# Physical Sciences Grade 12 Paper 1 NSC November 2012 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the NSC November Exam 2012, 12 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-sciences-p1-nov-2012-memo-afr-eng/ce0f1b88-3ac7-4821-bbaa-a0b7c294fcd0
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2012
- Exam period: NSC November Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 12

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Physical Sciences · Grade 12 · NSC November Exam · 2012. Memorandum, 12 pages. Read online or download the PDF.

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### Page 1

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                                      NATIONAL
                                 SENIOR CERTIFICATE
                                     NASIONALE
                                 SENIOR SERTIFIKAAT

                                         GRADE/GRAAD 12

                        PHYSICAL SCIENCES: PHYSICS (P1)
                        FISIESE WETENSKAPPE: FISIKA (V1)

                                        NOVEMBER 2012

                                        MEMORANDUM

MARKS/PUNTE: 150

                        This memorandum consists of 12 pages.
                      Hierdie memorandum bestaan uit 12 bladsye.

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### Page 2

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Physical Sciences P1/Fisiese Wetenskappe V1    2                         DBE/November 2012
                                     NSC/NSS – Memorandum

SECTION A

QUESTION 1/VRAAG 1

1.1       Frequency/Frekwensie                                                                   (1)

1.2       Capacitor/Kapasitor                                                                    (1)

1.3       Split ring commutator 
          Splitringkommutator                                                                     (1)

1.4       Photons/Fotone                                                                         (1)

1.5       Relative velocity/Relatiewe snelheid                                                   (1)
                                                                                                  [5]

QUESTION 2/VRAAG 2

2.1       D                                                                                     (2)

2.2       C                                                                                     (2)

2.3       D                                                                                     (2)

2.4       D                                                                                     (2)

2.5       A                                                                                     (2)

2.6       A                                                                                     (2)

2.7       D                                                                                     (2)

2.8       C                                                                                     (2)

2.9       C                                                                                     (2)

2.10      A                                                                                      (2)
                                                                                                  [20]

                                          TOTAL SECTION A/TOTAAL AFDELING A:                      25

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Physical Sciences P1/Fisiese Wetenskappe V1    3                         DBE/November 2012
                                     NSC/NSS – Memorandum

SECTION B/AFDELING B

QUESTION 3/VRAAG 3

3.1       Downward/afwaarts                                                                      (1)
3.2
3.2.1     Upwards positive/Opwaarts positief:
          v f = v i + aΔt 
             = 8 + (-9,8)(4) 
             = - 31,2 m∙s-1
          ∴ v f = 31,2 m∙s-1 

          Downwards positive/Afwaarts positief:
          v f = v i + aΔt
             = - 8 + (9,8)(4) 
          ∴ v f = 31,2 m∙s-1                                                                     (4)
3.2.2     OPTION 1/OPSIE 1
          Upwards positive/Opwaarts positief:
          Δy = v i Δt + ½aΔt2
             = (8)(4)  + ½(-9,8)(4)2 
             = -46,4 m

          Height of balcony/Hoogte van balkon:
          60 – 46,4 = 13,6 m 

          Downwards positive/Afwaarts positief:
          Δy = v i Δt + ½aΔt2
             = (-8)(4)  + ½(9,8)(4)2 
             = 46,4 m

          Height of balcony/Hoogte van balkon:
          60 – 46,4 = 13,6 m 

          OPTION 2/OPSIE 2
          Upwards positive/Opwaarts positief:
          Δy = v i Δt + ½ aΔt2
             = (27,13)  (6) + ½ (-9,8)(6)2
             = - 13,62 m

          Height of balcony/Hoogte van balkon:
          = 13,62 m 

          Downwards positive/Afwaarts positief:
          Δy = v i Δt + ½ aΔt2
             = (-27,13)  (6) + ½ (9,8)(6)2
             = 13,62 m

          Height of balcony/Hoogte van balkon:
          = 13,62 m                                                                              (5)
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Physical Sciences P1/Fisiese Wetenskappe V1    4                               DBE/November 2012
                                     NSC/NSS – Memorandum

3.3               OPTION 1/OPSIE 1
                  Upwards positive/Opwaarts positief:
                          27,13

        Velocity/snelheid (m∙s-1)
                                    8

                                    0            4                                  10    Time/tyd(s)

                              -31,2

                                                                                   Marks/
       Criteria for graph/Kriteria vir grafiek:
                                                                                   Punte
       Shape has two parallel lines with a gradient.
                                                                                      
       Vorm het twee ewewydige lyne met gradient.
       First part of graph starts at v = 8 m∙s-1 at t = 0 s
                                                                                      
       Eerste deel van grafiek begin by v = 8 m∙s-1 by t = 0 s.
       Positive marking from QUESTION 3.2.1:
       Positiewe nasien vanaf VRAAG 3.2.1:
       First part of the graph extends below the x axis until v = -31,2 m∙s-1
                                                                                      
       at t = 4 s.
       Eerste deel van die grafiek verleng onder x-as tot v = -31,2 m∙s-1
       by t = 4 s.
       Graph is discontinuous and object changes direction at 4 s.
                                                                                      
       Grafiek is nie kontinu nie en voorwerp verander van rigting by 4 s.
       Second part of graph starts at v = 27,13 m∙s-1 at t = 4 s.
                                                                                      
       Tweede deel van grafiek begin by v = 27,13 m∙s-1 by t = 4 s.
       Second part of graph extends below the x axis until t = 10 s.
                                                                                      
       Tweede deel van grafiek verleng onder x-as tot t =10 s.

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Physical Sciences P1/Fisiese Wetenskappe V1    5                                     DBE/November 2012
                                     NSC/NSS – Memorandum

                                  OPTION 2/OPSIE 2
                                  Upwards negative/Opwaarts negatief:

                                  31,2

      Velocity/snelheid (m∙s-1)
                                                                                          Time/tyd(s)
                                         0
                                                           4                        10

                                     -8

                          -27,13

                                                                                         Marks
              Criteria for graph/Kriteria vir grafiek:
                                                                                         Punte
              Correct shape as shown (two parallel lines).
                                                                                            
              Korrekte vorm soos aangetoon (twee ewewydige lyne).
              First part of graph starts at v = -8 m∙s-1 at t = 0 s
                                                                                            
              Eerste deel van grafiek begin by v = -8 m∙s-1 by t = 0 s
              Positive marking from QUESTION 3.2.1.
              Positiewe nasien vanaf VRAAG 3.2.1.
              First part of the graph extends above the x axis until v = 31,2 m∙s-1
                                                                                            
              at t = 4 s.
              Eerste deel van die grafiek verleng bokant x-as tot v = 31,2 m∙s-1
              by t = 4 s.
              Graph is discontinuous and object changes direction at 4 s.
                                                                                            
              Grafiek is nie kontinu en voorwerp verander van rigting by 4 s.
              Second part of graph starts at v = -27,13 m∙s-1 at t = 4 s.
                                                                                            
              Tweede deel van grafiek begin by v = -27,13 m∙s-1 by t = 4 s.
              Second part of graph extends above the x axis until t = 10 s.
                                                                                            
              Tweede deel van grafiek verleng bokant x-as tot t = 10 s.

                                             .                                                                 (6)
                                                                                                              [16]

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Physical Sciences P1/Fisiese Wetenskappe V1    6                            DBE/November 2012
                                     NSC/NSS – Memorandum

QUESTION 4/VRAAG 4

4.1       40 m∙s-1 east/oos                                                                        (2)

4.2       The total (linear) momentum remains constant/is conserved 
          in an isolated/a closed system/the absence of external forces/ if the impulse of
          external forces is zero. 

          Die totale (liniêre) momentum bly konstant/behoue 
          in 'n geïsoleerde sisteem/geslote sisteem/ die afwesigheid van eksterne
          kragte./ indien die impuls van eksterne kragte nul is.                                    (2)
4.3       East positive/Oos positief:
          Σp i = Σp f 
          m(20) + 2m(-20)  = (m + 2m)v f 
          ∴v f = -6,67 m·s-1
          ∴v f = 6,67 m·s-1  west /wes 
          East negative/Oos negatief:
          Σp i = Σp f 
          m(-20) + 2m(+20)  = (m + 2m)v f 
          ∴v f = 6,67 m·s-1  west /wes                                                             (6)

4.4
4.4.1     F
          Newton’s Third Law of motion/Newton se Derde Bewegingswet                                 (2)

4.4.2     -½ a / ½ a
                                        1
          (Same/Dieselfe F net ), a α     
                                        m                                                            (2)
4.4.3     Car driver 
          (Car - driver system) have greater acceleration. 
          (Car - driver system) have greater change in velocity /greater Δv.
          Motorbestuurder 
          (Motor -bestuurder sisteem) het groter versnelling. 
          (Motor -bestuurder sisteem) het groter verandering in snelheid / groter Δv.                (3)
                                                                                                     [17]

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Physical Sciences P1/Fisiese Wetenskappe V1    7                                DBE/November 2012
                                     NSC/NSS – Memorandum

QUESTION 5/VRAAG 5
5.1
                 N

                     F

             w
                                                                                                         (3)
5.2      The net (total) work (done on an object) is equal to 
         the change in kinetic energy (of the object.) 
         Die netto (totale) arbeid verrig (op 'n voorwerp) is gelyk aan 
         die verandering in kinetiese energie (van die voorwerp).                                       (2)

5.3
5.3.1     W net = ΔE k /ΔK  OR/OF F net Δxcosθ = ½ m(v f 2 – v i 2)
          F net (1,02)cos180° = ½ (1 200)(0 – 202) 
          F net = 235 294,12 N  (2,35 x 105 N)
                                                                                                         (4)
5.3.2     OPTION 1 /OPSIE 1                             OPTION 2/OPSIE 1
          F net Δt = mΔv                                      v + vf 
                                                        Δx =  i        Δt 
          ∴ (-235 294,12)Δt = (1 200)(0 - 20)                2 
          ∴ Δt = 0,1 s (0,102 s)                                  20 + 0 
                                                        1,02 =            Δt 
                                                                   2 
                                                        Δt = 0,1 s                                       (4)
                                                                                                         [13]
QUESTION 6/VRAAG 6
6.1       Frequency/Frekwensie                                                                          (1)

6.2       There is relative motion between the bird and the bird watcher. 
          Daar is relatiewe beweging tussen die voël en die voëlkyker nie.                              (1)

6.3       0,2 m                                                                                         (1)
6.4
6.4.1     v = fλ 
          340 = f(0,2) 
          ∴ f = 1 700 Hz                                                                                (3)
6.4.2         v ± vL                   v
          fL =       f s OR/OF fL =         fs 
              v ± vs                 v − vs
                           340
          ∴ 1 700  =               (1 650) 
                         340 − v s
             ∴ v s = 10 m∙s-1                                                                            (5)
                                                                                                         [11]
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Physical Sciences P1/Fisiese Wetenskappe V1    8                                DBE/November 2012
                                     NSC/NSS – Memorandum

QUESTION 7/VRAAG 7
7.1       Double slit/Dubbelspleet                                                                      (1)
7.2       (Alternate) dark and bright/blue bands. 
          Bright / blue bands of equal broadness (width). 
          (Afwissellende) donker en helder/blou bande. 
          Helder / blou bande van gelyke breedte.                                                       (2)
7.3
7.3.1              1 central band   1 sentraleband
          tan θ =   2              / 2
                 screen dis tan ce   skermafstand
                      1 (0,22) 
          ∴ tan θ = 2
                        1,4 
          ∴ θ = 4,49°                                                                                   (3)

7.3.2
         OPTION 1/OPSIE 1:                     OPTION 2/OPSIE 2:
                 mλ                                    mλ
         sin θ =                              sin θ =     
                  a                                     a
                                                                      
                   (1)( 470 × 10 −9 )                       ( −1)( 470 × 10 −9 )
         sin4,49 =                             sin (-4,49°) =
                           a                                           a
         ∴ a = 6 × 10 −6 m (6 003,67 nm)      ∴ a = 6 × 10 −6 m  (6 003,67 nm)                         (5)

7.4       λ red light > λ blue light 
          (Degree of) diffraction/sin θ / θ α wavelength ( λ ) 
          λ rooilig > λ bloulig 
          Diffraksie α golflengte ( λ )                                                                  (2)
                                                                                                         [13]

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Physical Sciences P1/Fisiese Wetenskappe V1    9                             DBE/November 2012
                                     NSC/NSS – Memorandum

QUESTION 8/VRAAG 8

8.1            V
          R=     
               I
                  12
          1 000 =    
                   I
          ∴I = 0,01 A                                                                                (3)
8.2       12 V                                                                                       (1)
8.3            Q
          C=     
               V
                      Q
          120 x 10-6 =   
                      12
          ∴Q = 1,44 x 10-3 C                                                                         (3)
8.4
8.4.1     Decreases/Verminder                                                                        (1)
8.4.2     Increases/Vermeerder                                                                       (1)
8.5
8.5.1
                                                                                          Marks/
+                                       Criteria for sketch:/Kriteria vir skets:
                                                                                          Punte
                                        Parallel lines equally spaced.
                                                                                             
                                        Parallelle lyne eweredig gespasieer.
                                        Direction from positive plate towards negative
                                        plate.(Polarity of plates must be indicated)
                                        Rigting vanaf positiewe plaat na negatiewe           
                                        plaat.(Polariteit van plate moet aangedui
 -                                      word)
                                        Field curved at the ends of the plates.
                                                                                             
                                        Veld gekrom aan einde van die plate.
                                                                                                      (3)
8.5.2
              V
          E=     
               d
                 12
           =              
              12 × 10 − 3
          ∴ E = 1 000 V∙m-1                                                                           (3)
                                                                                                      [15]

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Physical Sciences P1/Fisiese Wetenskappe V1   10                              DBE/November 2012
                                     NSC/NSS – Memorandum

QUESTION 9/VRAAG 9

9.1
9.1.1     1   1   1
            =   +   
          Rp R1 R 2
                 1   1
             =     +    
                60 60
          ∴ R p = 30 Ω                                                                                (3)

9.1.2     OPTION 1 / OPSIE 1                    OPTION 2 / OPSIE 2:
          R ext = 30 + 25 = 55 Ω               R tot = (30 + 25) +1,5 = 56,5 Ω
          Emf/emk = I(R + r)                   V = IR
          ∴12 = I(55 + 1,5)                   12 = I(56,5) 
          ∴I = 0,21 A                          ∴I = 0,21 A                                           (5)

9.1.3     OPTION 1/OPSIE 1                      OPTION 2/OPSIE 2
          V = IR                               V = IR 
            = (0,21)(30)                         = (0,105)(60) 
            = 6,3 V                              = 6,3 V 
                                                                                                       (3)

9.2
9.2.1     1,5 V                                                                                       (1)

9.2.2                    ∆V
          gradient/m =
                          ∆I
                         0,65 - 1,5 
                      =
                           1,0 - 0 
                      = - 0,85 Ω                                                                      (3)

9.2.3     Internal resistance 
          Interne weerstand                                                                            (2)

9.2.4     Decreases/Verminder 
          When I increases/Wanneer I toeneem:
          “Lost volts”/ Ir increases./“Verlore volts”/Ir neem toe. 
          V ext = emf – Ir decreases. /V ext = emk – Ir neem af.                                       (3)
                                                                                                       [20]

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Physical Sciences P1/Fisiese Wetenskappe V1   11                                    DBE/November 2012
                                     NSC/NSS – Memorandum

QUESTION 10/VRAAG 10
10.1      AC /WS                                                                                         (1)
10.2     V

                                             t

                                                                                  Marks
              Criteria for graph/Kriteria vir grafiek:
                                                                                  Punte
              Correct shape as shown; accept more than one cycle.
                                                                                    
              Korrekte vorm soos aangetoon; aanvaar meer as een siklus.
              If no/wrong labels: minus 1 mark
              Indien geen/verkeerde byskifte: minus 1 punt
                                                                                                          (2)
10.3      OPTION 1/OPSIE 1                               OPTION 2 / OPSIE 2
                       V                                 P ave = V rms Irms /P gem. = V wgk I wgk
          Vrms / wgk = max/ maks 
                           2                                            Vmax I rms Vmaks I wgk
                                                         P ave/gem. =             /            
                       30 × 10 3                                            2          2
                     =           
                           2
                                                                            (30 × 103 )Irms / wgk
                     = 2,12 x 104 V                      4,45 x 109  =                             
                                                                                      2
          P ave = V rms Irms /P gem. = V wgk I wgk        ∴I rms/wgk = 2,10 x 105 A 
                     9                   4
          4,45 x 10  = (2,12 x 10 )I rms/wgk
          ∴ Irms/wgk = 2,10 x 105 A                                                                      (5)
10.4      Less loss in (electrical) energy (as heat). 
          Minder verlies aan (elektriese) energie (as hitte).                                            (1)
                                                                                                          [9]

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Physical Sciences P1/Fisiese Wetenskappe V1   12                           DBE/November 2012
                                     NSC/NSS – Memorandum

QUESTION 11/VRAAG 11

11.1
11.1.1    Kinetic energy /Kinetiese energie (E k )                                            (1)
11.1.2    Frequency /Frekwensie(f)                                                            (1)
11.1.3    (Type of) metal
          (Soort) metaal                                                                      (1)
11.2      The minimum frequency needed to emit electrons 
          from (the surface of) a metal. 
          Die minimum frekwensie benodig om elektrone vry te stel
          vanaf (die oppervlak van) 'n metaal.                                                 (2)

11.3      9 x 1014 Hz                                                                         (1)

11.4
          E = W 0 + Ek
                             Any one /Enige een
          hf = hf 0 + E k
          (6,63 x 10-34)(14 x 1014)  = (6,63 x 10-34)(9 x 1014)  + E k
          ∴ E k =3,32 x 10-19 J  (3,31 x 10-19 J)
                                                                                               (4)

11.5      Remains the same/Bly dieselfde                                                       (1)
                                                                                               [11]
                                            TOTAL SECTION B/TOTAAL AFDELING B:                 125
                                                   GRAND TOTAL/GROOTTOTAAL:                    150

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