Physical Sciences Grade 10 P1 National November 2016 Memo

Downloaded from hlayiso.com “mie basic education wg VESraG Department: ] th Basic Education SZ REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 10 SITE RTT TT Tt IT TTL ee PHYSICAL SCIENCES: PHYSICS (P1) i FISIESE WETENSKAPPE: FISIKA (V1) " NOVEMBER 2016 A MEMORANDUM : eee 2 eee ee ee ne en ne ne nn ne nn nn ne ne nn ne ne ne en ee 2” MARKS/PUNTE: 150 This memorandum consists of 14 pages. Hierdie memorandum bestaan uit 14 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 2 DBE/November 2016 NSC/NSS —- Grade/Graad 10 — Memorandum QUESTION 1/VRAAG 1 14 Byy (2) 1.2 Cv (2) 1.3 Dvv (2) 14 Dvv (2) 15 Dvv (2) 16 Avv (2) 17 BYyY (2) 18 CYY (2) 1.9 DvY (2) 110 BY (2) [20] QUESTION 2/VRAAG 2 2.4 Mark allocation/Puntetoekenning: csv 1 x line AB: length, arrow, label 1 x lyn AB: lengte, rigting, benoem v¥ 1xline BC: length, arrow, label 1 x lyn BC: lengte, rigting, benoem (2) 2.2 2m VY west/to the leftv 2mv_ wes/na linksY (2) OR If the learner has done 2.2 on the vector diagram above, allocate 1 mark for the resultant from A to Cv and 1 mark for the label of the magnitude of 2m. As leerder 2.2 op die bostaande vektordiagram aangedui het, ken 1 punt vir die resultant van A tot C toe, en 1 punt vir die benoeming van 2m 2.3 Total distance/Totale afstand =5+7Vv =12mv (2) 2016 -1i- {1 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief he

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 3 DBE/November 2016 NSC/NSS — Grade/Graad 10 — Memorandum 2.4 For the total distance, the whole path length travelled is considered. For change in position, only the original position and final position“ of the man are considered. (2) Vir die totale afstand word die totale padiengte afgelé in berekening gebring, maar slegs die begin- en eindposisie word in berekening gebring vir verandering in posisie. 2.5 Velocity is the rate of change of displacement. ¥¥ Snetheid is die tempo waarteen verplasing (verandering in posisie) verander. (2) ACCEPT: Change in displacement over change in time AANVAAR: Verandering in verplasing gedeel deur verandering in tyd 2.6 POSITIVE MARKING FROM 2.2 POSITIEWE NASIEN VANAF 2.2 Ax =—Vv vet 2v ~ 20v =0,1m-s" west/to the left” wes/na links (4) [14] QUESTION 3/VRAAG 3 3.1 Acceleration is the rate of change of velocity. “vv” Versnelling is die tempo van snelheidsverandering. ORIOF Acceleration is the change in velocity per unit timev”. Versnelling is die verandering in snetheid per tydseenheid. (2) 3.2 Nov Nee (1) 3.3 NEGATIVE MARKING FROM 3.2 NEGATIEWE MERK VANAF 3.2 Velocity to the right, acceleration to the left ¥ 2016 -11- 11 Snetheid na regs, versnelling na links. OR/OF Taxi slowing down so acceleration is in opposite direction” to movement. Die taxi beweeg stadiger, dus is versnelling in die teenoorgestelde rigting van beweging. 1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief HE

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 4 DBE/November 2016 NSC/NSS — Grade/Graad 10 —- Memorandum 3.4 OPTION 1/OPS/E 1 OPTION 2/OPSIE 2 D (ve + vi) v =a or D=vxAtv Ax= > At ¥ D 25 +25 av = Tv =25x1v¥ = y¥x1Vv D =25mv =25mv =25mv OPTION 3/OPSIE 3 AX = vit + Zadt? =25x1¥ +%xOx tv =25mv (4) 2016 -i1- 11 Please turn over/Blaai om _ At Copyright reserved/Kopiereg voorbehou

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 5 NSCI/NSS — Grade/Graad 10 - Memorandum DBE/November 2016 3.5 POSITIVE MARKING FROM 3.4 POSITIEWE NASIEN VANAF 3.4 distance > 40 m nie, want die afstand is > 40 m OR / OF 0+25 = (025) oy 2 =25mv .. taxi will not stop at the traffic light as . die taxi sal nie V by verkeerslig stop .. distance available for braking time is 40 — 25 = 15 m, therefore the taxi is short of 10 m to stop. ” ..the taxi will not stop in time. ” ..die beskikbare afstand vir briektyd is 40-25 = 15m, dus sal die taxi10m te min hé om fe stop. -.dus sal die taxi nie betyds stop nie OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 ax =(““) av Vp =v; + aAtY 2 _ VE- Vi Only one mark ~ for either = (0425) x 2v at equation 7 _ (0 - 25) v Slegs een punt =25m a= 2 vir die enige een van die = “2 lykings. .. total distance/totale afstand = -12,5 ms — = 25+ 25V =50mv ve =v? + 2aAx 0 = 257 +2 x -12,5 x Ax¥ “Ax = 25m .. total distance/fotale afstand =25+25 = 50 mv ..taxi will not stop at the traffic light as distance > 40 mv .. die taxi sal nie by verkeerslig stop nie, want die afstand is > 40 m =25m . total distance/fofale afstand = 25+ 25 = 50 mv OPTION 3/OPSIE 3 = ty a At Only one mark for either = (0-25) y equation = 125 m-s? Slegs een punt vir enige een van Ax = vit+ Yadt? die vergelykings. =25x2 +%x-12,5x2?¥ -. taxi will not stop at the traffic light, as distance > 40 mY .. die taxi sal nie betyds stop nie, want die afstand is > 40 m (5) Copyright reserved/Kopiereg voorbehou Loe Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 6 DBE/November 2016 NSC/NSS — Grade/Graad 10 - Memorandum OPTION /OPSIE 4 _(t +Mi) Ay AtY 15 = yx aty At =1,2sv Actual time to reach traffic light is 1,2s which is shorter than the 2s. Therefore the taxi will not stop in time. ~ Ejintlike tyd of verkeerslig te bereik is 1,2s wat korter is as 2s. Dus sal die taxi nie betyds stop nie. 3.6 x (m) 50 25 | MARKING GUIDELINES/NASIENRIGL YNE ¥ Both axes correctly labelled with unit Beide asse korrek benoem met eenheid ¥ Straight line (t=O s and t = 1s) Reguitlyn (t =0 s ent = 7 s) ¥~Curve shape (t= 1s andt = 3) Kurwe (t =1 s ent = 3) A Y values for displacement and time indicated (A) Waardes vir verplasing en tyd aangedui [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 7 DBE/November 2016 NSC/NSS — Grade/Graad 10 - Memorandum QUESTION 4/VRAAG 4 4.1 4.2 4.3.1 4.3.2 441 4.4.2 Copyright reserved/Kopiereg voorbehou 5 m-s" north “ (accept range from 4,9 to 5,1) 5 m-s" noord (aanvaar vanaf 4,9 tot 5,1) 8,3m-s' vv (accept range from 8,2 to 8,4) 8,3 ms" (aanvaar vanaf 8,2 tot 8,4) The velocity is uniformly increasing. Velocity increases from 5 m-s" to 10 ms” in 150s. Pasitive acceleration. The girl is speeding up. The girl is uniformly accelerating Any ONE of the options VW” Enige EEN korrekte opsie Snelheid neem uniform toe. Snelheid neem van 5 m-s"' fot 10 m-s" in 150s foe. Positiewe versnelling. Die meisie se spoed neem toe. Die meisie se versnelling is uniform. Uniform/constant velocity Zero acceleration Any ONE of the options VY” Same speed / velocity Enige EEN korrekte opsie Uniforme/konstante snetheid Nil versnelling Dieselfde spoed/ snetheid OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Distance A to C Distance A to C Afstand A tot C Afstand A tot C =f{xb+%xbxhyv =lxb+lxb+%xbxhv =2125mv =2125mv =5 x 350 v+ %x 150 x Vv = 200 x 5 + 150 x 5V+% x 150 x 5¥ OPTION 3/OPSIE 3 Distance A to C Afstand A tot C ={xb+%(sum of parallel sides jh” ={xb+%(som van parallele sye )hY =5x 200 v+ %(5 + 10)(150) ¥ =2125mv Gradient of this graph is the acceleration ya-¥4) or Mi XQ - X4 tr-t _(-10)¥ ~ (65-0) ¥ =-0,15 m-s* + a=0,15m-s® Southv/ Suid gradient = (2) (2) (2) (2) (4) Ane (4) Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 8 DBE/November 2016 45 4.6 NSC/NSS - Grade/Graad 10 — Memorandum DtoE. vv DtotE (2) The change in speed from D to E is (-)10 m-s" v and that occurs over (65 s) a shorter period. W OR From B to CG, the change in speed is 5 m-s”' over a period of 150 s.vv OR Gradient is the steepest Die verandering is spoed van D tot E is (-)10m-s" »en die beweging gebeur oor 'n korter tydperk.(65 s) OF Vanaf B tot C is die verandering in spoed 5 m-s" oor 'n tydperk van 150 s. OF Gradient is die steilste (2) [20] QUESTION 5/VRAAG 5 5.1 5.2 5.3 5.4 3.600 =v 14x 1000 = 50,4 km-h'Y ORIOF 14x3,6v = 50,4 kmh"! ¥ 2) The energy an object possesses as a result of its motion. VV Die energie van 'n voorwerp as gevolg van die beweging daarvan. (2) E, = mgh¥ = 0,01 x 9,8 x Sv = 0,49 Jv (3) OPTION/OPSIE 1 (Ep + Ex)topibo = (Ep + Ex)bottomronder v mgh + %4mv? = mgh + %mv* (0,01)(9,8)(10) + OY = (0,01)(9,8)(5) +%x0,01xv ¥ v=9,89m-s? ¥ OPTION/OPSIE 2 (Ep + Etopibo = (Ep + Ex)bottom/onder v mgh + Ymv? = mgh + Ymv2 (0,01)(9,8)(10) +0 Vv =049+%x001xv v 049% 2x00Ixv v=9,89ms" ¥ ACCEPT/ AANVAAR: A { he. (Ep + Ex) = (Ep + Ex)s (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 9 DBE/November 2016 NSC/NSS — Grade/Graad 10 —- Memorandum 5.5 Equal tow’. Mechanical energy is conserved”, it is a closed system.” Gelyk aan. Meganiese energie word behou, dit is 'n gesiote stelsel. (3) [14] QUESTION 6/VRAAG 6 6.1 0,4mvv (2) 6.2.1 TroughyY Trog/buik (1) 6.2.2 CrestY Kruin (1) 6.3 Aand CY Aenc (1) 6.4 2uvv (2) 6.5 vafxav 0,4 =0,5 xAv “ N= 0,8 mY (3) 6.6 2% x 0,8¥ =2mv OR/OF _q vt 0,4 d v "5 [12] Copyright reserved/Kopiereg voorbehou Np Please turn over/Biaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 10 DBE/November 2016 NSC/NSS — Grade/Graad 10 —- Memorandum QUESTION 7/VRAAG 7 7.1 The wavelength of notes (played). “ Die golflengte van note (gespeel) (1) 7.2 The frequency. ¥ Die frekwensie (1) 7.3 120 100 — a 2 po og 56 s zs fe Le 0 5 10 15 Wavelength (m) Golflengte (m) MARK ALLOCATION: | | PUNTETOEKENNING: 7 | Y correct y-axis label and unit korrekte benoeming en eenheid op y-as_ (4) correct x-axis label and unit korrekte benoeming en eenheid op x-as Y¥ all points plotted correctly “ alle punte korrek gestip en beste lyn And best fit line drawn geteken v shape of graph ¥~ vorm van grafiek ¥ correct scales on axes vkorrekte skaal vir asse 74 Frequency and wavelength are inversely proportional “to each other. Frekwensie en golflengte is omgekeerd eweredig aan mekaar. rn Me 1 fas (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 11 NSC/NSS — Grade/Graad 10 - Memorandum 7.5 vafxav =55x6v = 330 m-s'!v QUESTION 8/VRAAG 8 8.1.1 8.1.2 8.1.3 8.2.1 8.2.2 8.3.1 8.3.2 8.3.3 Cv AY By Keeping food warm Remote controls Optical fibres v Animals like snakes which hunt Infrared scanners (for picking up heat) Hou voedse!l warm Afstandbeheerders Optiese vesels Sekere diere soos silange wat jag Infrarooi skandeerders (wat hitte optel) RADAR systems v RADAR speed traps Microwave ovens Telefoon- OF satelliet- OF selfoonkonneksies RADARsteisels RADARspoediokvaistelsels Mikrogolfoonde X-ray ¥ X-strale DBE/November 2016 2ni6 -tt 11 (10) (3) Any ONE v/Enige een (1) Telephone OR satellite OR cellphone connections Any ONE v/Enige een (1) (1) X-ray has a high frequency OR (high) penetration into soft tissues of humans, ¥ but not bones. X-strale het 'n hoé frekwensie OF sagte weefsel van mense (hoog)indring, maar nie been nie. X-rays can: e damage living tissue e cause cancer X-strale kan: « weefsel beskadig e kanker veroorsaak Copyright reserved/Kopiereg voorbehou (1) Any ONE ¥/Enige een eh (1) Please turn over/Biaai om asseblief

Downloaded from hlayiso.com DBE/November 2016 Physical Sciences/P1/Fisiese Weienskappe/Vi 12 NSCINSS - Grade/Graad 10 —- Memorandum 8.4 OPTION/OPSIE 1 OPTION/OPSIE 2 E=hiv f=£ a a = 6,63 x 10% x sity =u =6,63 x 107% J v =1x 10° Hz E = hf v (one mark for both equations/ een punt vir albei vergelykings) = 6,63 x 10%x 1 x 10° ¥ =6,63x 10%) v (4) QUESTION 9/VRAAG 9 9.1.1 A force exerted on an object without touching the object. vv OR A force exerted on an object that is at a distance. VW 'n Krag wat op ‘n voorwerp uitgeoefen word sonder om aan die voorwerp te raak. OF 'n Krag wat oor 'n afstand op ‘'n voorwerp uitgeoefen word. 9.1.2 e Gravity/Weight/Gravitational force e Electrostatic/Coulombic force ° Gravitasiekrag/Gewig/Gravitasie e Elektrostatiese/Coulomb-kragte 9.2 Attractive “ Aantrekkend 9.3 © North/NY e v Any ONE v/Enige een The direction of magnetic field lines is from north to southy Noord/ N Die rigting van magneetveldlyne is van noord na suid Copyright reserved/Kopiereg voorbehou [12] (2) (1) (1) (2) [6] by Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 13 DBE/November 2016 NSC/NSS — Grade/Graad 10 —- Memorandum QUESTION 10/VRAAG 10 10.1 BY (1) 10.2 B to AY Btot A (1) Qi+ Q: 10.3 Qhewinuut = = ZY _ (43x 10% + (2x10) y = 2 =5x10’Cv (3) 10.4 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 n=Sv n=fv e e — 5x 107-(-2x 10° dy = 5x 107-(+3 x 10°), 1,6 x 107° 1,6 x 10° = 1, 56 x 10° electrons” = 1, 56x 10" electrons” elektrone elektrone (3) [8] QUESTION 11/VRAAG 11 11.1.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 4 1 4 Ry xX Ro —— 1 —_— — Ry, = v Ry Ry * R2 v : Ry + Ro 4 4 - 8x 3, “5 *3" 6+3 2. Ry = 20 =20. ~Riotavtotaal = 4+ 2” ©. Reotaitotaal = 4 + 2% =6 Ov =6 a” (4) 11.1.2 [Ry : Reeves ACCEPT! AANVAAR: 20 : 40Vv ~ v2 as : : : Ran = 7% .. potential difference is also in ratio of ly Potensiaal verskil is ook in die verhouding 4= V2 v 2:40r1:2v 2 -. 12V +3 parts/dele = 4V V2=8VV¥ «. series = 2X 4 = BV ¥ (3) Copyright reserved/Kopiereg voorbehou 1. over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences/P1/Fisiese Wetenskappe/V1 14 NSC/NSS ~ Grade/Graad 10 - Memorandum 11.1.3 11.1.4 11.2 11.3 Copyright reserved/Kopiereg voorbehou DBE/November 2016 1 7 Ra yorin words: resistance is inversely proportional to current and -. ratio of resistors is 6 : 3V 2:1 «ratio of current is 1:2 “” A2: Ag lao = : x1 ola = 067A ¥ 1; R a; in woorde: weerstand is omgekeer eweredig aan stroom .. verhouding van resistors is 6 : 3¥ 2:1 .. verhouding van stroom is 1: 2¥ Ae2 i A3 2 wtlao = 3 x? wlap = 0,67 A v ACCEPT/IAANVAAR: v =av¥ R _ 12-8 6 =0,67AVY (3) Ai=2A Q=lAtY =2x120V =240CV Decrease ¥ Afneem NEGATIVE MARKING FROM 11.2 NEGATIEWE MERK VANAF 11.2 (3) (1) e Ifthe 6 Q resistor is removed, the resistance of the whole circuit increases V e SinceRa : ¥, if R increases, and V is constant V and I of the circuit decreases e Indien die 6 Q-resistor verwyder word, sal die totale weerstand van die stroombaan verhoog. °e R at ¥ , so indien R verhoog en V bly konstantv’, sal die stroom (I) verlaag. (3) 206 -1i- 11 [17] <yTOTAL/TOTAAL: 150 Me