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CBSEClass11physics
ImportantQuestions
Chapter5
LawsofMotion
3MarksQuestions
1.Atrainrunsalonganunbankedcircularbendofradius30mataspeedof54km/hr.
Themassofthetrainis106kg.Whatprovidesthenecessarycentripetalforcerequired
forthispurpose?Theengineortherails?Whatistheangleofbankingrequiredto
preventwearingoutoftherail?
Ans:(1)Thecentripetalforceisprovidedbythelateralforceactingduetorailsonthe
wheelsofthetrain.
(2)Outerrails
(3)
2.Ablockofmass15kgisplacedonalongtrolley.Thecoefficientofstaticfriction
betweentheblockandthetrolleyis0.18.Thetrolleyacceleratesfromrestwith0.5
for20sandthenmoveswithuniformvelocity.Discussthemotionoftheblockas
viewedby(a)astationaryobserverontheground,(b)anobservermovingwiththe
trolley.
Ans.
(a)Massoftheblock,m=15kg
Coefficientofstaticfriction, =0.18
Accelerationofthetrolley,a=
AsperNewton'ssecondlawofmotion,theforce(F)ontheblockcausedbythemotionofthe
trolleyisgivenbytherelation:
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F=ma= =7.5N
Thisforceisactedinthedirectionofmotionofthetrolley.
Forceofstaticfrictionbetweentheblockandthetrolley:
f=
=0.18x15x10=27N
Theforceofstaticfrictionbetweentheblockandthetrolleyisgreaterthantheapplied
externalforce.Hence,foranobserverontheground,theblockwillappeartobeatrest.
Whenthetrolleymoveswithuniformvelocitytherewillbenoappliedexternalforce.Only
theforceoffrictionwillactontheblockinthissituation.
(b)Anobserver,movingwiththetrolley,hassomeacceleration.Thisisthecaseofnon-
inertialframeofreference.Thefrictionalforce,actingonthetrolleybackward,isopposed
byapseudoforceofthesamemagnitude.However,thisforceactsintheoppositedirection.
Thus,thetrolleywillappeartobeatrestfortheobservermovingwiththetrolley.
3.Whatistheaccelerationoftheblocks?WhatisthenetforceontheblockP?What
forcedoesPapplyonQ.WhatforcedoesQapplyonR?
Ans:Ifaistheacceleration
ThenF=(3m)a
(1)NetforceonP
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(2)ForceappliedonQ
F2=(m+m)a
(3)ForceappliedonRbyQ
4.Howiscentripetalforceprovidedincaseofthefollowing?
(i)Motionofplanetaroundthesun,
(ii)Motionofmoonaroundtheearth.
(iii)Motionofanelectronaroundthenucleusinanatom.
Ans:(i)Gravitationalforceactingontheplanetandthesunprovidesthenecessary
centripetalforce.
(ii)Forceofgravityduetoearthonthemoonprovidescentripetalforce.
(iii)Electrostaticforceattractionbetweentheelectronandtheprotonprovidesthe
necessarycentripetalforce.
5.StateNewton’ssecond,lawofmotion.Expressitmathematicallyandhenceobtaina
relationbetweenforceandacceleration.
Ans:AccordingtoNewton’ssecondlawtherateofchangeofmomentumisdirectly
proportionaltotheforce.
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i.e.F rateofchangeofmomentum
(InS.I.unitK=1)
6.Arailwaycarofmass20tonnesmoveswithaninitialspeedof54km/hr.Onapplying
brakes,aconstantnegativeaccelerationof0.3m/s2isproduced.
(i)Whatisthebreakingforceactingonthecar?
(ii)Inwhattimeitwillstop?
(iii)Whatdistancewillbecoveredbythecarbeforeiffinallystops?
Ans:
(a)F=ma
F=-6000N
(b)
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t=50s
(c)
S=375m
7.Whatismeantbycoefficientoffrictionandangeloffriction?Establishtherelation
betweenthetwo?OR
Ablockofmass10kgisslidingonasurfaceinclinedataangleof30owiththe
horizontal.Calculatetheaccelerationoftheblock.Thecoefficientofkineticfriction
betweentheblockandthesurfaceis0.5
Ans:Angleoffrictionisthecontactbetweentheresultantoflimitingfrictionandnormal
reaction
withthenormalreaction
Coefficientofstaticfriction
Thelimitingvalueofstaticfrictionalforceisproportiontothenormalreactionis
Or
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From(1)&(2)
OR
Ablockofmass10kgisslidingonasurfaceinclinedataangleof30owiththehorizontal.
Calculatetheaccelerationoftheblock.Thecoefficientofkineticfrictionbetweentheblock
andthesurfaceis0.5
a=0.657m/s2
8.Stateandprovetheprincipleoflawofconservationoflinearmomentum?
Ans:Thelawofconservationoflinearmomentumstatesthatifnoexternalforceactsonthe
system.Thetotalmomentumofthesystemremainsunchanged.
i.e.if
Impulseexperiencedby
Impulseexperiencedby
AccordingtoNewton’sthirdlaw
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Thusmomentumgainedbyoneballislostbytheotherball.Hencelinearmomentum
remainsconserved.
9.Aparticleofmass0.40kgmovinginitiallywithconstantspeedof10m/stothenorthis
subjecttoaconstantforceof8.0Ndirectedtowardssouthfor30s.Takeatthatinstant,
theforceisappliedtobet=0,andthepositionoftheparticleatthattimetobex=0,
predictitspositionatt=-5s,25s,30s?
Ans.m=0.40kg
u=l0m/sdueNorth
F=-8.0N
(1)Att=-5s
x=-50m
(2)Att=25s
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x=-6000m
(3)Att=30s
(4)Att=30s
Motionfrom30sto100s
Totaldistancex=x1+x2
x=-50000m
10.Ablockofmass25kgisraisedbya50kgmanintwodifferentwaysasshowninFig.
5.19.Whatistheactiononthefloorbythemaninthetwocases?Iftheflooryieldstoa
normalforceof700N,whichmodeshouldthemanadopttolifttheblockwithoutthe
flooryielding?
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Ans.
750Nand250Nintherespectivecases;Method(b)
Massoftheblock,m=25kg
Massoftheman,M=50kg
Accelerationduetogravity,g=
Forceappliedontheblock,F=25x10=250N
Weightoftheman,W=50x10=500N
Case(a):Whenthemanliftstheblockdirectly
Inthiscase,themanappliesaforceintheupwarddirection.Thisincreaseshisapparent
weight.
∴Actiononthefloorbytheman=250+500=750N
Case(b):Whenthemanliftstheblockusingapulley
Inthiscase,themanappliesaforceinthedownwarddirection.Thisdecreaseshisapparent
weight.
∴Actiononthefloorbytheman=500-250=250N
Ifthefloorcanyieldtoanormalforceof700N,thenthemanshouldadoptthesecond
methodtoeasilylifttheblockbyapplyinglesserforce.
11.(a)Stateimpulse–momentumtheorem?
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(b)Aballofmass0.1kgisthrownagainstawall.Itstrikesthewallnormallywitha
velocityof30m/sandreboundswithavelocityof20m/s.calculatetheimpulseofthe
forceexertedbytheballonthewall.
Ans:(a)Itstatesthatimpulseismeasuredbythetotalchangeinlinearmomentumis
Impulse=
(b)m=0.1kgv=30m/s
Impulse=
Impulse= s
Impulse=m(-20–30)=-5Ns
12.Tenonerupeecoinsareputontopofoneanotheronatable.Eachcoinhasamassm
kg.Givethemagnitudeanddirectionof
(a)Theforceonthe7thcoin(countedfromthebottom)duetoallcoinsaboveit.
(b)Theforceonthe7thcoinbytheeighthcoinand
(c)Thereactionofthesixthcoinontheseventhcoin.
Ans.(a)Theforceon7thcoinisduetoweightofthethreecoinslyingaboveit.
Therefore,F=(3m)kgf=(3mg)N
Wheregisaccelerationduetogravity.Thisforceactsverticallydownwards.
(b)Theeighthcoinisalreadyundertheweightoftwocoinsaboveitandithasitsown
weighttoo.Henceforceon7thcoindueto8thcoinissumofthetwoforcesi.e.
F=2m+m=(3m)kgf=(3mg)N
Theforceactsverticallydownwards.
(c)Thesixthcoinsisundertheweightoffourcoinsaboveit
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Reaction,R=-F=-4m(kgf)=-(4mg)N
-vesignindicatesthatreactionactsverticallyupwards.
13.Givethemagnitudeanddirectionofthenetforceactingon
(a)adropofrainfallingdownwithaconstantspeed,
(b)acorkofmass10gfloatingonwater,
(c)akiteskillfullyheldstationaryinthesky,
(d)acarmovingwithaconstantvelocityof30km/honaroughroad,
(e)ahigh-speedelectroninspacefarfromallmaterialobjects,andfreeofelectricand
magneticfields.
Ans(a)Zeronetforce
Theraindropisfallingwithaconstantspeed.Hence,itaccelerationiszero.AsperNewton's
secondlawofmotion,thenetforceactingontheraindropiszero.
(b)Zeronetforce
Theweightofthecorkisactingdownward.Itisbalancedbythebuoyantforceexertedbythe
waterintheupwarddirection.Hence,nonetforceisactingonthefloatingcork.
(c)Zeronetforce
Thekiteisstationaryinthesky,i.e.,itisnotmovingatall.Hence,asperNewton'sfirstlawof
motion,nonetforceisactingonthekite.
(d)Zeronetforce
Thecarismovingonaroughroadwithaconstantvelocity.Hence,itsaccelerationiszero.As
perNewton'ssecondlawofmotion,nonetforceisactingonthecar.
(e)Zeronetforce
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Thehighspeedelectronisfreefromtheinfluenceofallfields.Hence,nonetforceisacting
ontheelectron.
14.Apebbleofmass0.05kgisthrownverticallyupwards.Givethedirectionand
magnitudeofthenetforceonthepebble,
(a)duringitsupwardmotion,
(b)duringitsdownwardmotion,
(c)atthehighestpointwhereitismomentarilyatrest.Doyouranswerschangeifthe
pebblewasthrownatanangleof45°withthehorizontaldirection?
Ignoreairresistance.
Ans.0.5N,inverticallydownwarddirection,inallcases
Accelerationduetogravity,irrespectiveofthedirectionofmotionofanobject,alwaysacts
downward.Thegravitationalforceistheonlyforcethatactsonthepebbleinallthreecases.
ItsmagnitudeisgivenbyNewton'ssecondlawofmotionas:
Where,
F=Netforce
m=Massofthepebble=0.05kg
a=g=
∴F= =0.5N
Thenetforceonthepebbleinallthreecasesis0.5Nandthisforceactsinthedownward
direction.
Ifthepebbleisthrownatanangleof45°withthehorizontal,itwillhaveboththehorizontal
andverticalcomponentsofvelocity.Atthehighestpoint,onlytheverticalcomponentof
velocitybecomeszero.However,thepebblewillhavethehorizontalcomponentofvelocity
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throughoutitsmotion.Thiscomponentofvelocityproducesnoeffectonthenetforceacting
onthepebble.
15.Aconstantforceactingonabodyofmass3.0kgchangesitsspeedfrom
to in25s.Thedirectionofthemotionofthebodyremainsunchanged.What
isthemagnitudeanddirectionoftheforce?
Ans.0.18N;inthedirectionofmotionofthebody
Massofthebody,m=3kg
Initialspeedofthebody,u=2m/s
Finalspeedofthebody,v=3.5m/s
Time,t=25s
Usingthefirstequationofmotion,theacceleration(a)producedinthebodycanbe
calculatedas:
v=u+at
AsperNewton'ssecondlawofmotion,forceisgivenas:
F=ma
= =0.18N
Sincetheapplicationofforcedoesnotchangethedirectionofthebody,thenetforceacting
onthebodyisinthedirectionofitsmotion.
16.Abodyofmass5kgisacteduponbytwoperpendicularforces8Nand6N.Givethe
magnitudeanddirectionoftheaccelerationofthebody.
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Ans. ,atanangleof37°withaforceof8N
Massofthebody,m=5kg
Thegivensituationcanberepresentedasfollows:
Theresultantoftwoforcesisgivenas:
istheanglemadebyRwiththeforceof8N
Thenegativesignindicatesthat isintheclockwisedirectionwithrespecttotheforceof
magnitude8N.
AsperNewton'ssecondlawofmotion,theacceleration(a)ofthebodyisgivenas:
F=ma
17.Astoneofmass0.25kgtiedtotheendofastringiswhirledroundinacircleof
radius1.5mwithaspeedof40rev./mininahorizontalplane.Whatisthetensioninthe
string?Whatisthemaximumspeedwithwhichthestonecanbewhirledaroundifthe
stringcanwithstandamaximumtensionof200N?
Ans.
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Massofthestone,m=0.25kg
Radiusofthecircle,r=1.5m
Numberofrevolutionpersecond,
Angularvelocity,
ThecentripetalforceforthestoneisprovidedbythetensionT,inthestring,i.e.,
Maximumtensioninthestring, =200N
Therefore,themaximumspeedofthestoneis34.64m/s.
18.Figure5.18showsamanstandingstationarywithrespecttoahorizontalconveyor
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beltthatisacceleratingwith .Whatisthenetforceontheman?Ifthe
coefficientofstaticfrictionbetweentheman'sshoesandthebeltis0.2,uptowhat
accelerationofthebeltcanthemancontinuetobestationaryrelativetothebelt?
(Massoftheman=65kg.)
Figure5.18
Ans.Massoftheman,m=65kg
Accelerationofthebelt,
Coefficientofstaticfriction, =0.2
ThenetforceF,actingonthemanisgivenbyNewton'ssecondlawofmotionas:
=65×1=65N
Themanwillcontinuetobestationarywithrespecttotheconveyorbeltuntilthenetforce
onthemanislessthanorequaltothefrictionalforce ,exertedbythebelt,i.e.,
Therefore,themaximumaccelerationofthebeltuptowhichthemancanstandstationaryis
.
19.Astreamofwaterflowinghorizontallywithaspeedof gushesoutofa
tubeofcross-sectionalarea ,andhitsaverticalwallnearby.Whatistheforce
exertedonthewallbytheimpactofwater,assumingitdoesnotrebound?
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Ans.Speedofthewaterstream,v=15m/s
Cross-sectionalareaofthetube,A=
Volumeofwatercomingoutfromthepipepersecond,
V=Av=
Densityofwater,
Massofwaterflowingoutthroughthepipepersecond= =150kg/s
Thewaterstrikesthewallanddoesnotrebound.Therefore,theforceexertedbythewater
onthewallisgivenbyNewton'ssecondlawofmotionas:
F=Rateofchangeofmomentum
20.Anaircraftexecutesahorizontalloopataspeedof720km/hwithitswingsbanked
at15°.Whatistheradiusoftheloop?
Ans.Speedoftheaircraft,v=720km/h
Accelerationduetogravity,g=
Angleofbanking,θ=15°
Forradiusr,oftheloop,wehavetherelation:
=14925.37m
=14.92km
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CBSEClass11physics
ImportantQuestions
Chapter5
LawsofMotion
4MarksQuestions
1.Givethemagnitudeanddirectionofthenetforceactingonastoneofmass0.1kg,
(a)justafteritisdroppedfromthewindowofastationarytrain,
(b)justafteritisdroppedfromthewindowofatrainrunningataconstantvelocityof
36km/h,
(c)justafteritisdroppedfromthewindowofatrainacceleratingwith1ms-2,
(d)lyingonthefloorofatrainwhichisacceleratingwith1ms-2,thestonebeingatrest
relativetothetrain.Neglectairresistancethroughout.
Ans.(a)1N;verticallydownward
Massofthestone,m=0.1kg
Accelerationofthestone,a=g=
AsperNewton'ssecondlawofmotion,thenetforceactingonthestone,
F=ma=mg
=0.1x10=1N
Accelerationduetogravityalwaysactsinthedownwarddirection.
(b)1N;verticallydownward
Thetrainismovingwithaconstantvelocity.Hence,itsaccelerationiszerointhedirectionof
itsmotion,i.e.,inthehorizontaldirection.Hence,noforceisactingonthestoneinthe
horizontaldirection.
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Thenetforceactingonthestoneisbecauseofaccelerationduetogravityanditalwaysacts
verticallydownward.Themagnitudeofthisforceis1N.
(c)1N;verticallydownward
Itisgiventhatthetrainisacceleratingattherateof .
Therefore,thenetforceactingonthestone,F'=ma=0.1x1=0.1N
Thisforceisactinginthehorizontaldirection.Now,whenthestoneisdropped,the
horizontalforceF,'stopsactingonthestone.Thisisbecauseofthefactthattheforceacting
onabodyataninstantdependsonthesituationatthatinstantandnotonearliersituations.
Therefore,thenetforceactingonthestoneisgivenonlybyaccelerationduetogravity.
F=mg=1N
Thisforceactsverticallydownward.
(d)0.1N;inthedirectionofmotionofthetrain
Theweightofthestoneisbalancedbythenormalreactionofthefloor.Theonlyacceleration
isprovidedbythehorizontalmotionofthetrain.
Accelerationofthetrain,
Thenetforceactingonthestonewillbeinthedirectionofmotionofthetrain.Itsmagnitude
isgivenby:
F=ma
=0.1x1=0.1N
2.Thedriverofathree-wheelermovingwithaspeedof36km/hseesachildstandingin
themiddleoftheroadandbringshisvehicletorestin4.0sjustintimetosavethe
child.Whatistheaverageretardingforceonthevehicle?Themassofthethree-
wheeleris400kgandthemassofthedriveris65kg.
Ans.Initialspeedofthethree-wheeler,u=36km/h=10m/s
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Finalspeedofthethree-wheeler,v=0m/s
Time,t=4s
Massofthethree-wheeler,m=400kg
Massofthedriver,m'=65kg
Totalmassofthesystem,M=400+65=465kg
Usingthefirstlawofmotion,theacceleration(a)ofthethree-wheelercanbecalculatedas:
v=u+at
Thenegativesignindicatesthatthevelocityofthethree-wheelerisdecreasingwithtime.
UsingNewton'ssecondlawofmotion,thenetforceactingonthethree-wheelercanbe
calculatedas:
F=Ma
=
Thenegativesignindicatesthattheforceisactingagainstthedirectionofmotionofthe
three-wheeler.
3.Abodyofmass0.40kgmovinginitiallywithaconstantspeedof10ms-1tothenorth
issubjecttoaconstantforceof8.0Ndirectedtowardsthesouthfor30s.Takethe
instanttheforceisappliedtobet=0,thepositionofthebodyatthattimetobex=0,
andpredictitspositionatt=-5s,25s,100s.
Ans.Massofthebody,m=0.40kg
Initialspeedofthebody,u=10m/sduenorth
Forceactingonthebody,F=
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Accelerationproducedinthebody,
(i)Att=-5s
Acceleration,a'=0andu=10m/s
=
(ii)Att=25s
Acceleration,a''= andu=10m/s
(iii)Att=100s
For
a=
u=10m/s
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For
Asperthefirstequationofmotion,fort=30s,finalvelocityisgivenas:
v=u+at
=
Velocityofthebodyafter30s=
Formotionbetween30sto100s,i.e.,in70s:
∴Totaldistance,
4.Atruckstartsfromrestandacceleratesuniformlyat .Att=10s,astoneis
droppedbyapersonstandingonthetopofthetruck(6mhighfromtheground).What
arethe(a)velocity,and(b)accelerationofthestoneatt=11s?(Neglectairresistance.)
Ans.(a)22.36m/s,atanangleof26.57°withthemotionofthetruck
(b)
(a)Initialvelocityofthetruck,u=0
Acceleration,a=
Time,t=10s
Asperthefirstequationofmotion,finalvelocityisgivenas:
v=u+at
=0+2×10=20m/s
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Thefinalvelocityofthetruckandhence,ofthestoneis20m/s.
Att=11s,thehorizontalcomponent( )ofvelocity,intheabsenceofairresistance,
remainsunchanged,i.e.,
=20m/s
Theverticalcomponent( )ofvelocityofthestoneisgivenbythefirstequationofmotion
as:
Where, = =1sand
Theresultantvelocity(v)ofthestoneisgivenas:
Letθbetheanglemadebytheresultantvelocitywiththehorizontalcomponentofvelocity,
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=26.57°
(b)Whenthestoneisdroppedfromthetruck,thehorizontalforceactingonitbecomeszero.
However,thestonecontinuestomoveundertheinfluenceofgravity.Hence,theacceleration
ofthestoneis anditactsverticallydownward.
5.Figure5.16showstheposition-timegraphofaparticleofmass4kg.Whatisthe(a)
forceontheparticlefort<0,t>4s,0<t<4s?(b)impulseatt=0andt=4s?(Consider
one-dimensionalmotiononly).
Figure5.16
Ans.
(a)Fort<0
Itcanbeobservedfromthegivengraphthatthepositionoftheparticleiscoincidentwith
thetimeaxis.Itindicatesthatthedisplacementoftheparticleinthistimeintervaliszero.
Hence,theforceactingontheparticleiszero.
Fort>4s
Itcanbeobservedfromthegivengraphthatthepositionoftheparticleisparalleltothetime
axis.Itindicatesthattheparticleisatrestatadistanceof
3mfromtheorigin.Hence,noforceisactingontheparticle.
For0<t<4
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Itcanbeobservedthatthegivenposition-timegraphhasaconstantslope.Hence,the
accelerationproducedintheparticleiszero.Therefore,theforceactingontheparticleis
zero.
(b)Att=0
Impulse=Changeinmomentum
=
Massoftheparticle,m=4kg
Initialvelocityoftheparticle,u=0
Finalvelocityoftheparticle,
∴Impulse
Att=4s
Initialvelocityoftheparticle,
Finalvelocityoftheparticle,v=0
∴Impulse
6.Twobodiesofmasses10kgand20kgrespectivelykeptonasmooth,horizontal
surfacearetiedtotheendsofalightstring.AhorizontalforceF=600Nisappliedto(i)
A,(ii)Balongthedirectionofstring.Whatisthetensioninthestringineachcase?
Ans.
Horizontalforce,F=600N
MassofbodyA, =10kg
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MassofbodyB, =20kg
Totalmassofthesystem, =30kg
UsingNewton'ssecondlawofmotion,theacceleration(a)producedinthesystemcanbe
calculatedas:
F=ma
WhenforceFisappliedonbodyA:
Theequationofmotioncanbewrittenas:
=400N…(i)
WhenforceFisappliedonbodyB:
Theequationofmotioncanbewrittenas:
…(ii)
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7.Abatsmandeflectsaballbyanangleof45°withoutchangingitsinitialspeedwhich
isequalto54km/h.Whatistheimpulseimpartedtotheball?(Massoftheballis0.15
kg.)
Ans.Thegivensituationcanberepresentedasshowninthefollowingfigure.
Where,
AO=Incidentpathoftheball
OB=Pathfollowedbytheballafterdeflection
∠AOB=Anglebetweentheincidentanddeflectedpathsoftheball=45°
∠AOP=∠BOP=22.5°=θ
Initialandfinalvelocitiesoftheball=v
Horizontalcomponentoftheinitialvelocity=vcos alongRO
Verticalcomponentoftheinitialvelocity=vsin alongPO
Horizontalcomponentofthefinalvelocity=vcosθalongOS
Verticalcomponentofthefinalvelocity=vsin alongOP
Thehorizontalcomponentsofvelocitiessuffernochange.Theverticalcomponentsof
velocitiesareintheoppositedirections.
∴Impulseimpartedtotheball=Changeinthelinearmomentumoftheball
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Massoftheball,m=0.15kg
Velocityoftheball,v=54km/h=15m/s
∴Impulse= =4.16kgm/s
8.Figure5.17showstheposition-timegraphofabodyofmass0.04kg.Suggesta
suitablephysicalcontextforthismotion.Whatisthetimebetweentwoconsecutive
impulsesreceivedbythebody?Whatisthemagnitudeofeachimpulse?
Figure5.17
Ans.
Aballreboundingbetweentwowallslocatedbetweenatx=0andx=2cm;afterevery2s,the
ballreceivesanimpulseofmagnitude fromthewalls
Thegivengraphshowsthatabodychangesitsdirectionofmotionafterevery2s.Physically,
thissituationcanbevisualizedasaballreboundingtoandfrobetweentwostationarywalls
situatedbetweenpositionsx=0andx=2cm.Sincetheslopeofthex-tgraphreversesafter
every2s,theballcollideswithawallafterevery2s.Therefore,ballreceivesanimpulse
afterevery2s.
Massoftheball,m=0.04kg
Theslopeofthegraphgivesthevelocityoftheball.Usingthegraph,wecancalculateinitial
velocity(u)as:
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Velocityoftheballbeforecollision,
Velocityoftheballaftercollision,
(Here,thenegativesignarisesastheballreversesitsdirectionofmotion.)
Magnitudeofimpulse=Changeinmomentum
9.AstoneofmassmtiedtotheendofastringrevolvesinaverticalcircleofradiusR.
Thenetforcesatthelowestandhighestpointsofthecircledirectedvertically
downwardsare:[Choosethecorrectalternative]
LowestPoint HighestPoint
(a)
(b)
(c)
(d)
Ans.(a)Thefreebodydiagramofthestoneatthelowestpointisshowninthefollowing
figure.
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AccordingtoNewton'ssecondlawofmotion,thenetforceactingonthestoneatthispointis
equaltothecentripetalforce,i.e.,
…(i)
Where, =Velocityatthelowestpoint
Thefreebodydiagramofthestoneatthehighestpointisshowninthefollowingfigure.
UsingNewton'ssecondlawofmotion,wehave:
…(ii)
Where, =Velocityatthehighestpoint
Itisclearfromequations(i)and(ii)thatthenetforceactingatthelowestandthehighest
pointsarerespectively and(T+mg).
10.Adiscrevolveswithaspeedof rev/min,andhasaradiusof15cm.Twocoins
areplacedat4cmand14cmawayfromthecentreoftherecord.Iftheco-efficientof
frictionbetweenthecoinsandtherecordis0.15,whichofthecoinswillrevolvewith
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therecord?
Ans.Coinplacedat4cmfromthecentre
Massofeachcoin=m
Radiusofthedisc,r=15cm=0.15m
Frequencyofrevolution, rev/min
Coefficientoffriction, =0.15
Inthegivensituation,thecoinhavingaforceoffrictiongreaterthanorequaltothe
centripetalforceprovidedbytherotationofthediscwillrevolvewiththedisc.Ifthisisnot
thecase,thenthecoinwillslipfromthedisc.
Coinplacedat4cm:
Radiusofrevolution,r'=4cm=0.04m
Angularfrequency,
Frictionalforce,f= mg=0.15×m×10=1.5mN
Centripetalforceonthecoin:
=0.49mN
Sincef> ,thecoinwillrevolvealongwiththerecord.
Coinplacedat14cm:
Radius, =14cm=0.14m
Angularfrequency,
Frictionalforce,f'=1.5mN
Centripetalforceisgivenas:
=1.7mN
Sincef< ,thecoinwillslipfromthesurfaceoftherecord.
11.A70kgmanstandsincontactagainsttheinnerwallofahollowcylindricaldrumof
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radius3mrotatingaboutitsverticalaxiswith200rev/min.Thecoefficientoffriction
betweenthewallandhisclothingis0.15.Whatistheminimumrotationalspeedofthe
cylindertoenablethemantoremainstucktothewall(withoutfalling)whenthefloor
issuddenlyremoved?
Ans.Massoftheman,m=70kg
Radiusofthedrum,r=3m
Coefficientoffriction, =0.15
Frequencyofrotation,v=200rev/min
Thenecessarycentripetalforcerequiredfortherotationofthemanisprovidedbythe
normalforce(FN).
Whenthefloorrevolves,themanstickstothewallofthedrum.Hence,theweightofthe
man
(mg)actingdownwardisbalancedbythefrictionalforce(f= )actingupward.
Hence,themanwillnotfalluntil:
mg<f
Theminimumangularspeedisgivenas:
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CBSEClass11physics
ImportantQuestions
Chapter5
LawsofMotion
5MarksQuestions
1.(a)Defineimpulse.StateitsS.I.unit?
(b)Stateandproveimpulsemomentumtheorem?
Ans:(a)Forcewhichareexertedoverashorttimeintervalsarecalledimpulsiveforces.
Impulse
Unit–NS
Impulseisavectorquantitydirectedalongtheaverageforce
(b)Impulseofaforceisequaltothechangeinmomentumofthebody.
AccordingtoNewton’ssecondlaw
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2.Amanofmass70kgstandsonaweighingscaleinaliftwhichismoving
(a)upwardswithauniformspeedof10 ,
(b)downwardswithauniformaccelerationof ,
(c)upwardswithauniformaccelerationof .
Whatwouldbethereadingsonthescaleineachcase?
(d)Whatwouldbethereadingiftheliftmechanismfailedandithurtleddownfreely
undergravity?
Ans.(a)Massoftheman,m=70kg
Acceleration,a=0
UsingNewton'ssecondlawofmotion,wecanwritetheequationofmotionas:
=ma
Where,maisthenetforceactingontheman.
Astheliftismovingatauniformspeed,accelerationa=0
∴R=mg
=70×10=700N
∴Readingontheweighingscale=
(b)Massoftheman,m=70kg
Acceleration,a= downward
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UsingNewton'ssecondlawofmotion,wecanwritetheequationofmotionas:
R+mg=ma
=
=350N
∴Readingontheweighingscale=
(c)Massoftheman,m=70kg
Acceleration, upward
UsingNewton'ssecondlawofmotion,wecanwritetheequationofmotionas:
R=m(g+a)
=70(10+5)=70×15
=1050N
∴Readingontheweighingscale=
(d)Whentheliftmovesfreelyundergravity,accelerationa=g
UsingNewton'ssecondlawofmotion,wecanwritetheequationofmotionas:
R+mg=ma
=0
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∴Readingontheweighingscale=
Themanwillbeinastateofweightlessness.
3.Twomasses8kgand12kgareconnectedatthetwoendsofalightinextensible
stringthatgoesoverafrictionlesspulley.Findtheaccelerationofthemasses,andthe
tensioninthestringwhenthemassesarereleased.
Ans.
Thegivensystemoftwomassesandapulleycanberepresentedasshowninthefollowing
figure:
Smallermass, =8kg
Largermass, =12kg
Tensioninthestring=T
Mass ,owingtoitsweight,movesdownwardwithaccelerationa,andmass moves
upward.
ApplyingNewton'ssecondlawofmotiontothesystemofeachmass:
Formass :
Theequationofmotioncanbewrittenas:
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Formass :
Theequationofmotioncanbewrittenas:
Addingequations(i)and(ii),weget:
Therefore,theaccelerationofthemassesis .
Substitutingthevalueofainequation(ii),weget:
Therefore,thetensioninthestringis96N.
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4.Explainwhy
(a)ahorsecannotpullacartandruninemptyspace,
(b)passengersarethrownforwardfromtheirseatswhenaspeedingbusstops
suddenly,
(c)itiseasiertopullalawnmowerthantopushit,
(d)acricketermoveshishandsbackwardswhileholdingacatch.
Ans.(a)Inordertopullacart,ahorsepushesthegroundbackwardwithsomeforce.The
groundinturnexertsanequalandoppositereactionforceuponthefeetofthehorse.This
reactionforcecausesthehorsetomoveforward.
Anemptyspaceisdevoidofanysuchreactionforce.Therefore,ahorsecannotpullacart
andruninemptyspace.
(b)Whenaspeedingbusstopssuddenly,thelowerportionofapassenger'sbody,whichisin
contactwiththeseat,suddenlycomestorest.However,theupperportiontendstoremainin
motion(asperthefirstlawofmotion).Asaresult,thepassenger'supperbodyisthrown
forwardinthedirectioninwhichthebuswasmoving.
(c)Whilepullingalawnmower,aforceatanangleθisappliedonit,asshowninthe
followingfigure.
Theverticalcomponentofthisappliedforceactsupward.Thisreducestheeffectiveweight
ofthemower.
Ontheotherhand,whilepushingalawnmower,aforceatanangleθisappliedonit,as
showninthefollowingfigure.
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Inthiscase,theverticalcomponentoftheappliedforceactsinthedirectionoftheweightof
themower.Thisincreasestheeffectiveweightofthemower.
Sincetheeffectiveweightofthelawnmowerislesserinthefirstcase,pullingthelawn
moweriseasierthanpushingit.
(d)AccordingtoNewton'ssecondlawofmotion,wehavetheequationofmotion:
Where,
F=Stoppingforceexperiencedbythecricketerashecatchestheball
m=Massoftheball
Δt=Timeofimpactoftheballwiththehand
Itcanbeinferredfromequation(i)thattheimpactforceisinverselyproportionaltothe
impacttime,i.e.,
Equation(ii)showsthattheforceexperiencedbythecricketerdecreasesifthetimeofimpact
increasesandviceversa.
Whiletakingacatch,acricketermoveshishandbackwardsoastoincreasethetimeof
impact(Δt).Thisisturnresultsinthedecreaseinthestoppingforce,therebypreventingthe
handsofthecricketerfromgettinghurt.
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5.Ahelicopterofmass1000kgriseswithaverticalaccelerationof .Thecrew
andthepassengersweigh300kg.Givethemagnitudeanddirectionofthe
(a)forceonthefloorbythecrewandpassengers,
(b)actionoftherotorofthehelicopteronthesurroundingair,
(c)forceonthehelicopterduetothesurroundingair.
Ans.(a)Massofthehelicopter, =1000kg
Massofthecrewandpassengers, =300kg
Totalmassofthesystem,m=1300kg
Accelerationofthehelicopter,
UsingNewton'ssecondlawofmotion,thereactionforceR,onthesystembythefloorcanbe
calculatedas:
=ma
=300(10+15)=300×25
=7500N
Sincethehelicopterisacceleratingverticallyupward,thereactionforcewillalsobedirected
upward.Therefore,asperNewton'sthirdlawofmotion,theforceonthefloorbythecrew
andpassengersis7500N,directeddownward.
(b)UsingNewton'ssecondlawofmotion,thereactionforceR',experiencedbythehelicopter
canbecalculatedas:
=m(g+a)
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=1300(10+15)=1300×25
=32500N
Thereactionforceexperiencedbythehelicopterfromthesurroundingairisactingupward.
Hence,asperNewton'sthirdlawofmotion,theactionoftherotoronthesurroundingair
willbe32500N,directeddownward.
(c)Theforceonthehelicopterduetothesurroundingairis32500N,directedupward.
6.Tenone-rupeecoinsareputontopofeachotheronatable.Eachcoinhasamassm.
Givethemagnitudeanddirectionof
(a)theforceonthe coin(countedfromthebottom)duetoallthecoinsonitstop,
(b)theforceonthe coinbytheeighthcoin,
(c)thereactionofthe coinonthe coin.
Ans.
(a)Forceontheseventhcoinisexertedbytheweightofthethreecoinsonitstop.
Weightofonecoin=mg
Weightofthreecoins=3mg
Hence,theforceexertedonthe coinbythethreecoinsonitstopis3mg.Thisforceacts
verticallydownward.
(b)Forceontheseventhcoinbytheeighthcoinisbecauseoftheweightoftheeighthcoin
andtheothertwocoins(ninthandtenth)onitstop.
Weightoftheeighthcoin=mg
Weightoftheninthcoin=mg
Weightofthetenthcoin=mg
Totalweightofthesethreecoins=3mg
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Hence,theforceexertedonthe coinbytheeighthcoinis3mg.Thisforceactsvertically
downward.
(c)The coinexperiencesadownwardforcebecauseoftheweightofthefourcoins(
)onitstop.
Therefore,thetotaldownwardforceexperiencedbythe coinis4mg.
AsperNewton'sthirdlawofmotion,the coinwillproduceanequalreactionforceon
the coin,butintheoppositedirection.Hence,thereactionforceofthe6thcoinonthe
coinisofmagnitude4mg.Thisforceactsintheupwarddirection.
7.Amonkeyofmass40kgclimbsonarope(Fig.5.20)whichcanstandamaximum
tensionof600N.Inwhichofthefollowingcaseswilltheropebreak:themonkey
(a)climbsupwithanaccelerationof
(b)climbsdownwithanaccelerationof
(c)climbsupwithauniformspeedof
(d)fallsdowntheropenearlyfreelyundergravity?
(Ignorethemassoftherope).
Fig.5.20
Ans.
Case(a)
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Massofthemonkey,m=40kg
Accelerationduetogravity,g=10m/s
Maximumtensionthattheropecanbear, =600N
Accelerationofthemonkey,a= upward
UsingNewton'ssecondlawofmotion,wecanwritetheequationofmotionas:
T-mg=ma
∴T=m(g+a)
=40(10+6)
=640N
SinceT> ,theropewillbreakinthiscase.
Case(b)
Accelerationofthemonkey,a= downward
UsingNewton'ssecondlawofmotion,wecanwritetheequationofmotionas:
mg-T=ma
∴T=m(g-a)
=40(10-4)
=240N
SinceT< ,theropewillnotbreakinthiscase.
Case(c)
Themonkeyisclimbingwithauniformspeedof5m/s.Therefore,itsaccelerationiszero,i.e.,
a=0.
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UsingNewton'ssecondlawofmotion,wecanwritetheequationofmotionas:
T-mg=ma
T-mg=0
∴T=mg
=400N
SinceT< theropewillnotbreakinthiscase.
Case(d)
Whenthemonkeyfallsfreelyundergravity,itswillaccelerationbecomeequaltothe
accelerationduetogravity,i.e.,a=g
UsingNewton'ssecondlawofmotion,wecanwritetheequationofmotionas:
mg-T=mg
∴T=m(g-g)=0
SinceT< ,theropewillnotbreakinthiscase.
8.TwobodiesAandBofmasses5kgand10kgincontactwitheachotherrestona
tableagainstarigidwall(Fig.5.21).Thecoefficientoffrictionbetweenthebodiesand
thetableis0.15.Aforceof200NisappliedhorizontallytoA.Whatare(a)thereaction
ofthepartition(b)theaction-reactionforcesbetweenAandB?Whathappenswhenthe
wallisremoved?Doestheanswerto(b)change,whenthebodiesareinmotion?Ignore
thedifferencebetween .
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Fig.5.21
Ans.
(a)MassofbodyA, =5kg
MassofbodyB, =10kg
Appliedforce,F=200N
Coefficientoffriction, =0.15
Theforceoffrictionisgivenbytherelation:
=0.15(5+10)×10
=1.5×15=22.5Nleftward
Netforceactingonthepartition= =177.5Nrightward
AsperNewton'sthirdlawofmotion,thereactionforceofthepartitionwillbeinthe
directionoppositetothenetappliedforce.
Hence,thereactionofthepartitionwillbe177.5N,intheleftwarddirection.
(b)ForceoffrictiononmassA:
=0.15×5×10=7.5Nleftward
NetforceexertedbymassAonmassB= =192.5Nrightward
AsperNewton'sthirdlawofmotion,anequalamountofreactionforcewillbeexertedby
massBonmassA,i.e.,192.5Nactingleftward.
Whenthewallisremoved,thetwobodieswillmoveinthedirectionoftheappliedforce.
Netforceactingonthemovingsystem=177.5N
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Theequationofmotionforthesystemofaccelerationa,canbewrittenas:
NetforcecausingmassAtomove:
=5×11.83=59.15N
NetforceexertedbymassAonmassB= =133.35N
Thisforcewillactinthedirectionofmotion.AsperNewton'sthirdlawofmotion,anequal
amountofforcewillbeexertedbymassBonmassA,i.e.,133.3N,actingoppositetothe
directionofmotion.
9.Therearsideofatruckisopenandaboxof40kgmassisplaced5mawayfromthe
openendasshowninFig.5.22.Thecoefficientoffrictionbetweentheboxandthe
surfacebelowitis0.15.Onastraightroad,thetruckstartsfromrestandaccelerates
with2ms-2.Atwhatdistancefromthestartingpointdoestheboxfalloffthetruck?
(Ignorethesizeofthebox).
Fig.5.22
Ans.Massofthebox,m=40kg
Coefficientoffriction, =0.15
Initialvelocity,u=0
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Acceleration,a=
Distanceoftheboxfromtheendofthetruck,s'=5m
AsperNewton'ssecondlawofmotion,theforceontheboxcausedbytheacceleratedmotion
ofthetruckisgivenby:
F=ma
=40×2=80N
AsperNewton'sthirdlawofmotion,areactionforceof80Nisactingontheboxinthe
backwarddirection.Thebackwardmotionoftheboxisopposedbytheforceoffrictionf,
actingbetweentheboxandthefloorofthetruck.Thisforceisgivenby:
f=
=
∴Netforceactingontheblock:
Thebackwardaccelerationproducedintheboxisgivenby:
Usingthesecondequationofmotion,timetcanbecalculatedas:
Hence,theboxwillfallfromthetruckafter fromstart.
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Thedistances,travelledbythetruckin isgivenbytherelation:
=20m
10.Youmayhaveseeninacircusamotorcyclistdrivinginverticalloopsinsidea
'death-well'(ahollowsphericalchamberwithholes,sothespectatorscanwatchfrom
outside).Explainclearlywhythemotorcyclistdoesnotdropdownwhenheisatthe
uppermostpoint,withnosupportfrombelow.Whatistheminimumspeedrequiredat
theuppermostpositiontoperformaverticalloopiftheradiusofthechamberis25m?
Ans.
Inadeath-well,amotorcyclistdoesnotfallatthetoppointofaverticalloopbecauseboth
theforceofnormalreactionandtheweightofthemotorcyclistactdownwardandare
balancedbythecentripetalforce.Thissituationisshowninthefollowingfigure.
Thenetforceactingonthemotorcyclististhesumofthenormalforce(FN)andtheforcedue
togravity( ).
Theequationofmotionforthecentripetalaccelerationac,canbewrittenas:
=mac
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Normalreactionisprovidedbythespeedofthemotorcyclist.Attheminimumspeed(
11.AthincircularloopofradiusRrotatesaboutitsverticaldiameterwithanangular
frequency .Showthatasmallbeadonthewireloopremainsatitslowermostpoint
for .Whatistheanglemadebytheradiusvectorjoiningthecentretothe
beadwiththeverticaldownwarddirectionfor ?Neglectfriction.
Ans.
Lettheradiusvectorjoiningthebeadwiththecentremakeanangleθ,withthevertical
downwarddirection.
OP=R=Radiusofthecircle
N=Normalreaction
Therespectiveverticalandhorizontalequationsofforcescanbewrittenas:
...(i)
…(ii)
InΔOPQ,wehave:
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l=R …(iii)
Substitutingequation(iii)inequation(ii),weget:
Substitutingequation(iv)inequation(i),weget:
...(v)
Sincecosθ≤1,thebeadwillremainatitslowermostpointfor ,i.e.,for
For or
Onequatingequations(v)and(vi),weget:
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CBSEClass11physics
ImportantQuestions
Chapter5
LawsofMotion
1MarksQuestions
1.Whatistheunitofcoefficientoffriction?
Ans:Ithasnounit.
2.Namethefactoronwhichcoefficientoffrictiondepends?
Ans:Coefficientoffriction dependsonthenatureofsurfacesincontactand
natureofmotion.
3.Whatprovidesthecentripetalforcetoacartakingaturnonalevelroad?
Ans:Centripetalforceisprovidedbytheforceoffrictionbetweenthetyresandtheroad.
4.Whyisitdesiredtoholdaguntighttoone’sshoulderwhenitisbeingfired?
Ans:Sincethegunrecoilsafterfiringsoitmustbeheldlightlyagainsttheshoulderbecause
gunandtheshoulderconstituteonesystemofgreatermasssothebackkickwillbeless.
5.Whydoesaswimmerpushthewaterbackwards?
Ans:Aswimmerpushesthewaterbackwardsbecauseduetoreactionofwaterheisableto
swimintheforwarddirection
6.Frictionisaselfadjustingforce.Justify.
Ans:Frictionisaselfadjustingforceasitsvaluevariesfromzerotothemaximumvalueto
limitingfriction.
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7.AthiefjumpsfromtheroofofahousewithaboxofweightWonhishead.Whatwill
betheweightoftheboxasexperiencedbythethiefduringjump?
Ans:WeightoftheboxW=m(g–a)=m(g–g)=0.
8.Whichofthefollowingisscalarquantity?Inertia,forceandlinearmomentum.
Ans:Inertiaandlinearmomentumismeasuredbymassofthebodyandisavectorquantity
andmassisascalarquantity.
9.Actionandreactionforcesdonotbalanceeachother.Why?
Ans:Actionandreactiondonotbalanceeachotherbecauseaforceofactionandreaction
actsalwaysontwodifferentbodies.
10.Ifforceisactingonamovingbodyperpendiculartothedirectionofmotion,then
whatwillbeitseffectonthespeedanddirectionofthebody?
Ans:Nochangeinspeed,buttherecanbechangeinthedirectionofmotion.
11.Thetwoendsofspring–balancearepulledeachbyaforceof10kg.wt.Whatwillbe
thereadingofthebalance?
Ans:Thereadingofthebalancewillbe10kgwt.
12.Aliftisacceleratedupward.Willtheapparentweightofapersoninsidethelift
increase,decreaseorremainthesamerelativetoitsrealweight?Iftheliftisgoingwith
uniformspeed,then?
Ans:Theapparentweightwillincrease.Iftheliftisgoingwithuniformspeed,thenthe
apparentweightwillremainthesameastherealweight.
13.Oneendofastringoflengthlisconnectedtoaparticleofmassmandtheothertoa
smallpegonasmoothhorizontaltable.Iftheparticlemovesinacirclewithspeedvthe
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netforceontheparticle(directedtowardsthecentre)is:
(i)T,
(ii) ,
(iii) ,
(iv)0
Tisthetensioninthestring.[Choosethecorrectalternative].
Ans.(i)Whenaparticleconnectedtoastringrevolvesinacircularpatharoundacentre,the
centripetalforceisprovidedbythetensionproducedinthestring.Hence,inthegivencase,
thenetforceontheparticleisthetensionT,i.e.,
F=T=
WhereFisthenetforceactingontheparticle.
14.If,inExercise5.21,thespeedofthestoneisincreasedbeyondthemaximum
permissiblevalue,andthestringbreakssuddenly,whichofthefollowingcorrectly
describesthetrajectoryofthestoneafterthestringbreaks:
(a)thestonemovesradiallyoutwards,
(b)thestonefliesofftangentiallyfromtheinstantthestringbreaks,
(c)thestonefliesoffatananglewiththetangentwhosemagnitudedependsonthe
speedoftheparticle?
Ans.(b)Whenthestringbreaks,thestonewillmoveinthedirectionofthevelocityatthat
instant.Accordingtothefirstlawofmotion,thedirectionofvelocityvectoristangentialto
thepathofthestoneatthatinstant.Hence,thestonewillflyofftangentiallyfromtheinstant
thestringbreaks.
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CBSEClass11physics
ImportantQuestions
Chapter5
LawsofMotion
2MarksQuestions
1.Givethemagnitudeanddirectionofthenetforceactingon
(a)Adropofrainfallingdownwithconstantspeed.
(b)Akiteskillfullyheldstationaryinthesky.
Ans:(1)AccordingtofirstlawofmotionF=0asa=0(particlemoveswithconstantspeed)
(2)Sincekiteisstationarynetforceonthekiteisalsozero.
2.Twoblocksofmassesm1,m2areconnectedbylightspringonasmoothhorizontal
surface.Thetwomassesarepulledapartandthenreleased.Provethattheratioof
theiraccelerationisinverselyproportionaltotheirmasses.
Ans:TheforcesF1andF2duetomassesm1andm2actsinoppositedirections
ThusF1+F2=0
m1a1+m2a2=0
m1a1=-m2a2
Henceproved.
3.Ashellofmass0.020kgisfiredbyagunofmass100kg.Ifthemuzzlespeedoftheshell
is80m/s,whatistherecoilspeedofthegun?
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Ans:Momentumbeforefiring=0
Momentumafterfiring=momentumof(bullet+gun)
Momentumafterfiring=mb b–mg
Accordingtolawofconservationoflinearmomentum
0=mb b–mg
mb b=mg
4.Aforceisbeingappliedonabodybutitcausesnoacceleration.Whatpossibilities
maybeconsideredtoexplaintheobservation?
Ans:(1)Iftheforceisdeformingforcethenitdoesnotproduceacceleration.
(2)Theforceisinternalforcewhichcannotcauseacceleration.
5.Forceof16Nand12Nareactingonamassof200kginmutuallyperpendicular
directions.Findthemagnitudeoftheaccelerationproduced?
Ans:
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6.Anelevatorweighs3000kg.Whatisitsaccelerationwhentheinthetension
supportingcableis33000N.Giventhatg=9.8m/s2.
Ans:Netupwardforceonthe
ElevatorF=T–mg
7.WritetwoconsequencesofNewton’ssecondlawofmotion?
Ans:(1)Itshowsthatthemotionisacceleratedonlywhenforceisapplied.
(2)Itgivesustheconceptofinertialmassofabody.
8.Abirdissittingonthefloorofawirecageandthecageisinthehandofaboy.The
birdstartsflyinginthecage.Willtheboyexperienceanychangeintheweightofthe
cage?
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Ans:Whenthebirdstartsflyinginsidethecagetheweightofbirdisnomoreexperiencedas
airinsideisinfreecontactwithatmosphericairhencethecagewillappearlighter.
9.Whydoesacyclistleantooneside,whilegoingalongcurve?Inwhatdirectiondoes
helean?
Ans:Acyclistleanswhilegoingalongcurvebecauseacomponentofnormalreactionofthe
groundprovideshimthecentripetalforceherequiresforturning.
Hehastoleaninwardsfromhisverticalpositioni.e.towardsthecentreofthecircularpath.
10.Howdoesbankingofroadsreducewearandtearofthetyres?
Ans:Whenacurvedroadisunbankedforceoffrictionbetweenthetyresandtheroad
providesthenecessarycentripetalforce.Frictionhastobeincreasedwhichwillcausewear
andtear.Butwhenthecurvedroadisbanked,acomponentofnormalreactionoftheground
providesthenecessarycentripetalforcewhichreducesthewearandtearofthetyres.
11.Amonkeyofmass40kgclimbsonaropewhichcanstandamaximumtension600N.
Inwhichofthefollowingcaseswilltheropebreak?Themonkey(a)climbsupwithan
accelerationof6m/s2(b)climbsdownwithanaccelerationof4m/s2(c)climbsupwitha
uniformseedof5m/s(d)fallsdowntheropefreelyundergravity.Takeg=10m/s2and
ignorethemassoftherope.
Ans:m=40kg,T=600N(maxtensionropecanhold)
Ropewillbreakifreaction(R)exceedsTension(T)
(a)a=6m/s2
R=m(g+a)=40(10+6)=640N(Ropewillbreak)
(b)a=4m/s2
R=m(g–a)=40(10–6)=240N(Ropewillnotbreak)
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(c) =5m/s(constant)a=0
R=mg=40 10=400N(Ropewillnotbreak)
(d)a=g;R=m(g–a)=m(g–g)
R=zero(Ropewillnotbreak)
12.Asodawaterbottleisfallingfreely.Willthebubblesofthegasriseinthewaterof
thebottle?
Ans:bubbleswillnotriseinwaterbecausewaterinfreelyfallingbottleisinthestateof
weight–lessenshencenoupthrustforceactsonthebubbles.
13.Twobilliardballseachofmass0.05kgmovinginoppositedirectionswithspeed6m/s
collideandreboundwiththesamespeed.Whatistheimpulseimpartedtoeachball
duetoother.
Ans:InitialmomentumtotheballA=0.05(6)=0.3kgm/s
Asthespeedisreversedoncollision,
finalmomentumofballA=0.05(-6)=-0.3kgm/s
ImpulseimpartedtoballA=changeinmomentumofballA=finalmomentum–initial
momentum=-0.3-0.3=-0.6kgm/s.
14.Anucleusisatrestinthelaboratoryframeofreference.Showthatifitdisintegrates
intotwosmallernuclei,theproductsmustbeemittedinoppositedirections.
Ans:Accordingtotheprincipleofconservationoflinearmomentum,totalmomentum
remainsconstant.
Beforedisintegrationlinearmomentum=zero
Afterdisintegrationlinearmomentum=
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15.Explainwhypassengersarethrownforwardformtheirseatswhenaspeedingbus
stopssuddenly.
Ans:Whenthespeedingbusstopssuddenly,lowerpartofthebodyincontactwiththeseat
comestorestbuttheupperpartofthebodyofthepassengerstendstomaintainitsuniform
motion.Hencethepassengersarethrownforward.
16.Arocketwithalift-offmass20,000kgisblastedupwardswithaninitialacceleration
of5.0m .Calculatetheinitialthrust(force)oftheblast.
Ans.Massoftherocket,m=20,000kg
Initialacceleration,a=5m/s2
Accelerationduetogravity,g=
UsingNewton'ssecondlawofmotion,thenetforce(thrust)actingontherocketisgivenby
therelation:
F-mg=ma
F=m(g+a)
=
=
17.Abobofmass0.1kghungfromtheceilingofaroombyastring2mlongissetinto
oscillation.Thespeedofthebobatitsmeanpositionis .Whatisthetrajectory
ofthebobifthestringiscutwhenthebobis(a)atoneofitsextremepositions,(b)atits
meanposition.
Ans.(a)Verticallydownward
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(b)Parabolicpath
(a)Attheextremeposition,thevelocityofthebobbecomeszero.Ifthestringiscutatthis
moment,thenthebobwillfallverticallyontheground.
(b)Atthemeanposition,thevelocityofthebobis1m/s.Thedirectionofthisvelocityis
tangentialtothearcformedbytheoscillatingbob.Ifthebobiscutatthemeanposition,then
itwilltraceaprojectilepathhavingthehorizontalcomponentofvelocityonly.Hence,itwill
followaparabolicpath.
18.Twobilliardballseachofmass0.05kgmovinginoppositedirectionswithspeed6
ms-1collideandreboundwiththesamespeed.Whatistheimpulseimpartedtoeach
ballduetotheother?
Ans.Massofeachball=0.05kg
Initialvelocityofeachball=6m/s
Magnitudeoftheinitialmomentumofeachball, =0.3kgm/s
Aftercollision,theballschangetheirdirectionsofmotionwithoutchangingthemagnitudes
oftheirvelocity.
Finalmomentumofeachball, =-0.3kgm/s
Impulseimpartedtoeachball=Changeinthemomentumofthesystem
=
=-0.3-0.3=-0.6kgm/s
Thenegativesignindicatesthattheimpulsesimpartedtotheballsareoppositeindirection.
19.Atrainrunsalonganunbankedcirculartrackofradius30mataspeedof54km/h.
Themassofthetrainis kg.Whatprovidesthecentripetalforcerequiredforthis
purpose-Theengineortherails?Whatistheangleofbankingrequiredtoprevent
wearingoutoftherail?
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Ans.Radiusofthecirculartrack,r=30m
Speedofthetrain,v=54km/h=15m/s
Massofthetrain,m= kg
Thecentripetalforceisprovidedbythelateralthrustoftherailonthewheel.Asper
Newton'sthirdlawofmotion,thewheelexertsanequalandoppositeforceontherail.This
reactionforceisresponsibleforthewearandrearoftherail
Theangleofbankingθ,isrelatedtotheradius(r)andspeed(v)bytherelation:
Therefore,theangleofbankingisabout36.87°.
20.Aconstantretardingforceof50Nisappliedtoabodyofmass20kgmovinginitially
withaspeedof .Howlongdoesthebodytaketostop?
Ans.Retardingforce,F=
Massofthebody,m=20kg
Initialvelocityofthebody,u=15m/s
Finalvelocityofthebody,v=0
UsingNewton'ssecondlawofmotion,theacceleration(a)producedinthebodycanbe
calculatedas:
F=ma
=20×a
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Usingthefirstequationofmotion,thetime(t)takenbythebodytocometorestcanbe
calculatedas:
v=u+at
=6s
21.Anucleusisatrestinthelaboratoryframeofreference.Showthatifit
disintegratesintotwosmallernucleitheproductsmustmoveinoppositedirections.
Ans.Letm, ,and betherespectivemassesoftheparentnucleusandthetwo
daughternuclei.Theparentnucleusisatrest.
Initialmomentumofthesystem(parentnucleus)=0
Let and betherespectivevelocitiesofthedaughternucleihavingmassesm1andm2.
Totallinearmomentumofthesystemafterdisintegration=
Accordingtothelawofconservationofmomentum:
Totalinitialmomentum=Totalfinalmomentum
Here,thenegativesignindicatesthatthefragmentsoftheparentnucleusmoveindirections
oppositetoeachother.
22.Ashellofmass0.020kgisfiredbyagunofmass100kg.Ifthemuzzlespeedofthe
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shellis80 ,whatistherecoilspeedofthegun?
Ans.Massofthegun,M=100kg
Massoftheshell,m=0.020kg
Muzzlespeedoftheshell,v=80m/s
Recoilspeedofthegun=V
Boththegunandtheshellareatrestinitially.
Initialmomentumofthesystem=0
Finalmomentumofthesystem=
Here,thenegativesignappearsbecausethedirectionsoftheshellandthegunareopposite
toeachother.
Accordingtothelawofconservationofmomentum:
Finalmomentum=Initialmomentum
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