Physical Sciences Grade 12 Paper 1 KZN June 2025 Memo

) KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE fh PHYSICAL SCIENCES P1 : 2025 JUNE EXAMINATION 1 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 11 pages. Copyright reserved Please turn over tS

2025 June examination QUESTION 1 1.1 Dvv (2) 1.2 Cvv (2) 1.3 AVY (2) 1.4 Byv (2) 1.5 Cvv (2) 1.6 Cvv (2) 4.7 AYY (2) 1.8 Dvv (2) 1.9 Cvv (2) 1.10 Byy (2) [20] QUESTION 2 251 Force that opposes the motion of a moving object relative to the surface. vv (2) 2:2 Accept force diagram: N N E F. ij f Ww w Accepted labels Fg/ Fw/ weight/gravitational forcev F./applied forceY (kinetic) friction/Fi/fky Fn /Normal/Fnormat ¥ Z\—>|\n/<=

2.3.1 2.3.2 2.4 2025 June examination Notes * Mark awarded for label and arrow. ¢ Do not penalise for length of arrows since drawing is not to scale. e Any other additional force(s): Max 3 4 e If everything correct, but no arrows: Max/Maks A Marking criteria If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark A body will remain in its state of rest or motion at constant velocity unless a non-zero resultant/net force acts on it.v ¥ Fret = ma Y Anyone Fcos© + mgsin © + fk = ma} i Fcos30° - (15)(9,8)sin30° v- 0,2v v[Fsin30° + (15)(9,8)cos30°)] v = Ov F=129,19N Vv OPTION 1 w=mg GMe 3800 = m(7,32) ¥ (2 m = 518,88 kg _ (6,67*107"')((5,98«10"4) (6,38x10°+1000«10°) w=mg 7,32 ms? = (518,88)(9,8) v = 5085,02 N v OPTION 2 re GmMg F R2 3800 r= oe * (R=1X106)2 3800 ( 6,38x10® + 1x10°)? = F(6,.38x108)2 v F=5084,58NY OPTION 3 Tor _ _GmMg 3800 = (R+1X10°)2 (6,67X10711)(m)}(5,98 x 1024) “ (6,38X 10° + 1X109)2 3800 = 1 | m = 518,88 kg w.=mg ¥ = (518,88)(9,8) ” = 5085,02 N v (4) (2) (6) (5) [19]

2025 June examination QUESTION 3 a 50 NY downwardsY 3.2.1 OPTION 1 (Accept if 0? is omitted) | OPTION 2 (Accept if 0? is omitted) UPWARDS AS POSITIVE UPWARDS AS NEGATIVE v2 = v2 + 2aly ¥ VP = v2 + 2aly ¥ = (0)? + (2)(-9,8)(-1,2) = (0)2 + (2)(9,8)(1,2) V vie = 4,85 m:s! v downwardsY vi = 4,85 m:s! Y¥ downwardsY 3.2.2 OPTION 1 OPTION 1 UPWARDS AS POSITIVE UPWARDS AS NEGATIVE POSITIVE MARKING FROM Q 3.2.1 | POSITIVE MARKING FROM Q 3.2.1 Ve = vi + aAt ¥ vr = vit alt -4,85 = 0 + (-9,8) At ]v 4,85 = 0 + (9,8) At Jv At=0,49s v At=0,49s v OPTION 2 OPTION 2 UPWARDS AS POSITIVE UPWARDS AS NEGATIVE POSITIVE MARKING FROM Q 3.2.1 | POSITIVE MARKING FROM Q 3.2.1 {7% _ (ity 7 + a = e ¥ = = v (1,2=(—S=)ag [1.2=(—)at) = v_ At=0.49 sv _ OPTION 3 OPTION 3 UPWARDS AS POSITIVE UPWARDS AS NEGATIVE Ay = viAt+ 3a At v Ay =viAt+ a ae v (£1,2=0+3(-9,8) Ary [1,2 = 0 +5(9,8) At] v At=049sv¥ At=0,49s v 32.3 OPTION 1 OPTION 1 UPWARDS AS POSITIVE v2 = vi2 + ZaAy ¥ 0 =v? + (2)(-9,8)(0,8) ¥ vi = 3,96 ms! ¥ UPWARDS AS NEGATIVE vi? = vi? + 2aly v 0 =v2 + (2)(9.8)(-0,8) ¥ vi =-3,96 ms vi = 3,96 ms? v OPTION 2 (U + K)top = (U + K)pbottom ~ (mgh + SMV"top = (mgh + SMV2)ootiom [(0,5)(9,8)(0,8) + 0 = 0 +(5)(0,5) vv v= 3,96 ms'v (2) (4) (3) (3)

3.2.4 2025 June examination FretAt = m(vr- vi) ¥ Frioor + Fg = m(vr- Vi) At = 0,098 s v POSITIVE MARKING FROM QUESTIONS 3.2.1 & 3.2.3 50 —(0,5)(9,8)At V = 0,5[3,96 — (-4,85)] v 3.3 POSITIVE MARKING FROM Q 3.2.2 i) 0,49 0,59 time (S) QUESTION 4 4.1 Marking criteria If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark In an isolated system the total linear momentum is conserved/remains constant. Vv 4.2 TO THE RIGHT AS POSITIVE 4 1,2 position (m) eeees 0.8 Criteria Correct shapev Graph starts at 1,2 m att = Ov Second maximum height at 0,8 mv Time at which ball reaches the floorv Time at which ball leaves the floory dpi = Tpr v= 6 mst ¥ MbVib + MbiVib| = (Mb + Mbi)vF } v Any one (0,006)(200) + 0 v = (0,006 + 0,194) vi vw TO THE RIGHT AS NEGATIVE Dpi = dpr vr = -6 ms? =6 ms ¥ MbVib + MbiVibI = (Mb + Moi)vs } “Any one (0,006)(-200) + 0 v = (0,006 + 0,194) viv (4) (5) [21] (2) (4)

2025 June examination 4.3 — OPTION 1 vie = vi2 + 2aAx 02v = (6)? + 2a(5) ¥ a= -3,6 m-s? = 3,6 ms”, to the left v I ___ OPTION 2 ' Whet = AK fAxcos@ = Yam(vr — vi?) f(5)(cos 180°) = %4(0,2)(02 -62) V #=0,72N -0,72 = 0,2av a=-3,6 = 3.6 m-s®, to the left ¥ 4.41 Ww Accepted labels W | Fo/ Fw/ weight/gravitational forcev f_| (kinetic) friction/Fi/fk~” N_| Fn/Normal/Frormai ¥ Notes e Mark awarded for label and arrow. ¢ Do not penalise for length of arrows since drawing is not to scale. e Any other additional force(s): Max * e If everything correct, but no arrows: Max/Maks 74 4.4.2. Normal forcev 443K =!mvv = 5 (60)(4)¥ = 480 Jv 4.44 Fro: = mgsin@ ~ f (60)(9.8)(sin25°) — 180 v 68,5 Nv ot (4) (3) (1) (3) (2)

2025 June examination 4.4.5 Marking criteria If any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark Work done on an object by a net force is equal to the change in object's kinetic energy. v ¥ (2) 44.6 POSITIVE MARKING FROM QUESTIONS 4.4.3 & 4.4.4 Whet = AK ¥ FnetAt = Kr - Ki (68,5)(15)cos0°V =[ 3(60)v?—- 480 Jv v=7,09 ms? v (4) [25] QUESTION 5 5.1. Rate at which work is done / energy is expended. vv (2) 52.4 p=“v At (0,85)(200) v = Zv W = 5100 Jv (4) 5.2.2 U=mghv = (90)(9,8)(3) ¥ = 2646 Jv (3) (5) [14]

2025 June examination QUESTION 6 6.1 Doppler Effect. v The change in frequency of sound detected by a listener because the sound source and the listener have different velocities relative to the medium of sound propagation. vv (3) 6.2 Doppler flow meter / Doppler ultrasound machinev (1) 6.3 flL=—— -f,v V-Vs __ 340 1695v = Wow (1500) v Vs = 39,12 m-s"! OPTION 1 a OPTION 2 \ _ ak __ (39,12 + 39,12 =a Ax = (722) (1,5) 39,12=5v = 58,68m Ax = 58,68 m ¥ (5) 4 64 £ (Hz) Criteria 4500 e Correct shapev ; e 1,5s and 1500 Hzv 15 te (2) [11]

2025 June examination QUESTION 7 ea Marking criteria lf any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark The (magnitude of the) electrostatic force exerted by one point charge on another point charge is directly proportional to the product of the (magnitudes of the) charges and inversely proportional to the_square of the distance between them. vv (2) 7.2 OPTION 1 OPTION 2 F = Srey Ea e 2 (9x10°)(3x10%)(9x10°) (9x10°)(3x10°) = v Exy= +,” (0,15)2 (0,15)? = 10,8N = 1,2%10° N.C" _ (9*10°)(6x10°)(9x10%) (9«10°)(6x10°) z= (0,1) Eyz= “one v =46,6.N = 5,4x10° NC" v, : n= ,6-10,8 = tv g Fret = 48,6~10,8 = 37, 8Ntothe right “|. 5,4x10° 1.20108 = 4,2x10° NC Fre = QE wv = (9x10°)(4,2x108v = 37,8 N to the righty (5) 7.3.1 Xto YY (1) 7.3.2 QgaMit® - (-3x10°)+(6x10°) / 2 = 1,5x10°C n=2¥ de _ (1,510) -(ax10*) ~ 1.61074 (4) = 2,81x10" (electrons) ” 7.3.3 es v (9x10°)(1,5x10°) | _ (9x10°)(9x10°) | r2 (0,15 - r)2 r=0,04mv (4) [16]

2025 June examination QUESTION 8 8.1 Marking criteria lf any of the underlined key words/phrases in the correct context is omitted, deduct 1 mark The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. Vv (2) 8.2.1 OPTION 1 OPTION 2 ee Rua Re Rp R, Ro P R,+Ro _1.1 _ 13*7 e “a7 “1347 Rp =4,550v = 4,55 Ov (3) 8.2.2 POSITIVE MARKING FROM QUESTION 8.2.1 €=I(Rtr) v 12 = 1(4,55 + 0,3)v |= 2,47 Av (3) 8.2.3 Vexternal = IR = (2,47)(4,55) v =11,24V I through 8 Q Vv l=av - 11,24 Vv 13 = 0,8646Av (4) 8.3 POSITIVE MARKING FROM QUESTION 8.2.3 Vsa=IR Isq = 2,47 — 0,8646 = 1,6054 = (0,8646)(5) v V30 = IR = 4,323 V = (1,6054)(3) v =4,816V No = 4,816 % 4,323 = 0,49 Vv (Range: 0,49 — 0,53 V) (4) 8.4 e Total resistance increases Vv e Total current decreases Vv © Vinternal (lost volts) decreases V (3)

8.5 Cost = energy x tariff 300 “ = (P x 8 x 30) x 2,59 P = 482,625 W v 482,625 = v no. of bulbs 80 =6v 2025 June examination TOTAL: (5) [24] 150