Physical Sciences Grade 12 Paper 1 Gauteng Mock 2019 Memo

Downloaded from hlayiso.com BEES, GAUTENG PROVINCE 3 *y' EDUCATION ‘wr REPUBLIC OF SOUTH AFRICA PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN 2019 MARKING GUIDELINES NASIENRIGLYNE 10841 PHYSICAL SCIENCES: PHYSICS (PAPER 1) / FISIESE WETENSKAPPE: FISIKA (VRAESTEL 1) 12 pages / bladsye

Downloaded from hlayiso.com 10841/19 GAUTENG DEPARTMENT OF EDUCATION / GAUTENGSE DEPARTEMENT VAN ONDERWYS PREPARATORY EXAMINATION / VOORBEREIDENDE EKSAMEN PHYSICAL SCIENCE: PHYSICS / FISIESE WETENSKAPPE: FISIKA (Paper 1 / Vraestel 1) MARKING GUIDELINES / NASIENRIGLYNE QUESTION / VRAAG 1 11 Dyvv 2 1.2 Dvv 2 1.3 Dvy A CORRECT 2 1.4 BYY = ANDC 2 1.5 Avv ANDB 2 1.6 Dvv 2 1.7 AVY 2 1.8 BYY 2 1.9 AVY 2 1.10 Dv¥v¥ 2 [20] 2 10841 Physical Science P1 Memo Bilcorrected (2)

Downloaded from hlayiso.com QUESTION / VRAAG 2 10841/19 2.1.1. When a net force acts on an object, the object will accelerate in the direction of the force” and the acceleration is directly proportional to the force and inversely T ¥ Tension/ Spanning 2.3 proportional to the mass~ of the object. OR The net (or resultant) force acting on an object is equal to the rate of change of momentum of the object in the direction of the net force. v¥ Wanneer n_netio krag op ‘n voorwerp inwerk sal die voorwerp versnel in die rigting van die netto krag. Die versneiling is direk eweredig aan die krag en omgekeerd eweredig aan die massa van die voorwerp. OF Die netto krag wat inwerk ap in voorwerp is gelyk aan die tempo van verandering van momentum van die voorwerp in die rigting van die netto krag. N/ Fx / Normal / Normaal ~ Ni Fu / Normal ” » ZONTFa 12N o Fag onaey Fo/w v¥ 3 kg block / 3 kg blok Fret = mav F,-T -f.= 3a 20 -T-1.2 = 3av 18,8 —T = 3a T=188-3a.... (1) 2 kg block | 2 kg blok Fret = ma 18,8 -3a = 2a a= 3,76 m:s* T = 2(3.76) v= 7,52 NY Set the two equations equal “ 3 10841 Physical Science P1 Memo Bilcorrected (2) Stel die twee vergelykings gelyk aan mekaar

Downloaded from hlayiso.com 10841/19 2.2.1 Each body in the universe attracts every other body with a force that is directly proportional to the product of their masses” and inversely proportional to the square of the distance between their centres. v Elke liggaam in die heelal trek elke ander liggaam aan met n krag wat direk eweredig is aan die produk van hul massas en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle middelpunte. 2 2.2.2 To the right as positive. Neem na regs as positief Force attracting mto earth,Fe = -G2%E (Ignore if the negative is missing.) Krag wat m aantrek na aarde for both correct expressions 9 ; mMy vir albei korrekte stellings Force attracting mto moon, Fu = Goat Krag wat m aantrek na die maan Net force on m, Fe + Fu = -GUe 4 ort = ov Netto krag op m - GUE + G mM yy 0 2. Ont 0; <a =0v 4 [20] QUESTION / VRAAG 3 3.1 The motion of an object in the force of gravity only. “” Die beweging van n voorwerp in gravitasiekrag alleen. 2 3.2 2/twiceY twee maal 1 3.3.1 OPTION / OPSIE 1 OPTION / OPSIE 2 OPTION / OPSIE 3 ve=vP + 2ahy ¥ V=vy + aAty vy=zyeo+ aAtv (107 v =O? + (2)-9,8)Ay ¥ 10 =0 + (-9,8)At ¥ ~10_= 0+ (-9,B)At ¥ “Ay = -5,1m ~At= 1,025 “At= 1,02 5 ieh=51m v oN, ~~ 0 10 Ay =viAt + % adt Ay = y= ( ) x1,02 ¥| ay= 0+ (A)-9,8)(4,02)~ —_s =-5,1m --5tm he5,1m¥ oh=51mv¥ 4 OPTION / OPSIE 4 OPTION / OPSIE 5 Woet = AK any one (U+K) =(U+K) } Woray = AK pow (mgh + % mv) =(mgh + % mv}, 2 mghcos® = % mv? -% mv? 2 (0,29,8)h(1) ¥ = (%4)\(0,2)(10"-0 ¥ ((0,2)(9,8)h +0) ¥ = (0 +(%X{0,2)(10)) ¥ uh=5,1mv ohe5,.mv 4 10841 Physical Science P1 Memo Bilcorrected (2)

3.3.2 3.4 Downloaded from hlayiso.com 10841/19 OPTION / OPSIE 1 ; OPTION / OPSIE 2 ve =v? + 2aAyv VYyavyieo+ aAt (OY = 7.5? + (2\(-9,8)Ay ¥ 0_=7.5 + (-9,8)At V v Ay =2.87m 4% - At = 0.775 ie. h=2.87m Ay=vAt + %adt? ay = 7.5 (0.77) _+ {(%\-9.8)0.77) ¥ = 2.87 mY « h=2.87m OPTION / OPSIE 3 Woe: = AK Woray = AK Y mghcos@ = % mv? -% mv? (0,2)(9,8)h(-1) = (4)(0,2)(0)" — (4)(0,27.5)? ¥ oh=2.87mv 5,1 mv (POSITIVE MARKING FROM 3.3.1) QUESTION / VRAAG 4 4.1 4.2.2 4.2.2 4.3 ¥ Isolated (system) or closed system Geisoleerde sisteem of gesiote sisteem Fret = Ap ¥ Fret (0,1) ¥ = 14 000 — 30 000 Fret =-160 000 N Fnet = 160 000 NY Xp; = Lp; } Any one/ Enige een | ¥ My¥4i F MV = MyVagt MoV2¢ 30 000 + (9000)(-1.5)“ = 14000 + (9000) ve) ¥ Ve =0.28m's" ¥ Foes Poet 20 20.1 20,2 203° t(s) on 20 adi 202 20,3 t{s) 5 10841 Physical Science Pi Memo Bilcorrected (2) [11] [12]

Downloaded from hlayiso.com 10841/19 QUESTION / VRAAG 5 5.1 5.1.1 A force for which the work done in moving an object between two points depends on the path taken. vv in Krag waarvoor die arbeid verrig om n voorwerp tussen twee punte te beweeg, is afhanklik van die roete wat gevolg word. 2 5.1.2 Wre= AEp + AEKY fAx cos 6 = AE, + AE, 18 x Ax cos 180° v= 5x9.8x(3-0) ¥ + % x 5x(0-9.907) Ax = 5.4458 (m) ¥ se 3 e= sin ; Gaal” 6 = 33.43°v 7 Include solution Wnet = Ack SEE LAST PAGE 5.2 5.2.1 The work done on an object by a net force” is equal to the change in the object's kinetic energy.“ OR The net work done on an object by a force” is equal to the change in the object's kinetic eneragy.~ OR The net work” done on an object is equal to the change in the object's kinetic energy. ~ Die netto arbeid verrig op ’n voorwerp deur 'n netto krag is gelyk aan die verandering in die voorwerp se kinetiese energie. Of Die arbeid verrig deur in netto krag is gelyk aan die verandering in die voorwerp se kinetiese energie. Of Die netto arbeid verrig op n voorwerp is gelyk aan die verandering in die voorwerp se kinetiese energie. 2 5.2.2 Wnet = AEkY Weg + Wea = FoAxcos® +Wapp = 2 Mp — 2 MVi2 (987)(9.8)(46) cos180° + Wapp = (9870.9) — OY Wea = 445339.34 JY 4 [45] 6 10841 Physical Science P1 Memo Bilcorrected (2)

Downloaded from hlayiso.com 1084119 QUESTION / VRAAG 6 6.1 The change in frequency (or pitch) of the sound detected by a listener,“ because the sound source and the listener have different velocities relative to the medium of sound propagation.” Die verandering in die waarneembare frekwensie van die klankgolwe as gevolg van die relatiewe beweging tussen die snelhede van die luisteraar en die klank Se voortplanting. 2 62, _ (v+v,) fs Vv * (vitv,)° gov = (940+9) sagv (340—v,) Vs = 41,099 m-s! v 5 63 63.1 910Hz~ 1 6.3.2 Positive marking from 6.2 (v+v,) = LS lo! £, (v+v,) 340-0 v = ——— 800V f 340+ 41000" = 713,73 Hz 4 6.4 800Hz 1 [13] 7 10841 Physical Science P1 Memo Bilcorrected (2)

Downloaded from hlayiso.com 10841/19 QUESTION / VRAAG 7 7.1 7.2 7.3 7.4 The magnitude of the electrostatic force exerted by one point charge (Q,) on another point charge (Q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them” if mass is used in definition 9/, Die grootte van die elektrostatiese krag uitgeoefen deur een puntiading op n ander puntlading is direk eweredig aan die produk van die grooites van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. Indien massas gebruik is in definsie o%, 2 v e Arrow and F correct labelv a8 © Pylen korrekte byskrif v Fa 2 kQ,Q 9x10°x4x10°x5x10° . F, = As v= ¥ = 4,5x104N 'Q2 due to O3 r? (20x10°)" kQ,Q 9x10°x2x10°x4x10° F ert = ¥ =3,2x104N Q2 due to 1 r2 a 5x1 0° y * Fret= + (4,5x10~4)? + (3,2x10-4)? V = 5,522 x 107 N ¥ 5 Take the right is positive / Regs is positief go -3 E-anc = KQy = x10" x2x10"" = 28 800 (N--C" to the left / na links) r (25x10) kQ 9x10°x4x10° 4 . Eanc= = = —— VY = 360 000(N-C"' to the right / na regs: 4nc 2 (10x10) ( ig gs) Enet = Eonc - Eanc Y This mark is for subtracting / Punt vir aftrek. = 360 000 - 28 000 = 331200 N-C" to the right / na regs v 5 [14] 8 10841 Physical Science P1 Memo Bilcorrected (2)

Downloaded from hlayiso.com 10841/19 QUESTION / VRAAG 8 8.1 Resistance within / offered by the cell” : 1 Weerstand gebied deur die sel self. 8.2 Temperature” 1 Temperatuur 8.3 Volts (V) V Potential difference / Potensiaal verskil 1 8.4 Internal resistance / interne weerstand VOR / OF r 1 85 7,2 1 8.6 slope / helling = AV/ Alv 3 =0-7.2“/08-0v =-9 r=9 Ov 8.7 POSITIVE MARKING FROM 8.5 4 7,2V [10] QUESTION / VRAAG 9 91 10VY (1) 9.2 OV OR/ OF zero (1) 93 141 + a RRs 10x5 Rp i 1 R,= RotRs = 10+5v=3,3330 = To + z v = Y= av Ry = 3,33 Rex = 3,333 +2 5,333 0 Rext = 3,33 +2 =5,330Y (3) 9 10841 Physical Science P1 Memo Bilcorrected (2)

9.4 9.5 Downloaded from hlayiso.com 10841/19 POSITIVE MARKING FROM 9.3 = Vext Rext = 1 v Vext _ _8 = Rext 5.333 =15(4) v emf = IRext + Ir y 10=8 + 1l5rv r=1330 Vv l= Option / Opsie 2 Vy = loRy “= 1,5 “x 3,333 = 4,9995 V Ve=1,5x2=3V Vext = VirtVs = 4,9995 +3 =7,9995 V Vint = & - Vex V= 10-7,9995V = 2,00 V Vint 2 Tint = =—= 13340 Vv in I 15 (5) increases ~ (Negative marking) vermeerder (negatiewe nasien) Circuit resistance increases v therefore (circuit) current decreasesv lost volts decreases / Vin decreases.” Stroombaan se weerstand vermeerder ¥ daarom sal die stroom afneem ¥ en die verlore volts sal verminder of die Vintern sal verminder. “ (4) [14] 10 10841 Physical Science P1 Memo Bilcorrected (2)

Downloaded from hlayiso.com 10841/19 QUESTION / VRAAG 10 10.1. 10.1.1 Slip rings ¥ Sleepringe (1) 10.1.2 coil/ solenoid ¥ spoe! / solenoid (1) 10.2 Connects the generator to the external circuit. OR Generated current exits the generator through C. “ OR Allows movement and conductivity between coil and external circuit. Verbind die generator aan die eksterne stroombaan OF Gegenereerde siroom veriaat generator deur D OF Laat beweging toe en geleiding tussen spoel en eksterne stroombaan (1) 10.3 parallel” (1) 10.4 = Imax/maks =25mA nx = v2 ¥ 25x10> v2" 17,678 x10 Av (3) lms 10.5 Positive marking from 10.4 Pay/gem = Inns J wok” xRv¥ = (17,678 x 40°)? x100 = 3,125x107 W v (3) [10] 11 10841 Physical Science P1 Memo Bilcorrected (2}

Downloaded from hlayiso.com 10841/19 QUESTION / VRAAG 11 11.1 11.2 11.3 11.4 11.5 Is the minimum energy that an electron in the metal needs to be emitted from the metal surface. “ ” Dit is die minimum energie wat n elekiron in die metaal benodig om vrygestel te word uit die metaal se opperviakte. Frequency of green light is below threshold frequency (f.). ““ OR Energy of green light lower than work function. OR Too low frequency. Frekwensie van groen lig is onder die drumpelfrekwensie OF Energie van groen lig is laer as die werksfunksie. Die frekwensie is te laag. Negative marking INCREASE ~“ Vermeerder Negative marking more photons sirike the metal per unit time’ hence more electrons ejected per unit timev’ thus rate of flow of charge increases hence increased current reading. Meer fotone tref die metaal per eenheid tyd daarom word meer elektrone vrygestel per eenheid tyd dus vermeerder die vioei van die lading dus word die stroom lesing verhoog. hf he Y any one / enige een 3.0x10° = p4yy, She vy 6,63 x 10 x 500x107 =3,98x 10? Uv E " 12 10841 Physical Science Pt Memo Bilcorrected (2) [11]