# Technical Sciences Grade 12 P1 GP Prelim 2021 Memo (English)

> Grade 12 Technical Sciences Paper 1 memorandum from the GP Prelim 2021 (English), 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/technical-sciences-p1-english-2021-memo/39e82137-9e36-4f04-996e-c5fc3078f3f5
- Updated: 2026-09-27
- Subject: Technical Sciences
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2021
- Exam period: GP Prelim
- Paper: 1
- Curriculum: CAPS
- Pages: 11

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## About this document

Technical Sciences · Grade 12 · GP Prelim · 2021 · English. Memorandum, 11 pages. Read online or download the PDF.

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- [Technical Sciences P1 (English) 2021 QP](https://www.hlayiso.com/library/technical-sciences-p1-english-2021-qp/ca8bec91-5842-46fc-9eaa-bc686d47ae51) — QP · 2021 · Eng

## Document text

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           GAUTENG DEPARTMENT OF
                  EDUCATION

      PREPARATORY EXAMINATION
                       2021
             MARKING GUIDELINES

                        11101
                 TECHNICAL SCIENCES
                      (PAPER 1)

11 pages

                          1

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QUESTION 1

1.1    C                                    (2)

1.2    B                                    (2)

1.3    A                                    (2)

1.4    A                                    (2)

1.5    B                                    (2)

1.6    A                                    (2)

1.7    C                                    (2)

1.8    C                                    (2)

1.9    B                                    (2)

1.10   C                                    (2)
                                              [20]

                           2

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QUESTION 2

2.1   Newton’s second law/Newton’s 2nd Law of motion 

      NEGATIVE MARKING FROM NAMING OF THE LAW

      When a resultant/net force acts on an object, the object will accelerate in the
      direction of the net force with an acceleration that is directly proportional to the
      force and inversely proportional to the mass of the object. 

      OR

      The (non-zero) resultant/net force acting on an object is equal to the rate of
      change of momentum of the object in the direction of the resultant/net
      force.  (2 or 0)

      ACCEPT
      Acceleration is directly proportional to the net force and inversely proportional
      to the mass of the object.

      NOTE:     If any of the underlined key words in the correct context is omitted
                deduct 1 mark                                                        (3)

2.2
                       N                                     N

                                   F                                     F
                   T                                     T

                           W                                     W

       Notes
           Mark is awarded for label and arrow.

              Do not penalise for length of arrows.

              If force(s) do not make contact with body: Max ¾.

              Deduct 1 mark for any additional force.

                 Accept the following symbol
       N         FN; Normal; + Normal force 
       F         F, FA, Applied force, 135 N
       W         Fg; mg; Weight; Fw ; Gravitational force
       T         Tension, FT,                                                               (4)

                                              3

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2.3   2.3.1   For 30 kg
              Fnet = ma 
              F – T = ma 
              135 – T = 30 … … (1)
              For 15 kg:
              Fnet = ma
              T = ma
              T = 15a … … (2)
              135 – 15a = 30a 
              a = 3 m . s –2                                                                   (3)

      2.3.2   Positive marking from 2.3.1

              Option 1:                                               Option 2:

              Fnet = ma                                                            
              F + (–T) = ma                                                               
              135 – T = 30 x3                                                       
              T = 135 – 90
              T = 45 N                                                                         (3)

2.4   2.4.1   For 30 kg:

              Fnet = ma                 
              F+T+ fk = ma
              135−T−(0,2)(30)(9.8)  = 30a 
              T = 76,2−30a ................................................. (1)

              For 15 kg:

              Fnet = ma
              T+ (–fk) = ma
              T−(0,1)(15)(9,8)  = 15a 
              T = 15a + 14,7 .................................................. (2)

              Solve for (1) and (2).
              76,2 − 30a = 15a + 14,7
              61,5 = 45a
              a = 1,37m∙s−2                                                                    (6)

      2.4.2   Decreased                                                                        (1)
                                                                                                [20]

                                                         4

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QUESTION 3

3.1   The total linear momentum in a closed system remains constant (is conserved). (2)

3.2   Left/Backwards/West                                                          (1)

3.3   p= mv 
      p = 1 x 10
      p = 10 kg. m·s-1 right                                                      (3)

3.4   Fnet. ∆t = ∆p 
      Fnet. 0,1 = 1 (10 – 0) 
      Fnet = 100 N right                                                           (4)
                                                                                    [10]

QUESTION 4

4.1   In a/an closed/isolated  system, the total  MECHANICAL ENERGY is
      conserved.                                                                    (2)

4.2   Em(top) = Em(bottom)
      (Ep + Ek)top = (Ep + Ek)bottom               
      (mgh + ½ mv2)top = (mgh + ½ mv2)bottom
      (1,5)(9,8)(2) + 0 = 0 + ½ (1,5)v2 
      v = 6,26 m∙s-1                                                               (4)

4.3   POSITIVE MARKING FROM 4.2
      m1vi1 + m2vi2 = v (m1 + m2) vf 
      1,5 x 6,26 + 2 x 0  = v (1,5 + 2) 
      9, 39 + 0 = 3,5v
             9,39
      v =
             3,5
        v = 2,68 m.s-1                                                             (4)

                                               5

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4.4   POSITIVE MARKING FROM 4.2 and 4.3
      Ek before collision = Ekp + Ekb
                         = ½ m1vi12 + ½ m1vi22
                         = 0,5 x 1,5 x 6,262 + 0,5 x 2,0 x 02 
                         = 29,39 J
      Ek after collision = Ekp + Ekb
                         = m1vf22 + ½ m2vf22
                         = 0,5 x 1,5 x 2,682 + 0,5 x 2,0 x 2,682 
                         = 12,57 J

      Ek before collision  Ek after collision 
       The collision is inelastic 
                                                                              (4)
                                                                              [14]

QUESTION 5

5.1              The internal restoring force per unit area of a body.      (2)

5.2   5.2.1      Area = l (length) × l (W) Width
                 = 4 × 10-3 × 4 × 10-3
                 = 1,6 × 10−5 m2 

                  σ=F 
                    A

                 = (40)(9,8) (Weight) 
                   1,6 × 10−5

                 = 245 000 000Pa              OR = 2,45 ×108 Pa              (4)

      5.2.2      ε=∆𝑙 
                    L

                 = 4 × 10−3 
                     4

                 = 1 × 10−3                                                  (4)

                                                       6

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      5.2.3   POSITIVE MARKING FROM 5.2.2 and 5.2.3

              K=σ 
                ε

              = 2,45 × 108     
                 1 × 10−3

              = 2,45 × 1011 N∙m−2                                                      (4)

5.3   5.3.1   𝐴 = 𝜋𝑟2 
              = π(0,001,5)2 
              = 7,07 × 10-6 m2                                                         (3)

      5.3.2   Ɛ=l
                   L
              = 0,7 ×10-3 
                  2,5

              = 2,8 × 10-4                                                             (3)
                                                                                        [20]

QUESTION 6

6.1   6.1.1   Viscosity is the property of a fluid to oppose relative motion between
              the two adjacent layers.                                             (2)

      6.1.2   When the temperature of the fluid increases, its viscosity decreases. 

              OR

              Viscosity increases with a decrease in temperature of the liquid.        (1)

6.2   6.2.1   Thrust is the normal force applied to a fluid at rest with the area with
              which it is in contact.                                                (2)

      6.2.2   Thrust                                                                   (1)

                                             7

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6.2.3   OPTION 1                                   OPTION 2

        F1 = F2                                   pA2= FA2
        A1 A2                                           AA2

                                                   = 4 000
             F1     = 4 000 
                 -3                                 1,9 ×10-2
        3,8 x 10      1,9 x 10-2
                                                   = 210 526,32 Pa
        ∴ F1 = 800 N 

                                                   pA1= FA1 
                                                        AA1

                                                   210 526,32 =     FA1       
                                                                   3,8 × 10 -3

                                                    FA1 = 800 N                         (3)

6.2.4   OPTION 1                                   OPTION 2
        pA1 = FA1                                 pA2 = FA2 
              AA1                                        AA2

        =      800                                = 4 000 
            3,8 × 10-3                              1,9 × 10-2

        = 210 526,32 Pa                           = 210 526,32 Pa                      (3)

6.2.5   Stays the same                                                                  (1)

6.2.6   ANY TWO 

        (Hydraulic) brakes; Car lifts; (Hydraulic) jacks; Forklifts; Dentist chairs;
        (Hydraulic) systems of a bulldozer; (Hydraulic) lifts used to lift heavy
        loads; Power assisted steering; (Hydraulic) beds; Jackhammer; Office
        chairs                                                                           (2)
                                                                                         [15]

                                          8

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QUESTION 7

7.1   Capacitance is the amount of charge that a capacitor can store per unit volt  (2)

7.2   Surface of the plates
      Distance between the plates
      Type of dielectric

      (Any 2)                                                                       (2)

7.3           

      = 6,5 x 10-6 
            10

      = 6.5 x 10-5 F                                                                 (3)
                                                                                      [7]

QUESTION 8

8.1   8.1.1   Rp = 1 + 1 
                   R1 R2

              =1 + 1 
               3   4

              = 1.71 Ω                                                              (3)

      8.1.2   V=IxR

              6 = I x (1.71 + 5) 

                  = 0.89 A                                                           (4)

      8.1.3   Option 1                 Option 2
              P = VI                        
              = 6 x 0,89 
                                            
              = 5,34 W 
                                       = 5,34 W                                      (3)
                                                                                      [10]

                                                 9

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QUESTION 9

9.1     Magnetic flux is the number of magnetic field lines perpendicular to the given
        surface area.                                                                (2)

9.2     A, has a magnetic flux, the coil is perpendicular to the field lines.                     (2)

9.3     Density of the field lines. 
        The strength of the magnet.                                                                (2)

9.4     9.4.1          B                                                                          (1)

        9.4.2          B has slip rings. Slip rings are used with the AC generator.               (2)
                                                                                                    [9]

QUESTION 10

10.1    When a magnet is moved into a coil of wire, changing the magnetic field
        and magnetic flux through the coil, a voltage will be generated in
        the coil according to Faraday's Law                                                        (1)

10.2    Np       Vp
                      
        Ns       Vs
        1 350 110
                 
         540   Vs
        Vs = 44V                                                                                   (4)
                                                                                                    [5]

QUESTION 11

11.1 11.1.1           (Phenomenon) whereby white light separates/breaks into its coloured
                      components.  (Penalise if there is no mention of white light and
                      coloured components)                                                           (2)

       11.1.2         Red. It has smaller frequency/has longer wavelength                         (3)

11.2 11.2.1           Angle of incidence must be greater than critical angle                       (2)

       11.2.2         When a light ray falls onto a plane surface it is reflected such that the
                      angle of reflection is equal to the angle of incidence.                      (2)

11.3 11.3.1           Normal line                                                                   (1)

       11.3.2         Diffused                                                                      (1)

                                                      10

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11.4      Image is upright 
          Image is virtual 
          Image size is equal to object size 
          Distance from object to the mirror is equal to distance from image to the
           mirror 
         Image is upright and inverted 
       (Accept any 2.)                                                                    (2)
                                                                                          [13]

QUESTION 12

12.1 12.1.1     c  fλ 

                3  108  = 89,7  106  

                  3.34 m                                                                  (4)

       12.1.2   Shorter than                                                                (1)

       12.1.3   Frequency is inversely proportional to the wavelength                       (2)
                                                                                             [7]

                                                                                 TOTAL:     150

                                               11

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