Physical Sciences Grade 11 P1 Gauteng June 2019 Memo (Afr)

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION / GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION / PROVINSIALE EKSAMEN JUNE / JUNIE 2019 GRADE / GRAAD 11 PHYSICAL SCIENCES / FISIESE WETENSKAPPE Physics / Fisika Paper 1 / Vraestel 1 MEMORANDUM 14 pages / bladsye

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 GAUTENG DEPARTMENT OF EDUCATION / GAUTENGSE DEPARTEMENT VAN ONDERWYS PREPARATORY EXAMINATION / VOORBEREIDENDE EKSAMEN PHYSICAL SCIENCES / FISIESE WETENSKAPPE (Paper / Vraestel 1) QUESTION 1 / VRAAG 1: 1.1 D  (2) 1.2 B  (2) 1.3 C  (2) 1.4 B  (2) 1.5 D  (2) 1.6 B  (2) 1.7 C  (2) 1.8 A  (2) 1.9 A  (2) 1.10 C  (2) [20] 2

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 QUESTION 2 / VRAAG 2: 2.1 The vector sum of all the vectors/forces acting on an object  OR The single vector that will have the same effect as the original vectors/forces taken together. Die vektorsom van al die vektore/kragte wat op ʼn voorwerp inwerk  OF Die enkele vektor wat dieselfde effek het as al die ander vektore/kragte saam. (2) 2.2 OR  F1y, and direction of arrow F1y, en rigting van pylpunt.  F1x and direction correct F1x, en rigting korrek  an angle shown 30˚ or 60˚ / ʼn hoek 30˚ of 60˚ aangetoon (3) 2.3.1 F1y = F x sin 30˚  F1y = F x cos 60˚  = 300 x sin 30˚ = 300 x cos 60˚ = 150 N at 0˚ / North/ upwards = 150 N at 0˚ / North/ upwards Noord / opwaarts Noord / opwaarts F1x = F x sin 60˚  F1x = F x cos 30˚  = 300 x sin 60˚ = 300 x cos 30˚ = 259,81 N at 90˚ / East/ right = 259,80... N at 90˚ / East/ right Oos/ regs Oos/ regs F2y = F x cos 70˚ F2y = F x sin 20˚ = 450 x cos 70˚ = 153,91 N at 180˚/South/ down = 450 x sin 20˚ Suid / afwaarts = 153,90... N at 180˚ / South/ down F1x = F x sin 70˚ Suid / afwaarts = 450 x sin 70˚ F1x = F x cos 20˚ = 422,86 N at 90˚ / East/ right = 450 x cos 20˚ Oos/ regs = 422,86... N at 90˚ / East/ right Oos/ regs 3

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 Fnet y = F1y + F1y  = + 150 – 153,91 = - 3,91 N = 3,91 N at 180˚ south/ downwards Suid / afwaarts  Fnet x = F1x + F1x = + 259,81 + 422,86 = 682,67 N at 90˚ / East/ right Oos/ regs  Fnet2 = Fnet y 2 + Fnet x2  = 3,912 + 682,672 Fnet = 682,68 N  (9) 2.3.2 OPTION 1: tan Ɵ =  Ɵ = tan -1 ( TB / Fg ) = tan-1 x (3,91 / 682,67)  = 0,328 ˚ OPTION 2: sin Ɵ = sin Ɵ = Fx / Fnet  = (3,91 / 682,68)  = 0,328 ˚ OPTION 3: cos Ɵ = cos Ɵ = Fy / Fnet  = (682,67 ÷ 682,68)  = 0,31 ˚ (3) [17] 4

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 QUESTION 3 / VRAAG 3: ACCEPT: 3.1 OR Criteria for Free body diagram / Kriteria vir Marks/ vryliggaam : Punte Fk – upwards /opwaarts  Fg/ W // Fg &Fg// – down towards centre of earth  - afwaarts na middel van aarde FN - upwards  to Fk / opwaarts  met Fk  (3) 3.2 The force that opposes the motion of an object and which acts parallel to the surface.  Is die krag wat die beweging van ʼn voorwerp teëstaan en wat parallel met die (2) oppervlak werk 3.3 OPTION 1: fk = µ k x N  = 0,112  x (70 x 9,8  x cos 20˚)  = 72, 20 N // up the slope  // op met helling OPTION 2: N = Fg x cos 20˚ = (70 x 9,8) x cos 20˚  = 644,63 N  fk = µ k x N  = 0,112 x 644,63  = 72, 20 N // up the slope  // op met helling (6) 5

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 3.4.1 DECREASE  VERMINDER (2) 3.4.2 Fg  will decrease  Normal F will decrease  - Fg  sal verminder as die helling steiler word  die Normale krag sal verminder.  Since fk  N, then fk will also decrease.  Aangesien fk  N, sal die fk ook verminder.  (2) 3.5.1 REMAIN THE SAME  - BLY DIESELFDE  (2) 3.5.2 As the mass increase, the normal will increase  fk will increase proportionally fk  N and µ remains the same for the surface.  As die massa verhoog, sal die Normaalkrag verhoog  fk sal proporsioneel verhoog fk  N en µ bly dieselfde vir die spesifieke oppervlak.  (2) OR / OF a  1/m  as m increase, a will decrease, but Fg// and fk will proportionally increase as well a  1/m  as m vermeerder sal a verminder, maar Fg// en fk sal proporsioneel verhoog [19] 6

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 QUESTION 4 / VRAAG 4: 4.1 An object will remain at rest or, if it is moving, will carry on moving in a straight line unless acted upon by a resultant/ net force.  ʼn Voorwerp sal in ʼn toestand van rus of konstante snelheid bly voortbeweeg tensy ʼn eksterne resultante/netto krag daarop inwerk.  (2) 4.2 The box will continue to move forward at 30 m.s-1 Die boks sal aanhou vorentoe beweeg teen 30 m.s-1 (2) 4.3 Fnet = m x a  = 20 x - 30  = - 600 N = 600 N against the motion / teen rigting van beweging  (4) 4.4 The back of the car seat has a much bigger area than the straps as pressure  1/ area  it will mean that the baby/toddler has a much bigger chance to survive as the pressure will decrease significantly. Die agterkant van die karstoeltjie het ʼn baie groter oppervlak as die veiligheids- gordels   as die druk 1/area  dit sal dus beteken dat die baba/kleuter ʼn baie groter kans het op oorlewing.  (3) [11] 7

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 QUESTION 5: VRAAG 5 5.1  Fg / w (lower case) and direction of arrow Fg / w (klein letter), rigting van pylpunt.  T and direction correct T en rigting korrek (2) 5.2 If a resultant force act 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 /net 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) 5.3 Fnet on A = m x a = - T + Fg F net on B = m x a = - Fg + T  0,375a = - T + (0,375 x 9,8) 0,225a = - (0,225 x 9,8) + T T = - 0,375a + 3,675…(1)  T = 0,225a + 2,205…. (2)  (1) = (2): - 0,375a + 3,675 = 0,225a + 2,205  0,6 a = 1,47 a = 2,45 m.s-2  (5) 5.4 Into (1) / Into (2) T = - 0,375a + 3,675…(1)  T = 0,225a + 2,205…… (2) T = (-0,375 x 2,45) + 3,675  T = (0,225 x 2,45) + 2,205 T = 2,76 N  T = 2,76 N (3) [12] 8

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 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 Gm1m2 F= r2  = 6,67 x 10 -11 x 5,98 x 1024 x 419725  3,64 x 106  r2 r = 6781804,029 m accept / aanvaar 6781804,03 m distance above the earth = r – radius of the earth = 6781804,029 – 6,38 x106  = 401804,029 m (5) = 401,80 km above the surface of the earth/ bokant die oppervlak van die aarde 6.3 Fg / w = m x g  650 = m x 9,8 m = 66,33 kg  (= 66,3265 kg) Mass remains constant / Massa is ʼn konstante  (3) 9

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 6.4 F=mxa = 419 725 x 0,6  = 251835 N  F net = Fapp - Fearth  251835 = Fapp - 3,64 x 106  = 3891835 N away from the earth. / Weg van die aarde af  (5) 6.5.1 DECREASE / VERMINDER  (2) 6.5.2 The gravitational force decreases as the satellite moves away from the earth.  g  1/r2  (2) Die gravitasiekrag verminder hoe verder die sateliet van die aarde af beweeg. g  1/r2  [19] QUESTION 7 / VRAAG 7: 7.1 The refractive index (n) is the ratio of the speed of light in air to the speed of light through another medium.  (2) Die brekingsindeks (n) is die verhouding van die spoed van lig in ʼn vakuum tot die spoed van lig in die medium  7.2 When light is moving from a less dense medium (air) to a more dense medium (water) it will be refracted towards the normal.  (2) Wanneer lig van ʼn minder digte medium (lug) na ʼn meer digte medium (water) beweeg word dit na die normaal toe gebreek.  7.3 n =  1,33 = 3 x 108  v (3) v = 2,26 x 108 m.s-1 7.4 ni sin θi = nr sin θr  1 x sin 42° = 1,33 sin r r = 30,21 °  (4) 10

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 7.5.1 Snell’s law / Snell se wet  (2) 7.5.2 The ratio of the sine of the angle of incidence in one medium to the sine of the angle of refraction in the other medium is constant.  Die verhouding van sin van die invalshoek in een medium tot sin van die brekingshoek in die ander medium is konstant.  (2) 7.6 Marking Grid: Merk gids The line is extrapolated straight into the water. Die lyn word reguit verleng in die water in.  The driver is seen higher up in the water. Die duiker word hoër op in die water gesien.  the diver is a bit further away from the normal. Die duiker word ‘n bietjie verder van die normaal af waargeneem. (3) 7.7 If the light from the fish hits the surface at 90˚  there will be no refraction of the light,  making it much easier for the bird to determine the exact position of the fish.  (3) As die lig wat van die vis af weerkaats die oppervlak teen 90˚ bereik, sal daar geen breking/ buiging van die lig plassvind nie. wat dit dus makliker maak vir die voël om die presiese posisie van die vis vas te stel. [21] QUESTION 8 / VRAAG 8 8.1 45 ˚  (2) 8.2 Is that angle of incidence that provides an angle of refraction of 90° Die invalshoek wat ʼn brekingshoek van 90° tot gevolg het.  (2) 8.3.1 Total internal reflextion.  Totale interne weerkaatsing.  (2) 8.3.2 Light must travel from an optical more dense medium to an optical less dense medium. Die invalstraal moet vanaf ʼn opties meer digte medium na ʼn opties minderdigte medium beweeg.  The angle of incidence must be bigger than the critical angle of that medium.  Die invalshoek moet groter as die grenshoek van die betrokke medium wees.  (4) 11

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 8.4 n =  = 3 x 108____  2,21 x 108 (3) = 1,36  8.5 Telescope / Microscope/ binoculars (Any One) Teleskoop/ mikroskoop/ verkyker (Enige een)  (2) 8.6 Signal move much faster Less interference / more data can be transferred Any 2  Very difficult to tap/ safer transfer of data (2) Sein beweeg baie vinniger Minder steurings / meer data kan versend word Enige 2  Dit is baie moeilik om inligting te steel. [17] QUESTION 9 / VRAAG 9: 9.1 All points on a wave front act like a point source. Each one of these point (2) sources (secondary sources) produces small circular waves moving forwards with the same speed as the wave. The new wave front is obtained by drawing a tangent to all the new little wave fronts.  Elke punt van ʼn golffront dien as ʼn puntbron van klein sferiese sekondêre golwe. Die golffront is dan ʼn raaklyn loodreg op die kleiner sekondêre golwe. 9.2.1 Diffraction  Diffraksie (2) 9.2.2 The ability of a wave to spread out in wave fronts as they pass through a small opening or around a sharp edge.  (2) Die vermoë van ʼn golf om uit te sprei in golffronte soos die golf deur ʼn klein opening of om ʼn skerp hoek voortplant.  12

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 9.3.1 Constructive interference  Konstruktiewe interferensie (2) 9.3.2 The angle of interference is directly proportional to the wavelength.  if the wavelength is longer (red light) the observed diffraction pattern is bigger  Die diffraksiehoek is direk eweredig aan die golflengte. (4)  hoe langer die golflengte(rooi lig) hoe groter die diffraksiepatroon wat waargeneem word. 9.3.3 Decrease the opening of the slit. (2) Maak die opening van die spleet kleiner  [14] TOTAL / TOTAAL: 150 13

Downloaded from hlayiso.com PHYSICAL SCIENCES / MEMORANDUM FISIESE WETENSKAPPE GRADE 11 / GRAAD 11 Taxonomy Grid Recall Comprehension Analysis Evaluation Q no: Mark Q no: Mark Q no: Mark Q no: Mark 1.1 2 1.3 2 1.7 2 2.3.2 3 1.2 2 1.4 2 1.8 2 3.5 4 2.1 2 1.5 2 1.9 2 4.6 4 3.2 2 1.6 2 1.10 2 6.5 3 4.1 2 2.2 4 2.3.1 9 7.7 3 5.1 2 3.1 4 3.3 6 5.2 2 4.2 2 3.4 4 6.1 2 5.4 3 4.3 4 7.1 2 6.3 3 4.4 3 7.6.2 2 7.2 2 5.3 5 8.2 2 7.3 4 6.2 4 9.1 2 7.4 3 6.4 4 8.1 1 7.5 4 8.3 6 7.6.1 2 8.4 2 8.5 2 8.6 4 9.3 6 9.2 4 9.4 2 16% 24 33,33% 50 40% 63 10,67% 17 Total mark 16% 24 33,33% 50 42% 63 11% 17 Total %/100% P1&2: 15% P1:35%/P2:40% P1:40%/P2:35% P1&2: 10% 14