# Physical Sciences Grade 12 Paper 1 NSC November 2018 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the NSC November Exam 2018, 29 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-sciences-p1-nov-2018-final-memo-afr-eng/fd1e0f54-fcfd-4b02-9e8c-9f770847620b
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: NSC November Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 29

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Physical Sciences · Grade 12 · NSC November Exam · 2018. Memorandum, 29 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 2018

                      MARKING GUIDELINES/NASIENRIGLYNE

MARKS/PUNTE: 150

                      These marking guidelines consist of 29 pages.
                      Hierdie nasien riglyne bestaan uit 29 bladsye.

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Physical Sciences P1/Fisiese Wetenskappe V1      2                         DBE/November 2018
                              NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 1/VRAAG 1

1.1       C /D                                                                               (2)

1.2       C                                                                                  (2)

1.3       C                                                                                  (2)

1.4       B                                                                                  (2)

1.5       B                                                                                  (2)

1.6       A                                                                                  (2)

1.7       A                                                                                  (2)

1.8       D                                                                                  (2)

1.9       D                                                                                  (2)

1.10      C                                                                                   (2)
                                                                                               [20]

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Physical Sciences P1/Fisiese Wetenskappe V1      3                               DBE/November 2018
                              NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 2/VRAAG 2

2.1       When a (non-zero) resultant/net force acts on an object, the object will
          accelerate in the direction of the force with an acceleration that is directly
          proportional to the force and inversely proportional to the mass of the object.
          
          Wanneer 'n (nie-nul) resultante/netto krag op 'n voorwerp inwerk, sal die
          voorwerp in die rigting van die krag versnel teen 'n versnelling wat direk
          eweredig is aan die (netto) krag en omgekeerd eweredig aan die massa van
          die voorwerp.

          OR/OF
          The (non-zero) resultant/net force acting on an object is equal to the rate of
          change of momentum of the object in the direction of the resultant/net force.
           (2 or 0)
          Die (nie-nul) netto krag wat op 'n voorwerp inwerk is gelyk aan die tempo van
          verandering van momentum.

          ACCEPT/AANVAAR
          Acceleration is directly proportional to the net force and inversely proportional
          to the mass of the object.
          Versnelling direk eweredig is aan die netto krag en omgekeerd eweredig aan
          die massa van die voorwerp.
          NOTE/LET WEL
          If any of the underlined key words in the correct context is omitted deduct 1
          mark.
          Indien enige van die onderstreepte sleutel woorde in die korrekte konteks
          uitgelaat is, trek 1 punt af.                                                                   (2)

2.2
                        N                                            N
                                   T                                             T
                    f                                           f

                             w                                             w
          Notes/Aantekeninge
          •   Mark is awarded for label and arrow
              Punt word toegeken vir byskrif en pyltjie
          • Do not penalise for length of arrows
              Moenie vir die lengte van die pyltjies penaliseer nie.
          •   If T is not shown but T || and T⊥ are shown, give 1 mark for both
              Indien T nie aangetoon is nie maat T || en T⊥ is getoon. Ken 1 punt toe vir beide.
          •   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 any additional force /Trek 1 punt af vir enige addisionele krag           (4)

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Physical Sciences P1/Fisiese Wetenskappe V1      4                                     DBE/November 2018
                              NSC/NSS – Marking Guidelines/Nasienriglyne

                   Accept the following symbols /Aanvaar die volgende simbole.
          N        F N ; Normal;Normal force /Normaal; Normaalkrag 
                   F f / f k / frictional force/wrywingskag/kinetic frictional force/ kinetiese
          f
                   wrywingskrag
                   F g ; mg; Weight;F Earth on block ;Fw / Gewig ;Gravitational force / Gravitasiekrag/
          w
                   78,4 N
          T        Tension/Spanning; F T /F A , F /16,96 N 

2.3.1     The 2/8 kg block /system is accelerating/Die 2/8 kg blok / sisteem is besig om
          te versnel 

          OR/OF
          The acceleration is not zero / a ≠ 0 (m·s-2) / a = 1,32 m.s-2 / Die versnelling is
          nie nul nie 

          OR/OF
          Velocity is /increasing/changing/not constant/Snelheid neem toe/ verander/is
          nie konstant nie 

          OR/OF
          F net is not equal to zero /F net is nie gelyk aan nul nie / F net ≠ 0 (N) 

          OR/OF
          The acceleration is changing / Die versnelling verander 

          Accept/Aanvaar
          An unbalanced force is acting on it / ‘n Ongebalanseerde krag werk in op die
          liggaam                                                                                            (1)

2.3.2     For 2 kg/Vir die 2 kg massa
                                                                Fnet = ma
          F net = ma  1 mark for any                           mg + T = ma
          mg -T= ma          1 punt vir
                                                                (2)(-9,8) + T = 2(-1,32) 
          (2)(9,8) – T = 2(1,32)                               T = 16,96 N 
          T = 16,96 N                                                                                        (3)

2.3.3     POSITIVE MARKING FROM 2.3.2/POSITIEWE NASIEN VANAF 2.3.2
          F net = ma
          Tcos15o - f = ma 

          T x = Tcos15o
              = 16,96 cos15o
              = 16,38 N (16,382 N)

          16,382 – f  = (8)(1,32) 
          f = 5,82 N (to the left/na links) 

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                              NSC/NSS – Marking Guidelines/Nasienriglyne

          OR/OF
          F net = ma
          Tcos15o + f = ma 

          T x = Tcos15o
              = 16,96 cos15o
              = 16,38 N (16,382 N)

          -16,382 + f  = (8)(-1,32) 
          f = 5,82 N (to the left/na links)                                                              (4)

2.4      ANY ONE/ENIGE EEN
         Normal force changes/decreases /Normaalkrag verander/neem af
         The angle (between string and horizontal) changes/increases. /Die hoek
         (tussen die toutjie en die horisontaal) verander/neem toe
         The vertical component of the tension changes/increases/Die vertikale
         komponent van die spanning verander / neem toe.                                                  (1)

2.5 -    Yes/Ja
         The frictional force (coefficient of friction) depends on the nature of the surfaces
         in contact. 
         Die wrywingskrag (wrywingskoëffisiënt) is afhanklik van die aard van die
         oppervlaktes in kontak met mekaar.

         ACCEPT/AANVAAR
         The nature of the surface changes / µ k changes
         Die aard van die oppervlakte verander / µk verander                                               (2)
                                                                                                          [17]

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Physical Sciences P1/Fisiese Wetenskappe V1      6                               DBE/November 2018
                              NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 3/VRAAG 3

3.1       Downwards/Afwaarts 
      -
          The only force acting on the object is the gravitational force/weight which acts
          downwards./Die enigste krag wat op die voorwerp inwerk is die
          gravitasiekrag/gewig wat afwaarts inwerk.

          ACCEPT/AANVAAR:
          The only force acting is gravitational/weight./Die enigste krag wat inwerk is
          gravitasie/gewig

          OR/OF
          Gravitational force/weight acts downwards./Gravitasiekrag/gewig werk
          afwaarts

          OR/OF
          The ball is in free-fall / Die bal in vry-val 

          OR/OF
          (Gravitational) acceleration is downwards/(Gravitasionele) versnelling is
          afwaarts                                                                                        (2)

3.2       OPTION 1/OPSIE 1
          Upward positive/Opwaarts positief
          v f = v i + aΔt 
          0 = 7,5 + (-9,8)Δt 
          Δt = 0,77 s 

          Downward positive/Afwaarts positief
          v f = v i + aΔt 
          0 = -7,5 + (9,8)Δt 
          Δt = 0,77 s 

          OPTION 2/OPSIE 2                             OPTION 2/OPSIE 2
          Upward positive                              Downward positive
          Opwaarts positief                            Afwaarts positief
          At highest point v f is zero                 At highest point v f is zero
          By hoogste punt is v f nul                   By hoogste punt is v f nul
             2     2                                      2     2
          v f = v i + 2a∆y                             v f = v i + 2a∆y
          0 = (7,5)2 + (2)(-9,8)∆y                     0 = (-7,5)2 + (2)(9,8)∆y
          ∆y = 2,87 (2,869) m                          ∆y = - 2,87 (-2,869) m

                v + vf                                      v + vf 
          Δy =  i       Δt                          Δy =  i        Δt 
                2                                           2 
                  7,5 + 0                                        - 7,5 + 0
          2,87 =           Δt                         - 2,87 =            Δt 
                     2                                               2

              Δt = 0,77 s                                  Δt = 0,77 s

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                              NSC/NSS – Marking Guidelines/Nasienriglyne

          OPTION 3/OPSIE 3                            OPTION 3/OPSIE 3
          Upward positive                             Downward positive
          Opwaarts positief                           Afwaarts positief
          F net Δt = m(v f – v i )                   F net Δt = m(v f – v i ) 
          mgΔt = m(v f – v i )                        mgΔt = m(v f – v i )
          (-9,8)Δt = 0 – 7,5                         (9,8)Δt = 0 –(- 7,5) 
          ∴ Δt = 0,76531 s (0,77 s)                  ∴ Δt = 0,76531 s (0,77 s)

          OPTION 4/OPSIE 4                            OPTION 4/OPSIE 4
          Upward positive                             Downward positive
          Opwaarts positief                           Afwaarts positief
          (Top to Bottom / Bo na onder)               (Top to Bottom /Bo na onder)
          v f = v i + aΔt                            v f = v i + Δt 
          -7,5 = 0 + (-9,8)Δt                        7,5 = 0 + (9,8)Δt
          ∴ Δt = 0,76531 s (0,77 s)                  ∴ Δt = 0,76531 s (0,77 s) 

          OPTION 5/OPSIE 5                            OPTION 5/OPSIE 5
          Upward positive                             Downward positive
          Opwaarts positief                           Afwaarts positief
          (Top to Bottom/ Bo na onder)                (Top to Bottom / Bo na onder)
          v f 2 = v i 2 + 2aΔy                        v f 2 = v i 2 + 2aΔy
          (7,5)2 = (0)2 + 2(-9,8)Δy                   (7,5)2 = (0)2 + 2(9,8)Δy
          Δy = -2,87 m                                Δy = 2,87 m

          Δy = v i Δt + ½ aΔt2                       Δy = v i Δt + ½ aΔt2 
          -2,87 = (0)Δt + ½ (-9,8)(Δt)2              2,87 = (0)Δt + ½ (9,8)Δt2 
          Δt = 0,765 s                               Δt = 0,765 s                                       (3)

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Physical Sciences P1/Fisiese Wetenskappe V1      8                               DBE/November 2018
                              NSC/NSS – Marking Guidelines/Nasienriglyne

NOTES for marking QUESTION 3.3
AANTEKENINGE vir merk van VRAAG 3.3
Formula mark/Formule punt                                                                  
Substitution mark /Vervangingspunt                                                         
Mark for height/distance / Punt vir hoogte/afstand                                         
Mark for comparison/Punt vir vergelyking                                                   
Mark for conclusion/Punt vir gevolgtrekking                                                

3.3       OPTION 1/OPSIE 1
          Upward positive/Opwaarts positief
          At highest point v f is zero/By hoogste punt is v f nul
             2     2
          v f = v i + 2a∆y 
          0 = (7,5)2 + (2)(-9,8)Δy 
          ∆y = 2,87 (2,869) m

          This is higher than height needed to reach point T (2,1 m)therefore the ball
          will pass point T.
          Dit is hoer as die hoogte benodig om punt T (2,1 m) te bereik dus sal die bal
          punt T verbygaan.

          Downward positive/Afwaarts positief
          At highest point v f is zero/By hoogste punt is v f nul
             2     2
          v f = v i + 2a∆y 
          0 = (-7,5)2 + (2)(9,8)Δy
          ∆y = - 2,87 (-2,869) m

          This is higher than height needed to reach point T (2,1 m)therefore the ball
          will pass the target. 
          Dit is hoer as die hoogte benodig om punt T (2,1 m) te bereik dus sal die bal
          punt T verbygaan.

          OPTION 2/OPSIE 2 (POSITIVE MARKING FROM 3.2)
          Upward positive/Opwaarts positief
          Δy = v i Δt + ½ aΔt2
          Δy = (7,5)(0,77)  + ½ (-9,8)(0,77)2
          Δy = 2,87 m (2,86 m)
          This is higher than height needed to reach point T (2,1 m)therefore the ball
          will pass point T . 
          Dit is hoer as die hoogte benodig om punt T (2,1 m) te bereik dus sal die bal
          punt T verbygaan.

          Downward positive/Afwaarts positief
          Δy = v i Δt + ½ aΔt2
          Δy = (-7,5)(0,77)  + ½ (9,8)(0,77)2
          Δy = -2,87 m (2,869 m)
          This is higher than the height needed to reach point T (2,1 m)therefore the
          ball will pass point T. 
          Dit is hoer as die hoogte benodig om punt T (2,1 m) te bereik dus sal die bal
          punt T verbygaan.

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                              NSC/NSS – Marking Guidelines/Nasienriglyne

          OPTION 3/OPSIE 3
          (E mech ) Top/Bo = (E mech ) Ground/Grond
                                                                      1 mark for any
          (E P +E K ) Top = (E P +E K ) Bottom/Onder
                                                                      1 punt vir enige
          (mgh + ½ mv2) Top/Bo = (mgh + ½ mv2) Bottom/Onder
          (9,8)(h) + 0 = 0 + (½ )(7,5)2 
          h = 2,87 m (2,869 m)

          This is higher than height needed to pass the target (2,1 m)therefore the ball
          will pass the target.
          Dit is hoer as die hoogte benodig om punt T (2,1 m) verby te gaan dus sal die
          bal punt T verbygaan.

          OPTION 4/OPSIE 4
          W net = ∆E K
          mg∆xcosθ = ½ mv f 2 - ½ mv i 2 
          (9,8)Δxcos180o = 0 - ½(7,5)2
          ∆x = 2,87 m (2,869 m)

          This is higher than point height needed to pass point T (2,1 m)therefore the
          ball will pass point T. 
          Dit is hoer as die hoogte benodig om punt T (2,1 m) verby te gaan dus sal die
          bal punt T verbygaan.

          OPTION 5/OPSIE 5
          Upward positive/Opwaarts positief
          If the highest point is y f then ∆y = (y f – y 1,6 ) At highest point v f is zero
          Indien die hoogste punt y f is, dan is ∆y = (y f – y 1,6 ). By hoogste punt is v f nul
          v f 2 = v i 2 + 2aΔy 
          0 = [(7,5)2 + (2)(-9,8)(y f – 1,6)]
          y f = 4,47 (4,469) m

          Yes 
          OR/OF
          This point (4,47m) is higher than point T  (or even the required height of
          2,1 m) therefore the ball will pass point T.
          Ja 
          Dit is hoer as die hoogte benodig om punt T (2,1 m) te bereik dus sal die bal
          punt T verbygaan.
          Downward positive/Afwaarts positief
          If the highest point is y f then Δy = (y f – y 1,6 ) At highest point v f is zero
          Indien die hoogste punt y f is, dan is Δy = (y f – y 1,6 ). By hoogste punt is v f nul
          v f 2 = v i 2 + 2aΔy 
          0 = [(-7,5)2 + (2)(9,8){y f – (-1,6)}] 
          y f = -4,47 (-4,469) m
          height is/hoogte is 4,47 m.
          This point (4,47 m) is higher than point T (or even the required height of
          2,1 m) therefore the ball will pass point T.
          Hierdie punt (4,47 m) is hoer as punt T (of selfs die benodigde hoogte van
          2,1 m ) dus sal die bal punt T verbygaan.

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          OPTION 6/OPSIE 6 (POSITIVE MARKING FROM 3.2)
          Upward positive/Opwaarts positief
          If the highest point is y f then Δy = (y f – y 1,6 ) At highest point v f is zero
          Indien die hoogste punt y f is, dan is Δy = (y f – y 1,6 ). By hoogste punt is v f nul
          Δy = v i Δt + ½ aΔt2
          (y f – 1,6) = (7,5)(0,77)  + ½ (-9,8)(0,77)2
          y f = 4,47 m (4,469 m)

          This point (4,47m) is higher than point T  (or even the required height
          of 2,1 m) therefore the ball will pass point T.
          Hierdie punt (4,47 m) is hoer as punt T (of selfs die benodigde hoogte van
          2,1 m ) dus sal die bal punt T verbygaan.

          Downward positive/Afwaarts positief
          If the highest point is y f then Δy = (y f – y 1,6 ) At highest point v f is zero
          Indien die hoogste punt y f is, dan is Δy = (y f – y 1,6 ). By hoogste punt is v f nul
          Δy = v i Δt + ½ aΔt2
          {y f – (-1,6)} = (-7,5)(0,765)  + ½ (9,8)(0,765)2
          y f = -4,47 m (-4,469 m)

          This point (4,47m) is higher than point T (or even the required height of
          (2,1m) therefore the ball will pass point T.
          Hierdie punt (4,47 m) is hoer as punt T (of selfs die benodigde hoogte van
          2,1 m ) dus sal die bal punt T verbygaan.

          OPTION 7/OPSIE 7 (POSITIVE                   OPTION 7/OPSIE 7 (POSITIVE
          MARKING FROM 3.2)
          Upward positive/Opwaarts positief            MARKING FROM 3.2
                v + vf                               Downward positive/Afwaarts
          Δy =  i       Δt                          positief
                2 
                                                             v + vf 
                0 + 7,5                              Δy =  i        Δt 
             =           (0,77 )                         2 
                2 
                                                              0 - 7,5
             = 2,89 m                                    =             (0,77) 
                                                                2
          This is higher than height needed to              = -2,89 m 
          pass the target (2,1 m)therefore the        Height /Hoogte is 2,89m
          ball will pass the target.                  This is higher than height needed to
          Dit is hoer as die hoogte benodig om         pass the target (2,1 m)therefore the
          die teiken verby te gaan (2,1 m) dus         ball will pass the target.
          sal die bal die teiken verbygaan.            Dit is hoer as die hoogte benodig om
                                                       die teiken verby te gaan (2,1 m) dus
                                                       sal die bal die teiken verbygaan.

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                              NSC/NSS – Marking Guidelines/Nasienriglyne

          OPTION 8/OPSIE 8
          Upward positive/Opwaarts positief
          At highest point v f is zero/By hoogste punt is v f nul
              2    2
          v f = v i + 2a∆y 
          0 = v i 2 – (2)(9,8)(2,1) 
          v i = 6,42 m∙s-1 

          This is the actual velocity needed to reach the target.
          The given velocity is greater than the actual velocity needed. 
          The ball will pass the target. 
          Dit is die werklike snelheid benodig is om die teiken te bereik
          Die gegewe snelheid is groter as die werklike snelheid benodig
          Die bal sal die teiken verbygaan.

          Downward positive/Afwaarts positief
          At highest point v f is zero
              2    2
          v f = v i + 2a∆y 
          0 = v i 2 + (2)(9,8)(-2,1) 
          v i = 6,42 m∙s-1 

          This is the actual velocity needed to pass the target.
          The given velocity is greater than the actual velocity needed. 
          The ball will reach the target. 
          Dit is die werklike snelheid benodig is om die teiken te verby te gaan.
          Die gegewe snelheid is groter as die werklike snelheid benodig
          Die bal sal die teiken verbygaan.

          OPTION 9/OPSIE 9
          W nc = ΔE p + ΔE k 
          0 = mgh f – mgh i + ½ mv f 2 – ½ mv i 2
          0 = (9,8)h f – (9,8)(1,6) + ½ (0) 2 – ½ (7,5)2
            0 = (9,8)h f – 43,805
          ∴ h f = 4,47 m 
          ∴ The ball will pass point T 
            Die bal sal punt T verbygaan.

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Physical Sciences P1/Fisiese Wetenskappe V1     12                               DBE/November 2018
                              NSC/NSS – Marking Guidelines/Nasienriglyne

          OPTION 10/OPSIE 10
          POSITIVE MARKING FROM 3.2 / POSITIEWE NASIEN VANAF 3.2
          Upward positive/Opwaarts positief
          Δt(max. height/maks. hoogte) = 0,77 s
          Δy = v i Δt + ½ aΔt2 
          2,1 = (7,5)Δt + ½ (-9,8)Δt2 
          ∴ Δt = 0,36 s 
          ∴ Δt (max height/maks.hoogte, 0,77 s) > Δt (to pass point T/ om T verby te
          gaan, 0,36 s) 
          ∴ The ball passed point T 
            Die bal het punt T verbygegaan.
          Downward positive/Afwaarts positief
          Δt (max height) = 0,77 s
          Δy = v i Δt + ½ a Δt2 
          2,1= (7,5)Δt + ½ (-9,8)Δt2 
          ∴ Δt = 0,36 s
          ∴ Δt (max height, 0,77 s) > Δt (to reach point T, 0,36 s) 
          ∴ The ball passed point T 
            Die bal het punt T verbygegaan

          OPTION 11/OPSIE 11
          Upward positive/Opwaarts positief
          ∆y = v i ∆t + ½ a∆t2
          (3,7 – 1,6) = 7,5 ∆t + ½ (- 9,8) ∆t2 
          ∆t = 0,375 s 
          The time to pass point T is less than time to reach maximum height . Ball
          will pass point T.
          Die tyd om punt T verby te gaan, is minder as tyd om maksimum hoogte te
          bereik.. Bal sal punt T verbygaan

          Downward positive/Afwaarts positief
          ∆y = v i ∆t + ½ a∆t2
          (3,7 – 1,6) = -7,5 ∆t + ½ (9,8) ∆t2 
          ∆t = 0,375 s 
          The time to reach point T is less than time to reach maximum height . Ball
          will pass point T.
          Die tyd om punt T verby te gaan, is minder as tyd om maksimum hoogte te
          bereik.. Bal sal punt T verbygaan

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Physical Sciences P1/Fisiese Wetenskappe V1     13                                   DBE/November 2018
                              NSC/NSS – Marking Guidelines/Nasienriglyne

          OPTION 12/OPSIE 12
          Upward positive/Opwaarts positief
          v f 2 = v i 2 + 2aΔy 
          v f 2 = (7,5)2 + 2(-9,8)(2,1) 
          v f = 3,88 m·s-1 

          Velocity at T is 3,88 m·s-1 therefore the ball still moving towards its maximum
          height 
          Snelheid by T is 3,88 ·m·s-1 dus beweeg die bal steeds opwaarts na
          maksimum hoogte
          Downward positive/Afwaarts positief
          v f 2 = v i 2 + 2aΔy 
          v f 2 = (-7,5)2 + 2(9,8)(-2,1)
          v f = -3,88 m·s-1

          Velocity at T is -3,88 m·s-1 therefore the ball is still moving towards its
          maximum height 
          Snelheid by T is -3,88 ·m·s-1 dus beweeg die bal steeds opwaarts na
          maksimum hoogte                                                                                     (6)

3.4       POSITIVE MARKING FROM 3.2 / POSITIEWE NASIEN VANAF 3.2
                                                                                     0,77 t(s)
                                                                0
                7,5

                                                            v (m∙s-1)

           v (m∙s-1)                                       - 7,5

                       0        0,77        t(s)
          Notes/Notas:
          Initial velocity and time for final velocity shown
                                                                                                 
          Beginsnelheid en tyd vir finale snelheid aangedui.
          Correct straight line (including orientation) drawn
                                                                                                 
          Korrekte reguitlyn (insluitend oriëntasie) geteken.                                                 (2)
                                                                                                              [13]

QUESTION 4/VRAAG 4

4.1       Momentum is the product of the mass of an object and its velocity 
          Momentum is die produk van die massa van 'n voorwerp en sy snelheid.
          [NOTE/LET WEL: 2 or/of 0]                                                                            (2)

4.2   -   To the left/Na links 
          Newton's third law/Newton se derde wet 

          ACCEPT/AANVAAR:
          Principle of conservation of linear momentum / law of action-reaction
          Beginsel van behoud van lineêre momentum/wet van aksie-reaksie
          Newton's third law and Newton’s second law/Newton se derde wet en
          Newton se tweede wet                                                                                 (2)
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Physical Sciences P1/Fisiese Wetenskappe V1     14                               DBE/November 2018
                              NSC/NSS – Marking Guidelines/Nasienriglyne

NOTE: For QUESTION 4.3 and 4.4 motion to the right has been taken as positive.
Candidates may use the opposite direction.
LET WEL: Vir VRAAG 4.3 en 4.4 word beweging na regs as positief geneem.
Kandidate mag die teenoorgestelde rigting gebruik.

4.3       OPTION 1/OPSIE 1
                      ∑p i = ∑p f                          Allocate mark if 0 is substituted on
          m 1 v 1i + m 2 v 2i = m 1 v 2f + m 2 v 2f         left hand side/Ken punt toe indien
          mass of girl is m / massa van meisie is m         0 aan linkerkant vervang is.

          {(m + 2)(0)} + {8(0)} = {(m + 2)(-0,6)} + (8)(4)
          m = 51,33 kg                                                NOTE: Penalise only
                                                                       once for the incorrect
          OPTION 2/OPSIE 2                                             sign of the 0,6.
          ∑p i = ∑p f                                                  LET WEL: Penaliseer
                                                                      slegs eenmaal die die
          m 1 v 1i + m 2 v 2i = m 1 v 2f + m 2 v 2f
                                                                       inkorrekte teken van
          0 = m 1 v 1f + m 2 v 2f                                      0,6
          0 = (8)(4)+ m 2 (-0,6) 
          ∴ m 2 = 53,33 kg
          ∴ m girl = 53,33 – 2
          ∴ m girl = 51,33 kg 

          OPTION 3/OPSIE 3
          Δp girl = -Δp parcel 
          m(v f – v i ) = -m(v f – v i )
          (m + 2)(-0,6 - 0)  = – 8(4 – 0) 
           m = 51,33 kg                                                                                  (5)

4.4       POSITIVE MARKING FROM 4.3/POSITIEWE NASIEN VANAF 4.3
          Impulse = Δp = m(v f – v i ) 
                 = (51,33 + 2)(-0,6 – 0) 
                 = -32 N·s / kg⋅m⋅s-1
          Magnitude of impulse/Grootte van die impuls is 32 N∙s /32 kg⋅m·s-1 
          OR/OF
          Impulse = Δp parcel/pakket = m(v f – v i ) 
                    Δp = (8)(4 – 0)  = 32 kg m∙s-1
                  ∴Δp girl/meisie = 32 kg m∙s-1                                                          (3)

4.5       POSITIVE MARKING FROM 4.4 /POSITIEWE NASIEN VANAF 4.4
          32 kg⋅m⋅s-1 / N·s  to the right/opposite direction /na regs /teenoorgestelde
          rigting                                                                                         (2)
                                                                                                          [14]

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Physical Sciences P1/Fisiese Wetenskappe V1     15                               DBE/November 2018
                              NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 5/VRAAG 5
5.1       A force is non-conservative if the work it does on an object which is moving
          between two points depends on the path taken.
          'n Krag is nie-konserwatief indien die arbeid wat dit verrig op 'n voorwerp wat
          tussen twee punte beweeg van die pad afhang.
          OR/OF
          A force is non-conservative if the work it does on an object depends on the
          path taken.
          'n Krag is nie-konserwatief indien die arbeid wat dit verrig afhang van die pad
          wat dit neem.

          OR/OF
          A force is non-conservative if the work it does in moving an object around a
          closed path is non-zero 
          'n Krag is nie-konserwatief indien die arbeid wat dit verrig om 'n voorwerp op
          'n geslote pad te beweeg nie-nul is.                                                            (2)
          NOTE/LET WEL
          If any of the underlined key words/phrases in the correct context is omitted
          deduct 1 mark. If the word work is omitted 0 marks
          Indien enige van die onderstreepte sleutel woorde/frases in die korrekte
          konteks uitgelaat is, trek 1 punt af. Indien die woord arbeid uitgelaat is 0
          punte.
5.2       No/Nee                                                                                         (1)

5.3       OPTION 1/ OPSIE 1
              W
          P=       
               Δt
               4,8 ×10 6
            =            
                 (90)
           = 53 333,33 W
           = 5,33 x 104 W (53,33 kW) 

          OPTION 2/OPSIE 2
                v + vi 
          Δx =  f       Δt
                2 
                 0 + 25 
              =          (90)
                 2 
              = 1 125 m

          W F = FΔxcosθ
          4,80 x 106 = F(1 125)cos0o
                     F = 4 266,667 N
          P ave = Fv ave 
                = (4 266,667)(12,5) 
                = 53 333, 33 W                                                                           (3)

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5.4       The net/total work done on an object is equal to the change in the object's
          kinetic energy
          Die netto/totale arbeid verrig op 'n voorwerp is gelyk aan die verandering in
          die voorwerp se kinetiese energie.
          OR/OF
          The work done on an object by a net force is equal to the change in the
          object's kinetic energy. 
          Die arbeid verrig op 'n voorwerp deur 'n netto krag is gelyk aan die
          verandering in die voorwerp se kinetiese energie.                                               (2)
          NOTE/LET WEL
          If any of the underlined key words/phrases in the correct context is omitted
          deduct 1 mark.
          Indien enige van die onderstreepte sleutel woorde/frases in die korrekte
          konteks uitgelaat is, trek 1 punt af.

5.5       OPTION 1/OPSIE 1
          W net = ΔK
          W w + W f + W F = ½ mv f 2 - ½ mv i 2
          mg∆xcosθ + W f + W F = ½ mv f 2 - ½ mv i 2
          (1 500)(9,8)200cos180o + W f + 4,8 x 106 = ½ (1 500)(252 – 0) 
          -2 940 000 + W f +4,8 x 106 = 468 750
          W f = -1 391 250 J
              = -1,39 x 106 J

          OR/OF
          W net = ΔK
          W w + W f + W F = ½ mv f 2 - ½ mv i 2
          -ΔEp + W f + W F = ½ mv f 2 - ½ mv i 2
          -(1 500)(9,8)(200 – 0) + W f + 4,8 x 106 = ½ (1 500)(252 – 0) 
          -2 940 000 + W f +4,8 x 106 = 468 750
          W f = -1 391 250 J
              = -1,39 x 106 J                                                                            (5)
          NOTE/LET WEL
          0 can be omitted in above substitutions.
          0 kan in bogenoemde vervangings weggelaat word.

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          OPTION 2/OPSIE 2
          W nc = ∆K + ∆U                               1 mark for any of
                        2          2
          W nc = ½ mv f - ½ mv i + mgh f - mgh i       these/
                          2    2                      1 punt vir enige
               = ½ m (v f - v i ) + mg(h f - h i )
          W nc = ½ mv f 2 + mgh f - ½ mv i 2 - mgh i   van hierdie
                                      2            2
                  W f + W F = ½ mv f - ½ mv i + mgh f - mgh i
          W f + 4,8 x 106 = [½ (1 500)(25)2 + -0] +[(1 500)(9,8)(200) - 0]
                       W f = - 1,39 x 106 J (-1,40 x 106J )

          OR/OF
                                                          1 mark for any of
          W nc = ∆K + ∆U                                  these/
          W nc = ½ mv f 2 - ½ mv i 2 + mgh f - mgh i
                                                      1 punt vir enige
               = ½ m (v f 2- v i 2) + mg(h f - h i )      van hierdie
          W nc = ½ mv f 2 + mgh f - ½ mv i 2 - mgh i
          W f + 4,8 x 106 = [½ (1500)(25)2 + (1500)(9,8)(200)] - [0 + 0]
                       W f = - 4,8 x 106 + 3,4 x 106
                             = - 1,39 x 106 J (-1,40 x 106J )                                            (5)

          ACCEPT THE FOLLOWING FOR: /AANVAAR DIE VOLGENDE VIR : 3                         ( 5)
          POSITIVE MARKING FROM 5.3/POSITIEWE NASIEN VANAF 5.3
          v f = v i + aΔt
          25 = 0 + a(90)
          a = 0,277…ms-2

          F net = ma
                =(1 500)(0,2777…) = 416,66… N

          F + (w || ) + (-f k ) = 416,666…
          4 266,6667 – 1 500(9,8)sinθ - f k = 416,666…
          f k = 1 236,6667 N

          W f = f k Δxcosθ 
              = (1 236,6667)(1 125)(cos180°)
              = - 1 391 250 J                                                                             (5)
                                                                                                          [13]

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QUESTION 6/VRAAG 6

6.1       The change in frequency (or pitch), of the sound detected by a listener because
          the sound source and the listener have different velocities relative to the
          medium of sound propagation. 
          Die verandering in frekwensie (of toonhoogte) (golflengte) van die klank
          waargeneem deur ‘n luisteraar omdat die klankbron en die luisteraar
          verskillende snelhede relatief tot die medium van klank voortplanting het.

          OR/OF
          An (apparent) change in observed/detected frequency (pitch), 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                                                                            (2)
          NOTE/LET WEL
          If any of the underlined key words/phrases in the correct context is omitted
          deduct 1 mark.
          Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks
          uitgelaat is, trek 1 punt af.

6.2       Away from/Weg vanaf 
          Observed frequency lower/Waargenome frekwensie is laer                                         (2)

6.3         v = fλ 
          340 = f (0,34) 
             f = 1 000 Hz                                                                                (3)

6.4       POSITIVE MARKING FROM 6.3/POSITIEWE NASIEN VANAF 6.3
               v ± vL                   v − vL
          fL =        fs      OR/OF fL =        fs
               v ± vs                       v

                              
               340 - v L
          950 =                   1 000 
                340 + 0

             v L = 17 m∙s-1

          distance/afstand x = vΔt
                             = (17)(10) 
                             = 170 m 

          OR/OF
               v − vL
          fL =        fs 
                  v
                           
                    340 - x 10
          950 =               (1000)
                    340 + 0

          distance/afstand x = 170 m 

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          ACCEPT/AANVAAR
          v L = Δfλ 
             = (50)(0,34)
             = 17 m·s-1

          distance/afstand x = vΔt
                             = (17)(10) 
                             = 170 m                                                                      (6)
                                                                                                          [13]

QUESTION 7/VRAAG 7

7.1                   Q1 + Q2 + Q3
          Q net/netto =
                            3
                        -15 ×10 - 9 + Q + 2 x10 -9
          -3 x 10-9 =                              
                                    3
          Q = + 4 x 10-9 C

          NOTE/LET WEL
           for addition of the three correct charges
           correct answer                                                                                (2)

7.2

          NOTES/NOTAS
          Correct shape /Korrekte vorm 
          Correct direction/Korrekte rigting 
          Lines must not cross and must touch spheres 
          Lyne moet nie kruis nie en moet die sfere raak                                                  (3)

7.3       The magnitude of the electrostatic force exerted by one point charge (Q 1 ) on
          another point charge (Q 2 ) is directly proportional to the product of the
          (magnitudes) of the charges and inversely proportional to the square of the
          distance (r) between them 
          Die grootte van die elektrostatiese krag uitgeoefen deur een puntlading (Q 1 )
          op 'n ander puntlading (Q 2 ) is direk eweredig aan die produk van die
          (groottes) van die ladings en omgekeerde eweredig aam die kwadraat van die
          afstand (r) tussen hulle.                                                                       (2)
          NOTE/LET WEL
          If any of the underlined key words/phrases in the correct context is omitted
          deduct 1 mark. If masses used (0/2)
          Indien enige van die onderstreepte sleutel woorde/frases in die korrekte
          konteks uitgelaat is, trek 1 punt af. Indien massas gebruik word, (0/2)

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7.4       OPTION 1/OPSIE 1
              kQ 1Q 2 
          F=
                 r2
                (9 ×10 9 )(3 ×10 -9 )(3 × 10 -9 ) 
          FSP =
                            (0,1)2
              = 8,1 x 10-6 N downwards/afwaarts

                   (9 ×109 )(3 ×10 -9 )(3 × 10 -9 )
          FTP =                                       
                                (0,3)2
                = 9 x 10-7 N left/links (0,9 X 10-6 N to the left/na links
          F net 2 = (F SP )2 + (F TP )2
          Fnet =     (F ) + (F )
                       SP
                            2
                                 TP
                                      2
                                                           for either
          Fnet =    (8,1×10 -6 ) 2 + (0,9 ×10 -6 )2
          F net = 8,15 x 10-6 N 

          OPTION 2/OPSIE 2
               kQ
          Es = 2 
                r
                  (9 × 10 9 )(3 × 10 -9 )
               =
                          (0,1) 2
              = 2 700 N.C-1
               kQ
          ET = 2                                                  both substitutions
                r
                                                                   beide vervangings
                 (9 × 10 9 )(3 × 10 -9 )
              =
                        (0,3) 2
             = 300 N.C-1

          E net = E S2 + E 2T
               = (2700 ) 2 + (30) 2 
               = 2 716,62 N.C-1

          F = Eq
            = (2 716,62)(3 x 10-9) 
            = 8,15 x 10-6 N                                                                                     (5)

7.5       POSITIVE MARKING FROM 7.4 / POSITIEWE NASIEN VANAF 7.4
          OPTION 1/OPSIE 1
            F
          E= 
             q
             8,15 ×10 -6
           =              
               3 × 10 - 9
            = 2,72 x 103 N.C-1                                                                                  (3)

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          OPTION 2/OPSIE 2
                 kQ
            Es = 2 
                 r
                (9 × 10 9 )(3 × 10 -9 )
              =                         
                       (0,1) 2
               = 2 700 N.C-1

                kQ
          ET =
                 r2
                 (9 × 10 9 )(3 × 10 -9 )
               =                         
                        (0,3) 2
               = 300 N.C-1

          E net = E S2 + E 2T
                = (2700 ) 2 + (30) 2
                = 2 716,62 N.C-1 

          NOTE/LET WEL
          Mark Allocation/Puntetoekenning
           correct formula/korrekte formula
           both subsitutions/beide vervangings
           correct answer/korrekte antwoord
          If calculation done in 7.4 award full marks for answer written here. / Indien
          berekening in 7.4 gedoen is, moet volle punte vir die antwoord wat hier
          geskryf is, toegeken word.

7.6.1     Sphere/Sfeer P or/of T                                                                         (1)

7.6.2     SPHERE P/SFEER P
               Q              Q
          ne =    or/of n e =
               qe             e
              − 15 x 10 −9
           =            −19
                             = 9,38 x 1010
             − 1,6 x 10
          mass gained/massa gewin = n e m e
          m gained /gewin= (9,38 x 1010)( 9,11 x 10-31 )
            = 8,55 x 10-20 kg 

          SPHERE T/SFEER T
               Q              Q
          ne =    or/of n e =
               qe             e
                − 5 x 10 −9
            =             −19
                               = 3,125 x 1010
              − 1,6 x 10
          mass gained/massa gewin = n e m e
          m gained /gewin= (3,125 x 1010)( 9,11 x 10-31 )
            = 2,85 x 10-20 kg                                                                             (3)
                                                                                                          [19]

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QUESTION 8/VRAAG 8

8.1       The battery supplies 12 J per coulomb/12 J per unit charge. 
          Die battery verskaf 12 J per coulomb lading

          OR /OF
          The potential difference of the battery in an open circuit is 12 V. 
          Die potensiaal verskil van die battery in 'n oop stroombaan is 12 V

          OR/OF
          The battery does 12 J of work per coulomb of charge.
          Die battery verrig 12 J arbeid per coulomb lading

          OR/OF
          Maximum work done by the battery per unit charge is 12 J
          Maksimum arbeid verrig deur die battery per eenheidslading is 12 J

          OR/OF
          Maximum energy supplied by the battery per unit charge is 12 J
          Maksimum energie verskaf deur die battery per eenheidslading is 12 J

          OR/OF
          The battery supplies 12 J of energy per coulomb/ 12 J of energy per unit
          charge
          Die battery verskaf 12 J energie per coulomb/12 J energie per eenheidslading

          OR/OF
          The greatest potential difference that can be generated by a battery is 12V
          Die grootste potensiaalveskil wat deur ‘n battery gelewer word, is 12 V

          OR/OF
          The total energy transferred by a battery to a unit electric charge is 12 J
          Die totale energie oorgedra deur die battery aan ‘n eenheid elektriese lading
          is 12 J

          OR/OF
          The total amount of electric energy supplied by the battery per coulomb/per
          unit charge is 12 J
          Die totale hoeveelheid elektriese energie verskaf deur die battery per
          coulomb/per eenheid lading is 12 J                                                              (2)
          NOTE/LET WEL
          If any of the underlined key words/phrases in the correct context is omitted
          deduct 1 mark.
          Indien enige van die onderstreepte sleutel woorde/frases in die korrekte
          konteks uitgelaat is, trek 1 punt af

8.2.1    OPTION 1/OPSIE 1
         V lost/verlore = I r 
                         = (2) (0,5)
                         =1V
         V ext/eks = Emf/emk – V lost/verlore
                      = (12 - 1) 
                      = 11 V
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Physical Sciences P1/Fisiese Wetenskappe V1     23                               DBE/November 2018
                              NSC/NSS – Marking Guidelines/Nasienriglyne

         OPTION 2/OPSIE 2
         ε = I(R + r) 
         12 = V ext/eks +(2)(0,5) 
         V ext/eks = 11 V

         OPTION 3/OPSIE 3
         ε = I(R + r) 
         12 = 2(R + 0,5)
         R = 5,5 Ω
         V = IR
           = 2(5,5)
           = 11 V                                                                                        (3)

8.2.2    POSITIVE MARKING FROM 8.2.1/POSITIEWE NASIEN VANAF 8.2.1
         OPTION 1/OPSIE 1
              V
         R =
              I
             11
           =     
              2
           = 5,5 Ω

         OPTION 2/OPSIE 2
         0,5:R
                    
           1:11
         R = 5,5 Ω

         OPTION 3/OPSIE 3
          1 11 
             =
         0,5 R
         R = 5,5 Ω

         OPTION 4/OPSIE 4
         V total = IR total
         12 = (2)R total
         R total = 6 Ω
         R = 6 – 0,5 
            = 5,5 Ω 

         OPTION 5/OPSIE 5
         ε = I(R + r)
         12 = 2(R + 0,5) 
         R = 5,5 Ω                                                                                       (2)

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Physical Sciences P1/Fisiese Wetenskappe V1     24                               DBE/November 2018
                              NSC/NSS – Marking Guidelines/Nasienriglyne

8.3 -    Decreases /Neem af 
         Total resistance decreases /Totale weerstand neem af
         Current increases/Stroom neem toe 
         "Lost volts" increases, (emf the same)/"Verlore volts" neem toe, (emk
         dieselfde)
         External potential difference decreases/eksterne potensiaal verskil neem af

         OR/OF
         Decreases /Neem af 
         Total resistance decreases /Totale weerstand neem af 
         Current increases/Stroom neem toe 
         ε = V ext/eks + Ir
         Ir increases/Ir neem toe 
         ε is constant/is konstant
         ∴V ext/eks decreases/neem af                                                                      (4)
                                                                                                          [11]

QUESTION 9/VRAAG 9

9.1       Temperature/Temperatuur                                                                        (1)

9.2.1     r = 3 Ω or/of 1,5 Ω 

          Accept for one mark only: /Aanvaar vir slegs een punt
          r = -3 Ω  or/of -1,5 Ω                                                                         (2)

9.2.2     ε = slope (gradient) of the graph/helling(gradiënt)van die grafiek 
              7,5 - (-3) 
          ε=                              Accept any correct values from the graph
                1, 5 - 0                  Aanvaar enige korrekte waardes vanaf die grafiek
             = 7 V

          OR/OF
          POSITIVE MARKING FROM 9.2.1 / POSITIEWE NASIEN VANAF 9.2.1
              ε
          R= -r           Accept any correct values on the line from the graph
              I
                           Aanvaar enige korrekte waardes op die lyn vanaf die grafiek
          7,5 = 1,5ε - 3 
          ε =7V

          OR/OF
          ε = I(R + r) 
            = 0,5(11 + 3) 
          ε =7V                                                                                          (3)
                                                                                                          [6]

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Physical Sciences P1/Fisiese Wetenskappe V1     25                                  DBE/November 2018
                              NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION 10/VRAAG 10

10.1.1    Y to/na X                                                                                         (1)

10.1.2    Faraday’s Law Electromagnetic Induction 
          Faraday se wet van Elektromagnetiese Induksie

          OR/OF
          Electromagnetic induction/Faraday's Law
          Elektromagnetiese induksie/Faraday se wet                                                          (1)

10.1.3    Mechanical (kinetic) energy to electrical energy 
          Meganiese (kinetiese) energie na elektriese energie                                                (2)

10.2.1    340 V 
                                                                                                             (1)
          Accept / Aanvaar
          -340 V

10.2.2    POSITIVE MARKING FROM 10.2.1/POSITIEWE NASIEN VANAF 10.2.1
                     V
          Vrms/wgk = max/maks 
                         2
                      340
                   =       
                        2
          V rms/wgk = 240,42 V                                                                              (3)

10.2.3   POSITIVE MARKING FROM
         10.2.2 / POSITIEWE NASIEN
         VANAF 10.2.3

         OPTION 1/OPSIE 1                       OPTION 2/ OPSIE 2
                          2
                        Vrms / wgk                                       V 2 max/maks
         Pave / gemid =
                            R                                   V 
                                                                2
                                                               rms / wgk        2        V 2 max/maks
                                                 Pave / gemid =         =              =
                         (240,42) 2                                R           R             2R
            1 600 =                  
                                R                                     2
                                                                 (340) 
                 R = 36,13 Ω (36,126 Ω)            1 600 =
                                                                   2R
                                                          R = 36,13 Ω (36,125 Ω) 
         OR/OF
         R = 36,12 Ω (36,124 Ω)

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