Physical Sciences Grade 12 P1 Free State Mock 2023 Paper & Memo

Downloaded from Stanmorephysics.com education Department of Education FREE STATE PROVINCE TIME: 3 HOURS This question paper consists of 15 pages and 3 data sheets. Copyright reserved Please turn over

Bewntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. INSTRUCTIONS AND INFORMATION dl, 10. uh 12. Write your name and other information in the appropriate spaces on the BOOK. StartEACH question on a NEW page in the ANSWER BOOK. Number the answers correctly according to the numbering system used in this question paper. Leave ONE line between two subquestions, for example between QUESTION 2.1 and QUESTION 2.2. You may use a non-programmable calculator. You may use appropriate mathematical instruments. Show ALL formulae and substitutions in ALL calculations. Round off your FINAL numerical answers to a minimum of TWO decimal places where applicable. Give brief motivations, discussions, et cetera where required. You are advised to use the attached DATA SHEETS. Write neatly and legibly. Copyright reserved Please turn over

Bewntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. QUESTION 1: MULTIPLE-CHOICE QUESTION Various options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the answer and write only the letter (A-D) nex ?the question numbers (1.1 to 1.10) in the ANSWER BOOK, e.g.1.41E 11 A bod: i$ moving at CONSTANT VELOCITY. The net force acting on it is ... A zero. B constant. C increasing. D decreasing. (2) 1.2 | Aball has a mechanical energy E when it is at rest h metres above the ground. The ball is then dropped from this rest position. What is the kinetic energy of the ball when it is ; h metres above the ground? A +E 3 B 7E 3 CE D se (2) 1.3. Acar of mass m is travelling at a constant velocity and has momentum p. The driver notices an object ahead of him and applies the brakes so that the momentum of the car changes to = p. 2 of the diagrams below correctly shows the relationship between ? A e pi Pr Ap Pr a Ap pr Ap (2) Copyright reserved Please turn over

Bewntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. 1.4 An object of mass m is accelerated from a velocity v to a velocity 2v, as shown in the diagram below. j] par F, Whee net work done on the object? A mv? B +m 2 Cc smv 2 D 2mv? (2) 1.5. A ball is thrown vertically upwards with velocity v, from the top of building, h metres above the ground. A, B and C are the positions of the ball during its motion as shown in the position-time graph below. Position ft time (Ss) Which ONE of the following is correct for the VELOCITY and DISPLACEMENT of the ball when it is at point C? Velocity Displacement (m-s*) (m) A v 2h B 0 Cc 2h D -V 0 (2) Copyright reserved Please turn over

Bowntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. 1.6 Astronomers observed that light from a distant star undergoes a red shift. Which ONE of the following combinations is correct for the OBSERVED WAVELENGTH and the OBSERVED FREQUENCY when compared to the wavelength and frequency of the light source? (OBSERVED WAVELENGTH | OBSERVED FREQUENCY Lower than Lower than Higher than Lower than Cc Lower than Higher than D Higher than Higher than (2) 1.7. Three small identical spheres A, B and C carry charges as shown in the diagram below. The distance between spheres A and B is x. +4Q -Q +Q For sphere B to experience a ZERO net electrostatic force, what must the distance between spheres B and C be? A tx 4 B ; x Cc 2x D 4x (2) 1.8 The batteries in CIRCUITS P and Q are identical. Resistors R1, Re and R3 Copyright reserved are identical. The ammeters and conducting wires have negligible resistance, the voltmeters have very high resistance, while the resistance of the batteries CANNOT be ignored. CIRCUIT P CIRCUIT Q © Ri ©) vi Please turn over

Bewntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. Which ONE of the following combinations is correct for the VOLTMETER READING and the AMMETER READING in circuits P and Q? — VOLTMETER READING AMMETER READING a "Nf Ve >Va Ap > Aa | By Vp < Va Ap > Aa ee Vp > Va Ap < Aa D Ve < Va Ap <Aa (2) 1.9 Which ONE of the energy conversions below takes place when a DC motor is in operation? A __ Kinetic energy to Electrical energy. B_ Electrical energy to Kinetic energy. C_ Potential energy to Electrical energy. D Electrical energy to Potential energy. (2) 1.10 Which ONE of the following graphs correctly illustrates the relationship between maximum kinetic energy (Ex(max)) of the emitted electrons from a metal surface and frequency (f) of the incident light? A, B Eximax) Eximax) 0 . f 0 f Cc D Eximax) Ekimax) Copyright reserved Please turn over

Bewntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. QUESTION 2 (Start on a new page.) A block mi of mass 8 kg is connected by a light inextensible string which runs over a light frictionless pulley to block mz also of mass 8 kg. Block m1 is moving up an incline as shown in-the diagram below. The coefficient of kinetic friction between block m1 and the inclifiélis 0,2. 7] | m2 2.1. State Newton's Second Law of motion in words. (2) 2.2. Drawa labelled free-body diagram showing ALL the forces acting on block m1. (4) 2.3. Consider the forces acting on each block separately, and calculate the magnitude of the acceleration of the system. (6) 2.4 According to Newton's Third law, which force is the reaction to the weight of block m2? (2) 2.5 While the blocks are moving, the string breaks. What is the direction of the acceleration of block m: immediately after the string breaks? Write only UP or DOWN the incline. (1) [15] Copyright reserved Please turn over

Bewntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. QUESTION 3 (Start on a new page.) A crane is lifting bricks on a wooden pallet at a constant velocity. When the pallet reaches a height of 17,2 m above the ground, on pallet. Ignore the ‘eff The position versus time graph shows the motion of the brick from the moment it falls from the pallet, until it strikes the ground. 3.1 3.2 3.3 3.4 3.5 the bricks falls from the t of friction. Bricks LY x x cereal ¢— Wooden 172 pallet ig F 0 Position (m) 0 07 1,4 27 Time(s) Explain what is meant by the term free fall. (2) Write down the time taken by the brick to reach its maximum height after falling off the pallet. (1) Calculate the velocity of the brick just before it falls off the pallet. (3) Calculate the distance covered by the brick during the last second of its free fall. (5) Draw an acceleration versus time graph for the motion of the brick from the moment it falls from the wooden pallet until it just reaches the ground. (2) [13] Copyright reserved Please turn over

Bewntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. QUESTION 4 (Start on a new page.) 41 4.2 A 400 g ball is approaching a baseball player horizontally in a westerly direction at a speed of 40 m-s" as indicated in the diagram. The baseball player exerts a forcésof 400 N on the ball in an easterly direction. The ballis in contact with the bat for 0,1 s. 4.1.1 Define impulse. (2) 4.1.2 Calculate the velocity of the ball immediately after contact with the bat. (4) When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, car A with mass 1 750 kg, travelling to the right at 1,5 m-s"™, collides with car B of mass 1 450 kg travelling to the left at 1,1 m-s”. Measurements show that the speed of car A after the collision is 0,25 m:s‘ in its original direction. Ignore the effects of friction during the collision. Before After A B A B < —_———> 1,5 m-s* 1,1 m-s* 0,25 m-s* 4.2.1 Calculate the velocity of car B immediately after théldallision. (5) 4.2.2 "When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage.’ Explain how flexible bumpers reduce damages during low-speed collisions. (3) [14] Copyright reserved Please turn over

Bewntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. QUESTION 5 (Start on a new page.) After an emergency landing at Bloemfontein Airport, a passenger of mass 70 kg evacuates the stationary aeroplane using the evacuation slide that makes an angle of 40° wi fi the horizontal. When the passenger reaches a height of 2,8 m above the ground, t s|bassenger has a speed of 0,35 m-s". 40° ie the term non-conservative force. tictional force acts on the passenger as he moves down the slide. 5.2. Draw a labelled free-body diagram showing all the forces exerted on the passenger. 5.3. Name a non-conservative force acting on the passenger. The coefficient of kinetic friction of the surface is 0,112. 5.4 Using energy principles, calculate the velocity of the passenger at the lowest point of the slide. 5.5 The angle between the horizontal and the slide is increased. How will the change influence the frictional force acting on the passenger? Write only INCREASES, DECREASES or STAYS THE SAME. Explain the answer. (2) (3) (1) (6) (3) [15] Copyright reserved Please turn over

Bewntoaded:from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. QUESTION 6 (Start on a new page.) A learner investigates the relationship between the ratio of the observed frequency to the source fri ncy (2) and the velocity with which the listener moves relative to the . The experiment is repeated by increasing the magnitude of the with which the observer moves relative to the same stationary source. The observed frequency (f.) is recorded and the ratio (2) is calculated for each experiment. The graph below shows the results obtained. 1,45 fi fs 1,10 - OF an 1,05 eH 1,00 @ 0 5 10 15 20 25 30 35 vi (m-s") 6.1. State in words the phenomenon described above. (2) Use the graph to answer the following questions. 6.2 Is the learner moving TOWARDS or AWAY FROM the stationary source? Explain your answer. (2) (1) (4) 6.3. What does the gradient of the graph represent? 6.4 Calculate the speed of sound in air during the experimen 6.5 An observation of the spectrum of a distant star shows that Tris moving away from the earth. Explain, in terms of the frequencies of the spectral line, how it is possible to conclude that the star is moving away from the earth. (2) [11] Copyright reserved Please turn over

Bowntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. QUESTION 7 (Start on a new page.) 7.1 2 Two point charges A and B, carrying charges of +1,2 nC and +1,5 nC ively are placed 5 cm apart along a straight line in a vacuum, as shown 7.1.1. State Coloumb's law in words. (2) 7.1.2 Calculate the net electrostatic force experienced by the electron. (5) P is a point in an electric field 0,03 m from charged sphere A, as shown in the diagram below. The electric field experienced at point P due to the charged sphere A is 4x10’ N-C" to the right. A OQ» 0,03 m 7.2.1 Calculate the magnitude of the charge on sphere A. (3) 7.2.2 \|s the charge on sphere A POSITIVE or NEGATIVE? (1) 7.2.3, A point charge X, carrying an excess of 938 electrons, is now placed at point P. Calculate the magnitude of the electrostatic force experienced by point charge X. (4) [15] Copyright reserved Please turn over

Bewntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. QUESTION 8 (Start on a new page.) The diagram below shows a circuit with a battery of emf 12 V and internal resistance r. The battery is connected to resistors of 25 Q, 20 Q, 5Q and resistor X of unknown resistance. Thelammeters, switch and the conducting wires have negligible resistance, while the gue have very high resistance. 250 On 5Q X 8.1. Explain what is meant by an emf of 12 V. (2) When switch S is closed, the reading on ammeter Az is 0,2 A and the reading on voltmeter V2 is 5,5 V. 8.2 Calculate the: 8.2.1 Reading on ammeter Ai (3) 8.2.2 Resistance of resistor X (3) 8.2.3 Total external resistance of the circuit (3) 8.2.4 Internal resistance (r) of the battery (3) 8.3. Resistor X is replaced by another ammeter. Will the reading on voltmeter Vi INCREASE, DECREASE on REMAIN THE SAME? Explain the answer. T (4) [18] Copyright reserved Please turn over

Bewntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. QUESTION 9 (Start on a new page.) 9.1 The diagram below is a simplified representation of a generator. The coil is rotated . clockwise. 9.1.1 Is the above a DC or an AC generator? (1) 9.1.2 Write down the name of component X. (1) 9.1.3 Write down the function of component Y. (1) 9.1.4 Is the direction of the induced current from P to Q or from Qto P? (1) 9.2 The graph below shows the output voltage from a household AC generator for the rotation of the coil for one complete cycle. 4 $+ 311,13 f pag 0,02 *t (s) Output voltage (V) - 311,13 v 9.2.1 The speed of rotation is doubled. Sketch the graph of the induced emf versus the time for this change for one cycle. Indicate the maximum voltage and relevant time values. (3) 9.2.2 An electrical device is connected to this generator. The-maximum current passing through the device is 8 A. Calculate the: (a) Resistance of the device (4) (b) Energy the device consumes in two hours (4) [15] Copyright reserved Please turn over

Bewntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. QUESTION 10 (Start on a new page.) A learner uses two photoelectric cells to determine the maximum kinetic energy of photoelectrons from caesium and sodium cathodes. Monochromatic light from the SAME SOURGE is shone on the cathode of each photoelectric cell, as shown in the diagram. Ultraviolet light The ammeter records a small current in each case. The learner compiles the following data for the two metals: MAXIMUM VELOCITY OF WORK FUNCTION METAL OF THE METAL (J) THE Pro ey RONS Caesium 3,36 x 10°19 7,14 x 10° Sodium 3,65 x 10°79 Vmax 10.1 Define the term work function of a metal. (2) 10.2 Use the information in the table to calculate the wavelength of the light used in the experiment. (5) 10.3 Calculate the maximum velocity of an ejected electron from sodium metal. (4) 10.4 The intensity of the incident light was INCREASED. How does the change affect the reading on the ammeter? Write down only INCREASE, DECREASE or REMAINS THE SAME. Explain the answer. (3) [14] TOTAL: 150 Copyright reserved

Bewntoaded: from Stanmorephysics.com Grade 12 Prep. Exam. FS/September 2023 DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 1 (PHYSICS) VRAESTEL 1 (FISIKA) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12 ICAL CONSTANTS / TABEL 1: FISIESE KONSTANTES NAME / NAAM SYMBOL / SIMBOOL | VALUE / WAARDE Acceleration due to gravity 3 Swaartekragversnelling 9 oe re Uni i itational tant niversal gravitational constan G 6,67 x 10" Nem2-kg2 Universele gravitasiekonstant Radius of the Earth 6 Radius van die Aarde Re a al Mass of the Earth M 5 98x 102k Massa van die Aarde e oe 9 Speed of light in a vacuum er Spoed van lig in 'n vakuum ° a nesendonien Planck's constant -34 |. Planck se konstante 5 SESe 14s Coulomb's constant k 9.0 x 10° N-m2-C2 Coulomb se konstante wx m Charge on electron 48 Lading op elektron © Aeon Electron mass 1-31 Elektronmassa me ari TOP Copyright reserved Please turn over

Bewntoaded: from Stanmorephysics.com FS/September 2023 Grade 12 Prep. Exam. TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING Vp =Vji + a Atay Ax =v,At+ tat? or/of Ay =v At+ tat? ve aye ee orlof v,* =v, +2ady | Ax= [“ = Jat orlof Ay -(* = Jat i —" FORCE/KRAG Fret =ma p=nwv f° =y,N f= U.N F_,At =Ap w= Ap= mv, - mv; a m. Freon orlof Freon 9 =c4t orlof 9 =GS WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING W =FAxcos 0 U=mgh orlof Ep=mgh 7 4 Wyet = AK orlof =W,., = AE, K=> mv? or/of E,= pnw’ AK=K,-K, oflof AE, =Ey —Ey Wye = AK +AU orfof Wy, = AE, +AE, p= Pav =Fvavy / Pgemia = FvVgemid WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG F = T=, v=fr f + aver gay E=hf orlof E=h® viv, ViV, x E = Wy + Excmax/maks) OF/Of E = Wy + Kmax/maxs Where/waar ae 7 ttf E=hf and/en w, =hf, and/en Excmax/maks) = 5 MV vax sia ce Syl K qnax/maks) = 2mv max/maks Copyright reserved Please turn over

Bowntoaded: from Stanmorephysics.com Grade 12 Prep. Exam. ELECTROSTATICS/ELEKTROSTATIKA ke 12 tH o|O I n= or/of Q n=— ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE emf (€ )= I(R +r) R= emk (€ )= I(R +r) R, =R,+R_ +... 1 1°41 q=lAt ee R, R, R, W=Vq pW. At W=VIAt P=VI W=FPRAt 5 P=PR w=“ _ve R ALTERNATING CURRENT/WISSELSTROOM mo 2 V v= Max m2 I ——maks we 2 V, Voge = Pires = V sired rae / Pyemiddela = Vuk Tce Pas =17,.R / Pryemidaeta = Fg v2 v2 fy Fu ~ R / Few = = Copyright reserved FS/September 2023

Copyright reserved/Kopiereg voorbehou Downloaded from Stanmorephysics.com education Department of Education FREE STATE PROVINCE PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) SEPTEMBER 2023 MARKS/PUNTE: 150 MARKING GUIDELINES NASIENRIGLYNE This marking guideline consists of 15 pages. Hierdie nasienriglyne bestaan uit 15 bladsye. Please turn over/Blaai om asseblief

Bewntoaded fsemvéiéanmevephysics.com FS/VSSeptember 2023 Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne (2) (2) (2) (2) (2) (2) (2) (2) (2) (2) [20] QUESTION/VRAAG 1 1.4 4 1.3 1.4 15 Dvv 16 By'¥ 17 Byv 18 Cvv 19 Byv 1:40) Dw QUESTION/VRAAG 2 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Net/Resultant force mentioned at least once. Indien enige van die onderstreepte sleutelwoorde/-frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase. Netto/resultante krag ten minste een keer genoem. 2.1. When a resultant/net force acts on an object, the object will accelerate in the direction of the net force at an acceleration that is directly proportional to the net/resultant force v and inversely proportional to the mass of the object. v Wanneer 'n resulterende/netto krag op 'n voorwerp inwerk, sal die voorwerp versnel in die rigting van die netto krag by 'n versnelling wat direk eweredig is aan die netto/resultante krag en omgekeerd eweredig aafdie massa van die voorwerp. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Bowntoaded fsemvéiéanmervepkysics.com FS/VSSeptember 2023 Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne Tv ACCEPT/AANVAAR v TY v Accept the following symbols/Aanvaar die volgende simbole T F1/Force on the rope/Krag op die tou Ww. Fg /weight/gewig N Fn/Normal/Normal force/Normaal/Normaalkrag fk NINES f/Friction/frictional force/kinetic frictional force f/Wrywing/Wywingskrag/Kinetiese wrywingskrag (4) NB: Allocate ONE mark for each correct arrow and label in first option. For the second option, give one mark for both components of weight, correctly drawn with arrows and labels. Deduct one mark for any additional forces. NB: Ken EEN punt toe vir elke korrekte pyltjie en etiket in eerste opsie. Vir die tweede opsie, gee een punt vir beide komponente van gewig, korrek geteken met pyle en byskrifte. Trek een punt af vir enige bykomende kragte. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Riowntoaded fremusiéanmervephysics.com FS/VSSeptember 2023 Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne 23 OPTION/OPSIE 1 Take the direction of motion of each block as positive Neempdie bewegingsrigting van elke blok as positief w+ (-T)=ma (8)(9,8) — TY = 8as Form: Any/Enige v Fret = ma T + (-wi) + (-fx) = 8a T-(8x9,8)sin30°V - 0,2(8)(9,8)cos30°V = 8a ....(2) 142: (8)(9,8) - (8x9,8)sin30° - 0,2(8)(9,8)cos30° = 16a a= 1,6 m-s?v OPTION/OPSIE 2 a Take the direction of motion of ea Neem die bewegingsrigting van elke’ iS negative negatief For/Vir m2 Fret = ma ¥ -w2+ T =-ma -(8)(9,8) + TY = -8a......... (1) For m1 Any/Enige Vv Fret = ma T + (-wi) + (-fx) = 8a -T+(8x9,8)sin30°v + 0,2(8)(9,8)cos30°V = -8a ....(2) T2: -(8)(9,8) + (8x9,8)sin30° + 0,2(8)(9,8)cos30° = -16a a=1.6 ms? ¥ (6) 2.4 — The force that block(mz) exerts on the earth. vv Die krag wat blok (m2) op die aarde uitoefen. (2) 2.5 Down (the incline)/Af (die helling) Vv (1) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Riowntoaded fremusiéanmervephysics.com Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne FS/VS/September 2023 QUESTION/VRAAG 3 3.1. Motion of an object under the influence of the gravitational force only.v ¥ Beweging van 'n voorwerp slegs onder die invloed van gravitasiekrag. (2) (1) 3.3 OPTION/OPSIE 1 OPTION/OPSIE 2 UPWARDS POSITIVE DOWNWARDS POSITIVE OPWAARTS POSITIEF | AFWAARTS POSITIEF vi = vi + gAtY Vt = vi + gAtY 0 = vi + (9,8)(0,7) ¥ 0 = vi t (-9,8)(0,7) ¥ Vi = —6,86 vi = 6,86 m-st¥ vi =6.86 ms? ¥ OPTION/OPSIE 3 OPTION/OPSIE 4 UPWARDS POSITIVE DOWNWARDS POSITIVE OPWAARTS POSITIEF | AFWAARTS POSITIEF Ay = viAt + ZgAryY = 2 Ay = MiAt + Vagaty’ -2,4 = vi + (9,8)(0,7)2¥ 2,4 = vi + (-9,8)(0,7)°¥ aa is vi = —6,86 vi = 6,86 m-s vi = 6,86 m-s? ¥ OPTION/OPSIE 5 OPTION/OPSIE 6 UPWARDS POSITIVE UPWARDS POSITIVE OPWAARTS POSITIEF OPWAARTS POSITIEF vp = vp + 2gAyY v7 = vp + 2gAyv 0 = v? + 2(9,8)(— 2,4)” 0 = v? + 2(—9,8)(2,4)¥ yV+v ay = (“aw vj = 6,86 m:s-t¥ iy = G86 v; = 6,86m:s iv OPTION/OPSIE 7 OPTION/OPSIE 8 UPWARDS POSITIVE UPWARDS POSITIVE OPWAARTS POSITIEF OPWAARTS POSITIEF vi + Y, ay = (7S 1) tw vp = 6,86m-s71¥ (mgh + sv”); = (mgh + 5mv*) v (gh + 30); = (gh + 5v), 9,8(2,4) +5(0) = 9,8(0) +5u7Y v,it+0 i +0 2,4 = (— v 24= e )o7v 2,4 ( )o7 - -1 Wr vy; = —6,86 vy, = 6,86 m:s~* down/af % = 6,86 m-s-* dewn/afv OPTION/OPSIE 9 OPTION/OPSIE10 Wnet = AKY mgAxcos6 =-sm{v; — v7) 9,8(2,4)(1) = 3(v} — 0) vp = 6,86m-s-t¥ Copyright reserved/Kopiereg voorbehou (3) Please turn over/Blaai om asseblief

Bewntoaded fsemvéiéanmervepkysics.com Grade/Graad 12 Prep.Exam/Voorb. Eksam. FS/VS/September 2023 Marking Guidelines/Nasienriglyne 3.4 POSITIVE MARKING FROMI/POSITIEWE NASIEN VANAF 3.3 OPTION/OPSIE 1 UPWARDS POSITIVE Ay = pile + 5ghtey = (-9,8)(1) + 3(-9,8)(1)? ~ =-14,7m Distance/Afstand = 14,7 mv OPTION/OPSIE 2 DOWNWARDS POSITIVE AFWAARTS POSITIEF Whole motion/Hele beweging ve = vj + gdt — 6,86 + (9,8)(1,7)¥ 98mst Ay 3 =v;At + 5gAtev = (9; Hay +5(9,8)(1)? ~ = +14,7m Distance/Afstand = 14,7 mY Wil UPWARDS POSITIVE OPWAARTS POSITIEF vy = y, + gAt = 6,86 + (—9,8)(1,7)¥ = -98ms"! Velocity striking the ground./ Snelheid as dit grond tref. (Whole motion/Hele beweging) ve = v + gdt = 6,86 + (—9,8)(2,7)” = -19,6m:s7} OR/OF (Maximum height/Maksimum hoogte) vy = Y, + gAt = 0 + (—9,8)(2) = -19,6m:s71 OR/OF (Point of launch/Punt waar vrygelaat) vy = YH + gdt = —6,86 + (—9,8)(1,3) = -19,6m-s7} OR/OF (Last one second/Duur een sekonde) Ve = yi + ght —9,8 + (—9,8)(1) —19,6m:s7! oil OPTION/OPSIE 3 UPWARDS POSITIVE OPWAARTS' POSITIEF vp = v} + 2ghyY (- ~19,6)? = (-9,8)? » + 2(—9,8) (Ay) ¥ Ay = -14,7m Distance/Afstand = 14,7 mY OPTION/OPSIE 4 DOWNWARDS POSITIVE AFWAARTS POSITIEF > ve = vp + 2gdyv (19,6)? = (9,8)? + 2(9,8)(Ay)“ Ay = +14,7m Distance/Afstand = 14,7 m¥ OPTION/OPSIE 5 UPWARDS POSITIVE OPWAARTS POSITIEF > Ay = 5) AtY _ care jay Ay = -14,7m Distance/Afstand = 14.7mv OPTION/OPSIE 6 DOWNWARDS POSITIVE AFWAARTS POS TIEF 1m xs (“ + say ee 19,6 2 av Ay = 14,7m Distance/Afstand = 14,7m~” Copyright reserved/Kopiereg voorbehou (5) Please turn over/Blaai om asseblief

Riowntoaded fremusiéanmervephysics.com Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne ua DOWNWARDS POSITIVE DOWNWARDS NEGATIVE AFWAARTS POSITIEF AFWAARTS POSITIEF vy a m-s? t(s) (m-s*) 9 t(s) -9,8 vv QUESTION/VRAAG 4 FS/VS/September 2023 (2) [13] 4.1.1. The product of the resultant/net force acting on an object and the time the resultant/net force acts on the object. VW Die produk van die resulterende/netto krag wat op 'n voorwerp inwerk (2/0) en die tyd wat die resulterende/netto krag op die voorwerp inwerk. Accept/Aanvaar: Impulse is the change in momentum of an object. (2/0) Impuls is die verandering in momentum van 'n voorwerp. OPTION/OPSIE 1 EAST AS POSITIVE OOS AS POSITIEF Fnet At = Ap ¥ FnetdAt = m(vr — vi) (400)(0,1) “= 0,4v1—-(0,4)(-40) ¥ vr = 60 m-s"', east/oos ¥ OPTION/OPSIE 2 EAST AS NEGATIVE OOS AS NEGATIEF Fnet At = Ap ¥ FnetAt = m(vr — vi) (-400)(0,1) ¥ = 0,4vr—(0,4)(40) ¥ vr = - 60 m-s*! *. ve = 60 m-s", east/oos ¥ OPTION/OPSIE 3 EAST AS POSITIVE OOS AS POSITIEF Fret = ma 400 = 04a v a= 1000 m-s? vf = vi + aAtv = -40 +(1000)(0,1)¥ = 60 m-s", east/oos ¥ OPTION/OPSIE 4 EAST AS NEGATIVE OOS AS NEGATIEF Fret = ma - 400 = 0,4a Vv a=-1000 m-s? vf = vi + aAtY = 40 +(-1000)(0,1)” =-60m = 60 m-s", east/oos ¥ Copyright reserved/Kopiereg voorbehou (2) (4) Please turn over/Blaai om asseblief

Riowntoaded fremusiéanmervepkysics.com FS/VSSeptember 2023 Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne 4.2.1 OPTION/OPSIE 1 RIGHT AS POSITIVE/REGS AS POSITIEF Zp(before/voor) = Xp(after/na) ~ mavai + MBVBi = MBVif + MBV2F a 750 5) + (1450)(-1,1)” = (1750)(0,25) + (1450)verr ver = 0,41 ms" to the right/na regs ¥ OPTION/OPSIE 2 LEFT AS POSITIVE/LINKS AS POSITIEF =p(before/voor) = Xp(after/na) ~ mavai + MBVBi = MaVvaf + MBVBF (1750)(-1,5) + (1450)(1,1)” = (1750)(-0,25) + (1450)varw Ver = - 0,41 m-s ver = 0,41 m-s"'v to the right/na regs V OPTION/OPSIE 3 RIGHT AS POSITIVE/REGS AS POSITIEF A pa =- A paY m(var — Vai) = - M(Ver — VBi) 1750(0,25 — 1,5)” = -1450(ver— (-1,1))” ver = 0,41 m-s"'v to the right/na regs¥ OPTION/OPSIE 4 LEFT AS POSITIVE/LINKS AS POSITIEF A pa =- A paY m(var — Vai) = -m(Ver — VBi) 1750(0,25 — 1,5)” = -1450(var— 1,1)” =-0,41 ms" ver = 0,41 m-s‘'v to the right/na regs ¥ (5) 4.2.2 Flexible bumpers reduce the net forcev by increasing the ti to bring about the same change in momentum. “ required Buigsame buffers verminder die netto krag deur die tyd wat-benodig word om dieselfde verandering in momentum teweeg te bring, te verhoog. (3) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Riowntoaded fremusiéanmervephysics.com FS/VSSeptember 2023 Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1. A force for which the work done (in moving an object between two points) ds-on the path taken. VV arvoor die arbeid verrig (om 'n voorwerp tussen twee punte te fhanklik is van die roete wat gevolg word. (2) NP 4 52 fv Accept the following symbols/Aanvaar die volgende simbole w Fg /weight/gewig v N Fn/Normal/Normal force/Normaal/normaalkrag | ¥ (3) wy f f/Friction/frictional force/kinetic frictional force | /, Wrywing/wrywingskrag/kinetiese wrywingskrag 5.3. Frictional force/Wrywingskrag ¥ (1) 5.4 OPTIO/OPSIE 1 x 2,8 2,8 x =n age ~ #956 Wnhet = 4K 1 = 2 2 Wegjj + We = zm(v7 -v?) v AnylEnige Fg//Axcos@ + fAxcos® = sm(vf — v?) v ) cos 0° + 0,112(70x9,8)cos 40° ( inae , % 1 2 _ Yaor2 70(9,8) sin 40 (= aa ) cos 1.85 = 5 (70) [v7 0,357] ve = 6,905 m-s! v sin 40° OPTION/OPSIE 2 Whee = AK mgAxcos@ + fAxcosé = 5m(v} - 70(9,8) 50° ( ae ye 0° + 0,112(70 x98) 40° ( co HIE sin4o) ©°S : ndeOs! sin40°. v Any/Enige vf) [nme v v ) cos1g0° = 5(70)[v? — 0,357) ame ve = 6,905 m:s tv OPTION/OPSIE 3 Ming: = Abs “BU v Any/Enige fxcoosd = m(v} — v7) + mg(hy — hy) weg , v v v 0,112(70 x 9,8)cos40° (=) cos180° = 3(70)[v?'— (0,35)2] + 9,8)(0 — 2,8) vp = 6,905 m:s7?” (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Bowntoaded fsemvéiéanmerveptyysics.com FS/VSSeptember 2023 Grade/Graad 12 Prep.Exam/Voorb. Eksam. 5.5 Marking Guidelines/Nasienriglyne Decreases ¥ The normal force decreases V (because cos@ decreases with increase of the size of the angle to the horizontal). Friction is directly proportional to the fetal / f =pN / (F< N). ¥ Die.normaalkrag neem af (omdat cos@ afneem met toename van die grootte van die hoek met die horisontaal). Wrywing is direk eweredig aan die normaal / f =iN / (f « N). (3) [15] QUESTION/VRAAG 6 6.1 6.2 6.3 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 (bespeur deur 'n luisteraar) omdat die klankbron en die luisteraar verskillende snelhede het relatief tot die medium van klankvoortplanting. OR/OF An (apparent) change in (observed/detected) frequency (pitch), (wavelength) Y because 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) Towards/Na ¥ & >1v OPTION/OPSIE 2 (2) fs £. increases as v, increases ¥ ‘s fu — Neem toe soos wat v, toeneem s (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Riowntoaded fremusiéanmervephysics.com Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne 6.4 Use any set of values from the graph. Gebruik enige ste! waardes uit die grafiek. FS/VS/September 2023 OPTION/OPSIE 1 OPTION/OPSIE 2 v dh VE v 20 v 1,06 =— +1 v v = 333,33m:s ty gradient = = 0,003 = 0,003 ssir 1,06¥ —1v 20-0 VY = 333,33m-s' y OPTION/OPSIE 3 A 1,06 ¥ = v(1,06 = v = 333,33m:s tv fi vt+uy, v v +20, v v + 20 (4) 6.5 The spectral lines of the star are shifted towards the lower frequency end Y which is the red end of the spectrum. “ Die spektraallyne van die ster word na die laer frekwensie-end verskuif wat die rooi-end van die spektrum is. Copyright reserved/Kopiereg voorbehou (2) [11] Please turn over/Blaai om asseblief

Bowntoaded fsemvéiéanmerveplysics.com FS/VSSeptember 2023 Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 7.1.1. The magnitude of the electrostatic force exerted by one point charge on another; int charge is directly proportional to the product of the magnitude of the chargesY and inversely proportional to the square of the si farice between them. “ Die piettte van die elektrostatiese krag wat deur een puntlading op na ‘puntlading uitgeoefen word, is direk eweredig aan die produk van die grootte van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. (2) TAQ _ kQiQ2 v Fae on r2 _ 9x 10° x (1,2 x 1079)(1,6 x 10719) ~ (4 x 1072)? v = 1,08 x 10715 N, left/links kQ1Q2 Fee = 72 9x 10° x (1,5 x 10°°)(1,6 x 10-1) v (1 x 10-2)2 = 2,16 x 10-"4N , right/regs Fret = Fae — Fee = 1,08x 10-15 — 2,16 x 10714 — 2,052 x 10714 2,052 x 10-14 N , right/regs ¥ (5) 72H E=k%v " 4x 107 =(9 x 109) pa Q = 4x10-cv (3) 7.2.2 Positive/Positief V (1) 7.2.3] Q = nage OPTION/OPSIE 2 = 938 x 1,6x 10-19 v pia Ko = 15 K 107% j a Q 4x10” =(9x10)° OPTION/OPSIE 1 (0,03)? Q = 4x10-°C F Bata Q F=K — ¥ 4x10? =———___ v d _ — 1,5 x 10-16 _ 9X 10°(1,5 x 10719)(4 x 107%) | F = 6x10-°NY ~ (0,03)? F =6x10-°NV (4) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Bewntoaded fsemvéiéanmerveptysics.com Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 8.1 821 8.2.2 POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 8.2.1 8.2.3 POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 8.2.2 Copyright reserved/Kopiereg voorbehou FS/VS/September 2023 12 J Vof energy is transferred per one coulomb of charge. 12 J “energie word per een coulomb lading oorgedra. OPTION/OPSIE 2 302 = V2oa Vs0q = IR 1399 (30) = Izo9 (20) = 0,2(25 + 5)v 0,2(30) = Ingq(20)¥ = OY = 034A V39 = Vaio = I, = 0,3 + 02% , = oem = 0,5AV =i lbp = 0,3A OPTION/OPSIE 3 1, = 0,34 + 0,24v ¥, = 0,2(30) = ia = 6VV 1 1 1 Rp 20° 25 +5 Rp = 12.2 Vy = IpRp 6 =1,(12)¥ Ip = 0,5A¥ Ve =—Vv “ 55 =v 0,5 = 110V OPTION/OPSIE 1 111 Ry Ri Ry 1 1 =— —vY 20 * 30 Ry = 120 Rent = Ry + Rg =12+11” = 230 OPTION/OPSIE 2 _ product/produk Pp sum/som _ 20x30 y ~ 20+30 =122 Rext = Rext +Rs = 12+11 =230 7% a a (2) (3) (3) (3) Please turn over/Blaai om asseblief

Riownhoaded fremusiéanmervephysics.com FS/VSSeptember 2023 Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne 8.2.4 OPTION/OPSIE 1 OPTION/OPSIE 2 e = TRi+ 1)” Viose = & — Vee 12 =105(23 + r) ~ ns 12— 115% r dinny = 0,5 pul Viose = Inv ge 0,5 = 0,5r (3) r=10V 8.3 Decrease/Afneem ¥ Total external resistance decreases/Totale eksterne weerstand neem af Total current increases/Totale stroom neem toe }v Viost increases/neem toe V Vext decreases/neem af V (Emr stays constant/bly konstant) (8 QUESTION/VRAAG 9 9.1.1 AC/WS v (1) 9.1.2 Slip rings/Sleepringe ¥ (1) 9.1.3. To ensure conductivity between the slip rings and the external circuit. Vv Om geleiding tussen sleepringe en die eksterne stroombaan te verseker. (1) OR/OF Allow for rotation of the slip rings while maintaining contact with the external circuit. Laat die sleepringe roteer terwyl kontak met die eksterne stroombaan behou word. 9.1.4 Pto/totQv (1) oe + 620k --,¥----------------- MARKING CRITERIA A NASIENKRITERIA 2 ! \ v | Shape/Vorm § ! 1 v | Amplitude +/-622V 3 Ya Y_| Half the period (0,01 s) 2 I 1/0,01s Helfte vat die tyd 3 0 t—_—1———*t (s) Jagr s 1 I Inn: z \ i eal 3 \ Lo 6 \! 3 - 622 -------- J----------- (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Riewntoaded fremusibarmervephysics.com Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne FS/VS/September 2023 9.2.2 (a) OPTION/OPSIE1 | OPTION/OPSIE 2 far - pal Vnax Vinax = ImaxR¥ Wu “7s | 311,12%= BRv 311, R = 38,870Y v2 =5,66A Yims = IrmsRY 220 “= 5,66RY R = 38,87 0Y (4) 9.2.2 (b) POSITIVE MARKING FROM/POSITIEWE NASIEN VANAF 9.2.2(a) OPTION/OPSIE 1 OPTION/OPSIE 2 OPTION/OPSIE 3 Pave = Vems!rms Pave. = TrmsR p= Vins = 220(5,66) = (5,66)?38,87 ave ~~ R- 1245,2W 1245,22W _ (220)? ~~ 38,87 pos p=“y 1245,18 W At ae af At on 1245,2=>v 1245,22 = Tv ity W = 8965440 Jv W = 8965611,16 JY v at 1 245,18 = —“-v 7200 W = 8965268,85 Jv OPTION/OPSIE 4 OPTION/OPSIE 5 OPTION/OPSIE 6 W = VIAty, W = PRAtv = 220(5,66}(7200)Y | = (5,66)2(38,87)(7200)¥ = 8965440 Jv = 8965611,16 JY Copyright reserved/Kopiereg voorbehou (4) [15] Please turn over/Blaai om asseblief

Howntoaded fsemvéiéanmervepteysics.com FS/VSSeptember 2023 Grade/Graad 12 Prep.Exam/Voorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 10.1 10.2 10.3 10.4 The work function of a metal is the minimum energy V that an electron (in thé;metal) needs to be emitted/ejected from the(metal)surface. ¥ ‘unksie van 'n metaal is die minimum energie wat 'n elektron taal) uit die (metaal)opperviak moet vrygestel. (2) E= Wo+ Kmax/maks AnviEnige 7 ny/Enige 1 he =Wo + 2 WV aaeprians (3 x 108)” vo 6,63 x 10-** = = 3,36 x'10-1 + 49,11 x 10-34)(7,14 x 105)?°v a A = 3,50 x 10-7 mY (5) Positive marking from/Positiewe nasien vanaf 10.2 E = Wo + Kmax/maks . c 1 Any/Enige v hy =Wo + 3 ™MVmax/maks te (3 X10) 3,50 Vmax/maks = 6,68 x 6,6,3 x 10734 1 v 65 x 10719 + 3 (9,11 % 10° VE a wimnniics 7 (4) Increase V More photo electrons strike the surface of the metal per unit time. ” More electrons are ejected per unit time. ¥ From the formula J = 2 (= ra ; | « n) ammeter reading increases. Verhoog Meer fotoelektrone tref die opperviak van die metaal per eenheid tyd. Meer elektrone word per eenheid tyd vrygestel. Vir die formule 1 = 2 (= “4 ; 1 « n) ammeterlesing neem toe. (3) TOTALITOTAAL: 150 ~~ es Copyright reserved/Kopiereg voorbehou