cbse class 9 science important...
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MaterialdownloadedfrommyCBSEguide.com. 1/17
CBSEClass9Science
ImportantQuestions
Chapter8
Motion
3MarksQuestions
1.Anobjecthasmovedthroughadistance.Canithavezerodisplacement?Ifyes,
supportyouranswerwithanexample.
Ans.Yes,ifanobjecthasmovedthroughadistanceitcanhavezerodisplacementbecause
displacementofanobjectistheactualchangeinitspositionwhenitmovesfromone
positiontotheother.SoifanobjecttravelsfrompointAtoBandthenreturnsbacktopoint
Aagain,thetotaldisplacementiszero.
2.Afarmermovesalongtheboundaryofasquarefieldofside10min40s.Whatwillbe
themagnitudeofdisplacementofthefarmerattheendof2minutes20seconds?
Ans.Distancecoveredbyfarmerin40seconds=
Speedofthefarmer=distance/time=40m/40s=1m/s.
Totaltimegiveninthequestion=2min20seconds=60+60+20=140seconds
Sincehecompletes1roundofthefieldin40secondssoinhewillcomplete3roundsin120
seconds(2mins)or120mdistanceiscoveredin2minutes.Inanother20secondswillcover
another20msototaldistancecoveredin2min20sec=120+20=140m.
Displacement = 200= (asperdiagram) =14.14m.
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3.Atrainstartingfromarailwaystationandmovingwithuniformaccelerationattains
aspeed40km in10minutes.Finditsacceleration.
Ans.Sincethetrainstartsfromrest(railwaystation)=u=zero
Finalvelocityoftrain=v=
=
=
time(t)=10min= =600seconds
Sincea=(v–u)/t
4.Whatcanyousayaboutthemotionofanobjectwhosedistance-timegraphisa
straight-lineparalleltothetimeaxis?
Ans.Iftheobject’sdistancetimegraphisastraightlineparalleltothetimeaxisindicates
thatwithincreasingtimethedistanceofthatobjectisnotincreasinghencetheobjectisat
resti.e.notmoving.
5.Whatcanyousayaboutthemotionofanobjectifitsspeedtimegraphisastraight
lineparalleltothetimeaxis?
Ans.Suchagraphindicatesthattheobjectistravellingwithuniformvelocity.
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6.Atrainistravellingataspeedof .Brakesareappliedsoastoproducea
uniformaccelerationof .Findhowfarthetrainwillgobeforeitisbrought
torest.
Ans.u= =
,v=0(trainisbroughttorest)
v=u+at=25+(-0.5)xt
0=25–0.5x
0.5t=25,ort=25/0.5=50seconds
=
=1250–625=625m
7.Astoneisthrowninaverticallyupwarddirectionwithavelocityof .Ifthe
accelerationofthestoneduringitsmotionis inthedownwarddirection,
whatwillbetheheightattainedbythestoneandhowmuchtimewillittaketoreach
there?
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Ans.
v=0(sinceatmaximumheightitsvelocitywillbezero)
v=u+at=
0=5–10t
10t=5,or,t=5/10=0.5second.
=
=2.5–1.25=1.25m
8.Derivethesecondequationofmotion graphically?
Ans.letattimeT=0bodymoveswithinitialvelocityuandattime‘t’bodyhasfinalvelocity
‘v’anduntime‘t’itcoversadistance’s.
AC=v,AB=u,OA=t,DB=OA=t,BC=AC-AB=V-u
Areaunderav-tcurvegivesdisplacementso,
S=Areaof DBC+AreaofrectangleOABD (i)
Areaof DBC= Base Height DB BC
= t (v-u) (ii)
AreaofrectangleOABD=length Breadth
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=OA BA
=t u (iii)
S=ut+ t (v-u)
S=ut+ t at( useV-u=at)
S=ut+ at2
9.Acarmovingwithacertainvelocitycomestoahaltiftheretardationwas ,
findtheinitialvelocityofthecar?
Ans.V=0(comestorest)V=finalvelocity
S=62.5m
(retardation)
U=?
From3rdequationofmotion,
=2 (-5) 62.5
=-10 62.5
,
u= [u=25m/s]
10.TwocarsAandBaremovingalonginastraightline.CarAismovingataspeedof
80KMphwhilecarBismovingataspeed50KMphinthesamedirection,findthe
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magnitudeanddirectionof
(a)tivevtherelativeofcarAwithrespecttoB
TherelativevelocityofcarBwithrespecttoA.
Ans.(a)VelocityofcarA=80KMph
VelocityofCarB=-50kmph
(-vesignindicatesthatCarBismovinginoppositedirectiontoCarA)
RelativevelocityofcarAwithrespecttoB
=velocityofcarA+(-velocityofcarB)
=80+(-(-50))
=80+50
=+130KMph
+130KMphshowsthatforapersonincarB,carAwillappeartomoveinthesamedirection
withspeedofsumoftheirindividualspeed.
(b)RelativevelocityofcarBwithrespecttoA
=velocityofcarB+(-velocityofcarA)
=-50+(-80)
=-130kmph
ItshowsthatcarBwillappeartomovewith130kmphinoppositedirectiontocarA
11.Aballstartsfromrestandrollsdown16mdownaninclinedplanein4s.
(a)Whatistheaccelerationoftheball?
(b)Whatisthevelocityoftheballatthebottomoftheincline?
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Ans.u=initialvelocity=0(bodystartsfromrest)
S=distance=16m
T=time=4s
(i)From,
16= a 16
(ii)From,v=u+at
v=0+2 4
[v=8m/s]
12.TwoboysAandB,travelalongthesamepath.Thedisplacement–timegraphfor
theirjourneyisgiveninthefollowingfigure.
(a)HowfardowntheroadhasBtravelledwhenAstartsthejourney?
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(b)Withoutcalculation,thespeed,statewhoistravelingfasterAorB?
(c)WhatisthespeedofA?
(d)WhatisthespeedofB?
(e)ArethespeedofAandBuniform?
(f)WhatdosepointXonthegraphrepresent?
(g)WhatisthespeedofapproachofAtowardsB?
WhatisthespeedofseparationofAfromB?
Ans.(a)WhenAstartshisjourneyat4sec,Bhasalreadycoveredadistanceof857m
(b)AtravelsfasterthanBbecauseAstartshisjourneylatebutcrossesBandcoversmore
distancethenBinthesametimeasB
(c)SpeedofA=
Letatt=12min,distancecovered=3500m
(d)SpeedofB=
(e)SpeedofapproachofAtowardsB=375m/min-214m/min
=161m/min
(f)SpeedofseparationofAfromB=161m/min.
13.Abodyisdroppedfromaheightof320m.Theaccelerationduetothegravityis
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?
(a)Howlongdoesittaketoreachtheground?
(b)Whatisthevelocitywithwhichitwillstriketheground?
Ans.Height=h
Distance=s=320m
Accelerationduetogravity=
Initialvelocity=u=0
(a)froms=ut+
(b)
14.Derivethirdequationofmotion numerically?
Ans.Weknow;
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………(i)
…….(ii)
Where,v=finalvelocity
u=initialvelocity
a=acceleration
t=time
s=distance
Fromequation(i)t=
Putthevalueoftinequation(ii)
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15.Thevelocitytimegraphofrunnerisgiveninthegraph.
(a)Whatisthetotaldistancecoveredbytherunnerin16s?
(b)Whatistheaccelerationoftherunneratt=11s?
Ans.(a)Weknowthatareaunderv-tgraphgivesdisplacement:
So,Area=distance=s=areaoftriangle+areaofrectangle
Areaoftriangle=
=
=30m
Areaofrectangle=length breadth
=(16-6) 10
=10 10
=100m
Totalarea=180m
Totaldistance=180m
(b)Sinceatt=11sec,particlestravelswithuniformvelocityso,thereisnochangeinvelocity
henceacceleration=zero.
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16.Aboythrowsastoneupwardwithavelocityof60m/s.
(a)Howlongwillittaketoreachthemaximumheight ?
(b)Whatisthemaximumheightreachedbytheball?
(c)Howlongwillittaketoreachtheground?
Ans.u=60m/s; ;v=0
(a)Thetimetoreachmaximumheightis;
(b)Themaximumheightis;
(c)Thetimetoreachtopisequaltotimetakentoreachbacktoground.Thus,timetoreach
thegroundafterreachingtopis6sOrthetimetoreachthegroundafterthrowingis6+6
=12s.
17.Thedisplacementxofaparticleinmetersalongthex-axiswithtime‘t’inseconds
accordingtotheequation-
(a)drawagraphifxversustfort=0andt=5sec
(b)Whatisthedisplacementcomeoutoftheparticlesinitially?
(c)Whatisslopeofthegraphobtained?
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Ans.X=20m+(12)t
(i)Att=0
X=20+12 0=12m
(ii)Att=1
X=20+12=32m
(iii)Att=2
X=20+24=44m
(iv)Att=5
X=20+12 5=72m
(a)
(b)AtT=0(initially)
Displacement=20m.
(c)Slope=
18.Thevelocityofabodyinmotionisrecordedeverysecondasshown-
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calculatethe–
(a)Acceleration
(b)distancetravelledanddrawthegraph.
Ans.(a)Acceleration=slopeofthevelocitytimegraph
a=
(b)Distance
=600-300=300m
(c)
19.Drawthegraphforuniformretardation–
(a)position–timegraph
(b)velocity–time
(c)Acceleration-time
Ans.(1)Position–time
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(2)Velocity–time
(3)Acceleration-time
20.Thedisplacement–timegraphforabodyisgiven.Statewhetherthevelocityand
accelerationofthebodyintheregionBC,CD,DEandEFarepositive,negativeorZero.
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Ans.(i)ForAB,thecurveisupwardstoppingi.e.slopeisincreasingsovelocityispositive
andremainssameso,V=+vebuta=0
(ii)ForBC,curvehasstillhas+veslopeso,V=+vebutvelocityisdecreasingwrttimeso,
a=negative
(iii)ForCD,bothvelocityandaccelerationareZerobecauseslopeisZero.
(iv)ForDE,velocityisthe(visincreasingwrttime)andsoisaccelerationis+ve.
(v)ForEF,velocityis+ve(positiveslopeofx-tgraph)butaccelerationisZerobecause
velocityremainssomewithtime.
21.Derivethethirdequationofmotion asgraphically?
Ans.Letattimet=0,bodymoveswithinitialvelocityuandtimeat‘t’hasfinalvelocity‘v’
andintime‘t’coversadistance‘s’
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Areaunderv-tgraphgivesdisplacement
S=Areaof DBC+AreaofrectangleOABD
S=
Now,v-u=at
Putthevalueof‘t’inequation(i)
thirdequationofmotion
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