# Physical Sciences Grade 12 Paper 1 NSC November 2017 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the NSC November Exam 2017, 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: 2017
- Exam period: NSC November Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 20

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Physical Sciences · Grade 12 · NSC November Exam · 2017. 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 2017

                     MARKING GUIDELINES/NASIENRIGLYNE

MARKS/PUNTE: 150

                      These marking guidelines consist of 20pages.
                      Hierdie nasienriglyne bestaan uit 20 bladsye.

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QUESTION 1 / VRAAG 1

1.1       B                                    (2)

1.2       D                                    (2)

1.3       A                                    (2)

1.4       C                                    (2)

1.5       B                                    (2)

1.6       A                                    (2)

1.7       D                                    (2)

1.8       B                                    (2)

1.9       C                                    (2)

1.10      D                                     (2)
                                                 [20]

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QUESTION 2 / VRAAG 2

2.1.1     An object continues in its state of rest or uniform motion (moving with
          constant velocity) unless it is acted upon by an unbalanced (resultant/net)
          force.
          OR
          A body will remain in its state of rest or motion at constant velocity unless a
          resultant/net force acts on it.
          OR
          A body will remain in its state of rest or of uniform motion in a straight line at
          constant velocity/speed unless a non-zero resultant/net force acts on it.

          'n Liggaam sal in sy toestand van rus of uniforme beweging (teen konstante
          snelheid) volhard tensy 'n ongebalanseerde (resulterende/netto) krag daarop
          inwerk.
          OF
          'n Liggaam sal in sy toestand van rus of beweging teen konstante snelheid bly
          tensy 'n resulterende/netto krag daarop inwerk
          OF
          'n Liggaam sal in sy toestand van rus of uniforme beweging in ‘n reguitlyn
          teen konstante snelheid/spoed volhard tensy 'n nie-nul resulterende/netto
          krag daarop inwerk.                                                                            (2)

2.1.3
                   N               F

                   fk

                               w

                                Accepted Labels/Aanvaarde benoemings
                  F g / F w /weight/mg /78,4 N/gravitational force
            w
                  F g / F w /gewig/mg/78,4 N/gravitasiekrag
                  F app /F A / applied force (Accept T / tension)
            F
                  F toegepas / F T / toegepaste krag (Aanvaar T / spanning)
            fk    (kinetic) Friction/F f /f/(kineties) wrywing/F w
            N     F N /Normal (force)/Normaal(krag)/ 67,9 N

                                                                                                         (4)

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2.1.3     F net = ma 
          F net = 0
          F+(-f k ) + (-F gll ) = ma
          F- (f k + F gll ) = ma
          F - 20,37 - (8)(9,8)sin30o = 0
          F = 59,57 N                                                                                    (5)

                         OPTION 1/OPSIE 1             OPTION 2/OPSIE 2
2.1.4     F net = ma                                  F net = ma
                                                                         
          (F g|| - f k ) = ma                         (f k - F gll ) = ma
          (8)(9,8)sin30o - 20,37 = 8a               20,37 + [-(8)(9,8)sin30o] = 8a 
          ∴magnitude/grootte: a = 2,35 m·s-2         ∴a = -2,35 m·s-2
                                                      ∴ magnitude/grootte: a = 2,35 m·s-2                (4)

2.2.1     Each body in the universe attracts every other body with a force that is
          directly proportional to the product of their masses  and inversely
          proportional to the square of the distance between their centres. 
          Elke liggaam in die heelal trek elke ander liggaam aan met 'n krag wat direk
          eweredig is aan die produk van hul massas en omgekeerd eweredig is aan
          die kwadraat van die afstand tussen hul middelpunte.

          OR/OF
          Every particle in the universe attracts every other particle with a force along a
          line joining them. The force is directly proportional to the product of the
          masses  of the particles and inversely proportional to the square of the
          distance between them. 
          Elke partikel in die heelal trek elke ander partikel aan met 'n krag wat direk
          eweredig is aan die produk van hul massas en omgekeerd eweredig is aan
          die kwadraat van die afstand tussen hulle.                                                     (2)

                   OPTION 1/OPSIE 1                           OPTION 2/OPSIE 2
2.2.2        GM                                            mm
          g=                                        F = G 12 2
              r2                                            r
           (6,67 x 10 -11 ) M                             GmM
          6=                                         mg = 2          
               (700 × 10 3 )2                              r
          M = 4,41 x 1022 kg                                  (6,67 ×10 -11 ) (200) M
                                                     (200)(6) =
                                                                    (700 × 103 )2 
                                                     M = 4,41 x 1022 kg                                   (4)
                                                                                                          [21]

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QUESTION 3 / VRAAG 3

3.1                                       OPTION 1/OPSIE 1
          Upwards positive                        Downwards positive
          Opwaarts positief:                      Afwaarts positief:
          v f = v i + aΔt                        v f = v i + aΔt
          0 = (12) + (-9,8)(Δt)                  0 = (-12) + (9,8)(Δt) 
          Δt = 1,22 s                            Δt = 1,22 s

                                          OPTION 2/OPSIE 2
          Upwards positive                        Downwards positive
          Opwaarts positief:                      Afwaarts positief:
          v f 2 = v i 2 + 2aΔy                    v f 2 = v i 2 + 2aΔy
          0 = 122 + 2(-9,8)Δy                    0 = (-12)2 + 2(9,8)Δy
          Δy = 7,35                               Δy = -7,35
          Δy = v i Δt + ½ aΔt2                   Δy = v i Δt + ½ aΔt2
          7,35 =12Δt + ½ (-9,8)Δt2                -7,35 = -12Δt + ½ (9,8) Δt2
          Δt = 1,22 s                            Δt = 1,22 s

                                          OPTION 3/OPSIE 3
          Upwards positive                        Downwards positive
          Opwaarts positief:                      Afwaarts positief:
          v f 2 = v i 2 + 2aΔy                    v f 2 = v i 2 + 2aΔy
          0 = 122 + 2(-9,8)Δy                    0 = (-12)2 + 2(9,8)Δy
          Δy = 7,35 m                             Δy = -7,35 m
                   v + vf                                v + vf 
          Δx =  i            Δt                Δx =  i            Δt 
                   2                                     2 
                    (12 + 0)                                    (- 12 + 0)
          7,35 =              Δt                     - 7,35 =                Δt
                         2                                      2
          Δt = 1,22 s                               Δt = 1,22 s

                                          OPTION 4/OPSIE 4
          (E mech ) A = (E mech ) top
          (½ mv2 + mgh) A = (½ mv2 + mgh) top
          ½m(12)2 + 0 = 0 + m(9,8)(h) 
          ∴h = Δy = 7,35 m
          OR/OF                                                        v + vf 
                                                                Δx =  i         Δt 
          W net = ΔE k                                                 2 
          F net Δycosθ = ½ m(v f 2 – v i 2)                             (12 + 0)
          m(9,8)Δycos 180o = ½ m(02 – (12)2)                   7,35 =           Δt
                                                                           2
          Δy = 7,35 m                                           Δt = 1,22 s
          OR/OF
          ΔE p + ΔE k = 0
          mg(h f – h i ) + ½ m(v f 2 – v i 2) = 0
          m(9,8)(h – 0) + ½(m)(0 - 122) = 0
          ∴h = Δy = 7,35 m
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                                          OPTION 5/OPSIE 5
          Upwards positive                      Downwards 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 – 12)                  (9,8)Δt = (0 – (-12)) 
          Δt = 1,2245 s                        Δt = 1,2245 s 

                                          OPTION 6/OPSIE 6
          Upwards positive                      Downwards positive
          Opwaarts positief:                    Afwaarts positief:
          Δy = v i Δt + ½ aΔt2                 Δy = v i Δt + ½ aΔt2 
          0 = 12Δt + ½ (-9,8)Δt2                0 = -12Δt + ½ (9,8)Δt2
          Δt = 2,4490 s                         Δt = 2,4490 s

          Δt = ½ (2,4490)                          Δt = ½ (2,4490) 
             = 1,2245 s                               = 1,2245 s                                       (3)

3.2                                  OPTION 1/OPSIE 1
          Upwards positive                   Downwards positive
          Opwaarts positief:                 Afwaarts positief:
          v f = v i + aΔt                   v f = v i + aΔt
          -3v = -v + (-9,8)(1,22)          3v = v + (9,8)(1,22) 
                         -1              -1
          v = 5,98 m∙s (5,978 – 6,03 m∙s )  v = 5,98 m∙s-1(5,978 – 6,03 m∙s-1)

                                        OPTION 2/OPSIE 2
          Upwards positive                      Downwards 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)(1,2245) = -3v – (-v)         (9,8)(1,2245) = 3v - v
          v = 6,00 m∙s-1                       v = 6,00 m∙s-1                                          (4)

3.3                                       OPTION 1/OPSIE 1
          Upwards positive                       Downwards positive
          Opwaarts positief:                     Afwaarts positief:
          Δy = v i Δt + ½aΔt2                   Δy = v i Δt + ½aΔt2
             = (-5,98)(2,44) + ½(-9,8)(2,44) 2         = (5,98)(2,44) + ½(9,8)(2,44) 2 
             = - 43,764                                 = 43,764
          ∴h = 43,76 m (43,764 – 44,08 m)          ∴h = 43,76 m(43,764 – 44,08)

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                                         OPTION 2/OPSIE 2
          Upwards positive                      Downwards positive
          Opwaarts positief                     Afwaarts positief:
          v f = v i + aΔt                       v f = v i + aΔt
          v f = - 5,98 + (-9,8)(2,44)           v f = 5,98 + 9,8(2,44)
          v f = - 29,892 m·s-1                      = 29,892 m·s-1
          v f 2 = v i 2 + 2aΔy                  v f 2 = v i 2 + 2aΔy 
          (-29,892)2 = (- 5,98)2 + 2(-9,8)Δy   (29,892)2 = (5,98)2 + 2(9,8)Δy
          Δy = - 43, 763 m                      Δy = 43,76 m
          ∴h = 43,76 m(43,764 – 44,08)                 ∴h = 43,76 m(43,764 – 44,08)

                                             OPTION 3/OPSIE 3
          Upwards positive                            Downwards positive
          Opwaarts positief                           Afwaarts positief:
          v f = v i + aΔt                             v f = v i + aΔt
          v f = - 5,98 + (-9,8)(2,44)                 v f = 5,98 + 9,8(2,44)
                                 -1
          v f = - 29,892 m·s                             = 29,892 m·s-1
                  v + vf                                    v + vf 
          Δx =  i          Δt                      Δx =  i           Δt 
                  2                                         2 
                    − 30 + ( −6,00)                         30 + 6,00            
               =                   (2,4490 )        =              (2,4490)  
                          2                                     2                
          Δx = - 44,082 m                             Δx = 44,082 m
          h = 44,082 m                               h = 44,082 m 

                                          OPTION 4/OPSIE 4
          Upwards positive                        Downwards positive
          Opwaarts positief                       Afwaarts positief:
          For A/ Vir A                            For A/ Vir A
          v f = v i + aΔt                         v f = v i + aΔt
          -12 = 12 + (-9,8)Δt                     12 = -12 + (9,8)Δt
          Δt = 2,45 s                             Δt = 2,45 s
          For B/ Vir B                            For B/ Vir B
                             2
          Δx = v i Δt + ½ aΔt                    Δx = v i Δt + ½ aΔt2 
              = (-5,98)(2,45) + ½ (-9,8)(2,45)2       = (5,98)(2,45) + ½ (9,8)(2,45)2 
               = - 44,06 m                             = 44,06 m
          h = 44,06 m                            h = 44,06 m                                                (3)

                                            OPTION 5/OPSIE 5
          Upwards positive                         Downwards positive
          Opwaarts positief                        Afwaarts positief:
          Δy = v i Δt + ½aΔt2                     Δy = v i Δt + ½aΔt2 
          Δy A = 12Δt + ½ aΔt2                     Δy A = -12Δt + ½ aΔt2
          Δy B = -6Δt + ½ aΔt2                     Δy B = 6Δt + ½ aΔt2
          Δy A – Δy B = 12Δt – (-6Δt)              Δy A – Δy B = 12Δt – (-6Δt)
          0 – Δy B = 18Δt                         0 – Δy B = -18Δt 
                    = 18(2,44)                               = -18(2,44)
                    = 43,92 m                                = 43,92 m
          h = 43,92 m                             h = 43,92 m                                               (3)

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                                        OPTION 6/OPSIE 6
          Upwards positive                     Downwards positive
          Opwaarts positief                    Afwaarts positief:
          W net = ΔE k                        W net = ΔE k 
          mgΔycosθ = ½ m(v f 2 – v i 2)        mgΔycosθ = ½ m(v f 2 – v i 2)
                       o        2        2
          (-9,8)hcos0 = ½ (-20) – ½ (-6)      (9,8)hcos0o = ½ (20)2 – ½ (6)2 
          h = 44,082 m                        h = 44,082 m 

                                                OPTION 7/OPSIE 7
          (E p + E k ) top/bo = (E p + E k ) bottom/onder
          mgh i + ½ mv i 2 = mgh f + ½ mv f 2             
          (9,8)h + ½ (6)2 = (9,8)(0) + ½ (30)2 
          h = 44,082 m                                                                                  (3)

3.4
          UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF

                         velocity (m∙s-1)   A

                         snelheid (m∙s-1)
                                                1,22

                                                              time (s)
                                -5,98
                                                              tyd (s)

                                                B

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          DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF

                          velocity (m∙s-1)

                          snelheid (m∙s-1)
                                               B

                                     5,98

                                                   1,22      time (s)
                                               A             tyd (s)

Criteria for graph/Kriteria vir grafiek                                       Marks/Punte
Time 1,22 s shown correctly/Tyd 1,22 s korrek getoon                               
Initial velocity for stone B at time t = 0 correctly shown with correct signs
                                                                                   
/ Aanvanklike snelheid vir klip B korrek met korrekte tekens getoon
Two sloping parallel lines with A crossing the time axis /Twee skuins
                                                                                   
parallelle lyne met A wat die tyd-as kruis
Straight line graph for A parallel to graph B, extending beyond the time
when B hits the ground/ Reguitlyn grafiek A parallel aan grafiek B                 
verleng verby die tyd wanneer B die grond tref

                                                                                                           (4)
                                                                                                          [14]

QUESTION 4 /VRAAG 4
4.1       The total linear momentum in an isolated/closed system is constant.
          Die totale liniêre momentum in 'n geïsoleerde (geslote) sisteem is konstant

          OR/OF
          In an isolated/closed system, total linear momentum before collision is equal
          to total linear momentum after collision. 
          In 'n geïsoleerde (geslote) sisteem is die totale liniêre momentum voor die
          botsing gelyk aan die totale momentum na die botsing.                                          (2)

4.2       Σp i = Σp f 
          m B v Bi + m b v bi = m B v Bf + m b v bf
          Δp bullet = -Δp block
          (0,015)(400) + 0 = (0,015)v Bf + 2(0,7)
          V Bf = 306,67 (306,666)m∙s-1                                                                   (4)

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4.3       OPTION 1/OPSIE 1
          F net Δt = Δp
          Δp = mv f - mv i 
          For bullet / Vir koeël                          For block / Vir blok
          Δp = (0,015)(306,666 – 400)                    Δp = (2)(0,7 - 0)
               = -1,4 kg∙m∙s-1                                 = 1,4 kg∙m∙s-1
          Fnet(0,002) = -1,4                              Fnet(0,002) = 1,4
          Fnet = -700 N                                   Fnet = 700 N
                                             W net = ΔE k
                                   F net Δxcosθ = ½ m(v f 2 – v i 2)
                          (700)Δxcos180o = ½ (0,015)(306,672 – 4002)
                                           Δx = 0,71 m 

          F net = ma               Δx = v i Δt + ½ aΔt2
          -700 = (0,015)a               = (400)(0,002) + ½(-46 666,67)(0,002)2
          OR/OF                         = 0,71 m (0,70667) m
          F met = ma               OR/OF
          700 = (0,015)a           v f 2 = v i 2 + 2aΔx
                                   (306, 67)2 = (400)2 + 2(-46 666,67)Δx
          a = - 46 666,67          Δx = 0,71 m (0,70667 m) 
          or/of 46 665 m·s-2

          OPTION 2/OPSIE 2
          v f = v i + aΔt
          306,666 = 400 + a (0,002) 
          a = -46 667 m·s-2

          v f 2 = v i 2 + 2aΔx
           (306,666)2 = 4002 + 2(-4667) Δx
          Δx = 0,71 m (0,706 m) 

          OPTION 3/OPSIE 3
                v + vf 
          Δx =  i       Δt 
                2 
                             
                400 + 306,666 
             =                  (0,002) 
                       2       
             = 0,71 m (0,707) m 

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          OPTION 4/OPSIE 4
          W net = ΔK/ΔE k 
          F net Δxcos θ = maΔxcosθ = ΔK/ΔE k

          v f = v i + aΔt
          306,666 = 400 + a (0,002) 
          a = -46 667 m·s-2

          W net = ΔK/ΔE k
          F net Δxcos θ = maΔxcosθ = ΔK/ΔE k
          (0,015)(46 667)Δxcos180o = ½(0,015)(306,6662 – 4002)
          Δx = 0,71 m (0,707) 
          OR/OF
          W nc = ΔE p + ΔE k
          (0,015)(46 667)Δxcos180o = ½(0,015)(306,6662 – 4002)
          Δx = 0,71 m (0,707)                                                                               (5)
                                                                                                            [11]

QUESTION 5/VRAAG 5

5.1      The net/total work done (on an object) is equal to the change in the object's
         kinetic energy.
         Die netto/totale arbeid wat (op 'n voorwerp) verrig is is gelyk aan die
         verandering in die voorwerp se kinetiese energie.

          OR/OF
         The work done on an object by a resultant/net force is equal to the change in
         the object's kinetic energy.
         Die arbeid verrig op in voorwerp deur die resultante/netto krag is gelyk aan
         die verandering in die voorwerp se kinetiese energie.                                              (2)

5.2                 T
                                            Accepted labels/Aanvaarde benoemings
                     ●            F g / F w / weight / mg/ 58,8N / gravitational force / F earth on block
                             w
                                  F g / F w / gewig / mg /58,8 N/ gravitasiekrag / F aarde op blok
                             T    F T / Tension/ spanning
                         w

5.3       W w = w∆xcosθ                              W w = - ∆E P 
              = mg∆xcosθ                                  = - mg(h f – h i )
              = (6)(9,8)(1,6)cos0o                       = - (6)(9,8)(0 – 1,6) 
          ∴W = 94,08 J                                   = 94,08 J                                        (3)

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5.4       OPTION 1/OPSIE 1
          W net = ∆E K /∆K = ½m(v f 2 – v i 2)
          W net = F net ∆xcosθ
          W net = W f + W g + W N
                = µ k NΔxcosθ + W g + W N

          W net = (0,4)(4)(9,8)(1,6)cos180o  + 94,08 + 0
                 = 68,992 J
          W net = ½m(v f 2 – v i 2)
          68,992= ½(4)(v f 2 – 0) + ½(6)(v f 2 – 0)
          v f = 3,71 m∙s-1

          OPTION 2/OPSIE 2
          W nc = ΔE p + ΔE k 
          fΔxcosθ = (m 1 gh f – m 1 gh i ) + (½m 1 v f 2 - ½m 1 v i 2) + (½m 2 v f 2 - ½m 2 v i 2)
          (0,4)(4)(9,8) (1,6) cos 180o = [0 - (6)(9,8)(1,6)]+ (½(6)v f 2 + ½(4)v f 2 - 0)
          68,992 = 5v f 2
          v f = 3,71 m∙s-1
          OPTION 3/OPSIE 3                                                 F net = ma
                                                                                   = (6)(4,312)
          f = µ N = (0,4)(4)(9,8) = 15,68 N                                        = 25,872
          k     k
          T – f k = ma
                                                                         W net = F net Δxcosθ
          w – T = ma
                                                                             = (25,872)(1,6)cos0o
          T – 15,68 = 4a …     .(i)                                          = 41,3952 J
          (6)(9,8) – T = 6a ……..(ii)
          ∴a = 4,312 m·s-2                                               W net = ΔE k 
                                                                         41,3952 = ½ m(v f 2 – v i 2
          ∴T = 32,928 N                                                  41,3952 = ½ (6)(vf 2 – 0)
                                                                         vf = 3,7146 m∙s-1

                                                                         Above calculations can be
                                                                         done with 4 kg or 10 kg /
                                                                         Bostaande berekeninge
                                                                         kan met 4 kg of 10 kg
                                                                         gedoen word
          4 kg block                                       6 kg block
          W net = ∆E K /∆K                                W net = ∆E K /∆ K
          W f + W T = ½m(v f 2 – v i 2)                    W w + W T = ½m(v f 2 – v i 2)

          fΔxcos180o + TΔxcos0o = ½(4)(v f 2 – 0)          mgΔxcos0o + TΔxcos180o = ½(6)(v f 2 – 0)
          (15,68)(1,6)(-1) + (32,928)(1,6)(1) = 2v f 2   (6)(9,8)(1,6)(1)+ (32,928)(1,6)(-1)= 3v f 2
          v f = 3,72 m·s-1                                v f = 3,72 m·s-1

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          OPTION 4/OPSIE 4
          W net = ∆E K /∆K 
          For the 4 kg mass / Vir die 4 kg massa
          T(1,6)cos0o + [(0,4)(9,8)(4)](1,6)cos180o = ½(4)v2 – 0)
          For the 6 kg mass/Vir die 6 kg massa
          (6)(9,8)(1,6)]cos0o+ T(1,6)cos180o = ½(6)(v2 – 0)
          Adding the two equations / Optel van twee vergelykings
          68,992 = ½ (4)v2 + ½ (6)v2 
          5v2 = 68,992
          v = 3,71 m·s-1

          OPTION 5/OPSIE 5
          W net = ΔE k 
          F net Δxcosθ = ½ m(vf 2 –v i 2)
          (F g – f)Δxcosθ = ½m(v f 2 –vi 2)
          [(6)(9,8) – (0,4)(4)(9,8)](1,6)cos0o = ½(10)(vf 2 - 0)
          v f = 3,71m.s-1                                                                                (5)
                                                                                                         [12]
QUESTION 6 / VRAAG 6
6.1       It is the (apparent) 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. 
          Dit is die verandering in frekwensie (of toonhoogte) van die klank
          (waargeneem deur 'n luisteraar) omdat die klankbron en die luisteraar
          verskillende snelhede relatief tot die medium van klankvoortplanting het.
          OR/OF
          An (apparent) change in (observed/detected) frequency (pitch),
          (wavelength) as a result of the relative motion between a source and an
          observer  (listener).
          'n       Skynbare       verandering      in      (waargenome)        frekwensie
          (toonhoogte),(golflengte) as gevolg van die relatiewe beweging tussen die
          bron en 'n waarnemer / luisteraar.                                                             (2)
6.2.1     170 Hz                                                                                        (1)
6.2.2     130 Hz                                                                                        (1)
6.3       POSITIVE MARKING FROM QUESTIONS 6.2.1 and 6.2.2/
          POSITIEWE NASIEN VANAF VRAAG 6.2.1 en 6.2.2
               v ± vL
          fL =        fs 
               v ± vs
            (340 + 0) 
          170 =              ×f   1
                 (340 - v s ) s

                          
             (340 - 0)
          130 =            × fs             2
                340 + v s
          v s = 45,33 m∙s-1  (45,33 – 45,45 m∙s-1)                                                       (6)
                                                                                                         [10]
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QUESTION 7 / VRAAG 7

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

7.2       OPTION 1/ OPSIE 1
                 QQ
          F = k 12 2 
                   r
             (9 × 10 9 )(6 × 10 -6 )(8 × 10 −6 )
          =
                          (0,2)2        
          = 10,8 N

          OPTION 2/ OPSIE 2
                                         kQ     (9 × 10 9 )(8 × 10 −6 )
                                     E=     2
                                              =              2
                                                                        = 1,8 x 104 N∙C-1
                                          r            (0,2)   
                                     F = Eq = (1,8 x 104)(6 x 10-6) = 10,8 N                                (4)
                             FXY
7.3

                    Fnet 

       FZY

                                                                                                              (3)

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7.4       OPTION 1 /OPSIE 1
           2      2     2
          Fnet = FXY + FZY
          15,20 2 = 10,8 2 + FZY
                              2

          F ZY = 10,696 N
                  Q Q
          F ZY = k Z 2 Y
                    r
                                   8 × 10 -6 × Q Z
          10,696 = 9 × 10 9 ×                     
                                      (0,30)2
          Q Z = 1,34 x 10-5 C

          OPTION 2 /OPSIE 2
                  10,8
          cosθ =                                                               FXY = 10,8 N
                  15,2                                                            θ
          θ = 44,72o

                                                                         Fnet = 15,20 N
                      F                     F
          sin 44,72 = ZY  OR/OF tan 44,72 = ZY
                     15,2                   FXY
          F ZY = 10,696 N

                     QZQ Y
          F ZY = k
                      r2
                               8 × 10 -6 × Q Z
                               9
          10,696 = 9 × 10 ×                   
                                  (0,30)2
          Q Z = 1,34 x 10-5 C                                                                             (4)
                                                                                                          [13]

QUESTION 8 / VRAAG 8

8.1       Electric field at a point is the force per unit positive charge placed at that
          point. 
          Elektriese veld by 'n punt is die krag per eenheids positiewe lading geplaas
          by daardie punt.                                                                                (2)

8.2             kQ
          E=        
                r2
          E net = (E A + E B )
                          (
                        9 1,5 × 10
                                      )
                                    -6 
                                                   (
                                                 9 2,0 × 10
                                                            -6
                                                               )
                = 9 ×10                 + 9 × 10
                              (0,4)2                 (0,3)2
                = 2,84 x 105 N∙C-1                                                                       (4)

8.3       OPTION 1/OPSIE 1
          F E = qE 
              = (3,0 x 10-9)( 2,84 x 105) 
              = 8,52 x 10-4 N 

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          OPTION 2/OPSIE 2
              kQ Q
          F= 12 2 
                 r
          F net = (F A + F B )
                    (9 × 10 9 )(3 × 10 −6 )(1,5 × 10 −6 ) (9 × 10 9 )(3 × 10 −6 )(2,0 × 10 −6 ) 
                =                                       +                                       
                                  (0,4)2                                (0,3)2                   
                = 8,53 x 10-4 N                                                                                    (3)
                                                                                                                    [9]

QUESTION 9 / VRAAG 9

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

9.1.2     (Equivalent) resistance/ (Ekwivalente) weerstand                                                         (1)

9.1.3                             ΔV
          Gradient/Helling =
                                  ΔI
                                   2-0
                                =          = 4 (Ω)
                                  0,5 - 0                                                                           (2)

9.1.4     OPTION 1/OPSIE 1
          In series R 1 + R 2 = 4 Ω .............(1)
                       R1R2
          In parallel          = 1 Ω ..........(2)
                      R1 + R2
          R1R2 = 4 Ω
          ∴R 1 = R 2 = 2 Ω 

          OPTION 2/OPSIE 2
          For graph X/Vir grafiek X:
          R 1 + R 2 = 4………………(1) 
          For graph Y/Vir grafiek Y
           1     1     1
              =     +
          R // R1 R 2
           1      1   1  
           +          =   ...........(2)
           1R    R  2      1
          R 12 - 4R 1 + 4 = 0
          R 1 = 2 Ω                                                                                                (4)

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9.2.1          V
          I=
               R
                   5
           =
               (RM + RN )

                 5
           =        
               ( 6)
           = 0,83 A                                                                                     (3)

9.2.2     OPTION 1/OPSIE 1                        OPTION 2/OPSIE 2
          Ɛ = I(R + r)                           Ɛ = (V s + V // + V r ) / V ext/eks + V int
                                                   = [5 + (0,833 x 1,5) + (0,9 x 0,833)] 
            = 0,83[(6 +1,5) + 0,9]
                                                    = 6,999 V
            = 6,997 V                               = 7(,00) V                (6,972 – 7,00 V)
            = 7(,00) V       (6,972 – 7,00 V)

9.2.3     The resistance R N will be 3 Ω 
          The voltage divides (proportionately) in a series circuit. Since the voltage
          across M is half the total voltage, it means the resistances of M and N are
          equal.

          Die weerstand R N sal 3 Ω wees.
          Die potensiaalverskil verdeel (eweredig) in 'n serie stroombaan. Aangesien
          die potensiaalverskil oor M die helfte is van die totale potensiaalverskil,
          beteken dit dat die weerstande van M en N gelyk is.                                             (2)
                                                                                                         [18]

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QUESTION 10 / VRAAG10
10.1
10.1.1            Mechanical to electrical / Meganies na elektries                                                      (1)

10.1.2

    Induced emf/           B           D                       Induced emf/                   B               D

    Geinduseerde emk                                           Geinduseerde emk

                       A        C             A        t (s)                        A                C                A t (s)

OR/OF

    Induced emf/                                                Induced emf/
                       A        C             A                                         A                C               A
                                                       t (s)                                                              t (s)

    Geinduseerde emk       B           D                        Geinduseerde emk               B                  D

    Criteria for graph/Kriteria vir grafiek                                                        Marks/Punte
    Correct DC shape, starting from zero/Korrekte GS vorm wat by nul begin                               
    Positions ABCDA correctly indicated on the graph/Posisies ABCDA of
                                                                                                         
    grafiek aangedui                                                                                                     (2)

10.2.1            20,5 Ω                                                                                                (1)

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10.2.2    OPTION 1/OPSIE 1

                    Vrms  25
          I rms =         =      
                     R      20,5
               = 1,22 (1,2195) A

          P ave = I rms 2 R                 P ave = I rms 2 R        W = I2 rms R∆t
               =(1,22)2(0,5)                     = (1,22)2(20)         = (1,22)2 (0,5)(1)
               = 0,74 W                          = 29,77 W             = 0,74 J

                 V 2                        OR/OF                             Vrms 2
          Pave = rms                                                  Pave =        
                   R                                                            R
                                            V rms/wgk device/toestel =
                 (25)2                       20 
                                                                              (25)2 
          Pave =                                  × 25 = 24,39 V      Pave =
                 20,5                       20,5                              20,5
          P ave = 30,49 W                                             P ave = 30,49 W
                                            P ave = V rms I rms 
          Actual energy delivered per                                 Actual energy delivered per
                                                  = (24,39)(1,22)     second(power) / Energie
          second(power) / Energie aan
          toestel gelewer per sekonde             = 29,76 W          aan toestel gelewer per
          (drywing)                                                   sekonde (drywing)
                = (30,49– 0,74)                                             = (30,49 – 0,74)
                = 29,75 W                                                  = 29,75 W 

          OPTION 2/OPSIE 2
                                20 
          V rms/wgk device/toestel =× 25 = 24,39 V
                               20,5
                 Vrms 2 (24,39)2
          Pave =        =         = 29,74 W
                  R         20                                                                           (5)
                                                                                                          [9]

QUESTION 11 / VRAAG 11

11.1.1    (Line) emission (spectrum) / (Lyn) emissiespektrum                                             (1)

11.1.2    (Line) absorption (spectrum ) / (Lyn) absorpsiespektrum                                        (1)

11.2.1    Emission / Emissie                                                                             (1)

11.2.2    Energy released in the transition from E 4 to E 2 = E 4 – E 2
          Energie vrygestel in die oorgang vanaf E 4 na E 2 = E 4 – E 2
          E 4 – E 2 = (2,044 x 10-18 – 1,635 x 10-18) = 4,09 x 10-19J

          E = hf
          4,09 x 10-19 = (6,63 x 10-34)f
          f = 6,17 x 1014 Hz                                                                             (4)

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11.2.3    E = W 0 + E k(max)
          hf = hf 0 + E k(max)                 Any one/Enige een
                            2
          hf = hf 0 + ½ m v max
                          2
          E = W 0 + ½ m v max

          4,09 x 10-19  = (6,63 x 10-34)(4,4 x 1014) + E k(max)
          E k(max) = 1,17 x 10-19 J
          OR/OF
          E k(max) = E light/lig – W o
                   = hf light/lig – hf o    Any one/Enige een

                    = (6,63 x 10-34)( 6,17 x1014) - (6,63 x 10-34)(4,4 x 1014)
                    = 1,17 x 10-19 J                                                                     (4)

11.2.4
          No / Nee
          The threshold frequency is greater than the frequency of the photon. 
          Die drumpelfrekwensie is groter as die frekwensie van die foton
          OR/OF
          The frequency of the photon is less than the threshold frequency
          Die frekwensie van die foton is minder as die drumpelfrekwensie
          OR/OF
          Energy of the photon is less than the work function of the metal
          Energie van foton is minder as die van die arbeidsfunksie van die metaal                         (2)
                                                                                                          [13]

                                                                               TOTAL/TOTAAL:              150

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