Physical Sciences Grade 12 Paper 1 Limpopo Mock 2021 Memo

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Downloaded from hlayiso.com Physical Sciences/Pl 2 Limpopo DoE/September 2021 NSC INSTRUCTIONS AND INFORMATION 1. Write your NAME in the appropriate space on the ANSWER BOOK. 2. This question paper consists of 10 questions. Answer QUESTION 10.4 on the attached graph paper. Answer ALL the other questions in the ANSWER BOOK 3. Start EACH question on a NEW page in the ANSWER BOOK. 4 Num�er the answers correctly according to the numbering syst�sed in this question paper. C.,• 5. Leave ONE line between two subquestions, !�,&,ample between QUESTION 2.1 and QUESTION 2.2. _ 6. You may use a non-programmable calculator. e<v. � 7. You may use appropriate mathematical i ents. 8. You are advised to use the attached �'i., �HEETS. 9. Show ALL formulae and substitu�� in ALL calculations. 10. Round off your FINAL t1t0-<S::a1 answers to a minimum of TWO decimal places. '\\ 11. Give brief motiv"Y...�iscussions, et cetera where required. 12. Write neatly a@lgibly. Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/Pl 3 Limpopo DoE/September 2021 NSC QUESTION 1: MULTIPLE-CHOICE QUESTIONS Various options are provided as possible answers to the following questions. Choose the answer and write only the letter (A-D) next to the question numbers (1.1 to 1.10) in the ANSWER BOOK, e.g. 1.11 E. Each question has only ONE correct answer. 1·1 Which ONE of the following physical quantities is represented by the ratio /iv ? lit A Acceleration B Displacement G0� C Momentum D Net force (2) �� 1.2 A satellite orbits Earth at a height where it exp rie e�a gravitational force four times less than that on the surface of the Earth. If Ea? ' radius is R, then the height of the satellite above Earth's surface is ... A 4R. B 3R. C 2R. D R. (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/Pl 4 Limpopo DoE/September 2021 NSC 1.3 A ball thrown vertically upwards from point X reaches its greatest height at point Y and returns to the same level as X at point Z, as shown in the diagram below. y X Z Neglect the effects of air resistance. Which ONE of the following statements is CORRECT? A The acceleration at point Y equals zero B The speed at X equals the speed at z C It takes longer time from X to Y than Y to Z D The velocity at X equals the velocity at Z (2) 1.4 The net force acting on a moving object is ZERO. Which statement about the momentum and kinetic energy of the object is CORRECT? MOMENTUM KINETIC ENERGY A Stays constant Changes Stays constant Stays constant C Changes Changes D Changes Stays constant (2) 1.5 Which ONE of the following is the BEST indication that mechanical energy is conserved? A Wnet + �Ek = 0 B Ep +fa= 0 (2) D �p + �E p = 0 Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/Pl 5 Limpopo DoE/September 2021 NSC 1.6 A sound source recedes (moves away) from a stationary observer at constant velocity. Which ONE of the following CORRECTLY describes how the observed frequency and wavelength differ from that of the sound source? FREQUENCY WAVELENGTH Greater than Greater than Greater than Less than C Less than Greater than D Less than Less than (2) ,, 1.7 Two id nUcal conducting spheres M nd N carry charges of +12, 8 x 10-19 C and -1 ,8 x 10·19 C, respectively. The sp eres are allowed to touch. ctrons does M gain wh n the spheres touch? A B C D (2) 1.8 Which ONE of the following graphs best represents the relationship between the current {I) and the power output (P) for an appliance with constant resistance? B p 0 C D p p 0 (2 ) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/Pl 6 Limpopo DoE/September 2021 NSC 1.9 . . of conventional A current-carrying, rectangular coil 1s 1rr-a rflagnel1cl1eltf. The d1rect1on . current in the coil is as shown in the diagram below. (Direction of rotation of coil) \.!/ I I I 1 t 2 - The coil is turning clockwise. Which combination correctly indicates the polarity of magnets 1 and 2? MAGNET 1 MAGNET 2 A South North B North North C South South D North South (2) 1.10 Ultra-violet light emits electrons from a Zn plate. How would the maximum kinetic energy, Ek(max), of the photoelectrons and the number of photoelectrons emitted per second be affected by using more intense ultra-violet light? Ek(max) Number of photoelectrons emitted per second A Decreased Increased B Unchanged Unchanged C Unchanged Increased D Increased Unchanged (2) [20] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/Pl 7 Limpopo DoE/September 2021 NSC A box of mass 65 kg slides down an incline, as shown in the diagram below. The incline makes an angle of 20° with the horizontal. The frictional force acting on the box is constant. 2.1 Define the term frictional force. (2 ) 2.2 Draw a labelled free-body diagram for the box as it slides down the incline. (3) The velocity versus time graph for the box as it moves down the incline is shown below. r , .......... ..-I ' _________________ I ________________ - 1 ' 'II) 6 _________________ L ________________ I -----------------�----------- ---- 1 I 55 -----------------�------ I I ---------O 1 � � ' I ·c::; 4 - - - - - - - - - - - - - - - - - - -1 I ------L-----------------1 0 0 I a3 3 O - - - - - - - - - - -1- - - - - - - - - - - - - - - - - - > r ' 1 ' 2 -----------------L----------------- - - - - - - - - - -- - - - - - -� - - -·- - - - - - - - -- - - - -: 1 1 - - -- - - - - --- ----------------- �� - - -- - - - - -- --- - - --: ----------------- o ,__-----....·------....' ----- 1 1;0 2:0 time (s ) 2.3 Use the data in the graph to show that the magnitude of the acceleration of the box is 1,75 m-s·2. (3) 2.4 State, in words, Newton's second Law of motion in terms of acceleration. (2) 2.5 Calculate the: 2.5.1 Resultant force acting on the box (3) 2.5.2 Kinetic fictional force acting on the box (4) [17] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/Pl 8 Limpopo DoE/September 2021 NSC QUESTION 3 (Start on a new page) A stone is thrown vertically from the ground. It passes a 2,0m high window whose sill is 10,0m above the ground. The stone takes 0,15 s to travel the 2,0m height of the window . Diagram is not drawn to scale. - ------- window . 2,0m -- ' -- - ,_ - ,------ 10,0m ' n� Stone Ground Neglect the effects of air resistance (i.e. the stone moves under the influence of gravitational force only). 3.1 Write down the term which describes the bracketed statement. (1) 3.2 USE EQUATIONS OF MOTION to calculate the: 3.2.1 The speed with which of the stone is thrown (5) 3.2.2 Maximum height above the ground that the stone reaches (4 ) 3.3 Sketch a velocity versus time graph to illustrate the motion of the stone from the moment it leaves the ground until it reaches its maximum height. (Take upwards as the positive direction). Clearly show the value of the initial velocity on the graph. (3) 3.4 Write down the magnitude of the acceleration of the stone at the maximum height. (1) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/Pl 9 Limpopo DoE/September 2021 NSC QUESTION 4 (Start on a new page) A bullet of mass 2 g travelling horizontally strikes a 1 kg wooden block which is at rest. Vbi c::> bullet The bullet emerges instantly on the opposite side of the block with a reduced velocity of 50 m-s-1. The frictional force acting on the block is 1,96 N. The block is observed to travel 0,09 m before stopping, as shown in the diagram above. 4.1 State, in words, Newton's second Law of motion in terms of momentum. (2) 4.2 Calculate the magnitude of the initial velocity of the bullet WITHOUT USING ENERGY PRINCIPLES. (7) 4.3 Draw a labelled vector diagram (not to scale) to illustrate the relationship among the initial momentum Pi, the final momentum pt and the change in momentum Ap for the bullet. (2) [11] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/Pl 10 Limpopo DoE/September 2021 NSC QUESTION 5 (Start on a new page) A block is released from rest at point A and slides down a curved frictionless track AB and then moves along a frictionless horizontal track BC, and finally moves up a ROUGH inclined plane CD. ,",, / ......... I I I h C B The coefficient of kinetic friction between the block and incline CD is 0,35. 5.1 State, in words, the principle of conservation of mechanical energy. (2) 5.2 Use the principle of conservation of mechanical energy to calculate the speed of the block at point B. (4) 5.3 Using energy principles, calculate the maximum height h reached by the block. (6) [12] QUESTION 6 (Start on a new page) When a sound source moves in an easterly direction towards a stationary learner at 15 m-s-1, the frequency observed by the listener is ti. When the source is stationary and the learner moves in a westerly direction towards the source at 25 m-s- 1, the frequency observed by the learner is fa. The learner notes that fa is greater than ti by the 37 Hz. 6.1 Write down the NAME of the phenomenon that the learner observes and state this phenomenon in words. (3) 6.2 Assuming that the speed of the sound in air is 340 m-s-1, calculate the frequency of the source. (6) 6.3 Calculate the wavelength of the sound observed by the STATIONARY learner. (3) 6.4 Write down ONE applications of the phenomenon referred to in QUESTION 6.1 in the medical field. (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/Pl 11 Limpopo DoE/September 2021 NSC nownloade.d ,ram !=itanmorenhvs. ·cs. com 6.5 7"f1e speclrallines from a aistane longer wavelength of the spectrum. Does this represent RED SHIFT or BLUE SHIFT? (1) [14] QUESTION 7 (Start on a new page) Two identical NEGATIVELY charged spheres, X and Y, having charges of the SAME MAGNITUDE, are placed 500 mm apart in a vacuum, as shown in the diagram below. The . magnitude of the electrostatic force that the one sphere exerts on the other is 0,288 N. -• I 500 m X ------------ y 7.1 State Coulomb's Law in words. (2) 7.2 Calculate the: 7.2.1 Magnitude of the charge on EACH sphere (4) 7.2.2 Number of EXCESS electrons on sphere Y (3) . Pis a point 800 mm from sphere Y, as shown in the diagram below. ----------------�p 1 500 mm 1 800 mm 1 x-------v 7.3 Define the term electric field at a point in words. (2 ) 7.4 What is the direction of the net electric field at P? Choose from LEFT or RIGHT. (1) 7.5 Calculate the number of electrons that must be removed from sphere Y so that the net electric field at P is 4 x 104 N-c- 1 to the right. (7) [19] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/Pl 12 Limpopo DoE/September 2021 NSC A battery with an emf of 12 V and internal resistance r is connected in series to two identical resistors, EACH with resistance X, as shown in circuit diagram 1 below. CIRCUIT DIAGRAM 1 ------------------------1 ------------------------1: t = 12 V X X The ammeter reading, A1, is 1,20 A. 8.1 Define the term internal resistance of a battery. (2) The same battery is now connected to the same two resistors in parallel, as shown in circuit diagram 2 below. CIRCUIT DIAGRAM 2 -------------------------, r _, - __: �-� I I � -- _€_::. )._� _y_ ----------- _: I I 3, OA X X The ammeter reading, A2, is 3,0 A. 8.2 Show that the total external resistance in circuit 1 is FOUR times that in circuit 2. (3) 8.3 Using the two circuits and the equation £ = l(R + r), Calculate the: 8.3.1 Resistance X (3) 8.3.2 Internal resistance of the battery r. (2) The resistors in circuits 1 and 2 are replaced by identical 12 V non-ohmic bulbs. Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 13 Limpopo DoE/September 2021 NSC 8.4 Rows nlpade d. crom esanmerep phy, si es Bhi in circuit 1 when the circuit (4) functions initially. 8.5 Explain why the resistance of a bulb in circuit 2 will not be the same as the resistance (3) of a bulb in circuit 1 after a while. [17] QUESTION 9 (Start on a new page) 9.1 An engineer has an electric drill, which operates from an alternating current supply. The drill is rated as follow: 2a0v QQ 1 800 W 50 Hz The current passing through the drill when operating is represented in the graph below Ka 10, 6f ----- ~~ 2 pee e ens nnn nnn nnn rrr nnn nnn nce o a x time (s) -10, 6 | --------------------------------------- = 9.1.1 Define the term root mean square (rms) for an alternating current. (2) Calculate the: 9.1.2 Time X indicated on the graph (3) 9.1.3 Root mean square (rms) current the drill draws when operating (3) 9.1.4 Resistance of the drill (3) 9.2 Explain why the transmission of the electrical energy is carried out using alternating currents and high voltages. 9.2.1 From the scientific point of view (1) 9.2.1 From the economic point of view co 13. Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 14 Limpopo DoE/September 2021 NSC Downloaded from Stanmorephysics.com QUESTION 10 (Start on a new page) Light with different wavelength (A) is incident on a metal surface. The variation of the maximum kinetic energy Eximax of electrons emitted from the metal surface with inverse of wavelength(~), is shown in the graph below. ry a = aol \ \ H : H \ H E ' ' ' ' ' ' ' H 30 Tt 10}-474 A 05 1,0 1,5 2,0 2,5 3,0 3,5 4,0 = (x 10° nm) 10.1 Define the term work function of a metal in words. (2) 10.2 Use the graph to determine the threshold frequency of the metal. (3) 10.3. Calculate the work function of the metal. (3) Caesium metal has a work function of 2, 48625 x 1079 J. 10.4 On the graph paper provided, draw a graph to show the variation of the maximum kinetic energy Ex(max for caesium metal with . Label your graph P. Show, by means of a suitable calculation, how you have arrived at the : -intercept (5) [13] TOTAL: 150 Copyright reserved Please turn over

Downloaded from hlayiso.com Limpopo DoE/September 2021 15 NSC Physical Sciences/P1 Downloaded from Stanmorephysics.com QUESTION 10.4 Name of leaner: ....... 0. eee ececce cece eeeee eee eeeeeeeeeeneeeeeeeeeeeeeeseeseeteeeseeseeeeeeeeeeeees (C gr-OT x) IE (x 106 m*) 1 a 05 1,0 1,5 2,0 2,5 3,0 3,5 4,0 Please turn over Copyright reserved

Downloaded from hlayiso.com Physical Sciences/P1 Downloaded from Stanmorephysics.com 16 NSC Limpopo DoE/September 2021 DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 1 (PHYSICS) TABLE 1: PHYSICAL CONSTANTS NAME SYMBOL VALUE Acceleration due to gravity g 9,8 m-s* Universal gravitational constant G 6,67 x 10°'' N-m*-kg? Speed of light in a vacuum c 3,0 x 10° m-s! Planck's constant h 6,63 x 10° J-s Coulomb's constant k 9,0 x 10°N-m?-C* Charge on electron e -1,6x10'C Electron mass me 9,11 x 10°! kg Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 17 Limpopo DoE/September 2021 NSC Downloaded from Stanmorephysics.com TABLE 2: FORMULAE MOTION Vy =Vitadt Ax = vAt+4adt? OR Ay =vAt+ tat? Vy =v? +2aax OR Vv, =v +2aay | AxX= (eta OR Ay ~( eat FORCE Fret =Ma p=mv an wemg WORK, ENERGY AND POWER W =FAxcos@ U=mgh OR E, =mgh ; ; W,.=AK OR W,, =AE, K=—mv? OR E,=—mv? 2 2 AK=K,-K, OR AE, =Ey —E, Wye =AK+AU OR W,, = AE, +AE, = Pave = FVave Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 WAVES, SOUND AND LIGHT 18 Limpopo DoE/September 2021 NSC Downloaded from Stanmorephysics.com v=fa T=- VtV, 4 - waver, E=hf OR E=h< viv, a E=W,+K,,, OR E=W,+E ‘k(max) 1 where E=hf and W,=hf, and Kin = DIV as OR Eximax) = FMV 2 ELECTROSTATICS kQ,Q kQ ef vw q q n=2 oR n-2 e a. ELECTRIC CIRCUITS R=v emf (E) =I(R +r) R, =R,+R> +... 1 91,1 q=IAt = tH + R, R, R, W=Vq Ww At W=VIAt P=VI W=PRAt P=rR _ Vat v2 R “R ALTERNATING CURRENT I Paverage = Vie lems Tine = \ Parerage =LinsR Ve = “8 Vie V2 Paverage = Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com —4 fas LIMPOPO 7 PROVINCIAL GOVERNMENT EX 43 REPUBLIC OF SOUTH AFRICA DEPARTMENT OF EDUCATION NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 MARKSI/PUNTE: 150 These marking guidelines consists of 16 pages./ Hierdie nasienriglyne bestaan uit 16 bladsye. Copyright Reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 NSC-Marking Guidelines LimpopoDoE/September 2021 2 BURFRTOLULUR FAG Stanmorephysics.com 1 Avy (2) 1.2 Dvv (2) 1.3 Byv (2) 1.4 BYv (2) 1.5 cyvy (2) 1.6 cyvv (2) 7 Dvv (2) 1.8 Avy (2) 1.9 Avy (2) 110 Cvv (2) [20] Copyright Reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 NSC-Marking Guidelines LimpopoDoE/September 2021 3 ROATIONGEL AGA Stanmorephysics.com 2.1 The force that opposes the motion of an objectv_and which acts parallel to the surface. ¥ (2) Die krag wat die beweging van 'n voorwerp teenwerk v en wat parallel met die opperviak inwerk. ~ 2.2 Fy/N Fi /f/Fr w/F, Accepted symbols Fuv | N/Normal force/ N/normaalkrag fv Ft /fx /frictional force/kinetic frictional force F /fk Awrywingskrag/kinetiese wrywingskrag wv | Fg/mg/ weight /Gravitational force/ Fearth on box Fg /mg/gewig/gravitasie krag/Faarde op blok NOTES/ Mark is awarded for label and arrow. Do not penalise for length of arrows. Deduct 1 mark for any additional force. If all forces are correctly drawn and labelled, but no arrows, deduct 1 mark. e If force(s) do not make contact with the dot: LET WEL Punt word toegeken vir byskrif en pylpunt. Moenie Straf vir die lengte van pyle nie. Trek 1 punt af vir enige bykomende krag. As alle kragte korrek getrek en gemerk word, maar geen pyle nie, trek 1 punt af. Indien krag(te) nie kontak met die kolletjie maak nie: maksimum. 2/3 (3) 2.3 OPTION 1:/OPSIE 1 OPTION 2:/ Marking Criteria:/ a = gradient OPSIE 2 nasien riglyne Av v Anyone |vw=vitadt ¥ . Av = at Enige een | 6=2,5v+a(2)v | Gradient/7— v _ 6-2,5 v a=1,75 m-s?@ Numerator as shown ¥ ~ 2,0-0% Denominator as shown ¥ (3) = 1,75 m-s2 Teller en noemer vgetoon Copyright Reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 NSC- Marking Guidelines LimpopoDoE/September 2021 DobwnlWa epee MPP Steep SAN object, the object accelerates in t Irection o Pet fet) force cum an acceleration which is 2.5.1 2.5.2 directly proportional to the (net ‘otee Y and inversely proportional to the mass of the object. ¥ OR The (non-zero) resultant/net force acting on an object is equal to the rate of change of momentum of the object_in the direction of the resultant force. Wanneer 'n (nie- nul) resultante / netto krag op 'n voorwerp inwerk, versnel die voorwerp in die rigting van die (netto) krag teen ‘n versnelling wat direk eweredig is aan die (netto) krag ven omgekeerd eweredig aan die massa van die voorwerp. ~ OF Die (nie-nul) resultante / netto krag wat op ‘n voorwerp inwerk, is gelyk aan die tempo van verandering van momentum van die voorwerp in die rigting van die resultante krag. (2) Fret = ma ¥ Marking criteria: Nasienriglyne: = (65) (1,75) V e Formula “/formuleY : e Substitution Y/vervangingy = 113,75 N, downhill ¢ Magnitude + direction V/grootte en afwaartsY rigting ~” (3) POSITIVE MARKING FROM QUESTION 2.5.1 POSITIEWE NASIEN VANAF VRAAG 2.5.1 Fret = ma wi + (-fx) = ma f ¥ Any one/enige een mgsiné - fk =ma +. (65) (9,8) (sin20°)¥ —fx = (65) (1,75) ¥ *. 217,8668 - fk = 113,75 +. 217,8668 — 113,75 = fx +. 104,1168 N = fk ¥ -. The kinetic frictional force/die kinetiese wrywingskrag is 104,1168 N (4) [17] Copyright Reserved Please turn over

Physical Sciences/P1 Downloaded from hlayiso.com NSC-Marking Guidelines 5 LimpopoDoE/September 2021 RO ATIONPOL AGH Stanmorephysics.com 3.1 3.2.1 3.2.2 Free fall W/vryvalv Consider the window (bottom to top) / Beskou die venster(onder tot bo) Ay = viAt + a(t)? ». 2,0 = vi(0,15) + + (-9,8)(0,15)? V + Vi = 14,06833 m-s* Now consider motion from ground to bottom of the window: Oorweeg nou beweging vanaf die grond tot onderkant die venster vit = vi? + 2ady ¥ + (14,06833)2 V = vi? + 2(-9,8)(10,0) ¥ + vi? = 393,918 “ Vi= +19,85 m-s :. The launch speed is/ Die lanseringsnelheid is 19,85 m-s? ¥ OPTION 1:/OPSIE 1 Consider motion from bottom of window OPTION 2:/OPSIE 2 Consider motion from up: Oorweeg beweging van onder venster boontoe: Vi? = vi? + 2adAy ¥ + (0)? = (14,06833)? + 2(-9,8)Ay ¥ . Ay = 10,0979 m v «. The height/die hoogte is 10,0 +¥ 10,0979 = 20,0979 m ground: Oorweeg beweging van die grond af: Vi? = vi? + 2ady ¥ V +. (0)? = (19,85)? + 2(-9,8)Ay +. Ay = 20,1032 m «. The height/die hoogte is 20,1032 m ¥ Copyright Reserved Please turn over (1) (5) (4)

Physical Sciences/P1 & 3 Ed Downloaded from hlayiso.com ed from Stanmorephysics.com NSC-Marking Guidelines LimpopoDoE/September 2021 6 Marking criteria: = Nasienriglyne 2 e Shape Y/vorm v sO » |¢ Launch velocity v/ 2 TimeTyd(s) lanseer snelheid ~ g e Velocity at max. height/ $ v=0v/ es} snelheid by max hoogte/ So os v=0~ > (3) 34 98ms?v (1) [14] QUESTION 4/VRAAG 4 4.1 The total (linear) momentum of an isolated system remains constant(is conserved) Vv ¥ Die totale (lineére) momentum van 'n geslote sisteem bly konstant (bly behoue) “~~ (2) 4.2 For the block: —_—. Vir die blok: fk = 1,96 N k Lpi = Zpr ¥ Fnet = ma (0,002)Vvbi + 0 = (1)(0,59397)+(0,002)(50) ¥ 1 -1,96Y = (1)av = sliv sa=-1,96 ms? Vbi = 346,985 m-s V2 = vi2 + 2aAx —— +. (0)? = vi2 + 2(-1,96)(0,09) Vv - 0 = vi? —0,3528 «vi = 0,59 397 ms? Pbiock = Mv = (1)(0,59 397) ¥ \ Copyright Reserved = 0,59 397 ems! —— “4 Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 NSC-Marking Guidelines LimpopoDoE/September 2021 7 Dowrihscttied "APSAY<S PAA RU epi ys RESscoReNtUM to the block / Die koeél het 0,59 397 kg-m-s*+ van sy momentum na die blok oorgedra -. Total initial momentum of the bullet,/ Totale aanvanklike momentum van die koeél, Pbullet/koeé! = MVbullet/koeé! + Pblock/blok = (0,002)(50) +v 0,59 397 = 0,69 397 kg-m-st But/maar pbi = MbVbi +. 0,69397 = (0,002)vbi ¥ _ 0,69397 ~ (0,002) o Vbi = 346,985 m-stv 4.3 Pi x >+ Py Ap ' Marking criteria/Nasienriglyne e Initial and final momenta correct v/ Aanvanklike en finale momentum korrek ~ e Change in momentum correctly drawn ¥/ Verandering in momentum korrek getrek ~” QUESTION S5/VRAAG 5 5.1 The total mechanical energy (sum of the gravitational potential energy and kinetic energy) in an isolated system remains constant (is conserved). Vv ¥ Die totale meganiese energie (som van die swaartekrag potensiéle energie en kinetiese energie) in 'n geslote sisteem bly konstant (bly behoue). “~~ 5.2 | (Ek + Ep)s = (Ek + Ep)a ; 1 1 +} v Any one/enige een G mv? + mgh)s = (mv? + mgh)a mv? + m(9,8)(0) v= 2m(0)2 + m(9,8)(5) ¥ 3v+0=0+49 “. V = 9,8995 m-st Vv :«. The speed of the block/ Die spoed van die blok is 9,8995 m-s+ Accept/aanvaar: 9,90 m-s* Copyright Reserved Please turn over (7) (2) [11] (2) (4)

Downloaded from hlayiso.com Physical Sciences/P1 NSC-Marking Guidelines LimpopoDoE/September 2021 8 DowMBEENa!9PS/E1S tanmorephysics.com Wret = AEk Ww + Wik + Wen = AE v Any one/enige een mgAxcosé + fkAx-cosB + 0 = smve - ; mvit mgAxcosé@ + [k.mgcosa:Ax: cosB = sve - ; mvi? (9,8)Axcos120°Y + (0,35)(9,8)cos30°:Axcos180°” = +(0)? - +(9,8995)2v ». (-4,9)Ax — (2,9705)Ax = -49 «. (-7,8705)Ax = -49 -. Ax = 6,2258 m h_.. But/maar — = sin30° v Ax «. h = (sin30°)(6,2258) ¥ ~h= 3,113 m NOTE:/LET WEL; Ww can also be calculated as follows/ W., kan ook soos volg bereken word: Ww=wiAxcosé = mgsina:Ax:cos@ = m(9,8)(sin30°)Ax:cosé ORIOF: Ww= -AEp = mghi - mghr = m(9,8)(0) — m(9,8)h = -m(9,8)h Copyright Reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 NSC-Marking Guidelines LimpopoDoE/September 2021 9 DowPS8gu2!'G?S2S tanmorephysics.com Whe = AEp + AEk fAx-cosB = mghi — mghi + smvee - ; mvi2 v Any one/enige een Llk.mgcosa-Ax: cosB = mghr — mghi + smve - > mv? (0,35)m(9,8)( cos30°)Ax(cos180°)” = m(9,8)h — m(9,8)(0) Vv + 0 - + m(9,8995)2v (-2,970467135)Ax-m = m(9,8)h - 49-m (-2,970467135)Ax = (9,8)h — 49 h -! v= - (-2,970467135)(——) v = (9,8)h - 49 (-5,94093427)h = (9,8)h — 49 49 = (9,8)h + (5,94093427)h 49 = (15,74093427)h ¥ +h =3,113m QUESTION 6/VRAAG 6 6.1 e Doppler effect v e The apparent 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. v VW Doppler effek ~” Die skynbare verandering in frekwensie (of 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. “~~ Copyright Reserved Please turn over (6) [12] (3)

Downloaded from hlayiso.com Physical Sciences/P1 NSC-Marking Guidelines LimpopoDoE/September 2021 10 Be = ynleatise) from Stanmorephysics.com v 340+0 f= Cronas) 68 v fi = (fs beeeeee (1) -. 1351,57 Hz = fs ¥ -. The frequency of the source/die frekwensie van die bron is (ignore 0 if omitted/ignoreer 0 indien uitgelaat) v 340+25 , ee . an fo = ( 34040 )fs (ignore 0 if omitted/ ignoreer 0 indien uitgelaat) _ 73 v p= Qe “....@) (1) - (2) 73 68 “fa-f= (Sts - (ts 73 68 “37 =fs (S- =) 121 “37 =f (70) 1351,57 Hz (6) 6.3 || POSITIVE MARKING FROM Q6.2/POSITIEWE NASIEN VANAF V 6.2 v=fav «. 340 = (1 351,57)A ¥ A= 0,2516 mv ] 3 -. The wavelength/die golflengte is 0,2516 m (3) 6.4 ANY ONE/ENIGE EEN: e Measuring the rate of blood flow in humans and animals. ¥ e Measuring the heartbeat of a foetus in the womb. ¥ e Ultrasound (scanning) ¥ Meet die tempo van bloedvioei in mense en diere. ~ Meet die hartklop van 'n fetus in die baarmoeder. “ Ultraklank (skandering) ~ (1) 6.5 Red shift/rooi verskuiwing ¥ (1) [14] Copyright Reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 NSC-Marking Guidelines LimpopoDoE/September 2021 11 ReAMIPOTHED 6m Stanmorephysics.com 7.1 7.2.1 7.2.2 7.3 7.4 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 deur een puntlading (Q1) 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 van die afstand(r) tussen hulle. ~ 7 ah, r2 9) 92 «2,88 x 1072 v= (9x10°)Q? | ‘ (0,5)? n Q =8~x 10722 « Q = 2,828 x 10° CV POSITIVE MARKING FROM QUESTION 7.2. / POSITIEWE NASIEN VANAF V 7.2: n=2 or/of n=2v e qe 2,828x1076 1,6x10-19 «n= 1,7675 x 10% Velectrons/elektrone The electrostatic force experienced per unit positive charge placed at that point. vv Die elektrostatiese krag per eenheidpositiewe-lading wat by daardie punt geplaas is.v¥ West/wes ¥ Copyright Reserved Please turn over (2) (4) (3) (2) (1)

Downloaded from hlayiso.com Physical Sciences/P1 NSC-Marking Guidelines LimpopoDoE/September 2021 12 Downk Ki p= <2, r2 Take ————~ as positive direction Enet = Ey + (-Ex) 9 10? v 9 -6 eax tots (9x10°)Qy¢ _ (9x 10°)(2,828x107®) y (0,8)? (1,3)? 4x 104 = (1,40625 x 10°)Qyr — 15060,35503 : = 6 * Qu = 3,9154 x 1o® C Marking criteria: AQy = Qyt = Qyi KQ AQy = 3,9154 x 106 _v (-2,828 x 10°) e 1mark for formula E = 7 AQy = 6,7434 x 10° C e¢ 1mark EACH for Q Q substitutions in Ey and Ex n=—orn=—Vv e 1mark for AQy e de Q e 1 mark for formula n = . -6 n= — v e 1 mark for substitution in ,6X107 «nN =4,215 x 1018 Velectrons formula . Range :4,206 X10"? to 4,215 x10 ¢ 1 mark for final answer QUESTION 8/VRAAG 8 8.1 8.2 The maximum energy/work done supplied by a _ battery per unit charge/coulomb of charge passing through it. vv Die maksimum energie/arbeid wat 'n battery lewer per eenheidslading/ coulomb van lading wat daardeur vloei. “~~” Rs _ Rit Ro V Rp (B1Rz_) Ry+ R2 _ X+xv Copyright Reserved Please turn over (7) [19] (2) (3)

Physical Downloaded from hlayiso.com Sciences/P1 13 NSC-Marking Guidelines LimpopoDoE/September 2021 Down 8.3.2 8.4 l6adéd*from Stanmorephysigs:-eem— Marking Criteria «12 = (1,2)(x + xr) YW “12 = (1,2)(2x + 1) © 10=2xX4+r (1) E€=1,(R+n .12= (3.96 +n v xX ~4=-4r 2 6B=xXt2r (2) From/vanaf (1):r=10-—2x ......... (3) Sub. (8) into (2): vervang (3) in (2) “xX + 2(10 — 2x) =8 “xX +20=4x-x - 12 = 3x ~x=4,00V 1 mark for substituting the emf equation for Circuit 1(series) 1 mark for substituting the emf equation in Circuit 2(parallel) Final Answer Nasienkriteria 1 punt vir die vervanging van die emk vergelyking vir stroombaan 1 1 punt vir die vervanging van die emk vergelyking in stroombaan 2 Finale Antwoord POSITIVE MARKING FROM 8.3.1:/ POSITIEWE NASIEN VANAF V 8.3.1: Sub./vervang x = 4,0 into/in (3): r=10-2(4)v =2QV POSITIVE MARKING FROM QUESTION 8.3: POSITIEWE NASIEN VANAF V 8.3: P=P?RV I?R Ratio/verhouding = Px _ (4,2)?x(4,0) ¥ ~ (4,5)2x(4,0) % = 0,64 = 0,64v Copyright Reserved Please turn over (3) (2) (4)

Downloaded from hlayiso.com Physical Sciences/P1 NSC-Marking Guidelines LimpopoDoE/September 2021 14 Po wnleaded fro Mist anmorephys ics.com potential difference/ current and the resistance across the other bulb remains the same. Vv GEEN VERANDERING v Wanneer een van die gloeilampe uitbrand, bly die potensiaalverskil / stroom en die weerstand oor die ander gloeilamp (3) dieselfde. v~ [17] QUESTION 9/VRAAG 9 9.1.1 The rms value of AC is the AC current which dissipates the same amount of energy as DC. Vv OR: The root mean square current is the value of the current in an AC circuit that will have the same heating effect as a DC circuit. vv Die wgk is die WS wat dieselfde hoeveelheid energie verbruik/oordra as ‘n ekwivalente GS. vw" OF: Die wgk stroom is die waarde van die stroom in 'n WS-stroombaan wat dieselfde verwarmingseffek as 'n GS-stroombaan sal hé. (2) 9.12 [ty f =v 50 «T=0,02s 0,02 exe (3) 2 “x=0,01sv 9.1.3 | OPTION 1/ OPSIE 1: OPTION 2/OPSIE 2: Pave = ImmsVims ¥ _ Irms Iims = TE v 2 1800 = Inms(240) ¥ 2 . = 10,6 lms = 7,50 A ¥ o Irms = v Irms = 7,4953 AY (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 NSC-Marking Guidelines LimpopoDoE/September 2021 15 Dbivn/GBAeN ¥ARS!EStanmorep, SIE 2: v?rms Pave = v Pave = |?rmsR ¥ “ = eR Vv 240)? 1800 = (7,4953)?R +. 1800 = (240)* | R= 32,040V R + 32,000 Vv OPTION 3/ OPSIE 3: OPTION 4/ OPSIE 4: R = Jims v R = vmax v I-ms Imax -2_, pe N2C240) 7,4953 “um 10,6 « R=32,020V « R=32,020V (3) 9.2.1 With AC, transformers can be used to step up voltage/step down current. ¥ (1) Met WS kan transformators gebruik word om spanning te verhoog/stroom te verlaag 9.2.2 Heat/energy loss by transmission cables is minimised by high voltage transmission. ¥ (1) Hitte/energie_verlies deur _transmissie_kabels_ word verminder deur 'n hoé spanning. ~ [13] QUESTION 10/VRAAG 10 10.1 The minimum energy that an electron in the metal needs to be emitted from the metal surface. VW Die minimum energie wat 'n elektron in die metaal benodig, om uit die metaalopperviak vrygestel te word. vv (2) . 1 10.2 — =1,75 x 10°v do But c = fodg + = fi ac Xo 0 «. (3 x 108)(1,75 x 10°) W= fo 3 « fo = 5,25 x 10!4 Hz ¥ (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 NSC-Marking Guidelines LimpopoDoE/September 2021 Bow 10.4 16 Wo = (6,63 x 10°%4)(5,25 x 1044) v Wo = 3,48 075 x 1079 J v (3) Wo = hfo he i v Any one/enige een Wo = x, 0 1 # 2,49625 x 10° v = (6,63 x 10°)(3 x 108). ~ v 0 1 “= = 12 50000 m+ Xo = 1,25 x 10° m+ . The graph p is parallel the original graph, but cuts the x — axis at 1,25(x 10°m™). / Die grafiek p is parallel met die oorspronklike grafiek, maar sny die x - as by 1,25(x 10°m7').(See graph below./ sien die grafiek hieronder) Graph of maximum kinetic energy versus the reciprocal of 4./ Grafiek van maksimum kinetiese energie teenoor die omgekeerde van A. vv 2 Ss Eximax) (x 10°19 J) BP N Nop -- any A 0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0 £6 108 m) Marking criteria: Formula /formuleY Left hand substitution /linkerhand vervangingY Right hand substitution/ regterhand vervanging ¥ x — intercept at 1,25(x 10°m7?) /x - sny by 1,25(x 10®m=1) ¥ Graph P parallel original graph/ Grafiek P parallel aan oorspronklike grafiek ~” (5) [13] GRAND TOTALIGROOT TOTAAL: [150] Copyright Reserved