# Physical Sciences Grade 12 P1 Western Cape Mock 2020 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the Western Cape Mock Exam 2020, 16 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/wc-2020-gr-12-sept-mced-physics-p1-common-cluster-exam-memorandum-2/b6743005-6dd1-403e-b52a-d4224e11cb5b - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2020 - Exam period: Western Cape Mock Exam - Paper: 1 - Curriculum: CAPS - Pages: 16 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 P1 Western Cape Mock 2020 Memo](https://www.hlayiso.com/library/wc-2020-gr-12-sept-mced-physics-p1-common-cluster-exam-memorandum-2/b6743005-6dd1-403e-b52a-d4224e11cb5b) ## About this document Physical Sciences · Grade 12 · Western Cape Mock Exam · 2020. Memorandum, 16 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 METRO CENTRAL EDUCATION DISTRICT GRADE 12 PHYSICAL SCIENCES PAPER 1 (PHYSICS) COMMON CLUSTER SEPT/OCT 2020 EXAMINATION MEMORANDUM / MARKING GUIDELINES MARKS: 150 TIME: 3 hours This MEMORANDUM consists of 15 pages including the cover sheet Copyright Reserved Page 1 of 16 Please turn over ### Page 2 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 QUESTION 1 1.1 A ✓✓ (2) 1.2 A ✓✓ (2) 1.3 D ✓✓ (2) 1.4 B ✓✓ (2) 1.5 C ✓✓ (2) 1.6 D ✓✓ (2) 1.7 D ✓✓ (2) 1.8 B ✓✓ (2) 1.9 A ✓✓ (2) 1.10 C ✓✓ (2) [20] QUESTION 2 2.1.1 When a resultant (net) force acts on an object, the object will accelerate in the direction of the force. This acceleration is directly proportional to the force ✓ and inversely proportional to the mass of the object. ✓ OR The net force acting on an object is equal to the rate of change of momentum ✓✓of the object (in the direction of the force). (2 or 0) (2) 2.1.2 (5) OR ACCEPT the labelled free-body diagram where the applied force is resolved into horizontal and vertical components (Fax and Fay) Copyright Reserved Page 2 of 16 Please turn over ### Page 3 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 2.1.3 For block A: N = Fg – Fay ✔ = (15)(9,8) - 120 sin300 ✔ = 147 - 60 = 87 N Ff = µN = (0,2)(87) ✔ = 17,4 N ✔ (4) 2.1.4 Fax = F cos300 = 120 cos300 = 103,923 N Fnet = Fax + (T) + (Ff ) ✔ (15)(2,08) ✔ = 103,923 - T - 17,4 ✔ T = 55,32 N ✔ (4) 2.2.1 Each object/body in the universe exerts a force of attraction on every other object/body. This force is directly proportional to the product of their masses ✔ and inversely proportional to square of the distance between their centres ✔ (2) #$%$& 2.2.2 𝐹= ✓ '& (6,67 X 10-11 )(5,98 X 1024 )(1200) = (6,38 X 106 + 3,6 X 107 ) 2 ✓✓ = 266,49 N ✓ (4) 2.2.3 EQUAL TO ✓ (1) [22] Copyright Reserved Page 3 of 16 Please turn over ### Page 4 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 QUESTION 3 3.1 Yes.✓ Only force of gravity acts on the tennis ball ✓ OR Weight is the only force acting on the tennis ball ✓ (2) 1 3.2.1 Δy = viΔt + 2 aΔt2 ✓ 1 - 21 = vi(3,77) + 2 (-9,8)(3,77)2 ✓ vi = 12,90 m·s-1 ✓ (3) 3.2.2 OPTION 1 POSITIVE MARKING FROM QUESTION 3.2.1 From point of projection to maximum height X 2 2 (a) v f = v i + 2aΔy ✓ 02 = (12,9)2 + 2(-9,8)Δy ✓ Δy = 8,49 m (b) Max height = 8,49 + 21 + 0,6 ✓ = 30,09 m ✓ (4) OPTION 2 POSITIVE MARKING FROM QUESTION 3.2.1 From point of projection to maximum height X (a) vf = vi + aΔt 0 = 12,9 + (-9,8)Δt Δt = 1,316 s ✔ 1 (b) Δy = viΔt + 2 aΔt2 1 = (12,9)(1,316) + 2 (-9,8)(1,316)2 ✓ = 8,49 m (c) Max height = 8,49 + 21 + 0,6 ✓ = 30,09 m ✓ (4) Copyright Reserved Page 4 of 16 Please turn over ### Page 5 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 OPTION 3 POSITIVE MARKING FROM QUESTION 3.2.1 From point of projection to maximum height X vf = vi + aΔt 0 = 12,9 + (-9,8)Δt Δt = 1,316 s ✓ 1 Δy = 2 (vi + vf)Δt 1 = 2 (12,9 + 0)(1,316)✓ = 8,49 m Max height = 8,49 + 21 + 0,6 ✓ = 30,09 m ✓ (4) OPTION 4 POSITIVE MARKING FROM QUESTION 3.2.1 From point of projection to roof of car 1 (a) Δy = viΔt + 2 aΔt2 1 = (12,9)(3,77) + 2 (-9,8)(3,77)2 ✓ = - 21,01 m = 21,01 m down (b) Max height above ground = 21,01 + 0,6 = 21,61 m Copyright Reserved Page 5 of 16 Please turn over ### Page 6 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 3.3 ✔ 25% of 24,04 = 6,01 m·s-1 75% of 24,04 = 18,03 m·s-1 (speed of the tennis ball 24,04 – 6,01 = 18,03 m·s-1 OR (speed of the tennis ball immediately after bouncing) immediately after bouncing) FnetΔt = Δp FnetΔt = m(vf – vi) ✓ Any ONE = 0,073[18,03 – (-24,04)] ✓ = 3,07 kg·m·s-1 = 3,07 N·s upwards ✓ (4) 3.4 Velocity 18,03 ------------------------------- (m∙s-1) 12,90 - 3,77 3,87 Time (s) -24,04 -------------------------- Criteria for graph Marks Time taken to reach the roof of the car and ✔ Time at which it leaves the roof of the car Velocities when thrown upwards, ✔ Velocity when it strikes the roof of the car ✔ Velocity when it bounces off the roof of the car straight lines ✔ Parallel lines ✔ (5) [18] Copyright Reserved Page 6 of 16 Please turn over ### Page 7 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 QUESTION 4 mg (h f - hi ) mg (0 - hi ) DE P 1 1 2 ghi m(v 2f - vi2 ) m(v 2f - 0) 2 4.1 gradient = DEK ✔ = 2 = 2 ✔ = - vf (2) 4.2 EK3 ✔ (1) 2 ghi 2 ghi 2 2 4.3 tan 1350 = - vf Accept: tan 450 = - v f (2)(9,8)hi -1 = - (2) 2 ✔ h = 0,20 m ✔ (2) 4.4 OPTION 1 Wg = -mg(hf – hi) ✔ 19,6 = -m(9,8)(0 – 0,2) ✔ m = 10 kg ✔ OPTION 2 Wnet = ΔEK 1 2 2 m( v f - vi ) 2 Wg = ✔ 1 19,6 = 2 m(22 – 02) ✔ m = 9,8 kg ✔ (3) 4.5 POSITIVE MARKING FROM QUESTION 4.4 Ʃpi = Ʃpf ✓ (mvi)1 + (mvi)2 = (mvf)1 + (mvf)2 (10)(2) + (2)(0)✓ = (10)(vf1) + (2)(0,3) ✓ vf1 = 1,94 m·s-1 ; right/original direction of motion✓(4) [12] Copyright Reserved Page 7 of 16 Please turn over ### Page 8 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 QUESTION 5 5.1 Trial 2 ✔ v2 cannot equal 0 ✔ or Net force must be equal to ZERO. OR Any other valid reason (2) 5.2 Wnet = ΔEK 1 FnetΔxCosθ = 2 m(Δv2) ✓ Δv2 of an object increases with increasing net force acting on an object provided the mass of the object, the displacement and the angle between the displacement and the force are kept constant. ✓ (2) 5.3 The net/total work done✔ on an object is equal to the change in the object’s kinetic energy✔. OR The work done on an object by a net force ✔ is equal to the change in the object’s kinetic energy. ✔ (2) 5.4 Wnet = ∆𝐸* 1 FnetΔxCosθ = 2 m(Δv2) ✔ In all case for this motion 𝜃 = 0- % 169,41Δx(1) = (m + 30)(5,76) ✔ & Δx = 0,017(m + 30) ---------------- (1) % 442,91Δx(1) = (m + 40)(13,69) ✔ ✔ & Δx = 0,01545 (m + 40) ------------- (2) (1) = (2): 0,017(m + 30) = 0,01545 (m + 40) ✔ M = 69,68 kg ✔ (6) [12] Copyright Reserved Page 8 of 16 Please turn over ### Page 9 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 QUESTION 6 6.1 The apparent/observed (change in) frequency (or pitch) of the sound detected by a listener because the sound source and listener have different velocities relative to the medium of sound propagation. ✓✓ OR The change in the (observed) frequency when there is relative motion between the source and the observer. ✓✓ (2) 6.2 NB: FL = 1,0625. FS 1204,16 = 1,0625. FS FS = 1133,33 Hz OPTION 1 v ± vL v fL = v ± vs fs ✓ OR fL = v - vs fs 340 + 0 1204,16 ✓ = ( 340 - vs )(1133,33) ✓ 1,0625(340 – vs) = 340 vs = 20 m.s-1 ✓ OPTION 2 v fL = v - vs fs ✓ fL v fs = v - vs 340 1,0625 ✓ = ( 340 - vs ) ✓ vs = 20 m.s-1 ✓ (4) 6.3 Waves in front of the moving source are compressed. The observed wavelength decreases ✓ For the same speed of sound,✓ a higher frequency or pitch will be observed. (2) 6.4 Increases ✓ (1) [9] Copyright Reserved Page 9 of 16 Please turn over ### Page 10 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 Copyright Reserved Page 10 of 16 Please turn over ### Page 11 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 QUESTION 7 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 (magnitude of the) charges ✓ and inversely proportional to the square of the distance between them. ✓ OR The force of attraction or repulsion between two (point) charges is directly proportional to the product of the (point) charges ✓ and inversely proportional to the square of the distance between (them) their centres. ✓ (2) 7.1.2 ✓ ✓ (2) ./0 /1 7.1.3 F = '1 ✓ Fnet = FPR + FSR 23 × %-6 72%,9 × %-:6 7 / 23 × %-6 72& × %-:6 7 / -1,27 x 10-6 ✓ = (-,<)1 ✓- (-,&)1 ✓ -1,27 x 10-6 = 150 Q - 450 Q Q = 4,23 x 10-9 C ✓ (5) 7.2.1 ✓ Correct shape. Lines must not cross and must touch sphere ✓ Correct direction (2) ./ 7.2.2 𝐸 = '1 ✓ 3×%-6 ×/ 3 x 107 = ✓ (-,9)1 QA = 8,33 x 10-14 C ✓ (3) Copyright Reserved Page 11 of 16 Please turn over ### Page 12 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 ✔ ./ (3 >%-6 )(?,<< > %-:0@ ) 7.2.3 (a) EAP = 1 = ✔ = 0,003 N·C-1 , left ' -,91 ./ (3 >%-6 )(%,A > %-:0@ ) (b) EBP = 1 = ✔ = 0,0036 N·C-1 , right ' -,&1 (c) Enet = EBP + EAP = 0,0036 – 0,003 ✔ = 0,0006 N·C-1 ✔, right (5) [19] Copyright Reserved Page 12 of 16 Please turn over ### Page 13 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 QUESTION 8 8.1.1 emf ( ε ) = IRext + Ir ✓ When the current increases, Ir (lost volts) increases ✓ emf ( ε ) is the same /constant ✓ ∴ IRext (terminal voltage/pd)( voltage of the load) decreases (3) 8.1.2 Group 2 ✓ (1) 8.1.3 ∆C ∆C gradient = - ü gradient = ✓ ∆D ∆D EF%& EF%& gradient = - EF- ü gradient = ✓ EF- =-2Ω = 2Ω ü -r = - 2 Ω r = 2Ω ü (3) 8.2.1 The battery supplies 18 J of energy✔ per coulomb of charge ✔ OR The battery supplies 18 J ✔ per unit charge ✔ OR 18 J of work is done ✔ in moving 1 C of charge ✔ through the battery OR ACCEPT: The potential difference of the battery in an open circuit is 18 V. ✓✓ OR Maximum work done by the battery per unit charge is 18 J. ✓✓ OR Maximum energy supplied by the battery per unit charge is 18 J. ✓✓ (2) 8.2.2 (a) VP = VR2 = IR ✓ Copyright Reserved Page 13 of 16 Please turn over ### Page 14 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 = (1,2)(10) ✓ = 12 V (b) PR1 = VI ✓ 6 = (12)IR1 ✓ IR1 = 0,5 A (c) V2 = IR 3,8 = (0,5 + 1,2)X ✓ X = 2,24 Ω ✓ (6) ACCEPT OTHER OPTIONS!!!!! 8.2.3. (a) INCREASES ✓ (1) (b) DECREASES ✓ (1) [17] Copyright Reserved Page 14 of 16 Please turn over ### Page 15 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 QUESTION 9 9.1.1 (Faraday’s Law) of Electromagnetic Induction ✓ (1) 9.1.2 Mechanical energy (kinetic) is converted to electrical energy ✓ (1) 9.1.3 Slip rings connected to brushes ✓ (1) 9.1.4 ● Shape ✓ ● Max emf for A, C & E and Zero emf for B & D ✓ (2) 9.2 Pave = VrmsIrms ✓ V ( max )( I rms ) = 2 35,36 ( ) = 2 ✓ (1, 22) ✓ = 30,50 W ✓ (4) [9] Copyright Reserved Page 15 of 16 Please turn over ### Page 16 Downloaded from hlayiso.com Physical Sciences/P1NSCMCED/SEPT 2020 QUESTION 10 10.1.1 Photoelectric effect ✔ (1) 10.1.2 UV light has a higher/greater frequency than the threshold frequency and energy to eject electrons from the zinc plate ✔ while the frequency of white light is less than the threshold frequency and will not eject photo-electrons ✔ (2) 10.2.1 Work function is the minimum amount of energy required by a metal, before electron are ejected. ✔✔ [2 or 0] (2) 10.2.2 𝑊- = ℎ𝑓- ✔ 3,36 × 10F%3 = 6,6 × 10F