# Physical Sciences Grade 12 Paper 1 NSC November 2014 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the NSC November Exam 2014, 20 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: 2014
- Exam period: NSC November Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 20

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Physical Sciences · Grade 12 · NSC November Exam · 2014. Memorandum, 20 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 2014

                                        MEMORANDUM

MARKS/PUNTE: 150

                        This memorandum consists of 20 pages.
                      Hierdie memorandum bestaan uit 20 bladsye.

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Physical Sciences P1/Fisiese Wetenskappe V1    2                             DBE/November 2014
                                     NSC/NSS – Memorandum

QUESTION 1/VRAAG 1

1.1       A                                                                                         (2)

1.2       A                                                                                         (2)

1.3       D                                                                                         (2)

1.4       C                                                                                         (2)

1.5       B                                                                                         (2)

1.6       C  (Accept/ Aanvaar R)                                                                    (2)

1.7       A                                                                                         (2)

1.8       D                                                                                         (2)

1.9       A                                                                                         (2)

1.10      C                                                                                         (2)
                                                                                                      [20]

QUESTION 2/VRAAG 2
2.1       When a resultant (net) force acts on an object, the object will accelerate in the
          direction of the force. This acceleration is directly proportional to the force and
          inversely proportional to the mass of the object.

          Wanneer 'n resulterende (netto) krag op 'n voorwerp inwerk, sal die voorwerp in
          die rigting van die krag versnel. Hierdie versnelling is direk eweredig aan die
          krag en omgekeerd eweredig aan die massa van die voorwerp.

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

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Physical Sciences P1/Fisiese Wetenskappe V1    3                         DBE/November 2014
                                     NSC/NSS – Memorandum

2.2

                T2 

                T1 

                        w
                                                                                                  (3)

2.3       OPTION 1/OPSIE 1
          F net = ma
          For 5 kg block/Vir 5 kg-blok
          T 2 + (-mg) + (-T 1 )= ma
          250 – (5)(9,8) – T 1 = 5 a 
          201 – T 1 = 5 a
           T 1 = 201 - 5a........(1)

          For 20 kg block/Vir 20 kg-blok
          T 1 + (-mg) = ma.......(2)
          T 1 + [-20(9,8)]  = 20a
          5 = 25 a
           a = 0,2 m∙s-2 upwards/opwaarts
          ∴ T 1 = 201 - 5(0,2) 
                = 200 N 
          OR/OF T 1 = 20(9,8) + 20(0,2) 
                       = 200 N 
                                                                                                  (6)

          OPTION 2 /OPSIE 2
          F net = ma
          For 5 kg block/Vir 5 kg-blok
          T 2 + (-mg) + (-T 1 )= ma
          250 - (5)(9,8) - T 1 = 5 a 
          201 - T 1 = 5a
           T 1 = 201 - 5a........(1)

          For 20 kg block/Vir 20 kg-blok ,
          T 1 + (-mg) = ma.......(2)
          T 1 + [-20(9,8)]  = 20a
          (1) x 4 : 4T 1 = 804 – 20a
          ∴T 1 - 196 = 804 – 4T 1 
          ∴5T 1 = 1 000
          ∴T 1 = 200 N 
                                                                                                  (6)

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Physical Sciences P1/Fisiese Wetenskappe V1    4                           DBE/November 2014
                                     NSC/NSS – Memorandum

          OPTION 3/OPSIE 3
          F net = ma
          For 5 kg block/Vir 5 kg-blok
          T 2 + (-mg) + (-T 1 )= ma
          250 - (5)(9,8) - T 1 = 5 a 
          201 - T 1 = 5 a
           T 1 = 201 - 5a........(1)
                   201 − T1
          ∴a=
                      5
          For 20 kg block/Vir 20 kg-blok ,
          T 1 + (-mg) = ma.......(2)
          T 1 + [-(20)(9,8)]  = 20a
                             201 − T1
          ∴T 1 - 196 = 20(            )
                                5
          ∴T 1 = 200 N                                                                             (6)
2.4       Q                                                                                         (1)
                                                                                                    [12]

QUESTION 3/VRAAG 3
3.1       An object moving / Motion under the influence of gravity / weight / gravitational
          force only (and there are no other forces such as friction). (2 or/of 0)
          ('n Voorwerp wat / Beweging slegs onder die invloed van swaartekrag / gewig /
          gravitasiekrag (en daar is geen ander kragte soos wrywing nie).                           (2)

3.2       OPTION 1/OPSIE 1
          Upwards positive/Opwaarts positief:      Downwards positive/Afwaarts positief:

          ∆y = v i ∆t + ½ a∆t2                    ∆y = v i ∆t + ½ a∆t2
          0 = 15 ∆t + ½ (-9,8) ∆t2               0 = -15 ∆t + ½ (9,8) ∆t2
          ∆t = 3,06 s                              ∆t = 3,06 s
          It takes/Dit neem 3,06 s                It takes/Dit neem 3,06 s
                                                                                                    (4)

          OPTION 2/OPSIE 2
          Upwards positive/Opwaarts positief:     Downwards positive/Afwaarts positief:

          v f = v i + a∆t                        v f = v i + a∆t
          0 = 15 +(-9,8)∆t                      0 = -15 +(9,8)∆t
          ∆t = 1,53 s                             ∆t = 1,53 s

          It takes (2)(1,53) = 3,06 s            It takes/Dit neem 3,06 s                         (4)

          OPTION 3 / OPSIE 3
          Upwards positive/Opwaarts positief:     Downwards positive/Afwaarts positief:

          v f = v i + a∆t                        v f = v i + a∆t
          -15 = 15 + (-9,8)∆t                   15 = -15 + (9,8)∆t
          ∆t = 3,06 s                            ∆t = 3,06 s                                      (4)
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Physical Sciences P1/Fisiese Wetenskappe V1    5                               DBE/November 2014
                                     NSC/NSS – Memorandum

          OPTION 4/OPSIE 4
          Upwards positive/Opwaarts positief:       Downwards positive /Afwaarts positief:

          F net ∆t = ∆p                            F net ∆t = ∆p
          mg ∆t = m (v f -v i )                     mg ∆t = m (v f -v i )
                (0 - 15)                                  0 − ( −15) 
          ∆t =                                      ∆t =
                   -9,8                                      9,8 
          ∆t = 1,53 s                               ∆t = 1,53 s
          It takes/Dit neem (2)(1,53s) = 3,06 s    It takes/Dit neem (2)(1,53s) = 3,06 s            (4)

          OPTION 5/OPSIE 5
          Upwards positive/Opwaarts positief:       Downwards positive/Afwaarts positief:

          F net ∆t = ∆p                            F net ∆t = ∆p
          mg ∆t = m (v f -v i )                     mg ∆t = m (v f -v i )
                 − 15 − (15)                             15 − ( −15)
          ∆t =                                      ∆t =
                    − 9,8                                    9,8 
            = 3,06 s                               ∆t = 3,06 s                                      (4)

          OPTION 5/OPSIE 6

          Upwards positive/Opwaarts positief:      Downwards positive/Afwaarts positief:
          v f = v i + 2aΔy 
            2     2
                                                   v f 2 = v i2 + 2aΔy 
          For ball A/Vir bal A                     For ball A/Vir bal A
          0 = (15)2 + 2 (-9,8)∆y                  0 = (-15)2 + 2 (9,8)∆y
          ∆y A = 11,48 m                           ∆y A = -11,48 m
                 v + vi                                   v + vi 
          Δy =  f       ∆t                       Δy =  f          ∆t
                 2                                        2 
                    15 + 0                                 − 15 + 0 
          11,48 =          ∆t                   -11,48 =           ∆t 
                    2                                         2     
          Δt = 1,53 s                              Δt = 1,53 s

          It takes/Dit neem (2)(1,53s) = 3,06 s It takes/Dit neem (2)(1,53s) = 3,06 s

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Physical Sciences P1/Fisiese Wetenskappe V1    6                           DBE/November 2014
                                     NSC/NSS – Memorandum

3.3       OPTION 1/OPSIE 1

          Upwards positive/Opwaarts positief:     Downwards positive/Afwaarts positief:
          v f = v i + 2aΔy 
            2    2
                                                  v f 2 = v i2 + 2aΔy 
          For ball A/Vir bal A                    For ball A/Vir bal A
          0 = (15)2 + 2 (-9,8)∆y                0 = (-15)2 + 2 (9,8)∆y
          ∆y A = 11,48 m                          ∆y A = -11,48 m

          When A is at highest point              When A is at highest point
          Wanneer A op hoogste punt is            Wanneer A op hoogste punt is

          ∆y B = v i ∆t + ½ a∆t2                  ∆y B = v i ∆t + ½ a∆t2
               = 0 + ½ (-9,8) (1,53)2               = 0 + ½ (9,8) (1,53)2
          ∆y B = -11,47 m                         ∆y B = 11,47 m
          ∆y B = 11,47 m downward/afwaarts        ∆y B = 11,47 m downward/afwaarts

          Distance/Afstand = y A + y B            Distance/Afstand = y A + y B
                           = 11,47 + 11,48                        = 11,48 + 11,47 
                           = 22,95 m                              = 22,95 m

          OPTION 2/OPSIE 2

          Upwards positive/Opwaarts positief:     Downwards positive/Afwaarts positief:

          At maximum height v f = 0:              At maximum height v f = 0:
          By maksimum hoogte v f = 0:             By maksimum hoogte v f = 0:

          Ball/Bal A                              Ball/Bal A
          ∆y A = v i ∆t + ½ a∆t2                 ∆y A = v i ∆t + ½ a∆t2
              = 15 (1,53)  + ½ (-9,8) (1,53)2        = (-15) (1,53) + ½ (9,8) (1,53)2
               = 11,48 m                               = -11,48 m

          When A is at highest/point              When A is at highest point
          Wanneer A op hoogste punt is            Wanneer A by hoogste punt is

          ∆y B = v i ∆t + ½ a∆t2                  ∆y B = v i ∆t + ½ a∆t2
              = 0 + ½ (-9,8) (1,53)2                = 0 + ½ (-9,8) (1,53)2
          ∆y B = -11,47 m                         ∆y B = -11,47 m
          ∆y B = 11,47 m downward/afwaarts        ∆y B = 11,47 m downward/afwaarts

          Distance/Afstand = y A + y B            Distance/Afstand = (y A + y B )
                           = 11,48 + 11,47                        = 11,48 + 11,47 
                           = 22,95 m                              = 22,95 m                       (7)

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Physical Sciences P1/Fisiese Wetenskappe V1    7                               DBE/November 2014
                                     NSC/NSS – Memorandum

          OPTION 3/OPSIE 3

          Upwards positive/Opwaarts positief:       Downwards positive/Afwaarts positief:

          Ball A/Bal A                              Ball A/Bal A
          ∆y A = v i ∆t + ½ a∆t2                   y A = v i ∆t + ½ a∆t2
          ∆y A = 15(1,53) + ½ (-9,8) (1,53)2      ∆y A = -15 (1,53)  + ½ (9,8) (1,53)2
                = 11,48 m                                 = -11,48 (m)
                                                          = 11,48 m upward/opwaarts
          For ball B/Vir bal B
          v f = v i + a∆t                           For ball B/Vir bal B
           v f = 0 + (-9,8)(1,53)                   v f = v i + a∆t
          v f = 14,99 m∙s-1                          v f = 0 + (9,8)(1,53)
                                                    v f = 14,99 m∙s-1
          v f = v i + 2a∆x
            2     2

                                                    v f = v i + 2a∆x
                                                      2     2
          14,992 = 0 + 2(-9,8) ∆y B 
          ∆y B = -11,47 (m)                         14,992 = 0 + 2(9,8) ∆y B 
               = 11,47 m downward/afwaarts          ∆y B = 11,47 (m)

          Distance/Afstand = (y A + y B )           Distance/Afstand = (y A + y B )
                             = 11,48 + 11,47                           = 11,48 + 11,47
                             = 22,95 m                                 = 22,95 m                      (7)

          OPTION 4/OPSIE 4

          Upwards positive/Opwaarts positief:       Downwards positive/Afwaarts positief:

          Ball A/Bal A                              Ball A/Bal A
                                                                              
                 vi + vf        (15 + 0)                    vi + vf        (-15 + 0)
          ∆y A =         Δt  =          (1,53)    ∆y A =          Δt  =           (1,53) 
                    2              2                           2               2
              = 11,48 m                                  = -11,48 (m)
                                                         = 11,48 m upwards/opwaarts
          For ball B/Vir bal B
          v f = v i + a∆t                           v f = v i ∆t + a∆t
             = 0 +( -9,8) (1,53)                        = 0 +( 9,8) (1,53)
             = -15 m∙s-1                                = 15 m∙s-1
                                                                             
                   vi + vf      (0 - 15) × 1,53              vi + vf      (0 + 15) × 1,53
          ∆y =             Δt =                    ∆y =             Δt =                 
                      2                2                         2              2
               = -11,47 m                                = 11,47m
               = 11,47 m downward/afwaarts
          Distance/Afstand = (y A + y B )           Distance/Afstand = y A + y B
                               = 11,48 + 11,47                      = 11,48 + 11,47
                               = 22,95 m                            = 22,95 m                         (7)

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Physical Sciences P1/Fisiese Wetenskappe V1    8                                        DBE/November 2014
                                     NSC/NSS – Memorandum

          OPTION 5/OPSIE 5

          Upwards positive/Opwaarts positief:           Downwards positive/Afwaarts positief:

          Ball A/Bal A                                  Ball A/Bal A
          W net = ∆K                                   W net = ∆K
          OR/OF                                         OR/OF
          ½ m( v2 f – v2 i ) = mg(h f – h i )cosθ       ½ m( v2 f – v2 i ) = mg(h f – h i )cosθ
                       2                            o
          ½ m(0 -15 ) = m(9,8)h f cos180              ½ m(0 -152) = m(9,8)h f cos180o 
          h = 11,48 m                                   h = 11,48 m
          OR/OF                                         OR/OF
          For Ball B when A is at highest point./       For Ball B when A is at highest point./
          Vir Bal B wanneer A by sy hoogste             Vir Bal B wanneer A by sy hoogste punt
          punt is.                                      is.

          v f = v i + a∆t                               v f = v i + a∆t
              = 0 +( -9,8) (1,53) = -15 m∙s-1               = 0 +(9,8) (1,53) = 15 m∙s-1
                                                                                
                   vi + vf      (0 - 15) × 1,53                  vi + vf      (0 + 15)(1,53)
          ∆y =             Δt =                        ∆y =             Δt =                
                      2                2                            2               2
               =-11,48 m                                     = 11,48 m downward/afwaarts
               = 11,48 m downward/afwaarts

          Distance/Afstand = y A + y B                  Distance/Afstand = y A + y B
                             = 11,48 + 11,48                              = 11,48 + 11,48
                             = 22,96 m                                    = 22,96 m                         (7)

          OPTION 7/OPSIE 7

          Upwards positive/Opwaarts positief:           Downwards positive/Afwaarts positief:
          Ball A                                        Ball A
          ½ m v2 i + mgh i = ½ m v2 f + mgh f          ½ m v2 i + mgh i = ½ m v2 f + mgh f 
          ½ m(152)  +0 = ½ m(0) + m(9,8)h             ½ m(152)  +0 = ½ m(0) + m(9,8)h 
          h = 11,48 m                                   h = 11,48 m
          OR/OF                                         OR/OF
          For Ball B when A is at highest point.        For Ball B when A is at highest point.
          Vir Bal B wanneer A by sy hoogste             Vir Bal B wanneer A by sy hoogste punt
          punt is.                                      is.

          v f = v i + a∆t                               v f = v i + a∆t
                = 0 +( -9,8) (1,53)                           = 0 +(9,8) (1,53)
                = -15 m∙s-1                                   = 15 m∙s-1
                   vi + vf                                       vi + vf
          ∆y =             Δt                           ∆y =             Δt
                      2                                             2
                                                                  
                (0 - 15)(1,53)                                (0 + 15)(1,53)
             =                                            =                 
                        2                                             2
              =-11,48 m                                     = 11,48 m downward/afwaarts
              = 11,48 m downward/afwaarts
          Distance/Afstand = y A + y B                  Distance/Afstand = y A + y B
                               = 11,48 + 11,48                          = 11,48 + 11,48
                               = 22,96 m                                = 22,96 m                           (7)
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Physical Sciences P1/Fisiese Wetenskappe V1    9                                       DBE/November 2014
                                     NSC/NSS – Memorandum

3.4

                                                15

                                    v (m·s-1)
                                                0
                                                        1,53          3,06     t (s)

                                            -15

                                                                                                               (4)
                                                                                                               [17]
          CONSIDER MOTION DOWNWARD AS POSITIVE/BESKOU BEWEGING
          AFWAARTS AS POSITIEF

                                    15

                        v (m·s-1)
                                        0
                                                     1,53      3,06

                                -15

                              Criteria/Kriteria                    Marks/Punte
           Graph starts at correct Initial velocity shown./Grafiek     
           begin by korrekte beginsnelheid aangetoon.
           Time for maximum height shown (1,53 s)./Tyd vir             
           maksimum hoogte aangetoon.(1,53 s)
           Time for return shown (3,06 s) /Tyd om terug te keer        
           (3,06) aangetoon.
           Shape/Vorm: Straight line extending beyond 3,06 s/          
           Reguitlyn wat verby 3,06 s strek.                                                                   (4)
                                                                                                               [17]

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Physical Sciences P1/Fisiese Wetenskappe V1    10                                 DBE/November 2014
                                     NSC/NSS – Memorandum

QUESTION 4/VRAAG 4

4.1       p = mv
            = 50(5)
            = 250 kg∙m∙s-1 (downward/afwaarts)

          OR/OF
          p = mv
            = 50(-5)
            = - 250 kg∙m∙s-1
            = 250 kg∙m∙s-1 (downward/afwaarts)                                                            (3)

4.2       The product of the (net) force and the time interval (during which the force
          acts) (2 or 0)
          Die produk van die (netto) krag en die tydinterval (waartydens die krag
          inwerk) (2 of 0).                                                                                (2)

4.3       OPTION 1/OPSIE 1
          ∆p = F net ∆t                ∆p = F net ∆t           ∆p = F net ∆t
          0 - 250 = F net (0,2)        250 - 0 = F net (0,2)   50(0 – (-5))= F net (0,2)
          F net = -1 250 N              F net = 1 250 N         F net = 1 250 N 
                = 1 250 N                                                                                 (3)

          OPTION 2/OPSIE 2
          m(v f – v i ) = F net ∆t              m(v f – v i ) = F net ∆t
          50 (0 - 5) = F net (0,2)              50 (5 - 0) = F net (0,2)
          F net = -1 250 N                       F net = 1 250 N 
                = 1 250 N                                                                                 (3)

          OPTION 3 /OPSIE 3
          v f = v i + a∆t                       v f = v i + a∆t
          0 = 5 + a(0,2)                       5 = 0 + a(0,2) 
          a = -25 m∙s-2                         a = 25 m∙s-2
          F net = ma                           F net = ma
                 = 50 (-25)                            = 50 (25)
                  = - 1 250 N                           = 1 250 N
                    = 1 250 N                                                                             (3)

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Physical Sciences P1/Fisiese Wetenskappe V1    11                              DBE/November 2014
                                     NSC/NSS – Memorandum

4.4      Greater than/Groter as                                                                        (1)

4.5     For the same momentum change, 
        the stopping time (contact time) will be smaller (less) 
        ∴ the (upward) force exerted (on her) is greater.
        Vir dieselfde verandering in momentum,
        sal die stilhoutyd (kontaktyd) kleiner wees
        ∴ die (opwaartse)krag wat (op haar) uitgeoefen word, sal groter wees.                            (3)
                                                                                                        [12]

QUESTION 5/VRAAG 5

5.1.1     In an isolated/closed system,  the total mechanical energy is conserved /
          remains constant 
          In 'n geïsoleerde/geslote sisteem bly die totale meganiese energie behoue / bly
          konstant.

          OR/OF
          The total mechanical energy of a system is conserved/ remains constant  in
          the absence of friction.
          Die totale meganiese energie van 'n sisteem bly behoue/bly konstant in die
          afwesigheid van wrywing.

          OR/OF
          The total mechanical energy of a system remains constant provided the net
          work done by external non conservative forces is zero.
          Die totale meganiese energie van 'n sisteem bly konstant, mits die arbeid verrig
          deur eksterne nie-konserwatiewe kragte, nul is.

          OR/OF
          In the absence of a non-conservative force, the total mechanical energy is
          conserved/remains constant
          In die afwesigheid van ‘n nie-konserwatiewe krag, bly die totale meganiese
          energie behoue / konstant

          OR/OF
          In an isolated/closed system,  the sum of kinetic and gravitational potential
          energy is conserved / remains constant 
          In 'n geïsoleerde/geslote sisteem bly som van kinetiese en gravitasionele
          potensiële energie behoue / bly konstant.

5.1.2     No/Nee                                                                                       (1)

5.1.3     OPTION 1/OPSIE 1

          Along AB/Langs AB                          Along AB/Langs AB

          E mechanical at A = E mechanical at B     W net = ΔE k 
          (E p + E k ) A = (E p + E k ) B          F g Δhcosθ = ½ m(v f 2 – v i 2)
          (mgh + ½ mv2) A = (mgh + ½ mv2) B         (10)(9,8)(4)cos0o = ½ (10)(v f 2 – 0) 
          (10)(9,8)(4) + 0 = 0 + ½ (10) v f 2      v f = 8,85 m∙s-1
          v f = 8,85 m∙s-1
                                                                                                        (6)
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Physical Sciences P1/Fisiese Wetenskappe V1    12                          DBE/November 2014
                                     NSC/NSS – Memorandum

          Along AB/Langs AB

          W nc = ΔK + ΔU 
          0 = ½ (10)(v f 2 – 0) + 10(9,8)(4 – 0) 
          v f = 8,85 m∙s-1
                        Substitute 8,85 m∙s-1 in one of the following options
                          Vervang 8,85 m∙s-1 in een van die volgende opsies
          Along BC/Langs BC                        Along BC/Langs BC
          W net = ∆K                             W nc = ∆K + ∆U 
          f∆x cosθ = ∆K                           f ∆xcosθ = ∆K + ∆U
          f(8)cos 180o = ½ (10)(0 – 8,852)      f(8)cos 180 = ½ (10)(0 - 8,852) + 0
          f = 48,95 N                            f = 48,95 N (Accept/ Aanvaar 49 N)

         OPTION 2/OPSIE 2
         Along AC/Langs AC

          W nc = ∆K + ∆U
          f∆xcosθ = ∆K + ∆U
          (f)(8)(cos180o) = (0 - 0)  + 10 (9,8)(0 - 4) 
          f = 49 N                                                                                  (6)

5.2.1     f k = µ k N
               = µ k mgcosθ
               = (0,19)(300)(9,8) cos 25o
               = 506,26 N                                                                           (3)

5.2.2

          OPTION 1/OPSIE 1
              F          FN
                  app

                                        f
                         θ
                             Fg
          F net = 0
                                         
          F app + (-F g sinθ) + (-f) = 0
          F app - (300)(9,8)sin 25o - 506,26 = 0
          F app = 1 748,76 N

          P ave = Fv ave 
               = 1748,76 x 0,5
               = 874,38 W                                                                          (6)

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Physical Sciences P1/Fisiese Wetenskappe V1    13                          DBE/November 2014
                                     NSC/NSS – Memorandum

          OPTION 2/OPSIE 2
          W f + W app + W N + W g = 0 
          F∆xcosθ + F app ∆x cosθ + 0 + F g ∆x cosθ = 0
          (506,26∆x cos180o) + (F app ∆x cos0) + 300(9,8)Δxcos115o= 0

          F app = 1748,76 N
          P ave = Fv ave 
          = (1748,76 ) (0,5)
          = 874,38 W                                                                               (6)

          OPTION 3/OPSIE 3
          W f + W app + W N + W g = 0
          F∆xcosθ + F app ∆x cosθ +0 + F g sinθ∆x cosθ = 0
          (506,26∆x cos 0)  + (F ap ∆x cos0) + 300 (9,8)sin 25o Δxcos180) = 0
          F app = 1 748,76 N

          P ave = Fv ave 
              = (1 748,76 )(0,5)                                                                   (6)
                 = 874,38 W                                                                        [18]

QUESTION 6/VRAAG 6

6.1.1     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.              (2)

6.1.2     Towards/Na

          Observed/detected frequency is greater than the actual frequency.
          Waargenome frekwensie is groter as die werklike frekwensie.                                (2)

6.1.3          v ± vL                   v
          fL =         f s OR/OF fL =        fs 
               v ± vs                 v − vs
                          343 
          (1200)=              1130 
                      343 - v s
          v s = 20,01 m∙s-1
          Accept/Aanvaar: (19,42 – 20,01 m∙s-1)                                                      (5)
6.2       The star is approaching the earth. 
          Die ster nader die aarde.
          OR/OF
          The earth and the star are approaching (moving towards) each other.
          Die aarde en die ster nader mekaar.

          The spectral lines in diagram 2 are shifted towards the blue end/blue shifted.            (2)
          Die spektrumlyne in diagram 2 het verskuif na die blou ent/blou verskuiwing               [11]

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Physical Sciences P1/Fisiese Wetenskappe V1    14                                     DBE/November 2014
                                     NSC/NSS – Memorandum

QUESTION 7/VRAAG 7
7.1       To ensure that charge does not leak to the ground/insulated. 
          Om te verseker dat die lading nie na die grond toe lek nie/isoleer.                                 (1)
          Notes/Aantekeninge
          Accept/Aanvaar
          In order retain original charge/To insulate the charges./ Om oorspronklike lading te
          behou/ Om lading te isoleer.

7.2                                       QR + Q S   + 8 + (-4)
          Net charge/Netto lading =                =            = 2 µC
                                             2           2                                                    (2)

7.3

                                                                                                    Marks/
                                                Criteria for sketch:/Kriteria vir skets:
                                                                                                    Punte
                                                Correct direction of field lines
                                                                                                       
                                                Korrekte rigting van veldlyne
                                                Shape of the electric field
                                                                                                       
                                                Vorm van elektrieseveld
                                                No field line crossing each other / No field
                                                lines inside the spheres/
                                                                                                       
                                                Geen veldlyne wat maekaar kruis nie / Geen
                                                veldlyne binne sfeer nie

                                                                                                              (3)
                                   T           
7.4
                         FS on T
                                    ●         F R on T                                                        (2)

7.5
          OPTION 1/OPSIE 1
                 QQ
          F = k 12 2 
                  r
                             (1× 10 -6 )(2 × 10 -6 ) 
          F ST = (9 × 10 9 )                         = 0,45 N / 4,5 x 10-1 N left/links
                                   (0,2)2 
          OR/OF
                  1          1
          F TS = F RT =         (1,8) = 0,45 N
                  4          4
                             (2 × 10 -6 )(1× 10 -6 )
          F RT = 9 × 10 9 ×                            = 1,8 N right/regs
                                    (0,1) 2
          OR/OF
          F RT = 4F ST = 4(0,45) = 1,8 N right /regs
          F net = F ST + F RT = 1,8 + (-0,45) 
                = 1,35 N or towards sphere S / na sfeer or/of right/regs S                                   (6)

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Physical Sciences P1/Fisiese Wetenskappe V1    15                                    DBE/November 2014
                                     NSC/NSS – Memorandum

          OPTION 2/OPSIE 2
              kQ    (9 × 10 9 )(2 × 10 −6 )
          ER = 2 =                          = 1,8 x 106 N.C-1 right/regs
               r           (0,1) 2
                 kQ     (9 × 10 9 )(2 × 10 −6 )
          Es =       =                          = 4,5 x 105 N.C-1 left/links
                 r2            (0,2) 2
          E net = 1,8 x 106 - 4,5 x 105 = 1,35 x 106 N.C-1right/regs

          F = EQ = (1,35 x 106)(1 x 10-6) 
                  = 1,35 N towards sphere S / na sfeer S right/regs                                          (6)

7.6       Force experienced per unit positive charge placed at that point.
          Krag ondervind per eenheid positiewe lading by daardie punt.                                        (2)

7.7

          OPTION 1/OPSIE 1
             F      1,35               6
          E= =          - 6 = 1,35 x 10  N∙C-1
             q     1× 10                                                                                      (3)

          OPTION 2/OPSIE 2
              kQ     (9 × 10 9 )(2 × 10 −6 )
          ER = 2  =              2
                                             = 1,8 x 106 N∙C-1 right/regs
               r            (0,1)
                 kQ      (9 × 10 9 )(2 × 10 −6 )
          Es =      2
                      =                2
                                                 = 4,5 x 105 N∙C-1 left/links
                  r              (0,2)
          E net = 1,8 x 10 - 4,5 x 105 = 1,35 x 106 N∙C-1
                            6
                                                                                                              (3)
          OPTION 3/OPSIE 3
                F         1,8 
          E= =                   = 1,8 x 106 N∙C-1
                q       1× 10 -6

                F      0,45
          E=        =           = 4,5 x 105 N∙C-1
                q 1× 10 -6
          E net = 1,8 x 106 - 4,5 x 105 = 1,35 x 106 N∙C-1
                                                                                                              (3)
                                                                                                              [19]

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Physical Sciences P1/Fisiese Wetenskappe V1    16                              DBE/November 2014
                                     NSC/NSS – Memorandum

QUESTION 8/VRAAG 8

8.1.1    Keep the temperature (of battery) constant.
         Hou die temperatuur (van battery) konstant                                                     (1)

8.1.2
                             Grafiek van potensiaalverskil teenoor stroom

                                                                                                Marks/
          Criteria for drawing line of best fit:/Kriteria vir teken van lyn van beste pas:
                                                                                                Punte
          ALL points correctly plotted (at least 4 points)
                                                                                                   
          ALLE punte korrek gestip (ten minste 4 punte)
          Correct line of best fit if all plotted points are used ( at least 3 point)
                                                                                                   
          Korrekte lyn van beste pas indien alle punte gebruik word (ten minste 3 punte)

                                                                                                         (3)

8.1.3     7,2 V
          (Accept any readings between 7,0 V and 7,4 V or the value of the y-intercept
          /Aanvaar enige lesing tussen 7,0 V en 7,4 V of die waarde van die y-afsnit                    (1)

8.1.4

                            ΔV
          Slope/Helling =
                            ΔI
                            0 - 7,2 
                          =         =-9
                            0,8 - 0 
                        r = 9 Ω                                                                        (3)

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Physical Sciences P1/Fisiese Wetenskappe V1    17                        DBE/November 2014
                                     NSC/NSS – Memorandum

8.2.1    OPTION 1/OPSIE 1
         P = VI 
         100 = 20(I)
         I=5A                                                                                     (3)

         OPTION 2/OPSIE 2
             V2
         P=     
              R
               (20) 2
         100 =
                 R
         R=4Ω

         V = IR
         20 = I(4) 
         I=5A                                                                                     (3)
         OPTION 3/OPSIE 3
              V2
         P=      
              R
                (20) 2
         100 =
                  R
         R=4Ω

         P=I2R
         100 = I2(4) 
         I=5A

8.2.2    OPTION 1/OPSIE 1
             V2
         P=     
              R
             (20) 2
         R=         
              150
          = 2,67 Ω                                                                               (3)

         OPTION 2/OPSIE 2
         P = VI
         150 = (20)I
         I = 7,5 A

         V = IR
         20 = (7,5)R 
         R = 2,67 Ω 

         OR/OF
         P = I2R
         150 = (7,5)2R
         R = 2,67 Ω                                                                              (3)

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Physical Sciences P1/Fisiese Wetenskappe V1    18                           DBE/November 2014
                                     NSC/NSS – Memorandum

         OPTION 3/OPSIE 3
         IX : IY
         5 : 7,5
         1 : 1,5

         RX : RY
         1,5 : 1
         4 : 2,67 Ω                                                                                (3)

8.2.3

         OPTION 1/OPSIE 1
         P = VI                           OR/OF      P = I2R
                  150                                          150
         I 150W =      = 7,5 A                      I150W =         = 7,5 A
                   20                                          2,67

         I tot = (5 + 7,5) 
         ε = I(R + r) 
         24 = 12,5(R + r)
         24 = V ext + V ir
         24 = 20 + 12,5(r ) 
         r = 0,32 Ω                                                                                  (5)

         OPTION 2/OPSIE 2
         V = Ir
         I tot = (5 + 7,5) 
         (24 - 20) = 12,5 r
                   4
         ∴r =
                  12,5
            r = 0,32 Ω                                                                               (5)

         OPTION 3/OPSIE 3
         1      1    1
             =     +
         R R1 R 2
          1      1    1                ( 4)(2,67)
               = +        OR/OF R // =
         R //    4 2,67                4 + 2,67
         ∴R || =1,6 Ω

                   20
         I tot =       = 12,5 A
                   1,6
         ε = I(R + r) 
         24 = 12,5(R + r)
         24 = V ext + V ir
          24 = 20 + 12,5(r ) 
         r = 0,32 Ω 
                                                                                                     (5)
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Physical Sciences P1/Fisiese Wetenskappe V1    19                         DBE/November 2014
                                     NSC/NSS – Memorandum

         OPTION 4/OPSIE 4
         P = VI
         250 = (20)I
         I = 12,5 A

         V = Ir
         4 = (12,5)r
         r = 0,32 Ω                                                                               (5)

8.2.4     Device Z is a voltmeter .
          Toestel Z is 'n voltmeter                                                                 (1)

8.2.5     Device Z should be a voltmeter (or a device with very high resistance) because
          it has a very high resistance and will draw very little current. 
          The current through X and Y will remain the same hence the device can operate
          as rated.
          Toestel Z moet 'n voltmeter wees (of 'n toestel met 'n baie hoë weerstand)
          omdat dit 'n baie hoë weerstand het en baie min sal stroom trek
          Die stroom deur X en Y sal dieselfde bly, gevolglik kan die toestel werk soos
          ontwerp.                                                                                  (2)
                                                                                                   [22]

QUESTION 9/VRAAG 9
9.1       Electromagnetic induction / Elektromagnetiese induksie                                  (1)
9.2       Rotate the coil faster/Increase the number of coils/ Increase the strength of
          the magnetic field.
          Roteer die spoel vinniger/Verhoog die aantal spoele / Verhoog die sterkte van
          die magneetveld.                                                                         (1)
9.3       Slip rings/Sleepringe                                                                   (1)

9.4.1     It is the value of the voltage in a DC circuit that will have the same heating
          effect as an AC circuit. 
          Dit is die waarde van die potensiaalverskil in 'n GS-stroombaan wat dieselfde
          verhittingseffek het as 'n WS-stroombaan                                       (2)

9.4.2               Vmax
          V rms =        
                      2
                    339,45
               =          
                      2                                                                            (3)
          V rms = 240,03 V                                                                        [8]

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QUESTION 10/VRAAG 10

10.1      The minimum frequency (of a photon/light) needed to emit electrons  from
          (the surface of) a metal. (substance) 
          Die minimum frekwensie (van 'n foton/lig) benodig om elektrone vanaf die
          (oppervlakte van)'n metaal (stof) vry te stel.                                                  (2)

10.2    OPTION 1/OPSIE 1
        E = Wo + E k(max)
                  1      2
        E = Wo + mv max            Any one / Enige een
                  2
            c          1    2
          h = hf0 + mv max
            λ          2
        (6,63 × 10-34 )(3 × 108 )                                      1
                                  = (6,63 × 10-34 )(5,548 × 1014 )  + (9,11 × 10-31 )(5,33 × 105 )2 
                   λ                                                   2
        λ = 4 x 10 m
                   -7
                                                                                                           (5)
        OPTION 2/OPSIE 2
        E = Wo + E k(max)
                   1   2
        E = Wo + mv max           Any one / Enige een
                  2
                     1
          hf = hf0 + mv max
                         2

                     2
                                                               1
        (6,63 x 10-34)f = (6,63 × 10-34 )(5,548 × 1014 )  +     (9,11 × 10-31 )(5,33 × 105 )2 
                                                               2
        f = 7,5 x 1014 Hz

        c = fλ
        3 x 108 = (7,5 x 1014)λ
        λ = 4 x 10-7 m                                                                                    (5)

10.3      Smaller (less) than 
          Kleiner (minder) as                                                                             (1)

10.4      The wavelength/frequency/energy of the incident light (photon/hf) is constant.
          Die golflengte/frekwensie/energie van die invallende lig (foton/hf) is konstant

          Since the speed is larger, the kinetic energy is larger                           the work
          function/W 0 /threshold frequency smaller. 

          Aangesien die spoed vergroot, is die kinetiese energie groter, is die
          arbeidsfunksie / W 0 / drumpel frekwensie kleiner                                                (3)
                                                                                                          [11]

                                                          GRAND TOTAL/GROOTTOTAAL:                        150

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