chapter 6 work energy and power 1 marks questions 1.if two

Chapter6

WorkEnergyandPower

1MarksQuestions

1.Iftwobodiessticktogetheraftercollisionwillthecollisionbeelasticorinelastic?

Ans:Inelasticcollision.

2.Whenanairbubblerisesinwater,whathappenstoitspotentialenergy?

Ans:Potentialenergyofanairbubbledecreasesbecauseworkisdonebyupthrustonthe

bubble.

3.Aspringiskeptcompressedbypressingitsendstogetherlightly.Itisthenplacedina

strongacid,andreleased.Whathappenstoitsstoredpotentialenergy?

Ans:Thelossinpotentialenergyappearsaskineticenergyofthemoleculesofthecid.

4.Definetriplepointofwater?

Ans.Triplepointofwaterrepresentsthevaluesofpressureandtemperatureatwhichwater

co-existsinequilibriuminallthethreestatesofmatter.

5.StateDulongandpetitlaw?

Ans.Acc.tothislaw,thespecificheatofallthesolidsisconstantatroomtemperatureandis

equalto3R.

6.Whytheclockpendulumsaremadeofinvar,amaterialoflowvalueofcoefficientof

linearexpansion?

Ans.TheclockpendulumsaremadeofInverbecauseithaslowvalueofα(co-efficientof

linearexpansion)i.e.forasmallchangeintemperature,thelengthofpendulumwillnot

changemuch.

7.Whyismercuryusedinmakingthermometers?

Ans.Mercuryisusedinmakingthermometersbecauseithaswideandusefultemperature

rangeandhasauniformrateofexpansion.

8.Howwouldathermometerbedifferentifglassexpandedmorewithincreasing

temperaturethanmercury?

Ans.Ifglassexpandedmorewithincreasingtemperaturethanmercury,thescaleofthe

thermometerwouldbeupsidedown.

9.Showthevariationofspecificheatatconstantpressurewithtemperature?

Ans.

10.Twothermometersareconstructedinthesamewayexceptthatonehasaspherical

bulbandtheotheranelongatedcylindricalbulb.Whichonewillresponsequicklyto

temperaturechange?

Ans.Thethermometerwithcylindricalbulbwillrespondquicklytotemperaturechanges

becausethesurfaceareaofcylindricalbulbisgreaterthantheofsphericalbulb.

11.Abodyconstrainedtomovealongthez-axisofacoordinatesystemissubjecttoa

constantforceFgivenby

Where areunitvectorsalongthe axisofthesystemrespectively.

Whatistheworkdonebythisforceinmovingthebodyadistanceof4malongthe

axis?

Ans.Forceexertedonthebody,

Displacement,s= m

Workdone,W=

Hence,12Jofworkisdonebytheforceonthebody.

12.Amoleculeinagascontainerhitsahorizontalwallwithspeed andangle

30°withthenormal,andreboundswiththesamespeed.Ismomentumconservedinthe

collision?Isthecollisionelasticorinelastic?

Ans.Yes;Collisioniselastic

Themomentumofthegasmoleculeremainsconservedwhetherthecollisioniselasticor

inelastic.

Thegasmoleculemoveswithavelocityof200m/sandstrikesthestationarywallofthe

container,reboundingwiththesamespeed.

Itshowsthatthereboundvelocityofthewallremainszero.Hence,thetotalkineticenergyof

themoleculeremainsconservedduringthecollision.Thegivencollisionisanexampleofan

elasticcollision.

13.ThebobAofapendulumreleasedfrom30°totheverticalhitsanotherbobBofthe

samemassatrestonatableasshowninFig.6.15.HowhighdoesthebobAriseafter

thecollision?Neglectthesizeofthebobsandassumethecollisiontobeelastic.

Ans.BobAwillnotriseatall

Inanelasticcollisionbetweentwoequalmassesinwhichoneisstationary,whiletheotheris

movingwithsomevelocity,thestationarymassacquiresthesamevelocity,whilethemoving

massimmediatelycomestorestaftercollision.Inthiscase,acompletetransferof

momentumtakesplacefromthemovingmasstothestationarymass.

Hence,bobAofmassm,aftercollidingwithbobBofequalmass,willcometorest,whilebob

BwillmovewiththevelocityofbobAattheinstantofcollision.

14.Atrolleyofmass300kgcarryingasandbagof25kgismovinguniformlywitha

speedof27km/honafrictionlesstrack.Afterawhile,sandstartsleakingoutofahole

onthefloorofthetrolleyattherateof0.05 .Whatisthespeedofthetrolley

aftertheentiresandbagisempty?

Ans.Thesandbagisplacedonatrolleythatismovingwithauniformspeedof27km/h.The

externalforcesactingonthesystemofthesandbagandthetrolleyiszero.Whenthesand

startsleakingfromthebag,therewillbenochangeinthevelocityofthetrolley.Thisis

becausetheleakingactiondoesnotproduceanyexternalforceonthesystem.Thisisin

accordancewithNewton'sfirstlawofmotion.Hence,thespeedofthetrolleywillremain27

km/h.

15.WhichofthefollowingpotentialenergycurvesinFig.6.18cannotpossiblydescribe

theelasticcollisionoftwobilliardballs?Hereristhedistancebetweencentresofthe

balls.

Ans.(i),(ii),(iii),(iv),and(vi)

Thepotentialenergyofasystemoftwomassesisinverselyproportionaltotheseparation

betweenthem.Inthegivencase,thepotentialenergyofthesystemofthetwoballswill

decreaseastheycomeclosertoeachother.Itwillbecomezero(i.e.,V(r)=0)whenthetwo

ballstoucheachother,i.e.,atr=2R,whereRistheradiusofeachbilliardball.Thepotential

energycurvesgiveninfigures(i),(ii),(iii),(iv),and(vi)donotsatisfythesetwoconditions.

Hence,theydonotdescribetheelasticcollisionsbetweenthem.

2MarksQuestions

1.AbodyismovingalongZ–axisofaco–ordinatesystemissubjectedtoaconstant

forceFisgivenby

Where areunitvectoralongthex,yandz–axisofthesystemrespectivelywhat

istheworkdonebythisforceinmovingthebodyadistanceof4malongtheZ–axis?

Ans:

W=12J

2.Aballisdroppedfromtheheighth1andifrebouncestoaheighth2.Findthevalueof

coefficientofrestitution?

Ans:Velocityofapproach

(Balldropsformheighth1)

Velocityofseparation

(Ballreboundstoheighth2)

Coefficientofrestitution

3.Stateandproveworkenergytheoremanalytically?

Ans:Itstatesthatworkdonebyforceactingonabodyisequaltothechangeproducedinits

kineticenergy.

If forceisappliedtomoveanobjectthroughadistancedS

Then

HenceW=Kf–KiWhereKfandKiarefinalandinitialkineticenergy.

4.Anobjectofmass0.4kgmovingwithavelocityof4m/scollideswithanotherobjectof

mass0.6kgmovinginsamedirectionwithavelocityof2m/s.Ifthecollisionisperfectly

inelastic,whatisthelossofK.E.duetoimpact?

Ans:m1=0.4kg,u1=4m/s,m2=0.6kgu2=2m/s.

TotalK.E.beforecollision

Sincecollisionisperfectlyinelastic

TotalK.E.aftercollision

LossinK.E. =Ki–Kf=4.4–3.92=0.48J

5.Whydoesthedensityofsolid|liquiddecreaseswithriseintemperature?

Ans.LetP=Densityofsolid|liquidattemperatureT

P1=Densityofsolid|liquidatTemperatureT+∆T

SinceDensity=

So,P= →(1)P1= (2)

V1=VolumeofsolidattemperatureT+∆T

V=VolumeofsolidattemperatureT

Sinceonincreasingthetemperature,solids|liquidsexpandthatistheirvolumesincreases,

sobyequation

i)&2)Densityisinverselyproportionaltovolumes,soifvolumeincreasesonincreasingthe

temperature,Densitywilldecrease.

6.TwobodiesatdifferenttemperaturesT1,andT2arebroughtinthermalcontactdo

notnecessarilysettledowntothemeantemperatureofT1andT2?

Ans.TwobodiesatdifftemperaturesT1andT2wheninthermalcontactdonotsettlealways

attheirmeantemperaturebecausethethermalcapacitiesoftwobodiesmaynotbealways

equal.

7.Theresistanceofcertainplatinumresistancethermometerisfoundtobe2.56Ωat00c

and3.56Ωat1000c.Whenthethermometerisimmersedinagivenliquid,itsresistance

isobservedto5.06Ω.Determinethetemperatureofliquid?

Ans.Ro=Resistanceat00c=2.56Ω

Rt=ResistanceattemperatureT=1000c=3.56Ω100

Rt=Resistanceatunknowntemperaturet;

Rt=5.06Ω

Since,

t=

t=2500c

8.Aballisdroppedonafloorfromaheightof2cm.Afterthecollision,itrisesuptoa

heightof1.5m.Assumingthat40%ofmechanicalenergylostgoestothermalenergy

intotheball.Calculatetheriseintemperatureoftheballinthecollision.Specificheat

capacityoftheballis800J/k.Takeg=10m/s2

Ans.Initialheight=h1=2m

Finalheight=h2=1.5m

Sincepotentialenergy=mechanicalenergyforabodyatrestasK.E=0

Mechanicalenergylost=

=

=

=5J

Now(mechanicalenergylost)×40%=heatgainedbyball

∆T=2.5×10-30C

9.Athermometerhaswrongcalibration.Itreadsthemeltingpointoficeas–100C.It

reads600Cinplaceof500C.Whatisthetemperatureofboilingpointofwateronthe

scale?

Ans.Lowerfixedpointonthewrongscale=-100C.

Let‘n’=no.divisionsbetweenupperandlowerfixedpointsonthisscale.IfQ=readingon

thisscale,then

Now,C=IncorrectReading=600C

Q=CorrectReading=500C

So,

n=140

Now,

On,theCelsiusscale,Boilingpointofwateris1000C

So,

Q=1300C

10.Writetheadvantagesanddisadvantagesofplatinumresistancethermometer?

Ans.AdvantagesofPlatinumResistancethermometer:-

1)Highaccuracyofmeasurement

2)Measurementsoftemperaturecanbemadeoverawiderangeoftemperaturei.e.from–

2600Cto12000C.

→DisadvantagesofPlatinumResistancethermometer:-

1)HighCost

2)Requiresadditionalequipmentsuchasbridgecircuit,Powersupplyetc.

11.Ifthevolumeofblockofmetalchangesby0.12%whenitisheatedthrough200C.

Whatistheco-efficientoflinearexpansionofthemetal?

Ans.07.Theco-efficientofcubicalexpansionyofthemetalisgivenby:-

Here,

∴Co-efficientoflinearexpansionofthemetalis:-

12.Thedensityofasolidat00Cand5000Cisintheratio1.027:1.Findtheco-efficientof

linearexpansionofthesolid?

Ans.Densityat00C=SO

Densityat5000C=S500

Now,SO=S500

Where,Y=Co-efficientofvolumeexpansion

∆T=Changeintemperature

∆T=Changeintemperature

∆T=FinalTemperature–Initialtemperature

∆T=500-00C

∆T=5000C

Or

Now,Co-efficientoflinearexpansion(α)isrelatedtoco-efficientofvolumeexpansion(Y)as

:-

13.IfoneMoleofamonatomicgasismixedwith3molesofadiatomicgas.Whatisthe

molecularspecificheatofthemixtureatconstantvolume?

Ans.09.For,amonatomicgas,Specificheatatconsentvolume=CV1= ;R=Universal

GasConstant

No.ofmolesofmonatomicgas=n1=1mole

No.ofmolesofdiatomicgas=n2=3moles.

For,diatomicgas,specificheatatconstantvolume

Applying,conservationofenergy.

LetCV=Specificheatofthemixture;

R=UniversalGasconstant

14.Calculatethedifferencebetweentwoprincipalspecificheatsof1gofheliumgasat

N.T.P.GivenMolecularweightofHelium=4andJ=4.186J/calandUniversalGas

constant,R=8.314J/mole/K?

Ans.10.MolecularweightofHelium=M=4

UniversalGasConstant,R=8.31J|mole|K

CP=specificheatatconstantPressure

CV=specificheatatconstantVolume

Now, for1moleofgas.

WhereR=UniversalGasConstant=8.31J|mole|K

J=4.186J|cal

M=MolecularweightofHelium=4

15.Whydoesheatflowfromabodyathighertemperaturetoabodyatlower

temperature?

Ans.11.Whenabodyathighertemperatureisincontactwithabodyatlowertemperature,

moleculewithmorekineticenergythatareincontactwithlessenergeticmoleculesgiveup

someoftheirkineticenergytothelessenergeticones.

16.Aoneliterflaskcontainssomemercury.ITisfoundthatatdifferenttemperatures,

thenvolumeofairinsidetheflaskremainsthesame.Whatisthevolumeofmercuryin

theflask?Giventheco-efficientoflinearexpansionofglass=9×10-6/0Candco-

efficientofvolumeexpansionofmercury=1.8×10-4/0C

Ans.Itisgiventhatvolumeofairintheflaskremainsthesameatdifferenttemperature.This

ispossibleonlywhentheexpansionofglassisexactlyequaltotheexpansionofmercury,

Co-efficientofcubicalexpansionofglassis:-

Co-efficientofcubicalexpansionofmercuryis:→

Volumeofflask,V=1liter=1000cm3.

LetVmCm3bethevolumeofmercuryintheflask.

Expansionofflask=ExpansionofMercury

∴VolumeofMercury,

17.Thepotentialenergyfunctionforaparticleexecutinglinearsimpleharmonic

motionisgivenbyV(x)= /2,wherekistheforceconstantoftheoscillator.Fork=

0.5N ,thegraphofV(x)versusxisshowninFig.6.12.Showthataparticleoftotal

energy1Jmovingunderthispotentialmust'turnback'whenitreachesx=±2m.

Ans:Totalenergyoftheparticle,E=1J

Forceconstant,k=0.5N

Kineticenergyoftheparticle,K=

Accordingtotheconservationlaw:

E=V+K

Atthemomentof‘turnback',velocity(andhenceK)becomeszero.

Hence,theparticleturnsbackwhenitreachesx= m.

18.Stateifeachofthefollowingstatementsistrueorfalse.GivereasonsforyourAns..

(a)Inanelasticcollisionoftwobodies,themomentumandenergyofeachbodyis

conserved.

(b)Totalenergyofasystemisalwaysconserved,nomatterwhatinternalandexternal

forcesonthebodyarepresent.

(c)Workdoneinthemotionofabodyoveraclosedloopiszeroforeveryforcein

nature.

(d)Inaninelasticcollision,thefinalkineticenergyisalwayslessthantheinitialkinetic

energyofthesystem.

Ans.(a)False

(b)False

(c)False

(d)True

Explanation:(a)Inanelasticcollision,thetotalenergyandmomentumofboththebodies,

andnotofeachindividualbody,isconserved.

(b)Althoughinternalforcesarebalanced,theycausenoworktobedoneonabody.Itisthe

externalforcesthathavetheabilitytodowork.Hence,externalforcesareabletochangethe

energyofasystem.

(c)Theworkdoneinthemotionofabodyoveraclosedloopiszeroforaconservationforce

only.

(d)Inaninelasticcollision,thefinalkineticenergyisalwayslessthantheinitialkinetic

energyofthesystem.Thisisbecauseinsuchcollisions,thereisalwaysalossofenergyinthe

formofheat,sound,etc.

19.Abodyisinitiallyatrest.Itundergoesone-dimensionalmotionwithconstant

acceleration.Thepowerdeliveredtoitattimetisproportionalto

(i) (ii)t(iii) (iv)

Ans.(ii)t

Massofthebody=m

Accelerationofthebody=a

UsingNewton'ssecondlawofmotion,theforceexperiencedbythebodyisgivenbythe

equation:

F=ma

Bothmandaareconstants.Hence,forceFwillalsobeaconstant.

F=ma=Constant…(i)

Forvelocityv,accelerationisgivenas,

Where, isanotherconstant

Powerisgivenbytherelation:

P=F.v

Usingequations(i)and(iii),wehave:

Hence,powerisdirectlyproportionaltotime.

20.Abodyismovingunidirectionallyundertheinfluenceofasourceofconstant

power.Itsdisplacementintimetisproportionalto

(i) (ii)t(iii) (iv)

Ans.(iii)

Powerisgivenbytherelation:

P=Fv

Integratingbothsides:

Fordisplacement ofthebody,wehave:

Where Newconstant

Onintegratingbothsides,weget:

21.Apumponthegroundfloorofabuildingcanpumpupwatertofillatankofvolume

30 in15min.Ifthetankis40mabovetheground,andtheefficiencyofthepumpis

30%,howmuchelectricpowerisconsumedbythepump?

Ans.Volumeofthetank,V=

Timeofoperation,t=15min=15×60=900s

Heightofthetank,h=40m

Efficiencyofthepump, =30%

Densityofwater,

Massofwater,m=

Outputpowercanbeobtainedas:

Forinputpower ,efficiency isgivenbytherelation:

22.Abodyofmass0.5kgtravelsinastraightlinewithvelocity where

.Whatistheworkdonebythenetforceduringitsdisplacementfromx=

0tox=2m?

Ans.Massofthebody,m=0.5kg

Velocityofthebodyisgovernedbytheequation,

Initialvelocity,u(atx=0)=0

Finalvelocityv(atx=2m)

Workdone,W=Changeinkineticenergy

23.Afamilyuses8kWofpower.(a)Directsolarenergyisincidentonthehorizontal

surfaceatanaveragerateof200Wpersquaremeter.If20%ofthisenergycanbe

convertedtousefulelectricalenergy,howlargeanareaisneededtosupply8kW?(b)

Comparethisareatothatoftheroofofatypicalhouse.

Ans.

(a)

(a)Powerusedbythefamily,P=8kW= W

Solarenergyreceivedpersquaremetre=200W

Efficiencyofconversionfromsolartoelectricityenergy=20%

Arearequiredtogeneratethedesiredelectricity=A

Aspertheinformationgiveninthequestion,wehave:

(b)Theareaofasolarplaterequiredtogenerate8kWofelectricityisalmostequivalentto

theareaoftheroofofabuildinghavingdimensions14m×14m.

24.Aboltofmass0.3kgfallsfromtheceilingofanelevatormovingdownwithan

uniformspeedof7 .Ithitstheflooroftheelevator(lengthoftheelevator=3m)

anddoesnotrebound.Whatistheheatproducedbytheimpact?WouldyourAns.be

differentiftheelevatorwerestationary?

Ans.

Massofthebolt,m=0.3kg

Speedoftheelevator=7m/s

Height,h=3m

Sincetherelativevelocityoftheboltwithrespecttotheliftiszero,atthetimeofimpact,

potentialenergygetsconvertedintoheatenergy.

Heatproduced=Lossofpotentialenergy

=mgh= =8.82J

Theheatproducedwillremainthesameeveniftheliftisstationary.Thisisbecauseofthe

factthattherelativevelocityoftheboltwithrespecttotheliftwillremainzero.

25.Considerthedecayofafreeneutronatrest:n→p+e-

Showthatthetwo-bodydecayofthistypemustnecessarilygiveanelectronoffixed

energyand,therefore,cannotaccountfortheobservedcontinuousenergydistribution

inthe decayofaneutronoranucleus(Fig.6.19).

[Note:Thesimpleresultofthisexercisewasoneamongtheseveralarguments

advancedbyW.Paulitopredicttheexistenceofathirdparticleinthedecayproducts

of decay.Thisparticleisknownasneutrino.Wenowknowthatitisaparticleof

intrinsicspin½(likee-,porn),butisneutral,andeithermasslessorhavingan

extremelysmallmass(comparedtothemassofelectron)andwhichinteractsvery

weaklywithmatter.Thecorrectdecayprocessofneutronis:n→p+e-+v]

Ans.

Thedecayprocessoffreeneutronatrestisgivenas:

FromEinstein'smass-energyrelation,wehavetheenergyofelectronas

Where,

Δm=Massdefect=Massofneutron-(Massofproton+Massofelectron)

c=Speedoflight

Δmandcareconstants.Hence,thegiventwo-bodydecayisunabletoexplainthecontinuous

energydistributioninthe decayofaneutronoranucleus.Thepresenceofneutrinovon

theLHSofthedecaycorrectlyexplainsthecontinuousenergydistribution.

3MarksQuestions

1.Provethatinanelasticcollisioninonedimensiontherelativevelocityofapproach

beforeimpactisequaltotherelativevelocityofseparationafterimpact?

Ans.

Accordingtolawofconservationoflinearmomentum

K.E.alsoremainsconserved.

Dividing(2)by(1)

I.e.Relativevelocityofapproach=Relativevelocityofseparation.

2.CalculateCpforair,giventhatCv=0.162calg-1k-1anddensityairatN.T.Pis0.001293

g|cm3?

Ans.Specificheatatconstantpressure=Cp=?

Specificheatatconstantvolume=Cv=0.162Calg-1k-1

Now,Cp–Cv=

OrCP–Cv=

Cp–Cv=

=

=

=6.8×10-4+2

Cp–Cv=0.068

Cp=0.162+0.068

Cp=0.23Calg-1k-1

3.Developarelationbetweentheco-efficientoflinearexpansion,co-efficient

superficialexpansionandcoefficientofcubicalexpansionofasolid?

Ans.Since,co-efficientoflinearexpansion=α=

∆L=changeinlength

L=length

∆T=changeintemperature

Similarly,co-efficientofsuperficialexpansion=β=

∆S=changeinarea

S=originalarea

∆T=changeintemperature

Co-efficientofcubicalexpansion,=Y=

∆V=changeinvolume

V=originalvolume

∆T=changeintemperature.

Now,∆L=αL∆T

L+∆L=L+αL∆T

L+∆L=L(1+α∆T)→(1)

SimilarlyV+∆V=V(1+Y∆T)→(2)

AndS+∆S=S(1+β∆T)→(3)

Also,(V+∆V)=(L+∆L)3

V+∆V=

V+∆V=L3

Sinceα2,α3arenegligible,so,

V+γV∆T=V(1+3α∆T)[asL3=V]

So,V+γV∆T=V+V3α∆T

γV∆T=3α∆T

Y=3α

Similarly,β=2α[usingL2=S(Area)]

So,

4.Calculatetheamountofheatrequiredtoconvert1.00kgoficeat–100cintosteamat

1000catnormalpressure.Specificheatofice=2100J|kg|k.Latentheatoffusionofice=

3.36x105J|kg,specificheatofwater=4200J|kg|k.Latentheatofvaporizationofwater=

2.25x106J|kg?

Ans.(1)Here,heatisrequiredtoraisethetemperatureoficefrom–100cto00c.

So,changeintemperature=∆T=T2-T1=0-(-10)=100c

So,∆Q1=cm∆T

C=specificheatofice

M=Massofice

∆T=100c

∆Q1=2100×1×10=21000J

(2)Heatrequiredtomelttheiceto00cwater:-

∆Q2=mL

L=Latentheatoffusionofice=3.36×105J/kg

m=Massofice

∆Q2=1×3.36×105J/kg

∆Q2=3.36×105J

∆Q2=336000J

(3)Heatrequiredtoraisethetemperatureofwaterfrom00cto1000c:-

∆T=T2-T1=100-0=1000c

∆Q3=cm∆Tc=specificheatofwater

=4200×1×100

=420,000J

(4)Heatrequiredtoconvert1000cwatertosteamat1000c

∆Q4=mLL=Latentheatofvapourisation=2.25×106J/kg

∆Q4=1×2.25×106J|kg

∆Q4=2250000J

∴TotalHeatrequired=∆Q1+∆Q2+∆Q3+∆Q4

∆Qtotal=21000+336000+420000+2250000

∆Qtotal=3027000J

∆Qtotal=3.027x106J

5.Thesignofworkdonebyaforceonabodyisimportanttounderstand.State

carefullyifthefollowingquantitiesarepositiveornegative:

(a)Workdonebyamaninliftingabucketoutofawellbymeansofaropetiedtothe

bucket.

(b)Workdonebygravitationalforceintheabovecase,

(c)Workdonebyfrictiononabodyslidingdownaninclinedplane,

(d)Workdonebyanappliedforceonabodymovingonaroughhorizontalplanewith

uniformvelocity,

(e)Workdonebytheresistiveforceofaironavibratingpenduluminbringingitto

rest.

Ans.(a)Positive

Inthegivencase,forceanddisplacementareinthesamedirection.Hence,thesignofwork

doneispositive.Inthiscase,theworkisdoneonthebucket.

(b)Negative

Inthegivencase,thedirectionofforce(verticallydownward)anddisplacement(vertically

upward)areoppositetoeachother.Hence,thesignofworkdoneisnegative.

(c)Negative

Sincethedirectionoffrictionalforceisoppositetothedirectionofmotion,theworkdoneby

frictionalforceisnegativeinthiscase.

(d)Positive

Herethebodyismovingonaroughhorizontalplane.Frictionalforceopposesthemotionof

thebody.Therefore,inordertomaintainauniformvelocity,auniformforcemustbe

appliedtothebody.Sincetheappliedforceactsinthedirectionofmotionofthebody,the

workdoneispositive.

(e)Negative

Theresistiveforceofairactsinthedirectionoppositetothedirectionofmotionofthe

pendulum.Hence,theworkdoneisnegativeinthiscase.

6.Underlinethecorrectalternative:

(a)Whenaconservativeforcedoespositiveworkonabody,thepotentialenergyofthe

bodyincreases/decreases/remainsunaltered.

(b)Workdonebyabodyagainstfrictionalwaysresultsinalossofitskinetic/potential

energy.

(c)Therateofchangeoftotalmomentumofamany-particlesystemisproportionalto

theexternalforce/sumoftheinternalforcesonthesystem.

(d)Inaninelasticcollisionoftwobodies,thequantitieswhichdonotchangeafterthe

collisionarethetotalkineticenergy/totallinearmomentum/totalenergyofthesystem

oftwobodies.

Ans.(a)Decreases

(b)Kineticenergy

(c)Externalforce

(d)Totallinearmomentum

Explanation:

(a)Aconservativeforcedoesapositiveworkonabodywhenitdisplacesthebodyinthe

directionofforce.Asaresult,thebodyadvancestowardthecentreofforce.Itdecreasesthe

separationbetweenthetwo,therebydecreasingthepotentialenergyofthebody.

(b)Theworkdoneagainstthedirectionoffrictionreducesthevelocityofabody.Hence,

thereisalossofkineticenergyofthebody.

(c)Internalforces,irrespectiveoftheirdirection,cannotproduceanychangeinthetotal

momentumofabody.Hence,thetotalmomentumofamany-particlesystemisproportional

totheexternalforcesactingonthesystem.

(d)Thetotallinearmomentumalwaysremainsconservedwhetheritisanelasticcollisionor

aninelasticcollision.

7.Answercarefully,withreasons:

(a)Inanelasticcollisionoftwobilliardballs,isthetotalkineticenergyconserved

duringtheshorttimeofcollisionoftheballs(i.e.whentheyareincontact)?

(b)Isthetotallinearmomentumconservedduringtheshorttimeofanelasticcollision

oftwoballs?

(c)Whataretheanswersto(a)and(b)foraninelasticcollision?

(d)Ifthepotentialenergyoftwobilliardballsdependsonlyontheseparationdistance

betweentheircentres,isthecollisionelasticorinelastic?(Note,wearetalkinghereof

potentialenergycorrespondingtotheforceduringcollision,notgravitationalpotential

energy).

Ans.(a)No

Inanelasticcollision,thetotalinitialkineticenergyoftheballswillbeequaltothetotalfinal

kineticenergyoftheballs.Thiskineticenergyisnotconservedattheinstantthetwoballs

areincontactwitheachother.Infact,atthetimeofcollision,thekineticenergyoftheballs

willgetconvertedintopotentialenergy.

(b)Yes

Inanelasticcollision,thetotallinearmomentumofthesystemalwaysremainsconserved.

(c)No;Yes

Inaninelasticcollision,thereisalwaysalossofkineticenergy,i.e.,thetotalkineticenergyof

thebilliardballsbeforecollisionwillalwaysbegreaterthanthataftercollision.

Thetotallinearmomentumofthesystemofbilliardsballswillremainconservedeveninthe

caseofaninelasticcollision.

(d)Elastic

Inthegivencase,theforcesinvolvedareconservation.Thisisbecausetheydependonthe

separationbetweenthecentresofthebilliardballs.Hence,thecollisioniselastic.

8.Apersontryingtoloseweight(dieter)liftsa10kgmass,onethousandtimes,toa

heightof0.5meachtime.Assumethatthepotentialenergylosteachtimeshelowers

themassisdissipated.(a)Howmuchworkdoesshedoagainstthegravitationalforce?

(b)Fatsupplies Jofenergyperkilogramwhichisconvertedtomechanical

energywitha20%efficiencyrate.Howmuchfatwillthedieteruseup?

Ans.(a)Massoftheweight,m=10kg

Heighttowhichthepersonliftstheweight,h=0.5m

Numberoftimestheweightislifted,n=1000

∴Workdoneagainstgravitationalforce:

(b)Energyequivalentof1kgoffat= J

Efficiencyrate=20%

Mechanicalenergysuppliedbytheperson'sbody:

Equivalentmassoffatlostbythedieter:

4MarksQuestions

1.Abodyofmass2kginitiallyatrestmovesundertheactionofanappliedhorizontal

forceof7Nonatablewithcoefficientofkineticfriction=0.1.

Computethefollowing:

(a)Workdonebytheappliedforcein10s,

(b)Workdonebyfrictionin10s,

(c)Workdonebythenetforceonthebodyin10s,

(d)Changeinkineticenergyofthebodyin10s,

andinterpretyourresults.

Ans.Massofthebody,m=2kg

Appliedforce,F=7N

Coefficientofkineticfriction, =0.1

Initialvelocity,u=0

Time,t=10s

TheaccelerationproducedinthebodybytheappliedforceisgivenbyNewton'ssecondlaw

ofmotionas:

Frictionalforceisgivenas:

=

heaccelerationproducedbythefrictionalforce:

Totalaccelerationofthebody:

Thedistancetravelledbythebodyisgivenbytheequationofmotion:

(a)Workdonebytheappliedforce, F×s=7×126=882J

(b)Workdonebythefrictionalforce,

(c)Netforce= =5.04N

Workdonebythenetforce, =5.04×126=635J

(d)Fromthefirstequationofmotion,finalvelocitycanbecalculatedas:

v=u+at

=0+2.52×10=25.2m/s

Changeinkineticenergy

2.Atrolleyofmass200kgmoveswithauniformspeedof36km/honafrictionless

track.Achildofmass20kgrunsonthetrolleyfromoneendtotheother(10maway)

withaspeedof4 relativetothetrolleyinadirectionoppositetotheitsmotion,

andjumpsoutofthetrolley.Whatisthefinalspeedofthetrolley?Howmuchhasthe

trolleymovedfromthetimethechildbeginstorun?

Ans.Massofthetrolley,M=200kg

Speedofthetrolley,v=36km/h=10m/s

Massoftheboy,m=20kg

Initialmomentumofthesystemoftheboyandthetrolley

=(M+m)v

=(200+20)×10

=2200kgm/s

Letv'bethefinalvelocityofthetrolleywithrespecttotheground.

Finalvelocityoftheboywithrespecttotheground

Finalmomentum

Asperthelawofconservationofmomentum:

Initialmomentum=Finalmomentum

2200=

Lengthofthetrolley,l=10m

Speedoftheboy,v''=4m/s

Timetakenbytheboytorun,

∴Distancemovedbythetrolley

3.A1kgblocksituatedonaroughinclineisconnectedtoaspringofspringconstant

100 Nm-1asshowninFig.6.17.Theblockisreleasedfromrestwiththespringin

theunstretchedposition.Theblockmoves10cmdowntheinclinebeforecomingto

rest.Findthecoefficientoffrictionbetweentheblockandtheincline.Assumethatthe

springhasanegligiblemassandthepulleyisfrictionless.

Ans.Massoftheblock,m=1kg

Springconstant,k=

Displacementintheblock,x=10cm=0.1m

Thegivensituationcanbeshownasinthefollowingfigure.

Atequilibrium:

Normalreaction,R=mgcos37°

Frictionalforce,f R=mgsin37°

Where, isthecoefficientoffriction

Netforceactingontheblock=

=

Atequilibrium,theworkdonebytheblockisequaltothepotentialenergyofthespring,i.e.,

4.GiveninFig.6.11areexamplesofsomepotentialenergyfunctionsinonedimension.

Thetotalenergyoftheparticleisindicatedbyacrossontheordinateaxis.Ineachcase,

specifytheregions,ifany,inwhichtheparticlecannotbefoundforthegivenenergy.

Also,indicatetheminimumtotalenergytheparticlemusthaveineachcase.Thinkof

simplephysicalcontextsforwhichthesepotentialenergyshapesarerelevant.

Ans.(a)x>a;0

Totalenergyofasystemisgivenbytherelation:

E=P.E.+K.E.

Kineticenergyofabodyisapositivequantity.Itcannotbenegative.Therefore,theparticle

willnotexistinaregionwhereK.E.becomesnegative.

Inthegivencase,thepotentialenergy( )oftheparticlebecomesgreaterthantotalenergy

(E)forx>a.Hence,kineticenergybecomesnegativeinthisregion.Therefore,theparticle

willnotexististhisregion.Theminimumtotalenergyoftheparticleiszero.

(b)Allregions

Inthegivencase,thepotentialenergy( )isgreaterthantotalenergy(E)inallregions.

Hence,theparticlewillnotexistinthisregion.

(c

Inthegivencase,theconditionregardingthepositivityofK.E.issatisfiedonlyintheregion

betweenx>aandx<b.

Theminimumpotentialenergyinthiscaseis .Therefore, =E+V1.

Therefore,forthepositivityofthekineticenergy,thetotalenergyoftheparticlemustbe

greaterthan .So,theminimumtotalenergytheparticlemusthaveis .

(d)

Inthegivencase,thepotentialenergy( )oftheparticlebecomesgreaterthanthetotal

energy(E)for .Therefore,theparticlewillnotexistinthese

regions.

Theminimumpotentialenergyinthiscaseis .Therefore,K.E.= .

Therefore,forthepositivityofthekineticenergy,thetotalenergyoftheparticlemustbe

greaterthan .So,theminimumtotalenergytheparticlemusthaveis .

5.Anelectronandaprotonaredetectedinacosmicrayexperiment,thefirstwith

kineticenergy10keV,andthesecondwith100keV.Whichisfaster,theelectronorthe

proton?Obtaintheratiooftheirspeeds.(electronmass= kg,protonmass=

kg,1eV= J).

Ans.Electronicsfaster;Ratioofspeedsis13.54:1

Massoftheelectron, kg

Massoftheproton,

Kineticenergyoftheelectron, =10keV= eV

=

=

Kineticenergyoftheproton,

Forthevelocityofanelectron ,itskineticenergyisgivenbytherelation:

Forthevelocityofaproton ,itskineticenergyisgivenbytherelation:

Hence,theelectronismovingfasterthantheproton.

Theratiooftheirspeeds:

6.Thebobofapendulumisreleasedfromahorizontalposition.Ifthelengthofthe

pendulumis1.5m,whatisthespeedwithwhichthebobarrivesatthelowermostpoint,

giventhatitdissipated5%ofitsinitialenergyagainstairresistance?

Ans.Lengthofthependulum,l=1.5m

Massofthebob=m

Energydissipated=5%

Accordingtothelawofconservationofenergy,thetotalenergyofthesystemremains

constant.

Atthehorizontalposition:

Potentialenergyofthebob,EP=mgl

Kineticenergyofthebob,EK=0

Totalenergy=mgl…(i)

Atthelowermostpoint(meanposition):

Potentialenergyofthebob,EP=0

Kineticenergyofthebob,

Totalenergy …(ii)

Asthebobmovesfromthehorizontalpositiontothelowermostpoint,5%ofitsenergygets

dissipated.

Thetotalenergyatthelowermostpointisequalto95%ofthetotalenergyatthehorizontal

point,i.e.,

7.ThebladesofawindmillsweepoutacircleofareaA.(a)Ifthewindflowsata

velocityvperpendiculartothecircle,whatisthemassoftheairpassingthroughitin

timet?(b)Whatisthekineticenergyoftheair?(c)Assumethatthewindmillconverts

25%ofthewind'senergyintoelectricalenergy,andthatA= ,v=36km/handthe

densityofairis1.2kg .Whatistheelectricalpowerproduced?

Ans.Areaofthecirclesweptbythewindmill=A

Velocityofthewind=v

Densityofair=p

(a)Volumeofthewindflowingthroughthewindmillpersec=Av

Massofthewindflowingthroughthewindmillpersec=pAv

Massm,ofthewindflowingthroughthewindmillintimet=pAvt

(b)Kineticenergyofair

(c)Areaofthecirclesweptbythewindmill=A=30m2

Velocityofthewind=v=36km/h

Densityofair,

Electricenergyproduced=25%ofthewindenergy

Electricalpower

5MarksQuestions

1.(a)Definepotentialenergy.Giveexamples.

(b)Drawagraphshowingvariationofpotentialenergy,kineticenergyandthetotal

energyofabodyfreelyfallingonearthfromaheighth?

Ans:(a)Potentialenergyistheenergypossessedbyabodybyvirtueofitspositioninafield

orduetochangeinitsconfigurationexample–Agascompressedinacylinder,Awound

springofawater,waterraisedtotheoverheadtankinahouseetc.

(i)Gravitationalpotentialenergydecreasesasthebodyfallsdownwardsandiszeroatthe

earth

(ii)Kineticenergyincreasesasthebodyfallsdownwardsandismaximumwhenthebody

juststrikestheground.

(iii)Accordingtolawofconservationofenergytotalmechanical(KE+PE)energyremains

constant.

2.Answerthefollowing:

(a)Thecasingofarocketinflightburnsupduetofriction.Atwhoseexpenseistheheat

energyrequiredforburningobtained?Therocketortheatmosphere?

(b)Cometsmovearoundthesuninhighlyellipticalorbits.Thegravitationalforceon

thecometduetothesunisnotnormaltothecomet'svelocityingeneral.Yetthework

donebythegravitationalforceovereverycompleteorbitofthecometiszero.Why?

(c)Anartificialsatelliteorbitingtheearthinverythinatmospherelosesitsenergy

graduallyduetodissipationagainstatmosphericresistance,howeversmall.Whythen

doesitsspeedincreaseprogressivelyasitcomescloserandclosertotheearth?

(d)InFig.6.13(i)themanwalks2mcarryingamassof15kgonhishands.InFig.

6.13(ii),hewalksthesamedistancepullingtheropebehindhim.Theropegoesovera

pulley,andamassof15kghangsatitsotherend.Inwhichcaseistheworkdone

greater?

Fig.6.13

Ans.(a)Rocket

Theburningofthecasingofarocketinflight(duetofriction)resultsinthereductionofthe

massoftherocket.

Accordingtotheconservationofenergy:

TotalEnergy(T.E.)=Potentialenegry(P.E.)+Kineticenergy(K.E.)

Thereductionintherocket'smasscausesadropinthetotalenergy.Therefore,theheat

energyrequiredfortheburningisobtainedfromtherocket.

(b)Gravitationalforceisaconservativeforce.Sincetheworkdonebyaconservativeforce

overaclosedpathiszero,theworkdonebythegravitationalforceovereverycompleteorbit

ofacometiszero.

(c)Whenanartificialsatellite,orbitingaroundearth,movesclosertoearth,itspotential

energydecreasesbecauseofthereductionintheheight.Sincethetotalenergyofthesystem

remainsconstant,thereductioninP.E.resultsinanincreaseinK.E.Hence,thevelocityof

thesatelliteincreases.However,duetoatmosphericfriction,thetotalenergyofthesatellite

decreasesbyasmallamount.

(d)Inthesecondcase

Case(i)

Mass,m=15kg

Displacement,s=2m

Workdone,W

Where, =Anglebetweenforceanddisplacement

Case(ii)

Mass,m=15kg

Displacement,s=2m

Here,thedirectionoftheforceappliedontheropeandthedirectionofthedisplacementof

theropearesame.

Therefore,theanglebetweenthem,

Since

Workdone,

=

Hence,moreworkisdoneinthesecondcase.

3.Araindropofradius2mmfallsfromaheightof500mabovetheground.Itfalls

withdecreasingacceleration(duetoviscousresistanceoftheair)untilathalfits

originalheight,itattainsitsmaximum(terminal)speed,andmoveswithuniformspeed

thereafter.Whatistheworkdonebythegravitationalforceonthedropinthefirstand

secondhalfofitsjourney?Whatistheworkdonebytheresistiveforceintheentire

journeyifitsspeedonreachingthegroundis ?

Ans.Radiusoftheraindrop,r=2mm= m

Volumeoftheraindrop,

Densityofwater,

Massoftheraindrop,

=

Gravitationalforce,F=mg

=

Theworkdonebythegravitationalforceonthedropinthefirsthalfofitsjourney:

=

=0.082J

Thisamountofworkisequaltotheworkdonebythegravitationalforceonthedropinthe

secondhalfofitsjourney,i.e., ,=0.082J

Asperthelawofconservationofenergy,ifnoresistiveforceispresent,thenthetotalenergy

oftheraindropwillremainthesame.

∴Totalenergyatthetop:

=

=0.164J

Duetothepresenceofaresistiveforce,thedrophitsthegroundwithavelocityof10m/s.

∴Totalenergyattheground:

∴Resistiveforce=

4.Twoidenticalballbearingsincontactwitheachotherandrestingonafrictionless

tablearehithead-onbyanotherballbearingofthesamemassmovinginitiallywitha

speedV.Ifthecollisioniselastic,whichofthefollowingfigureisapossibleresultafter

collision?

Ans.Case(ii)

Itcanbeobservedthatthetotalmomentumbeforeandaftercollisionineachcaseis

constant.

Foranelasticcollision,thetotalkineticenergyofasystemremainsconservedbeforeand

aftercollision.

Formassofeachballbearingm,wecanwrite:

Totalkineticenergyofthesystembeforecollision:

Case(i)

Totalkineticenergyofthesystemaftercollision:

Hence,thekineticenergyofthesystemisnotconservedincase(i).

Case(ii)

Totalkineticenergyofthesystemaftercollision:

Hence,thekineticenergyofthesystemisconservedincase(ii).

Case(iii)

Totalkineticenergyofthesystemaftercollision:

Hence,thekineticenergyofthesystemisnotconservedincase(iii).

5.Abulletofmass0.012kgandhorizontalspeed strikesablockofwoodof

mass0.4kgandinstantlycomestorestwithrespecttotheblock.Theblockis

suspendedfromtheceilingbymeansofthinwires.Calculatetheheighttowhichthe

blockrises.Also,estimatetheamountofheatproducedintheblock.

Ans.

Massofthebullet,m=0.012kg

Initialspeedofthebullet, =70m/s

Massofthewoodenblock,M=0.4kg

Initialspeedofthewoodenblock, =0

Finalspeedofthesystemofthebulletandtheblock=v

Applyingthelawofconservationofmomentum:

Forthesystemofthebulletandthewoodenblock:

Massofthesystem,m'=0.412kg

Velocityofthesystem=2.04m/s

Heightuptowhichthesystemrises=h

Applyingthelawofconservationofenergytothissystem:

Potentialenergyatthehighestpoint=Kineticenergyatthelowestpoint

Thewoodenblockwillrisetoaheightof0.2123m.

Heatproduced=Kineticenergyofthebullet-Kineticenergyofthesystem

6.Twoinclinedfrictionlesstracks,onegradualandtheothersteepmeetatAfrom

wheretwostonesareallowedtoslidedownfromrest,oneoneachtrack(Fig.6.16).Will

thestonesreachthebottomatthesametime?Willtheyreachtherewiththesame

speed?Explain.Given =30°, =60°,andh=10m,whatarethespeedsandtimes

takenbythetwostones?

Ans.

No;thestonemovingdownthesteepplanewillreachthebottomfirst

Yes;thestoneswillreachthebottomwiththesamespeed

=14m/s

=2.86s; =1.65s

Thegivensituationcanbeshownasinthefollowingfigure:

Here,theinitialheight(AD)forboththestonesisthesame(h).Hence,bothwillhavethe

samepotentialenergyatpointA.

Asperthelawofconservationofenergy,thekineticenergyofthestonesatpointsBandC

willalsobethesame,i.e.,

,say

Where,

m=Massofeachstone

v=SpeedofeachstoneatpointsBandC

Hence,bothstoneswillreachthebottomwiththesamespeed,v.

ForstoneI:

Netforceactingonthisstoneisgivenby:

ForstoneII:

Usingthefirstequationofmotion,thetimeofslidecanbeobtainedas:

ForstoneI:

ForstoneII:

Hence,thestonemovingdownthesteepplanewillreachthebottomfirst.

Thespeed(v)ofeachstoneatpointsBandCisgivenbytherelationobtainedfromthelawof

conservationofenergy.

Thetimesaregivenas: