# Physical Sciences Grade 12 P1 NSC Supplementary 2018 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the NSC Supplementary Exam 2018, 14 pages. Read online or download the free PDF.

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- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: NSC Supplementary Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 14

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Physical Sciences · Grade 12 · NSC Supplementary Exam · 2018. Memorandum, 14 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)

             FEBRUARY/MARCH/FEBRUARIE/MAART 2018

                     MARKING GUIDELINES/NASIENRIGLYNE

MARKS/PUNTE: 150

                      These marking guidelines consist of 13 pages.
                       Hierdie nasienriglyne bestaan uit 13 bladsye.

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QUESTION 1/VRAAG 1

1.1       A                                                                                           (2)

1.2       B                                                                                           (2)

1.3       C                                                                                           (2)

1.4       B                                                                                           (2)

1.5       A                                                                                           (2)

1.6       B                                                                                           (2)

1.7       C                                                                                           (2)

1.8       C                                                                                           (2)

1.9       D                                                                                           (2)

1.10      B                                                                                            (2)
                                                                                                        [20]

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QUESTION 2/VRAAG 2

2.1.      A body will remain in its state of rest or motion at constant velocity unless a
          non-zero resultant/net force acts on it. 
          'n Liggaam sal in sy toestand van rus of beweging teen konstante snelheid in
          'n reguitlyn volhard tensy 'n nie-nul resulterende/netto krag daarop inwerk.                     (2)

2.2
           Accepted labels/Aanvaarde benoemings
              Fg / Fw/weight / mg / gravitation force
           w
              Fg / Fw / gewig / mg / gravitasiekrag
              FT/tension
           T
              Fs/spanning
           fk (Kinetic) Friction / Ff / 4 N / f / wrywing / Fw
           N FNormal / Normal / FN

                                                   N

                                        fk
                                                                T

                                                       w
                                                                                                           (4)

          Notes/Aantekeninge
           Mark awarded for label and arrow/Punt toegeken vir benoeming en pyltjie
           Do not penalise for length of arrows since drawing is not to scale./Moenie
             vir die lengte van die pyltjies penaliseer nie aangesien die tekening nie
             volgens skaal is nie
           Any other additional force(s)/Enige ander addisionele krag(te) Max/Maks
             3
               4
           If force(s) do not make contact with body/Indien krag(te) nie met die
             voorwerp kontak maak nie: Max/Maks: 3 4

2. 3      Object Q/Voorwerp Q

          Fnet = ma
          Fnet = 0               
          T + (fk) = ma
          T–3 =0
          T=3N

          Object P/Voorwerp Q
          Fnet = ma
          Fhor – (fk +T) = ma
          (Fcos 30o) – 5 – 3 = 0 
          F = 9,24 N (9,238 N)                                                                            (6)

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2.4       3 s                                                                                           (1)

2.5       Y
          Graph Y represents the motion of Q after the string breaks.
          The graph Y shows a decreasing velocity with a negative acceleration. 
          This is because the net force (friction) acting on Q is in the opposite direction
          to its motion,  (accept: only frictional force acts on Q).

          Grafiek Y verteenwoordig die beweging van Q na die toutjie breek
          Die versnelling is negatief vir grafiek Y en toon 'n afnemende snelheid
          Dit is omdat wrywing op Q inwerk                                                                (4)
                                                                                                         [17]

QUESTION 3/VRAAG 3

3.1       10 m∙s-1                                                                                      (1)

3.2       The gradient represents the acceleration due to gravity (g) which is constant
          for free fall. 
          Die helling verteenwordig die versnelling as gevolg van gravitasie (g) wat
          konstant vir vry-val is.

          [The graphs represent free fall/Die grafieke verteenwoordig vryval]                            (1)

3.3.1     POSITIVE MARKING FROM QUESTION 3.1
          POSITIEWE NASIEN VANAF VRAAG 3.1
          OPTION 1/OPSIE 1
          Δy = viΔt + ½ aΔt2           Δy = viΔt + ½ aΔt2
             = (10)(2) + ½ (9,8)( 22)     = (-10)(2) + ½ (-9,8)( 22) 
             = 39,6 m                      = -39,6 m
          Height/Hoogte= 39,6 m        Height/Hoogte = 39,6 m 
          OPTION 2/OPSIE 2              OPTION 3/OPSIE 3
                                                                        
                                
                                                                                      

                                                                   

                           
          OPTION 4/OPSIE 4
          Height = Area under the graph
                                             Any one/Enige een
          Hoogte = Area onder die grafiek
                  =Area of/van +Area of/van 
                  = (10)(2)+(½)(2)(19,6) 
                  = 39,6 m 
          OPTION 5/OPSIE 5                                                                              (3)
          Height = Area of trapezium = Hoogte = Area van trapesium 
          = ½ (10 +29,6) x 2  = 39,6 m 

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3.3.2      OPTION 1/OPSIE 1                         OPTION 2/OPSIE 2
          vf = vi + aΔt                            vf = vi + aΔt
           0 = -25 + (9,8)( Δt )                    0 = 25 + (-9,8)( Δt ) 
          Δt = 2,55 s                               Δt = 2,55 s
          Total time T/Totale tyd = 8 +2,55        Total time T/Totale tyd = 8 +2,55 
                                   = 10,55 s                                = 10,55 s
          OPTION 3/OPSIE 3                          OPTION 4/OPSIE 4

                                              
                                                                                             

                                  

          Total time T/Totale tyd = 8 +2,55 
                                   = 10,55 s

                                                                           

                                                    Total time T/Totale tyd = 8 +2,55 
                                                                             = 10,55 s
          OPTION 5/OPSIE 5                          If values of vi and vf are swopped                  (4)
                                                   around,and a negative time is
          Slope of graph = 9,8                      obtained, give 1 mark for formula and
                            0  ( 25)             1 mark for adding calculated time to
                         =                
                              T8                   8 s, (max 2/4).

          Total time T/Totale tyd
          = 10,55 s 

3.4.1     0,2 s                                                                                         (1)

3.4.2     4,955 s                                                                                      (2)

3.4.3     - 27 (ms-1) [Must include the negative/Moet negatief insluit]                                (1)

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3.5       Inelastic./Onelasties
          The speeds at which it strikes and leaves the ground are not the same/The
          kinetic energies will not be the same 
          Die spoed waarmee dit die grond tref en die grond verlaat is nie dieselfde
          nie./Die kinetiese energieë sal nie dieselfde wees nie                                               (2)
                                                                                                              [16]

QUESTION 4/VRAAG 4
4.1       The total linear momentum of a closed (isolated) system remains constant (is
          conserved). 
          Die totale lineêre momentum in 'n geslote sisteem bly konstant (bly behoue)

           OR/OF
          In an isolated system, the total linear momentum before collision is equal to
          the total linear momentum after collision 
          In 'n geïsoleerde sisteem is die totale lineêre momentum voor botsing gelyk
          aan die totale lineêre momentum na botsing.
          (If key words isolated and total missing -1 mark for each.)                                         (2)
4.2       ∑pi = ∑p f
                                                 Any one/
          m1v1i + m2v2i = m1v1f + m2v2f          Enige een 

          For the system cat-skate board A/Vir die sisteem kat-skaatsplank
          (3,5)(0) + (2,6)(0)  = (3,5)vskateboard/skaatsplank + (2,6)(3) 

          vskateboard/skaatsplank = 2,23 m∙s-1 to the left/na links

          ACCEPT/AANVAAR v = -2,23 m∙s-1                                                                    (5)

4.3       Fnet Δt = Δp = mvf - mvi
                       = (3,5)(1,28 - 0) 
                       = 4,48 N∙s (4,48 kg∙ms-1) 

          OR/OF
          Fnet Δt = Δp = mvf - mvi
                       = (2,6)(1,28 - 3) 
                       = -4,48 N∙s (4,48 kg∙ms-1)
           Δp = 4,48 Ns                                                                                     (3)
                                                                                                              [10]

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QUESTION 5/VRAAG 5
5.1       The total mechanical energy/sum of kinetic and gravitational potential energy
          in a closed/isolated system is constant (conserved).
          Die totale meganiese energie/some van kinetiese en gravitasionele potensiële
          energie in 'n geslote sisteem bly behoue.
          (If key words isolated and total missing -1 mark for each.)                                              (2)
5.2       E MECH P = E MECH Q
          (EP + EK)P = (EP + E K )Q
                                                          1 mark for any of the three/
                                                          1 punt vir enige van die drie

                   1                1
          (mgh +     mv2)P = (mgh + mv2)Q
                   2                2
          50 (9,8)(3) + 0 = 0 + ½ (50)v2

          v = 7,67 m∙s-1  (7,668 m∙s-1 )                                                                          (4)
5.3           Accepted labels/Aanvaarde benoemings
                   Fg / Fw / weight / mg / gravitational force
              w    Fg/ Fw / gewig / mg / gravitasiekrag                                             
              N    FN                                                                               
              fk   Ff/friction/f                                                                    
                                                                                                                   (3)

                        N                                          N

                                           fk                                         fk
                                                                   

                         w                                          w

          Notes/Aantekeninge
               Mark awarded for label and arrow/Punt toegeken vir benoeming en pyltjie
               Do not penalise for length of arrows since drawing is not to scale./Moenie vir die lengte
                van die pyltjies penaliseer nie aangesien die tekening nie volgens skaal is nie
               Any other additional force(s)/Enige ander addisionele krag(te) Max/Maks 2
                                                                                             3
               If force(s) do not make contact with body/Indien krag(te) nie met die voorwerp kontak
                maak nie: Max/Maks: 2
                                          3

5.4       fk = μkN
             = µk (mgcosθ)              Any one/Enige een

               = 0,08 (50 x 9,8) cos 30o
               = 33,95 (33, 948) N                                                                                (3)

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NOTE/LET WEL:
IN ALL THE OPTIONS FOR QUESTION 5.5 BELOW, ACCEPT THE
SUBSTITUTION:/IN AL DIE OPSIES VIR VRAAG 5.5 HIERONDER, AANVAAR DIE
VERVANGING
5 cos 60o IN PLACE OF/IN PLAAS VAN 5 sin 30o

5.5       OPTION 1/OPSIE 1
          POSITIVE MARKING FROM QUESTION 5.4/POSITIEWE NASIEN VANAF
          VRAAG 5.4
          W = Fnet∆x cosθ
          W =W +W +W                       1 mark for any one/
            net     f     w     N
                                                           1 punt vir enige van die drie
          Wnet = W f + (-∆ EP) + WN
          Wnet = fk ∆x cos180 o + mgsinθ ∆xcos0 + 0
          Wnet = ∆ EK/∆ K

          Wnet = [33,948)(5)(-1)] + [(50)(9,8) (5)sin 30 o + 0]
                = 1055, 26 (1055,259)
          1055,259 = ½ (50) (vf 2 – 7,6682) 
          vf = 10,05 m∙s-1

          OPTION 2/OPSIE 2
          POSITIVE MARKING FROM QUESTION 5.2/POSITIEWE NASIEN VANAF
          VRAAG 5.3
          Wnc = ΔEp + ΔEk                             1 mark for any of the two/
                                           2       2  1 punt vir enige van die twee
          fΔxcosθ = (mghf – mghi) + ( ½ mvf – ½ mvi )

          μmg∆x cos180 o = [ 0 - (mg∆x sin 30 o ] + ½ m(vf2 – vi2)
          [33,948)(5)(-1)] = [0 - 50(9,8) (5) sin 30 o ] + ½ (50) (vf 2 – 7,6682 )
           vf = 10,05 m∙s-1

          OPTION 3/OPSIE
          POSITIVE MARKING FROM QUESTION 5.2 AND 5.4/POSITIEWE NASIEN
          VANAF VRAAG 5.2 EN 5.4                            1 mark for any of the two/
          Wnc = ΔEp + ΔEk                                   1 punt vir enige van die twee
          f Δxcosθ = (mghf – mghi) + ( ½ mvf2 – ½ mvi2)
          (33,95)(5) cos 180 = [(50)(9,8) {0 - 5 sin30o}] + ½ (50)( vf2 - 7,6682) 
          vf = 10,05 m∙s-1                                                                                (5)
                                                                                                          [17]

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QUESTION 6/VRAAG 6
6.1       An (apparent) change in the observed frequency (pitch), (wavelength) as a
          result of the relative motion between a source and an observer (listener).
          'n (Waarskynlike) verandering in die waargenome frekwensie (toonhoogte)
          (golflengte) aas gevolg van die relatiewe beweging tussen bron en
          waarnemer (luisteraar)                                                                        (2)

6.2       Towards A./Na A
          Recorded frequency higher./Aangetekende frekwensie is hoër                                   (2)

6.3          v  vL
          fL       fs 
             v  vs
          FOR A/VIR A

                           
                   340
          690              fs
                  340 - v s              (1)

          FOR B/VIR B:

                  340 
          610              fs
                  340  v s              (2)

          690 340  v s
             
          610 340 - v s

          1,131 (340 - vs) = 340 + vs
          vs = 20,90 m.s-1 (20.90 to 20.92 m.s-1)                                                      (6)

6.4       ANY ONE/ENIGE EEN
          Doppler flow meter/Doppler-vloeimeter 
          Measuring foetal heartbeat/Meet van fetale hartslag
          Measure speed of blood flow
          Ultra sound/Ultraklank
          Sonar
          Radar (for speeding/vir jaag)                                                                  (1)
                                                                                                        [11]

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QUESTION 7/VRAAG 7

7.1       The magnitude of the electrostatic force exerted by one point charge on
          another point charge is directly proportional to the product (of the magnitudes)
          of the charges and inversely proportional to the square of the distance
          between them. 
          Die grootte van die elektrostatiese krag uitgeoefen deur een puntlading op 'n
          ander puntlading is direk eweredig aan die produk (van die groottes) van die
          ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen
          hulle                                                                                            (2)

7.2

                                               ●               FRP
                                        FRS    R
          NOTE/LET WEL:                                                                                    (2)
          One mark for each force, correctly shown./Een punt vir elke krag korrek
          getoon.

7.3       Taking right as positive/Neem regs as positief
                Q1Q 2
          F =k 2         
                 r       c
          FnetR = FPR + FvSR

                   kQ 1Q 2       kQ 1Q 2
          Fnet =         2   +
                     r             r2
                                                                 
                          (9  10 )(1,5c 10 )(Q) (9  10 )(2 c10 -9 )(Q) 
                                    9         -9           9
          - 1,27c  10  
                      -6
                                                                            
                                     (0,3v)
                                            2
                                                             (0,2) 2v        
                v
                                            c
          - 1,27 × 10-6 = 150Q   - 450Q v (for subtraction/vir aftrekking)
                                  c
          Q = 4,23x10-9 C v                                                                                (7)
          Accept answers where left is taken as positive.                                                  [11]

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QUESTION 8/VRAAG 8

8.1

                                                      Shape (radial)/Vorm (radiaal)
                                                      Polarity (sign) of A/Polariteit (Teken) van A
                                                      Do not penalize for incorrect direction/Moet
                                        _             nie penaliseer vir verkeerde rigting nie

                                                                                                        (2)

8.2           kQ
          E      
              r2
                   (9  10 9 )(Q)
          3 x107 =
                       (0,5) 2
          Q= 8,33 x 10-4 C                                                                             (3)

8.3       Q = ne
            = (105)(1,6x10-19)                      (Positive marking from 8.2 for this
                                                     option)
             = 1,6 x10-14 C
                                                          Q1Q 2
               F                                     F =k 2 
          E=                                               r
               Q
                                                                  (
                                                               9 8,33 ×10
                                                                          -4
                                                                             )(1,6 ×10-14 )
                         F                              (
                                                     F = 9 ×10  )                           
          3 x 107 =                                                    (0,5)2
                    1,6x10 -14
                                                       = 4,8 x 10-7 N Right/Regs
          F = 4,8 x10-7 N Right/Regs                                                                   (6)
                                                                                                        [11]

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QUESTION 9/VRAAG 9

9.1.1     The potential difference across a conductor is directly proportional to the
          current in the conductor at constant temperature..
          Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die
          geleier by konstante temperatuur.

          OR/OF
          The current in a conductor is directly proportional to the potential difference
          across the conductor at constant temperature. 
          Die stroom in 'n geleier is direk eweredig aan die potensiaalverskil oor die
          geleier by konstante temperatuur                                                                (2)

9.1.2     Graph X./Grafiek X
          Graph X is a straight line (passing through the origin) therefore potential
          difference is directly proportional to current. 
          /Grafiek X is lineêr
                                                                                                          (2)

9.2.1
           1    1   1                                          R10 × R15
              =   +                                   R // =
          R // R10 R15                                         R10 + R15
           1     1   1
               =   +   
          R // 10 15                                       10 × 15
          R//= 6 Ω                                     R // =      
                                                             25
                                                         =6Ω
          R = (10 + 6 + 2)(for the addition/vir optelling)
             = 18 Ω
                                                   Rext = (10 + 6) = 16 Ω
             V                                      ℇ = I (R + r)
          R=                                        6 = I(16 + 2)
              I
                                                      I = 0,33 A
              6
          I=    
             18
           = 0,33 A                                                                                      (5)

9.2.2     Decrease. 
          The total resistance of the circuit increases.
          Afneem
          Die totale weerstand van die stroombaan neem toe.                                               (2)

9.2.3     Increase/Neem toe                                                                              (1)

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9.2.4     The total resistance in the external circuit increases./Die totale weerstand in
          die eksterne stroombaan neem toe
          Current decreases/Stroom neem af
          ''Lost'' volts decreases/''Verlore'' volts neem af

          OR/OF

          The total resistance in the external circuit increases./Die totale weerstand in
          die eksterne stroombaan neem toe 
          V α R for constant/vir konstante I
          Therefore V increases./Dus neem V toe                                                           (3)
                                                                                                         [15]
QUESTION 10/VRAAG10
10.1      ANY THREE/ENIGE DRIE
             I. Permanent magnets/Permanente magnete
            II. coils (armature)/spoel
           III. commutator/kommutator
          IV. brushes/borsels
            V. power supply/battery/kragbron                                                             (3)
10.2.1    The rms voltage of AC is the potential difference which dissipates the same
          amount of energy as the equivalent DC potential difference. 
          Die wgk spanning/potensiaalverskil van WS is die potensiaalverskil wat
          dieselfde aantal energie verkwis as GS.
          Accept formula for Vrms as 1 mark.                                                             (2)
10.2.2    OPTION 1/OPSIE 1                            OPTION 2/OPSIE 2
                                                                                         
          Vrms = Irms R                                         Vmax
                                                       Vrms 
                    Vrms                                       2
           Irms =
                     R                                Vmax = (240) 2 
                   240                                     = 339,41
          Irms =         
                    11
                = 21,82 A                             Vmax = Imax R
                  I                                         339,41 
          I rms  max
                     2                                Imax = 11
              Imax = (21,82) 2 
              Imax = 30,86 A                         Imax = 30,86 A
          OPTION 3/OPSIE 3                                                                               (4)

                            

                                                                                               
                                                      
                                                                                     
                                                                                                         [9]
                                                          

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### Page 14

Downloaded from hlayiso.com
Physical Sciences P1/Fisiese Wetenskappe V1         14                 DBE/Feb.–Mar./Feb.–Mrt. 2018
                                  NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 11/VRAAG 11
11.1.1    Greater than/Groter as

          Electrons are ejected from the metal plate./Elektrone word vrygestel vanaf die
          metaalplaat  Accept: a current is registered on the ammeter.                                (2)
11.1.2    Increase in intensity means that (for the same frequency) the number of
          photons per second increases (ammeter reading increases)but the energy
          of the photons stays the same (Therefore the statement is incorrect).
          OR
          An increase in the energy of the photons only increases the kinetic energy of
          the photoelectrons and not the number of photoelectrons, thus the ammeter
          reading will not change.
          Toename in intensiteit beteken dat (vir dieselfde frekwensie) die aantal fotone
          neem toe (ammeterlesing neem toe) maar die energie van die fotone bly
          dieselfde. (Dus is die stelling verkeerd)                                                    (2)
11.1.3    Light has a particle nature/Lig het 'n deeltjieaard
          Accept light energy is quantized/Aanvaar ligenergie is gekwantiseer                         (1)
11.2.1    The minimum frequency needed for the emission of electrons (from a metal
          surface).
          Die minimum energie benodig vir die vrystelling van elektrone (vanaf die
          metaaloppervlak)                                                                             (2)
11.2.2    W0 = hf0 
             = (6,63 x 10-34)(5,73 x 1014)
             = 3,8 x 10-19J [3,799 x 10-19 J]                                                         (3)
11.2.3    POSITIVE MARKING FROM QUESTION 11.2.2
          POSITIEWE NASIE VANAF VRAAG 11.2.2
          OPTION 1/OPSIE 1
          E = W 0 + Ek(max/maks)
          hf = hf0 + Ek(max/maks)
          hf = hf0 + ½ mv2         Any one/Enige een
                            2
          E = W 0 + ½ mv

          (6,63 x 10-34)f = 3,8 x 10-19 + [½(9,11 x 10-31)(4,19 x 105)2]
          f = 6,94 x 1014 Hz            [7 x 1014 Hz]
          OPTION 2/OPSIE 2
          E = W 0 + Ek(max/maks)
          hf = hf0 + Ek(max/maks)
           hf = hf0 + ½ mv2             Any one/Enige een
           E = W 0 + ½ mv2

          E = 3,8 x 10-19 + [½(9,11 x 10-31)(4,19 x 105)2]
          hf = 4,599 x 10-19
          (6,63 x 10-34)f = 4,599 x 10-19
          f = 6,94 x 1014 Hz         [7 x 1014 Hz]                                                   (3)
                                                                                                      [13]
                                                                           TOTAL/TOTAAL:               150

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