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CBSEClass11physics

ImportantQuestions

Chapter5

LawsofMotion

3MarksQuestions

1.Atrainrunsalonganunbankedcircularbendofradius30mataspeedof54km/hr.

Themassofthetrainis106kg.Whatprovidesthenecessarycentripetalforcerequired

forthispurpose?Theengineortherails?Whatistheangleofbankingrequiredto

preventwearingoutoftherail?

Ans:(1)Thecentripetalforceisprovidedbythelateralforceactingduetorailsonthe

wheelsofthetrain.

(2)Outerrails

2.Ablockofmass15kgisplacedonalongtrolley.Thecoefficientofstaticfriction

betweentheblockandthetrolleyis0.18.Thetrolleyacceleratesfromrestwith0.5

for20sandthenmoveswithuniformvelocity.Discussthemotionoftheblockas

viewedby(a)astationaryobserverontheground,(b)anobservermovingwiththe

trolley.

(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:

=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?

(a)F=ma

F=-6000N

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S=375m

7.Whatismeantbycoefficientoffrictionandangeloffriction?Establishtherelation

betweenthetwo?OR

Ablockofmass10kgisslidingonasurfaceinclinedataangleof30owiththe

horizontal.Calculatetheaccelerationoftheblock.Thecoefficientofkineticfriction

betweentheblockandthesurfaceis0.5

Ans:Angleoffrictionisthecontactbetweentheresultantoflimitingfrictionandnormal

reaction

withthenormalreaction

Coefficientofstaticfriction

Thelimitingvalueofstaticfrictionalforceisproportiontothenormalreactionis

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From(1)&(2)

Ablockofmass10kgisslidingonasurfaceinclinedataangleof30owiththehorizontal.

Calculatetheaccelerationoftheblock.Thecoefficientofkineticfrictionbetweentheblock

andthesurfaceis0.5

a=0.657m/s2

8.Stateandprovetheprincipleoflawofconservationoflinearmomentum?

Ans:Thelawofconservationoflinearmomentumstatesthatifnoexternalforceactsonthe

system.Thetotalmomentumofthesystemremainsunchanged.

i.e.if

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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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

∴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:

= =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:

17.Astoneofmass0.25kgtiedtotheendofastringiswhirledroundinacircleof

radius1.5mwithaspeedof40rev./mininahorizontalplane.Whatisthetensioninthe

string?Whatisthemaximumspeedwithwhichthestonecanbewhirledaroundifthe

stringcanwithstandamaximumtensionof200N?

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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,

Densityofwater,

Massofwaterflowingoutthroughthepipepersecond= =150kg/s

Thewaterstrikesthewallanddoesnotrebound.Therefore,theforceexertedbythewater

onthewallisgivenbyNewton'ssecondlawofmotionas:

F=Rateofchangeofmomentum

20.Anaircraftexecutesahorizontalloopataspeedof720km/hwithitswingsbanked

at15°.Whatistheradiusoftheloop?

Ans.Speedoftheaircraft,v=720km/h

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:

=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:

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

u=10m/s

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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

(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

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

(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

(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?

Horizontalforce,F=600N

MassofbodyA, =10kg

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MassofbodyB, =20kg

Totalmassofthesystem, =30kg

UsingNewton'ssecondlawofmotion,theacceleration(a)producedinthesystemcanbe

calculatedas:

WhenforceFisappliedonbodyA:

Theequationofmotioncanbewrittenas:

=400N…(i)

WhenforceFisappliedonbodyB:

…(ii)

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7.Abatsmandeflectsaballbyanangleof45°withoutchangingitsinitialspeedwhich

isequalto54km/h.Whatistheimpulseimpartedtotheball?(Massoftheballis0.15

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

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

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

Frequencyofrotation,v=200rev/min

Thenecessarycentripetalforcerequiredfortherotationofthemanisprovidedbythe

normalforce(FN).

Whenthefloorrevolves,themanstickstothewallofthedrum.Hence,theweightofthe

(mg)actingdownwardisbalancedbythefrictionalforce(f= )actingupward.

Hence,themanwillnotfalluntil:

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:

Where,maisthenetforceactingontheman.

Astheliftismovingatauniformspeed,accelerationa=0

∴R=mg

=70×10=700N

∴Readingontheweighingscale=

(b)Massoftheman,m=70kg

Acceleration,a= downward

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R+mg=ma

(c)Massoftheman,m=70kg

Acceleration, upward

R=m(g+a)

=70(10+5)=70×15

=1050N

(d)Whentheliftmovesfreelyundergravity,accelerationa=g

R+mg=ma

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Themanwillbeinastateofweightlessness.

3.Twomasses8kgand12kgareconnectedatthetwoendsofalightinextensible

stringthatgoesoverafrictionlesspulley.Findtheaccelerationofthemasses,andthe

tensioninthestringwhenthemassesarereleased.

Thegivensystemoftwomassesandapulleycanberepresentedasshowninthefollowing

figure:

Smallermass, =8kg

Largermass, =12kg

Tensioninthestring=T

Mass ,owingtoitsweight,movesdownwardwithaccelerationa,andmass moves

upward.

ApplyingNewton'ssecondlawofmotiontothesystemofeachmass:

Formass :

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Formass :

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:

=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.

(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

Case(a)

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Massofthemonkey,m=40kg

Accelerationduetogravity,g=10m/s

Maximumtensionthattheropecanbear, =600N

Accelerationofthemonkey,a= upward

T-mg=ma

∴T=m(g+a)

=40(10+6)

SinceT> ,theropewillbreakinthiscase.

Case(b)

Accelerationofthemonkey,a= downward

mg-T=ma

∴T=m(g-a)

=40(10-4)

SinceT< ,theropewillnotbreakinthiscase.

Case(c)

Themonkeyisclimbingwithauniformspeedof5m/s.Therefore,itsaccelerationiszero,i.e.,

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T-mg=ma

T-mg=0

∴T=mg

SinceT< theropewillnotbreakinthiscase.

Case(d)

Whenthemonkeyfallsfreelyundergravity,itswillaccelerationbecomeequaltothe

accelerationduetogravity,i.e.,a=g

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

(a)MassofbodyA, =5kg

MassofbodyB, =10kg

Appliedforce,F=200N

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

Initialvelocity,u=0

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Acceleration,a=

Distanceoftheboxfromtheendofthetruck,s'=5m

AsperNewton'ssecondlawofmotion,theforceontheboxcausedbytheacceleratedmotion

ofthetruckisgivenby:

=40×2=80N

AsperNewton'sthirdlawofmotion,areactionforceof80Nisactingontheboxinthe

backwarddirection.Thebackwardmotionoftheboxisopposedbytheforceoffrictionf,

actingbetweentheboxandthefloorofthetruck.Thisforceisgivenby:

∴Netforceactingontheblock:

Thebackwardaccelerationproducedintheboxisgivenby:

Usingthesecondequationofmotion,timetcanbecalculatedas:

Hence,theboxwillfallfromthetruckafter fromstart.

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Thedistances,travelledbythetruckin isgivenbytherelation:

10.Youmayhaveseeninacircusamotorcyclistdrivinginverticalloopsinsidea

'death-well'(ahollowsphericalchamberwithholes,sothespectatorscanwatchfrom

outside).Explainclearlywhythemotorcyclistdoesnotdropdownwhenheisatthe

uppermostpoint,withnosupportfrombelow.Whatistheminimumspeedrequiredat

theuppermostpositiontoperformaverticalloopiftheradiusofthechamberis25m?

Inadeath-well,amotorcyclistdoesnotfallatthetoppointofaverticalloopbecauseboth

theforceofnormalreactionandtheweightofthemotorcyclistactdownwardandare

balancedbythecentripetalforce.Thissituationisshowninthefollowingfigure.

Thenetforceactingonthemotorcyclististhesumofthenormalforce(FN)andtheforcedue

togravity( ).

Theequationofmotionforthecentripetalaccelerationac,canbewrittenas:

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Normalreactionisprovidedbythespeedofthemotorcyclist.Attheminimumspeed(

11.AthincircularloopofradiusRrotatesaboutitsverticaldiameterwithanangular

frequency .Showthatasmallbeadonthewireloopremainsatitslowermostpoint

for .Whatistheanglemadebytheradiusvectorjoiningthecentretothe

beadwiththeverticaldownwarddirectionfor ?Neglectfriction.

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:

(ii) ,

(iii) ,

Tisthetensioninthestring.[Choosethecorrectalternative].

Ans.(i)Whenaparticleconnectedtoastringrevolvesinacircularpatharoundacentre,the

centripetalforceisprovidedbythetensionproducedinthestring.Hence,inthegivencase,

thenetforceontheparticleisthetensionT,i.e.,

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?

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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

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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

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:

=20×a

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Usingthefirstequationofmotion,thetime(t)takenbythebodytocometorestcanbe

calculatedas:

v=u+at

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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osbincbse.com CBSE Class 11 physics Chapter 5 Laws of Motion · Material downloaded from...

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