Physical Sciences Grade 12 P1 NSC June 2025 Memo (Afrikaans)

SENIOR CERTIFICATE EXAMINATIONS/ NATIONAL SENIOR CERTIFICATE EXAMINATIONS/ SENIORSERTIFIKAAT-EKSAMEN/ NASIONALE SENIORSERTIFIKAAT-EKSAMEN PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) MAY/JUNE 2025/MEI/JUNIE 2025 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 20 pages. Hierdie nasienriglyne bestaan uit 20 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 D  (2) 1.2 A  (2) 1.3 B  Accept Q / Aanvaar Q (2) 1.4 D  (2) 1.5 D  (2) 1.6 C (2) 1.7 A  (2) 1.8 C  (2) 1.9 C  (2) 1.10 C  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. A body will remain in its state of rest or motion at constant velocity unless a non-zero resultant/net force acts on it.  'n Liggaam sal in sy toestand van rus of beweging teen konstante snelheid volhard, tensy 'n nie-nul resulterende/netto krag daarop inwerk. (2) 2.2 ACCEPT/AANVAAR F N F N fk fk w w Accepted labels/Aanvaarde benoemings N FN/Normal force/Fnormal/ 212,18 N / Fnormaal/Normaalkrag fk (kinetic) friction/Ff/ f / 40 N / (kinetiese) wrywing/Fw F Applied force/ Fapplied /Fa / Toegepaste krag/Ftoegepas w Fg/Fw/mg/245 N/weight/gravitational force/gewig/gravitasiekrag 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. • If arrows do not touch the dot/Indien pyle nie die kolletjie raak nie: Max/Maks 3�4 • Any other additional force(s)/Enige ander addisionele krag(te): Max/Maks 3�4 • If everything correct, but no arrows/Indien alles korrek, maar geen pyltjies: Max/Maks 3�4 • If components are drawn/Indien komponente geteken: max/maks 3�4 (4) 2.3 OPTION 1/OPSIE 1 DOWNWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ AFWAARTS AS POSITIEF OPWAARTS AS POSITIEF Fnet = 0 Fnet = 0 Fnet = ma Any one/ Fnet = ma Any one/ Fgll – F – fk = ma Enige een – Fgll + F + fk = ma Enige een mgsinθ – F – fk = ma –mgsinθ +F + fk = ma 25(9,8)sin30° – F – 40 = 25(0)  –25(9,8)sin30° + F + 40 = 25(0)  F = 82,5 N  F = 82,5 N OPTION 2/OPSIE 2 W net = ∆Ek Any one/ W F + W f + W Fg = 0 Enige een F∆xcosθ + f∆xcosθ + mgsinΦ∆xcosθ = 0 Fcos180o + 40cos180°+ (25)(9,8)sin30°cos0° = 0  F = 82,5 N  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 2.4 Fnet = 0 NOTE: mark only awarded with Fg// = f Any one/Enige een correct substitution of Fgll NOTA:punt toegeken slegs indien Fg// - f = Fnet = 0 korrekte vervanging van Fgll OR/OF f = μkN  mgsinθ = µk mgcosθ  OR/OF µk = mgcosθ  mgsinθ  sinθ µk = cosθ (3) 2.5 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 tanθ = 0,19 tanθ = 0,19 θ = 10,76° θ = 10,76° fk = µkmgcos θ fk = Fgll = mgsin θ = (0,19)(25)(9,8) cos10,76° = (25)(9,8) sin10,76°  = 45,73 N  = 45,73 N  (Range/Gebied: 45,73 – 45,74 N) (Range/Gebied: 45,73 – 45,74 N) (3) [15] QUESTION 3/VRAAG 3 3.1 Motion under the influence of gravitational force only.  Accept weight for gravitational force. Beweging slegs onder die invloed van gravitasiekrag. Aanvaar swaartekrag/gewig vir gravitasiekrag. (2 or/of 0) OR/OF Motion in which the only force acting is gravitational force.  Accept weight for gravitational force. Beweging waar die enigste krag wat inwerk, gravitasiekrag is. Aanvaar swaartekrag/gewig vir gravitasiekrag. (2 or/of 0) NOTE: If projectile is defined: 0�2 LET WEL: Indien projektiel gedefinieer is: 0�2 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 3.2 Marking criteria /Nasienkriteria • Correct formula for Δy = viΔt + ½aΔt2 / Korrekte formule vir Δy = viΔt + ½aΔt2  • Correct substitution into formula for A-B./Korrekte vervanging in formule vir A-B.  • Correct substitution into formula for B-C or A-C./ Korrekte vervanging in formule vir B-C of A-C.  • Equating vf at B and vi at B / Gelykstelling van vf by B en vi by B  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF OPTION 1/OPSIE 1 A-B NOTE/NOTA: A Δy = viΔt + ½aΔt 2 If candidate uses t = 3 s or t = 1 s h Δt h = (0) + ½(9,8)(t)2  Indien kandidaat t = 3 s of t = 1 s h = 4,9 t2 gebruik B A-C max/maks 1�4 Δy = viΔt + ½aΔt 2 9h = (0)(t + 2)+ ½(9,8)(t + 2)2  8h 2s 9h = (4,9)(t + 2)2 9t2 = (t + 2)2 3t = (t + 2) C t=1s OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 A-B A-B Δy = viΔt + ½aΔt2  vf2 = vi2 + 2aΔy h = (0) t + ½(9,8) t2  = (0) + 2(9,8)(h)  h = 4,9 t2 vB = �19,6h B-C Δy = viΔt + ½aΔt2 B-C 8h = vi(2)+ ½(9,8)(2)2  Δy = viΔt + ½aΔt2  = 2vi +19,6 8h = vB(2)+ ½(9,8)(2)2  (8h − 19,6) 8h = 2(�19,6h ) + 19,6 vi = 2 8h - 2�19,6h = 19,6 A-B & B-C vf = vi + aΔt h = 4,9 m vf = 0 + (9,8) t vi(B-C) = vf(A-B) Any one / A-B (8h − 19,6) Enige een Δy = viΔt + ½aΔt2 = (9,8) t 2 4,9 = (0) + ½(9,8)(t)2  8(4,9t2) = (19,6) t + 19,6 t=1s 2t2 −t – 1 = 0 t=1s Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 4/OPSIE 4 OPTION 5/OPSIE 5 A-B Fnet∆t = m(vf – vi) B-C ∆x = vi∆t + ½a∆t2  mg∆t = m(vf – vi) 8h = vi(2) + ½(9,8)(2)2  vf = 9,8t …(1) vi = 4h – 9,8 ……..(1) B-C A-B vf = vi + a∆t ∆y = vi∆t + ½a∆t2  4h – 9,8 = 0 + 9,8t 8h = 9,8t(2) + ½(9,8)(2)2  4h t = 9,8 - 1 …..(2) 8h = 19,6t + 19,6 …(2) A-B A-B ∆x = vi∆t + ½ a∆t2 ∆y = vi∆t + ½a∆t2 h = 0 + ½(9,8)t2  h = 0t + ½(9,8)t2  = 4,9t2 ……..(3) h = 4,9t2 …(3) 4(4,9t2 ) 8(4,9t2) = 19,6t + 19,6  (3) into (2) (3) into (2): t = 9,8 -1 39,2t2 – 19,6t + 19,6 = 0 2t2 - t - 1 = 0 t=1s t=1s UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF OPTION 1/OPSIE 1 A A-B Δy = viΔt + ½aΔt2  NOTE/NOTA: −h = (0) + ½(−9,8)(t)  2 If candidate uses t = 3 s or t = 1 s h Δt h = 4,9 t2 Indien kandidaat t = 3 s of t = 1 s A-C gebruik B Δy = viΔt + ½aΔt2 max/maks 1�4 −9h = (0)(t + 2)+ ½(−9,8)(t + 2)2  8h 2s 9h = (4,9)(t + 2)2 9t2 = (t + 2)2  3t = (t + 2) t=1s C OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 A-B Δy = viΔt + ½aΔt2  A-B −h = (0)t + ½(−9,8) t2  vf2 = vi2 + 2aΔy −h = −4,9 t2 = (0) + 2(-9,8)(h)  B-C Δy = viΔt + ½aΔt2 vB = −�19,6h −8h = vi(2)+ ½(−9,8)(2)2  B-C -8h = 2vi −19,6 Δy = viΔt + ½aΔt2 (−8h + 19,6) vi = −8h = vB(2)+ ½(−9,8)(2)2  2 A-B & B-C −8h = 2(−�19,6h ) – 19,6 vf = vi + aΔt 8h + 2�19,6h = 19,6 vf = 0 + (−9,8) t h = 4,9 vi(B-C) = vf(A-B) Any one / (−8h + 19,6) Enige een A-B = (−9,8) t 2 Δy = viΔt + ½aΔt2  −8(4,9Δt2) = (−19,6)Δt – 19,6 − 4,9 = (0) + ½(−9,8)(t)2  2t2 −t – 1 = 0 t=1s t=1s Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 4/OPSIE 4 OPTION 5/OPSIE 5 A-B B-C: ∆x = vi∆t + ½a∆t2  Fnet∆t = m(vf – vi) -8h = vi(2) + ½(-9,8)(2)2  mg∆t = m(vf – vi) vi = -4h + 9,8 …(1) vf = 9,8t …(1) A-B: vf = vi + a∆t B-C -4h + 9,8 = 0 + (-9,8)t ∆y = vi∆t + ½a∆t2  4h t = 9,8 - 1 ...(2) -8h = -9,8t(2) + ½(-9,8)(2)2  8h = 19,6t + 19,6 …(2) A-B: ∆x = vi∆t + ½ a∆t2 A-B -h = 0 + ½(-9,8)t2  ∆y = vi∆t + ½a∆t2 h = 4,9t2 ...(3) -h = 0t + ½(-9,8)t2  h = 4,9t2 …(3) 4(4,9t2 ) (3) into (2) t = 9,8 -1 8(4,9t2) = 19,6t + 19,6  (3) into (2) t=1s t=1s (4) 3.3 Marking criteria/Nasienkriteria • Formula to calculate vf/Formule om vf te bereken  • Correct substitution/Korrekte vervanging  • Correct final answer/Korrekte finale antwoord: 29,4 m·s-1  DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 A-C A-B vf = vi + aΔt  vf = vi + aΔt A = 0 + (9,8)(3)  = 0 + (9,8)(1) 1s vf = 29,4 m·s-1 vf = 9,8 m·s-1 B-C B vf = vi + aΔt  vf = 9,8 + (9,8)(2)  vf = 29,4 m·s-1 2s OPTION 3/OPSIE 3 A-C A-C Δy = viΔt + ½aΔt 2 vf2 = vi2 + 2aΔy  = (0)(3) + ½(9,8)(3)2 vf2 = (0)2 + (2)(9,8)(44,1) Δy = 44,1 m vf = 29,4 m·s-1 C OPTION 4/OPSIE 4 A-C vi + vf ∆y = � �∆t  2 0 + vf 44,1 = � �(3)  2 vf = 29,4 m·s-1 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 5/OPSIE 5 A-B B-C vf = vi + aΔt vf2 = vi2 + 2aΔy  = 0 + (9,8)(1) vf2 = (9,8)2 + (2)(9,8)(39,2)  vf = 9,8 m·s-1 vf = 29,4 m·s-1 B-C OPTION 6/OPSIE 6 Δy = viΔt + ½aΔt2 B-C = (9,8)(2)+ ½(9,8)(2)2 vi + vf Δy = 39,2 m ∆y = � �∆t  2 9,8 + vf 39,2 = � �(2)  2 vf = 29,4 m·s-1 UPWARDS AS POSITIVE/ OPWAARTS AS POSITIEF OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 A-C A-B vf = vi + aΔt  vf = vi + aΔt = 0 + (−9,8)(3)  = 0 + (−9,8)(1) A vf = −29,4 vf = −9,8 m·s-1 1s vf = 29,4 m·s-1 B-C vf = vi + aΔt  vf = −9,8 + (−9,8)(2)  B vf = −29,4 vf = 29,4 m·s-1 2s OPTION 3/OPSIE 3 A-C A-C Δy = viΔt + ½aΔt2 vf2 = vi2 + 2aΔy  = (0)(3) + ½(−9,8)(3)2 vf2 = (0)2 + (2)(−9,8)( −44,1)  Δy = −44,1 m vf = −29,4 C vf = 29,4 m·s-1 OPTION 4/OPSIE 4 A-C vi + vf ∆y = � �∆t  2 0 + vf −44,1 = � �(3)  2 vf = −29,4 vf = 29,4 m·s-1 OPTION 5/OPSIE 5 A-B B-C vf = vi + aΔt vf2 = vi2 + 2aΔy  = 0 + (−9,8)(1) vf2 = (−9,8)2 + (2)( −9,8)( −39,2)  vf = −9,8 m·s-1 vf = −29,4 B-C vf = 29,4 m·s-1 Δy = viΔt + ½aΔt 2 OPTION 6/OPSIE 6 = (−9,8)(2)+ ½(−9,8)(2)2 B-C Δy = −39,2 m vi + vf ∆y = � �∆t  2 −9,8 + vf −39,2 = � �(2)  2 vf = −29,4 vf = 29,4 m·s-1 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 3.4.1 POSITIVE MARKING FROM QUESTION 3.2 and 3.3 POSITIEWE NASIEN VANAF VRAAG 3.2 en 3.3 Note/Let wel: If 9h = 44,1 m already calculated in 3.2 or 3.3, award 2 marks for answer./Indien h = 44,1 m alreeds in 3.2 of 3.3 bereken, ken 2 punte toe vir antwoord. DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF OPTION 1/OPSIE 1 h = 4,9 OR/OF h = (4,9)(1)2 OR/OF Δy = viΔt + ½aΔt2 h = (0) + ½(9,8)(1)2 9h = 9(4,9)  = 44,1 m OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 Δy = 4,9 + 39,2  Δy = viΔt + ½aΔt2 = 44,1 m = (0)(3) + ½(9,8)(3)2  Δy = 44,1 m  OPTION 4/OPSIE 4 OPTION 5/OPSIE 5 vi + vf vf2 = vi2 + 2aΔy ∆y = � �∆t (29,4)2 = (0)2 + (2)(9,8)Δy  2 0 + 29,4 Δy = 44,1 m =� �(3)  2 Δy = 44,1 m  OPTION 6/OPSIE 6 (Emech)i = (Emech)f 1 0 + m(9,8)h = 2m(29,4)2 + 0  h = 44,1 m  DOWNWARDS AS NEGATIVE/AFWAARTS AS NEGATIEF OPTION 1/OPSIE 1 h = 4,9 OR/OF h = (4,9)(1)2 OR/OF Δy = viΔt + ½aΔt2 h = (0) + ½(-9,8)(1)2 9h = 9(4,9)  = 44,1 m OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 Δy = 4,9 + 39,2  Δy = viΔt + ½aΔt2 = 44,1 m = (0)(3)+ ½(-9,8)(3)2  Δy = −44,1 Δy = 44,1 m  OPTION 4/OPSIE 4 OPTION 5/OPSIE 5 vi + vf vf2 = vi2 + 2aΔy ∆y = � �∆t (-29,4)2 = (0)2 + (2)(-9,8)Δy  2 0 + (-29,4) Δy = −44,1 =� �(3)  Δy = 44,1 m 2 Δy = −44,1 Δy = 44,1 m  OPTION 6/OPSIE 6 (Emech)i = (Emech)f 1 0 + m(9,8)h = 2m(-29,4)2 + 0  h = 44,1 m  (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 3.4.2 ΔyBC = 39,2 m (1) 3.5 POSITIVE MARKING FROM QUESTIONS 3.3 and 3.4 POSITIEWE NASIEN VANAF VRAAG 3.3 en 3.4 DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF Marking criteria: • Initial position -44,1 m AND ball strikes the 1 ground at 3 s  3 position (m) 0 • Position at 1s is -39,2 m Time (s) • Correct shape  • If ground is not used as zero max 3�4 • If axis incorrectly/not labelled max 3�4 Nasienkriteria: • Aanvanklike posisie -44,1 m EN bal tref die grond by 3 s  −39,2 • Posisie na 1s is -39,2 m −44,1 • Korrekte vorm  • Indien grond nie as nul gebruik is nie maks 3�4 • Indien asse verkeerd/nie benoem maks 3�4 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF Marking criteria: • Initial position 44,1 m AND ball strikes the 44,1 ground at 3 s  39,2 • Position at 1s is 39,2 m • Correct shape  position (m) • If ground is not used as zero max 3�4 • If axis incorrectly/not labelled max 3�4 Nasienkriteria: • Aanvanklike posisie 44,1 m EN bal tref die grond by 3 s  0 1 3 • Posisie na 1s is 39,2 m Time (s) • Korrekte vorm  • Indien grond nie as nul gebruik is nie maks 3�4 • Indien asse verkeerd/nie benoem maks 3�4 (4) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. IF: closed system -1 mark INDIEN: geslote sisteem – 1 punt The total (linear) momentum in an isolated system is conserved/remains constant.  Die totale (lineêre) momentum in 'n geïsoleerde sisteem bly behoue/konstant. Accept for 1 mark/Aanvaar vir 1 punt In a isolated system the total momentum before a collision is equal to the total momentum after a collision.  In 'n geïsoleerde sisteem is die totale momentum voor 'n botsing gelyk aan die totale momentum na 'n botsing. (2) 4.2 Δx = viΔt + ½a∆t2 0,8 = vi(2)  vi = 0,4 m·s-1 RIGHT AS +/REGS AS + LEFT AS +/LINKS AS + OPTION 1/OPSIE 1 ∑pi = ∑pf ∑pi = ∑pf Any one/ m1v1i + m2v2i = m1v1f + m2v2f Enige een m1v1i + m2v2i = m1v1f + m2v2f Any one/ Enige een m1v1i + m2v2i = (m1 + m2)vf m1v1i + m2v2i = (m1 + m2)vf (2)(v1i) + 0 = (2 + 1,5)(−0,4) (2)v1i + 0  = (2 + 1,5)(0,4) V1i = -0,7 vi1 = 0,7 m·s-1  v1i = 0,7 m·s-1  OPTION 2 /OPSIE 2 Any one/ Δp1 = − Δp2 Any one/ − Δp1 = Δp2 Enige een m(v1f – v1i) = − m(v2f – v2i) Enige een − m(v1f – v1i) = m(v2f – v2i) 2(0,4 − v1i) = − 1,5(0,4 − 0)  −2(0,4 − v1i) = 1,5(−0,4 − 0)  v1i = 0,7 m·s-1  v1i = 0,7 m·s-1  (5) 4.3 Inelastic / Onelasties (1) 4.4 DECREASES/NEEM AF  Δp 1 FnetΔt = Δp OR/OF Fnet = Δt OR/OF FnetΔt = mΔv OR/OF Fnet α ∆t  (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. IF: The word 'work' is omitted: 0 marks. INDIEN: Die woord 'arbeid' uitgelaat is: 0 punte. A non-conservative force is a force for which the work done (in moving an object between two points) is dependent on the path taken.  'n Nie-konserwatiewe krag is 'n krag waarvoor die arbeid wat verrig is (om 'n voorwerp tussen twee punte te beweeg) afhanklik is van die pad wat gevat word. (2) 5.2 ACCEPT/AANVAAR Fpump Fpump w w Accepted labels/Aanvaarde benoemings Fpump F/FP/Felectric pump/F/1 960 N/Fapplied/Fup/Fp/Felectriese pomp/Ftoegepas/Fop w Fg/Fw/weight/mg/gravitational force/1 960 N/gewig/gravitasiekrag Notes/Aantekeninge • Mark is awarded for label and arrow./Punt word toegeken vir byskrif en pyltjie. • Do not penalise for length of arrows./Geen penalisasie vir die lengte van pyltjies • If arrows do not touch the dot/Indien pyle nie die kolletjie raak nie: Max/Maks 1�2 • Any other additional force(s)/Enige ander addisionele krag(te): Max/Maks 1�2 • If everything correct, but no arrows/Indien alles korrek, maar geen pyltjies: Max/Maks 1�2 (2) 5.3.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 W nc = ∆Ep + ∆Ek Any one/ Fpump = w = 1960 N Wpump = mg∆h / mgh Enige een Wpump = FpumpΔxcos θ  = 200(9,8)(12,5)  = (1960)(12,5)(1)  = 24 500 J = 24 500 J OPTION 3/OPSIE 3 W net = ∆Ek Any one/ Wpump + WFg = 0 Enige een Wpump + 200(9,8)(12,5)cos 180° = 0  Wpump = 24 500 J  (3) 5.3.2 NOTE: Q5.3.3 can be done first for Q5.3.2. Credit the candidate NOTA: Q5.3.3 kan eerste gedoen word vir V5.3.2. Ken punte toe OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 200 200 ∆t = 2,5  = 80 s ∆t = 2,5  = 80 s ∆x v = (v + v ) ∆t ∆x = i f ∆t 12,5 2 =  2v 80 12,5 = (80)  2 = 0,16 m∙s-1 (0,156 m∙s-1) v = 0,16 m∙s-1 (0,156 m∙s-1) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 5.3.3 POSITIVE MARKING FROM Q5.3.1 and Q5.3.2 / POSITIEWE NASIEN VANAF V5.3.1 en V5.3.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 W P = ∆t  Pave = Fvave  24500 = (200)(9,8)(0,16)  = 80  = 313,60 W  = 306,25 W  (Range/Gebied: 305,76 W - 313,60 W) (3) [13] QUESTION 6/VRAAG 6 6.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. Doppler effect/ Doppler effek The change in frequency (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 frekwensie (toonhoogte) van die klank waargeneem deur 'n luisteraar omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium waarin die klank voortgeplant word, het. OR/OF A change in observed/detected frequency (pitch), as a result of the relative motion between a source and an observer (listener). 'n Verandering in waargenome frekwensie (toonhoogte), as gevolg van die relatiewe beweging tussen die bron en 'n waarnemer (luisteraar). (3) 6.2 v = fL λ 340 = fL x 0,016  fL = 21 250 Hz v ± vL v v ± vL fL = fS  OR/OF fL = fS OR/OF fL = fS v ± vs v − vs v ± vB 340 + 0 21 250 = � �fS  340 – vB fS = 21 250 – 62,5vB (4) 6.3 POSITIVE MARKING FROM Q6.2 for 21 250 Hz for fs POSITIEWE NASIEN VANAF Q6.2 vir 21 250 Hz vir fs fLꞋ = 21 250 – 62,5vB + 850 NOTE/NOTA: Any one/Enige een = 22 100 – 62,5vB fLꞋ is the frequency detected by the bat after reflection v ± vL v + vL off the cliff. fLꞋ = fS OR/OF fLꞋ = fS v ± vs v fLꞋ is die frekwensie   waargeneem deur die 340 + vB vlermuis na weerkaatsing [22 100 - 62,5vB] = (21 250)  vanaf die krans 340 - 0 vB = 6,8 m·s-1  (5) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1 + Criteria for sketch/Kriteria vir skets: Correct shape (straight field lines)./Korrekte vorm (reguit veldlyne).  Correct direction away from the charge./Korrekte rigting weg van die  lading. Notes/Aantekeninge: • Sphere with field lines inside OR If electric field lines cross or touch/ Sfeer met veldlyne binne-in OF Indien elektriese veldlyne kruis of raak: Max/Maks 2�3 • If negative sign shown/Indien negatiewe teken getoon: Max/Maks 2�3 • If more than one charge or combined field drawn/Indien meer as een lading of gekombineerde veld geteken is: 0�3 (3) 7.2.1 Marking criteria/Nasienkriteria: 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 word/frase. 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 charges  and inversely proportional to the square of the distance (r) between them.  Die grootte van die elektrostatiese krag wat een puntlading (Q1) op 'n ander puntlading (Q2) uitoefen, is direk eweredig aan die produk van die ladings en omgekeerd eweredig aan die kwadraat van die afstand (r) tussen hulle. OR/OF The magnitude of the electrostatic force exerted by one charge (Q1) on another charge (Q2) is directly proportional to the product of the (magnitudes) of the charges  and inversely proportional to the square of the distance (r) between their centres.  Die grootte van die elektrostatiese krag wat een lading (Q1) op 'n ander lading (Q2) uitoefen, is direk eweredig aan die produk van die ladings en omgekeerd eweredig aan die kwadraat van die afstand (r) tussen hulle middelpunte. NOTE: If masses are mentioned instead of charges: 0�2 LET WEL: Indien massas in pleks van ladings genoem word: 0�2 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 7.2.2 Accept/Aanvaar T T FE FE w w Accepted labels/Aanvaarde byskrifte w Fw/Fg/mg/1,47 x 10-3 N/gravitational force/gravitasiekrag/weight/gewig F/Electrostatic force/Coulomb force/FE Field/FR on S / FEveld/FR op S / FR FE  Elektrostatiese krag/Coulomb-krag T  FT/Fstring/tension/spanning/Ftou Notes/Aantekeninge: • Mark awarded for label and arrow./Punt toegeken vir benoeming en pyltjie. • Do not penalise for length of arrows since drawing is not to scale./Moenie vir die lengte van die pyltjies penaliseer nie aangesien die tekening nie volgens skaal is nie. • Any other additional force(s)/Enige ander addisionele krag(te): Max/Maks 2�3 • If everything is correct, but no arrows/Indien alles korrek is, maar geen pyltjies: Max/Maks 2�3 • If force(s) do not make contact with the dot /Indien krag(te) nie met die kolletjie kontak maak nie: Max/Maks 2�3 (3) 7.2.3 kQ1 Q2 F=  r2 9 -9 -9 � 9 x 10 �� 6 x 10 �� 3 x 10 � =  0,032 =1,8 x 10-4 N w = mg = (1,5 x 10-4)(9,8)  = 1,47 x 10-3 N FE 2 2 2 tanθ = OR/OF T = (FE ) + (Fg ) mg -4 = (1,47 x 10-3)2 + (1,8 x 10-4)2 1,8 × 10 = -3  1,47 × 10 T = 1,481 x 10-3 N θ = 6,98° 1,47 x 10-3 1,8 x 10-4 θ T cos θ = -3  OR/OF sin θ = Fg 1,481 x 10 1,481 x 10-3 θ = 6,98°  FE (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The potential difference (voltage) across a conductor is directly proportional to the current in the conductor at constant temperature.  Die potensiaalverskil (spanning) oor ‘n geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur. OR/OF The current in a conductor is directly proportional to the potential difference (voltage) across the conductor if the temperature is constant.  Die stroom in ‘n geleier is direk eweredig aan die potensiaalverskil (spanning) oor die geleier indien die temperatuur konstant is. OR/OF The ratio of potential difference to current for a conductor is constant provided the temperature is constant.  Die verhouding van potensiaalverskil tot stroom vir ‘n geleier is konstant indien die temperatuur konstant bly. (2) 8.2.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 V = IR V13 = VL V = 0,8(13)  0,8(13) = IL(9,5)  = 10,4 V IL = 1,09 A  V = IR 10,4 = IL(9,5)  IL = 1,09 A  OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 R IR13 IRL = IR13 � R13� (9,5) = 0,8  L 22,5 IR13 = 1,89 A 13  IRL = (0,8)� � 9,5  IL = 1,89 – 0,8  = 1,09 A  = 1,09 A  (3) 8.2.2 POSITIVE MARKING FROM QUESTION 8.2.1 / POSITIEWE NASIEN VANAF VRAAG 8.2.1 OPTION /OPSIE 1 OPTION 2/OPSIE 2 Vp = V9Ω OR/OF Vp = V9Ω R 10,4= I9Ω(9)  I1 R13 = I9ΩR9Ω IR9Ω = IR � 13 � 13 R9 I9Ω = 1,16 A (0,8)(13) = I9Ω(9)   I9Ω = 1,16 A 13 IR9Ω = (0,8)� �  9 = 1,16 A Itotal = I13Ω + IL + I9Ω l = 0,8 + 1,09 +1,16  = 3,05 A  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3 R1 R2 13(9,5) Rp1 = � �= V = IR R1 + R2 13 + 9,5 V = 0,8(13)  Rp1= 5,49 Ω = 10,4 V V = IR Vp = V9 10,4 = I(5,49) 10,4 = I(9)  I = 1,89 A I = 1,156 A Itotal = 1,156 + 1,89  = 3,05 A  (4) 8.2.3 POSITIVE MARKING FROM Q 8.2.2. / POSITIEWE NASIEN VANAF V 8.2.2. OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 ε = I(R + r) ε = I(R + r)  OR/OF V = IR ε = Vext + Vint  Any one/ 24 = 3,05(Rext + 0,3)  ε = Vext + Ir enige een Rext = 7,57 Ω V = IR 1 1 1 1 R1 R2 24 = Vext + 3,05(0,3)  = + + OR/OF Rp1 = � � Rp R1 R2 R3 R1 + R2 Vext = 23,09 V 13(9,5) Vext = Vp + VR 1 1 1 1 = = + +  13 + 9,5 23,09 = 10,40 + VR Rp 13 9,5 9 Rp1= 5,49 Ω VR = 12,69 V Rp = 3,41 Ω 5,49(9) Rp2= 5,49 + 9  V = IR 12,69 = (3,05)R Rp1= 3,41 Ω R = 4,16 Ω R =R +R ext p (Range: 4,157 Ω – 4,161 Ω) 7,57 = 3,41+ R R = 4,16 Ω (5) 8.3 Decreases/Neem af  The total/external resistance increases/Totale/eksterne weerstand neem toe (2) 8.4 Increases  • VR / IR decreases  thus • VLost / Ir decreases AND VL increases  (and emf stays constant) V2 • RL constant AND P α V2 / P = R  therefore L is brighter OR • RL constant AND IL increases AND P α I2 / P = I2R therefore L is brighter Neem toe • VR / IR neem af, daarom • Vverlore / Ir neem af EN VL neem toe (terwyl emk konstant bly) V2 • RL konstant EN P α V2 / P = R daarom is L helderder OF • RL konstant EN IL neem toe EN P α I2 / P = I2R daarom is L helderder Accept for this exam ONLY: calculation of P/ Aanvaar SLEGS vir hierdie eksamen: berekening van P (4) 8.5 C, D and/en F OR/OF A1, A2 and V9Ω Mark first three answers /Merk eerste drie antwoorde (3) [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1.1 North/Noord  (2) 9.1.2 ANY ONE of the following/ENIGE EEN van die volgende • Increase speed of the rotation/Verhoog spoed van rotasie ACCEPT/AANVAAR • Increase number of coils (turns)/Meer windings • Use magnet with stronger field/stronger magnet/increase magnetic flux/electromagnets/ insert soft iron core/Gebruik magneet met sterker veld/sterker magneet/verhoog magnetiese vloedkoppeling/ elektromagneet/voeg sagte ysterkern by (1) 9.1.3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Induced current(A) Induced current(A) 0 0 Time (s) Time (s) Criteria for graph/Kriteria vir grafiek AC-graph starts with maximum current/ WS-grafiek begin met maksimum  stroom. Graph drawn for one cycle/one complete wavelength./  Grafiek geteken vir een siklus/een volledige golflengte. Correct shape (cosine graph)./Korrekte vorm (cosinus grafiek)  If one cycle of a sin-graph drawn/Indien sin-grafiek geteken: max/maks 1�3 (3) 9.2.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. Root-mean-square current is the alternating current (AC) that dissipates the same amount of energy/heating effect as an equivalent direct current (DC)  Die wortelgemiddeldekwadraat-stroom is die wisselstroom (WS) wat dieselfde hoeveelheid energie verbruik/verhittingseffek het as 'n ekwivalente gelykstroom (GS) het. NOTE: If learner writes DC first / If energy or heating effect is omitted / If learner defines potential difference: 0�2 LET WEL: Indien leerder GS eerste skryf / energie of verhittingseffek uitgelaat is / Indien leerder potensiaalverskil definieer : 0/2 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 19 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 9.2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 Vmax I ×V Pave = max/maks 2 max/maks Vmax Vrms = Vrms = √2 Imax/maks × 340 √2 340 1200 =  340 =  2 =  √2 Imax/maks = 7,06 A √2 = 240,42 V Imax = 240,42 V Irms =  √2 V2rms Pave = VrmsIrms  7,06 Pave = R =  1200 = (240,42) Irms √2 240,422 1200 = R Irms = 4,99 A  = 4,99 A  R = 48,17 Ω V = IR  240,42 = I(48,17)  I = 4,99 A  (4) NOTE: Penalise for subscripts not shown/ LET WEL: Penaliseer indien voetskrifte nie getoon nie 9.2.3 POSITIVE MARKING FROM QUESTION 9.2.2 / POSITIEWE NASIEN VANAF VRAAG 9.2.2. OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 NOTE: Vrms Vmax Options 1 and 2 subscripts R=  R=  not necassary Irms Imax NOTA: 240,42 340 Opsies 1 en 2 voetskrifte =  =  4,99 7,06 nie nodig nie R = 48,18 Ω R = 48,18 Ω  OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 NOTE: 2 Vrms Pave = Irms2R If R calculated in Q9.2.2 Pave =  1200 = (4,99) R 2 (option 3) give credit. R R = 48,19 Ω NOTA: 240,422 Indien R reeds bereken in 1200 =  R V9.2.2 (opsie 3) gee krediet R = 48,17 Ω (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P1/Fisiese Wetenskappe V1 20 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The minimum energy (of incident photons) that can eject electrons from a metal/surface.  Die minimum energie (van invallende fotone) wat elektrone kan vrystel vanuit 'n metaal/oppervlak. NOTE: If reference to frequency: 0�2 LET WEL: Indien na frekwensie verwys word: 0�2 (2) 10.2 1  (2) 10.3.1 1 Ek(max) = 2 mv2max  2,99 x 10-19 = ½(9,11 x 10-31)v2max  vmax = 8,10 x 105 m∙s-1  (3) 10.3.2 OPTION 1/OPSIE 1 POSITIVE MARKING FROM Q10.2 E = W 0 + Ek(max) POSITIEWE NASIEN VANAF Q10.2  Any one/Enige een OPTION 2/OPSIE 2 hf = W 0 + Ek(max) ∆Ek(max) = 3,68 x 10-19 + 2,99 x 10-19  gradient =  ∆E E = 6,67 x 10-19 (J)  -19 2,99 x 10 -0  1= Y - 3,68 x 10-19  Y = 6,67 x 10-19  (4) 10.4 Remains the same/Bly dieselfde  The energy of the incident photons is not changing./Die energie van die invallende fotone verander nie  OR/OF (Kinetic energy is affected by/depends on the frequency of the photons and) the frequency remains the same/(Kinetiese energie word beïnvloed deur/ hang af van die frekwensie van die fotone en) die frekwensie bly dieselfde (2) [13] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou