Physical Sciences Grade 12 P1 Free StateNovember 2014 Memo

_ Downloaded from hlayiso.com education Department of Education FREE STATE PROVINCE PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN GRADE 12/GRAAD 12 PHYSICAL SCIENCES P1 (PHYSICS) FISIESE WETENSKAPPE V1 (FISIKA) SEPTEMBER 2014 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 11 pages. Hierdie memorandum bestaan uit 11 bladsye. Copyright reserved/Kopiereg voorbehou

Downloaded from hlayiso.com Physical Sciences P1 (Physics) 2 FS/VS/September 2014 Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum SECTION A QUESTION 1/VRAAG 7 11 Ccyvy (2) 12 Dvwy (2) 1.3 Cvy (2) 14. Cvv (2) 15 Dvv (2) 16 Dvv (2) 17 Byv (2) 18 By (2) 19 Dvv (2) 110 Dvv (2) [20] TOTAL SECTION A/TOTAAL AFDELING A: 20 Copyright reserved/Kopiereg voorbehou

Downloaded from hlayiso.com Physical Sciences P1 (Physics) 3 Grade/Graad 12 Prep.Exam./Voorb.Eksam SECTION B/AFDELING B QUESTION 2/VRAAG 2 2.1.1 2.1.2 2.2 FS/VS/September 2014 Memorandum Upwards positive/Opwaarts positief: Vp =VjtaAty 0 =30v +-9,8At At = 3,06 sv Downwards positive/Afwaarts Ve =Vj taAty 0 =-30V +9,8At At = 3,06 sv Upwards positive/Opwaarts positief: Ay =v At+taat? Vv = 30(4) v + 4(-9,8)(4)°v =41,.6m Height /Hoogte= 41,6 mv” Downwards positive/Afwaarts positief: Ay =v At+taat? v = - 30(4) v + 14(9,8)(4)2v =-41,6m Height/Hoogte = 41,6 mv OPTION 1/OPSIE 1 Upwards positive/Opwaarts positief: A > Time/tyd(s) Criteria for graph/Kriteria vir grafiek: pom Shape of first part of graph up till maximum height. v Vorm van eerste deel van grafiek tot by maksimum hoogte. Shape of second part of graph and above zero height at t = 4s. v Vorm van tweede deel van grafiek en bokant zero hoogte by t= 4s Ground not zero position (provided everything else is correct): (2) Grond nie zero posisie (indien die res korrek is): 4% Copyright reserved/Kopiereg voorbehou (3) (4)

Downloaded from hlayiso.com Physical Sciences P1 (Physics) 4 FS/VS/September 2014 Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum OPTION 2/OPSIE 2 Downwards positive/Afwaarts positief: 4 E zo a ap ! Time/tyd(s) v Criteria for graph/Kriteria vir grafiek: pe Shape of first part of graph up till maximum height. v Vorm van eerste deel van grafiek tot by maksimum hoogte. Shape of second part of graph and above zero height at t = 4 s. v Vorm van tweede deel van grafiek en bokant zero hoogte byt = 4s Ground not zero position (provided everything else is correct): 14 (2) Grond nie zero posisie (indien die res korrek is): 4% 2.3 Positive marking from 2.1.2/ Positiewe merk vanaf 2.1.2. Upwards positive/Opwaarts positief: Vv,’ =v,? + 2ady ¥ 0_= (24)? + 2(-9,8)Ay~” Ay = 29,39 m Total dispalcement = -41,6 +_29,39 Totale verplasing Positive marking from 2.1.2/ Positiewe merk vanaf 2.1.2 Downwards positive/Afwaarts positief: VP =v, + 2ady O = (-24)? + 2(9.8)Ay~ Ay = -29,39 m Total dispalcement = 41,6 - Vv 29,39 =-12.21m Totale verplasing = 12.21 mv =12,21mv downwards” /afwaarts downwardsv /afwaarts Copyright reserved/Kopiereg voorbehou (8)

Downloaded from hlayiso.com Physical Sciences P1 (Physics) 5 FS/VS/September 2014 Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum 2.4 InelasticY The velocity/speed with which the ball bounces from the ground is less than/differs from the velocity/speed at which the ball strikes the ground. ¥ The kinetic eneray is not conserved/changes. Y Onelasties Die snelheid/spoed waarmee die bal vanaf die grond bons is minder as/verskillend van die snelheid waarmee dit die grond tref. Die kinetiese energie word dus nie behou nie/verander. (3) [17] QUESTION 3/VRAAG 3 3.1 3.2 3.3 3.4 3.5 Newton’s First Law of motion. Vv A body will remains in its state of rest or motion at constant velocity” unless a non-zero resultant/net force acts on it.” Newton se Eerste bewegingswet. 'n Liggaam sal in sy toestand van rus of konstante snelheid beweging volhard tensy ‘n nie-nul resultante/netto krag daarop inwerk. (3) A system in which no net external forces are acting on the spaceship.v” 'n Sisteem waarin geen netto eksteme kragte op ruimtetuig inwerk nie. (2) F,At =Ap F,.At=mv, —mv, (12 000)(30) v= 3500(v;~ vi) v Av = 102,86 m-s” (3) TF fF triction/friction” /Frywing, WryWing Fg/wiweight/force of gravity” /gewig, gravitasiekrag (2) (3500)(9,8) Vv +f= as00(300—600, vv f = -156 800 N ..f = 156 800 N ¥ (5) [15] Copyright reserved/Kopiereg voorbehou

Downloaded from hlayiso.com Physical Sciences P1 (Physics) 6 FS/VS/September 2014 Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum QUESTION 4/VRAAG 4 41 Newton's Second Law of motion. “ When a resultant/net force acts on an object, the object will accelarates in the direction of the net force at an accelaration which is directly proportional to v the force and inversly proportional to the mass of the object. v Newton se Tweede bewegingswet. Indien ‘n resulterende/netto krag op 'n voorwerp inwerk, versnel die voorwerp in die rigting van die netto krag met 'n versnelling wat direk eweredig is aan die krag en omgekeerd eweredig aan die massa van die voorwerp. (3) 4.2 fs(max) = UsFnV = 0,35(500)(9,8)v” = 1715 NV (3) 4.3 Fappiied/ 10000 NY f/F dF riction/friction” <= ,_e=_, Ty (3) 4.4 Left positive: /links positief: } Car: Fre = ma ¥ T+f= ma T v - (0,2)(500)(9,8) “= 500(3,04)v T= 2500 NY OR/OF Truck/Trok: Fret = ma js Fappiied/Ftoegepas + T+f= ma 10 000 v + T - (0,2)(1500)(9,8) v= 1 500(3,04)v T = -2500 N “.T = 2500 NY (5) [14] Copyright reserved/Kopiereg voorbehou

Downloaded from hlayiso.com Physical Sciences P1 (Physics) 7 FS/VS/September 2014 Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum QUESTION 5/VRAAG 5 5.1 Yes, ¥ no other forces other than the gravitational force is acting on the block.” Ja, geen ander krag behalwe gravitasiekrag werk op die blok in nie. 5.2 Whet = AK ¥ = %4(3)(6)-0v = 54JV 5.3 | OPTION 4/ OPSIE 4 OPTION 2/ OPSIE 2 (U + Kya = (U + K)a ¥ Wret = AE/AK (3)(9.8V(6) + Ov = (3)(9,8)h + (316) || (3)(9,8)Aycos0® v= 14(3)(62) - Ov h= 4,16 mv Ay = 1,84m “h=6-1,84 = 4,16 mv Positive marking from 5.3/ Positiewe merk vanaf 5.3 5.4 OPTION 1/ OPSIE 4 Wre= AU + AK Y/Wre= AE, + AEx (f)(5)cos 180° = (3)(9,8)(1) - (3)(9,8)(4,16) VY + % (3)(2)? - % (3)(6)? V f= 28,18NV OPTION 2/ OPSIE 2 Whet = AEWAK W, + Wy = AK (f\(5)cos180°V + (3)(9,8)(4,16 — 1)cos 0° v= % (3)(2)? - % (3)(6)? ¥ f= 28,18NV QUESTION 6/VRAAG 6 6.1 Red shift is a shift in the spectra of distant stars ~ towards the longer wave lengths/red end of the spectrum.” Rooi verskuiwing is ‘n verskuiwing in die spektra van verafgeleé sterre na die langer golflengte/rooi kant van die spektrum. 6.2 Moving away from the earth’ The wavelength of the light coming from the distant star has increased/ shift towards longer wavelengths. ” Beweeg weg van die aarde af. Die goifiengte van die lig komende van die ster het toegeneem/verskuif na die langer golglengtes. Copyright reserved/Kopiereg voorbehou (2) (3) (4) (5) [14] (2) (2)

Downloaded from hlayiso.com Physical Sciences P1 (Physics) 8 FS/VS/September 2014 Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum 6.3.1 The difference between the speed of the car and the speed of light is very large.Y The frequency shift/fincrease is too small to observe.” Die verskil tussen die spoed van die kar en die van lig is baie groot. Die frekwensieskuif toename is dus te klein om waar te neem. (2) 6.3.2 32 4, =YEMLt onioF f -——4, v viv, -V, 340 of, = ————-v (1200) v t * 340-77,78 © (120) = 1555,95 Hzv (4) [10] QUESTION 7/VRAAG 7 7A Electrostatic force experiencedv per unit positive chargev’ at that point./Die elektrostatiese krag ervaar per eenheid positiewe lading by daardie punt. (2) 7.2 E, -¢ v 9x10" x12x10° Vv 0,3? =1,2x 10° N.C1” (3) 7.3 Fry” ; Note: No deduction of Fy” marks for labels. *. Let wel: Geen punte aftrek vir byskrifte nie. (2) 7.4 kQ,Q Fy = 72 zy 9x10? x8x 107 x10x10° 7 0,4? =4,5x10°N Fret? = Fxy? + Fy? = (9,6 x 10%)? + (4,5 x 10%)? Fyz Fret = 1,06 X 10°N S ety ny s 4,5x10 Tan@ = ——__v 9,6x10° Fxy © = 25,11° ©. Fret = 1,06 X 10°N»% 295,11°/ 25,11° North of West/Noord van wes 25,11° upwards from negative x-axis »/opwaarts vanaf negatiewe x - as (6) [13] Copyright reserved/Kopiereg voorbehou

Downloaded from hlayiso.com Physical Sciences P1 (Physics) 9 FS/VS/September 2014 Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum QUESTION 8/VRAAG 8 8.1 21 J of work done /energy transfered” per coulomb/unit charge that is moved between two points in an electric field./ 21 J arbeid verrig/energie oorgedra per coulomb/eenheid lading wat tussen twee punte in ‘n elektriese veld beweeg. 8.2.1 ma v T Vi= 24-21 Je ee 37 a 1 =3A V=IR = (3)(3) ¥ = OV V2 = 21-9 = 12Vwv 8.2.2 | Positive marking from 8.2.1/ OPTION 2/ OPSIE 2 Positiewe merk vanaf 8.2.1 OPTION 1/ OPSIE 7 E=\(R+9 v 24= 3(R + 1) Vv R=— R=79 Rp =7-3v -V =40 =70 eee ee Rp =7-3¥ R, R, R, =40 1,.41,1, Lie aipeniay. 4° 6'R R, R, R, “R=12Q0 1 11 Ratio of current in parallel v= —4V branches:2:1/Verhouding van stroom in 4 6R a parallelle vertakking: 2:1 R=120 Ie=1A Vp = 21-9 p= IR? nee = (1) ¥ (12) ¥ pi =12 WY ‘ v 12? P= 12 ¥ = 12 Wv 8.3 Increasev /vergroot Total resistance in circuit decrease.v/Totale weerstand in stroombaan verklein. Total current increasev (and through 3 resistor)./Totale stroom vegroot Therefore: Power increase according to P= I?R for R (30) staying constant.v/ Daarom: Drywing vergroot ooreenstemmend met P = FR vir R (3 Q) bly konstant,. Copyright reserved/Kopiereg voorbehou (2) (4) (8) (4) [18]

Downloaded from hlayiso.com Physical Sciences P1 (Physics) 10 FS/VS/September 2014 Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum QUESTION 9/VRAAG 9 9.1.1 Electromagnetic Inductionv /elektromagnetiese induksie (1) 9.1.2 | Maximumv/ maksimum (1) 9.1.3 Av The rate of change in the magnetic flux” is a maximumv at position C./ Die tempo van verandering in die magnetiese vioed is ‘n maksimum by punt C. (3) 9.1.4 v vw ' Mirror image of graph / 4 also correct. s BC Ts) Spieélbeeld van grafiek ‘= ook korrek. o (4) 9.2.1 AC voltages can be stepped up or down by making use of transformers for different users. OR Electrical energy can be transmitted over long distances” with low current and high voltage. (ANY ONE) WS spanning kan m.b.v transformators verhoog of verlaag word vir verskillende vebruikers OF Elektriese energie kan oor lang afstande oorgedra word deur lae stroom en hoé spanning. (ENIGE EEN) (1) 9.2.2 Pmax = Viaxlmax 300 VY = 80(Imax) ¥ (Imax) = 3,75 A = 2,65 AV (5) 9.2.3 (3) [18] Copyright reserved/Kopiereg voorbehou

Downloaded from hlayiso.com Physical Sciences P1 (Physics) 1 FS/VS/September 2014 Grade/Graad 12 Prep.Exam./Voorb.Eksam Memorandum QUESTION 10/VRAAG 10 10.1.1 Line emission (spectrum) /Lynemissie(spektrum) (1) 10.1.2 Driffraction gratingy / Diffraksierooster Prism / Prisma (Any one/ Enige een) (1) 10.2 The frequency of red vlight is smaller than the threshold frequency’ of the metal used in photo cell. / Die frekwensie van rooi lig is kleiner as die drumpelfrekwensie van die metaal gebruik in fotosel. OR/OF The energy of the photon of red light is lower than the work function v of the metal used in photo cell./ Die energie van die foton van rooi lig is minder as die werksfunksie vir die metaal gebruik in fotosel. (2) 10.3.1 The minimum energy” needed for an electron to be set free from the surface of a metal.v/ Die minimum energie nodig om ‘n elektron uit die opperviak van ‘n metaal vry te stel. (2) 10.3.2 | OPTION 1/ OPSIE 1 OPTION 1/ OPSIE 1 E=hfv = hiv = (6,63 x 10°)(1,1 x 10%)v 3x 108V =x 10° (f v =7.29x1079 J f = 6.67 x 10 Eeh® fo= 1,1 x 10° (given/gegee) ~) No, ¥ fohoton/f foton< fo (cut-off f) _ 8,63 x10 x3x10° 450x107 =4,42x1079 J No, ¥ Ephoton/Etoton < Work (5) Function’ /Werksfunksie [11] TOTAL SECTION B/ITOTAAL AFDELING B: 130 GRAND TOTALIGROOTTOTAAL: 150 Copyright reserved/Kopiereg voorbehou