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

> Grade 12 Physical Sciences Paper 1 memorandum from the Gauteng Mock Exam 2014 (Afrikaans), 18 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2014afr-gau-physical-sciences-p1-memo/db77042f-0bdc-4e06-8f6b-ce760a7e4148
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2014
- Exam period: Gauteng Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 18

Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 P1 Gauteng Mock 2014 Memo (Afr)](https://www.hlayiso.com/library/2014afr-gau-physical-sciences-p1-memo/db77042f-0bdc-4e06-8f6b-ce760a7e4148)

## About this document

Physical Sciences · Grade 12 · Gauteng Mock Exam · 2014. Memorandum, 18 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

Downloaded from hlayiso.com

     PREPARATORY EXAMINATION/
      VOORBEREIDENDE EKSAMEN
                     2014
               MEMORANDUM

                 PHYSICAL SCIENCES: PHYSICS (P1)/
SUBJECT/VAK:     FISIESE WETENSKAPPE: FISIKA (V1)
                 (10841)

### Page 2

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

### Page 3

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

### Page 4

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

### Page 5

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

### Page 6

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

### Page 7

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

### Page 8

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

### Page 9

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

### Page 10

Downloaded from hlayiso.com
                              10841/14

                                  [10]

             9

### Page 11

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

### Page 12

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

### Page 13

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

### Page 14

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

### Page 15

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

### Page 16

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

### Page 17

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

### Page 18

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

## Related documents

- [2014ENG gau physical sciences p1 memo](https://www.hlayiso.com/library/2014eng-gau-physical-sciences-p1-memo/d8eca56a-f776-4202-84cd-3e2e7e299ae3) — Physical Sciences · Memo · 2014
- [2014ENG gau physical science p2 eng memo](https://www.hlayiso.com/library/2014eng-gau-physical-science-p2-eng-memo/5fca6597-743c-4f85-8f26-3dea294e1dfe) — Physical Sciences · Memo · 2014 · Eng
- [PHYS SC P1 MEMO ENGAFR](https://www.hlayiso.com/library/phys-sc-p1-memo-engafr/da9b9ae0-718d-442a-95bc-aa0bfbf308f3) — Physical Sciences · Memo · 2014
- [Physical Sciences P1 Nov 2014 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2014-memo-afr-eng/44924678-c33b-42e1-8937-fcfda16ea7b3) — Physical Sciences · Memo · 2014 · Eng
- [wkp kwod physc p1 sept 2014 eng memo](https://www.hlayiso.com/library/wkp-kwod-physc-p1-sept-2014-eng-memo/bb06a52a-4a49-454d-ae6e-5128be28cc24) — Physical Sciences · Memo · 2014 · Eng
- [2019 AFR ENG Gauteng Paper 1 Memo](https://www.hlayiso.com/library/2019-afr-eng-gauteng-paper-1-memo/7e7c9cfc-f2f5-4700-a132-61eae8192020) — Physical Sciences · Memo · 2019 · Eng

## More from Grade 12 Physical Sciences

- [PHYSICAL SCIENCE P1 QP JCT 2025 AFR D watermark](https://www.hlayiso.com/library/physical-science-p1-qp-jct-2025-afr-d-watermark/b9429ac5-acbb-4fae-989d-af9e1478a96b) — QP · 2025 · Afr
- [PHYSICAL SC P2 QP GR12 SEPTEMBER 2025 AFR1 watermark](https://www.hlayiso.com/library/physical-sc-p2-qp-gr12-september-2025-afr1-watermark/37cff2db-7c1a-4ec8-8ac5-07ec5bcf87a9) — QP · 2025
- [Physics Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/physics-grade-12-nsc-p2-qp-september-2025-gauteng/4d3cbea6-10ec-47ff-98ee-50646165ae9f) — QP · 2025
- [Physics Grade 12 NSC P1 QP September 2025 KZN](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-kzn/6d225dc8-24c6-4c28-a181-df906d5a2c0e) — QP · 2025
- [Physical Sciences Physics Grade 12 PS01 QP March 2025 Limpopo](https://www.hlayiso.com/library/physical-sciences-physics-grade-12-ps01-qp-march-2025-limpopo/482ee9af-e88a-4818-9120-11ab9e045a1c) — QP · 2025
- [Physics Grade 12 NSC P1 QP September 2025 Limpopo](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-limpopo/b65e9832-a6be-4312-8f5a-57425d647121) — QP · 2025

## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

---

Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.
