# Physical Sciences Grade 12 Paper 1 KZN Prelim 2024 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the KZN Prelim Exam 2024, 12 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-sciences-p1-grade-12-september-2024-mg-final/71489f8b-b54b-4313-9f55-78d497c13e5e
- Updated: 2026-10-29
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2024
- Exam period: KZN Prelim Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 12

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### Page 1

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                                      FINAL

                                   NATIONAL
                              SENIOR CERTIFICATE

                                       GRADE 12

                      PHYSICAL SCIENCES P1 (PHYSICS)

                         PREPARATORY EXAMINATION

                                 SEPTEMBER 2024

                            MARKING GUIDELINES

 MARKS: 150

                       These Marking Guidelines consist of 12 pages.

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Physical Sciences P1                          2                September 2024 Preparatory Examination
                                       NSC – Marking Guideline

 QUESTION 1: MULTIPLE-CHOICE QUESTIONS

 1.1       B

 1.2       B

 1.3       D

 1.4       D

 1.5       A

 1.6       C

 1.7       B

 1.8       C

 1.9       A

 1.10      C
                                                                                                  [20]

 QUESTION 2

 2.1

 2.1.1   When a resultant/net force acts on an object, the object will accelerate in the
         direction of the force. The acceleration is directly proportional to the force  and
         inversely proportional to the mass of the object.                                        (2)

 2.1.2

           Accepted labels
            W     Fg/Fw/Weight/Gravitational force
             fk   f/friction/Kinetic friction
             N    FN/Normal force
             F    FApplied force / 40N
             T    FT/ Tension
                                                                                                   (5)
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Physical Sciences P1                           3                September 2024 Preparatory Examination
                                        NSC – Marking Guideline

 2.1.3   MARKING CRITERIA
            Mark for formula Fnet = ma. 

                Mark for substitution of acceleration of 0,5 in Fnet = ma. 

                Mark for calculation of the kinetic friction. 

                Left hand side substitution. , one for each block.

                Mark for the answer.
         UP THE INCLINE AS POSITIVE:

         For 3 kg block:
         Fnet = ma
         T + FgII + (−fk ) + (−F) = ma

         [T + (3)(9.8) sin 3 00 − (0,2)[(3)(9.8) cos 3 00 ] − 40 ]= 3(0.5) 

         T = 31,89N

         For block X:

         Fnet = ma

         F + (−T) = ma

         X((9.8) − 31.89)= X(0.5)

         X = 3.43kg                                                                                (6)

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Physical Sciences P1                               4                September 2024 Preparatory Examination
                                            NSC – Marking Guideline

 2.2

 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.                                                 (2)

 2.2.2
              Gm1 m2
         F=            
                r2

                           (𝟔.𝟔𝟕×𝟏𝟎−𝟏𝟏 )(𝟓.𝟗𝟖×𝟏𝟎𝟐𝟒 )(𝟕.𝟑𝟓×𝟏𝟎𝟐𝟐 )
         1.91 × 1020                                         
                               (𝟔.𝟑𝟖×𝟏𝟎𝟔 +𝟏.𝟐×𝟏𝟎𝟖 +𝐑 𝐦 )𝟐 

         R m = 2.65 × 108 m                                                                            (5)
                                                                                                        [20]

 QUESTION 3

 3.1

 3.1.1   An object which has been given an initial velocity and then it moves under the
         influence of the gravitational force only.

         NB: Object will only receive a mark if used in the correct contexts of vertical
         motion.                                                                                        (2)

 3.1.2

                                                                                                        (1)

 3.1.3    Zero/ 0 (J)                                                                                  (1)

 3.2.1

 Upwards Positive                                      Upwards Negative

 𝑣𝑓 = 𝑣𝑖 + 𝑎𝛥𝑡                                        𝑣𝑓 = 𝑣𝑖 + 𝑎𝛥𝑡 
 0 = 10 + (−9,8)𝛥𝑡                                    0 = −10 + (9,8)𝛥𝑡 
 𝛥𝑡 = 1,02𝑠                                           𝛥𝑡 = 1,02𝑠 
                                                                                                        (3)

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Physical Sciences P1                          5                September 2024 Preparatory Examination
                                       NSC – Marking Guideline

3.2.2
         DOWNWARD POSITIVE:

         BALL A:
                       1
         Δy = vi Δt + 2 aΔt 2
                           1
         yA = (−10)t + 2 (9.8)t 2 
         yA = −10t + 4.9t 2

         BALL B:
                             1
         yB = (0)(t − 0.5) + 2 (9.8)(t − 0.5)2
         8 + yA = yB

         8+= −10𝑡 + 4.9𝑡 2 = 4,9(t − 0.5)2
         −10t + 4.9t 2 = 4.9t 2 − 4,9t + 1,225 − 8

         ∴ t = 1.33s       

         DOWNWARD NEGATIVE:

         BALL A:

                     1       
         Δy = vi Δt + aΔt 2
                     2
                      1
         yA = (10)t + 2 (−9.8)t 2 
         yA = 10t − 4.9t 2

         BALL B:

                               1
         yB = (0)(t − 0.5) + (−9.8)(t − 0.5)2 
                               2
         8 + yB = yA
         8 − 4.9t 2 + 4,9t − 1,225 = 10t − 4.9t 2

         ∴ t = 1.33s       
                                                                                                   (4)

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Physical Sciences P1                              6                September 2024 Preparatory Examination
                                           NSC – Marking Guideline

 3.3.2    POSITIVE MARKING FROM 3.3.

                    8

             Position (m)

                            0        0,5        1,02       1,33        t (s)

           Starting times (0.5 s; 8 m) for Ball B, (0s ; 0 m) for Ball A         
           1.33s the time at which balls meet                                                         (3)
           Correct shapes for the graphs                                                              [14]

 QUESTION 4

 4.1     The total linear momentum of an isolated system remains constant (is
         conserved)
         NB: If closed system is used award 1 mark.                                                    (2)

 4.2              1
         𝐸𝐾 = 2 𝑚𝑣 2 
                                1
         258750 = 2 (2300)𝑣 2 
         𝑣 = 15m.s-1
                                                                                                       (3)

 4.3
         ∑ pbefore = ∑ pafter
                                          
         m1 vi1 + m2 vi2 = (m1 + m2 )vf
         mT (20) + (2300 − mT )(−10) = (2300)(15) 
         mT = 1916,67kg                                                                               (4)

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Physical Sciences P1                         7                September 2024 Preparatory Examination
                                      NSC – Marking Guideline

 4.4    POSITIVE MARKING FROM 4.3
        Δp = m(vf − vi ) 
        Δp = 1916.67(15 − 20) 
        Δp = −9583.35 kg.m.s-1

        ∴Impulse = 9583.35 N.s                                                                   (3)
                                                                                                  [12]

 QUESTION 5

 5.1    No. Presence of kinetic frictional force/rough surface                                  (2)
 5.2    The work done on an object by a net force is equal to the change in the object's
        kinetic energy.                                                                         (2)
 5. 3   𝑊𝑛𝑒𝑡 = 𝛥𝐸𝑘 
                           1
        𝑚𝑔𝛥𝑦𝑐𝑜𝑠𝜃 + 𝑊𝑓𝑘 = 𝑚(𝑣𝑓 2 − 𝑣𝑖 2 )
                           2
                                  1
        (2)(9.8)(2)𝑐𝑜𝑠00 +𝑊𝑓𝑘 = 2 (2)(52 − 02 ) 
        𝑊𝑓𝑘 = −14.20 J
        Option 2:
        𝑊𝑛𝑐 = 𝛥𝐸𝑘 + 𝛥𝐸𝑝 
              1
        𝑊𝑓𝑘 = 𝑚(𝑣𝑓 2 − 𝑣𝑖 2 ) + 𝑚𝑔(ℎ𝑓 − ℎ𝑖 )
              2
               1
        𝑊𝑓𝑘 = 2 (2)(52 − 02 ) +(2)(9.8)(0 − 2) 

        𝑊𝑓𝑘 = −14.20 J                                                                           (6)
                                                                                                  [10]

 QUESTION 6
 6.1    Doppler effect is the change in frequency (or pitch) of the sound detected by
        a listener, because the sound source and the listener have different velocities
        relative to the medium of sound propagation.                                            (2)
 6.2
        𝑣 = 𝑓×𝜆 

        340 = 𝑓(0.24) 

        𝑓 = 1416.67 Hz                                                                           (3)
6.3     POSITIVE MARKING FROM 6.2
             𝐯±𝐯
        𝐟𝐋 = 𝐯±𝐯𝐋 𝐟𝐬 
                   𝐬

                      
                  340+0
        1466.67 = 340−v (1416.67) 
                         s

        ∴ vs = 11.59 m.s-1
                                                                                                  (6)

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Physical Sciences P1                                     8                September 2024 Preparatory Examination
                                                  NSC – Marking Guideline

              Δx
         v=
              Δt
                     x
         11.59 = 15

         x = 173.85m
 6.4     Doppler flowmeter. 
         Or
         Determine heartbeat of foetus.                                                                       (1)
                                                                                                              [12]

 QUESTION 7

 7.1

 7.1.1    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 between them. (2)

 7.1.2               T        

                         FE   

                                                                                                              (3)

                         Fg       

 7.1.3    Fnet = ma
                                      kQ1 Q2             
          mg + (−T) + (−                     ) = ma
                                        r2
                                  (9×109 )(50×10−9 )(20×10−9 )
          [m(9.8) − 8 −                                         ]  = 0
                                            (4×10−2 )2
          m = 0.82 kg 

          ms = 0.82 × 2 = 1.63kg                                                                             (6)
 7.2

 7.2.1    The electric field at a point is the electrostatic force experienced per unit positive
          charge placed at that point.                                                                      (2)

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Physical Sciences P1                                           9                September 2024 Preparatory Examination
                                                        NSC – Marking Guideline

 7.2.2    B: The field from both charges are in the same direction.
                   1
          C:𝐸𝛼 𝑟 2 , C is closer to q2 than q1, and q2 > q1
          Or

             kq 2         𝑘𝑞1
                −2 2
                     >
          (5𝑋10 )      (65𝑋10−2 )2                                                                                  (2)

 7.2.3
          Enet = E1 + (−E2 )
                 kQ        kQ
          Enet = r22 +(− r21 )
                       (9×109 )(30×10−6 )           (9×109 )(90×10−6 )
          Enet =                            −                          
                          (20×10−2 )2                     (80×10−2 )2
          ∴ Enet   = 5484375N.C-1 to the left                                                                     (6)
                                                                                                                    [21]

 QUESTION 8.

 8.1     The potential difference across a conductor is directly proportional to the current
         in the conductor at constant temperature.                                                                (2)

 8.2

       8.2.1
                                       1     1    1
                                          =     +
                                       𝑅𝑝 𝑅1 𝑅2
                                       1     1 1
                                          =     + 
                                       𝑅𝑝 18 6
                                     𝑅𝑝 = 4,5 𝛺

                                            𝑉      
                                     𝑅=
                                            𝐼
                                                 𝑉
                                     4,5 =          
                                                0,4

                                     𝑉 = 1,8𝑉                                                                      (4)

       8.2.2   POSITIVE MARKING FROM 8.2.1

                          V2 Δt
               W=                    
                           R
                               
                           (1,8)2 (300)
               W=(                       )
                                  (18)
               W = 54 J

               OR
                                                                                                                    (4)
                  V
               R=
                  I
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Physical Sciences P1                           10                 September 2024 Preparatory Examination
                                        NSC – Marking Guideline

                       1,8
                 18 =
                        I
                 I = 0,1A
                                                        OR
                                                        W = VI△t
                 W = I2R△t 
                        
                                                        W = [(1,8 X0,1 ) x 300]
                 W = [(0,1)2 (18) (300) ] 
                                                                       
                                                        W = 54J
                 W = 54J
                                                                              

        8.2.3    POSITIVE MARKING FROM 8.2.1 AND 8.2.2

                 𝑉𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 = 𝐼𝑟 
                          
                 5-1,8 = 0,4r
                 r=8Ω               If learners end here award full marks
                 r = 4 Ω

                 OR

                 ε = I(R + r) 
                   
                 5 = (0,4)(4,5 + r) 
                 r=8Ω
                 r = 4 Ω                                                                            (4)

 8.3      Increases.                                                                                (1)
                                                                                                     [15]

 QUESTION 9

 9.1

       9.1.1    DC                                                                                  (1)

       9.1.2    Mechanical (energy) is converted to Electrical (energy).                            (1)

       9.1.3    Y.                                                                                  (1)

       9.1.4    Maximum.                                                                            (1)

 9.2
       9.2.1            𝐼
                𝐼𝑟𝑚𝑠 = 𝑚𝑎𝑥 
                            √2
                                                                                                      (5)

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Physical Sciences P1                                    11                    September 2024 Preparatory Examination
                                                 NSC – Marking Guideline

                            6
                   𝐼𝑟𝑚𝑠 =                                   OR
                            √2
                                                             Vmax = Vrms√2
                                                             Vmax = 210√2
                   𝐼𝑟𝑚𝑠 = 4.24A                              Vmax = 296,98V
                   𝑉𝑟𝑚𝑠 = 𝐼𝑟𝑚𝑠 𝑅 
                                                                     V
                                                              𝑅 = I max 
                   210 = 4.24𝑅                                       max

                                                                     296,98
                   𝑅 = 49.53𝛺                               R= 6 
                                                             R = 49, 50 Ω

     9.2.2         POSITIVE MARKING FROM 9.2.1

                                                           2                                  V2rms
                   Pave = Vrms . I                Pave = Irms .R                   Pave =           
                                                                                               R
                   Pave = (210)(4.24)                           2
                                                   Pave = (4,24) (49,53) 
                                                                                              (210)2
                                                   Pave = 890.43W                    Pave = (49.53)
                   Pave = 890.40W
                                                                                     Pave = 890,37W

                        W
                   P = Δt
                              W
                   890.40 =
                            7200
                   W = 6,41 X 106 J                                                                                  (5)
                                                                                                                      [14]
 QUESTION 10

 10.1 The process whereby electrons are ejected from a metal surface when light of
      suitable frequency is incident on that surface.                                                               (2)

 10.2
        At Ek(max) = 0 J:
                    1
        f0 = c(λ) 
        f0 = (3 × 108 )(3.6 × 106 ) 
        f0 = 1.08 × 1015 Hz                                                                                          (3)

 10.3    hc
                 = hf0 + Ek(max) 
             λ

             (6,63×10−34 )(3,0×108 )
         (                             = (6,63X10-34) (1,08X1015)  + Ek(max)
                     5,6×106 

         Ek(max) = 3,98X10-19J 
                                                                                                                      (5)

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Physical Sciences P1                   12                 September 2024 Preparatory Examination
                                NSC – Marking Guideline

 10.4

        10.4.1   REMAIN THE SAME                                                            (1)

        10.4.2   INCREASES                                                                  (1)
                                                                                             [12]
                                                                               TOTAL:       150

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