# Physical Sciences Grade 12 P1 KwaZulu Natal Mock 2018 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the KwaZulu Natal Mock Exam 2018, 11 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/kzn-2018-paper-1-memorandum/72eeb2b6-43d3-432b-be5a-bcd5a5b882e1 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2018 - Exam period: KwaZulu Natal Mock Exam - Paper: 1 - Curriculum: CAPS - Pages: 11 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 KwaZulu Natal Mock 2018 Memo](https://www.hlayiso.com/library/kzn-2018-paper-1-memorandum/72eeb2b6-43d3-432b-be5a-bcd5a5b882e1) ## About this document Physical Sciences · Grade 12 · KwaZulu Natal Mock Exam · 2018. Memorandum, 11 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 (P1) 1 Preparatory Examination September 2018 NSC – Marking Guideline NATIONAL SENIOR CERTIFICATE GRADE 12     PHYSICAL SCIENCES P1 MARKING GUIDELINE PREPARATORY EXAMINATION SEPTEMBER 2018 This marking guideline consists of 11 pages. Copyright Reserved Please Turn Over ### Page 2 Downloaded from hlayiso.com Physical Sciences (P1) 2 Preparatory Examination September 2018 NSC – Marking Guideline QUESTION 1 1.1 Cüü 1.2 Cüü 1.3 Aüü 1.4 Büü 1.5 Düü 1.6 Cüü 1.7 Aüü 1.8 Düü 1.9 Cüü 1.10 Cüü 10x 2 = [20] QUESTION 2 2.1 When a resultant/net force acts on an object, the object will accelerate in the direction of the force at an acceleration directly proportional to the forceü and inversely proportional to the mass of the object. ü OR The 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. üü (2) 2.2 Notes : Accepted labels Mark w Weight / Fg / FG ü T Tension in string ü f Frictional force ü N Normal force ü (4) 2.3 Box Object Fnet = maü Fnet = ma T + (-f) = ma w + (-T) = ma T – 6 = 9a………..(1) ü (2.5)(9,8) – T = 2.5a………(2) ü Substituting (1) into (2): -6 + (2.5)(9.8) = 11.5aü a = 1.61 m.s-2ü (5) Copyright Reserved Please Turn Over ### Page 3 Downloaded from hlayiso.com Physical Sciences (P1) 3 Preparatory Examination September 2018 NSC – Marking Guideline M1M2 2.4 F=G P r2 But FE&M = FS&M P (net zero gravitational force) (!.!"×!"!!! )(!,!"×!"!! )(!,!!×!"!" ) 1,997×10!"   P =   P !! r = 2,21 x 1011m P (5) [16] QUESTION 3 3.1 An object upon which the only force acting is the force of gravity. ü ü (2) 3.2 OPTION 1 Upward is positive Ball Y Ball X Δy1 = vi Δt + ½ a Δt2ü Δy2 = vi Δt + ½ a Δt2 = 20t + ½(-9.8) Δt2 -(50- Δy1) = (0)( Δt) + ½(-9.8) Δt2 = 20t – 4.9 t2……….(1)  ü Δy1 = -4.9t2 + 50………..(2)  ü Now, 20t – 4.9 t2 = -4.9t2 + 50 ü 20t = 50 t = 2.5 sü (5) Upward is negative Ball Y Ball X Δy1 = vi Δt + ½ a Δt2ü Δy2 = vi Δt + ½ a Δt2 -Δy1 = -20t + ½(9.8) Δt2 (50 – Δy1) = (0)( Δt) + ½(9.8) Δt2 Δy1 = 20t - 4.9 t2……….(1)  ü Δy1 = -4.9t2 + 50………..(2)  ü Solving (1) and (2): 20t = 50ü t = 2.5 sü OPTION 2 The balls are approaching each other (relative velocity increases) Δx (vx + vy) = ü Δt 50 (0ü + 20ü)   = ü (5) Δt Δt = 2.5 s ü Copyright Reserved Please Turn Over ### Page 4 Downloaded from hlayiso.com Physical Sciences (P1) 4 Preparatory Examination September 2018 NSC – Marking Guideline 3.3 OPTION 1 OPTION 2 Δy = vi Δt + ½ a Δt2 ü Δy = vi Δt + ½ a Δt2ü = (20)(2.5) ü + ½(-9.8)(2.5)2ü = (0)(2.5) + ½(-9.8)(2.5)2ü = 19.375 m = -30.625 h = 19.375 mü h = 50 –ü 30.625 = 19.375 mü OPTION 3 Ball X vf = vi + aΔt = 0 + (-9.8)(2.5) ü vf = -24.50 m.s-1 Em at top = Em at meeting point (mgh + ½mv2)at top = mgh + ½mv2)at meeting pointü m(9.8)(50) + 0 = m(9.8)Δy + ½m(-24.50)2ü Δy = 19.375 m h = 19.375 mü (4) 3.4 Mark positively from 3.2 and 3.3 OPTION 1 (downward is positive) Criteria • Shapeü • Starts at 50 mü • Ends at 2.5 sü • Ends at 19.375mü OPTION 2 (downward is negative) Criteria • Shapeü • Starts at -50 mü • Ends at 2.5 sü • Ends at -19.375mü (4) Copyright Reserved Please Turn Over ### Page 5 Downloaded from hlayiso.com Physical Sciences (P1) 5 Preparatory Examination September 2018 NSC – Marking Guideline OPTION 3 Can shift the lower graph to start at the origin, and the rest will remain the same. (4) [15] QUESTION 4 4.1 The product of mass and velocity of the object. üü (2) 4.2 0 kg.m.s-1 (Accept 0 / Zero) ü (1) 4.3 To the right is positive To the right is negative p = mvü p = mvü = (1)(10) ü = (1)(-10) ü = 10 kg.m.s-1 to the rightü = -10 kg.m.s-1 = 10 kg.m.s-1 (to the right)ü (3) 4.4 To the right is positive To the right is negative Σpbefore = Σpafter ü Σpbefore = Σpafter ü 0ü = 100vgb ü+ 10ü 0ü = 100vgb ü+ (-10)ü vfi = -0.1 m.s-1 vfi = 0.1 m.s-1 to the leftü vfi = 0.1 m.s-1to the leftü (5) [11] QUESTION 5 5.1 The work done by a net force on an object is equal to üthe change in the kinetic energy of the object. ü OR Net work done on an object is equal to üthe change in the kinetic energyof the object. ü (2) 5.2 Fg = gravitational force or weightü FN / N= normal forceü Ff / f = frictional forceü   (3) NB : Ignore the relative lengths the forces Copyright Reserved Please Turn Over ### Page 6 Downloaded from hlayiso.com Physical Sciences (P1) 6 Preparatory Examination September 2018 NSC – Marking Guideline 5.3 OPTION 1 Wnet = FnetΔx cosθü = Wf + WFg// = (190)(10)(cos180°)ü + (50)(9.8)(sin30°)(10)(cos0°)ü = -1900 + 2450 = 550 Jü (4) OPTION 2 Fnet = Fg// - Ff = (50)(9.8)(sin 30º) – 190 ü = 55 N Fnet = ma 55 = 50a OPTION 3 a = 1.10 m.s-2 Wnet = Wf + WFg// ü vf2 = vi2 + 2aΔx = (50)(9,8)(10)cos60 ü -1900 ü = 550 J ü = 22 + 2(1.10)(10) ü vf = 5.10 m.s-1 Wnet = ½mvf2 - ½mvi2 ü = ½(50)(5.10)2 -½(50)(2)2 = 550 Jü 5.4 POSITIVE MARKING FROM QUESTION 5.3 OPTION 1 Wnet = ΔK ü OPTION 2 = ½ mvf2 – ½ mvi2 550ü= ½ (50)(vf2 – 22)ü Wnc =ΔK +ΔUü vf = 5.10 m.s-1 ü -1900 = ½(50)(v2 – 22) + (50)(9.8)(hY-hX) -1900ü = ½(50)(v2 – 22) + (50)(9.8)(-10sin30º) ü v = 5.10 m.s-1ü (4) [13] Copyright Reserved Please Turn Over ### Page 7 Downloaded from hlayiso.com Physical Sciences (P1) 7 Preparatory Examination September 2018 NSC – Marking Guideline QUESTION 6 6.1 14 x 10−4 s PP (2) 6.2 POSITIVE MARKING FROM QUESTION 6.1 1 f = P T 1 = P 14 x 10 − 4 = 714,29 Hz P (3) 6.3 It 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. PP OR It is the change in the observed frequency of a sound wave when the source of sound is moving relative to the listener.PP (2) 6.4 POSITIVE MARKING FROM QUESTION 6.1 !!!  ! f! = f P !!!! ! !"# (714.29)P = P(600)P !"#!!! v! =  54.40  m.s-­‐1  P (accept 54.28 to 54.40  m.s-­‐1) (5) [12] QUESTION 7 !!   !  !! 7.1 Qnew = ! !×!"!! P!(!!×!"!! )P = ! = -1.5×10!!  CP (3) 7.2.1 The magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge (Q2) is directly proportional to the product of the magnitudes of the chargesP and inversely proportional to the square of the distance (r) between them. P OR The force of attraction or repulsion between two point charges is directly proportional to the product of the charges ü and inversely proportional to the square of the distance between them. ü (2) Copyright Reserved Please Turn Over ### Page 8 Downloaded from hlayiso.com Physical Sciences (P1) 8 Preparatory Examination September 2018 NSC – Marking Guideline 7.2.2 Criteria for marking Marks the diagram Correct shape ü   Correct direction ü   Field lines not touching each ü   other or entering the spheres (3) 7.2.3 Tx = (0.07)cos60ºP = 0.035 NP (2) !!!   !! 7.2.4 F =   P !! (!×!"! )(!.!×!"!! )(!.!×!"!! ) 0.035P = P !! r = 0.012  m x = 0.012  mP OR ! !"   E =   ! !.!"# either 𝐸 = !×!"!! P = 729166.67 N.C-1 P   ! 𝐸 = 𝑘 !!  !.!×!"!! 729166.67=  𝑘 P !! r = 0.012 mP (4) [15] QUESTION 8 8.1 The potential difference across the ends of the conductor is directly proportional to the current flowing to the conductor at constant temperature. üü (2) Copyright Reserved Please Turn Over ### Page 9 Downloaded from hlayiso.com Physical Sciences (P1) 9 Preparatory Examination September 2018 NSC – Marking Guideline 8.2.1 OPTION 1 OPTION 2 ! ! R !Ω   = ! P R !Ω   = ! P ! ! 8= 8= !.! !.! V = 2.4  V  V = 2.4  V ! !×! R !Ω   = R !     =   = 2.67  ΩP ! !!! !.! ! 4   = ! R           = ! !.! I     = 0.6  AP 2.67 = P ! I!"!   = (0.3 + 0.6)P = 0.9  AP I         = 0.9  AP OPTION 3 Current through 4 Ω = 2 x 0,3 P   = 0,6A P   ∴ A1 = (0,6 + 0,3) P= 0,9A P (4) 8.2.2 Positive marking from Question 8.2.2 ! R !   = P ! !".! = P !.! = 17.33  ΩP (3) 8.2.3 Positive marking from Question 8.2.2 !×! R ! =   !!! = 2.67  ΩP R ! = (2.67 + 17.33)  P = 20  ΩP (3) 8.2.4 Positive marking from Questions 8.2.1 and 8.2.3 𝑒𝑚𝑓 = 𝐼 𝑅 + 𝑟 P 19P = 0.9 20 + r P 𝑟             = 1.11 ΩP (4) 8.3.1 DecreaseP (1) 8.3.2 Total resistance of the circuit decreases,  P therefore current increases.  P More volts will be lost,  P and Vext decreases (Vext = ε – Ir)P (4) [21] Copyright Reserved Please Turn Over ### Page 10 Downloaded from hlayiso.com Physical Sciences (P1) 10 Preparatory Examination September 2018 NSC – Marking Guideline QUESTION 9 9.1 Slip ringsP (1) 9.2 Criteria • Starts at 50 VP • Ends at 50 VP • Ends at 0.08X10-3 sP • 2 cyclesP (4) 9.3 Increase the rotation speed of the coil. PP (2) I max 9.4.1 I rms = P 2 6.27 I rms = P 2 I rms = 4.43 AP (3) 9.4.2 OPTION 1   OPTION 2 V ! R = max PP 𝐼!"# =   !"# P I max ! Vmax V P 52.6 = P 4.43 = rms P 6.27 52.6 Vmax = 329.80V P Vrms = 233.018V Vmax Vrms = P 2 Vmax 233.018 = P 2 Vmax = 329.54V P   (5) [15] Copyright Reserved Please Turn Over ### Page 11 Downloaded from hlayiso.com Physical Sciences (P1) 11 Preparatory Examination September 2018 NSC – Marking Guideline QUESTION 10 10.1 The energy of the photons of red light is greaterP than the work function of the metal in the photocell.P OR The frequency of red light is higherP than the threshold/cut-off frequency of the metal in the photocell. P (2) 10.2.1 IncreaseP (1) 10.2.2  Stays the same P Since intensity is unchangedP, same number of photons reach cathode. P Therefore number of photo electrons emitted is unchanged. P (4) 10.3 E = Wo + Ek (max) P                     c h = hfo + ½mv2                               λ   6,63 x 10 -34 × 3 × 10 8 P  6,63 x 10-34(f ) P + ½(9,11 x 10-31)(5,06 x 105)2P   = o       4,5 × 10 −7 P     fo = 6,67 x 1014 Hz P (6)         [13] GRAND TOTAL: 150     Copyright Reserved Please Turn Over ## Related documents - [GP 2018 Paper 1 Memorandum](https://www.hlayiso.com/library/gp-2018-paper-1-memorandum/23750489-743e-47ce-ab89-fb6303430e77) — Physical Sciences · Memo · 2018 - [Physical Sciences P1 Feb March 2018 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-feb-march-2018-memo-afr-eng/4d0a2b36-b101-4443-aa19-d5c8b0d293a0) — Physical Sciences · Memo · 2018 · Eng - [Physical Sciences P1 May June 2018 Eng](https://www.hlayiso.com/library/physical-sciences-p1-may-june-2018-eng/4bbee28e-11a5-458f-8d92-6c6f1d7103d3) — Physical Sciences · QP · 2018 · Eng - [WP 2018 Paper 1](https://www.hlayiso.com/library/wp-2018-paper-1/7745da05-6836-4240-9b30-1cc144cc0c0d) — Physical Sciences · QP · 2018 - [GP 2018 Paper 2 Memorandum](https://www.hlayiso.com/library/gp-2018-paper-2-memorandum/b76b99ee-4834-4d11-bd2c-3650944f8134) — Physical Sciences · Memo · 2018 - [Gr 12 Physical Sciences P1 (English and Afrikaans) June 2024 Possible Answers](https://www.hlayiso.com/library/gr-12-physical-sciences-p1-english-and-afrikaans-june-2024-possible-answers/f78ca708-f7f0-4bc5-8ff9-36594b1b1052) — Physical Sciences · Memo · 2024 · Afr ## More from Grade 12 Physical Sciences - [PHYSICAL SCIENCE P1 QP JCT 2025 AFR D watermark](https://www.hlayiso.com/library/physical-science-p1-qp-jct-2025-afr-d-watermark/b9429ac5-acbb-4fae-989d-af9e1478a96b) — QP · 2025 · Afr - [PHYSICAL SC P2 QP GR12 SEPTEMBER 2025 AFR1 watermark](https://www.hlayiso.com/library/physical-sc-p2-qp-gr12-september-2025-afr1-watermark/37cff2db-7c1a-4ec8-8ac5-07ec5bcf87a9) — QP · 2025 - [Physics Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/physics-grade-12-nsc-p2-qp-september-2025-gauteng/4d3cbea6-10ec-47ff-98ee-50646165ae9f) — QP · 2025 - [Physics Grade 12 NSC P1 QP September 2025 KZN](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-kzn/6d225dc8-24c6-4c28-a181-df906d5a2c0e) — QP · 2025 - [Physical Sciences Physics Grade 12 PS01 QP March 2025 Limpopo](https://www.hlayiso.com/library/physical-sciences-physics-grade-12-ps01-qp-march-2025-limpopo/482ee9af-e88a-4818-9120-11ab9e045a1c) — QP · 2025 - [Physics Grade 12 NSC P1 QP September 2025 Limpopo](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-limpopo/b65e9832-a6be-4312-8f5a-57425d647121) — QP · 2025 ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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