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