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

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION 2021 MARKING GUIDELINES 11101 TECHNICAL SCIENCES (PAPER 1) 11 pages 1

Downloaded from hlayiso.com 1101/21 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

Downloaded from hlayiso.com 11101/21 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

Downloaded from hlayiso.com 11101/21 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

Downloaded from hlayiso.com 11101/21 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

Downloaded from hlayiso.com 11101/21 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

Downloaded from hlayiso.com 11101/21 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

Downloaded from hlayiso.com 11101/21 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

Downloaded from hlayiso.com 11101/21 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

Downloaded from hlayiso.com 11101/21 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

Downloaded from hlayiso.com 11101/21 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