

vmvm

FdtF

t

=+

==

Imp

Impa orce are

force*time.

Units for the impulse of a

The final momentum of the particle can be

obtained by adding vectorially its initial

force aresmkgssmkgsN 2 ==


momen um an e mpu se o e orce ur ng

the time interval.

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g ynth

on


SOLUTION:

Apply the principle of impulse and

momentum. The impulse is equal to the

product of the constant forces and thetime interval.

An automobile weighing 4000 lb is

driven down a 5o

incline at a speed of60 mi/h when the brakes are applied,

causing a constant total braking force of

1500 lb.

Determine the time required for theautomobile to come to a stop.


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g ynth

on


SOLUTION:

A l the rinci le of im ulse and

momentum in terms of horizontal and

vertical component equations.

velocity of 80 ft/s. After the ball is hit

by the bat, it has a velocity of 120 ft/s

in the direction shown. If the bat and

ball are in contact for 0.015 s,determine the average impulsive force

exerted on the ball during the impact.


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th

n


SOLUTION:

Apply the principle of impulse and

momen um o e pac age-car sys em

to determine the final velocity. Apply the same principle to the package

A 10 kg package drops from a chute

into a 24 kg cart with a velocity of 3

a one o e erm ne e mpu se exer e

on it from the change in its momentum.

m/s. Knowing that the cart is initially at

rest and can roll freely, determine (a)

the final velocity of the cart, (b) the

impulse exerted by the cart on the

package, and (c) the fraction of the

initial energy lost in the impact.



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th

n


Apply the principle of impulse and momentum to the package-cart

system to determine the final velocity.

y

x

2211 vmmvm cpp

+=+ Imp

=

( )( ) ( ) 2kg25kg1030cosm/s3kg10 v

cpp

+=

m/s742.02=v


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th

n


Apply the same principle to the package alone to determine the impulse

exerted on it from the change in its momentum.

x

y

2211 vmvm pp

=+ Imp

x components:

( )( ) ( ) 2

21

kg1030cosm/s3kg10

cos

vtF

vmtvm

x

pxp

=+

=+

sN56.18 =tFx

com onents: 030sin =+ tFvm

( )( ) 030sinm/s3kg10 =+ tFy sN15 =tFy


s.ss.21 =+== ttmp


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th

n


To determine the fraction of energy lost,

( )( ) J45sm3kg102

212

121

1 === vmT p

( ) ( )( ) J63.9sm742.0kg25kg10 2212

221

1 =+=+= vmmT cp

786.0J45

.

1

21 =

=

T



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h

Impact

Impact: Collision between two bodies which

occurs during a small time interval and during

which the bodies exert large forces on each other.

Line of Impact: Common normal to the surfacesin contact during impact.

Central Impact: Impact for which the mass

centers of the two bodies lie on the line of impact;

otherwise, it is an eccentric impact..


Direct Impact: Impact for which the velocities of

the two bodies are directed along the line of

mpac .

Oblique Impact: Impact for which one or both of



impact.


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h


Bodies moving in the same straight line,

vA > vB .

period of deformation, at the end of which,

they are in contact and moving at acommon velocit .

Aperiod of restitution follows during

which the bodies either regain their

deformed.

Wish to determine the final velocities of the

.

two body system is preserved,

BBBBBBAA vmvmvmvm +=+


A second relation between the final

velocities is required.


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h


= nrestitutiooftcoefficiene AAA

10

==

e

uv

vu

Pdt A

A

Period of restitution: AAA vmRdtum =

B uv

e

=Bvu

Combining the relations leads to the desired ( )BAAB vvevv = .

Perfectly plastic impact, e = 0: vvv AB == ( )vmmvmvm BABBAA +=+


Perfectly elastic impact, e = 1:

Total energy and total momentum conserved.BAAB vvvv =


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h


Final velocities are

unknown in magnitude

.

equations are required.

No tangential impulse component;

tangential component of momentum

( ) ( ) ( ) ( )tBtBtAtA vvvv ==

.

Normal component of total

momentum of the two particles is

( ) ( ) ( ) ( )nBBnAAnBBnAA vmvmvmvm +=+

conserve .

Normal components of relative

velocities before and after impact( ) ( ) ( ) ( )

nBnAnAnB vvevv =


are related by the coefficient of

restitution.


Edition

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Block constrained to move along horizontal

sur ace.

Impulses from internal forces FF and

exerted by horizontal surface and directed

along the vertical to the surface.

ext

Final velocity of ball unknown in direction and

magnitude and unknown final block velocity

. .



Edition

Obli C t l I t

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Tangential momentum of ball is

conserved.

( ) ( )tBtB vv =

=and ball is conserved.

xx

Normal component of relative ( ) ( ) ( ) ( )nBnAnAnB vvevv =

by coefficient of restitution.


Note: Validity of last expression does not follow from previous relation for

the coefficient of restitution. A similar but separate derivation is required.


Edition

P bl I l i E d M t

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Problems Involving Energy and Momentum

Three methods for the analysis of kinetics problems:

- Direct application of Newtons second law

- e o o wor an energy

- Method of impulse and momentum

Select the method best suited for the problem or part of a problem

under consideration.



Edition

Sample Problem 13 14

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SOLUTION:

Resolve ball velocity into components

normal and tan ential to wall.

Impulse exerted by the wall is normalto the wall. Component of ball

momen um angen a o wa s

conserved.

Assume that the wall has infinite mass,vertical wall. Immediately before the

ball strikes the wall, its velocity has a

ma nitude v and forms an le of 30o

so that wall velocity before and after

impact is zero. Apply coefficient of

restitution relation to find change in

with the horizontal. Knowing thate = 0.90, determine the magnitude and

direction of the velocity of the ball as

norma re a ve ve oc y e ween wa

and ball, i.e., the normal ball velocity.


it rebounds from the wall.


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SOLUTION: Resolve ball velocity into components parallel and

perpendicular to wall.

Component of ball momentum tangential to wall is conserved.

vvvvvv tn 500.030sin866.030cos ====

vvv tt .==

Apply coefficient of restitution relation with zero wall

t

ve ocity.

( )

( ) vvv

vev

n

nn

779.0866.09.0

00

==

=n

+= 500.0779.0 vvv tn


=

= 7.32500.0

.tan926.0 1vv


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SOLUTION:

Resolve the ball velocities into components

normal and tangential to the contact plane.

Tangential component of momentum for

The magnitude and direction of the

velocities of two identical

.

Total normal component of the momentum

of the two ball system is conserved.frictionless balls before they strike

each other are as shown. Assuming

e = 0.9, determine the magnitude

The normal relative velocities of the

balls are related by the coefficient of

an rect on o t e ve oc ty o eac

ball after the impact.

.

Solve the last two equations simultaneously

for the normal velocities of the balls after


the impact.

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Edition


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The normal relative velocities of the balls are related by the

coefficient of restitution.

( ) ( ) ( ) ( )= nBnAnBnA vvevv

( )[ ] 4.410.200.2690.0 ==

Solve the last two equations simultaneously for the normal

velocities of the balls after the impact.

( ) sft7.17= nAv ( ) sft7.23= nBv

+=

0.15

0.157.17

1

ntAv

n

+=

=

=

6.347.23

.

7.17

ans.

ntB

A

v

v


=

= 6.557.23

.tansft9.41Bv


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SOLUTION:

Determine orientation of impact line of

The momentum component of ballAtangential to the contact plane is

conserve .

The total horizontal momentum of the

two ball system is conserved.BallB is hanging from an inextensible

cord. An identical ballA is released

from rest when it is just touching the

The relative velocities along the line of

action before and after the impact are

cord and acquires a velocity v0before

striking ballB. Assuming perfectly

elastic impact (e = 1) and no friction,

.

Solve the last two expressions for the

velocity of ballA along the line of action


eterm ne t e ve oc ty o eac a

immediately after impact.

an t e ve oc ty o a w c s

horizontal.

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The relative velocities along the line of action beforeand after the impact are related by the coefficient of

restitution.

( )( ) 0

0

866.05.0030cos30sin

vvvvvv

vvevv

AB

nAB

nBnAnAnB

==

=

Solve the last two expressions for the velocity of ball

A along the line of action and the velocity of ballB

.

( ) 00 693.0520.0 vvvv BnA ==

=

==

=

10

00

1.465.0

52.0tan721.0

..

vv

vvv

A

ntA


=

==

0693.0

..

vvB


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SOLUTION:

Apply the principle of conservation of

block at the instant of impact.

Since the impact is perfectly plastic, the

oc an pan move toget er at t e same

velocity after impact. Determine that

velocity from the requirement that the

A 30 kg block is dropped from a height

of 2 m onto the the 10 kg pan of a

conserved.

Apply the principle of conservation of.

perfectly plastic, determine themaximum deflection of the pan. The

constant of the s rin is k= 20 kN/m.

energy to determine the maximum

deflection of the spring.


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Edition


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p

Apply the principle of conservation of energy todetermine the maximum deflection of the spring.

( ) ( )( )21231

3 J4427.41030vmmT BA =+=+=

( )( )233212

321

3

J241.01091.410200 kx

VVV eg

==+=

+=

Initial spring deflection due to ( )( )242

14

4 0

kxhWWVVV

T

BAeg ++=+=

=

( )( )m1091.4

1020

81.910 333

=

==k

Wx B ( ) ( ) 24321

34

4234

10201091.4392

1020392

xx

xxx

+=

+=

( ) ( )m230.0

10201091.43920241.0442 243

213

4

4433

=

+=+

+=+

x

xx


m1091.4m230.0 334== xxh m225.0=h

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ES 12 Chapter 13