Physical Sciences Grade 10 P1 2016 Memo

basic education Department: Basic Education REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 10 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 2016 MEMORANDUM 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

Physical Sciences/P1/Fisiese Wetenskappe/V1 2 DBE/November 2016 NSC/NSS —- Grade/Graad 10 — Memorandum QUESTION 1/VRAAG 1 14 Byv (2) 1.2 Cv (2) 13 Dyvv (2) 1.4 Dvv (2) 15 Dvv (2) 1.6 AvY (2) 1.7 BYY (2) 1.8 Cvy (2) 1.9 DvY (2) 110 BYv (2) [20] QUESTION 2/VRAAG 2 2.1 Mark allocation/Puntetoekenning: Cc —® Iv 1 xiine ab: length, arrow, label 1 x lyn AB: lengte, rigting, benoem ¥ 1xline BC: length, arrow, label 1x 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

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

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 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om _ Ae

Physical Sciences/P1/Fisiese Wetenskappe/V1 5 NSCI/NSS — Grade/Graad 10 - Memorandum 3.5 POSITIVE MARKING FROM 3.4 POSITIEWE NASIEN VANAF 3.4 DBE/November 2016 .. taxi will not stop at the traffic light as distance > 40 m . die taxi sal nie V by verkeerslig stop nie, want die afstand is > 40 m OR / OF - oe ov =25mv .. 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 O = 257 +2 x -12,5 x AxY “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 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?¥ =25m . total distance/fofale afstand = 25+ 25 = 50 mv -. 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 Copyright reserved/Kopiereg voorbehou Loe Please turn over/Blaai om asseblief (5)

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

Physical Sciences/P1/Fisiese Wetenskappe/V1 7 NSC/NSS — Grade/Graad 10 - Memorandum QUESTION 4/VRAAG 4 4.1 4.2 4.3.1 4.3.2 441 4.4.2 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" 8,3 ms" (aanvaar vanaf 8,2 tot 8,4) The velocity is uniformly increasing. Pasitive acceleration. The girl is speeding up. The girl is uniformly accelerating Snelheid neem uniform toe. Positiewe versnelling. Die meisie se spoed neem toe. Die meisie se versnelling is uniform. Uniform/constant velocity Zero acceleration Same speed / velocity Uniforme/konstante snetheid Nil versnelling Dieselfde spoed/ snetheid ¥¥ (accept range from 8,2 to 8,4) DBE/November 2016 (2) (2) Velocity increases from 5 m-s" to 10 ms” in 150s. Any ONE of the options VW” Enige EEN korrekte opsie Snelheid neem van 5m-s" tot 10 m-s" in 150s toe. (2) Any ONE of the options v “ Enige EEN korrekte opsie (2) OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Distance A to C Afstand A tot C ={xb+%xbxhv = 5x 350 w+ “4x 150 x 5” =2125mv Distance A to C Afstand A tot C =lxb+lxb+%xbxhv = 200 x5 + 150 x 5v+% x 150 x 5¥ =2125mv 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 y2-Y4 VE Vi XQ - X4 tr-t _(-10)¥ ~ (65-0) ¥ =-0,15 m-s* + a=0,15m-s® Southv/ Suid gradient = Yor Copyright reserved/Kopiereg voorbehou (4) Ane (4) Please turn over/Blaai om asseblief

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

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 v=ftxay 0.4=0,5xAv “= 0,8 mY (3) 6.6 2% x 0,8¥ =2mv ORIOF = vt oaetv "5 [12] Copyright reserved/Kopiereg voorbehou Np Please turn over/Biaai om asseblief

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 =P ze po og 5 5 s zs es Le 0 5 10 15 Wavelength (m) Golflengte (m) MARK ALLOCATION: | | PUNTETOEKENNING: | 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

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

Physical Sciences/P1/Fisiese Weienskappe/Vi 12 DBE/November 2016 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

Physical Sciences/P1/Fisiese Wetenskappe/V1 13 NSC/NSS - Grade/Graad 10 —- Memorandum QUESTION 10/VRAAG 10 10.1 10.2 10.3 10.4 11.1.1 11.1.2 DBE/November 2016 By (1) Bto Av Btot A (1) Qi+ Q: Qhew/nuut = 5 ay _ (43x 10% + (2x10) y = 3 =5x107CV (3) 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 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 1 1 1 Ry x Ro —— 1 —_— — Ry, = v Ry Ra * Ro ” me Rt + Ro 4 4 = 8x 3, “5*3” 6+3 2. Ry = 20 =22 --Riotavtotaal = 4 + 2 ©. Riotaitoteal = 4 + 2¥” =6 OY =6 OY (4) Ry : Reeries ACCEPT/ AANVAAR: 20 : 40v _Y2 an . . . Ran = 7% .. potential difference is also in ratio of ly Potensiaal verskil is ook in die verhouding 4= V2 v 2: 407 1:24 2 -. 12V +3 parts/dele = 4V V2=8VV¥ -. Veeries = 2x 4 = BV 4 (3) Copyright reserved/Kopiereg voorbehou 1. over/Blaai om asseblief

Physical Sciences/P1/Fisiese Wetenskappe/V1 14 NSC/NSS ~ Grade/Graad 10 - Memorandum DBE/November 2016 11.1.3. | Ro orin words: ACCEPT/IAANVAAR: I . Vv resistance is inversely proportional to =R v current and _ 12-8 .. ratio of resistors ia =0,67AY «ratio of current is 1:2 “” Az: Ag lao = : x1 wb2=067A v 1, R azin woorde: weerstand is omgekeer eweredig aan 2:1 Ae2 i A3 stroom .. verhouding van resistors is 6 : 3¥ .. verhouding van stroom is 1: 2¥ wtlao = z x? wlap = 0, 67A v 11.1.4 Ai=2A Q=lAtY =2x120V =240CV 11.2 Decrease ¥ Afneem 11.3 NEGATIVE MARKING FROM 11.2 NEGATIEWE MERK VANAF 11.2 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. (3) (3) (1) °e R at ¥ , so indien R verhoog en V bly konstantv’, sal die stroom (I) verlaag. (3) Copyright reserved/Kopiereg voorbehou 206 -1i- 11 [17] <yTOTAL/TOTAAL: 150 Me