# Physical Sciences Grade 12 Paper 1 NSC June 2019 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the NSC June Exam 2019, 28 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: 2019
- Exam period: NSC June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 28

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Physical Sciences · Grade 12 · NSC June Exam · 2019. Memorandum, 28 pages. Read online or download the PDF.

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### Page 1

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             SENIOR CERTIFICATE EXAMINATIONS/
         NATIONAL SENIOR CERTIFICATE EXAMINATIONS
               SENIORSERTIFIKAAT-EKSAMEN/
          NASIONALE SENIORSERTIFIKAAT-EKSAMEN

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

                                          2019

                      MARKING GUIDELINES/NASIENRIGLYNE

                                                                 MARKS/PUNTE: 150

                     These marking guidelines consist of 28 pages./
                      Hierdie nasienriglyne bestaan uit 28 bladsye.

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Physical Sciences/P1/Fisiese Wetenskappe/V1       2                                       DBE/2019
                              SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION 1/VRAAG 1

1.1       A                                                                                             (2)

1.2       B                                                                                             (2)

1.3       D                                                                                             (2)

1.4       C                                                                                             (2)

1.5       C                                                                                             (2)

1.6       C                                                                                             (2)

1.7       D                                                                                             (2)

1.8       D                                                                                             (2)

1.9       C                                                                                             (2)

1.10      A                                                                                          (2)
                                                                                                      [20]

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Physical Sciences/P1/Fisiese Wetenskappe/V1       3                                               DBE/2019
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QUESTION 2/VRAAG 2

2.1.1
                                                       N

                                          f
                                                   ●
                                                                   F
                                                   w

                                        ACCEPT/
                                        AANVAAR                N                                  N
                             N
                                                                                       F

                     f
                            ●                 Fx       f                                f
                                                                                               ●
                                 Fy
                                                                            F
                           w                               w                                  w

           Accepted labels/Aanvaarde benoemings
           F F A /90 N/F 90
              F g / F w/ weight / mg / gravitational force
           w
              F g / F w / gewig / mg / gravitasiekrag
           f (Kinetic) Friction / F f / / f k / wrywing / F w
           N F Normal / Normal/Normaal / F N

         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
         • Deduct 1 mark for an arrow/arrows omitted / trek 1 punt af indien pyl/pyle
            weggelaat                                                                                        (4)

2.1.2     It is moving at constant speed in a straight line/, the acceleration is zero/ the
          net force (resultant) acting on it is zero/it is moving at constant velocity 
          Dit beweeg teen konstante spoed in ‘n reguit lyn / versnelling is nul / netto
          krag (resultant) wat daarop inwerk is nul/ dit beweeg teen konstante snelheid                      (1)

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2.1.3
            Fnet = ma
            Fnet = 0
            Fx= f                 any one
            Fx – f = 0             enige een
            Fcos 40o- f = 0

             90 cos 40o – f = 0
            f = 68,94 N

          OR/OF
           Fnet = ma                                   Fnet = ma
                                                                             any one
           Fnet = 0                                    Fnet = 0
           Fx= f                  any one             Fx = f                 enige een
           Fx – f = 0              enige een           Fx – f = 0
           Fcos 320o- f = 0                            Fsin 50o – f = 0
           90 cos 320o – f = 0                        90 sin 50o – f = 0
           f = 68,94 N                                f = 68,94 N
                                                                                                       (3)
          NOTE:
          1 mark for formula/equation, 1 mark substitution with zero, 1 mark answer.
          LW:1 punt vir formule/vergelyking, 1 punt substitusie, 1 punt antwoord.

2.1.4     POSITIVE MARKING FROM 2.1.3 /POSITIEWE NASIEN VANAF 2.1.3
          OPTION 1/OPSIE 1
          v f = v i + a∆t                          Fnet = ma
           2 = 0 + a(3)                          Fcos 320o – f = 15(0,67)
          a = 0,67 m∙s-2                           F cos 320o – 68,94 = 15(0,67)
                                                   F = 103,11N
          F net = ma
          F cos 40o – 68,94  = 15 (0,67)

          F = 103,11N (103,05 N – 103,11N)

          POSITIVE MARKING FROM 2.1.3 /POSITIEWE NASIEN VANAF 2.1.3
          OPTION 2/OPSIE 2
          F net .Δt = Δp
          Fcos 40º - (68,94) (3) = 15(2 – 0) 
          F = 103,11 N

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          POSITIVE MARKING FROM 2.1.3 /POSITIEWE NASIEN VANAF 2.1.3
          OPTION 3/OPSIE 3
          F net = ma                any one
          F x – f k = ma             enige een

                                  (2 - 0) 
          F x – 68,94 =     15
                                    3
          F x = 78,94 N
                  Fy
          tan θ =
                  Fx
                      Fy
          tan 40 o =
                     78,94
          F y = 66,24 N

          F2 = Fx2 + Fy2
          F2 = (78,94)2 + (66,24)2
          F = 103,05 N

          POSITIVE MARKING FROM 2.1.3 /POSITIEWE NASIEN VANAF 2.1.3
          OPTION 4/OPSIE 4
                  vi + v f
          Δx =             Δt
                     2
                  (2 + 0)  
              =            (3)
                    2
          ∆x = 3 m
          W net = ∆K
          W F + W f = ∆K 
          F∆x cos θ + f∆x cos θ = ∆K
          F(3)cos40o + 68,94(3)cos180o = ½ (15)(22) – ½ (15)(0)2
          F = 103.06 N
                                                                                                       (6)

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2.2       OPTION 1/OPSIE 1
                 mm
          F = G 12 2 
                   r
                               m planet (10)
          20= (6,67 ×10 -11 )
                               (6 ×10 5 )2 
          m planet = 1,08 x 1022 kg

          OPTION 2/OPSIE 2
          w = mg
          20 = (10)(g)             Any one
          g = 2 m·s  -2             Enige een
              GM
          g= 2
               R
              (6,67 × 10 -11 ) M
          2=                     
                 (6 × 10 5 )2
          M = 1,08 x 1022 kg                                                                           (4)
                                                                                                       [18]

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QUESTION 3/VRAAG 3
3.1       Motion of an object under the influence of gravity/gravitational force (weight)
          only.
          Beweging van ʼn voorwerp slegs onder die invloed van gravitasie/gravitasie
          krag (gewig).
          OR/OF
          Motion in which the only force acting on the object is gravity/weight. 
          Beweging waar die enigste krag wat op die voorwerp inwerk, gravitasie/gewig
          is.
          ACCEPT/AANVAAR
          Vertical motion in which friction/air resistance is absent. 
          Vertikale beweging waar wrywing/lugweerstand afwesig is.
          Motion in air with an acceleration of 9,8 m∙s-2. 
          Beweging in lug met ‘n versnelling van 9,8 m∙s-2.                                            (2)
          NOTE: 2 OR ZERO/ 2 of nul
3.2.1     OPTION 1/OPSIE 1
          Upwards positive/Opwaarts positief:
            v f = v i +aΔt 
            0 = v i +(-9,8) (1,53) 
          ∴ v i = 14,99 m∙s-1 (15 m∙s-1) 
          Downwards positive/Afwaarts positief
              v f = v i +aΔt 
            0 = v i +(9,8) (1,53) 
          ∴ v i = -14,99 m∙s-1
               v i = 14,99 m∙s-1 (15 m∙s-1) 
          OPTION 2/OPSIE 2
          F net = ma
                = 9,8 (m)
          F net Δt = mΔv
          (9,8)(m)(1,53) = (m)(v f – 0) 
          v f = 14,99 m·s-1 (15 m·s-1) 
          OPTION 3/OPSIE 3
          Upwards positive/Opwaarts positief:
          Δy = v i Δt + ½aΔt2 
             0 = v i (3,06) + ½ (- 9,8)(3,06)2
              v i = 14,99 m∙s-1(15 m∙s-1)

          Downwards positive/Afwaarts positief
          Δy = v i Δt + ½aΔt2 
           0 = v i (3,06) + ½ (9,8)(3,06)2
            v i = 14,99 m∙s-1 (15 m∙s-1)
          NOTE: initial and final velocities can be swopped if starting from top, as long
          as sign of g is changed accordingly.
          LW: v f en v i kan omgeruil word indien van bopunt begin, solank teken van g
          dienooreenkomstig verander word.                                                             (3)
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3.2.2     OPTION 1/OPSIE 1
          POSITIVE MARKING FROM 3.2.1/ Positiewe nasien vanaf 3.2.1
          Upwards positive/Opwaarts positief:
          Δy = v i Δt + ½aΔt2 
             = 14,99 (1,53) + ½ (- 9,8)(1,53)2
             = 11,47 m(11,46-11,48)
          Maximum height is/Maksimum hoogte is 11,47 m

          Downwards positive/Afwaarts positief
          Δy = v i Δt + ½aΔt2 
             = -14,99 (1,53) + ½ (9,8)(1,53)2 
             = -11,47 m (11,46-11,48)
          Maximum height is /Maksimum hoogte is 11,47 m

          OPTION 2/OPSIE 2
          POSITIVE MARKING FROM 3.2.1/ Positiewe nasien vanaf 3.2.1
          Upwards positive/Opwaarts positief:
          v f 2 = v i 2 + 2aΔy
          0 = (14,99)2 + 2(-9,8)(Δy)
          Δy= 11,47 m∙(11,46-11,48)
          Maximum height reached is/Maksimum hoogte bereik is 11,47 m
          Downwards positive/Afwaarts positief:
          v f 2 = v i 2 + 2aΔy
          0 = (-14,99)2 + 2(9,8)(Δy)
          Δy= -11,47 m∙(11,46-11,48)
          Maximum height reached is/Maksimum hoogte bereik is 11,47 m
          OPTION 3/OPSIE 3
          POSITIVE MARKING FROM 3.2.1/ Positiewe nasien vanaf 3.2.1
          Upwards positive/Opwaarts positief:
                v + vf 
          Δy =  i       Δt 
                2 
                (14,99 + 0 ) 
              =               (1,53 ) 
                     2       
          ∆y = 11,47 m
          Maximum height reached is /Maksimum hoogte bereik is 11,47 m

          Downwards positive/Afwaarts positief:

                v + vf  
          Δy =  i       Δt
                2 
                 (- 14,99 + 0 ) 
              =                 (1,53 )
                       2        
          ∆y = -11,47 m (11,46-11,48)
          Maximum height reached is /Maksimum hoogte bereik is 11,47 m

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          OPTION 4/OPSIE 4
          POSITIVE MARKING FROM 3.2.1

          ∆E = ∆K + ∆U                          1 mark for any
          ½ mv i 2 + mgh i = ½ mv f 2 + mgh f   1 punt vir enige

          ½ (14,994)2 + (9,8)(0) = 0 + 9,8 h f 
          h f = 11,47 m (11,46-11,48)
          Maximum height reached is /Maksimum hoogte bereik is 11,47 m

          OR/OF
          ∆K = - ∆U
          ½ m(v f 2 – v i 2) = - mg(h f – h i )
              ½ (0 - 14,992) = - 9,8( h f - 0) 
          h f = 11,47 m (11,46-11,48)
          Maximum height reached is /Maksimum hoogte bereik is 11,47 m                                  (3)

3.3       OPTION 1/OPSIE 1
          POSITIVE MARKING FROM 3.2.1
          Upwards positive/Opwaarts positief:
          Δy = v i Δt + ½aΔt2 
             = (14,99) (4) + ½ (- 9,8)(4)2
             = -18,4 m
          Position is 18,4 m downwards (below the edge of the roof)  / Posisie is
          18,4 m afwaarts (onder die kant van die dak).

          Downwards positive/Afwaarts positief
          Δy = v i Δt + ½aΔt2 
             = (-14,99) (4) + ½ (9,8)(4)2
             = 18,4 m
          Position is 18,4 m downwards (below the edge of the roof) / Posisie is
          18,4 m afwaarts (onder die kant van die dak)

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          OPTION 2/OPSIE 2
          POSITIVE MARKING FROM 3.2.1
          Upwards positive/Opwaarts positief:
          v f = v i +aΔt
              = (14,99) + (-9,8) (4)
            = - 24,2 m∙s-1
          v f 2 = v i 2 + 2aΔy
          (-24,2)2 = (14,99)2 + 2(-9,8)(Δy)
          Δy= - 18,4 m∙
          Ball is 18,4 m downwards (below the edge of the roof) / Bal is 18,4 m afwaarts
          (onder die kant van die dak)
          Downwards positive/Afwaarts positief:
          v f = v i +aΔt
              = (-14,99) + (9,8) (4)
            = 24,2 m∙s-1
          v f 2 = v i 2 + 2aΔy
          (24,2)2 = (-14,99)2 + 2(9,8)(Δy)
          Δy= 18,4 m∙
          Ball is 18,4 m downwards (below the edge of the roof)/ Bal is 18,4 m
          afwaarts (onder die kant van die dak)

          OPTION 3/OPSIE 3
          POSITIVE MARKING FROM 3.2.1
          Upwards positive/Opwaarts positief:
                 v + vf                        vf = vi +aΔt
          Δy =  i        Δt 
                   2                              = (14,99) + (-9,8) (4)
                 (14,99 - 24,2 )               = - 24,2 m∙s-1
              =                   (4 )
                       2         
          ∆y =- 18,4 m
          Ball is 18,4 m downwards (below the edge of the roof)  / Bal is 18,4 m
          afwaarts (onder die kant van die dak).

          Downwards positive/Afwaarts positief:
                                                        vf = vi +aΔt
                v + vf                                  = (-14,99) + (9,8) (4)
          Δy =  i       Δt
                2                                      = 24,2 m∙s-1
                                    
                 (- 14,99 + 24,2 ) 
              =                    (4 )
                        2          
          ∆y = 18,4 m
          Ball is 18,4 m downwards (below the edge of the roof)  / Bal is 18,4 m
          afwaarts (onder die kant van die dak.)

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          OPTION 4/OPSIE 4
          Total time to return to starting point/totale tyd terug na beginpunt
           = 2(1,53) = 3,06 s
          ∴ time from reference point to ground/ tyd vanaf verwysingspunt tot by grond
          = (4 – 3,06) = 0,94 s
          ∆y = v i ∆t + ½ (g)∆t2 
             = (14,99)(0,94) + ½(9,8)(0,94)2 
              = 18,43 m downwards (below the edge of the roof) /afwaarts (onder die
          kant van die dak.                                                                            (3)

3.4       No/Nee
       - The motion of the ball is only dependent on its initial velocity/the initial
          velocity depends on the time taken to reach maximum height.
          Die beweging van die bal is slegs afhanklik van sy beginsnelheid./die
          aanvanklike snelheid hang af van die tyd wat dit neem om maksimum hoogte
          te bereik.

          ACCEPT for 1 mark/ AANVAAR vir 1 punt:

          The ball will still be in the air.
          Die bal sal nog steeds in die lug wees.

          OR/OF
          The ball is still falling.
          Die bal is steeds besig om te val.

          OR/OF
          The ball would not have reached the ground.
          Die bal sal nog nie die grond bereik het nie.

          OR/OF
          The motion of the ball is independent of the height of the building. 
          Die beweging van die bal is onafhanklik van die hoogte van die gebou.

          NOTE: If learners gave separate answers for 3.2 and 3.3, mark them
          together. Thus, if one answer is correct and the other incorrect 0/3                          (3)
          LW: Indien leerders twee afsonderlike antwoorde gee vir 3.2 en 3.3, sien as                  [14]
          geheel na. Dus, indien een verkeerd is, 0/3

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QUESTION 4/VRAAG 4

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

          OR/OF
          In an isolated/closed system the total momentum before a collision is equal to
          the total momentum after the collision. 
          In 'n geslote/geïsoleerde sisteem is die totale momentum voor die botsing
          gelyk aan die totale momentum na die botsing.

          NOTE/LET WEL:
          -1 for each key word/phrase omitted.
          -1 vir elke sleutel woorde/frase weggelaat.
          Take the whole statement in context /Vat die hele stelling in konteks.                               (2)

4.2       OPTION 1/OPSIE 1
          ∑p i = ∑p f
                                                      1 mark for any
          m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f
                                                      1 punt vir enige
          m 1 v 1i + m 2 v 2i = (m 1 +m 2 )v f

          {0,45(9) + 0,20(0)} = (0,45 + 0,20)v
          v = 6,23 m⋅s-1
          OR
          ∆p ball/bal = - ∆p cont/houer                If – sign omitted from formula 0/4
          0,45(v – 9)  = - 0,2(v – 0)                 Indien – teken weggelaat uit
          v = 6,23 m⋅s-1                               formula 0/4
          OPTION 2/OPSIE 2

          ∑p i = ∑p f           1 mark for any
          p f Total = p i Total 1 punt vir enige
          (Thus change in total momentum = 0 /Dus verandering in momentum is=0))
          0 = (0,65v f ) – (9)(0,45)
          v f = 6,23 m∙s-1                                                                                    (4)

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4.3       POSITIVE MARKING FROM 4.2/POSITIEWE NASIEN VANAF 4.2
          K = ½ mv2 (or E K = ½ mv2)
          Total kinetic energy before collision:/Totale kinetiese energie voor botsing:
          ½ (0,45)(9)2 + 0= 18,225J

          Total kinetic energy after collision:/ Totale kinetiese energie na botsing:
          ½ (0,45 + 0,20)(6,23)2 = 12,614J
          ∑K before/voor ≠ ∑K after/na
          Collision is inelastic. /Botsing is onelasties
          If start with/ indien begin met ∑E Ki =∑E Kf 4/5 max/maks
          No calculation/ geen berekening: 0
          Do not accept a conclusion of inelastic collision based on any other
          calculation (such as that of momentum or mechanical energy)./ Moet geen
          afleiding van ‘ n onelastiese botsing aanvaar wat op enige ander berekening
          gebaseer is nie (soos byvoorbeeld momentum of meganiese energie).                             (5)
                                                                                                       [11]

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                              SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION 5/VRAAG 5

5.1         Tension/Spanning                                                                                (1)

5.2         There is friction/ tension in the system     OR/OF
            Daar is wrywing/spanning in die sisteem       The internal energy increases
            OR/OF                                         because of friction 
            Friction/tension is a non-conservative        Die interne energie neem toe as
            force                                        gevolg van wrywing.
            Wrywing/spanning 'n 'n nie-                   OR
            konserwatiewekrag                             The applied force is non-
            OR/OF                                         conservative
            The system is not isolated because            Die toegepaste krag is nie-
            there is friction/tension                    konservatief
            Die sisteem in nie geïsoleerd nie omdat       OR
            daar wrywing/spanning is                      It is not an isolated system                      (1)

5.3
                                                                 ACCEPT/AANVAAR
           N                                                     N
                               T                                                             T

      f        •                                            f

                w                                                          w
                                                                N
                                                                                            T
          NOTE: maximum ¾ if friction
          and tension are not on a
          straight line
          LW: maksimum ¾ indien
          wrywing en spanning nie in ‘n
                                                           f             •
                                                            w||/Fg//
          reguitlyn nie.

                                                                                   w⊥/Fg⊥

                                                         NOTE: maximum ¾ if N and
                                                         w ⊥ are not on a straight line
                                                         LW: maksimum ¾ indien N en
                                                         w ⊥ nie in ‘n reguitlyn nie.

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           Accepted labels/Aanvaarde benoemings
              F g /F w /weight/mg/gravitational force
           w                                                               
              F g /F w /gewig/mg/gravitasiekrag
           f Friction/F f /f k /178,22 N/wrywing/F w                       
           N Normal (force)/F normaL /F N /F normaal /F reaction/reaksie   
           T F T /F A /F applied/toegepas /700 N/Tension                   

          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
          Deduct 1 mark for an arrow/arrows omitted / trek 1 punt af indien pyl/pyle
          weggelaat                                                                                                  (4)

5.4       W = F∆xcosθ
          W f = [178,22(4)cos180o] 
              = - 712,88 J                                                                                          (3)

5.5       OPTION 1/OPSIE 1
          POSITIVE MARKING FROM QUESTIONS 5.4 POSITIEWE NASIEN VANAF
          VRAE 5.4
          W net = ∆E K                  1 mark for any one/
          W f + W g + W T = ∆K          1 punt vir enige een
          W f + mgsinθ∆xcosθ + W T = ∆K

          -712,88 + (70)(9,8) (sin 30o)(4) cos 180o + (700 x 4 x cos 0o) = ½ 70(v f 2 – 0)
          v f = 4,52 m⋅s-1

          NOTE: W g can be obtained using any of the following formulae:
          LW: w g kan verkry word deur enige een van die volgende formules:
          W gravity/gravitasie = mg∆xcosθ
                                = (70)(9,8)(4).(cos 120o)

          ∴
               -712,88 + (70)(9,8)(4)cos 120o) + (700 x 4 x cos 0o) = ½ 70(vf2 – 0)
               vf = 4,52 m⋅s-1
          W gravity/gravitasie = -∆ mgh = - mg(h f – h 0)
                               = mg∆ycosθ
                                = ((70)(9,8) 4(sin 30o).cos 180o
           W gravity/gravitasie = mgsinθ∆xcosθ
                               = (70)(9,8) (sin 30o)(4).cos 180o

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          OPTION 2/OPSIE 2
          POSITIVE MARKING FROM 5.4 / POSITIEWE NASIEN VANAF 5.4
          W nc = ∆E K + ∆E p 
          W T + W f = ∆E K + ∆E p
          (700)(4) cos 0o) + (-712,88) = [(70)(9,8) 4(sin 30o). - 0 ]+ ½ 70(v f 2 – 0)
          v f = 4,52 m⋅s-1
          OPTION 3
          F net = F T – [mgsinθ + f k ]
                 = 700 – [(70 x 9,8 sin 30o) + 178,22] 
                 = 178,78 N
          W net = ∆E K 
          F net . ∆xcosθ =∆E K
          (178,78)(4)cos0 o =.½ 70(v f 2 – 0)
          v f = 4,52 m⋅s-1                                                                               (5)

5.5       WHERE EQUATIONS OF MOTION ARE USED:/ WAAR
          BEWEGINGSvERGELYKING GEBRUIK: MAX/MAKS 1
                                                                            5
          F net = ma
          F T – [mgsinθ + f k ] = ma
          700 – [(70 x 9,8 sin 30o) + 178,22] = 70a
          a = 2,554 ms-2
          v f 2 = v i 2 + 2a∆x
               = 0 + 2(2,554)(4)
          v f = 4,52 m⋅s-1

5.6       POSITIVE MARKING FROM 5.4/POSITIEWE NASIEN VANAF 5.4

          2(-712,88) = -1425,76 J

          OR/OF
          Double the answer (in question 5.4). 
          Dubbel die antwoord (in vraag 5.4)                                                               (1)
                                                                                                          [15]

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QUESTION 6/VRAAG 6
6.1.1     ∆x = v i ∆t + ½ a∆t2
          300 = v i (10)                    d 300 
                                        v=     =    =30m∙s-1
          v i = 30 m∙s-1                    t   10
          NOTE/LET WEL:
          Accept/Aanvaar ∆x = v i ∆t                                                                   (2)
6.1.2     The change in frequency (or pitch) (of the sound) detected by a listener
          because the source and the listener have different velocities relative to the
          medium of sound propagation. 
          Die verandering in die frekwensie (of toonhoogte) (van die klank)
          waargeneem deur ‘n luisteraar omdat die bron en die luisteraar verskillende
          snelhede relatief tot die voortplantingsmedium het.
          OR/OF
          An (apparent) change in observed/detected frequency (pitch), (wavelength)
          as a result of the relative motion between a source and an
          observer (listener). 
          'n Skynbare verandering in waargenome frekwensie (toonhoogte),(golflengte)
          as gevolg van die relatiewe beweging tussen die bron en 'n waarnemer/
          luisteraar.
          NOTE/LET WEL:
          -1 for each key word/phrase omitted.
          -1 vir elke sleutel woorde/frase weggelaat                                                   (2)
6.1.3     Car/source (just) passes observer 
          Motor beweeg net verby die waarnemer
          Accept:
          Car moves away from observer 
          No relative motion between car and observer 
          Car and observer at the same place/position

         Aanvaar:
         Motor beweeg verby waarnemer
         Geen relatiewe beweging tussen motor en waarnemer
         Motor en waarnemer by dieselfde plek/posisie.                                                 (2)

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6.1.4     POSITIVE MARKING FROM 6.1.1/POSITIEWE NAISEN VANF 6.1.1
               v ± vL                v
          fL =        fs OR/OF fL =        fs
               v ± vs               v − vs
                             
                 340
          932 =               fs
                340 - 30

            f s = 849,76 Hz                                                                           (4)

          Notes/Aantekeninge:
          • Any other Doppler formula, e.g. /Enige ander Doppler formula b.v.
                    v − vL
             fL =          f - Max/Maks. 3
                    v − vs s               4
          Marking rule 1.5: No penalisation if zero substitutions are omitted.
          Nasienreël 1.5. Geen penalisering indein nul vervangings uitgelaat word

6.2       Doppler / Blood flow meter
          Dopplervloeimeter/ bloedvloeimeter
          Measuring the heartbeat of a foetus             Any 2 
          Meting van hartklop van 'n fetus                Enige 2
          Radar
          Sonar
          Used to determine whether stars are receding or approaching earth/
          Gebruik om te bepaal of sterre na of weg van die aarde beweeg                                 (2)
                                                                                                       [12]

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QUESTION 7/VRAAG 7

7.1          The electric field at a point is the electrostatic force experienced per unit
             positive charge placed at that point. 
             Die elektriese veld by 'n punt is die elektrostatiese krag wat per eenheid
             positiewe lading wat by daardie punt geplaas word, ervaar word. 

             NOTE/LET WEL:
             -1 for each key word/phrase omitted. If definition of electric field: 0/2
             -1 vir elke sleutel woorde/frase weggelaat. Indien definisie van elektriese veld
             0/2                                                                                         (2)

7.2          q 2 is positive 
          - The electric field due to q points to the right because q is negative.  Since
                                        1                                1
             the net field is zero, field due to q 2 must point to the left away from q 2 , 
             hence q 2 is positive.

             q 2 is positief
             Die elektriese veld as gevolg van q 1 is na regs gerig omdat q 1 negaitief is.
             Aangesien die net veld nul is, moet die veld as gevolg van q 2 na links weg
             van q 2 wees.

             OR/OF
             q 2 is positive 
      -      Since E net is zero,E 1 and E 2 are in opposite directions therefore q 1 and q 2
             are oppositely charged. 

             q 2 is positief
             Omdat E net nul is, is E 1 en E 2 in teenoorgestelde rigtings daarom is q 1 en
             q 2 teenoorgesteld gelaai.                                                                 (3)

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7.3                Q
          E= k 2                                       1 mark for formula
                   r
          E net = 0                                     1 mark for equating the two fields
                                                        1 mark for both substitutions
                q    q                                  1 mark for answer
          ∴ k 21 = k 22 OR
                r1   r2                                 1 punt vir vergelyking
                                                        1 punt vir twee velde gelyk gestel
          q1 q 2
               =                                        1 punt vir altwee substitusies
          r12 r2 2                                      1 punt vir antwoord

          (9 × 10 9 )(3 ×10 -9 )  (9 × 10 9 )q 2   
                                 =
                 (0,1)2               (0,4)2

          q 2 = + 4,8 x 10-8 C                                                                         (4)
7.4       The electrostatic force (of attraction/repulsion) between two point charges is
          directly proportional to the product of the charges and inversely proportional
          to the square of the distance between them. 
          Die elektrostatiese krag(aantekking/afstotend) tussen twee puntladings is                     (2)
          direk eweredig aan die produk van die ladings en omgekeerd eweredig aan
          die kwadraat van die afstand tussen hulle.

          NOTE/LET WEL:
          -1 for each key word/phrase omitted.If masses used instead of charges 0
          -1 vir elke sleutel woorde/frase weggelaat Indien massas gebruik 0

7.5       POSITIVE MARKING FROM 7.3/POSITIEWE NASIEN VANAF 7.3
             kQ 1Q 2 
          F=
                r2
              (9 ×109 )(3 ×10-9 )(4,8 × 10-8 ) 
          F=
                          (0,3)2
            = 1,44 x 10-5 N                                                                            (3)

7.6       POSITIVE MARKING FROM 7.2 AND 7.3/POSITIEWE NASIEN van 7.2 en
          7.3
          YES/JA
          Both charges are equal and positive 
          Beide ladings is gelyk en positief

          Accept calculation which shows charges the same and positive/ Aanvaar
          berekening wat toon dat ladings dieselfde en positief is.
          If the answer is YES, mark according to the memo, if NO check 7.2 first for
          sign of charge. If stated NEGATIVE at 7.2, then answer is:
          No, the direction is incorrect. 
          Positiewe nasien vanaf 7.2: Indien antwoord vir 7.2 NEGATIEF, dan is hierdie
          antwoord: Nee , die rigting is verkeerd. 
                                                                                                         (2)
                                                                                                        [16]

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QUESTION 8/VRAAG 8

8.1.1     The rate at which (electrical) energy is converted (to other forms) (in a circuit)
          The rate at which energy is used/Energy used per second
          The rate at which work is done 
          (2 or zero)
          Die tempo waarteen elektriese energie omgesit word (in ander vorms) in 'n
          stroombaan.
          Die tempo waarteen energie verbruik word.
          Die tempo waarteen arbeid verrig word.
          (2 of nul)                                                                                     (2)

8.1.2         V2                                         P = VI                 P = VI
          P=                                           6 = (12)(I)            6 = (12)(I)
              R
                                        (12)2 (1)        ∴I = 0,5 A             ∴I = 0,5 A
             (12)2                                      P = I2R               V = IR
          6=                      6=
               R                        R                6 = (0,5)2 R          12 = (0,5)R
                                   R=24 Ω               R = 24 Ω              R = 24 Ω
          R = 24 Ω                                                                                      (3)

8.1.3     POSITIVE MARKING FROM 8.1.2/POSITIEWE NASIEN VANAF 8.1.2
          OPTION 1/OPSIE 1

                                                                          R1R 2
           1 1 1                                        R tot = (R s +            )
               = +                                                       R1 + R 2
          R// R1 R2                                                      (24)(24)  
                  1   1                                 R tot = 24 +
               =    +                                                     48
                 24 24                                      = 36 Ω
          R // = 12 Ω

          R ext = (R s + R // )
          R ext = (24 +12) 
              = 36 Ω
          V = IR
          OR                     any one
          ε = I(R + r)          Enige 1
          12 = I(36 + 2) 
          I = 0,32 A(0,316 A)

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          POSITIVE MARKING FROM 8.1.2/POSITIEWE NASIEN VANAF 8.1.2
          OPTION 2/OPSIE 2

          Rext = (Rs + R//)                                              R1R 2
                                                        R ext = R s +
                                                                        R1 + R 2
           1 1 1
             = +                                                        (24)(24)  
          R// R1 R2                                     R ext = 24 +
                                                                          48
               1   1
                 +                                            = 36 Ω
           = 24 24 
          R// = 12 Ω
          Rext = (24 + 12)
             = 36 Ω                                           V2
                                                  I2R I2R =
                                                              R
               2    V2                            I2R2 = V2
          P=I R =                                V = IR
                     R
                                                  12 = I(38) 
                         (12 )2
               2
                 (    )
              I 36 + 2 =                         I = 0,316 A
                           38                     =
          I = 0,32 A(0,316)                                                                            (5)

8.1.4     POSITIVE MARKING FROM 8.1.3              POSITIVE MARKING FROM 8.1.3
          POSITIEWE NASIEN VANAF 8.1.3             POSITIEWE NASIEN VANAF 8.1.3
          OPTION 1/OPSIE 1                         OPTION 2/OPSIE 2
          V = IR                                   V = IR
          V = I(R A +r)                            For the parallel portion (or from 8.1.3):
             = 0,316(26)                          Vir die parallel gedeelte (of vanaf
             = 8,216 V (8,32 V)                    8..1.3)
                                                    1 1 1                  RR
                                                      = +      OR R = 1 2
          V // = (12 – 8,216)                     R R1 R 2              (R1 + R 2 )
               = 3,784 V(3,68 V)
                                                          (24)(24)
          ∴V C = 3,78 V (3,68 V)                   R =
                                                              48
                                                        = 12 Ω                                           (3)
                                                   V // = V C 
                                                   V = IR //
                                                   = (0,316)(12) 
                                                        = 3.79 V (3,84 V)
          POSITIVE MARKING FROM 8.1.3
          POSITIEWE NASIEN VANAF 8.1.3
          OPTION 3/OPSIE 3
          IA = IB + IC
              = 2 IB
          0,316 = 2I B 
          I B = 0,158 A
          V = 0,158 (24) 
             = 3,79 V                                                                                   (3)

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8.1.5     OPTION 1/OPSIE 1
          The power rating (output voltage) of the bulb is 6 W, 12 V. /Die gloeilamp is
          gemerk 6W ; 12 V
               V2
          P=
                R
           [For a given resistance, power is directly proportional to V2] 
           [of Vir ‘n gegewe resistor is drywing direk eweredig aan V2]
          Since the potential difference across light bulb C is less than the operating
          voltage, the output/power will be less, / Omdat die potensiaalverskil oor
          gloeilamp C minder is as die benodigde spanning sal die uitset/drywing
          minder wees.

           OPTION 2/OPSIE 2
              V2
          P=
               R
          The potential difference across light bulb C is less than the operating voltage.
           Thus for the same resistance,  brightness decreases.
          Die potensiaalverskil oor gloeilamp C is minder as die benodigde
          potensiaalverskil. Dus vir dieselfde weerstand, sal die helderheid afneem.

          OPTION 3/OPSIE 3
          P = I2 R
          For a given resistance, power is directly proportional to I2 Since current
          decreases, brightness decreases.]
           [vir ‘n gegewe resistor is drywing direk eweredig aan I2 Omdat stroom
          afneem sal die helderheid afneem]

          OPTION 4/OPSIE 4
          P = I2 R
          In the circuit, the total current in light bulb C is less than the optimum current
          required (0,5 A).  Thus for the same resistance,  the power will be less
          hence brightness will decrease.
          In die stroombaan is die totale stroom in gloeilamp C minder as die optimum
          stroom benodig (0,5 A). Dus vir dieselfde weerstand, is die drywing minder en
          die helderheid sal afneem.

          OPTION 5/OPSIE 5
          P = IV [Power is directly proportional/equal to product of V and I. 
          Since current decreases, brightnes decreases/
                     drywing is direk eweredig/gelyk aan produk van V en I. Omdat
          stroom afneem sal die helderheid afneem]]
          OR/OF
          The voltage across light bulb C, as well as the current in the bulb are all less
          than the optimum values hence power is less and brightness is less.
          Die potensiaalverskil oor gloeilamp C sowel as die stroom in die gloeilamp is
          almal minder as die optimum waardes, dus is die drywing minder en die
          helderheid minder.                                                                           (3)
          NOTE: No mark if only equation is given.

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8.2.1     The total current passes through resistor A.  For the parallel portion, the
          current branches, therefore only a portion of the total current passes through
          resistor C. 
          Die totale stroom vloei deur resistor A. Vir die parallelle gedeelte verdeel die
          stroom, dus vloei slegs 'n gedeelte van die stroom deur resistor C.                          (2)
          ACCEPT for 1 mark: Resistor C is connected parallel to resistors B and D
          together. Current is dividing at the junction.
          AANVAAR vir 1 punt: Resistor C is in parallel geskakel met B en D saam. Die
          stroom breek op  by die koppeling.

8.2.2     The current in B is equal to the current in A. The circuit becomes a series
          circuit. 
          Die stroom in B is gelyk aan die stroom in A. Die stroombaan word 'n serie
          stroombaan.                                                                                   (2)
                                                                                                       [21]

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QUESTION 9/VRAAG 9

9.1       Slip rings/Sleep ringe                                                                      (1)

9.2       B                                                                                           (1)

9.3                   Vmax/maks 
          Vrms/wgk =
                          2
                     312 
                   =
                       2
                   = 220,62 V                                                                         (3)

9.4.1     POSITIVE MARKING FROM 9.3/POSITIEWE NASIEN VANAF 9.3
          OPTION 1/OPSIE 1
                              2
                            Vrms / wgk
          Paver / gemid =          
                              R
                            (220,62)2 
                       =
                                40
                       = 1216,83 W 
          POSITIVE MARKING FROM 9.3/POSITIEWE NASIEN VANAF 9.3
          OPTION 2/OPSIE 2
                   Vrms / wgk
          I rms =
                      R
                   (220,62)              for any/ vir enige
                =
                       40
                = 5,515
          P ave = I2 rms R
                       = (5,515)2(40) 
                       = 1216,61 W 
          OR
          P ave = V rms I rms
                = (220,62)(5,515) 
                = 1216,72 W

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          OPTION 3/OPSIE 3
                 Vmax
          Imax =
                   R
                  312                       for any/ vir enige
               =
                   40
                = 7,80 A
                   Imax Vmax
          Pave =
                        2
                    (7,8)(312) 
                 =
                         2
                 = 1216,80 W                                                                            (3)

9.4.2     OPTION 1/OPSIE 1
                        V                                      V
          I max/maks = max/maks        Accept/ Aanvaar: I =
                           R                                   R
                       312 
                     =
                        40 
                     = 7,8 A

          POSITIVE MARKING FROM 9.3 AND 9.4.1/POSITIEWE NASIEN VANAF
          9.3 EN 9.4.1
          OPTION 2/OPSIE 2
          P ave/gemid = V rms/wgk I rms/wgk
           1 216,83 = 220,62 I rms/wgk 
                                             for any/ vir enige
            I rms/wgk = 5,515 A
                        I
            I rms/wgk = max/maks
                            2
                        I
             5,515 = max/maks 
                            2
             I max/maks = 7,8 A

          OPTION 3/OPSIE 3
          P ave/gemid = I2 rms/wgk R
          1 216,83 = I2 rms/wgk (40)
            I rms/wgk = 5,515 A                     for any/ vir enige
                         I
              Irms/wgk = max/maks
                             2
                         I
             5,515 = max/maks 
                             2
             I max/maks = 7,8 A 
                                                                                                          (4)
                                                                                                         [12]

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Physical Sciences/P1/Fisiese Wetenskappe/V1      27                                        DBE/2019
                              SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION 10/VRAAG 10
10.1      The minimum frequency of light needed to eject electrons from a metal
          (surface).
          Minimum frekwensie van lig benodig om elektrone vanaf ʼn metaal
          (oppervlak) vry te stel.

          NOTE/LET WEL:
          -1 for each key word/phrase omitted.
          -1 vir elke sleutel woorde/frase weggelaat.                                                  (2)

10.2      OPTION 1/OPSIE 1                          OPTION 2/OPSIE 2
               c                                    c = fλ
          E= h                                                                    Both equations
               λ                                    3 x 108 = f(5 x 10-7)
                                                                                    Beide vergelykings
              (6,63 × 10 -34 )(3 × 10 8 )           f = 6 x 1014 Hz
           =                              
                     5 × 10 -7                   E = hf
           = 3,98 x 10-19 J                       = (6,63 x 10-34)(6 x 1014) 
                                                   = 3,98 x 10-19 J                                   (3)
          NOTE: do not penalise if v is used in place of c.

10.3      OPTION 1/OPSIE 2
          POSITIVE MARKING FROM QUESTION 10.2/POSITIEWE NASIEN VANAF
          VRAAG 10.2
          E = W 0 + E kmax
                                       1 mark any one/1 punt vir
                   1     2
          hf = W0 + mv max             enige
                   2                

                        c
                    h       = W 0 + E K(max/maks)
                        λ
                     c
                    h     = hf 0 + E K(max/maks)
                     λ
          3,98 x 10-19 = (6,63 x 10-34)(5,55 x1014) + E K(max/maks) 
          E K(max/maks) = 3,0 x10-20 J

          E K(max/maks > 0 

          (The electrons emitted from the metal plate have kinetic energy to move
          between the plates, hence the ammeter registers a reading.
          Die elektrone vrygestel vanaf die metaalplaat het kinetiese energie om
          tussen die plate te beweeg en gevolglik registreer die ammeter ʼn lesing)

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Physical Sciences/P1/Fisiese Wetenskappe/V1      28                                 DBE/2019
                              SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

          OPTION 2/OPSIE 2
          POSITIVE MARKING FROM                 QUESTION       10.2/POSITIEWE    NASIEN
          VANAF VRAAG 10.2
          W o = hf o 
             = (6,63 x 10-34)(5,55 x1014) 
             = 3,68 x 10-19 J

          E photon > W o 
          (The energy of the incident photon is greater than the work function of
          potassium. From the equation hf = W 0 + E Kmax , the ejected photoelectrons
          will move between the plates,  hence the ammeter registers a reading.
          Die energie van die invallende foton is hoër as die arbeidsfunksie van kalium.
          Vanaf die vergelyking hf = hf 0 + E K(maks) , sal die vrygestelde foto-elektrone
          tussen die plate te beweeg en gevolglik registreer die ammeter ʼn lesing.)

          OPTION 3/OPSIE 3
          c = fλ 
          3 x 108 = f(5 x 10-7) 
          f = 6 x 1014 Hz

          f > f0 
          The frequency of the incident photon is higher than the threshold frequency.
          From the equation hf = hf 0 + E K(max) , the ejected photoelectrons will be able
          to move between the plates  (for the given frequency), hence the ammeter
          registers a reading.
          Die frekwensie van die invallende foton is hoër as die drumpelfrekwensie.
          Vanaf die vergelyking hf = hf 0 + E K(maks) , sal die vrygestelde foto-elektrone
          tussen die plate kan beweeg en gevolglik registreer die ammeter ʼn lesing.)           (4)
10.4      The increase in intensity increases the number of photons per second.
          Soos die intensiteit toeneem, neem die aantal fotone per sekonde toe.

          Since each photon releases one electron the number of ejected electrons
          per second increases.
          Aangesien elke foton een elektron vrystel, neem die aantal vrygesteld
          e elektrone per sekonde toe.
          ACCEPT: Flow of electrons per unit time increases  (1 mark)
          AANVAAR: vloei van elektrone per eenheidstyd neem toe (1 punt)

          This causes the current /ammeter reading to increase.
          Dit veroorsaak dat die stroom/ammeterlesing toeneem.                                  (3)
                                                                                               [12]

                                                                           TOTAL/TOTAAL:       150

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