PHYSICAL SCIENCES Grade 11 TERM 1 RESOURCE PACK
[Pages:124]PHYSICAL SCIENCES Grade 11
TERM 1
RESOURCE
PACK
Gr11_Physical_Science_Term1_Resource_Book.indb 1
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Contents
Worksheets
3
Topic 1: Vectors in Two Dimensions
4
Topic 2: Newton's Laws and Application of Newton's Laws
13
Topic 3: Atomic Combinations (Molecular Structure)
33
Topic 4: Intermolecular Forces and Chemistry of Water
48
Formal Experiment
59
Technical Instructions
61
Newton's Second Law of Motion
64
Marking Guidelines
74
Assessments
81
Topic 1: Vectors in Two Dimensions
82
Topic 2: Newton's Laws and Application of Newton's Laws
89
Topic 3: Atomic Combinations (Molecular Structure)
101
Topic 4: Intermolecular Forces and Chemistry of Water
113
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WORKSHEETS
Gr11_Physical_Science_Term1_Resource_Book.indb 3
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RESOURCE PACK
Topic 1 ? Vectors in Two Dimensions
WORKSHEET
1. A forward horizontal force of 50 N is applied to a crate. A second horizontal force of
180 N is applied to the crate in the opposite direction. Determine the magnitude and
direction of the resultant force acting on the crate.
(2)
2. The athlete at point A runs 150 m east, then 70 m west and then 100 m east. Determine
the resultant displacement of the athlete relative to point A.
(2)
3. Three forces act on an object in the vertical plane. Two forces of 500 N and 300 N act
vertically upwards and the third force of 600 N acts vertically downwards. Determine
the resultant force acting on the object.
(2)
4. An aircraft flies 8 km north from an airport and then 12 km east.
4.1 Use the tail-to-head method to draw a neat labelled displacement vector diagram.
Draw in the resultant displacement vector.
(3)
4.2 Determine the magnitude and direction of the resultant displacement of the aircraft
relative to the airport.
(5)
5. Draw the following force vectors (not necessarily to scale) in the Cartesian plane acting from the origin:
5.1 F1 = 20 N acts to the right in the horizontal plane.
5.2 F2 = 50 N acts upwards in the vertical plane.
5.3 F3 = 80 N acts to the left in the horizontal plane.
5.4 F4 = 90 N acts downwards in the vertical plane.
(4)
6. Consider the four forces in question 5.
6.1 Determine the resultant horizontal force and the resultant vertical force.
(4)
6.2 Use the tail-to-tail method to calculate the magnitude and direction of the resultant
of the two forces in Question 6.1. Draw a neat labelled vector diagram to support
your calculation.
(7)
6.3 Now use the tail-to-head method to draw a neat labelled vector diagram.
(3)
7. A drone at point P is flown 2 km north, then 5 km east and then 8 km south. Find the
magnitude and direction of the resultant displacement of the drone from point P. Draw
a neat labelled displacement vector diagram to support your calculation.
(8)
4Grade 11 Physical Sciences
Gr11_Physical_Science_Term1_Resource_Book.indb 4
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WORKSHEETS
TOPIC 1: VECTORS IN TWO DIMENSIONS
8. Resolve each of the following vectors into their perpendicular components. In each case draw a neat, labelled vector diagram (not necessarily to scale).
8.1 A 100 N force which acts at 30? above the horizontal.
(6)
8.2 A displacement of 280 m on a bearing of 070?.
(6)
8.3 A 400 N force which acts at 50? to the vertical.
(6)
8.4 A 500 N force which acts at 40? bTeolopwitche1h:oVrizeocnttaol.rs in two dimensio(6n)s
9. Three forces T, F and W act on an object as shown in the diagram below. 9. Three forces T, F and W act on an object as shown in the diagram below.
F = 300 N
T = 100 N
30?
60?
W = 100 N
9.1 Determine the perpendicular components of force F.
(6)
9.2 Determine th9e.1peDrpeetenrdmicinuelarthceompepropneenndticsuolfafrocrocme Tp.onents of force F.
(6)
9.3 Determine th9e.2maDgenteitrumdienaentdhediprecrptieonndoicfutlhaer rceosmulptaonntefnotrscoefafcotricneg Ton the object. Draw a neat l9a.b3elDleedtevermctionredtihaegrmamagtnoitusudpepaonrdt ydoiruercctiaolncuolaf ttihoen.resultant force ac(8ti)ng on the object.
Draw a neat labelled vector diagram to support your calculation.
Gr11_Physical_Science_Term1_Resource_Book.indb 5
Term 15
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RESOURCE PACK
CONSOLIDATION EXERCISE
TOPIC 2: NewTotopnic's1:LVaewcstoarnsdInTAOTpwTpAolLic:D4a0itmiMoeAnnRoKsSfioNneswton's Laws
1. Define a resultanGtTRvOeAcDtPoEIrC.11:2C:ONNSeOwTLoItDopAniTcI'Os1N:LVaQUewEcsStToaIrOnsNdSInATpwpolicDaitmioennosfioN(2n)eswtoTOnT'AsLL: a4w0 MsARK
2. A block is puGlleR1dA.bDyEtDw1e1o:fihnCeoOraNizrSoenOstuLalIltDafoAnrtTcveIOescNotfo1Qr.5UNESaTnIdO2N5SN to the left, and by two
TOTAL: 40 MARKS
horizontal f1o.rce2s.Doef f1iAn0ebNlaoacrnkedsisu3l0ptauNnlltetvdoetbchytoetrwr.igohhto. rizontal forces of 15 N and 25 N to the left, and by two
(2
2.1 Draw th2e. se fAoubrlhofoocrkricziesovnpetuaclltleofodrsrcbienysttwhoefo1Ch0aorNrtizeaosinnadtna3lp0floaNrnceetos(atochftei1nr5gigNahtta.thnedo2r5igNint)o.the lef(t2, )and by two
hori2z.o1ntDalrafowrctehsesoef 1fo0uNr faonrcde3v0ecNtotorsthinethrieghCt.artesian plane (acting at the origin).
2.2
Find
the
mag2n.1it2uD.d2reaaFwnintdhddetsihreeecfmotiuoarngfnooirftuctedheeveraecnstduorldtsairnientctfthoioerncCeo.af rttheesiarenspulatannet
(foarcctien.g
at
(2) the
origin).
(2
3. A learner hikes38.2k.m2 AFnsointrudtdhtehanentdmh1iak2genksimt8udkwemesaNtn. odUrstdhiinraegncttdhioetnhteaoif1l-t2thoek-mhreeWsadueltmsatne. ttUhfosoirdnc,ged.tehteertmaiil-ntoe-head method,
(2
the
magnitude 3.
anAdsdtduiredetecetnrimtohninikoeef stthh8ee
kmlemaargNnnoeitrrut'shdreaesnaudnltdtahndetir1de2icstkpiomlancWoemfetshetne.tUs.tsuidnegntth'se
rteasilu-tlota-hnet(5ad)dispmlaectheomde, nt.
4. Three forces, A,4B.deanteTdrhmCreinaeectftoohrneceamsn,aoAgbn,jiBetucadt neadsasCnhdoawdcitnreoicnntiatohnneoodbfijatehgcertasamtsu.dsDheeontwte'srnmrienisntuehltetahndetiadgisrapmla.cDemeteenrmt. ine the
(5
magnitude a4n. d dTirhercemtieoanfgonorcifteutshd,eeAra,ensBduladtnairndetcCftoioarnccetooafcntthianengroeobsnujetltchatenaotsbfosjerhccoet.wanctiinngthoendtihaegroabmje. cDt.etermine the
magnitude and direction of theBr=es4u0ltaNnt force acting on the object.
B = 40 N
A = 60 N A = 60 N
C = 10 N
C = 10 N
5. Consider the three forces acting on an object as shown below. (6)
(6
5. Consider th5e.threCeofnosrcideesratchteintghoreneafnorocbejsecatcatsinsghoown naninotbhTjee=cdt5iaa0gs0rsaNhmowbenlobwe.low.
F = 300 N
T = 500 N
F = 300 N
20?
45?
20?
45?
W = 200 N
W = 200 N
5.1 Calculate the horizontal and vertical components of force T.
5.1 Calculate the5.h1o5rCi.z2aolncCutaalallctauenlatdhtevethrhtoeircihazlocrnioztamolnpatoannldaevnnetdsrtviocefarfltoicroacmel cTpo.omnpeonntsenotfsfoorfcfeorTc.e F.(4)
(4
5.2 5C.3alcDuelatetermthineehtohreizmonatganl iatunddevaenrtdicdailrceocmtiopnoonfetnhtes oref sfourltcaentF.horizontal force.
(4
5.2 Calculate the horizontal and vertical components of force F.
(4)
5.3 5D.4eteDremteinrme itnhe tmheagmnaitgundietuadnedadnidredcitrioenctioofnthoef trheesureltasunlttahnotrivzeorntitcaal lfoforcrcee. .
(2
5.3 Determine th5e.4m5Da.5genteUitrsumidninegeatnhthedetdamiirle-atcgotn-itoiatnuildomefetahtnhedordde,siruderlctaatwinotnahonofertiahztoelnartbeaeslulfleoltdarcnfeot.rvceertdicialg(fr2oa)rmcea. nd determine
(2
5.5 Usinthgethmeatganili-ttuod-teaial nmdedthiroedc,tidornawof athneeraetslualbtaenllet dfofrocreceacdtiinaggroanmthaendobdjeetcetr.mine
the magnitude and direction of the resultant force acting on the object.
(5
6Grade 11 Physical Sciences
4 Gr11_Physical_Science_Term1_Resource_Book.indb 6
Grade 11 PHYSICAL SCIENCES Term 1
2018/12/24 8:21:36 AM
WORKSHEETS
TOPIC 1: VECTORS IN TWO DIMENSIONS
5.4 Determine the magnitude and direction of the resultant vertical force.
(2)
5.5 Using the tail-to-tail method, draw a neat labelled force diagram and determine the
magnitude and direction of the resultant force acting on the object.
(5)
PHYSICAL SCIENCES RESOURCE PACK
6. The diagram below shows a top view of a tug boat that is pulled by two forces. North is GRADE 11 shown. TERM 1 Use the component method to find the magnitude and direction of the resultant force
WORKSHEETS acting on the tugboat. State the direction as a bearing, clockwise from north.
PAGE 7 DIAGRAM FOR QUESTION 6.
(6) PAGE 8 QUESTIO N 4.2 [40]
x2 = 82 + 122
x = 208 = 14,42 km
tan
=
12 8
= tan-1 182 = 56,31o
x = 14,42 km on a bearing of 056,31o
PAGE 9 DIAGRAM FOR QUESTION 7
Gr11_Physical_Science_Term1_Resource_Book.indb 7
Term 17
2018/12/24 8:21:37 AM
RESOURCE PACK
Topic 1: Vectors In Two Dimensions
MARKING GUGIRDAEDLEI1N1: EWSORKSHEETTMoEpMicOR1A:NVDeUcMtors In Two Dimensions
1. CThheono:seRfo=rw+a5r0d1+a.s^t-hG1eR8pCT0AohhhsDeoi=tnEoi-:vs1eeR11:(f3+oW0=)rw=dOa+irR1red53Kc00atSisoNH+ntEh.^Ee-Tp1TobM8soa0icEthpkiMvwe=iOca(rR+-d1A)s:1dN3iVrD0e eUc=tMciotn1o.3r0sN In TbwacokwDar(d2ims) ensions
2. 3. 4.
CTCT4.hhhh1eeoonnoo::sseeTReuxap=s=wt+a+ar5sd10ts5h03240ae...+1s+p312t3oh^...0s-ei0tCTCTp127i+vhhGo0..hheees^hooRCTCTCTi-(nnoo+tA+hhhhhhi::ss6veee)ooo1Deee0DRnnnCTCT0ooodE0ue(:::xhh0sssihh+hrpa=eeeeeoo1DRRe=)=ws1=nnooc:ufextdo::ta+ss1pa==iaWirreeo+r8DRwsw=ds25en0tfesOxa10ca++to.ahat=mr50r+Raridsde55sw=0os+K=st1001tnaa+pht5a0S+a.3hsr+o22e0s+d5sH0+es0(t10ph0ti-a0teEkth+p53h^oieasEv+emo-N00+e7ss(eTpstt0i0-p1^htpio+i(-)t^o8ve+Moi+s7-v es0(si)+60pe^tEi-h1iit(0d-)toviuM1i+8(vv0ies7=+pr0e0)+6eiheO0)wt0(h0dic()+1vR(d-0+=ait+re)0=ii=Arhe)+ro)10dde(cndN+3d+si1c=t-1r.i0ii=Drte)0ro82ieeo1c0dUn00+c =cn3t1i.m0itMrto0.i=82eioon100c=nn.=3mt01.i.0o82ENn1=00.3am00s2NEtN0a0sNbtEaacuskpb((twwa22ua))capkrrdwwdssaarrddss
4.1 3. Choose upwards as the positive (+) direction.
4.81 k8mTkhmen:R = + 51002+1kx32m00kx+m^- 600h = + 200 = 200N(3) upwards
4.2 tanxiixx2 ===== t8118a4222n{,0 4-+45812b.. 12k=182m4221 .{l24o{,n4txxx 4a=.22a2n==5k=tbxxxa6m=e21n,a==834=2{rt2xx11,i=a084+21nc28n882=4g2=12k1,=k082o4=+=m28fm228t12ak0b=2112=0mn5e8+422a86t-,a1r1b,41=F(3ine22=42n1a1-2k=8,tg12rc4am51(i{25)n0n41k68-Ng,2=m1,435()21165k8=?2,6m3,)135?=16?, 3x516?, 31?
(5)
5.
5.
x = 14, 42km bearing 56, 31?
F3 = 80 N
FF21==5200N
5.
F3 = 80 N
F21 = 520 N
F3 = 80 N F4 = 90 N
F1 = 20
6.1 Rx = + 20 + ^- 80h =F4 =-906N0 = 60N left
6.1
Ry
= + 50 + ^- 90h = - 40 = 40N downwards Rx = + 20 + ^- 80h =F4 -= 9600N = 60N left
(4)
6.
6.1
Rx
=
+
20
+
^6-.280h
Ry =6-.160
==R6x0+=N506l+0e+fN2t^0{-+9^0-h
= - 40 80h = -
= 60
40N downwards = 60N left
Ry
=+
50
+
^- 960.h2=
-
40
40
N=R4y0 =N
d+oR5w600n+Nw^a4-r0dN9s0{h = - 40 = 40N downwards(4)
40 N 60 N
40 N
6R.22 =
R
602 + 402
60 N
R
=
5200
40=N
60 N
72, 11NR
40 N
tan
R2 Ri
= ==
56462000020=+=4t0a72n2-6,101(1N64N00
)=
33, 69?
R =tan72iRR, 12==1=N6456a0020t0230=+3,=4t6a09n72?2-b1,e(1lo614w0N0 )th=e h3o3ri,z6o9n?tal
8Grade 11 Physical Sciences
R
=ta7n2i,
1=1N
4a0t 60
3=3,
6t9a?nb-1e(lo64w00
t)he=ho3r3i,z6o9n?tal
6
R = 72, 11N at 33,G6r9ad?ebe11lowPHthYeSIhCoArLizSoCnIEtaNlCES Term 1
6
Gr11_Physical_Science_Term1_Resource_Book.indb 8
Grade 11 PHYSICAL SCIENCES Term 1 2018/12/24 8:21:42 AM
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