Technical Sciences Grade 12 P1 Gauteng Mock 2020 Memo (Eng)

PREPARATORY EXAMINATION 2020 MARKING GUIDELINES TECHNICAL SCIENCES (PAPER 1) 11101 10 pages 1

11101/20 QUESTION 1 1.1 A (2) 1.2 B (2) 1.3 D (2) 1.4 B (2) 1.5 C (2) 1.6 D (2) 1.7 A (2) 1.8 C (2) 1.9 B (2) 1.10 B (2) [20] 2

11101/20 QUESTION 2 2.1 When a net force (Fnet) is applied on an object of mass (m), the object will accelerate in the direction of the net force. The acceleration is directly proportional to the net force and inversely proportional to the mass of the object.  (2) 2.2.1 Accepted labels w Fg/Fw/weight/mg/gravitational force T FT/tension N FN/FNormal f Ff/Fw/friction/µN/frictional force F FA/ Fapplied Notes • Mark awarded for label and arrow • Do not penalise for length of arrows since drawing is not to scale. • Any other additional force(s) Minus 1 (-1) mark • If force(s) do(es) not make contact with body, Minus 1 (-1) mark • Minus 1 mark if all arrows are omitted but correctly labelled. (5) 2.2.2 Ff = µN = 0.6(15)(9.8)  = 88,2 N (3) 2.2.3 + marking from 2.2.2 Fnet = 0✓ For Block B T – Fk = 0 T = Fk = 88,2 N✓ For Block A Fnet = 0 FA cosɵ - T – Fk =0 FA cos140✓ – 88,2 – 0,6(10)(9,8) ✓ = 0✓ FA = 151,5N✓ (6) 3

11101/20 2.3 + marking from 2.2.2 Fnet = ma✓ 88,2 = 15a✓ a = 5,88 m.s-2✓ (3) 2.4.1 Newton’s 3rd law✓ (1) 2.4.2 When object A exerts a force on object B, object B simultaneously exerts an opposing force of equal magnitude on object A. ✓✓ (2) [22] QUESTION 3 3.1 The total (linear) momentum remains constant/is conserved in an isolated/ a closed system/the absence of external forces. (2) 3.2 To the right as positive Σ pbefore = Σ pafter✓ (2)(5) + (9)(0)✓ = (2)vf1 + (9)(1)✓ ∴ vf1 = 0,5 m.s-1 right ✓ (4) 3.3 Option 1: (Wooden block) FnetΔt = mΔv OR FnetΔt = Δp✓ Fnet(0,6) = ✓2(0,5 – 5)✓ Fnet = -15 N ∴ magnitude of Fnet = 15 N✓ (4) 3.4 Impulse = FnetΔt✓ = 15 x 0,6✓ = 9 Ns✓ (3) [13] 4

11101/20 QUESTION 4 4.1 The product of the force applied on an object ✓ and the displacement in the direction of the force ✓ (2) 4.2 Scalar ✓ (1) 4.3 The total energy of an isolated system is constant. ✓✓ (2) 4.4.1 Ep = mgh ✓ = 500 x 9,8 x 48 ✓ = 235 200 J ✓ (3) 4.4.2 Em = Ep + Ek ✓ = mgh + 1 2 mv 2 = 235 200 + (250✓ x (1,97)2) ✓ = 235 200 J ✓ (4) [12] QUESTION 5 5.1 The maximum force that can be applied to a body so that the body regains its original form completely on removal of the force (2) 5.2.1 r  2 m2 =   2 2  0,0032 =    2  = 2,56 x 10-6 m2  (2) 5.2.2 F σ=  A = 67 000/2,56 x 10-6 m2  = 2,62 x 1010 N.m-2  (3) 5.2.3 Δl ε=  L 0,26 = 305 = 8,52 x 10-8  (2) 5.2.4  = K ε   = 2,62 x 10 /8.52 x 10-8 10 = 3,08 x 1017 N.m  (4) [13] 5

11101/20 QUESTION 6 6.1 Viscosity is the property of a fluid to oppose relative motion between the two adjacent layers. (2) 6.2 High viscosity (1) 6.3 The high viscosity will keep the toothpaste on the toothbrush, it will not fall off/Toothpaste does not flow (like water). (1) 6.4.1 Honey being heated Dynamic viscosity (mPa.s) Temperature (°C) Marking guidelines: Labelling of X-axis  Labelling of Y-axis  Heading of graph  Shape of graph  (5) 6.4.2 Honey has a high viscosity. As it gets warmer the kinetic energy between particles of honey increases,  giving them more energy hence it flows easily, thereby decreasing its viscosity.  (2) [11] 6

11101/20 QUESTION 7 7.1 Depth below the surface/height Density of the liquid  (2) 7.2 Top Hole Bottom hole P = pgh  P = pgh = 1 000 x 9,8 x 0,4 = 1 000 x 9,8 x 0,25 = 392 Pa = 2 450 Pa  The pressure at the bottom.  Pressure at the bottom hole is 2 450 Pa and 392 Pa at the top. (6) [8] QUESTION 8 8.1 In a continuous liquid at equilibrium, the pressure applied at any point is transmitted equally to other parts of the liquid.  (2) 8.2 Hydraulic lifts  Hydraulic brakes  Hydraulic steering  Hydraulic jacks Heavy equipment Airplanes Hydraulic shock absorbers ANY THREE (3) 8.3 F F =  A A F 2 000 =  0,5 = 1 000 Pa  (3) [8] 7

11101/20 QUESTION 9 9.1 A device for storing an electrical charge (2) 9.2.1 50A C= ✓ d (8,85 x 10−12 )(1) = ✓✓ 0,002 = 4 425 x 10-12F ✓ (4) 9.2.2 C=Q✓ V Q 4 425 x 10-12 = ✓ 250 Q = 110,6 x 10-8 C✓ (3) 9.2.3 Increase the potential difference. ✓✓ (2) 9.2.4 Dielectric✓ (1) 9.3.1 B✓ (1) 9.3.2 It is reverse bias with the battery.✓✓ (2) [15] 8

11101/20 QUESTION 10 1 1 1 10.1 = + ✓ R11 R1 R2 1 1 = + 3 6 = 2Ω ✓ Rtotal = Rs + Rll ✓ =2+3 = 3Ω ✓ (4) P3 = I3 R3 2 10.2 ✓ ✓ ✓ V 12 I3 = I = = = 4A R 3 OR P3 = V3I3 V 12 I3 = I = = = 4A R 3 OR 2 V3 P3 = R V 12 I3 = I = = = 4A R 3 (3) 2 ✓ ✓ 10.3 P3 = I3 R3 = (4)2  1 = 16W ✓ OR V3 = I3R3 = 4 1 = 4V P3 = V3I3 = 4  4 = 16W OR V3 = I3R3 = 4 1 = 4V 2 V3 (4)2 P3 = = = 16 W R 1 (3) ✓ ✓ 10.4 E3 = P3 t = 16  20 = 320 J ✓ (3) [13] 9

11101/20 QUESTION 11 11.1.1 Step-down There are more turns on the primary coil than on the secondary coil.  (2) 11.1.2 VP NP = VS NS N Any one VP = P × VS NS 1200 VP = × 24  60 VP = 480 V  (3) 11.2 An AC generator has slip rings while a DC generator has a split ring commutator. ✓✓ (2) 11.3.1 Irms = Imax/√2✓ Irms = 5,9✓/0,7✓ = 835 A✓ (4) 11.3.2 Vmax = ImaxR ✓ = 5,9✓ X 46✓ = 271 V✓ (4) [15] TOTAL: 150 10