Physical Sciences Grade 12 P1 Gauteng Mock 2014 Memo (Afr)

Downloaded from hlayiso.com PREPARATORY EXAMINATION/ VOORBEREIDENDE EKSAMEN 2014 MEMORANDUM PHYSICAL SCIENCES: PHYSICS (P1)/ SUBJECT/VAK: FISIESE WETENSKAPPE: FISIKA (V1) (10841)

Downloaded from hlayiso.com 10841/14 GAUTENG DEPARTMENT OF EDUCATION/ GAUTENGSE DEPARTEMENT VAN ONDERWYS PREPARATORY EXAMINATION/ VOORBEREIDENDE EKSAMEN PHYSICAL SCIENCES: PHYSICS (P1)/ FISIESE WETENSKAPPE: FISIKA (V1) MEMORANDUM QUESTION 1/VRAAG 1 1.1 B  (2) 1.2 D  (2) 1.3 B  (2) 1.4 B  (2) 1.5 C  (2) 1.6 A  (2) 1.7 D  (2) 1.8 C  (2) 1.9 A  (2) 1.10 C  (2) [20] 1

Downloaded from hlayiso.com 10841/14 QUESTION 2/VRAAG 2 2.1 Acceleration is the rate of change in velocity /  Versnelling is die tempo van verandering in snelheid  Acceleration is the change in velocity per unit time /  Versnelling is die verandering in snelheid per eenheid tyd (2) 2.2 Accepted Labels/Aanvaarde benoemings w Fg/Fw/force of Earth on block/weight/49 N/mg/gravitational force Fg/Fw/krag van Aarde op blok/gewig/49 N/mg/gravitasiekrag N FN/normal FN/normaal f Friction/Ff Wrywing/Ff  N f Note/Nota  One mark for correct arrow and label. If any other forces shown max. 2 3  w Een punt vir die korrekte pyl en benaming Indien enige ander krag getoon word maks. 2 3 (3) 2.3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Upward positive: Upward negative: Opwaarts positief: Opwaarts negatief: For/Vir 2 kg For/Vir 2 kg Fnet  ma Fnet  ma  T  FG  ma FG  T  ma  21 19,6  2a   19,6  21  2a  a  0,7 m  s -2 a  0,7 m  s -2  a  0,7 m  s -2 (3) 2

Downloaded from hlayiso.com 10841/14 2.4 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Parallel down positive: Parallel down negative: Parallel af positief: Parallel af negatief: For/Vir 5 kg For/Vir 5 kg Fnet  ma Fnet  ma FG sin  T  f  ma  FG sin  T  f  ma   78,4 sin20  21 f  8(0,7)   78,4 sin20  21 f  8( 0,7)  f  0,21N  f  0,21N Note/Nota  Any one of the  f  0,21N two answers. Enige een van die twee antwoorde (4) [11] 3

Downloaded from hlayiso.com 10841/14 QUESTION 3/VRAAG 3 3.1 9,8 m⋅s-2  downward/to the centre of the earth /  9,8 m⋅s-2  afwaarts / na die middelpunt van die aarde  (2) 3.2 OPTION 1/ OPSIE 1 OPTION 2 / OPSIE 2 Downward positive: Downward negative: Afwaarts positief: Afwaarts negatief: From t1 to t2 / Vanaf t1 na t2 From t1 to t2 / Vanaf t1 na t2 y  v i t  21 at 2  y  v i t  21 at 2  0,9  0  21 (9,8)t 2  0,9  0  1 ( 9,8)t 2  2 t  0,43 s  t  0,43 s  OPTION 3 / OPSIE 3 OPTION 4 / OPSIE 4 Downward positive: Downward negative: Afwaarts positief: Afwaarts negatief: From 0 to t1 / Vanaf 0 na t1 From 0 to t1 / Vanaf 0 na t1 v 2f  v i2  2ay v 2f  v i2  2ay 0  v i2  2(9,8)( 0,9)  0  v i2  2( 9,8)(0,9)  v i  4,2 m  s -1 v i  4,2 m  s -1  v i  4,2 m  s -1  v i  4,2 m  s -1 v f  v i  at  v f  v i  at  0  4,2  (9,8)t  0  4,2  ( 9,8)t  t  0,43 s  t  0,43 s  4

Downloaded from hlayiso.com 10841/14 OPTION 5 / OPSIE 5 OPTION 6 / OPSIE 6 Downward positive: Downward negative: Afwaarts positief: Afwaarts negatief: From 0 to t1 / Vanaf 0 na t1 From 0 to t1 / Vanaf 0 na t1 v 2f  v i2  2ay v 2f  v i2  2ay 0  v i2  2(9,8)( 0,9) 0  v i2  2( 9,8)(0,9) v i  4,2 m  s -1 v i  4,2 m  s -1  v i  4,2 m  s -1  v i  4,2 m  s -1 vi  v f vi  v f y  t  y  t  2 2  4,2  0  4,2  0  0,9  t 0,9  t  2 2 t  0,43 s  t  0,43 s  (4) 3.3 POSITIVE MARKING FROM QUESTION 3.2 POSITIEWE NASIEN VANAF VRAAG 3.2 OPTION 1 / OPSIE 1 OPTION 1 / OPSIE 1 Downward positive: Downward negative: Afwaarts positief: Afwaarts negatief: From 0 to t1 / Vanaf 0 na t1 From 0 to t1 / Vanaf 0 na t1 v 2f  v i2  2ay  v 2f  v i2  2ay  0  v i2  2(9,8)( 0,9)  0  v i2  2( 9,8)(0,9)  v i  4,2 m  s -1 v i  4,2 m  s -1   v i  4,2 m  s -1  v i  4,2 m  s -1 upward/opwaarts   v i  4,2 m  s -1 upward/opwaarts OPTION 3 / OPSIE 3 OPTION 4 / OPSIE 4 Downward positive: Downward negative: Afwaarts positief: Afwaarts negatief: From 0 to t2 / Vanaf 0 na t2 From 0 to t2 / Vanaf 0 na t2 y  v i t  21 at 2  y  v i t  21 at 2      0,9  v i (0,43)  21 (9,8)(0,43) 2  0,9  v i (0,43)  21 (9,8)(0,43) 2 v i  4,2 m  s -1 v i  4,2 m  s -1    v i  4,2 m  s -1 upward/opwaarts  v i  4,2 m  s -1 upward/opwaarts (4) 5

Downloaded from hlayiso.com 10841/14 3.4 POSITIVE MARKING FROM QUESTION 3.2 AND 3.3 POSITIEWE NASIEN VANAF VRAAG 3.2 EN 3.3 -1 v (m·s ) Upward positive/Opwaarts positief Criteria for graph/Kriteria vir Marks/ grafiek: Punte A Straight line, crossing x-axis 4,2  Reguitlyn, kruis die x-as B AB is shorter than BC 0  t1 t (s) AB is korter as BC vi = ± 4,2 m⋅s-1  C t1 = 0,43 s  Downward positive/Afwaarts positief -1 v (m·s ) Note/Nota C if no/incorrect labels on axis: max. as geen/verkeerde benamings vir asse dan: maks. B t (s) t1 - 4,2 A (4) [14] 6

Downloaded from hlayiso.com 10841/14 QUESTION 4/VRAAG 4 4.1 Gm ZmX  FG  d2  (6,67  1011)(5,98  1024 )(1,99  1019 ) FG  (1,496  106 )2  FG  3,52  1021 N attractive / aantrekkend  OR/OF FG  3,55  1021 N towards X / na X (4) 4.2 (3,55 × 1021) × 8 = 2,84 × 1022 N towards moon Y / na maan Y (2) 4.3 2 Fnet  FMX 2  FMY 2   3,55  1021   2.84  1022  2 2 2 Fnet Fnet  2,86  1022 N  (2) [8] QUESTION 5/VRAAG 5 5.1 Impulse  area between graph and x - axis Impuls  oppervlakte tussen grafiek en x - as Impulse  1 bh  1 bh  lb  2 2 Impulse  (2)(10)  12 (2)( 10)  (6)( 10)  1 2 Impulse  60 N  s   Impulse  60 N  s in oppositedirection/ west  Impuls  60 N  s in teenoorges telde rigting/we s (3) 5.2 Greater than / Groter as  (1) 5.3 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 East positive: West positive: Oos positief: Wes positief: Impulse  mv f  mvi  Impulse  mv f  mvi    60  (12)v  (12)(5,5)    60  (12)v f  (12)(5,5)  f v f  0,5 m  s east / oos  -1 v f  0,5 m  s -1 east / oos  (4) [8] 7

Downloaded from hlayiso.com 10841/14 QUESTION 6/VRAAG 6 6.1 W  Fx cos   W  (20)(32) cos 10  W  630,28 J  (3) 6.2 OPTION 1 / OPSIE 1 Notes/Aantekeninge: 2 If “net” left out for Fnet : Max. Fnet  F  FG sin   f 5 Fnet  20  59,78 sin 10  11  As “net” uitgelaat is Fnet  1,38 N 2 Fnet  1,38 N backwards/agteruit vir Fnet: Maks. 5 Wnet  E K  Fnet x cos   m( v  v ) 1 2 2  2 f i (1,38)(32) cos180  21 (6,1)( v 2f  5 2 ) v 2f  10,51  v f  3,24 m  s -1 (5) OPTION 2 / OPSIE 2 Wnet  E K  Fx cos   fx cos   Fg x cos   12 m(v 2f  vi2 )    (20)(32) cos 0  (11)(32) cos 180  (59,78)(32) cos 10  12 (6,1)v 2f  12 (6,1)(5 2 ) v 2f  10,51 v f  3,24 m  s -1  OPTION 3 / OPSIE 3 Wnet  EK   Fx cos   fx cos   Fgx cos   21 m( v 2f  v i2 )   (20)(32) cos 0  (11)(32) cos 180  (59,78 sin10)(32) cos 180  21 (6,1)v 2f  21 (6,1)(52 ) v 2f  10,51 v f  3,24 m  s-1  6.3 The net work done by the non-conservative force / frictional force depends on the path the object travelled  / Die netto arbeid verrig deur ’n nie-konserwatiewe krag / wrywingskrag is onafhanklik van die roete geneem  OR/OF The mechanical energy is not constant for a non-conservative force / Die meganiese energie is nie konstant vir ’n nie-konserwatiewe krag nie (2) 8

Downloaded from hlayiso.com 10841/14 [10] 9

Downloaded from hlayiso.com 10841/14 QUESTION 7/VRAAG 7 7.1.1 ΣE m 56m  ΣE m 18m EP  EK 56m  EP  EK 18m  mgh  0  mgh  21 mv 2 9,8(56) (9,8)(18)  1 v2 2 -1 v  27,29 m  s  (4) 7.1.2 POSITIVE MARKING FROM QUESTION 7.1 POSITIEWE NASIEN VANAF VRAAG 7.1 v  vL Notes: fL  fs  Any other Doppler formula, e.g. v  vs v  vL OR/OF fL  fs - Max. 3 for calculation v v  vs 4 fL  fs  v  vs  Marking rule 1.5: No penalisation if zero substitutions are omitted.  f   340  497  10 3  Aantekeninge: L  340 - 27,29   Enige ander Dopplerformule, bv. f  5,4  10 Hz 5  v  vL L fL  fs - Maks. 3 vir berekening v  vs 4  Nasienreël 1.5: Geen punte word verbeur indien nul substitusies weggelaat word nie. (4) 7.1.3 The phone is moving away from Keenan who is stationary, the phone is increasing the distance between each wavefront and therefore the wavelength increases. This causes the frequency to decrease since for the same speed of sound  / Die selfoon beweeg weg van Keenan af, die foon vergroot dus die grootte van die afstand tussen die golffronte en die golflengte vergroot.  Dit veroorsaak dat die frekwensie afneem  aangesien vir dieselfde spoed van klank. (3) 7.1.4 HIGHER;  / HOËR  (1) 7.2 7.2.1 The observed frequency is the same as the original frequency of the source. / the car is moving perpendicular to the camera / Die waargenome frekwensie is dieselfde as die oorspronklike frekwensie van die bron / die motor beweeg loodreg ten opsigte van die kamera. (1) 7.2.2 3 × 108 m⋅s-1  It is an electromagnetic wave. / Dit is ’n elektromagnetiese golf.  (2) [15] 10

Downloaded from hlayiso.com 10841/14 QUESTION 8/VRAAG 8 8.1 Criteria for sketch Marks Kriteria vir skets Punte Correct shape as shown.  Korrekte vorm soos getoon Direction from positive to negative.  Rigting van positief na negatief. Field lines start on spheres and do  not cross. Veldlyne begin op elke sfeer en kruis nie. (3) 8.2 kQ A Q B  FB on A  r2 FB on A      9  10 9 4,5  10 5 9,5  10 5  0,352  FB on A  314,08 N to the right/ na regs  (4) 8.3 POSITIVE MARKING FROM QUESTION 8.2 POSITIEWE NASIEN VANAF VRAAG 8.2 Option 1/Opsie 1 kQ EA  2A  r EA    9  10 9 4,5  10 5  0,352 E A  3,31 10 6 N  C -1 to the right / na regs Option 2/Opsie 2 FA op B  EA  qB 314,08 EA   9,5  10 -5 E A  3,31 10 6 N  C -1 to the right / na regs  (3) [10] 11

Downloaded from hlayiso.com 10841/14 QUESTION 9 / VRAAG 9 9.1 Criteria for hypothesis /Kriteria vir hipotese: Mark/ Punt The dependent and independent variables are stated.  Die afhanklike en onafhanklike veranderlikes is genoem. - Makes a statement about the relationship between dependent and independent variables.  Maak ’n stelling oor die verwantskap tussen die afhanklike en onafhanklike veranderlikes. Independent variable/Onafhanklike veranderlike: Notes/ Amount of potato cells / Aantal aartappelselle Aantekeninge: A statement that does Dependent variable/Afhanklike veranderlike: not contain a Bulb glows or not /Gloeilampie gloei of nie relationship: Max ½ Or/of ’n Stelling wat geen Current (strength) /Stroom(sterkte) verwantskap bevat nie: Maks. ½ Examples/Voorbeelde: More potato cells will increase the brightness of the bulb/ Example/ Meer aartappelselle sal die helderheid van die gloeilampie laat Voorbeeld: toeneem. Three potato cells are needed to make penlight bulb glow. ½ More potato cells will increase the current through the bulb/ Meer aartappelselle sal die grootte van die stroom deur die Drie aartappelselle is gloeilampie laat toeneem. nodig om die gloei- lampie te laat gloei. ½ (2) 9.2 Dependent variable/Afhanklike veranderlike: Bulb glows or not /Gloeilampie gloei of nie  Or/of Current (strength) /Stroom(sterkte)  (1) 9.3 1,6 V  (1) 9.4 The battery has an internal resistance /  Die battery het ’n interne weerstand  (1) 9.5 V2  P R (0,02) 2  P 2 P  2  10  4 W  (3) [8] 12

Downloaded from hlayiso.com 10841/14 QUESTION 10/VRAAG 10 10.1 The potential difference is directly proportional to the current  through the resistor if the temperature stays the same.  Die potensiaalverskil is direk eweredig aan die stroomsterkte  deur die resistor, mits die temperatuur konstant bly  vir die resistor. (2) 10.2 V  R1  I 12,9 R1   1,5 R 1  8,6   (3) 10.3 POSITIVE MARKING FROM QUESTION 11.2 POSITIEWE NASIEN VAN VRAAG 11.2 1 1 1   R tot R1 R series  1 1 1    R tot 8,6 ( 4  5) 1 88  R tot 387 R tot  4,4   (3) 10.4 POSITIVE MARKING FROM QUESTION 11.3 POSITIEWE NASIEN VAN VRAAG 11.3 Option 1/Opsie 1: Option 2/Opsie 2: V V I total   I total   R total R total 12,9 12,9 I total   I total  4,4 4,4 I total  2,93 A I total  2,93 A Vlost   IR  Ir r I 15  (2,93)( 4,4)  (2,93)r  15  12,9  r  0,72   r 2,93 r  0,72   13

Downloaded from hlayiso.com 10841/14 Option 3/Opsie 3: Option 4/Opsie 4: V V I serie  I serie  R serie   R serie 12,9 12,9 I serie   I serie   9 9 I serie  1,43 A I serie  1,43 A I total  1,5  1,43 I total  1,5  1,43 I total  2,93 A I total  2,93 A Vlost   IR  Ir r I 15  (2,93)( 4,4)  (2,93)r 15  12,9  r  0,72   r 2,93 r  0,72   (4) [12] QUESTION 11/VRAAG 11 11.1 11.1.1 DC;  It is powered by a DC-source  GS;  Dit word aangedryf deur ʼn GS-bron  (2) 11.1.2  Increase the emf of the source  Increase the amount or turns on the coil  Increase the strength of the magnet  [any two]  Verhoog die emk van die bron  Vermeerder die aantal windings op die spoel  Gebruik ’n sterker magneet  [enige twee] (2) 11.1.3 X: South pole / Suidpool Y: North pole / Noordpool  (2) 14

Downloaded from hlayiso.com 10841/14 11.2 11.2.1 Option 1/Opsie 1: Option 2/Opsie 2: I max I rms   Vrms  I rms R 2 80 I I rms   Vrms  max R 2 2 I rms  56,57 A 80 Vrms  ( ) )(510  2 Vrms  I rms R  Vrms  2,89  10 4 V Vrms  (56,57)(510)  Vrms  2,89  10 4 V  (5) 11.2.2 Option 1/Opsie 1: Option 2/Opsie 2: Pave  I rms 2 R 2 Vrms Pave   Pave  (56,57) 2 (510)  R Pave  1,63  10 6 W  (2,89  10 4 ) 2  Pave  510  Pave  1,63  10 6 W (3) Option 3/Opsie 3: Pave  Vrms I rms  Pave  (2,89  10 4 )(56,57)  Pave  1,63  10 6 W  11.2.3 Criteria for graph/Kriteria vir Marks/ I (A) grafiek: Punte 80 Sinus or cosines graph  Sinus of kosinus grafiek Two cycles  Twee siklusse t (s) Max and min I-values indicated  Maks en min I-waardes getoon -80 Note/Nota if no/incorrect labels on y-axis: max. as geen/verkeerde benamings vir y-as dan: maks. (4) [18] 15

Downloaded from hlayiso.com 10841/14 QUESTION 12/VRAAG 12 12.1 12.1.1 Workfunction is the minimum energy of a photon needed to set an electron free from the surface of a metal / Werksfunksie is die minimum energie wat ’n foton moet hê om ’n elektron vry te stel uit die oppervlak van ’n metaal. (2) 12.1.2 43,9 × 10-19 Hz  (2) 12.1.3 POSITIVE MARKING FROM QUESTION 13.1.2 POSITIEWE NASIEN VAN VRAAG 13.1.2 Option 1/Opsie 1: E  W0  E K hf  hf0  21 mv 2  Any one/Enige een    (6,63  10 34 )(100  10 19 )  (6,63  10 34 )( 43,9  10 19 )  21 (9,11 x 10 -31 )v 2 v 2  8,17  10 21 v  9,04  10 11 m  s -1  Option 2/Opsie 2: E  hf  E  (6,63  10 34 )(100  10 19 ) E  6,63  10 51 J W0  hf0  W0  (6,63  10 34 )( 43,9  10 19 ) W0  2,91 10 51 J  E K  21 mv 2  E  W0  21 (9,11 x 10 -31 )v 2 6,63  10 51  2,91 10 51  21 (9,11 x 10 -31 )v 2 v 2  8,17  10 21 v  9,04  10 11 m  s -1  (5) 12.1.4 The data points are not connected directly, an average straight line connecting most of the points are drawn / Die data punte is nie direk verbind nie, ’n gemiddelde reguitlyn wat deur so veel as moontlik punte gaan word geteken.  (1) 16

Downloaded from hlayiso.com 10841/14 12.2 12.2.1 Emission spectra occurs when a light source gives off light and absorption spectra occurs when white light is observed through a cold gas/ Emissiespektra vorm as ’n ligbron lig afgee terwyl absorpsiespektra vorm as die wit lig waargeneem word deur ’n koue gas.  (2) 12.2.2 RED  it has the longest wavelength of all the visible colours/ (2) ROOI  Dit het die langste golflengte van al die sigbare kleure.  [14] [125] 150 17