Physical Sciences Grade 11 P1 2016 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 2016 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 14 pages. Hierdie memorandum bestaan uit 14 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 2 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION 1/VRAAG 1 1.1 D  (2) 1.2 B  (2) 1.3 C  (2) 1.4 A  (2) 1.5 A  (2) 1.6 C  (2) 1.7 B  (2) 1.8 D  (2) 1.9 A  (2) 1.10 C  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 3 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION 2/VRAAG 2 2.1 The sum of two or more vectors  Die som van twee of meer vektore OR/OF The single vector having the same effect as two or more vectors together. Die enkele vektor wat dieselfde effek as twee of meer vektore saam het. (2) 2.2 The object is lifted/moved upwards. There will be a resultant/net force not equal to zero. Die voorwerp word opgelig/beweeg opwaarts Dit sal 'n resultante/netto krag wees wat nie nul is nie. (2) 2.3.1 F y = F A sin 70˚ = 730 sin 70˚  = 685,98 N (2) 2.3.2 F x = F A cos70˚ = (730)cos70˚  = 249,67 N  (2) 2.4 F y = F B cos10˚ = (1 440)cos10˚ = 1418,12 N  OR/OF F y = F B sin 80˚ = (1 440)sin80˚ = 1 418,12 N Maximum/Maksimum w = F YA + F YB = 685,98 + 1418,12  = 2 104,1 N  (4) 2.5 If the distance x increases, the vertical components of the applied forces will decrease  and then the system will possibly not be able to pick up the weight.  Indien die afstand x vergroot, sal die vertikale komponente van die toegepaste krag verklein en dan sal die stelsel moontlik nie die gewig kan optel nie. (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 4 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION 3/VRAAG 3 3.1 A body will remain in its state of rest or motion at constant/uniform velocity  unless a non-zero resultant/net force acts on it.  'n Liggaam sal in sy toestand van rus of beweging teen konstante/uniforme snelheid bly tensy 'n nie-nul resulterende/netto krag daarop inwerk. (2) 3.2 T Fblades w (3) Notes: Accepted Labels/Aanvaarbare Byskrifte Mark/Punt w weight/F G /F g  gewig/gravitasiekrag/swaartekrag T Tension/F T /F applied /F A  Spanning/F toegepas F blades F of the blades/f/Air friction/Downward force of the blades  F van die lemme/f/Lugweerstand/Afwaartse krag van lemme Any additional force: deduct 1 mark (maximum ⅔) Enige addisionele krag: trek 1 punt af (maksimum ⅔) Lines must touch object otherwise (maximum ⅔ ) Lyne moet voorwerp raak anders (maksimum ⅔) 3.3 There is an extra downward force (on the container created by the blades of the helicopter) Daar is 'n ekstra afwaartse krag (op die houer as gevolg van die lemme van die helikopter) (2) 3.4 w = mg 1 960 = m(9,8)  m = 200 kg  (2) 3.5 POSITIVE MARKING FROM 3.4 POSITIEWE NASIEN VAN 3.4 F net = ma T – F g – F blades = ma  OR/OF T – 2 100 = ma T – 1 960 – 140  = (200)(0,13)  T – 2 100 = (200)(0,13) T = 2 126 N  T = 2 126 N (2) 3.6 2 100 N  (1) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 5 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION 4/VRAAG 4 4.1 When a net force acts on an object, it will accelerate in the direction of the net force. The acceleration is directly proportional to the net force and inversely proportional to the mass of the object.  Wanneer 'n netto krag op 'n voorwerp inwerk, sal dit in die rigting van die netto krag versnel. Die versnelling is direk eweredig aan die netto krag en omgekeerd eweredig aan die massa van die voorwerp. OR/OF 1 Fnet α a and a α m 1 Fnet α a en a α m (2) 4.2 T N f w Notes: Accepted Labels/Aanvaarbare byskrifte MARK/PUNT N Normal force/ F N  Normaalkrag w Weight/F G /F g  Gewig/Gravitasiekrag/Swaartekrag T Tension/F T  Spanning f Friction/F f  Wrywingskrag Any additional force: deduct 1 mark maximum (maximum ¾) Enige addisionele krag: trek 1 punt af (maksimum ¾) Lines must touch dot otherwise (maximum ¾ ) Lyne moet kolletjie raak anders (maksimum ¾) Do not penalise if angle is shown/not shown Moenie penaliseer as hoek getoon/nie getoon is nie. (4) 4.3 f k = µk N  f k = 0,2(4)(9,8)cos 40°  f k = 6,01 N  up the slope/teen die helling op (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 6 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum 4.4 POSITIVE MARKING FROM 4.3 POSITIEWE NASIEN VAN 4.3 F net = ma For the 8 kg box: For the 4 kg box: Vir die 8 kg-blok: Vir die 4 kg-blok: T – f = ma F gll – T – f = ma T – 0,2(8)(9,8)  = 8a  4(9,8) sin 40˚ - T – 6,01 = 4a  T = 8a + 15,68 ….(1) 25,2 – 6,01 – T = 4a 19,19 – 4a = T ….(2) Combining equations (1) and (2): Kombineer vergelykings (1) en (2) 4(9,8)sin 40˚ – 6,01 - 4a = 0,2(8)(9,8) + 8a  a = 0,29 m.s-2  OR/OF 8a + 15,68 = 19,19 – 4a a = 0,29 m.s-2 (6) 4.5 Greater than  Groter as Explanation:/Verduideliking: The component parallel to the slope increases  The tension in the rope stays the same  The net force increases  Die komponent parallel (ewewydig) aan die helling vergroot Die spanning in die tou bly dieselfde Die netto krag vergroot OR/OF For the 8 kg box: For the 4 kg box: Vir die 8 kg-blok: Vir die 4 kg-blok: 8(9,8) sin 40º – (0,2)(8)(9,8) cos 40º - T = 8a  T – 0,2(4)(9,8) = 4a  50,39 – 12,01 – T = 8a …(1) T – 7,84 = 4a …(2) 50,39 – 12,01 – 8a = 4a + 7,84 a = 2,545 m.s-2  (4) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 7 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION 5/VRAAG 5 5.1 Every 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.  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 is aan die kwadraat van die afstand tussen hul middelpunte. (2) 5.2 Weightlessness is the sensation experienced when all contact forces are removed.  Gewigloosheid is die sensasie wat ervaar word wanneer alle kontakkragte verwyder word. (2) 5.3 m1m 2 M m F= G 2 OR/OF F = G E 2  r RE = 6, 67 10 -11 × (5,98 × 10 )(3800 )  = 1 539,23 N 24 [6.38 × 10 6 + 25 × 10 6 ] 2  (4) 5.4 OPTION 1/OPSIE 1 Greater than  the mass is greater  and for the same force  the distance must also be greater  (because the product of the masses is directly proportional to the square of the distance between the centres.) Groter as Die massa is groter en vir dieselfde krag moet die afstand ook groter wees (omdat die produk van die massas direk eweredig is aan die kwadraat van die afstand tussen die middelpunte) OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 POSITIVE MARKING FROM 5.3 Greater than  POSITIEWE NASIEN VAN 5.3 Groter as Greater than  mm mm 2 Groter as G 21 = G 2 R1 R2 mm F = G 12 2 m1 m R 2 = 22 -11 1539,23 = 6, 67 10 × ( ) 5,98 × 10 24 (4500 ) R1 3800 4500 R2 R2 2 = 2  R1 R2  4500 2 2 7 R2 = R1 R = 3,41 x 10 m 3800 2 R 2 = 1,0882...( 25 × 10 6 + 6,38 × 10 6 ) Distance above the surface of the Earth Afstand bo oppervlak van Aarde R 2 = 3,41 x 107 D = 3,41 x 107 – 6,38 x 106  D = 3,41 x 107 – 6,38 x 106  D = 2,78 x 107 m  (or 27 768 214,93 m) D = 2,78 x 107 m  OR/OF D = 1,0882...(25 x 106) D = 2,78 x 107 m (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 8 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION 6/VRAAG 6 6.1 The refractive index is the ratio between the speed of light in a vacuum  and the speed of light in a medium.  Die brekingsindeks is die verhouding tussen die spoed van lig in 'n vakuum en die spoed van lig in 'n materiaal/medium. (2) 6.2 c n=  v 3 × 10 8 1,47 =  v -1 v = 2.04 x 108 m . s  (3) 6.3 ni sin θi = nr sin θr  1,33 sin 37°  = 1,47 sin θ  θ = 32,99°  (4) 6.4 Olive oil has a higher optical density than water  Olyfolie het 'n groter optiese digtheid as water (2) 6.5.1 Away from the normal  (1) Weg van die normale 6.5.2 The speed of light in air is faster than the speed of light in oil  because the refractive index in oil is higher than in air.  (When a light ray speeds up it refracts away from the normal.) Die spoed van lig in lug is vinniger as die spoed van lig in olie omdat die brekingsindeks in olie groter is as in lug. (Wanneer die ligstraal vinniger (2) beweeg breek dit weg van die normale.) 6.6.1 Oil and air (1) Olie en lug 6.6.2 One of the conditions for total internal reflection is that the light has to travel from a medium with high optical density  to a medium of lower optical density.  Een van die voorwaardes vir totale interne weerkaatsing is dat die lig van 'n medium met 'n hoë optiese digtheid na 'n medium met 'n laer optiese (2) digtheid moet beweeg. 6.7 Wavelength  (1) Golflengte [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 9 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION 7/VRAAG 7 7.1 Every point of a wave front serves as a point source of spherical, secondary waves.  The positions of the new wave front will be that of the surface tangent to the secondary waves. Elke punt van 'n golffront dien as 'n puntbron van sferiese, sekondêre golwe. Die posisies van die nuwe golffront sal dié van die oppervlakraaklyn tot die sekondêre golwe wees.. (2) 7.2.1 Slit width  Spleetwydte (1) 7.2.2 Degree of diffraction  Mate van diffraksie (1) 7.2.3 Wavelength/Frequency/Colour of light  Golflengte/Frekwensie/Kleur van lig (1) 7.3 The greater the width of the slit, the less the degree of diffraction  Hoe groter die wydte van die spleet, hoe kleiner die mate van diffraksie OR/OF The smaller the width of the slit, the greater the degree of diffraction  Hoe kleiner die spleetwydte, hoe groter die mate van diffraksie (2) 7.4.1 The bright, central band will increase  The coloured bands will be red instead of green  Die helder, sentrale band sal vergroot/toeneem Die gekleurde bande sal rooi in plaas van groen wees (2) 7.4.2 If the wavelength increases, the degree of diffraction will increase.  Indien die golflengte vergroot/toeneem, sal die mate van diffraksie vergroot/toeneem OR/OF Degree of diffraction is directly proportional to wavelength.  Die mate van diffraksie is direk eweredig aan die golflengte (1) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 10 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION 8/VRAAG 8 8.1 + - Shape/Vorm  Direction/Rigting  Touching the charge, no crossing lines, etc.  Raak aan die lading/geen lyne wat kruis, ens. (3) 8.2 kQ 1Q 2 F=  r2 F= ( )( 9 × 10 9 5 × 10 -6 5 × 10 -6 )  (0,04 ) 2  F Y on X = 140,63 N right/regs  (4) 8.3.1 No  Nee (1) 8.3.2 The electric field is stronger closer to the charges (not a uniform field) and therefore not constant  which means the force will not be constant  Die elektriese veld is sterker nader aan die ladings (nie 'n uniforme veld nie) en daarom nie konstant nie wat beteken die krag sal ook nie konstant wees nie (2) 8.4 POSITIVE MARKING FROM 8.2 POSITIEWE NASIEN VAN 8.2 kQ 1Q 2 F= r2 F= ( )( 9 × 10 9 4 × 10 -6 5 × 10 -6  ) (0,03 )2 F Z on X = 200 N upwards  (F net )2 = (F Y on X )2 + (F Z on Y )2 (F net )2 = 140,632 + 2002  F net = 140,63 2 + 200 2 F net = 244,49 N  (4) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 11 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION 9/VRAAG 9 9.1 Φ = BA cos θ  Φ = (3,2)π(0,04)2cos 0˚  Φ = 0,016 Wb  (or 0,02 Wb) (3) 9.2 POSITIVE MARKING FROM 9.1 POSITIEWE NASIEN VAN 9.1 NΔΦ ε=  Δt (  − 250 0,016 cos 25  − 0,016 cos 0  2,8 = )  ∆t ∆t = 0,13 s  (0,17 s if 0,02 Wb was used/ gebruik was) (4) 9.3 Faraday's law.  The magnitude of the induced emf across the ends of a conductor is directly proportional to the rate of change in the magnetic flux linkage with the conductor.  Faraday se wet. Die grootte van die geïnduseerde emk oor die ente van 'n geleier is direk eweredig aan die tempo van verandering in die magnetiese vloedkoppeling met die geleier. (2) 9.4.1 Smaller  Kleiner as (2) 9.4.2 The area of a square is smaller than the area of a circle with the radius equal to the side length of the square  Die oppervlakte van 'n vierkant is kleiner as die oppervlakte van 'n sirkel met die radius gelyk aan die sylengte van die vierkant. OR/OF 0,042 < π x 0,042 area of square is smaller than area of circle. 0,042 < π x 0,042 oppervlakte van vierkant is kleiner as oppervlakte van sirkel. OR/OF ε directly proportional to A √ ε direk eweredig aan A (2) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 12 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION 10/VRAAG 10 10.1 Ohm's law Ohm se wet (1) 10.2 Graph/Grafiek Graph of Potential difference versus Current Grafiek van Potensiaalverskil teenoor Stroom 4 3.5 3 Potential difference (V) 2.5 2 1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 -0.5 Current (A) Marking criteria for graph Nasienkriteria vir grafiek Axes with correct/appropriate scale  Asse met korrekte en toepaslike skaal 5 or more of the 6 coordinates correctly plotted  (3–4 one mark only) 5 of meer van die 6 koördinate korrek gestip (3–4 slegs een punt) Drawing a line of best fit  Teken 'n lyn van beste passing (4) 10.3 Resistance of the parallel connection  Weerstand van die parallel kombinasie (1) 10.4 Stay the same  Bly dieselfde (1) 10.5 Increase  Toeneem (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 13 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum 10.6 OPTION 1/OPSIE 1 1 1 1 = +  R p R1 R 2 1 1 1 = +  Rp 4 6 R p = 2,4 Ω V R tot = I 6 R tot =  0,8 R tot = 7,5 Ω R = R tot - R par = 7,5 – 2,4 = 5,1 Ω  (5) OPTION 2/OPSIE 2 V tot = 6 V V R = V tot – V 2 = 6 – 1,9  = 4,1 V  V R =  I 4,1 R =  0,8 R = 5,1 Ω  (5) 10.7 V R = 6 – 2,4 = 3,6 V  W = (3,6)(1)(10)  W = 36 J  (3) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 14 DBE/November 2016 CAPS/KABV – Grade/Graad 11 – Memorandum ANSWER SHEET FOR QUESTION 10.2/ANTWOORDBLAD VIR VRAAG 10.2 HAND IN THIS ANSWER SHEET TOGETHER WITH THE ANSWER BOOK./ LEWER HIERDIE ANTWOORDBLAD SAAM MET DIE ANTWOORDEBOEK IN. Graph of potential difference versus current Grafiek van potensiaalverskil teenoor stroom V (V) 4 3,5 3 2,5 2 1,5 1 0,5 0 0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 I (A) TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou