﻿
2|OneDimensionalMotion
67
2.5Equilibrium
Atapointofequilibrium,thetotalforceonthemassiszero,andinonedimensionthat’sallyouneed.
F
x
dU=dx
,soapointof
U
atwhichtheslopevanishesisanequilibriumpoint. Now,isitstable
orunstable? Stable e equilibrium means thatifthe mass is disturbed slightly from the critical point,
then theforce pushesitback toward theequilibrium position. . Ifbeinga a slightdistance away y from
equilibriumcausesittobepushedevenfartherawaythenitisanunstableequilibrium.
U
x
(2
:
36)
Therstofthesecurveshasapotentialmaximumat
dU=dx
=0,anunstableequilibrium. Tothe
rightofthemaximumtheforce,
dU=dx
,ispositive|towardtheright;toitsleftit’stotheleft.
Thesecondcurvehasapotentialminimumas
dU=dx
=0,astableequilibriumbecausetotheleft
theforceistotherightandontherightitistotheleft.
Thethird hasapotential minimumeventhoughthederivativeisn’tzerothere. . In n factitdoesn’t
evenexistthere. Itisastablepointhowever. . Justlookatthedirectionsoftheforcesontheleftand
theright.
Thefourthgraphhasapointwithzeroderivative,butitisnotastableequilibriumwhenyouconsider
motionsinbothdirections. Ontheleftthepushisbacktotheright,butontherightthepushisalso
totheright.
And of f course e the
x
-axis itself f graphs a potential l function n that has zero derivative everywhere.
Neutralequilibrium.
Seewhichofthesecurvescorrespondtowhichregionsinthegraphonpage61.
Thegraph of
U
onpage61showsseveralequilibria. Betweenc c andC and between aand A
andbetweengandG thereareminimumpointsof
U
. Thosearestableequilibriumpoints. Moveto
therightfromanyofthemand
dU=dx
becomespositive,making
F
x
dU=dx
negative.Thereare
unstableequilibriabetweenCanddandbetweenFandgandjusttotherightofG.Iftheenergyis
justabovethatofastableequilibrium,the motion nearthatpointwillbe anoscillationaround the
equilibrium.Thenextchapter,onharmonicmotionwilllookatthatmotionquantitatively.
bindingtheatomstogetherisafunctionoftheirseparationdistance. Fortheexampleofthemolecule
:
3
A(0
:
13nm),
howeverthereisnosimpleexactexpressionforthepotentialenergyofthemolecule asafunctionof
theatomicseparation. Thereishoweveraverygoodapproximatefunctionrepresentingthisenergy;it
wasdevelopedbyPhilipMorse.Youcanseefromthegraphofthisfunctionthatwhentheinteratomic
distanceshrinks,theforce(
dU=dr
)pushingthemapartbecomesvery large. . When n the atomsare
fartherapartthantheequilibriumpositiontheyarepulledbacktogetherunless theyaretoofarapart
andthentheattractiveforcedropstozeroastheslopeofthegraphof
U
becomes at.This attening
representsthefactthatthemoleculecanbreakapartifyougiveittoomuchenergy|dissociation.
U
(
r
)=
B
e
(
r
r
0
)
2
B
r
0
MorsePotential
(2
:
37)
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2|OneDimensionalMotion
68
The three parameters
B
,
, and
r
0
canbe tto thedatafor r various s diatomicmolecules. . Forthis
exampleofHCltheyare(datafromT.Zielinski)
r
0
=0
:
127nm
;
B
=7
:
3910
19
J=4
:
61eV
;
=1
:
8110
10
m
1
=18
:
1nm
1
(2
:
38)
Morse’s function is acleverone for r two reasons. . First, , it ts s theenergy data a quite well with h only
threeparameters. Second,whenyougetintoquantummechanicsandstudymolecules,thischoiceof
2.6ConservationofEnergy
Whenenergyisconserveditissometimeseasiertostartfromthatconservationlawthanitistostart
from
~
F
=
m~a
. DierentiateEq.(2.22)withrespecttotimeanduse
dE=dt
=0tostatethatenergy
isconserved.Usethechainruleacoupleoftimes.
dE
dt
=
d
dt
1
2
mv
2
x
+
U
(
x
)
=
1
2
m
dv
2
x
dv
x
dv
x
dt
+
dU
(
x
)
dx
dx
dt
=
mv
x
a
x
+
dU
(
x
)
dx
v
x
=0
The
v
x
factors cancelandyou have
ma
x
dU=dx
=
F
x
,Newton’s equation of motion. . When
togetthroughthediculties.
Example
As aninstanceof thisconsidertheAtwoodmachine,hangtwomassesoverapulleyandndthe
accelerationofeithermass.Thisisveryeasytosetup,andifyoulookbacktochapteroneinEqs.(1.17)
itbecomeshardjusttowritetheequationsofmotion. Thegureonpage 45remainsthesame,but
that’sall.
Ignore the mass s of the pulley y and ignore e friction. . The e linearmass density y of the rope e is
(=
dm=d‘
). Thelengthoftheropeis
L
,and
y
isthecoordinateofthetopofeachmass,measured
upfor
m
2
anddownfor
m
1
. Takethepotentialenergytobemeasuredfrom
y
=0,thenthepotential
energiesof
m
1
and of
m
2
are easy. . Fortheropeyouhavetodivide e itintothree pieces. . The e two
verticalsegmentsandthesemicircleoverthetop.
potentialenergyofmass
m
2
:
m
2
gy
lengthofropeabove
m
2
when
y
=0:
h
=(
L
R
)
=
2
midpointofropeabove
m
2
when
y
6=0:(
h
+
y
)
=
2
massofropeabove
m
2
when
y
6=0:
(
h
y
)
potentialenergyropeabove
m
2
:
(
h
y
)
g
(
h
+
y
)
=
2
for
m
1
,justchangethesignsof
y
.
y
m
2
(
h
y
)
=
2
y
=
h
m
1
Thetotalenergyis
E
=
1
2
m
1
v
2
y
+
1
2
m
2
v
2
y
+
1
2
Lv
2
y
+
m
2
gy
m
1
gy
+
g
(
h
y
)(
h
+
y
)
=
2+
g
(
h
+
y
)(
h
y
)
=
2
Thekineticenergyisclear,comingfromthetwomasses andtherope. . Thepotentialenergyforthe
twomasses
m
1
and
m
2
areclear. Fortherope,thelength
R
atthetopdoesn’tchangeitsposition
soitdoesn’tmatter.Forthetwoverticalsegmentsofropeyouhavefor
m
1
and
m
2
respectively
mgy
cm
=
(
h
y
)
.
g
.
(
h
+
y
)
=
2
or
(
h
+
y
)
.
g
.
(
h
y
)
=
2
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2|OneDimensionalMotion
69
Ifyouset
=0anddierentiatethiswithrespectto
t
yougetthesameresultasEq.(1.18).
dE
dt
=(
m
1
+
m
2
)
v
y
dv
y
dt
+(
m
2
m
1
)
gv
y
=0
;
or
a
y
=
m
1
m
2
m
1
+
m
2
g
(2
:
39)
Forthemorecomplexcaseinwhichyoudon’tneglectthemassoftherope,dothesamething.
dE
dt
=(
m
1
+
m
2
+
L
)
v
y
dv
y
dt
+(
m
2
m
1
)
gv
y
2
gyv
y
=0
or
(
m
1
+
m
2
+
L
)
d
2
y
dt
2
2
gy
=(
m
1
m
2
)
g
(2
:
40)
I’llleavethesolutionofthisequationforchapterthree,problem3.72. Ifyouwouldliketotrysetting
upthisequationstartingfrom
~
F
=
d~p=dt
(not
~
F
=
m~a
),you’llseethatitisnotsoeasytodo.Notice
thatevenif
m
1
=
m
2
=0,sothattherearenomasseshangingontheends,youstillgetacceleration.
Doesthisequationputitinthecorrectdirectionatleast?
Exercises
WhathappeninEqs.(2.10)and(2.11)if
T
1
=0? Isthisplausible? Whatif
F
2
=0?
DothedimensionsinEq.(2.14)agree? Forboth
x
and
v
x
.
Starting at t Eq. . (2.24) I started to be more careful to distinguish between n integration variables
(dummies) and parameters. . Whatwouldequations s (2.15)through(2.18)become with morecare in
writing?Asitsaysonpage60,\goback
:::
andcorrectmynotation."
InEq.(2.3)thesolution
x
(
t
)startedoas
t
3
forsmalltime. Whatisthebehaviorof
x
(
t
)nearthe
pointwhere
!t
=2
andwhereisthispointinthegraphFig.2.1?
StartfromtheresultEq.(2.31)andanalyzethevelocityforsmallfriction.Thatisforlargeterminal
speed
v
t
. Carry y the analysis atleast through terms linear r in
b
. Howdoes s this analysis dierfrom
InEq.(2.26)thezeropointofpotentialenergywasatinnity;theconstantofintegrationinnding
U
U
(
R
)=0.
thepotentialenergyofthemassasafunctionoftheanglefromthevertical? Andgraphitofcourse.
(b)Usethistodescribeallthequalitativelydierentsortsofmotionthismasscanhave.
SketchagraphoftheforcefunctionfortheMorsepotentialenergy,Eq.(2.37).
Writethetotalenergyforamassfallingfreelyundergravity.Dierentiateitwithrespectto
t
.
10 In Eq. (2.40), what is s the e acceleration for
= 0? ? More e important: : Analyze e this s answer for
plausibility.
11 With the parametersstated,does theequation (2.37)really agreewith thegraph drawn there?
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2|OneDimensionalMotion
70
Problems
2.1 ForthethreesetsofconditionsspeciedinEq.(2.1),(b)solvefor
C
and
D
andgetthecorre-
sponding
y
and
v
y
,verifying(ornot)theresultsstatedintheequationfollowingthere. (a)Butrst,
aretheresultsstatedthereplausible?
2.2_Thesolutionforapurelytime-dependentforce,Eq.(2.2),canbewrittenasasingleintegral.
x
(
t
)=
x
0
+
v
x
0
t
+
1
m
Z
t
0
dt
0
(
t
t
0
)
F
x
(
t
0
)
Dierentiatethistwicetoverifythatitworks. Thegeneralframeworkandthereasonwhythisiscorrect
can waituntil section 3.11andproblem 3.42,though derivingthisparticularresult is problem 2.54.
Especially, repeat the analysis s of the solution and compare the results here to those with the sine.
2.1?
sine. Canyouanticipatewhatatleastsomepiecesofthesolutionshouldbeevenbeforeyoudoany
integrals?
2.5 Aforcealongthe
x
-directionisgiven tobe
F
0
forthetimebetween zeroand
T
. Itthendrops
tozeroinastraightlineforthetimefrom
T
to2
T
. (a)Writethisfunction
F
x
usingthenotationof
Eqs.(0.27)and(2.4). Thengraphit. . (b)Amass
m
startsattheorigin fromrestandis subjectto
thisforce.Thepositionandvelocityequationsareproposedtobe
v
x
(
t
)=
F
0
m
t
x
(
t
)=
F
0
2
m
t
2
(0
<t<T
)
v
x
=
F
0
m
T
+
F
0
mT
T
(
t
T
1
2
(
t
T
)
2
x
=
F
0
2
m
T
2
+
F
0
mT
T
2
(
t
T
)+
1
2
T
(
t
T
)
2
1
6
(
t
T
)
3
(
T<t<
2
T
)
Determine if youshouldbelieve these. . Try y to showthatthey’re wrong, , and if youfail then maybe
they’reright. Firsttrysketching whatyouexpectthegraphofvelocityversustimeshouldbe. . Then
sketchthegraphofthisequationandcompareittoyours. Thenexaminethesetoseeiftheybehave
properly.(Youdidcheckthattheyhavethecorrectunitsdidn’tyou.)
2.6_Derivetheequationsintheprecedingproblem,unlessofcourseyouhaveconcludedthattheyare
wrong.Inthatcase,derivethecorrectequations. Also,assumingthattheforceiszeroaftertime2
T
,
whatare
x
(
t
)and
v
x
(
t
)then?
t>T
2
theapplied
forceiszero. Computethevelocityandpositionforthisthirdcaseandnishthegraphshownthere.
2.8 Fortheconstantforce
F
y
mg
andgraphicalanalysisasintheexamplefollowingit,where
U
=
kx
2
=
knowtheanswer. Useallthe toolsinthat second analysistoverify theplausibility of this solution.
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2|OneDimensionalMotion
71
Stoppingpoints,force,kineticenergy,
:::
Fromthis,ndthestoppingpointundertheinitialconditions
v
y
(0)=
v
0
.
F
x
bv
x
.Startingfrom
x
(0)=0and
v
x
(0)=
v
0
,themotionisclaimedtobe
(let
=1
,
=2
)
v
x
=
v
0
bt=m
1
=
x
=(
m=b
)
h
v
0
v
0
bt=m
=
i
Analyzethis proposed solution to seeifitisplausible. . Examinedimensions, , smalltime,largetime,
various ranges of
’s. Toanalyze e thespecial case
=1,recallthe basic limitfortheexponential
function: lim
n
!1
(1+
x=n
)
n
=
e
x
.
2.10Derivetheresultclaimedintheprecedingproblem. (Assume
6=1
;
2forreasonsthatshould
becomeclearinthemiddleofthecalculation.)
2.11 Amassisinitiallymovingatavelocity
v
x
(0)=
v
0
>
0,startingfromtheorigin. Thefrictional
forceis
F
0
e
kv
(
v>
0).Thepositionisclaimedtobe
x
=
v
0
t
(1+
t
)ln(1+
t
t
=k
where
=
e
kv
0
kF
0
=m
. Analyzetheresults(dimensions,smalltime,largetime,smallandlarge
k
)to
seeiftheclaimisplausible.
2.12_Derivetheresultclaimedintheprecedingproblem.
2.13 StartfromtheformofEq.(2.26)andsolvefor
dt
. Setuptheintegraltondtherelationbetween
r
and
t
,thencarryouttheintegralinthespecialcasethat
E
caseandyoucanspendyourtimeanalyzingtheresult.
2.14 You thought I spent enough time e on n analyzing g the results s of Eq. . (2.30) and (2.32) but t no.
(a)Findthemaximumheighttowhichtheprojectilerises.(b)Forsmallairresistanceyourresultlooks
likeitisgoingtoinnity. Finditsbehaviorforsmall
b
andseeifitisplausible.(c)Approximatelywhat
isthisresultforlargeairresistance?Thatis,whattermsdominate?
Ans:
y
max
v2
t
g
ln
1+
v
0
v
t
+
v
t
v
0
g
(
v
2
0
=
2
g
) (
bv
3
0
=
3
mg
2
)+
2.15 AfterEq.(2.23) thereisaqualitativeanalysisofthepotentialenergy
U
(
x
)=
kx
2
=
2. Nowdo
theintegralinEq.(2.23)toget
t
intermsof
x
(atrigsubstitution)andthensolvefor
x
(
t
). Compare
yoursolutiontothequalitativedescriptioninthetext.
downrst. Now
U
(
x
)=
kx
2
=
2. (b)Fortheexampleinthetext,Ididnotconsiderthecasesfor
which
E<
thisproblem,seeproblem2.56.
2.18 Aparticleofmass
m
movesalongthe
x
-axis.Itspotentialenergyisgivenby
U
(
x
)=
ax
3
bx
,
where
a>
0and
b>
0. (a)Whatistheforce
F
x
onthisparticle?(b)Draw
U
. (c)Findallequilibria
and say if they are stable orunstable (and of course, why). . (d) If the mass is atthe minimum of
potentialenergy,whatkineticenergymustithavetobecomeunbound?
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2|OneDimensionalMotion
72
2.19_Aparticleofmass
m
issubjecttotheforceasspeciedbythepotentialenergy
U
(
x
)=
U
0
j
x
j
=a
where
a
is a(constant) length and
U
0
is apositiveconstanthavingthedimensionsofenergy. . The
masshasatotalenergy
E
. Whatistheperiodofoscillation? Theamplitude? Howdoesthisperiod
varywiththeamplitudeoftheoscillation? Sketchagraphof
T
versus
E
. Suggestion: : Write e j
x
jas
twocases:
x>
0and
x<
0. Ans:
T
=8(
a=U
0
)
p
mE=
2
2.20_When you brakeanautomobile to ahalt, none of the velocity-dependentforms of friction in
as in Eq.(2.12),butyou vary theforce withyour r foot. . Asmooth h stoprequires no bigjerk atthe
end|that’s
j
x
=
da
x
=dt
. Youaregoingfrom
v
0
L
. Deviseaforcefunction,
F
x
(
t
)=
ma
x
thatdoesthiswithnobigjerkattheend.Alsonobigjerkinthemiddle. (It’s\big"if
youtrytodierentiateastepfunction.) Suggestion: : Graphs,lotsofgraphs. Figureoutwhatsortof
graphwouldaccomplishthisandthyouenconstructafunctionthatmatchesit.
2.21 Aparticlewithmass
m
issubjectedtoaforce
F
x
(
x
)=
F
0
x
3
=a
3
.Attime
t
=0theparticleis
atrestatposition
x
0
(a)Find
v
x
(
x
),drawthepotentialenergygraphintheprocess. (b)Setupthe
integralthatyouwouldhavetodotogofromtheresultin(a)togettherelationbetween
x
and
t
.If
youndthatyoucandotheintegraleasily,gobackandndyourmistake,startingwiththeunitsof
course.
2.22 Whenamass
m
isoscillatingwithinapotentialenergyfunction
kx
2
=
2,asonpage61,youcan
tryassumingthesimplestfunctionalformforthemotion,
x
(
t
)=
A
cos
!t
. Donotassumethatyou
haveanyknowledgeof
A
or
!
.Computethetotalmechanicalenergy
mv
2
=
2+
kx
2
=
2forthisassumed
x
.This
E
(
t
)willbeafunctionoftime,butitisnotsupposedtobe. Findthevalueof
!
sothat
E
is
independentof
t
2.23 If an elastic ball is bouncingon the  oor, the potential energy curve is like the third one at
Eq.(2.36). Astraightline
mgy
for
y >
0andjumpingalmost straightupat
y
=0. Assumethe
ball’s totalenergyis
E
and usethemethods of section 2.3to ndthe total timebetween bounces.
Ans:(1
=g
)
p
2
E=m
2.24 Find
x
(
t
)if
F
x
=
F
0
e
ct
,given
x
(0)=0and
v
x
(0)=0. (Isitnecessarytosaythatyoushould
analyzetheresult?) Ans:
F
0
mc
1
c
e
ct
1
t
2.25 Derivetheequation(2.11)usingtheresultofproblem2.2.
F
2
isturnedoaftertime
T
2
. Whatare
v
x
and
x
foralltimesafterthat?
2.27_FortheforceinEq.(2.19),withinitialvelocity
v
0
,ndthe velocityandpositionasafunction
oftime. Analyzetheresultsandcomparethisanalysistothatdoneintheparagraphsjustafterthat
equation.
2.28 Duringabungeejump,whatisthemaximumforcethatyouwillfeel?Youwillhavetomake(and
explain) some plausible assumptions that you must makeaboutthe bungeecord. . Itisquiteelastic,
withaforcewhenstretchedthatisproportionaltothestretch.Isthecordtiedsothatitexertsaforce
cordisstraightenedout?CheckoutWikipediaforinformationonbungeejumping.
2|OneDimensionalMotion
73
airresistanceis
bv
2
y
onthewayup,so
ma
y
mg
bv
2
y
.Aresultclaimedtobetrueforthevelocity
duringtheupwardjourneyis
v
y
(
t
)=
v
t
v
0
v
t
tan(
gt=v
t
)
v
t
+
v
0
tan(
gt=v
t
)
where
v
t
istheterminalspeed,
mg
=
bv
2
t
.(a)Analyzethisclaimedresulttodetermineitsplausibility:
smalltime,largetime,dimensions,graphs,etc.(b)Showthatif
v
t
=100m/s(typicalfora30caliber
bullet)then nomatterhowbigthe initialspeedisthemass willstop inlessthan about 16seconds.
DoesthesamestatementapplytoEq.(2.30)?
2.30_Derivetheresultclaimedintheprecedingproblem.
2.31_Inproblem2.29,if themassismovingdownthentheairresistanceis
F
y
=+
bv
2
y
,notminus.
Youhavetohandlethiscaseseparatelybecausetheformoftheequationisdierent,andthere’snota
neatanalyticalwaydobothcasesatonce. Solveforthevelocityifthemassstartsfromrestanddrops.
2.32 A boat t is slowed d by y a a frictional l force
F
x
(
v
). Its s speed decreases s according to the equation
v
(
t
)=
c
2
(
t
t
1
)
2
where
c
isaconstantand
t
1
isthetimeatwhichitstops.Findtheforce
F
x
(
v
)as
afunctionof
v
. (
t<t
1
ofcourse.) Ans: 2
mcv
1
=
2
2.33_Ifamassisbouncingupanddownelasticallyonarigid oor,theprobabilityofseeingthe
masswithintheinterval
y
to
y
+
y
isproportionaltotheamountoftimeitspendsinthatinterval.
(Ifyou lookatitonly occasionally,the probabilityof seeingitsomewhereisproportionaltothe
timeitspendswhereyoulook.) Statedmathematically,theprobability
dP
toseeitintheinterval
dy
is
Cdt
,where
dt
isthetimespentthereand
C
issomeconstant. Theprobabilitydensityis
denedas
dP=dy
,sothisis
Cdt=dy
=
C=v
upto
y
max
.(a)Computethisasafunctionof
y
for
abouncingballthathastotalenergy
E
=
mgy
max
,andthe oorisat
y
=0.Thetotalprobability
mustbeone:
R
dP
=1. Thisletsyouevaluate
C
andthengraph
dP=dy
versus
y
. Whydoes
thegraphhavetheshapethatitdoes,anddoesitmakeanysense?Dothedimensionsof
dP=dy
make sense? ? Examine e thesimulationonthe rightandcompareitto yourresult. . Andwhen n in
doubt,rememberthatphysicsisanexperimentalsciencesoyoumaywanttotrybouncingatennis
ballyourself. (b)Justasthetotalprobabilityisone
R
dP
=1
themean(oraverage)valueof
theheightis
R
ydP
=
R
y
max
0
y
(
dP=dy
)
dy:
Evaluatethis. (c)Also,whatisthemeanvalueof1?
Ans:(b)
2
3
y
max
Cf.thevideoat
2.34_Intheprecedingproblem,you have100ballsallbouncingup anddownrandomly(butall
withthe sameenergy). . (a) At any onetime, , if you takea photograph of all of them m atonce,
approximately howmany of the ballswould beat aheight from theground up to one-halfthe
maximumheight? (b)Themedianheightisthatheightsothatone-halftheballsarebelowitand
one-halfareabove.Whatisthatmedianforthesebouncingballsandhowdoesitcomparetothe
meanvaluefoundintheprecedingproblem? Thereare100dotsinthesimulationontheright,so
y
med
=
3
4
y
max
2.35 Aforceactson
m
,with
F
x
=
Av
2
=x
.Find
x
(
t
)if
x
(0)=
x
0
>
0and
v
x
(0)=
v
0
>
0.
2.36 Aparticleofmass
m
isinitiallyatrestatacoordinate
x
=
x
0
. Itisrepelledfromtheoriginby
aforce
F
x
=
A=x
3
.Solvefor
x
(
t
)and
v
x
(
t
).Ans:
x
=
x
2
0
+
At
2
=mx
2
0
1
=
2
2.37 Amass
m
moves inonedimension andsubjecttoavarietyof forcesandinitialconditions. . In
eachcase,doesthemassstopinanitetimeordistanceasthecasemaybe?
[a]
F
x
=t
for
t>
0.Attime
t
=
t
0
,thevelocityis
v
x
=
v
0
>
0;nd
v
x
(
t
).
2|OneDimensionalMotion
74
[b]
F
x
=v
x
for
v
x
>
0.Attime
t
=
t
0
,thevelocityis
v
x
=
v
0
>
0;nd
v
x
(
t
).
[c]
F
x
=x
for
x>
0. Atposition
x
=
x
0
,thevelocityis
v
x
=
v
0
>
0;nd
v
x
(
x
).
2.38 Theforceonamass
m
is
F
x
=
kv
x
x
where
k
isapositiveconstant. Attime
t
=0themass
has
x
(0)=0and
v
x
(0)=
v
0
. Find
x
(
t
).
2.39 Thecoecientoffrictionbetween amass
m
andthehorizontaltable onwhichitrests is
k
.
Theairalsoprovidesafrictionalforceproportionaltovelocity.Themassisgivenaninitialvelocity
v
0
.
Find
v
x
(
t
).Find
x
(
t
).
2.40_Airresistancedependsonthedensityoftheair,andthereislessatgreateraltitude.Assumethat
airresistanceis governedbyEq.(2.29),
bv
2
,andthat
b
isproportional toairdensity. . Areasonable
modelforairdensityis
(
y
)=
0
e
y=h
,where
0
isthedensityatsealeveland
h
youjump fromaplaneinfreefallyoureachterminalspeedfairlyfast(
mg
=
bv
2
t
). (a)Whatisthe
terminalspeedasafunctionof
y
,expressedintermsoftheterminalspeedatsealevel,
v
0
? (b)Assume
thatyou are fallingat thisvarying terminalspeedalltheway down,then whatisthe timetoreach
sealevelgiven thatyou start atheight
H
? (c)Someonejumpsfrom m 36kmupanddoesn’topen a
parachuteuntilalmost atthesurface,howmuchtimedoes ittake? ? Terminal l speed atlowaltitude
isabout200km/hrifyou arefallingprone. . (d) Howmuchtimewouldithavetakenifyou usedthe
sea-levelterminalspeedfortheentiredrop? (e)Explainwhytheapproximationshereareunreasonable
allthewaytothetopforthis36kmfall. (f)Whattimedoyougetfora10kmfall? ? Ans:(a)
v
0
e
y=
2
h
,
(c)(2
h=v
0
)
e
H=
2
h
2.41 Apotentialenergyusgivenby
U
(
x
)=(
a=x
2
) (
b=x
),where
a
,
b
,and
x
arepositive.Analyze
thepossiblemotionsqualitatively. Forwhatrangeofenergyisthemotionperiodic? Forwhatrangeis
itnot?
2.42_Forthepotentialenergyof theEarth,asdescribedin Eqs.(2.26)-(2.27),howmuchtimedoes
ittaketogofromthesurface,
r
=
R
,tothestoppingpoint. Asusual,whatisthisresultinlimiting
cases: escapeorsmall(notzero)
v
0
.
R
is
(
t
)=
At
3
where
A
isaconstantand
thatwasatthebottom attime
t
=0hasgonearound
N
timesandthen comes uptotheposition
directlytotherightoftheaxis,asmallpieceofthewheelatthatpointbreaksoand iesstraightup.
Usetheaxisofthewheelastheoriginandndthedisplacementvectorfromtheorigintotheposition
ofthispieceatthetimeitreachesitsmaximumheight.
2.44 Themoregeneralform ofNewton’sequationofmotionis
F
x
=
d
(
mv
x
)
=dt
,not=
ma
x
. An
M
istravelinginastraightline,startingwithaninitialvelocity
v
0
. Itrains
straight down,causing waterto ll the carat arate
dm=dt
=
. Find d the car’s velocity and the
is:Whyistheinitialaccelerationproportionalto
_
M
? Ans:
x
=(
Mv
0
=
)ln
1+
t=M
2.45Auniformchainishungverticallywithitsbottompointjusttouchingatable.Releasethechain,
lettingitdropontothetable. Computetheforcethechainexertsonthetableasthishappens. Show
onthetable. Thereareexactlytwothingsactingonthechain: gravityandthetable. Oneapproach
istogureoutwhatthemotionwillbeandthentakethetimederivativeofthechain’smomentum.
Anotherapproachistodrawcarefulpicturesatthetimes
t
and
t
+
t
andtoseewhathashappened
betweenthetwotimes. Remember:
~
F
6=
m~a
;it=
d~p=dt
.]
2|OneDimensionalMotion
75
2.46 Inchapternineyouwillndananalysisofarelativisticeectthatcausesdustparticlesorbiting
r
(
t
)=
p
r
2
0
2
t
where
isaparameterwhosevalue,
expressedinAstronomicalUnits(AU)andyearsisnumerically1
=
2800.
r
0
(This is fora1
m radius particle.) ) AnAstronomicalUnitisvery y closetothedistance from Sun to
Earth (1AU=15010
9
km). (a)Whataretheunitsof
? (b)Whatistheinward d componentof
velocityasafunctionof
t
? of
r
?(c)Startingfromanorbitaldistanceof1AU,howmuchtimewillit
taketohitthesun?
2.47 Amass
m
isdroppedwithinitialvelocityzero,andtheforceson
m
arefromgravityandsome
formofairresistance. Measurethecoordinate
y
positivedownwardfromthestartingposition. Ifyou
gettheresult
y
(
t
)=
v
2
t
g
ln
sinh
gt
v
t

(a)Whatisthispositionfunctionforlargetime?Isthebehaviorcorrect? (b)Whatisthisfunctionfor
smalltime,keepingenoughtermsthatyoumaybeabletosurmisewhattheassumedformfortheair
resistanceforceis?Aretheseresultsplausible?
y
(
t
)=
v
2
t
g
ln
cosh
gt
v
t

Inthesameway,alsoanalyzethissolutionforlargeandforsmalltimes,andtrytodeterminewhatthe
airresistanceforceis. Determineifthissolutionismoreorlessplausiblethantheothersolution.
2.49 Forthepotentialenergyfunction
U
(
x
)=
U
0
x
2
=a
2
cos(
x=a
)
,analyzethepossiblemotions.
(
U
0
>
0.) Forexample,if
issmall(0
:
1)whatmotioncanoccur? Orifit’sbig(10)? ? Without
lotsofgraphsketching,thisishopeless.Withcarefulsketching,it’seasy.
2.50 AmoleculemorecomplicatedthanHClhasapotentialenergyfunctionmorecomplicatedthan
theMorsepotentialofEq.(2.37).Itis
U
(
r
)=
B
e
(
r
r
0
)
2
Be
2
2
(
r
r
1
)
2
where
=0
:
05and
r
1
=3
r
0
.Describethebehaviorandpossibleequilibriaofthismolecule. Forthese
equilibria,andusingthenumbersinEq.(2.38),howmuchenergywouldittaketokickthemoleculeout
ofoneequilibriumstateandintotheother?Howmuchtokickitback?Lookuptheterm\metastable".
Ifyoundthisproblemhard,thatmeansthatyoudidn’tsketchacarefulgraph.
2.51 Asimplependulumisapointmass
m
ontheendofastringoflength
. Thepotentialenergy
ofthe massis
mgh
measured from whateverheightyou choose. . Write e the totalmechanicalenergy
and compute
dE=dt
,settingitequal to zeroto derivethe dierentialequation ofmotion. . Express
everything in terms of theangularcoordinate. . And d howdoesyour r resultchange e if youchange the
originfromwhichyoumeasureheight?
2.52_Usethesamefrictional force asin problem 2.12,butnowthemass starts from restand drops
becauseofgravity.Assume
F
0
<mg
.
2.53 Fireanobjectstraightupatinitialspeed
v
0
,andassumethattheairresistanceisproportional
tothesquareofthespeed. UsethemethodofEq.(2.28)tond(a)howhightheobjectwillgoand
(b)itsspeedwhenitreturnstotheoriginalheight. Expresstheresultsusing
v
t
,theterminalspeed,