# Physical Sciences Grade 12 Paper 1 NSC November 2020 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the NSC November Exam 2020, 19 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/physical-sciences-p1-nov-2020-eng-afr-memo/36887c96-549a-48ef-8172-dbbab7bf9add - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2020 - Exam period: NSC November Exam - Paper: 1 - Curriculum: CAPS - Pages: 19 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 Paper 1 NSC November 2020 Memo](https://www.hlayiso.com/library/physical-sciences-p1-nov-2020-eng-afr-memo/36887c96-549a-48ef-8172-dbbab7bf9add) ## About this document Physical Sciences · Grade 12 · NSC November Exam · 2020. Memorandum, 19 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 SENIOR CERTIFICATE/SENIOR SERTIFIKAAT NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 2020 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 19 pages./ Hierdie nasienriglyne bestaan uit 19 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 2 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 B  (2) 1.2 D  (2) 1.3 C  (2) 1.4 C  (2) 1.5 C  (2) 1.6 A  (2) 1.7 A  (2) 1.8 D  (2) 1.9 A  (2) 1.10 B  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 3 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1 Marking criteria/Nasienriglyne If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase The perpendicular force exerted by a surface on an object in contact with the surface.  Die loodregte krag deur 'n oppervlak uitgeoefen op 'n voorwerp wat daarmee in kontak is. (2) 2.2 Fapplied N f T w Accepted symbols/Aanvaarde simbole N FN/Normal/Normal force/173,5N /Normaal/Normaalkrag Ff /fk /frictional force/wrywingskag/kinetic frictional force/kinetiese f wrywingskrag/5 N Fg /mg/Weight/FEarth on block/Fw/Gewig/Gravitational force/ w Gravitasiekrag/196 N T Tension/Spanning/FT Fapplied  F/Applied force/35 N/Toegepaste krag/ FA Ftoegepas Notes/Aantekeninge  Mark is awarded for label and arrow./Punt word toegeken vir byskrif en pyltjie.  Do not penalise for length of arrows./Moenie vir die lengte van die pyltjies penaliseer nie.  Deduct 1 mark for any additional force./Trek 1 punt af vir enige addisionele krag.  If all forces are correctly drawn and labelled, but no arrows, deduct 1 mark. / Indien all kragte korrek geteken en benoem is, maar geen lyne nie, trek 1punt af. (5) 2.3 For the/Vir die 20 kg: Marking criteria/Nasienriglyne Fnet = ma  Formula for 20 kg or m kg/Formule vir 20 kg of  m kg / Fnet = ma  T – f – FAx = ma T – 5 – 35 cos40°  = 0   Substitution of zero into either formula  T = 31,81 N Vervanging van nul in een van die formules  All substitutions into Fnet for 20 kg as shown  For/vir m: Alle vervanging in Fnet for 20 kg soos getoon  Substitution of value of T in eqn for m Fnet = ma /Substitusie van waarde vir T in vgl vir m  mg – T = ma  Final answer/finale antwoord: 3,25 kg  m(9,8) – 31,81  = 0 m = 3,25 kg  (5) 2.4.1 Decreases/Neem af  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 4 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 2.4.2 POSITIVE MARKING FROM QUESTION 2.3 POSITIEWE NASIEN VANAF VRAAG 2.3 Moving to the right/Beweeg na regs Velocity decreases/snelheid neem af  Accelerates/Net force to left /Versnelling/netto krag na links OR/OF As the tension force decreases, the net force/acceleration acts in the opposite direction of motion /to the left.  Soos die spanning afneem, is daar ‘n netto krag/versnelling in die teenoorgestelde rigting / na links Moving to the left/Beweeg na links Velocity increases/snelheid neem toe  Accelerates/Net force to left /Versnelling/netto krag na links (3) [16] QUESTION 3/VRAAG 3 3.1 (Motion of an object) under the influence of gravity (weight) only.  (2 or 0) (Beweging van ʼn voorwerp) slegs onder die invloed van gravitasie (gewig). OR/OF (Motion in which) the only force acting on the object is gravity (weight). (Beweging waar) die enigste krag wat op die voorwerp inwerk, gravitasie (gewig) is. (2) 3.2.1 Δt = 0,67 – 0,64 = 0,03 s  (2) 3.2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 (1,90  0,67)  Δx = viΔt + ½ aΔt2 Δt = (-1,85) = 0 + ½ (-9,8)Δt2  2 = 0,62 s  (0,615 s) Δt = 0,61 s  (0,6145 s) OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 (1,90  0,67) vf2 = vi2 + 2aΔx Δt = = 1,285 s 0 = vi2 + 2(-9,8)(1,85) 2 Δt = 1,285 – 0,67  vi = 6,02 m·s-1 = 0,62 s  (0,615 s) vf = vi + aΔt 0 = 6,02 + (-9,8)Δt  Δt = 0,61 s  (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 5 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 3.2.3 POSITIVE MARKING FROM QUESTION 3.2.2 POSITIEWE NASIEN VANAF VRAAG 3.2.2 Marking Criteria/Nasienriglyne  Any appropriate formula/Enige geskikte formule   Correct substitution/Korrekte vervanging   Final answer/Finale antwoord: 5,94 to 6,08 m·s-1  OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Upwards positive/Opwaarts positief Upwards positive/Opwaarts positief vf = vi + aΔt  Δy = viΔt + ½ aΔt2 0 = vi + (-9,8)(0,62)  1,85 = vi (0,62) + ½ (-9,8) (0,62)2  vi = 6,08 m·s-1 (6,076 m·s-1 )  vi = 6,02 m·s-1 (6,022 m·s-1 )  Downwards positive/Afwaarts positief Downwards positive/Afwaarts vf = vi + aΔt  positief 0 = vi + (9,8)(0,62)  Δy = viΔt + ½ aΔt2  vi = -6,08 1,85 = vi (0,62) + ½ (9,8) (0,62)2   6,08 m·s-1 (6,076 m·s-1 )  vi = -6,02  vi = 6,02 m·s-1 (6,022 m·s-1)  OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 Motion from top to bottom / Upwards positive/Opwaarts positief Beweging vanaf bo na onder Δy = viΔt + ½ aΔt2  Downwards positive/Afwaarts positief 0 = vi(1,23) + ½ (-9,8)(1,23)2  v f  v i  2ay  2 2 vi = 6,03 m·s-1  vf2 = 0 + 2(9,8)(1,85)  vf = 6,02 m·s-1  Downwards positive/Afwaarts 2 -1 positief Δy = viΔt + ½ aΔt  initial velocity/beginsnelheid=6,02 m·s 0 = vi(1,23) + ½ (9,8)(1,23)2  Upwards positive/Opwaarts positief vi = - 6,03 m·s-1 speed/spoed = 6,03 m·s-1 v f  v i  2ay  2 2 vf2 = 0 + 2(-9,8)(-1,85)  OPTION 5/OPSIE 5 vf = 6,02 m·s  -1  v f  vi  -1 Δy =   Δt  initial velocity/beginsnelheid=6,02 m·s  2   0  vi  Motion from bottom to top 1,85 =   (0,62) Beweging vanaf onder na bo  2  Downwards positive/Afwaarts positief vi = 5,97 m·s-1  v f  v i  2ay  2 2 OPTION 6/OPSIE 6 FnetΔt = mΔv 02 = vi2 + 2(9,8)(-1,85)  FnetΔt = m(vf – vi)  vi = 6,02 m·s-1  m(9,8)(0,62) = m(0 – vi)  vi = 6,08 m·s-1  Upwards positive/Opwaarts positief v f  v i  2ay  2 2 0 = vi2 + 2(-9,8)(1,85)  vi = 6,02 m·s-1  OPTION 7/OPSIE 7 (Ep + Ek)floor/vloer = (Ep + Ek)top/bo  (mgh + ½ mv2)floor/vloer = (mgh + ½ mv2)top/bo 0 + ½ v2 = (9,8)(1,85) + 0  v = 6,02 m·s-1  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 6 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 3.2.4 OPTION/OPSIE 1, 2, 3, 4: Marking criteria/Nasienriglyne Calculate initial velocity: Calculate/Bereken Δt: Bereken aanvanklike snelheid:  Appropriate formula/Geskikte formule   Appropriate formula/Geskikte formule   Substitution/Vervanging   Substitution/Vervanging   1,97 s + Δt   Fin answer/Fin antwoord: 2,95 – 2,97 s  Calculate initial velocity: Calculate time Δt Bereken beginsnelheid Bereken tyd Δt OPTION 1/OPSIE 1 Upwards positive Downwards positive/Afwaarts positief Opwaarts positief v f  v i  2ay  2 2 Δy = viΔt + ½ aΔt2 0 = vi2 + 2(9,8)(-1,2)  1,2 = (4,85)Δt + ½(-9,8)Δt 2  vi = - 4,85 m·s-1 Δt = 0,4898 s / 0,5 s t = 1,97 + 2(0,4898)  Upwards positive/Opwaarts positief = 2,95 s / 2,97 s  OR/OF v f  v i  2ay  2 2 Δy = viΔt + ½ aΔt2  0 = vi2 + 2(-9,8)(1,2)  0 = (4,85)Δt + ½(-9,8)Δt2  vi = 4,85 m·s-1 Δt = 0,9898 s (or Δt = 0) OPTION 2/OPSIE 2 t = 1,97 + 0,9898  = 2,96 s  (Emech)top = (Emech)bot/ond Any one/ (Ep +Ek)top = (Ep +Ek)Bot/Ond Enige een Downwards positive 2 2 (mgh + ½mv )top = (mgh + ½mv )Bot/Ond Afwaarts positief (9,8)(1,2) + 0 = 0 + (½)v2  Δy = viΔt + ½aΔt2 vi = 4,85 m·s-1 upwards /opwaarts 1,2 = (-4,85)Δt + ½(9,8)Δt2  OPTION 3/OPSIE 3 Δt = 0,4898 s / 0,5 s Wnc = ΔEp + ΔEk t = 1,97 + 2(0,4898)  Any one/ = 2,95 s / 2,97 s  0 = (0 – mgh) + ½m( v 2f  v i2 ) Enige een OR/OF 0 = -(9,8)(1,2) + ½vi2  Δy = viΔt + ½aΔt2  vi = 4,85 m·s-1 upwards /opwaarts 0 = (4,85)Δt + ½(9,8)Δt 2  OPTION 4/OPSIE 4 Δt = 0,9898 s (or Δt = 0) Wnet = ΔEk Any one/ t = 1,97 + 0,9898  = 2,96 s  wΔxcos180° = ½m( v 2f  v i2 ) Enige een OR/OF (9,8)(1,2)cos180° = ½vi2  vf = vi + aΔt  vi = - 4,85 m·s-1 -4,85 = 4,85 + (-9,8)Δt  Δt = 0,9898 s Δt = 1,97 + 0,9898 = 2,96 s OR/OF Upwards positive Opwaarts positief vf = vi + aΔt  0 = 4,85 + (-9,8)Δt  Δt = 0,4949 s Δt = 1,97 + (2)(0,4949)  = 2,96 s  OR/OF  v  vf  Δy =  i  Δt   2   0  4,85  1,2 =   ΔtΔt = 0,4948s  2  Δttotal = 2(0,4948) = 0,99 s Δt = 1,97 + 0,99 = 2,96 s  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 7 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 5/OPSIE 5 OPTION 5: Marking criteria/ Downwards positive/Afwaarts positief OPSIE 5: Nasienriglyne Δy = viΔt + ½ aΔt2   Formula /Formule 1,2 = 0 + ½(9,8) Δt2  Substitution/Vervanging Δt = 0,49 s Δy = 1,2  t = 1,97 +  2(0,49)   Substitution/Vervanging 0 + ½(9,8) Δt2 = 2,96 s   1,97 s +  Upwards positive/Opwaarts positief  2 Δt  Δy = viΔt + ½ aΔt2  -1,2 = 0 + ½(-9,8)Δt2  Final answer/Finale antwoord: 2,95 - 2,97 s  Δt = 0,49 s t = 1,97 +  2(0,49)  = 2,96 s  (6) [15] QUESTION 4/VRAAG 4 4.1 (Linear) momentum (of an object) is the product of mass and velocity.  (Liniêre) momentum (van 'n voorwerp) is die produk van massa en snelheid. (2 or/of 0) (2) 4.2.1 OPTION 1/OPSIE 1 East as positive/Oos as positief ∑pi = ∑pf Any one/Enige een mpvpi + mQvQi = mpvpf + mQvQf (0,16)(10) + (0,2)(-15) = (0,16)(-5) +(0,2)vQf  vQf = -3 m∙s-1 vQf = 3 m∙s-1  west/wes  OPTION 2/OPSIE 2 West as positive/Wes as positief ∑pi = ∑pf Any one/Enige een mpvpi + mQvQi = mpvpf + mQvQf (0,16)(-10) + (0,2)(15) = (0,16)(5) +(0,2)QNf  vQf = 3 m∙s-1  west/wes  OPTION 3/OPSIE 3 Δpp = -ΔpQ  (0,16)(-5 – 10) = -(0,2)(v – (-15)) v = -3 m·s-1 = 3 m·s-1 west/wes  (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 8 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 4.2.2 For ball/Vir bal P: POSITIVE MARKING FROM West as negative/Wes as negatief QUESTION 4.2.1 /POSITIEWE Impulse = Δp NASIEN VANAF VRAAG 4.2.1 FnetΔt = Δp Any one/ For ball/Vir bal Q: Δp = m(vPf – vPi) Enige een West as negative/Wes as negatief = 0,16(-5 – 10)  Impulse = Δp = - 2,4 FnetΔt = Δp  Any one/  2,4 N∙s  (2,4 kg∙m∙s-1) = m(vQf – vQi) Enige een = 0,2[-3 – (-15)]  OR/OF = 2,4 N∙s  (2,4 kg∙m∙s-1) West as positive /Wes as positief Impulse = Δp OR/OF FnetΔt = Δp Any one/ West as positive /Wes as positief = m(vPf – vPi) Enige een Impulse = Δp = 0,16(5 – (-10))  FnetΔt = Δp Any one/ = 2,4 N·s  = m(vQf – vQi) Enige een = 0,16(3 – (15))  = - 2,4 N·s  2,4 N∙s  (2,4 kg∙m∙s-1) (3) [10] QUESTION 5/VRAAG 5 5.1 Marking criteria/Nasienriglyne If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. However, IF: The word “work” is omitted 0 marks Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase. Maar, INDIEN: Die woord “arbeid” uitgelaat is, 0 punte A force is non-conservative if the work it does on an object (which is moving between two points) depends on the path taken.  'n Krag is nie-konserwatief indien die arbeid wat dit verrig (op 'n voorwerp wat tussen twee punte beweeg) afhanklik is van die pad. OR/OF A force is non-conservative if the work it does on an object depends on the path taken.  'n Krag is nie-konserwatief indien die arbeid wat dit verrig afhanklik is van die pad. OR/OF A force is non-conservative if the work it does in moving an object around a closed path is non-zero.  'n Krag is nie-konserwatief indien die arbeid wat dit verrig om 'n voorwerp op 'n geslote pad te beweeg, nie-nul is nie. (2) 5.2 K = ½ mv2 /Ek = ½ mv2 ΔK = Kf - Ki Any one /Enige een K = ½mvf2 - ½mvi2 = ½m(vf2- vi2) =½(200)(22 - 42) K = - 1 200 J  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 9 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 5.3 POSITIVE MARKING FROM QUESTION 5.2. POSITIEWE NASIEN VANAF VRAAG 5.2. Marking criteria/Nasienriglyne  Appropriate formula/Geskikte formule   Substitution into appropriate formula together with/Vervanging in geskikte formule saam met -3,40 × 103   Final answer/Finale antwoord: 8,88 m  OPTION 1/OPSIE 1 Wnc = K + U Wnc = ½ mvf2 - ½ mvi2 + mghf - mghi Any one/Enige een = ½ m (vf2- vi2) + mg(hf - hi) -3,40 × 103  = -1 200 + 200(9,8)(hf - 10)  h = 8,88 m  (8,87765 m) OPTION 2/OPSIE 2 E(mech/meg)A + W f = E(mech)B (Ep +Ek)A + W f = (Ep +Ek) B Any one/Enige een 2 2 (mgh + ½mv ) A + W f = (mgh + ½mv ) B 200(9,8)(10) + ½(200)(42) - 3,40 × 103  = 200(9,8)(h) + ½(200)(2)2  h = 8,88 m  (8,87755) OPTION 3/OPSIE 3 Wnet = K Wf + W w = ½mvf2 - ½mvi2 Any one/Enige een Wf – ΔEp = ½mvf2 - ½mvi2 Wf - mg(hf - hi) = ½m(vf2- vi2) -3,40 × 103 - 200(9,8)(h-10) = -1 200  h = 8,88 m  (8,87755 m) (4) 5.4 OPTION 1 AND 2/OPSIE 1 EN 2: Marking criteria /Nasienriglyne  Appropriate formula/Geskikte formule   Work done by friction/Arbeid verrig deur wrywing  Substitution of/Vervanging van (200)(9,8)(13,12)   Appropriate formula/Geskikte formule  Substitution into power formula/Vervanging in drywingformule  Final answer /Finale antwoord: 1 814,35 W OPTION 1/OPSIE 1 Wnc = K + U Wengine + Wf = ½mvf2 - ½ mvi2 + mghf - mghi Any one/Enige een 2 2 = ½m(vf - vi ) + mg(hf - hi) Wengine + (50)(15)(2)cos180°  = 0 + 200(9,8)(22 – 8,88) Wengine = 27 215,20 J Wengine Pengine = t 27 215,20 = 15 = 1 814,35 W Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 10 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 2/OPSIE 2 Wnet = K WN + W engine + W w + Wf = 0 Any one/Enige een WN + W engine - ΔEp + Wf = 0 0 + W engine - (200)(9,8)(13,12) + (50)(2)(15)cos180°  = 0 Wengine = 27 215,20 J OR/OF Wnet = K  WN + W engine + W w|| + Wf = 0 WN + W engine +mgsinθΔxcos180o + Wf = 0  13,12  0 + W engine - (200)(9,8)   Δx(-1) + (50)(2)(15)cos180°  = 0  x  Wengine = 27 215,20 J Wengine Pengine = t 25 215,20 = 15 = 1 814,35 W OPTION/OPSIE 3: Marking criteria/Nasienriglyne Opsie 3  Appropriate formula/Geskikte formule   Substitution of/Vervanging van - 50   Substitution of/Vervanging van (-200)(9,8)(0,4373) or/of (-200)(9,8)(0,44)  Appropriate formula/Geskikte formule  Substitution into/Vervanging in Pave = Fvave  Final answer/Finale antwoord: 1 814,35 W - 1 824,8 W OPTION 3/OPSIE 3 Fnet = ma Any one/Enige een Fengine + Ffriction + Fg// = 0 Fengine + (-50)  + (-200)(9,8)(0,4373) = 0 Fengine = 906,52 N (906,52 – 912,4) Pave = Fvave h sinθ = Pave = (908,52)(2) x = 1 813,04 W (1 824,8 W) 13,12 = 2(15) OR/OF W = FengineΔxcosθ = 0,4373 = (906,52)(30)cos0o = 27 195,6 J (27 372 W) W 27195,6 P= = = 1 813,04 W (1 824,8 W) t 15 (5) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 11 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1 Marking criteria/Nasienriglyne If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase The change in frequency (or pitch) (of the sound) detected by a listener because the source and the listener have different velocities relative to the medium of propagation.  Die verandering in die frekwensie (of toonhoogte) (van die klank) waargeneem deur 'n luisteraar omdat die bron en die luisteraar verskillende snelhede relatief tot die voortplantingsmedium het. OR/OF An (apparent) change in (observed/detected) frequency (pitch), as a result of the relative motion between a source and an observer (listener). 'n (Skynbare) verandering in (waargenome) frekwensie (toonhoogte), as gevolg van die relatiewe beweging tussen die bron en 'n waarnemer/ luisteraar. (2) 6.2 Towards/Nader  (1) 6.3 v  vL v v fL  fs  fL  fs fL  fs v  v s OR/OF v  vs OR/OF v  vs    340  0  340  0 3148  fs 2073  fs 340  v s 340  v s 3148(340  v s ) 2073(340  v s ) = 340  0 340  0 vs = 70 m·s-1 (69,95 - 70,16 m·s-1) (6) 6.4 POSITIVE MARKING FROM QUESTION 6.3 POSITIEWE NASIEN VANAF VRAAG 6.3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 x Δx = viΔt + ½ aΔt2  v  vf  t  350 = 70Δt + 0  Δx   i  Δt v  2  350  Δt = 5 s  t   70  70  70 350    Δt  t = 5 s   2  Δt = 5 s  (2) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 12 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1 Q n=  e ( ) 4  10 -6 =  ( ) 1,6  10 -19 = 2,5 x 1013  (3) 7.2 Electrostatic force on B due to A:/Elektrostatiese krag op B a.g.v. A: kQ1Q 2 FAB =  r2  9  10 9 (4  10 -6 )(3  10 -6 )  Ignore negative signs =    Ignoreer negatiewe tekens  0,2 2  doeleindes = 2,7 N  (3) 7.3 Marking criteria/Nasienriglyne If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase Electric field is a region (in space) where (in which) an (electric) charge experiences a (electric) force.  Elektriese veld is 'n gebied (in die ruimte) waarin 'n (elektriese) lading 'n (elektriese) krag ondervind. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 13 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 7.4 Marking criteria/Nasienriglyne  Appropriate formula/Geskikte formule   Correct substitution for A and B/Korrekte vervanging van A en B   Subtraction of electric fields/Aftrek van elektriesevelde  Final answer/Finale antwoord: 2,3 x106 N∙C-1 OPTION 1/OPSIE 1 Electric field at M due to / Elektriese veld by M as gevolg van: -4 x10-6 C Q EAM = k 2  r (4 x 10 -6 ) = 9 10 9  (0,3) 2 = 4,0 x 105 N∙C-1 (to left /links) Electric field at M due to / Elektriese veld by M as gevolg van: +3 x10-6 C, Q EBM = k 2 r (3 x10 -6 ) = 9 x10 9  (0,1) 2 = 2,7 x106 N∙C-1 (to right /regs) Net electric field at M /Netto elektrieseveld by M Enet = EBM + EAM = 4,0 x105 – 2,7 x106  = 2,3 x106 N∙C-1 (right/regs) OR/OF Net electric field at M /Netto elektrieseveld by M Enet = EBM + EAM = -4,0 x105 + 2,7 x106  = - 2,3 x106 N∙C-1 = 2,3 x106 N∙C-1  (right) OPTION 2/OPSIE 2 kQ1Q 2 (9  109 )(4  10 6 )Q FAM = 2 = 2  = 4 x 105Q N r (0,3) kQ1Q 2 (9  10 )(3  10 6 )Q 9 FBM = 2 = 2  = 2,7 x 106Q N r (0,1) Fnet = 2,7 x 10 Q + (-4 x 105Q) = 2,3 x 106Q 6 F 2,3  10 6 Q E= = = 2,3 x 106 N·C-1 (right/regs) q Q (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 14 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 7.5 Positive/Positief  (1) 7.6 POSITIVE MARKING FROM 7.2/POSITIEWE NASIEN VANAF 7.2 Marking criteria/Nasienriglyne  Correct substitution into Pythagoras’s equation/Korrekte vervanging in Pythagoras se vergelyking   Correct substitution into Coulomb’s Law/Korrekte vervanging in Coulomb se wet   Correct answer/Korrekte antwoord  (Fnet)2 = (FAD)2 + (FAB)2 (7,69)2 = (FAD)2 + (2,7)2  FAD = 7,2 N kQ1Q 2 FAD = r2 (9  109 )( 4 x 10-6 )Q 7,2 =  (0,15)2 QD = 4,5 x 10-6 C  OR/OF Q1Q 2 FAD = k r2 ( 4 x 10 -6 )Q = 9 x 10 9  0,15 2 = 1,6 x 106 Q Fnet = FAB  FAD FAB  FAD 2 2 2 2 2 OR/OF Fnet = 7,69 = 2,7 2  (1,6x10 6 Q) 2  Q = 4,50 x10-6 C  (3) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 15 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 Marking criteria/Nasienriglyne If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase (Maximum) energy provided (work done) by a battery per coulomb/unit charge passing through it.  (Maksimum) energie verskaf (arbeid verrig) deur 'n battery per coulomb/ eenheidslading wat daardeur beweeg. Work done by the battery to move a unit coulomb of charge across the circuit./Arbeid verrig deur die battery om 'n eenheidslading oor die stroombaan te beweeg. (2) 8.2 Energy (per coulomb of charge) is converted to heat in the battery due to the internal resistance.  Energie (per coulomb lading) word na hitte omskep binne-in die battery a.g.v. interne weerstand. (2) 8.3.1 V I  R 1,5 I  0,5 =3A (3) 8.3.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 1 1 1 R R = +  Rp  1 2  RP R1 R 2 R1  R 2 1 1 1 (25)(15)    Rp   RP 25 15 25  15 Rp = 9,375  Rp = 9,375  Rext = 9,375 + 4  = 13,38   Rext = 9,375 + 4 = 13,38   (13,375 ) (13,375 ) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 16 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 8.3.3 POSITIVE MARKING FROM QUESTIONS 8.3.1 AND 8.3.2. POSITIEWE NASIEN VANAF VRAAG 8.3.1 EN 8.3.2. OPTION 1/OPSIE 1 Ɛ = I(R + r)  = 3(13,38 + 0,5)  = 41,64 V  (Range/Gebied: 41,625 – 41,64) OPTION 2/OPSIE 2 Ɛ = Vext/eks + Vint  = (3)(13,38) + 1,5  = 41,64 V  (Range/Gebied: 41,625 – 41,64) (3) 8.4 Yes. /Ja For the same voltage/potential difference,  V a larger current will flow through a smaller resistor ( I  ) R Vir dieselfde spanning/ potensiaalverskil V sal 'n groter stroom deur die kleiner weerstand vloei ( I  ). R OR/OF 1 Iα , V = constant /konstant R I is inversely proportional to R and V is constant. I is omgekeerd eweredig aan R en V is konstant. OR/OF V|| = IR = (3)(9,38) = 28,14 V V 28,14 IR2 = = = 1,13 A  R 25 V 28,14 IR3 = = = 1,88 A  R 15 OR/OF V is the same / V is dieselfde  25 I15Ω = I 40 15  I25Ω = I 40 (3) 8.5 Remains the same/Bly dieselfde (1) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 17 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1.1 (DC) motor/(GS-)motor  (1) 9.1.2 POSITIVE MARKING FROM QUESTION 9.1.1 POSITIEWE NASIEN VANAF VRAAG 9.1.1 Electrical to mechanical /kinetic (energy)  (2 or 0) Elektriese na meganiese/kinetiese (energie) (2 of 0) (2) 9.1.3 Split ring/commutator/Splitring/kommutator  (1) 9.1.4 Anticlockwise/antikloksgewys  (2) 9.2.1 (The rms voltage/value of AC is) the AC voltage/potential difference which dissipates the same amount of energy/heat/power as an equivalent DC voltage/potential difference.  (2 or 0) (Die wgk-waarde van WS is) die WS-potensiaalverskil/spanning wat dieselfde hoeveelheid energie/hitte/drywing verbruik as 'n ekwivalente GS-spanning/ potensiaalverskil. (2 of 0) ACCEPT/AANVAAR The rms voltage/value of AC is the DC potential difference which dissipates the same amount of energy/heat/power as AC. Die wgk-waarde van WS is die GS-potensiaalverskil wat dieselfde hoeveelheid energie/hitte/drywing verbruik as die WS. (2) 9.2.2 Marking criteria/Nasienriglyne  Appropriate formula for Pave/Geskikte formule vir Pave   Substitution to calculate/Vervanging vir berekening van R   Final answer/Finale antwoord: 242 Ω  OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 V2 Pave  VrmsIrms  Pave  VrmsIrms  Pave  rms  R 200 = Irms (220) 200 = Irms (220) 2202 Irms = 0,909 A (0,91) Irms = 0,909 A(0,91) 200 =  V V R R  rms or/of R  Pave = Irms2R R = 242 Ω  Irms I 200 = (0,909)2R  220 R   R = 242 Ω  0,909 (241,52 Ω) R = 242 Ω  (241,76 Ω) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 18 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 9.2.3 Marking criteria for options 1,2 and 3 /Nasienriglyne vir opsies 1,2 en 3  Appropriate formula to calculate P or Irms /Geskikte formule om P of Irms te bereken   Substitution/Vervanging   Formula for P or W containing t/Formule vir P of W wat t bevat   Substitution/Vervanging   Final answer/Finale antwoord: 55 785,12 J  POSITIVE MARKING FROM QUESTION 9.2.2. POSITIEWE NASIEN VANAF VRAAG 9.2.2. OPTION 1/OPSIE 1 Marking criteria / Nasienriglyne  Appropriate formula for W containing V/Geskikte formule vir W wat V bevat   Substitution/Vervanging   Final answer/Finale antwoord: 55 785,12 J  V 2 t W   R (1502 ) (10 x 60)   242  = 55 785,12 J  OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 2 Vrms V V V V Pave =  R  rms / R  R  rms / R  R Irms I Irms I 2 150 150 150 =  242   242   242 Irms Irms Pav = 92,975 W Irms = 0,620 A Irms = 0,620 A W P  Pave  Irms Vrms Δt W = I2R  t  = (0,62)(150) W = (0,62)2(242)(10)(60)  92,975   = 92,97 W (93 W) (10)(60) W = 55 814,88 J  W = 55 785,12 J  P   (55785,12 – 55896 J) Δt (55785,12 – 55896 J) W 92,975   OR/OF (10)(60) W = VIΔt W = 55 785,12 J  = (150)(0,62)(600) (55785,12 – 55896 J) = 55 800 J OPTION 5/OPSIE 5 2 Vrms 1502 Pave = = = 92,975 W R 242 Pave= Irms2R 92,975 = Irms2(242) Irms = 0,6198 A W = I2RΔt  = (0,6198)2(242)(10)(60)  = 55 778,88 J  (5) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 19 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1 Photoelectric effect/Fotoëlektriese effek  (1) 10.2 Work function (of potassium)/Werksfunksie/Arbeidsfunksie (van kalium)  (1) 10.3 Potassium/Kalium  It has the lowest work function / threshold frequency / highest threshold wavelength.  Dit het die laagste arbeidsfunksie / drumpelfrekwensie / hoogste drumpel golflengte. (2) 10.4 Marking criteria/Nasienriglyne If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase The work function of a metal is the minimum energy that an electron (in the metal) needs to be emitted/ejected from the metal / surface.  Die werkfunksie/arbeidsfunskie van 'n metaal is die minimum energie benodig om 'n elektron vanaf ‘n oppervlak / metaal vry te stel. (2) 10.5.1 Wo= hfo  = (6,63 x 10-34)(1,75 x 1015)  = 1,160 x 10-18 J  OR/OF E = W o + Ek(max)  Any one / Enigeen hf = W o + Ek(max) (6,63 x 10-34)(1,75 x 1015) = W o + 0  Wo = 1,160 x 10-18 J  (3) 10.5.2 POSITIVE MARKING FROM QUESTION 10.5.1. POSITIEWE NASIEN VANAF VRAAG 10.5.1. E  W0  Ek (max) 1 Any one/Enige een hf  hfo  2 mvmax 2 1 (6,63 x 10-34)f  = 1,160 x 10-18 + (9,11 x 10-31) (5,60 x 105)2 2  f = 1,97 x 1015 Hz  (4) [13] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou ## Related documents - [WC 2020 Gr 12 SEPT MCED Physics P1 COMMON CLUSTER EXAM MEMORANDUM 2](https://www.hlayiso.com/library/wc-2020-gr-12-sept-mced-physics-p1-common-cluster-exam-memorandum-2/b6743005-6dd1-403e-b52a-d4224e11cb5b) — Physical Sciences · Memo · 2020 - [Physical Sciences P1 Nov 2019 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2019-memo-afr-eng/6250ed76-71a4-4f9b-bfe7-95a1f244df41) — Physical Sciences · Memo · 2019 · Eng - [Physical Sciences P1 Nov 2015 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2015-memo-afr-eng/140f25b7-f3c1-406a-a99d-6549fa7ea9e0) — Physical Sciences · Memo · 2015 · Eng - [Physical Sciences P1 Nov 2014 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2014-memo-afr-eng/44924678-c33b-42e1-8937-fcfda16ea7b3) — Physical Sciences · Memo · 2014 · Eng - [Physical Sciences P1 Nov 2013 Memo Afr Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2013-memo-afr-eng/b8f92a42-9c73-4898-adeb-03f8dd258340) — Physical Sciences · Memo · 2013 · Eng - [Physical Sciences P2 Nov 2020 Afr](https://www.hlayiso.com/library/physical-sciences-p2-nov-2020-afr/798fcbd8-dd69-4871-95a6-d42a4af1a841) — Physical Sciences · QP · 2020 · 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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