# Physical Sciences Grade 12 Free StateNovember 2016 Memo

> Grade 12 Physical Sciences memorandum from the Free StateNovember 2016, 14 pages. Read online or download the free PDF.

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- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2016
- Exam period: Free StateNovember
- Curriculum: CAPS
- Pages: 14

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Physical Sciences · Grade 12 · Free StateNovember · 2016. Memorandum, 14 pages. Read online or download the PDF.

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### Page 1

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                PREPARATORY EXAMINATION
                VOORBEREIDENDE EKSAMEN

                         GRADE 12/GRAAD 12

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

                             SEPTEMBER 2016

                               MEMORANDUM

                           MARKS/PUNTE: 150

                      This memorandum consists of 14 pages.
                    Hierdie memorandum bestaan uit 14 bladsye.

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Physical Sciences P1/Fisiese Wetenskappe V1  2                      FS/VS/September 2016
Grade/Graad 12 Prep. Exam./Voorb. Eksam Memorandum

QUESTION 1/VRAAG 1

1.1     C                                                                              (2)

1.2     C                                                                              (2)

1.3     C                                                                              (2)

1.4     C                                                                              (2)

1.5     C                                                                              (2)

1.6     B                                                                              (2)

1.7     D                                                                              (2)

1.8     B                                                                              (2)

1.9     A                                                                              (2)

1.10    B                                                                              (2)
                                                                                        [20]

QUESTION 2/VRAAG 2

2.1     It is the force that opposes the motion of an object and which acts
        parallel to the surface
        Dit is die krag wat die beweging van 'n voorwerp teenstaan en wat
        parallel met die oppervlak inwerk.                                               (2)

2.2.1              Fv
        tan 20o =
                   FH
              Fv = (tan 20o)(38)
                 = 13,83 N                                                              (2)

2.2.2   POSITIVE MARKING FROM 2.2.1
        POSITIEWE NASIEN VANAF 2.2.1

        FN = Fg - Fv
           = (5)(9,8) - 13,83
           = 35,17 N                                                                    (3)

2.3     When a resultant force acts on an object the object will accelerate in
        the direction of the force at acceleration directly proportional to the
        force and inversely proportional to the mass of the object.
        Indien 'n resulterende krag op 'n voorwerp inwerk sal die voorwerp
        in die rigting van die krag versnel met 'n versnelling direk eweredig
        aan die grootte van die krag en omgekeerd eweredig aan die massa
        van die voorwerp.                                                                (2)

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Physical Sciences P1/Fisiese Wetenskappe V1  3                              FS/VS/September 2016
Grade/Graad 12 Prep. Exam./Voorb. Eksam Memorandum

2.4
                T/FT /Tension/Spanning

               ●
                   w/Fg/Fw/weight/gravitational force/29,4N/gewig/gravitasiekrag
                                                                                                 (2)

2.5     POSITIVE MARKING FROM 2.2.2
        POSITIEWE NASIEN VANAF 2.2.2

        Option 1/Opsie 1
        Left/upwards as positive:/ Links/opwaarts as positief

        5 kg block:             Fnet = ma
                         -T + FH - f = ma
               -T + 38 - (0,2)(35,17)  = 5a              1

        3 kg block :             -Fg + T     = ma
                            -(3)(9,8) + T    = 3a         2

        Substitute 2 into 1:
                                          a = 0,196 m·s-2
        Substitute a into 2:
                               -29,4 + T = (3)(0,196) 
                                       T = 29,99 N

        Option 2/Opsie 2
        Right/downwards as positive:/ Regs/afwaarts as positief

        5 kg block:             Fnet = ma
                        T - FH + f = ma
               T - 38 + (0,2)(35,17)  = -5a              1

        3 kg block :             Fg - T = ma
                             (3)(9,8) - T = -3a           2

        Substitute 2 into 1:
                                          a = 0,196 m·s-2
        Substitute a into 2:
                                 29,4 - T = -(3)(0,196) 
                                        T = 29,99 N
                                                                                                 (6)
                                                                                                [17]

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Physical Sciences P1/Fisiese Wetenskappe V1  4                     FS/VS/September 2016
Grade/Graad 12 Prep. Exam./Voorb. Eksam Memorandum

QUESTION 3/VRAAG 3

3.1       The motion of an object if the only force acting on it is the force of
          gravity.
          Die beweging van 'n voorwerp indien die enigste krag wat daarop
          inwerk gravitasiekrag is.                                                     (1)

3.2       9,8 m·s-2 Downwards/ 9,8 m·s-2 Afwaarts
                                                                                        (1)
3.3       Upwards positive/Opwaarts         Downwards positive/Afwaarts
          positief:                         positief:
          v f = v i + a Δt                 v f = vi + a ∆t 
          -20 = 20 + (-9,8)(∆t)           20 = -20 + (9,8)(∆t) 
           ∆t = 4,08 s                      ∆t = 4,08 s
                                                                                        (4)

3.4     Upwards positive/Opwaarts           Downwards positive/Afwaarts
        positief:                           positief:
        Height of cliff:                    Height of cliff:
        Δy = v i Δt + 21 a∆t 2             Δy = v i Δt + 21 a∆t 2 
            = 20(6) + ½(-9,8)(6)2              = -20(6) + ½(9,8)(6)2
            = - 56,4 m                          = 56,4 m
        Height /Hoogte= 56,4 m              Height /Hoogte= 56,4 m
        Ball B:                             Ball B:
        Δy = v i Δt + 21 a∆t 2              Δy = v i Δt + 21 a∆t 2
            = 20(2,8) + ½(-9,8)(2,8)2          = -20(2,8) + ½(9,8)(2,8)2
            = 17,584 m                          = -17,584 m
        Distance = 56,4 + 17,584           Distance = 56,4 + 17,584
                    = 73,98 m                          = 73,98 m                      (5)

        OPTION 1/OPSIE 1
        Upwards positive/Opwaarts positief:

3.5

                     B

        ∆y (m)   0
                                    6s
                                                  Time/tyd(s)

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Physical Sciences P1/Fisiese Wetenskappe V1  5                    FS/VS/September 2016
Grade/Graad 12 Prep. Exam./Voorb. Eksam Memorandum

                                                                             Marks/
   Criteria for graph/Kriteria vir grafiek:
                                                                             Punte
   Shape till the ball reaches the ground.
                                                                                 
   Vorm totdat die bal grond bereik.
   Zero position at 6 s.
                                                                                 
   Zero posisie by 6s.
   Shape for bounce up till maximum height.
                                                                                 
   Vorm vir bons tot by maksimum hoogte.
   Ground not zero position (provided everything else is correct): 2/3
                                                                                (3)
   Grond nie zero posisie nie (op voorwaarde die res is korrek) : 2/3

        OPTION 2/OPSIE 2
        Upwards negative/ Opwaarts negatief:

                                        6s

        ∆y (m)
                 0
                                                  Time/tyd(s)

                                                                             Marks/
   Criteria for graph/Kriteria vir grafiek:
                                                                             Punte
   Shape till the ball reaches the ground.
                                                                                 
   Vorm totdat die bal grond bereik.
   Zero position at 6 s.
                                                                                 
   Zero posisie by 6s.
   Shape for bounce up till maximum height.
                                                                                 
   Vorm vir bons tot by maksimum hoogte.
   Ground not zero position (provided everything else is correct): 2/3
   Grond nie zero posisie nie (op voorwaarde die res is korrek) : 2/3           (3)

                                                                                      [14]

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Physical Sciences P1/Fisiese Wetenskappe V1  6                     FS/VS/September 2016
Grade/Graad 12 Prep. Exam./Voorb. Eksam Memorandum

QUESTION 4/VRAAG 4

4.1     Impulse is the product of the net force acting on an object and the
        time the force is acting on the object. 
        Impuls is die produk van die netto krag wat op 'n voorwerp inwerk
        en die tyd wat die krag op die voorwerp inwerk.
                                     OR
        Impulse is the change in momentum of an object.
        Impuls is die verandering in die momentum van 'n voorwerp.                      (2)

4.2          Σ pbefore = Σ pafter 
          (0,04)(800) = (8,04)vf 
                   vf = 3,98 m·s-1
                  ∆p = m(vf - vi)
                       = 8 ( 3,98 – 0) 
                       = 31,84 kg·m·s-1, to the right/east/na regs/oos
                                                                                        (5)

4.3      POSITIVE MARKING FROM 4.2
         POSITIEWE NASIEN VANAF 4.2

         OPTION 1/OPSIE 1
                 W net = ∆K    
              f∆xcosӨ = ½mvf2 - ½mvi2
          f(4)cos1800 = ½(8,04)0 - ½(8,04)(3,98)2
                      f = 15,92 N

         OPTION 2/OPSIE 2
                   W nc = ΔU + ΔK 
          f(4)cos1800 = 0 + ½(8,04)0 - ½(8,04)(3,98)2
                     f = 15,92 N                                                        (4)
                                                                                        [11]

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Physical Sciences P1/Fisiese Wetenskappe V1  7                     FS/VS/September 2016
Grade/Graad 12 Prep. Exam./Voorb. Eksam Memorandum

QUESTION 5/VRAAG 5

5.1      The force which a surface excerts on an object with which it is in
         contact and which is perpendicular to the surface.
         Die krag wat 'n oppervlak uitoefen op 'n voorwerp waarmee dit in
         kontak is en wat loodreg op die voorwerp is.                                   (2)

5.2      Fnet = ma            
         Fapplied – Fg// = ma
         600 – m(9,8)(sin 250)  = 0
                                m = 144,87 kg                                          (5)

5.3      Total mechanical energy in a closed system remains constant. 
         Totale meganiese energie in 'n geslote sisteem bly konstant.                   (2)

5.4.1   POSITIVE MARKING FROM 5.2/POSITIEWE NASIEN VANAF
        5.2

        OPTION 1/ OPSIE 1
         (Ep + Ek)X = (Ep + Ek)B
         (144,87)(9,8)h  + 0 = 0 + ½ (144,87)(6)2
                             h = 1,84 m
          OPTION 2/ OPSIE 2
          Wnc = ΔU + ΔK / Wnc = ΔEp + ΔEk
          0   =    0 - (144,87)(9,8)(5)h  + ½ (144,87)(6)2 - 0                        (4)
          h   =    1,84 m

5.4.2   POSITIVE MARKING FROM 5.2/POSITIEWE NASIEN VANAF 5.2

        OPTION 1/ OPSIE 1
                     Wnet = ∆K 
                       W f = ∆K
         μk(144,87)(9,8)(5)cos1800= ½(144,87)22 -½(144,87)(62) 
                         μk = 0,33 

          OPTION 2/ OPSIE 2
          Wnc = ΔU + ΔK 
          Wf = ΔU + ΔK
          μk(144,87)(9,8)(5)cos1800= 0 + ½(144,87)22 -½(144,87)(62) 
                                   μk= 0,33                                            (5)
                                                                                       [18]

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Physical Sciences P1/Fisiese Wetenskappe V1  8                        FS/VS/September 2016
Grade/Graad 12 Prep. Exam./Voorb. Eksam Memorandum

QUESTION 6/VRAAG 6

6.1.1   The (apparent) change in the frequency/pitch of the sound
        detected by a listener because the sound source and listener
        have different velocities relative to the medium of sound
        propagation.
        Dit is die (skynbare) verandering in die frekwensie/toonhoogte van
        die klank waargeneem deur deur die luisteraar omdat die klankbron
        en die luisteraar verskillende snelhede relatief tot die medium
        waarin die klank voortgeplant word, het.                                          (2)

6.1.2   B. There is an increase in the detected/observed frequencyby
        the listener.
        B. Daar is 'n toename in die waargenome frekwensie deur die
        luisteraar.                                                                       (2)

6.1.3         v ± vL            v − vL                 v ± vL
         fL =        fs OR fL =        fs  OR/OF fL =        fs OR
              v ± vs              v                    v ± vs
              v + vL                                            
         fL =        fs
                 v
                340 − v L                              340 + v L
          620 =           650                     680 =          650
                  340                                     340         
          
           vL = 15,69 m.s-1                                       -1
                                                    vL = 15,69 m.s 
                                                                                          (5)

6.1.4   The listener is moving towards the source with a constant
        accelaration/constant increase in velocity. 
        Die luisteraar nader die bron met 'n konstante versnelling/konstante
        toename in snelheid.                                                              (2)
6.1.5   Measurement of foetal heart beat.OR
        Measurement and monitoring of blood flow.
        Meting van hartklop van fetus. OF
        Meting en waarneming van bloedvloei.                                              (1)
                                                                                         [12]

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Grade/Graad 12 Prep. Exam./Voorb. Eksam Memorandum

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 magnitude of the charges and inversely
        proportional to the square of the distance between them. 

        Die grootte van die elektrostatiese krag uitgeoefen deur een
        puntlading op 'n ander puntlading is direk eweredig aan die produk
        van die grootte van die ladings en omgekeerd eweredig aan die
        kwadraat van die afstand tussen die ladings.                                          (2)

                                        QR + Q S - 6 + 2          
7.2     Net Charge/netto lading =               =        = -2 μC or (-2 x 10 -6 C)
                                           2        2
                                                                                              (1)

7.3      POSITIVE MARKING FROM 7.2/POSITIEWE NASIEN VANAF 7.2

               kQRQS
         FSR =          
                  r2
               9 × 109 × 2 × 10 −6 × 2 × 10 −6
             =                                 
                            0,22
             = 0,9 N
               kQ Q
         FTR = R2 T
                 r
               9 × 10 9 × 2 × 10 −6 × 4 × 10 −6 
             =
                            0,3 2
             = 0,8 N

         Fnet2 = FSR2 + FTR2
               = (0,9)2 + (0,8)2
         Fnet = 1,20 N
                                                                            Fnet     Ө FTR
                  0,9
         Tan Ө =      
                  0,8                                              FSR
             Ө = 48,37o
         ∴ Fnet = 1,20 N, 228,37o/48,37o West of South/ 41,63o downwards
                  from negative x-axis/48,37o Wes van Suid/41,63o
                  afwaarts van negatiewe x-as
                                                                                              (7)
                                                                                             [10]

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Grade/Graad 12 Prep. Exam./Voorb. Eksam Memorandum

QUESTION 8/VRAAG 8

8.1.1   To change/vary the external resistancein the circuit in order to
        change/vary the (total) current through the battery.

        Om die eksterne weerstand in die stroombaan te verander/varieer
        om sodoende die (totale) stroom deur die battery te
        verander/varieer.                                                                  (2)

8.1.2   Current/ stroom                                                                   (1)

8.1.3   6V                                                                                (1)

8.1.4   Vi = 6 – 2 = 4V                                                                  (2)

8.2.1   It is an indication of the maximum amount of energygiven per
        colomb of charge passing through the battery. 

        Dit is 'n aanduiding van die maksimum hoeveelheid energie wat aan
        elke coulomb lading gegee word wat deur die battery beweeg.                        (2)

8.2.2      P = VI
          12 = V(2)
         V12w = 6 V

          V6Ω = IR
              = (2)(6)
              = 12V
           Vp = 12 + 6 = 18 V

                  V2
               P=
                   R
                    62
              12 =     
                    R                                 1           1           1
                                                              =        +
            R12W = 3 Ω                                R           R        R
                                                          p        1           2
        RATIO: R9Ω : R6Ω                                          1       1
                                                                               
                                 I6Ω = V/R                    =       +
                  3 : 2                                           9       6
                 I9Ω : I6Ω           = 18/6      OR
                                     = 3A                RP = 3,6
                  2 : 3        
                                                                RP
               I6Ω = 3 A                                  I9Ω =     I
                                                                 R
                                                                 3,6 
                  It   = 3 + 2 = 5 A                         2=      I
                                                                  9
                                                              I =5A
                 Vs = IR
                       = (5)(2) = 10 V
                 Vekst = Vp + Vs
                       = 18 + 10
                       = 28 V                                                           (8)

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8.2.3   POSITIVE MARKING FROM 8.2.2
        POSITIEWE NASIEN VANAF 8.2.2

        OPTION 1/ OPSIE 1                         OPTION 2/ OPSIE 2

                  Vi                              emf/emk ( ε ) = I(R + r) 
              ri =     
                  I                                   30 = 5(5,6 + r) 
                  30 - 28                               r = 0,4 Ω
                =         
                     5
              I = 0,4 A                                                                (3)

8.2.4   Increase
        Total resistance in circuit increase and the total current decrease. 
        Vinternal will decrease. 
        Therefore: Vexternal will increase because emf stays constant. 

        Toeneem
        Totale weerstand in stroombaan neem toe en die totale stroom neem
        af.
        Vintern sal afneem.
        Dus: Vekstern sal toeneem omdat die emk konstant bly.
                                                                                        (4)
                                                                                       [23]

QUESTION 9/VRAAG 9

9.1.1   Anticlockwise/anti-kloksgewys                                                  (1)

9.1.2   The rate of change in the magnetic flux is a maximum at
        this position.
        Die tempo van verandering in die magnetiese vloed is 'n maksimum
        by hierdie posisie .                                                            (2)

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Grade/Graad 12 Prep. Exam./Voorb. Eksam Memorandum

9.1.3     OPTION 1/OPSIE 1

            V(V)                               t(s)

          OPTION 2/OPSIE 2

             V(V)                                  t(s)

                                                                                         (2)

                                                                           Marks/
        Criteria for graph/Kriteria vir grafiek:
                                                                           Punte
        Shape for one complete rotation of coil.
                                                                               
        Vorm vir een volledige rotasie van spoel.
        Starting at maximum/minimum emf value.
                                                                               
        Begin by maksimum/minimum waarde van emk.
                                                                               (2)

9.1.4     Increase the rotation speed of the coil. 
          Verhoog die rotasiespoed van die spoel.                                       (1)

9.1.5     Electrical energy can be transmitted over long distances (with the use of
          transformers).  OR
          The voltage can easily be adapted for different needs (by means of
          transformers.)

          Elektriese energie kan oor lang afstande oorgedra word (m.b.v.
          transformators) OF
          Die spanning kan maklik aangepas word vir verskillende behoeftes
          (m.b.v. transformators.)                                                      (1)

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Grade/Graad 12 Prep. Exam./Voorb. Eksam Memorandum

9.2                      I
                 I rms = max 
                            2
                           7
                       =      
                            2
                Irms = 4,95 A

              Pave = I2rmsR 
                80 = 4,952(R) 
                R = 3,26 A 
                                                                                              (5)
                                                                                             [12]

QUESTION 10/VRAAG 10

10.1    The process wereby electrons are ejected from the surface of a metal
        when irradiated by light of a suitable frequency.
        Die proses waardeur elektrone uit 'n metaaloppervlak vrygestel word
        wanneer lig van geskikte frekwensie invallend op die oppervlak is.                    (2)

10.2    OPTION 1/OPSIE 1

             E = Wo + Ek (max)    
             c
         h     = hfo + ½mv2
             λ
              6,63 x 10 -34 × 3 × 108
                                      =   6,63 x 10-34(6,45 x 1014)  +
                        λ                 ½(9,11 x 10-31)(2,78 x 105)2

                               λ = 4,30 x 10-7 m·s-1
        Ultaviolet light is used/ Ultaviolet lig is gebruik
                                                                                              (5)

10.3.1 Increase/Verhoog                                                                       (1)

10.3.2 Stays the same/Bly dieselfde                                                           (1)

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Grade/Graad 12 Prep. Exam./Voorb. Eksam Memorandum

10.3.3 The intensity of the light only has an influence on the number of
       photo electrons emitted per time unit. 
       The intensity of the light has no influence on the kinetic energy of the
       photo electrons. 
       OR
       Only the frequency of the light has an influence on the energy of the
       photon and the kinetic energy of the photoelectron.
       The frequency of the photons stay the same.

         Die intensitiet van lig het slegs 'n invloed op die aantal foto-elektrone
         wat per tydeenheid vrygestel word.
         Die intensitiet van lig het geen invloed op die kinetiese energie van
         die foto-elektrone nie.
         OF
         Slegs die frekwensie van die lig het 'n invloed op die energie van die
         foton en die kinetiese energie van die fotoelektron.
         Die frekwensie van die fotone bly dieselfde.                                     (2)

10.4    Electrons from atoms in the excited state fall back to the ground
        state/lower energy state.Energy is radiated as light.
        Elektrone van atome in die opgewekte toestand val terug na die
        grondtoestand/laer energietoestand. Energie word uitgestraal as lig.              (2)
                                                                                         [13]

                                             GRAND TOTAL/GROOTTOTAAL:                    150

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