Gr11 Phy P1 (Bilingual) June 2018 Possible Answers_hlayiso.com_.pdf

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION / GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION / PROVINSIALE EKSAMEN JUNE / JUNIE 2018 GRADE / GRAAD 11 PHYSICAL SCIENCES / FISIESE WETENSKAPPE Physics / Fisika Paper / Vraestel 1 MEMORANDUM TIME / TYD: 3 hrs / uur 180 min MARKS / PUNTE: 150 16 pages / bladsye

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 2 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) GAUTENG DEPARTMENT OF EDUCATION / GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION / PROVINSIALE EKSAMEN PHYSICAL SCIENCES: (Physics) / FISIESE WETENSKAPPE: (Fisika) Paper 1 / Vraestel 1 MEMORANDUM QUESTION 1 / VRAAG 1: 1.1 A (2) 1.2 C (2) 1.3 D (2) 1.4 B (2) 1.5 B (2) 1.6 D (2) 1.7 A (2) 1.8 B (2) 1.9 C (2) 1.10 D (2) [20] 2

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 3 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) QUESTION 2 / VRAAG 2: 2.1  Fg / w (lower case) and direction of arrow Fg / w (klein letter), rigting en pylpunt.  TA and direction correct TA en rigting korrek  TB and direction correct TB en rigting korrek  an angle shown / ʼn hoek OR / OF aangetoon (4) 2.2 F net = 0 N  (2) 2.3 Forces are in equilibrium  Newton’s 1st law.  Kragte is in ewewig  Newton se Eerste wet (2) 2.4 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 Fg2 = TA2 +TB2  OR / OF cos  =  (m x 9,8)2 = 7,512 + 62  cos 40˚  = 7,51 / m x 9,8 m = 9,6125  m x 9,8 = 7,51 / cos 40˚ 9,8 m = 1 kg  m = 0,981 kg  (5) 2.5 Decrease  Verminder  (2) [15] 3

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 4 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) QUESTION 3 / VRAAG 3: 3.1 Frictional force: is a contact force  that develops between two surfaces to oppose the motion.  Wrywingskrag: is ʼn kontakkrag wat ontstaan tussen twee oppervlaktes om die beweging teen te staan.  (2) 3.2  Fg / w (lower case) and direction of arrow Fg / w (klein letter), rigting en pylpunt.  N and direction correct N en rigting korrek  f and direction correct f en rigting korrek an angle shown / ʼn hoek aangetoon (3) 3.3 3.3.1 Fnet  = 0 = +N – Fg    N = (80 x 9,8) x cos 4,76˚  = 781,30 N  up the slope  /  opwaarts  (3) 3.3.2 fs = µ s x N  = 0,1 x 781,30 = 78,13 N // up the slope  / op met helling  (2) 3.3.3 Fg// = Fg x sin   = (80 x 9,8) x sin 4,75˚  = 65,06 N // down slope.  / af teen helling  Fg// < fs  object will remain stationary   voorwerp beweeg nie (4) 4

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 5 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) 3.4 3.4.1 Decrease  Verminder  (1) 3.4.2 Remain the same  Bly dieselfde  (1) 3.4.3 The Normal will increase as the incline decrease and the mass will remain the same  fk  N fk increase  Die Normale krag sal vermeerder as die helling verminder en die massa van die voorwerp sal konstant bly.  fk  N fk word groter (3) [19] 5

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 6 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) QUESTION 4 / VRAAG 4: 4.1 If a resultant force acts on a body, it causes the body to accelerate in the direction of the force  and the acceleration is directly proportional to the resultant force and indirectly proportional to the mass of the body.  Indien ʼn resulterende /netto krag op ʼn voorwerp inwerk, sal die voorwerp versnel in the rigting van die resulterende krag.  Die versnelling is direk eweredig aan die krag en omgekeerd eweredig aan die massa van die voorwerp.  (2) 4.2 4.2.1 Horizontal forces on A Horizontal forces on B Horisontale kragte op A Horisontale kragte op B Fnet on / op A = m x a = T – fk  F net on / op B = m x a = Fx - T - fk  15 a = T - 11 15 a = (50 x cos25˚) - T - 11 T = 15 a + 11 ……(1)  T = - 15 a + 45,32……. (2)  (1) + (2) 15 a + 11 = - 15 a + 45,32  30 a = 45.32 a = 1,15 m.s-2  Into (1) / In (2) T = 15 a + 11 ……(1)  T = - 15 a + 215,58 ……. (2) T = (15 x 1,15) + 11 T = (-15 x 1,15) + 45,32 T = 28,25 N  T = 28,07 N (8) 6

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 7 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) 4.2.2 Fnet y = 0 = N + Fy - Fg  N = (- 50 x cos 15°) + (15 x 9,8) = 101,68 N upwards  / opwaarts µ =  = 11 101,68 = 0,10817 / 0,108  (4) 4.3 Trolley B will experience more friction as 2 kg mass was added to it, thus it will slow down.  Trolley A will still experience a frictional force of 11 N and will collide into the back of trolley B.  Trollie B sal ʼn groter wrywingskrag ondervind as die 2 kg massa  daarop geplaas word en sal dus stadiger beweeg.  Trollie A sal steeds ʼn wrywingskrag van 11 N ondervind en sal dus teen die agterkant van trollie B bots.  (3) [17] 7

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 8 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) QUESTION 5 / VRAAG 5: 5.1 Criteria for Free body diagram / Marks/ Kriteria vir vryliggaamdiagram : Punte T – upwards /opwaarts  Fg – down towards centre of earth /  - afwaarts na middel van aarde Fair friction / lug wrywing – downwards /  afwaarts / lug wrywing - 1 for any extra forces / - 1 vir enige ekstra kragte. (max 2/3 ) (3) 5.2 There is the force of gravity and air friction down to balance the tension in the rope upwards.  Resultant force is zero.  Die gravitasiekrag en die krag van lugweerstand afwaarts wat die spanning in die tou opwaarts balanseer.  Resulterende krag is nul.  (2) 5.3 Fnet = 0 = - Fg - fair / lug + T 0 = (- 80 x 9,8) - f air / lug + 920 fair/ lug = 136 N downwards / afwaarts  OPTION 1 / OPSIE 1 (up as +) ( op is +) Fnet = m x a = - Fg - fair / lug + T  - 80 x 0,18  = (- 80 x 9,8) - 136 + T  T = 905,60 N upwards / opwaarts  OPTION 2 / OPSIE 2 (up as -) ( op is -) Fnet = m x a = Fg + fair / lug - T  80 x 0,18  = ( 80 x 9,8) + 136 - T  = -905,60 N  T = 905,60 N upwards / opwaarts  (5) 8

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 9 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) 5.4 If an object A exerts a force on object B, then B will exert a force equal in magnitude, but opposite in direction on object A.  Indien voorwerp A ʼn krag op voorwerp B uitoefen, dan sal voorwerp B ʼn krag, van dieselfde grootte maar in die teenoorgestelde rigting op voorwerp A uitoefen. (2) 5.5 The rope on Bale and Bale on the rope  Rope on the helicopter and the helicopter on the rope Earth on Bale and Bale on the earth Earth on the helicopter and the helicopter on the earth (any one) Die tou op Bale en Bale op die tou Die tou op die helikopter en die helikopter op die tou Die aarde op Bale en Bale op die aarde Die aarde op die helikopter en die helikopter op die aarde (enige een) (2) 5.6 905,60 N downwards / afwaarts  (1) [15] 9

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 10 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) QUESTION 6 / VRAAG 6: 6.1 Every particle in the universe exerts a force of gravitational attraction on every other particle. The force between the two particles is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.  Elke voorwerp in die heelal trek elke ander voorwerp aan met ʼn krag wat direk eweredig is aan die produk van die massas van die voorwerpe  en omqekeerd ewerediq is aan die kwadraat van die afstand tussen die massa-middelpunte van die twee voorwerpe. (2) 6.2 The force of the earth on Apollo 11 would be equal to the force of the moon on Apollo 11 at that point, acting in opposite directions Die krag van die aarde op Apollo 11 sal gelyk wees aan die krag van die maan op Apollo 11 op daardie punt, maar in teenoorgestelde rigtings.  (2) 6.3 FEA = GmEmA  r2 = 6,67 x 10-11 x 6,02 x 1024 x 300 (193620 x 103+ 6,38 x 106) 2 = 3,012 N  (5) 6.4 The mass of the astraunaut is so small and he is so far away from any planet (earth/ moon) thus the force will be very small and he appears weightless.  Die massa van die ruimtevaarde is so klein en hy is so vêr weg van enige planeet dat die aantrekkingskragte wat hy sal ervaar so klein is dat dit lyk of hy gewigloos is.  (2) 6.5 Fg = Gmmmc  r2 = 6,67 x 10-11 x 600 x 1021 x 899 (1737 x 103) 2 = 11924,42 N  (4) 10

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 11 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) 6.6 Fnew = Gmmmc r2 = 1x½ x1 ( 3)2 1 Fn = 18F  or /of 0,0556 F or/of 662,47N (2) [17] QUESTION 7 / VRAAG 7: 7.1 Refraction is the change of the path of a light ray when it moves from one optical medium to another optical medium.  Die verandering van rigting van ʼn ligstraal a.g.v die verandering in spoed van een medium na die volgende.  (2) 7.2 A = incident ray / invallende straal  B = reflected ray / weerkaatste straal  C = refracted ray / gebreekte straal  (3) 7.3 To ensure a fair test and obtain more accurate results / reduce the factor of human error.  Om ʼn meer geloofwaardige en akkurate resultate te verkry / die effek van menslike foute te verminder.  (2) 11

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 12 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) 7.4 6 Points plotted correct  6 Punte korrek geplot. Close fit line drawn  Naaste paslyn getrek. Correct lables on x and y-axis Korrekte opskrifte op beide asse Applicable calibration  Gepaste kalibrering (5) (5) 7.5 According to graph is sin i  sin r which is Snell’s law  and the gradient of graph is equal to the refractive index Volgens die grafiek is sin i  sin r  wat Snell se wet verteenwoordig.  en die helling van die grafiek gee die brekingsindeks (3) 7.6 y  = 0,94 – 0,50  = 1,42  x 0,64 - 0,33  (4) [19] 12

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 13 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) QUESTION 8 / VRAAG 8: 8.1 Towards the Normal / towards the centre of the core  Na die normale toe/ in die rigting van die middel van die kern  (2) 8.2 ni sin θi = nr sin θr  1 x sin 39° = 1,33 sin r r = 28,24 ° (4) 8.3 Criteria for RAY diagram / Kriteria vir Marks / Punte STRAALDIAGRAM At A, break towards normal  By A, breek na die normale toe. At C: bend towards core (remains inside)  By C: buig terug na die middel van die kern Arrows indicate direction of movement  Pyltjies dui die rigting van beweging aan (3) 8.4 When light moves from an optical more dense to an optical less dense medium  and the light ray is refracted back into the optic more dense medium (angle of refraction bigger than 90˚). Wanneer ʼn ligstraal van ʼn opties digter medium na ʼn opties minder digte medium beweeg  en die ligstraal binne in die digter en medium weerkaats word.  (brekingshoek groter as 90˚) (2) 8.5 8.5.1 ni sin θi = nr sin θr 1,56 x sin  = 1 sin 90˚  = sin-1 (1 x sin 90 / 1,56) r = 39,87 ° (3) 13

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 14 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) 8.5.2 ni sin θi = nr sin θr 1,56 x sin  = 1,49 sin 90˚ = sin-1 (1,49 x sin 90 / 1,56) r = 72,77 ° (2) 8.6 Because the critical angle is now so much bigger,  the light can travel for longer distances before undergoing TIR, thus it can travel faster.  As gevolg van ʼn baie groter grenshoek,  kan die ligstraal baie verder in ʼn reguit lyn beweeg voordat dit totale interne weerkaatsing ondergaan.  (2) [18] QUESTION 9 / VRAAG 9: 9.1 Every point on a wave front is a source of a secondary wavelets. These wavelets spread out in the forward direction, at the same speed as the source wave. Elke punt op ʼn golffront reageer as die bron van sekondêre golfies wat in alle rigtings met dieselfde spoed as die golf uitsprei.  (2) 9.2 Moving a straight stick or a piece of wood/object up and down in the water.  Deur ʼn reguit stok of stuk hout. / voorwerp op en af in die water te beweeg.  (2) 9.3 The wavelength can be shortened by moving the object up and down faster  Die golflengte kan verkort word deur die voorwerp vinniger op en af te beweeg in die water.  (2) 14

Downloaded from hlayiso.com PHYSICAL SCIENCES: (Physics) / 15 FISIESE WETENSKAPPE: (Fisika) (Paper 1 / Vraestel 1) 9.4 Criteria for diagram / Marks/ Kriteria vir diagram Punte Small slit = big diffraction (round waves)  Klein opening = groter diffraksie (ronde golwe) Wavelength remains the same  Golflengte bly dieselfde (2) 9.5 DECREASE  VERMINDER  (2) [10] TOTAL / TOTAAL: 150 15