Physical Sciences Grade 12 P1 Free State November 2021 Memo

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Downloaded from hlayiso.com FS/September 2021 Grade 12 Prep.Exam. INSTRUCTIONS AND INFORMATION 1. Write your name and other information in the appropriate spaces on the ANSWER BOOK. 2. This question paper consists of 10 questions. Answer ALL questions in the ANSWER BOOK. 3. Start EACH question on a NEW page in the ANSWER BOOK. 4. Number the answers correctly according to the numbering system used in this question paper. 5. Leave one line between two sub-questions, for example between r&sTION .._ 2.1 and QUESTION 2.2. 6. You may use a non-programmable pocket calculator. 7. You may use appropriate mathematical instrumen � 8. You are advised to use the attached DATA SH� i- 9. Show ALL formulae and substitutions in �\culations. 10. Round off your FINAL numerical ans1 ,.. k�to a minimum of TWO decimal places where applicable. 11. Give brief motivations, discu��s, et cetera where required. Q 12. Write neatly and legibly \\ Copyright reserved Please turn over

Downloaded from hlayiso.com FS/September 2021 QUESTION 1 Various options are provided as possible answers to the following questions. Choose the answer and write down only the letter ( A - D) next to the question number (1.1- 1.10) in your ANSWER BOOK, e.g. 1.11 E. 1.1 A car is moving at a CONSTANT velocity on an incline. Which ONE of the following statements about the forces acting on the car is CORRECT? A There are no forces acting on the car. B The net force acting on the car is zero. C D �f There is a non-zero net force acting on the car. The weight of the car is equal to the normal (2) 1. 2 A ball of mass m strikes a wall perpendicularly q.)�peed 2v. Immediately after the collision the ball moves in the opposi ��frection at a speed v, as shown in the diagram below. The magnitude of the impulse of the t,a.� ... m -, A 111 \ ,_., m I B :v C 2mt\O D �◊ (2) 1.3 A Ot�ne is dropped from a certain height above the ground. Which ONE of the following combinations is TRUE while the stone is moving downwards? Ignore the effects of air resistance. NET FORCE ACTING ON MECHANICAL ENERGY OF THE STONE THE STONE A Remain the same Increases B Decreases Remains the same C Remain the same Remain the same D (2) Increases Decreases Copyright reserved Please turn over

Downloaded from hlayiso.com FS/September 2021 1.4 Which ONE of the following forces is an example for conservative force? A Gravitational force. B Tension in a chord. C Frictional force. D Normal force. (2) 1.5 Which ONE of the following combinations is TRUE for an elastic collision? TOTAL LINEAR MOMENTUM TOTAL KINETIC ENERGY A Conserved Conserved B Not conserved Not conserved C Conserved Not conserved D Not conserved Conserved (2) 1.6 SPECTRUM A represents the visible light spectrum recorded on earth. SPECTRA B and C are recorded from two stars Y and Z respectively '-------'---II -----'-'I_------'----_ Blue Red SPECTRUM A Blue Red I SPEC TRUM B � ____... ____._l.,_I ______ �II �I�� __ _ ..__ _ Blue Red SPECTRUMC Which ONE of the following combinations most correctly describes the movement of the stars relative to earth? A Star Y is moving away from the earth and star Z is moving towards the earth. B Star Y is moving towards the earth and star Z is moving away from the earth. C Both Star Y and star Z are moving towards the earth. D Both Star Y and star Z are moving away from the earth. (2) Copyright reserved Please turn over

Downloaded from hlayiso.com FS/September 2021 1.7 A stone starting from rest and rolls down a frictionless track, reaches the bottom of the track with a speed v. What change must be effected on the track for the stone to reach a speed of 2v at the bottom of the track? Stone A Double the height of the track. B Increase the height by a factor ✓2 . C Increase the height by four times. D keep the height the same, but make the track twice steeper. (2) 1.8 Two identical charged spheres kept at a distance d apart. The force exerted on one charge by th other is F. What is the new force, when the charge on one sphere i OOUBLED? (2) Copyright reserved Please turn over

Downloaded from hlayiso.com FS/September 2021 1.9 5 0 and 10 0 resistors are connected in parallel as shown in the diagram. Ii 5n 12 10.0 A : I Ii The relationship between current Ii and current 12 can be expressed as: A B 1 C 12 = - Ii 3 D 12 = -11 (2) 2 1.10 When an electron moves from a higher energy level to a lower energy level a specific ... A emission line in an emission spectrum is observed. B absorption line in an emission spectrum is observed. C emission line in an absorption spectrum is observed. D absorption line in an absorption spectrum is observed. (2) [20] Copyright reserved Please turn over

Downloaded from hlayiso.com FS/September 2021 QUESTION 2 A block of mass 100 kg is sliding down a rough inclined surface at a CONSTANT speed when a force F is applied parallel to the inclined plane as indicated in the diagram below. The coefficient of kinetic friction for the surfaces is 0,29. 2.1 Define the term normal force in words. (2) 2.2 Draw a labelled free-body diagram for the block. (4) 2.3 Calculate the magnitude of the applied force, F. (5) 2.4 Name the force that will form an action-reaction pair with the normal force in this case. (1 ) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com FS/September 2021 QUESTION 3 The position - time graph below, not according to scale, represents the motion of a ball thrown vertically downwards with a speed of 3,28 m-s- 1 from the edge of the roof of a 35 m high building. The ball hits the ground and is in contact with the ground for 0,1 s. The ball then bounces off to a maximum height x above the ground. Ignore the effects of air resistance. 35 '- -'- time (s) Q ---t 1--12---'--- 3,86 3.1 Explain the term projectile. (2) 3.2 Calculate the: 3.2.1 Speed of the ball at time ti (3) 3.2.2 Time, ti, it takes the ball to hit the ground (3) 3.2.3 Height, x, reached by the ball after it bounces off the ground (7) 3.3 Draw the velocity-time graph for the ball from the moment it is thrown until it reaches the maximum height x after the first bounce. Indicate: (i) The values of time h and 12 (ii) The velocities of the ball at time ti and 12 (3) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com FS/September 2021 QUESTION 4 Car A of mass 1 085 kg travelling east at a speed of 33 m-s-1, collides head on with car B of mass 3 450 kg travelling west at a speed of 28 m-s·1. Immediately after the collision, car A moves west at a speed of 5 m-s·1. Ignore friction. 33 m-s·1 28 m-s·1 ) Car A CarB mass - 1085 kg mass - 3450 kg s 4.1 State Newton's second law of motion in terms of momentum in words. (2) 4.2 Calculate the: 4.2.1 Velocity of car B after collision (5) 4.2.2 Magnitude of the impulse on car B during the collision (3) 4.3 Airbags serves as a safety feature during collision. 4.3.1 Use the relevant laws of physics to explain the purpose of air bags in cars. ( 3) 4.3.2 Which driver sustains more injuries during above collision? Choose from CAR A or CAR B (1) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com FS/September 2021 QUESTION 5 A 70 kg crate, starts from rest at point A (top of slope) and slides down a slope inclined at 25° to the horizontal. The distance from point A to point B is 135 m. 135 m B The total kinetic friction between the crate and the surface is 150 N. 5.1 Define the term non-conservative force in words. (2) 5.2 Calculate the net work done on the crate, as it slides from point A to point B. (4) 5.3 Calculate the time it takes the crate to slide from point A to point B. (6) (12] QUESTION 6 An ambulance moving at a constant speed along a straight horizontal road, emits sound with a constant frequency. When the ambulance approaches a detector mounted next to the road a frequency of f is recorded. After passing the detector, a frequency of 0,55 f is recorded. 6.1 State the Doppler effect in words. (2) 6. 2 Calculate the speed of the ambulance. Take the speed of sound in air as 340 m-s· 1. (6) 6.3 State TWO applications of the Doppler effect. (2) (10] Copyright reserved Please turn over

Downloaded from hlayiso.com FS/September 2021 QUESTION 7 x is a point 0,2 m from point charge, Qi, carrying charge -3x10-5 C. 0,2 m >I •---------------------• Qi X 7 .1 Define electric field at a point in words. (2) 7.2 Draw the electric field pattern due to the charge Qi. (2) Now another point charge Q2, carrying a charge of +12x10-5 C is placed at a distance of 0,35 m from Qi as shown below. N ----- 0,35 m----- ,-----------------------------------. I< 0,2 m >I Q2 Qi X 7.3 Calculate the net electric field at point x due to the two charges. (6) 7.4 State Coloumb 's law in words. (2) 7.5 Calculate the magnitude of the electrostatic force exerted by charge Q2 on charge Qi. (3) 7.6 Charge Q2 is moved further west. How will this change affect the net electrostatic force at this new distance? Choose from INCREASE, DECREASE or REMAIN THE SAME. (1 ) [16] Copyright reserved Please turn over

Downloaded from hlayiso.com FS/September 2021 QUESTION 8 A battery has an emf(£) of 12 V and an internal resistance of 0,2 0. The resistances of the connecting wires are negligible, as shown in the circuit below. 10 0 R 15 0 s E= 12V 0,2 0 8.1 What is the meaning of the underlined words? (2) When switch S is CLOSED, the voltmeter reading decreases from 12 V and the ammeter reading is 0,9 A. 8.2 Explain why the voltmeter reading decreases. (2) Calculate the: 8.3 Total current in the circuit (4) 8.4 Resistance of resistor R (5) 8.5 The total power dissipated by the battery (3) 8.6 Will the ammeter reading INCREASE, DECREASE or REMAIN THE SAME if the 15 n resistor is removed? Give a reason for the answer. (2 ) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com FS/September 2021 QUESTION 9 9.1 The diagram below represents a simplified electric machine. The coil abed rotates at 20 cycles per second. "T • The direction of the current in the coil is from a to b. com N 9.1.1 What type of electric machine is represented by the above diagram? (1) 9.1.2 Name the principle on which the machine operates. (1) 9.1.3 In which direction is the coil rotating? Choose from CLOCKWISE or ANTICLOCKWISE. (1) 9.1.4 How long does it take for the coil abed to complete one rotation? (3) 9.1.5 State ONE way in which this machine can produce a lower output voltage without changing its construction. (1) 9.2 An electrical device is rated 220 V, 1850 W. 9.2.1 Define the term rms voltage in words. (2) Calculate the: 9.2.2 Rms current passing through the appliance when it is operating. (3) 9.2.3 Resistance of the device. (3) [15] Copyright reserved Please turn over

Downloaded from hlayiso.com FS/September 2021 QUESTION 10 An experiment is conducted to investigate the relationship between the frequency of light incident on a metal cathode and the maximum kinetic energy of the emitted electrons from the surface of a metal cathode of a photocell. The graph represents the results obtained. Graph of maximum kinetic energy versus frequency of light > 0 Cl) C Cl) u ·.;::::; Cl) C 0, 5 10 >5' 2,0 f (x1015 Hz) , _-�· -X , 10.1 For this investigation, write down the following: 10.1.1 Controlled variable (1) 10.1.2 Dependent variable (1) 10.2 Name the phenomenon on which this experiment is based. (1) 10.3 Define the term work function in words. (2) 10.4 Calculate the: 10.4.1 Value of x as shown on the graph (3) 10.4.2 Frequency of the incident light that will emit electrons from the surface with a maximum speed of 5,23x105 m•s- 1 (4) 10.5 The intensity of the light is now increased. State how this change affects the current in the photocell . Choose from INCREASES, DECREASES or REMAINS THE SAME. Explain your answer. (3) [15] GRAND TOTAL: 150 Copyright reserved Please turn over

Downloaded from hlayiso.com FS/SEPTEMBER/ 2021 DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 1 (PHYSICS) GEGEWENS VIR FIS/ESE WETENSKAPPE GRAAD 12 VRAESTEL 1 (FIS/KA) TABLE 1: PHYSICAL CONSTANTS/ TABEL 1: FIS/ESE KONSTANTES NAME/ NAAM SYMBOL/ S/MBOOL VALUE/ WAARDE Acceleration due to gravity g 9,8 m-s-1 Swaartekragversnelling Universal gravitational constant G 6,67 x 10-11 N-m 2-kg·2 Universe/e gravitasiekonstant Speed of light in a vacuum C 3,0 x 108 m-s· 1 Spoed van Jig in 'n vakuum Planck's constant h 6,63 X 10-34 J·S Planck se konstante Charge on electron e -1,6 X 10-19 C Lading op elektron Electron mass me 9,11 X 10-31 kg Elektronmassa Mass of the Earth M 5,98 X 1024 kg Massa van die Aarde Radius of the Earth RE 6,38 X 106 m Radius van die Aarde Copyright reserved Please turn over

Downloaded from hlayiso.com FS/SEPTEMBER/2021 TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEG/NG 2 V1 =V; +2MX Oro If V 1 =V; +2My 2 2 2 FORCE/KRAG Fn et = ma P= ITW f max s =µ s N fk µk N = Fnet l'.lt = L'.lp W=l'TY:J Llp = mv 1 mv i Mf F=G-2 - F=G-2 - g =G- m1m m1m g =G 2 2 or/of or/of c 2 d r r WORK, ENERGY AND POWER/ARBE/D, ENERGIE EN DRYWING W=F�xcose U=mgh or/of Ep = mgh Wnet = �K or/of wnet = b.Ek mv 2 1 K=_! 2 or/of Ek =-rnv 2 2 �K = K, -K; or/of �Ek = Ekf -Eki W Wnc = �K + �U or/of wnc =�Ek + �Ep P= b.t Pav = FVav / Pgemid = FVgemid WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG V=fA T=� f V±VL v v f= f f = ± L f b E= hf or /of E=h� L v+ - vs s L v v ± b "- E = Wo + E k(max) or/of E =WO + K max where/waar 1 -1 E =hf and/en Wo = hfo and/en Ek (max) = rnv �ax or/of K max - rnv 2max 2 2 Copyright reserved Please turn over

Downloaded from hlayiso.com ELECTROSTATICS/ELEKTROSTA TIKA k kQ F= Q l2Q 2 E= 2 r r W F V= E= q q n= Q or/of n= _2_ qe ELECTRIC CIRCUITS/ELEKTR/ESE STROOMBANE emf ( £)= l(R + r) V R =- emk ( £)= l( R + r) R s =R1 +R 2 + ... 1 1 1 q=I fit -=-+-+ ... RP Rl R2 W=Vq W P= Lit W =VI fit P =VI W= 1 2Rfit 2 P =l R 2 w---- V Lit R v P=- 2 R ALTERNATING CURRENT/W/SSELSTROOM Pa1.e =Vrms l rms I pgemiddeld = Vwgkl wgk _ l r ms = ! max ✓2 I I wgk ✓ - lma2ks pa1.e = l �sR I pgemiddeld = l�kR vmax vmaks v,ms = 2 ✓ I Vwgk - - ✓2 vr�s V1gk pa1.e = R I pgem iddeld = R Copyright reserved

Downloaded from hlayiso.com �1- �� �..� . \•", l. � ' ' Department of Education FREE STATE PROVI CE PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (Pl) FIS/ESE WETENSKAPPE: FIS/KA (V1) SEPTEMBER 2021 MARKS/PUNTE: 150 MARKING GUIDELINES NASIENRIGL YNE This marking guidelines consists of 17 pages. Hierdie nasienriglyne bestaan uit 17 bladsye. Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 1.1 B ✓✓ (2) 1.2 o ✓✓ (2) 1.3 c ✓✓ (2) 1.4 A✓✓ (2) 1.5 A✓✓ (2) 1.6 A✓✓ (2) 1.7 C✓✓ (2) 1.8 A✓✓ (2) 1.9 B ✓✓ (2) 1.10 A✓✓ (2) [20] Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne QUESTIONNRAAG 2 Marking criteria/Nasienkriteria: If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase As een van die onderstreepte s/eutelwoorde/frases in die regte konteks weggelaat word: - 1 punt per woordlfrase 2.1 The force or the component of a force which a surface exerts on an object with which it is in contact, and which is perpendicular to the surface. ✓ ✓ Die kraq of die komponent van 'n kraq wat 'n oppervlak op 'n voorwerp uitoefen waarmee dit in kontak is, en wat /oodreq op die oppervlak is. (2) 2.2 N ACCEPTIAANVAAR F Fg11 Accept the following symbols/Aanvaar die volgende simbole F Fapplied / Force applied / FA ✓ Froeaeoas I Krag toegepas I FA Fg w / Fw / weight/ gewig ✓ N FN / Normal / Normal force ✓ FN I Normaal I Normaalkrag f FtI fk / friction / frictional force/kinetic frictional force ✓ Ft/ fk / wrywing / wrywingskrag I kinetiese wrywingskrag NOTE/LET WEL: • Mark awarded for label and arrow/Punt toegeken vir byskrif en pyl • Do not penalise for length of arrow since drawing is not to scale. Moenie penaliseer vir die lengte van die pyl nie, want die tekening is nie volgens skaal nie. 3 • Any additional force(s)!Enige bykomende krag(te) 4 • If force(s) do not make contact with body/As krag(te) nie kontak met liggaam maak nie � • No labels/Geen byskrifte 4 Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne 2.3 Marking criteria / Nasienkriteria: • Showing Fnet/netto = O ✓ • • Substitution of first term0 ✓ • Substitution of second term • Final answer of/ Finale antwoord van 157,48 N✓ Range (152,98N to/tot 157,45N) OPTION / OPSIE 1: Downwards as positive/Afwaarts positief: Fnet = ma or Fnet = O -F - fk + F911 = 0 } Any ✓ -F - µkmgcos0 + mgsin0 = 0 -F - O,289x(1OOx9,8cos25 °) ✓ + (1OOx9,8)sin25° ✓ = O ✓ F = 157,48 N ✓ Downwards as negative / Afwaarts positief Fnet = ma or Fnet = 0 F + fk - Fgtt = 0 F + µkmgcos0 - mgsin0 = 0 } F + O,289x(1OOx9,8cos25 ) ✓ - (1OOx9,8)sin25° ✓ = 0 ✓ ° F = 157,48 N✓ (5) 2.4 Perpendicular force tha the water tank exerts on the surface. ✓ Loodreg krag wat die watertenk op die oppervlak uitgeoefen. ACCEPT/AANVAAR mgcose (1) [12] Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 3.1 An object upon which the only force acting is the force of gravity (2 or 0) ✓ ✓ 'n Voorwerp waarop die eniqste kraq wat daarop inwerk, swaartekraq is. (2) 3.2.1 Marking criteria / Nasinkriteria: Appropriate formula/ Toepas/ike formu/e ✓ Substitution of/ Vervanging van 3,28 & 35 ✓ Final answer of/ Finale antwoord van 26,396 m-·1 ✓ Range (26,397 to 26,41) Downwards as positive/ Upwards as positive/ Afwaarts positief Opwaarts posiitief OPTION 1 / OPSIE 1 v 21 = v 2i + 2a�y ✓ v: = v� + 2a�y✓ = ( ,28 )2 + 2 (9,8 )( 5) ✓ 3 3 = ( ,28 )2 + 2(9,8 )( 5) ✓ 3 3 = 26, 96 m • s- 1 ✓ v t = 26, 96 m · s- ✓ 1 3 3 Vr OPTION 2/ OPSIE 2 v 1 = v i + a�t ✓ v, = v i + a�t ✓ = ,28 + (9,8 X2, 59) ✓ 3 3 = - ,2s+ (-9,sX2, 59)✓ 3 3 v 1 = 26, 98 m · s-1✓ 3 v 1 = 26, 98 m · s-1 ✓ 3 OPTION 3/ OPSIE 3 Vr+V Vt +V- �y = ( )�t✓ I �y = ( I ) �t ✓ 2 2 Vf + 3,28 = (Vf \ •28 35 = ( 3 )( 2,359)✓ 35 )(2,359) ✓ 2 v 1 = 26, 98 m · s-1 ✓ 3 v 1 = 26, 98 m · s-1 ✓ 3 ( ) 3 Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne 3 .2.2 POSITIVE MARKING FROM QUESTION 3.2.1 / POSIT/EWE NASIEN ANAF 3.2.1 Downwards as positive/ Upwards as positive/ Afwaarts positief Opwaarts posiitief OPTION 1/0PS/E 1 V 1 = V; + aL'it ✓ V 1 = V; + aL'it ✓ 26 ,396 = 3,28 + ( 9, 8 )L'i t ✓ -26,396 = 3,28 + (-9, 8 )L'it ✓ L'i t = 2359 s ✓ ' L'it = 2,359 s ✓ OPTION 2/0PS/E 2 L'iy = V;L'it + ½a!it2 ✓ L'iy = V;L'it + ½a!it2 ✓ 1 2 3 5=3,28L'it+ !( 9, 8 )L'it ✓ - 35=- 328L'i' t+-(-9 8 )L'it2 ✓ 2 2 ' L'i t = 2'359 s ✓ ✓ L'i t = 2,359 s OPTION 3/OPS/E 3 OPTION 3/OPS/E 3 v1 v; V V; L'iy=( ; ) L'it ✓ liy =( 1 ; )lit ✓ = 26,39 + 3,28 = -26,396 (- 3,28 ) 35 ( � )L'it ✓ 35 ( ; )L'it ✓ L'it = 2,359 s ✓ L'it = 2,359 s ✓ (3) Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne 3.2.3 Calculate velocity at tz/ Bereken sne/heid by ti 12 = 2,359 + 0.1✓ = 2,459 s Time to max(x) = 3,86 - 2.459 ✓ = 1,40 s V r =V;+ai"1t ✓ 0 = V; +9,8(1,40) ✓ V =-1373m•S-1 I , Downward as positive/ Upward as positive/ Afwaarts positief Opwaarts positief OPTION 1/ OPSIE 1 v2 - 2 - v 1 +2a6..y ✓ v1 = v� +2a6..y ✓ f 2 0=(-13, 73) +2(9,8)Lly ✓ 2 0 = (13,73) + 2(- 9,8)Lly ✓ 6..y = -9,62 m 6,.y = 9,62 m :. Height/ Hoogte = 9,62 m ✓ :. Height/ Hoogte = 9,62 m ✓ OPTION 2/ OPSIE 2 6..y = V 6..t + 2.la6..t2 ✓ I 6..y =V;6..t + -½a6..t2✓ 2 2 =(-1 3 ,73 X1,4)+ 1 (9,sX1A) ✓ = (1 3 ,73 X1,4 )+ 1 (- 9,8 X1 ,4 )✓ 2 2 6,.y = -9,62 m 6,.y = 9,62 m :. Height/ Hoogte = 9,62 m ✓ :. Height/ Hoogte = 9,62 m ✓ OPTION 3/ OPSIE 3 V; V V; V 6,.y = ( ; f )6,.t✓ 6,.y = ( ; f )6,.t✓ +(- 3 , 73 ) = ( (o + - 3 , 73 ( � ))}1,4 ✓ = ((o � ) )1.4) ✓ ) 6..y = -9,62 m 6..y = -9,62 m :. Height/ Hoogte = 9,62 m ✓ :. Height/ Hoogte = 9,62 m ✓ (7) Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne 3.3 POSITIVE MARKING OF QUESTION 3.2.1 AND 3.3.2 I POSIT/EWE NASIEN VAN 3.2.1 EN 3.2.2 OPTION 1: Downward Positive OPSIE 1: Afwaarts Positief Marking criteria/ Nasienkriteria: • Velocity striking the ground and leaving the ground at hand t2. ✓ 26 ,39 ---------------------- Snelheid waarteen grand getref en grand verlaat word by t1 en t2 • Contact time hand t2 I Kontaktyd t1 en t2 ✓ :"S;! (l) 3,28 • Lines parallel / Paralelle lyne ✓ :S (l) 2,4 6 3, 86 >, :!:: 2,36 u Time(s)/Tyd 0 -13, 76 --------------------------------- OPTION 2: Upwards positive OPS/E 2: Opwaarts positief ............. 13, 76 -- --------------------------------- ---------------, I I :"S;! :S (l) 2 36 >, 2,46 3,86 Time(s)/Tyd ·g -3,28 > Q) -26 , 39 ------------------------ (3) [18] Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 4.1 The resultant/net force acting on an object is equal to the rate of change of momentum of the object in the direction of the net/resultant force. ✓ ✓ (2 or 0) Die resulterendelnetto kraq wat op 'n voorwerp inwerk is qelyk aan die tempo van veranderinq van momentum van die voorwerp in die rigting van die nettolresulterende krag. (2 of 0) (2) 4.2.1 OPTION 1 / OPSIE 1 Motion to the east(+)/ Beweging na o s(+) r Any!Enige ✓ " _ L..P L-P t - " j m 1 v 1 i + m 2 v 2i,/'° m 1v 11 + m 2v 21 ✓ 1085 (+33) + 3450 (-28) = 1 085 (-5) + 3450v 21 v 21 = - 15, 05 ✓ west / forward ✓ 1 = 16,05 m · s- wes I voorwaarts OPTION 2/ OPSIE 2 Motion to the east(-)/ Beweging na oos(-) LP, - LP, m,v,i + m 2 v 2i = m 1 v 11 + m 2 V 21 J l Any/Enige ✓ 1 085 (-33) + 34 50 ( +28)✓-= 1 085 ( +5 ) + 3450v 21✓ V 21 =+16,05 = 16,05 m · s- ✓ west / forward ✓ 1 wes I voorwaarts (5) Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne 4.2.2 POSITIVE MARKING FROM QUESTION 4.2.1/ POSIT/EWE NASIEN VANAF 4,2,1 OPTION 1: CAR B OPTION 2: CAR B Motion to the west(+) Motion to the west(-) OPSIE 1: KAR B OPSIE 2: KAR B Beweging na wes (+) Beweging na wes (-) Impulse= �P = m�v✓ Impulse= �P = m�v✓ = 3450(16,05 - 28)✓ = 3450(-16,05 - ( - 28))✓ = - 41227,5 = 41227,5 N·s ✓ = 41227,5 N-s✓ OPTION 3: CAR A OPTION 4: CAR A Motion to the west(+) Motion to the west(-) OPSIE 3: KAR A OPSIE 4: KAR A Beweging na wes (+) Beweging na wes (-) Impulse= �P = m�v ✓ Impulse= �P = m�v✓ = 1085(5 -( - 3 3))✓ = 3450(-5 - ( + 3 3)) ✓ = 412 30 N-s✓ = - 412 30 = 412 30 N-s ✓ (3) 4.3.1 According to Newton's second law ✓ I Fnec �p M Air bag causes an increase in collision time ✓ which results in a lesser / smaller force ✓ ( for the driver/passenger) (3) 4.3.2 Car A / Kar A ✓ (1) [14] QUESTION/VRAAG 5 5.1 A force for which the work done in moving an object between two points is dependent of the path taken.✓✓ 'n Krag waarvan die werk wat verriq word om 'n voorwerp tussen twee punte te skuif, afhanklik is van die pad wat qevolq word. (2) 5.2 Wnet = W1 + WFg// = Mxcose + mgsin8�xcos8 ✓ } = 150(135)C0Sl80 ✓+ 70(9,8)sin25 °(135)COS0° ✓ ° = -20 250 + 39 138, 677 = 18 888,677 J ✓ (4) Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne 5.3 POSITIVE MARKING FROM QUESTION 5.2/ POSIT/EWE NASIEN VANAF 5.2 OPTION 1/0PS/E 1 Wnet = L'iK 1 2 2 }Anv ✓ v = m(v t - i ) 2 18 888,677✓= !(70)(v; -0) ✓ 2 v 1 = 23 ,?3 m · s-1 '' ' L'iy = (V f +''V\) L'it ✓ 2 23 , 2 0)L'i 135 = ( �+ t ✓ L'it = 11,625 s ✓ OPTION 2/ OPSIE 2 - Wnc = L'iK + L'iU 1 - Any ✓ faycos0=-m(v; -v;}+mg(h 1 -h i ) 2 150( 135)cos180 ° ✓ = !(70 Xv; -o) +(70X9,8Xo -sin 25° x 135) ✓ 2 v 1 = 23,2� m · s-1 '' I L'iy = (v + � )L'it ✓ 1 2 \ 23, 2 + o)L'it 135 = ( � ✓ L'it = 11,625 s✓ Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne OPTION 3/ OPSIE 3 Wnet = Fnet 6.xcos0 ✓ 18 888,68 = Fnei (135)cosO ° ✓ Fnet = 139,92 N Fnet = ma 139,92 = 7Oa ✓ a= 2 m-s-2 1 6.y = V;6.t + -a6.t 2 ✓ 2 135 = 0 + !(2)6.t ✓ 2 2 6.t = 1 1,625 s ✓ (6) [12] QUESTION/VRAAG 6 6.1 It is 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. ✓ Dit is die (skynbare) verandering in frekwensie (of toonhoogte) van die klank (waargeneem deur 'n /uisteraar) omdat die klankbron en die luisteraar verskillende snelhede het in verhouding tot die medium van klankverspreiding. OR/OF An (apparent) change in (observed/detected) frequency (pitch), (wavelength) ✓ as a result of the relative motion between a source and an observer ✓ (listener). 'n (Skynbare) verandering in (waargenome) frekwensie (toonhoogte), (golflengte) as gevolg van die relatiewe beweging tussen 'n bron en 'n waarnemer (luisteraar). (2) Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne 6.2 Towards: fl = ( v-vv Jt s 340 tV: ( 340- Jf ·•·equation 1 Vs ✓ 5 Away : fl = ( +v Jt V V5 s 0,55 f = ( 340340 Jt5✓ ... equation 2 +Vs Equating-eqn 1 & eqn 2 153f = 1,55v J V 5 =98,71m•S- 1 ✓ (6) 6.3 ANY TWO ✓✓ Doppler /Blood flow meter Measuring the heart beat of a foetus Radar Sonar Use to determine whether stars are receding or approaching earth (2) (10] QUESTION/VRAAG 7 7 .1 The electric field at a point is the electrostatic force ✓ experienced per unit positive charge (placed at that point). ✓ Die elektrieseveld by 'n punt is die elektrostatiese krag ervaar per eenheid positiewe lading (op daardie punt geplaas). (2) Note: If electric field defined ( ½) / Let we/: As elektriese veld gedefinieer (½) Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne 7.2 Markin criteria/Nasienkriteria: Correct direction towards the sphere ✓ Korrekte ri tin na die steer Field lines not crossing and radial ✓ Veld/ ne wat nie die radial kruis nie (2) 7.3 EXQl = kQ r2 ✓ 9 X 109 (3 X 10-5) =----- -✓ (0,2)2 =6,75xl0 6 N-c-1 west EXQ2 = kQ r2 _ 9 X 109 (12 X 10- 5) ✓ - 2 (0,15) = 4' 8 x10 7 N • c-1 west ✓ E net = EXQl + EXQ2 = 6,75xl0 6 + 4,8xl0 7 = 5,475 x10 7 N-c-1✓ west✓ (6) 7.4 The electrostatic force(of attraction/repulsion) between two point charges is directly proportional to the product of the charges and inversely proportional to the squareof the distance between them. ✓✓ (2) 7.5 = k0 , 0 2 F ✓ r2 9 5 5 9 X 1 0 ( 3 X 1 0 - 0( 12 X 1 0 - =--------- ) 2 ✓ (0,35 ) = 264,49 N ✓ (3) 7.6 Increase / Neem aan ✓ (1) [16] Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 8.1 The battery supplies 12J ✓of energy per coulomb of/per unit charge. ✓ Die battery voorsien 12J per coulomb van/per eenheidsladinq. (2 ) 8.2 Due to the internal resistance ✓ energy is dissipated (per coulomb of charge ) in the battery. ✓ As gevolg van die interne weerstand, word energie per coulomb lading in die battery omgeskakel. (2 ) 8.3 Marking criteria/ Nasienkriteria: Appropriate f ormula/ Toepaslike formule V = IR ✓ Calculating current in/ Bereken stroom in 15 n ✓ Adding two currents/✓ Final answer /Finale antwoord van 6Q ✓ V= IR✓ ACCEPT V1on = V1sn lion: lisn =15 :10 ✓ lionV1on = lisnV1sn 0,9(10) = lisn{15) ✓ 0,9 0,6 ✓ ITOT =0,9+0,6✓ lisn = 0,6 A =1, 5 A ✓ IToT = lion+lisn =09+06 ✓ I I =15A ✓ I (4) Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne 8.4 POSITIVE MARKING FROM QUESTION 8.3 / POSIT/EWE NASIEN VANAF 8.3 OPTION 1 / OR/OF OR/OF OPS/El R, xR2 1 1 1 R= -=-+- P R1 + R2 RP Rl R2 10 X 15 _, = 1 1 1 1 10 +15 ( + ) =-+- ✓ = 10 15 10 15 =6Q =6n ... R P = 6Q OPTION 2 / OPSIE 2 c = I (R+Rp+ r) ✓ ✓ 12 = 1,5 (R+ 6 + 0,2) ✓ :.R = 1,8Q ✓ =- 1,5 =80 R = R+R + r T P ✓ 8 = R + 6 + O,� R= 1,8 Q ✓ (5) 8.5 OPTION 1/ OPSIE 1 OPTION 2 / OPSIE 2 OPTION / OPSIE 3 ✓ P =VI✓ P = 1 2R✓ v2 = 12(1,5)✓ = (1,5)2(8)✓ P=- R ✓ = 18W✓ = 18W✓ (12)2 8 =18 W✓ (3) 8.6 Decreases✓ The total resistance increases. ✓ (2) (18] Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Grade/Graad 12 Prep.ExamNoorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 (1) 9.1.1 AC Generators/ WS-generatorslopwekkers ✓ (1) 9.1.2 Electromagnetic induction/ Elektromagnetiese induksie ✓ 9.1.3 Anticlockwise/ Antikloksgewys (1) T =! ✓ 9.1.4 =- ✓ 20 = 0,05 s ✓ (3) 9.1.5 Reduce rotation speed (to less than 20Hz). ✓ Verminder rotasiespoed (na minder as 20Hz). (1) 9.2.1 The rms value of AC is the DC potential difference which dissipates the same amount of energy as AC. ✓ ✓ Die wgk-waarde van WS is die GS-potensiaalverskil wat diese/fde hoeveelheid energie as WS versprei. (2) 9.2.2 Pave = Vrms lrms ✓ 1850 = (220) I rms ✓ lrms = 8,41 A ✓ (3) 9.2.3 POSITIVE MARKING FROM 9.2.2 / POSIT/EWE NASIEN VANAF 9.2.2 OPTION / OPS/E 1 OPTION / OPSIE 2 OPTION 3/OPSIE 3 Pave = 1 2rms R✓ = (V ,m,) 2 1850 (8,41) R✓ V,m., ✓ R- = 2 -- ✓ pave R R = 26,160 ✓ J,ms 220 ( 220? ✓ =- ✓ 1850 = 8,41 R ✓ = 26,16 n ✓ :.R=26,16Q (3) [15] Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief

Downloaded from hlayiso.com FSNS/September 2021 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 10.1.1 Metal (cathode)/ Metaal (katode)lwork function/threshold frequency I Wo /fo✓ (1) 10.1.2 (Maximum) Kinetic energy✓ (1) 10.2 Photoelectric effect✓ (1) 10.3 The minimum energy ✓ that an electron in the metal needs to be emitted from the metal surface ✓ Die minimum enerqie wat 'n e/ektron in die metaa/ vanaf die metaa/- oppervlak moet uitstraal (2) 10.4.1 W = hf ✓ o o =(6,63x10-34 )(1,75x10 15 ) ✓ = 1 , 16 X 1 o-lB J ✓ (3) 10.4.2 POSITIVE MARKING FROM QUESTION 10,4.1 / POSIT/EWE NASIEN VANAF 10.4.1 E = WO + Ek(max) hf= hf�+ ½�v,�1., ✓ (6,63 X10-34 )f = 1,J 6 X 10-IS + ½(9,11 X 10-)I )(5,23 X 10 5 ) ✓ 2 15 f = 1,94x 10 Hz ✓ (4) 10.5 Increases/ Neem toe ✓ The number of photons per unit time increases ✓ when the intensity of the light increase. Therefore, the number of photo-electrons per unit time increase. ✓ (Thus the current (I =_g__) increases.) Die aantal fotone per eenheidstyd neem toe wanneer die intensiteit van die Jig toeneem. Daarom sa/ die aantal foto-elektrone per eenheidstyd toeneem. (Dus neem die stroom (I =;j toe.) (3) [15] TOTAL/TOTAAL: 150 Copyright reserved/kopiereg voorbehou Please turn over/B/aai om asseblief