NW NSC GR 12 PS P1 memo 21 hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) SEPTEMBER 2021 MARKING GUIDELINES/NASIENRIGLYNE MARKS/ PUNTE: 150 These marking guidelines consist of 17 pages including 2 pages with the cognitive grid./ Hierdie nasienriglyne bestaan uit 17 bladsye nwat 2 bladsye met die kognitiewe tabel insluit. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 2 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 D  (2) 1.2 C  (2) 1.3 A  (2) 1.4 C  (2) 1.5 B  (2) 1.6 C  (2) 1.7 D  (2) 1.8 C  (2) 1.9 B  (2) 1.10 D  (2) o [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 3 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne o QUESTION 2 /VRAAG 2 o 2.1.1 The force or the component of a force which a surface exerts on an object with which it is in contact, and which is perpendicular to the surface.  /Die krag of die komponent van die krag wat `n voorwerp op `n oppervlakte uitoefen waarmee dit in kontak is, en wat loodreg op die oppervlakte is. 2 or/of 0 (2) 2.1.2 Accept the following symbols N FN/Normal/Normal force Normaal/Normaalkrag f Ff/frictional force / Wrywingskrag w Fg/mg/weight/Fearth on suitcase/gravitational force Gewig/F aarde op tas/gravitasiekrag (3) 2.1.3 Ff = Fg‖ Ff = mg sinθ Any one / Enige een Ff = 32 x 9,8 x sin 30 o  Ff = 156,8 N (3) 2.1.4 fsmax = µs N  156,8 = µs x 32 x 9,8 cos 30 o  µs = 0,58 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 4 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 2.2.1 Marking Criteria / Nasien kriteria mg sin θ 5 Substitution of / Substitusie van =  mg cos θ 3  Any appropriate formula for Fnet / Enige aanvaarbare formule vir Fnet   All substitutions into Fnet for 30 kg / Alle substitusies in Fnet vir 30 kg  All substitutions into Fnet for 50 kg / Alle substitusies in Fnet vir 50 kg   Substitutions for ‘ma’ in any one of the equations / Substitusies vir ‘ma’ in enige van die vergelykings  Final answer / Finale antwoord  Note: System approach maximum 4/6 marks Sisteem benader maksimum 4/6 punte Accept positive final answer: Range 332,1 N -332,3 N Aanaar positiewe finale antwoord: Interval 332,1 N -332,3 N OPTION 1 / OPSIE 1 mg sin  5 𝑦 5   or/of 𝑡𝑎𝑛𝜃 = 𝑥 = 3 mg cos  3 θ = 59,04̊ Fnet = ma any one / enige een  T – Fg‖ – Ff = ma For 50 kg: FA – Fg‖ – Ff – T = ma 500 – (50 x 9,8 x sin 59,04̊ ) – Ff(50) – T = 50 x 2 – T = 20,19 + Ff(50) ……………..(1) For 30 kg: Any one/enige een T – (30 x 9,8 x sin 59,04̊ ) – Ff(30)  = 30 x 2 T= 312,11 + Ff(30)…………….(2) combine equations (1) and (2) / kombineer vergelykings (1) en (2) 0 =312,11 + Ff(30) + 20,19 + Ff(50) Ff(30)+ Ff(50) = - 332,3 N Total frictional force/ Totale weerstandskrag = - 332,3 N OPTION 2 / OPSIE 2 mg sin θ 5 =  mg cos θ 3 θ = 59,04̊ Fnet = ma Any one / enige een  T – Fg‖ – Ff (30) = ma For the 50 kg: FA – Fg‖ – Ff – T = ma 500 – (50 x 9,8 x sin 59,04̊ ) – Ff(50) – T = 50 x 2 – T = 20,19 + Ff(50) ……………..(1) For the 30 kg: Any one  T–(30 x 9,8 x sin 59,04̊ )–3/5 Ff(50) = 30 x 2 T= 312,11 + 3/5 Ff(50) …………….(2) combine equations (1) and (2) 0 =312,11 + 3/5 Ff(50) + 20,19 + Ff(50) 8/5 Ff(50) = - 332,3 N (Ff ) 50 = - 207,69 N Total frictional force = - 332,3 N / Totale weerstandskrag = - 332,3 N  (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 5 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 3 / OPSIE 3 System approach / Sisteem benadering mg sin θ 5 =  mg cos θ 3 θ = 59,04̊ Fnet = ma Any one  T – Fg‖ – Ff (80) = ma 500 – (80)(9,8)(sin 59,04) – Ff (80)  = (80)(2)  500 – 672,3 - Ff (80) = 160 Ff (80) = - 332,3 N Maximum marks/ Maksimum punte 4/6 2.2.2 POSITIVE MARKING FROM QUESTION 2.2.1 / POSITIEWE NASIEN VANAF 2.2.1 Ff(Tot) = 332,3 N OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 5 3 Ff(50) = (8) - 332,3 N Ff(30) =(8) - 332,3 N Ff(50) = - 207,69 N Ff(30 = -124,613 N Substitute in equation (1) Substitute in equation (2) Substitusie in vergelyking (1) Substitusie in vergelyking (2) – T = 20,19 + Ff(50) T= 312,11 + 3/5 Ff(30) – T = 20,19 + -207,69  T =312,11 + -124,613  T = 187,5 N  T = 187,5 N  (2) [19] QUESTION 3/ VRAAG 3 3.1 o Motion during which the only force acting on an object is the force of gravity. Beweging waar die enigste krag wat op die voorwerp inwerk gravitadsiekrag is. 2 or/of 0 marks (2) o 3.2 o 5 m·s-1 upwards / opwaarts (1) 3.3 Marking Criteria / Nasien kriteria  Any appropriate formula / Enige aanvaarbare formule   All substitutions to calculate the value of Δ𝑦 / Alle substitusies om die waarde van Δ𝑦 te bereken   Addition of 60 + Δy / Som van 60 + Δy   Final answer / Finale antwoord Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 6 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne UPWARDS AS POSITIVE / OPWAARTS AS POSITIEF OPTION 1 / OPSIE 1 vf2 = vi2 + 2aΔ𝑦  (0)2 = 52 + 2 (-9,8) Δ𝑦  Δ𝑦 = 1,28 m The ball will reach a maximum height of (60 +1,28) = 61,28 m  above the ground /Die bal sal `n maksimum hoogte van (60 + 1,28) = 61,28 m bo die grond bereik DOWNWARDS AS POSITIVE / AFWAARTS AS POSITIEF vf2 = vi2 + 2aΔ𝑦  (0)2 = - 52 + 2 (9,8) Δ𝑦  Δ𝑦 = -1,28 m Height = 1,28 m The ball will reach a maximum height of (60 +1,28) = 61,28 m  above the ground /Die bal sal `n maksimum hoogte van (60 + 1,28) = 61,28 m bo die grond bereik OPTION 2 / OPSIE 2 vf = vi + aΔt 0 = 5 + (-9,8) Δ𝑡 Δ𝑡 = 0,51 s Any one / Enige een  v vf  Δy   i  Δt  2  5+0 ( ) 0,51 2 = 1,28 m The ball will reach a maximum height of (60 +1,28)  = 61,28 m above the ground Die bal sal `n maksimum hoogte van (60 + 1,28) = 61,28 m bo die grond bereik OPTION 3 / OPSIE 3 vf = vi + aΔt 0 = 5 + (-9,8) Δ𝑡 Any one / Enige een  Δ𝑡 = 0,51 s ∆y = vi∆t + ½ a∆t2 ∆y = 5 x 0,51 + ½ (-9,8 x 0,512)  = 1,28 m The ball will reach a maximum height of (60 +1,28) = 61,28 m  above the ground Die bal sal `n maksimum hoogte van (60 + 1,28) = 61,28 m bo die grond bereik (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 7 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 3.4 OPTION 1 / OPSIE 1 The hot- air balloon moved upwards at a constant velocity. /Die warmlugballon het opwaarts beweeg teen `n konstante snelheid. ∆y = vi∆t + ½ a∆t2  ∆y = (5)(3) + 0  ∆y = 15 m After 3 s the hot- air balloon will be 15 m above the starting point. /Na 3 s sal die warmlugballon 15 m bo die beginpunt wees. The distance travelled by the ball after 3s / Afstand deur bal beweeg na 3s. ∆y = vi∆t + ½ a∆t2 ∆y = (5)(3) + ½ (-9,8) (3)2  ∆y = -29,1 m  The ball is 29,1 m below the point from where it was released. After 3 s the hot air balloon and the ball will be (15 + 29,1) = 44,1 m apart / Die bal is 29,1 m onder die punt vanwaar dit laat val is. Na 3s sal die bal en die warmlugballon (15 + 29,1) = 44,1 m van mekaar wees. OPTION 2 / OPSIE 2 Vave = ∆y  ∆t ∆y = (5)(3)  15 m vf = vi + aΔt = 5 +(-9,8)(3) = -24,4 m·s-1  v vf  Δy   i  Δt  2  −24,4 +5 = ( 2 )(3) = -29,1 m The ball is 29,1 m below the point from where it was released. After 3 s the hot air balloon and the ball will be (15 + 29,1) = 44,1 m apart / Die bal is 29,1 m onder die punt vanwaar dit laat val is. Na 3s sal die bal en die warmlugballon (15 + 29,1) = 44,1 m van mekaar wees. (6) 3.5 o ANY ONE o Some of the ball’s kinetic energy is converted into heat and sound energy. o / Sommige van die energie van die bal word omgeskakel in hitte en klank energie.  o o OR/OF o o The collision between the ball and the ground is inelastic.  o / Die botsing tussen die bal en die grond is onelasties (2) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 8 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 4 / VRAAG 4 4.1 The total linear momentum of an isolated system is conserved both in (2) magnitude and direction Die totale liniêre momentum van `n geslote sisteem bly behoue in grootte en rigting. (2 or/of 0) 4.2 ΣPbefore = ΣPafter  (15) (VR )+ 0 = (15 +13,5)( 4,4) VR = 8,36 m·s-1. (3) 4.3 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2 vf = vi + a∆t  Fnet ∆t = ∆P = m(vf – vi ) 0 = 4,4 + (a) (3) Fnet (3) = 28,5 (0-4,4)  a = - 1,47 m·s-2 Fnet = Ff Ff = Fnet = ma Fnet = - 41,8 N Ff = (15 + 13,5) (-1,47) Ff = 41,8 N Ff = 41,9 N  Accept – 41,8 N (4) Accept – 41,9 N [9] QUESTION 5 / VRAAG 5 5.1 The net/total work done on an object is equal to the change in the object’s kinetic energy.  / Die netto werk verrig op `n voorwerp is gelyk aan sy verandering in kinetiese energie. OR The work done on an object by a resultant/net force is equal to the change in the objects kinetic energy.  / Die werk verrig op `n voorwerp deur `n netto/resultante krag is gelyk aan sy verandering in kinetiese energie. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 9 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 5.2 FA  FN  Ff Fg  Accept the following symbols / Aanvaar die volgende simbole FA  Applied Force /Force by engine/ Toegepaste krag/krag van engin N FN/Normal/Normal force / Normaal / Normaalkrag f Ff/frictional force / Wrywingskrag w Fg/mg/weight/Fearth on car/gravitational force Gewig/F aarde op tas/gravitasiekrag (4) 5.3 OPTION 1 / OPSIE 1 Marking Criteria for Option 1  Any appropriate formula / Enige aanvaarbare formule   Substitution of 0,1 in equation of Wnet  / Substitusie van 0,1 in vergelyking van Wnet   All substitutions to calculate Wnet  / Alle substitusies om Wnet te bereken  Substituting Wnet = 0  / Vervanging van Wnet = 0   Calculation of the power required / Berekenig van die drywing benodig  Calculation of the power of the engine  / Berekening van die drywing van die motor  Stating the car has enough power  / Staaf dat die motor genoeg drywing het Wnet = ∆K Wg‖ + W Ff + W FA = ∆K Any one / Enige een mg sinθ ∆x cosθ + Ff ∆x cos θ + FA ∆x cos θ = ∆K Wnet = (1200) (9,8) (0,1) (100) (cos180̊ ) + (820)(100)(cos180̊ ) + + FA (100)(cos0̊ ) = 0  sinθ = 10/100 FA = 1996 N upward /opwaarts sinθ = 0,1 Power = FA v Fnet = Fg‖ + Ff + FA . Prequired = (1996)(6)  Fg‖ = mg sinθ = 11976 W =11,976 kW Accept range P/ Aanvaar interval van = 11,972 kW to 11,977 kW 83 Real power of the engine / Werklike drywing van die motor = (62)( ) 100 PEngine = 51,460 kW Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 10 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne or PEngine = 51460 W PEngine > Prequired The car has enough power / Die motor het genoegsame drywing OPTION 2 Marking Criteria for Option 1  Any appropriate formula   Substitution of 0,1 in equation of ∆K   All substitutions to calculate ∆K   Substituting ∆K = 0   Calculation of the power required   Calculation of the power of the engine   Stating the car has enough power   Enige aanvaarbare formule   Substitusie van 0,1 in vergelyking ∆K   Alle substitusies om ∆K te bereken  Vervanging van ∆K = 0   Berekenig van die drywing benodig  Berekening van die drywing van die motor  Staaf dat die motor genoeg drywing het Wnet = ∆K Wg‖ + W Ff + W FA = ∆K Any one / Enige een mg sinθ ∆x cosθ + Ff ∆x cos θ + W FA = ∆K (1200) (9,8) (0,1) (100) (cos180̊ ) + (820)(100)(cos180̊ + W FA  = 0 W FA = 199600 J Δx v= Δt 100 6= Δt ∆t = 16,67 s W P t 199600 Prequired =  16,67 P = 11973,61 W Accept range P = 11972 W-11977 W Aanvaarbare interval P = 11972 W-11977 W Prequired = 11,974 kW 83 Real power of the engine/Werklike drywing van motor = (62000) ( ) 100 PEngine = 51460 W PEngine = 51,460 kW Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 11 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne PEngine > Prequired The car has enough power /Die motor het genoeg drywing (7) [13] QUESTION 6 / VRAAG 6 6.1 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.  / Die verandering in die frekwensie (toonhoogte) van die waargenome klank deur die luisteraar agv die klankbron en die luisteraar wat verskillende snelhede realtief tot mekaar het. (2) 6.2 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2   910 = (340+vL) 850  790 = (340-vL) 850  (340 -0)  ( 340 + 0)  vL = 24 m·s-1  vL = 24 m·s-1  (5) 6.3 OPTION 1/ OPSIE 1 OPTION 2 / OPSIE 2 Δx  v i Δt  21 at 2  ∆x = v ∆t  ∆x = (24)(6) + ½ x0 x 62 ∆x = 24 x 6  ∆x = 144 m  ∆x = 144 m  (3) [10] QUESTION 7 / VRAAG 7 7.1.1 The magnitude of the electrostatic force exerted by one point charge (Q 1) on another point charge (Q2) is directly proportional to the product of the magnitudes of their charges and inversely proportional to the square of the distance (r) between them  / Die grootte van die elektrostatiese krag wat deur een puntlading (Q 1) op 'n ander puntlading (Q2) uitgeoefen word, is direk eweredig aan die produk van die groottes van die ladings en omgekeerd eweredig aan die kwadraat (2) van die afstand (r) tussen hulle Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 12 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 7.1.2 OPTION 1 OPTION 2 Before contact/Voor kontak kQ1Q 2 F=  kQ1Q 2 r2 F=  r2 F = k q x 3q  F = k ( q) (3q) d2 d2 = 3 k q2  F = 3 kq2 d 2 d2 kQ1Q 2 After contact/Na kontak F new = r2 New Charge qnew = Q1 + Q2 = kq2  2 d2 = -q + 3q Fnew = F  2  3 =q Fnew = kqq d2 = kq2  d2 Fnew = F  3 (5) 7.2.1 kQ E  r2 ‘E’ at ‘X’ due to Q1 E =( 9,0 x 109) (6 x 10-9 ) 22 =13,5 N·C-1 to the left/na links ‘E’ at x due to Q2 kQ E r2 E =( 9,0 x 109) (8 x 10-9 ) 12 =72,0 N·C-1 to the right/na regs Net electric field at X / Netto elektrieseveld by X =72,0 -13,5 = 58,5 N·C-1  (4) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 13 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 8 / VRAAG 8 8.1 Guideline for allocating marks/Riglyne vir toekenning van punte Arrows point outwards Pyle uitwaarts gerig Correct shape Korrekte vorm (2) 8.2 E = F/Q  E = (3,23 x 10-5)/ (4,8 x 10-9)  E = 6729,17 N.C-1  (3) 8.3 OPTION 1 OPSIE 1 kQ1Q 2 F=  r2 3,23 x 10-5 = (9 x 109 x 4,8 x 10-9 x 4,8 x 10-9)  r2 r = 0,08 m OPTION 2: POSITIVE MARKING FROM QUESTION 8.2 OPSIE 2: POSITIEWE NASIEN VANAF VRAAG 8.2 kQ E  r2 6729,17 = 9 x 109 x 4,8 x 10-9  r2 r = 0,08 m (3) [8] QUESTION 9 / VRAAG 9 9.1.1 o V = IRT  o 14 = 1,4 x RT  RT = 10 Ω (3) 9.1.2 o POSITIVE MARKING FROM QUESTION 9.1.1 o POSITIEWE NASIEN VANAF VRAAG 9.1.1 o RT = (Rext +r) o 10 = (6+3,6) +( r ) r = 0,4 Ω  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 14 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 9.1.3 W = I2R∆t  W = 1,42 x 6 x 3 x 60  = 2116,8 J (3) 9.2 1 1 1 OPTION 1    ... V = IR R p R1 R 2 Vp = 1,88 x 3,43  1 1 1 = 6,46 V = +  Rp 8 6 I8Ω = 6,45  Rp = 3,43 Ω 8 emf ( ε )  I(R + r)  /emk ( ε )  I(R + r) = 0,81 A 14  I(3,43 + 3,6 + 0,4)  OPTION 2 I = 1,88 A I 8Ω = 6 x 1,88  14  = 0,81 A  (6) 9.3 Increases / energy transfer from the battery increases.  Toeneem / energie oorgedra van die battery neem toe.  OR / OF External resistance decreases because the parallel combination is eliminated by the short circuit. Die eksterne weerstand neem af omdat die paralelle kombinasie uitgesluit word duer die kortsluiting. (2) [17] QUESTION 10 / VRAAG 10 10.1.1 AC generator/WS generator (1) 10.1.2 Q- Carbon brush / Koolstofborseltjies R- Slip ring / Sleepring (2) 10.2.1 Graph A represents direct current.  Grafiek A verteenwoordig gelykstoom Graph B represents alternating current.  Grafiek B verteenwoordig wisselstroom (2) 10.2.2 no of ocillations aantal ossilasies f= / time tyd = 1,5/0,03 = 50 Hz (2) 10.3.1 V max Vrms =  2 V max 200 =  2 Vmax = 282,84 V  (3) 10.3.2 V rms = I rms x R 200 = I rms x 10 I rms = 20 A (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe/V1 15 NW/September 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 10.3.3 OPTION 1 / OPSIE 1 OPTION 3 / OPSIE 3 Pave = IrmsVrms Pave  I rms 2 R  = 20 x 200 =(20)2(10)  = 4000 W = 4000 W  OPTION 2 / OPSIE 2 V2 Pave  rms  R 2002 =  10 = 4000 W  (3) [16] QUESTION 11 / VRAAG 11 11.1 The work function of a metal is the minimum energy that an electron needs to be emitted from the metal surface  2 or 0/ Die werksfunksie van `n metaal is die minimum hoeveelheid energie benodig om elektrone uit die oppervlakte van die metaal vry te stel  2 or 0 (2) 11.2.1 c = fλ 3 x 108 = f (200 x 10-9)  f = 1,5 x 1015 Hz  (2) 11.2.2 W o = hf0 2,3 x 10-19 = 6,63 x 10-34 x f0  f0 = 3,47 x 1014 Hz  (3) 11.2.3 POSITIVE MARKING FROM QUESTION 11.2.1 /POSITIEWE NASEIN VANAF 11.2.1 E = W o + Ek(max)  (6,63 x 10-34) (1,5 x 1015 ) = (2,3 x 10-19 ) + Ek  Ek = 7,65 X 10-19 J  (4) 11.3 Decrease/Verminder (1) [12] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou

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