Physical Sciences Grade 12 Paper 1 Gauteng Mock 2015 Memo

Downloaded from hlayiso.com S GAUTENG PROVINCE EDUCATION > REPUSLIC OF SOUTH AFRICA PREPARATORY EXAMINATION 2015 MEMORANDUM SUBJECT: PHYSICAL SCIENCES: PHYSICS P1 (10841) FISIESE WETENSKAPPE: FISIKA V1 (10841)

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION — 2015 10841/15 PHYSICAL SCIENCES: PHYSICS/ FISIESE WETENSKAPPE: FISIKA (First Paper/Eerste Vraestel) MEMORANDUM QUESTION 1/VRAAG 1 14 . 4.2 1.3 1.4 1.5 1.6 V7 1.8 1.9 1.10 Avy AvY ByY AVY BvYY evv AVY Cyvy DvY Avy

Downloaded from hlayiso.com , : 1084115 QUESTION 2/VRAAG 2 24 2.2 2.3 The net force acting on an object is equal to the rate of change in the momentum of the object. Vv” Die netto krag wat op 'n voorwerp inwerk is gelyk aan die tempo van verandering in momentum vir die voorwerp. ¥ ” OR/OF When a resultant/net force acts on an object, the object will accelerate in the direction of the force at an acceleration directly proportional to the force “and inversely proportional to the mass of the object. / Wanneer ‘'n resultante/netio krag op ‘n voorwerp inwerk sal die voorwerp in die rigting van die krag versnel met 'n versnelling direk eweredig aan die krag Y en omgekeerd eweredig aan die massa.” (2) Accepted Labels/Aanvaarde benoemings w F,/F,/force of Earth on block/weight/49 N/mg/gravitational force F,/Fy/krag van Aarde op blok/gewig/49 N/mg/gravitasiekrag N Fy/normal F,/normaal f Friction/Fy Wrywing/Fr T Tension/Force in string/Fr Spanning/Trekkrag (van die tou)/ Fr F Fapo/15 N Fro 15.N Note/Nota NY FY One mark for correct arrow and label. TY lf any other forces shown max. % iv Een punt vir die korrekte py! en benaming wy Indien enige ander krag getoon word maks. % (5) fe =H,N © r +6) f, = (0,35)(49-15sin30°). ~ (oa X49 — LS = v aa (3)

Downloaded from hlayiso.com 24 OPTION 1/OPSIE 1 Right positive: Regs positief: For/Vir 5 kg Fy.=ma v ” Foos6+T -f =m(2,2) ~ 15cos30°+T—14,53'= 5(2,2)” T = 14,53 +15c0s30°+11 T =3352N Upward positive: Opwaarts positief: For/Vir B Fa =ma - 3392 +F, =m(22) ~ZBSU- 9.8m'= m(2.2) 39,52 =Thn m= 5S 0Tkg¥ 10841/15 OPTION 2/OPSIE 2 Upward positive: Opwaarts positief: For/Vir B Fig = ma v ~T4Fy = m(22) “ -T+98mS m(22) T =7,bm (A) Paralle! down positive: Ewewydig af positief: For/Vir5 kg Fi = ma ~Fcos0 +T —f = m(2,2) ~15cos 30° +T - 14,53 v= 5(2,2) T = 14,85 + 15cos 30° + 11 T= 3)SN (B) Set/stel (A) = (B) Thm =3%, 5 m=S0ikg v 6) [15]

Downloaded from hlayiso.com 10841/15 3.5 Upward positive/Opwaarts positief y (m) B Criteria for graph/Kriteria vir Marks/ tafiek: Punte Parabolic shape from A to C v A Paraboliese vorm van A tot C E t(s) Shape from C to D 4 : Vorm van C tot D Shape from D to E v D Vorm van D tot E Combination of negative parabola Vv Kombinasie van negatiewe parabole A siarts at zero Y A begin vanaf nul Downward positive/Afwaarts positief Note/NVota if no/incorrect labels on axis: max. 4 y(m) s as geen/verkeerde benamings vir asse dan: maks. : t (s) (5) [13]

Downloaded from hlayiso.com 10841/15 QUESTION 3/VRAAG 3 3.1 9,8 m-s® Vv downward/to the centre of the earth ¥ 9,8 m-s* “ afwaarts / na die middelpunt van die aarde “ (2) 3.2 Once/one time “ Een keer “ (1) 3.3 Upward “ Opwaarts”” qd) 3.4 OPTION 1/ OPSIE 7 OPTION 2/ OPSIE 2 Upward positive: Upward positive: Opwaarts positiet: Opwaarts positief. From t; to tp / Vanaf t; na te vi =v? + 2ady v — 3,2°"= 6? + 2(-9,8)Ay v Vv, =v, + adt , Ay = 131 m ¥ =~ 32 = 6 + (-9,8)At : At = 0,94s Ay = v,At + tadt’v Ay = (6)(0,94) + 3 (-9,8)(0,94)* ¥ Ay = 131m % OPTION 3 / OPSIE 3 OPTION 4/ OPSIE 4 Upward positive: Upward negative: Opwaarts positief. Opwaarts negatief: Vv, =V, +adt All the options 1 to 3 with all the ~32=6+(-98)At ¥ applicable signs changed. Maximum ae = ores hy At = 0,94s Al die opsies 1 tot 3 met die toepasiike tekens omgekeerd. ij 3, Ay = Yet ay Maksimum % Ay = 73.246 (0,94) v Ay =131m % (4)

Downloaded from hlayiso.com ; ; 10841/15 QUESTION 4/VRAAG 4 41 4.2 Each body in the universe attracts every other body with a force that js directly proportional tg the product of their masses “and inversely proportional to the square of the distance between their centres.“ Elke voorwerp in die heelal trek elke ander voorwerp aan met 'n krag wat direk eweredig is aan die produk van hul massas V en omgekeerd eweredig is aan die kwadraat van die afstand tussen hul middelpunte. “ (2) Gm.m. Fa = z 2 57 1% 10 = (6,67 x10-"1)(6,39 x10%)(108x10%) da? d? = 8,06 x10" d=898x10°m y -h=8,98x10° —3390x10° (6) h=5,59x108m v [8] QUESTION 5/VRAAG 5 5.1 5.2 The total finear momentum of a closed system remains constant (is conserved). vv Die totale line6re momentum in ‘n gesiote stelsel bly konstant (bly behoue). v OR/OF In a closed/isolated system the total momentum before collision is equal to the total momentum after collison. Vv“ In ’n geslote/geisoleerde sisteem is die totale momentum voor 'n botsing gelyk aan die totale momentum na 'n botsing. Vv (2) OPTION 1/ OPSIE 1 East positive: Oos positief. Zp=Zp, ¥ Myo) = MV + MV oy (54 +15) (0,08) 54v,,415(10.2) v Vyq=—0,2m-s7 Vy= 0,2m-s" west/wes “ OPTION 2/ OPSIE 2 West positive: Wes positief: =p, = Zp, MioVy = MV + MV or (54+ 15)(-0,08)= 54v,4.+15(-10,2) v Vy=0,2m-s" westtwes % (4)

Downloaded from hlayiso.com . 10841/15 5.3 OPTION 1/ OPSIE 1 East positive: Oos positief: FieAt=mv,—mv, v F,,, (4,2) = (1,5)(10,2) — (4.5)(0,08) ” Fist =12,65Neast/oos “ OPTION 2/ OPSIE 2 West positive: Wes positief- Fi.At=mv,-mv, v Fre (1,2) = (1,5)(-10,2) — (15)(-0,08) % Fret = —12,65 N Fg, =12,65Neast/oos % OPTION 3/ OPSIE 3 East positive: Oos positief: Vv, =v, tadt 10,2 = 0,08 + a(1,2) a=843m-s* Fis =Ma v Fre = 15(8,43) “ Fret = 12,65 N east / oos ¥ OPTION 3/ OPSIE 3 West positive: Wes positief. Vv, =V,+adt —10,2 = —0,08 + a(1,2) a=-843m- sg? Fi = ma v > Fret = 1.5(-8,43) ~ Fi, =-12,65N (3) F, = 12,65 N east / oos “ [9]

Downloaded from hlayiso.com 10841/15 QUESTION 6/VRAAG 6 6.1 6.2 6.3 6.4 A force for which the net work done in any closed path is dependent on the path the object travelled. vv OR/OF ‘n Krag waarvoor die netto arbeid verrig langs enige geslote pad. _afhanklik is van die roete geneem Vv OR/OF The net work done by the non-conservative force / frictional force depends on the path the object travelled vv / Die netto arbeid verrig deur 'n nie-konserwatiewe krag / wrywingskrag is afhanklik van die roete geneem vv OR/OF The mechanical energy is not constant v for a non-conservative force / Die meganiese energie is nie konstant vv vir ‘n nie-konserwatiewe krag nie (2) i =nN } Any one/enige een ¥ f= Fa f, =0,22%(869,8)xcos25° i, = 168,04N The net/total work done v’ on an object is equal to the change in the object's kinetic energy. “ OR The work done on an object by a resultant/net force is equal to the change in the object's kinetic energy. Die neito/totale arbeid V wat op 'n voorwerp verrig word, is gelyk aan die verandering in die kine in die kinetiese energie van die voorwerp. V OF Die arbeid verrig op 'n - voorwerp deur 'n resultante/netto krag is gelyk aan die verandering in die kinetiese energie van die voorwerp. (2) Accepted Labels/Aanvaarde benoemings w F,/F,/force of Earth on block/weight/842,8 N/ma/gravitational force F,fF,/krag van Aarde op blok/gewig/842,8 N/mg/gravitasiekrag N Fy/normal Fy/normaal f Friction/F; Wrywing/F; F Fapp/15 N Fiooo/15 N v N Fenan wy (4)

Downloaded from hlayiso.com . 10841/15 6.5 OPTION 1/ OPSIE 7 Parallel to the plane down is positive Ewewydig aan die viak af is positief Vee Ve + 2aAx v2 —v? = 2(1,54)(0,9) Ve -V) = 2,772 W,, = AE, ¥ Won + Weray + Wes 4tm(vi-v}) Woran + (86%9,8)sin25°(0,9)cos0° + 168,04(0,9)cos180°= £(86)(2,772) 7 Wan + 320,56 — 151,236 = 119,196 W.,, =-50,13] “ OPTION 2 / OPSIE 2 Parallet to the plane up is positive Ewewydig aan die viak op is positief Vz = Vi +2aAx vi -v? = 2(-1,54)(-0,9) ¥ viv? = 2,772 Wr = ABK v Wrun + Weay + Wy =im(; - vi) W,,, + (86 X9,8)sin25°(0,9)cos0° + 168,04(0,9)cos 180° $(86)(2,772) Wen + 320,56 — 151,236 = 119,196 Woe =—50,133 “ OPTION 3 / OPSIE 3 Parallel to the plane up is positive Ewewydig aan die viak op is positief Ra =ma Foon ~ Fsin€ +f = ma E,,,, —(86X9,8)sin25° + 168,04 = 86(-1,54) ” Faan = 356,18 — 168,04 — 132,44 Fug, = 55,7 N Woon = (55,79%(0,9)cosi80° 10

Downloaded from hlayiso.com 10841/15 OPTION 4/ OPSIE 4 Parallel to the plane down is positive Ewewydig aan die viak af is positief FP, =ma ~ F,, + F,sin€ +f = ma —E,,, + (86X9,8)sin25° — 168,04 = 86(1,54) Fi, = —356,18 + 168,04 + 132,44 Fun = 55,7 N Wun = Fain Wn = (55.7) ¥(0,9)c0s180° “ Wren = -50,13 Iv (5) [15] 14

Downloaded from hlayiso.com 10841/15 QUESTION 7/VRAAG 7 7.1 7.2 7.3 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.v Die verandering in die frekwensie(of toonhoogte) van die klank waargeneem deur ‘n luisteraar omdat die klankbron en die luisteraar verskillende snelhede het relatief t'die medium van voortplanting. V (2) OPTION 1 / OPSIE 1 viv f=” = “Vay, f, OR/OF f, NV, f, f= “eo Y¥ 430 v 335 -25 f, = 464,68 Hz viv R= f _ 335 464,68 X= 0,72m ¥ OPTION 2 / OPSIE 2 + f= Yt, Y ORVOF f,=—Y4, vey, v-V, Vv = (aes v 430” ay 335 - 25 335v_ ( 335 430 AL 335 - 25 X= 072m v (6) THE SAME; v The relative velocity of the train driver and the train is zerov’, There will therefore not be an observed change ¥ in the frequency heard by the driver. DIESELFDE V Die relatiewe snetheid van die treindrywer ten opsigte van die trein is nul. “ Daar sal dus nie 'n verandering in die waargenome frekwensie V (3) soos gehoor deur die treindrywer wees nie, - [11] 12

Downloaded from hlayiso.com 10841/15 QUESTION 8/VRAAG 8 8.1 Negative OR - v The direction of E, on the diagram is to the left “which is opposite to that of E2 which indicates the field for a positive charge. “ Negatief of — ¥ Die rigting van E; in die diagram is na links V wat in die teenoorgestelde rigting is as Ez wat die veld vir'n positiewe lading is. “ (3) 8.2 POSITIVE MARKING FROM QUESTION 8.1 POSITIEWE NASIEN VANAF VRAAG 8.1 Option 1/Opsie 1 Fret = Ee Ey Any one/enige een v - «E, =E,~ EB, ~ re ‘Enet (ex10°R,” (x10? \12x10°)% 600 “ (a6y (03h v ~ 25x10°Q, =1200- 600 Q, = 2,4 x1o%o ¥ 3 Option 2/Opsie 2 kQ, ¥ E, = re 9x10°\12x10") ¥ E,= 2 (0,3) E, = 1200 N-C" to the right / na regs Const, -E Any one/enige een” “E,=&, ~ Ene E, =1200 ~ 600°“ _ E, = 600 N-C" to the left/na links kQ E,= = bove exi0°}Q. “ Q, =4tkx10 FC (7) 13

Downloaded from hlayiso.com 10841/15 8.3 ceFy F = (146x107) (1600) ” F=256x10"" N tothe tight /naregs v ey 14

Downloaded from hlayiso.com 1084115 QUESTION 9/ VRAAG 9 9.1 Independent variable/Onafhanklike veranderlike: Total current in the circuit ~ Totale stroom(sterkte) in die stroombaan’ (1) 9.2 Cell; switch rheostat correctly drawn lh Ss and placed in series iF Sel, skakelaar en reostaat korrek jeteken en in serie met mekaar. pal ~ Ammeter in series with theostat. (7 / VA} Ammeter in serie met die reostaat. 7 Voltmeter in parallel with the Vv LN theostat and ammeter or cell. v Voltmeter in parallel met die YQ p teostaat en die ammeter of sel, (3) 9.3 . Ay dient =—*+ gracien AX Accept any legitimate coordinates . 10-8” taken from the graph gradient = Oni” Aanvaar antwoorde enige geldige gradient =—2 ko6rdinate afgelees vanaf die grafiek nf=20 7 (3) 9.4 GREATER THAN; “ (1) GROTERAS v [8] 15,

Downloaded from hlayiso.com 10841/15 QUESTION 10/VRAAG 10 10.1.1 RA, =R,+R,v R, = 3,7+34 R, = 7,1 QY (2) 10.12 e=IR+Ir vv 20= 1(71+1) vv 1=247A ¥ (3) 10.2 INCREASES; “ When the total resistance decreases v; the total current increases when emf and intemal resistance stay the same. “ NEEM TOE; V Wanneer die totale weerstand afneem, v sal die totale stroom toeneem as die EMK en die interne weerstand dieselfde bly. “ (3) 10.3 Option 1/Opsie 1; Option 2/Opsie 2: e=IR+Ir e=IR+Ir 20 = (2,31)R, +(2,31)(1} ¥ 20 = (2,31)R, + (2.31) R, =7,66 @ R, = 7,66 Q V=IR P=PR ¥ V = (2,31)(7,66) P = 2.31°(7,66) ¥ V =17,69V P = 4087Wv v2 P=— v Accept answers between R Aanvaar antwoorde tussen p — 1769" v 40,86 W en 40,88 W 7,66 (4) P= 40,86W ¥ [12] 16

Downloaded from hlayiso.com 10841/15 QUESTION 11/VRAAG 17 11.1. AC; ¥ The magnetic field changes continuously Vv which causes the current through the globe to alternate direction.“ WS; ~ Die magneetveld word kontinu omgekeer V wat veroorsaak dat die stroom deur die gloeilamp van rigting verander, V 11.2 The globe will light up/glow/bum. “ Die gloeilamp sal brand/ gloei, Vv (1) 11.3 The strength of the magnetic field/magnetic flux linking with the copper coil changes continuously. “ This causes an induced emf in the copper coil. “ A current is able to flow through the globe because the coil is part of that circuit” Die sierkte van die magneetveld/magnetiese fluks wat met die koperspoel skakel verander voortdurend. v Dit veroorsaak 'n geinduseerde EMK in die koperspoel. Vv ‘n Stroom kan deur die gloeilamp vioei omdat die spoe! deel vorm van daardie stroombaan. “ (3) 11.4 « Increase the speed/tempo at which the wheel/crank is turned. Increase the number or turns on the coil e Use a stronger magnet vv fany two] Verhoog die tempo waarteen die wiel/kruk gedraai word. Vermeerder die aantal windings op die spoel Gebruik ‘n sterker magneet ¥¥ Tenige twee} (2) 11.5 Option 1/Opsie 1: Option 2/Opsie 2: = 2 Pace = Vooslins P,, = Vin y 6 = (12)lins ¥ F Tne = 0,5 A 6 = (2) . R R=249 ¥ Tins =! Lex V2 =0 v2 ms = i 0 = a v2 2 7 O7A = Trax (4) [13] 17

Downloaded from hlayiso.com 10841/15 QUESTION 12/VRAAG 12 12.1 12.2 12.3 Workfunction is the minimum energy of a photon “needed to set an electron free from the surface of a metalv’ / Werksfunksie is die minimum energie wat ’n foton “ moet hé om 'n elektran vry te stel uit die opperviak van ’n metaal.v The wavelength is then inversely proportional to the maximum kinetic energy “ 1 (Ao ). E, max Wavelength is inversely proportional to the frequency (hot) of the radiation. “ The frequency is directly proportional to the maximum kinetic energy(foE,,..): “ if the cathode remains the same. “ Die golflengte is omgekeerd eweredig aan die maksimum_kinetiese energie” 1 ered } E k max Die gotflengte is onqekeerd eweredig aan die frekwensieY (uot \van die straling. Die frekwensie is direk eweredig aan die maksimum kinetiese energie” (fOE na) mits die katode dieselfde bly. “ STAYS THE SAME BLY DIESELFDE 18 ) (4) (1)

Downloaded from hlayiso.com 12.4 Qption 1/Opsie 1: E=W, +E, . he he ¥ Any one/Enige een 12 2 sagt 0 -34 3 34 3a Wa (6,63 x10 (8x10 )¥, (6,63x10 18x10 M2911 10°%,,? 160x10 255107 . 124x10 =78x10-° +.4555x10 "Vv, ? Vinx = 1.009x 107 Vinex = 401X108 mst ¥ Option 2/Opsie 2: g= he nr (6,63x 10 )(3x10°) 160x10° F=124xi0" J v W, = hf, (6,63x10™)(3x 10°) v W, =e 285x107 W, =7,8x1078 J K E-Wy =2(9,11x10v,,.? v 12410" —7,8x10°? =4(9,11x10" jv? Vinee = 1-009 10” Vinx = 1401x108 m-st % 10841/15 6) [12] GRAND TOTAL/GROOTTOTAAL: 150 19