Physical Sciences Grade 12 P1 Northern Cape Prelim 2024 Memo

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com ————————————————————— DEPARTMENT OF EDUCATION ————————————————————— DEPARTEMENT VAN ONDERWYS ————————————————————— LEFAPHA LA THUTO ————————————————————— ISEBE LEZEMFUNDO ————————————————————— PROVINSIALE VOORBEREIDENDE EKSAMEN/ PROVINCIAL PREPARATORY EXAMINATION GRAAD/GRADE 12 FISIESE WETENSKAPPE/PHYSICAL SCIENCES VRAESTEL/PAPER 1 FISIKA/PHYSICS (FISIKA/PHYSICS) SEPTEMBER 2024 PUNTE/MARKS: 150 TYD/TIME: 3 uur/hours Hierdie vraestel bestaan uit 16 bladsye en 3 gegewensblaaie./ This question paper consists of 16 pages and 3 data sheets. Kopiereg voorbehou/Copyright reserved Blaai om asseblief/Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 2 com September 2024 INSTRUCTIONS AND INFORMATION 1. Write your name on the ANSWER BOOK. 2. This question paper consists of 10 questions. Answer ALL the 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 subquestions, e.g. between QUESTION 2.1 and QUESTION 2.2. 6. You may use a non-programmable calculator. 7. You may use appropriate mathematical instruments. 8. Show ALL formulae and substitutions in ALL calculations. 9. Round off your FINAL numerical answers to a minimum of TWO decimal places. 10. Give brief motivations, discussions, etc. where required. 11. You are advised to use the attached DATA SHEETS. 12. Write neatly and legibly. Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 3 com September 2024 QUESTION 1: MULTIPLE-CHOICE QUESTIONS 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) next to the question numbers (1.1 to 1.10) in the ANSWER BOOK, e.g. 1.11 E. 1.1 Which ONE of the following physical quantities is defined as the rate of change of velocity? A Impulse B Net force C Inertia D Acceleration (2) 1.2 According to Newton's Third Law of Motion, the reaction force to the weight of a book lying on a table is the … A normal force. B force of the book on Earth. C force of Earth on the book. D force of the book on the table. (2) 1.3 A 300 kg bag of cement and a 30 kg iron sphere are dropped simultaneously from the roof of a building. Ignore the effects of air friction. When they are 9 m above the ground, they will have the same … A momentum. B acceleration. C kinetic energy. D potential energy. (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 4 com September 2024 1.4 A ball of mass m strikes a wall perpendicularly at a speed v. After the collision, the ball moves in the opposite direction at the same speed v, as shown in the diagram below. wall v m v m Which ONE of the following represents the magnitude of the change in momentum of the ball? A 0 B mv C 2mv D 2mv2 (2) 1.5 The kinetic energy of a car moving at a constant velocity of 2v is K. The velocity of the car then changes to a constant velocity of 4v. Which ONE of the following represents the kinetic energy of the car at a constant velocity of 4v? A ⅛K B ¼K C 4K D 8K (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 5 com September 2024 1.6 The siren of a police car travelling at 20 m·s-1 emits sound waves of frequency f Hz. The frequency observed by a pedestrian standing on the road when the police car moves towards him, will be … A 0 Hz. B smaller than f Hz. C equal to f Hz. D greater than f Hz. (2) 1.7 The diagram below represents a positive point charge. Points R, S and T are at different distances from the charge. R ● T ● + ● S The magnitude of the electric field at a point is the … A greatest at point R. B greatest at point S. C greatest at point T. D same at point R, S and T. (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 6 com September 2024 1.8 In the circuit represented below, the resistance of a variable resistor is INCREASED. rr V A How would this increase affect the readings on the voltmeter and ammeter? VOLTMETER READING AMMETER READING A Increases Decreases B Decreases Increases C Increases Increases D Decreases Decreases (2) 1.9 Which ONE of the energy conversions below takes place when an AC motor is in operation? A Electrical energy to mechanical energy B Mechanical energy to electrical energy C Kinetic energy to mechanical energy D Heat energy to mechanical energy (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 7 com September 2024 1.10 When a clean metal plate is irradiated with red light of sufficient energy, photoelectrons are emitted. The red light is now replaced with blue light of the same intensity. This change will … A increase the number of photoelectrons emitted per second. B decrease the number of photoelectrons emitted per second. C increase the kinetic energy of the emitted photoelectrons. D decrease the kinetic energy of the emitted photoelectrons. (2) [20] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 8 com September 2024 QUESTION 2 (Start on a new page.) Two blocks of masses 9 kg and 4 kg are connected by a light, inextensible string. A constant force, F, acting at 75° to the horizontal, is applied to the 9 kg block as shown in the diagram below. The string runs over a frictionless pulley. F 4 m·s-1 75o 9 kg 4 kg The 9 kg block moves at a CONSTANT velocity of 4 m·s-1 while experiencing a frictional force of 2,5 N. 2.1 State Newton's First Law of Motion in words. (2) 2.2 Draw a labelled free-body diagram for the 9 kg block. (5) 2.3 Calculate the magnitude of the force F acting on the 9 kg block. (5) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 9 com September 2024 QUESTION 3 (Start on a new page.) Object A is projected vertically upwards from the ground. It takes 0,8 s to pass a window which is 1,5 m high and 8 m above the ground. It strikes a balcony at a velocity of 1,4 m·s-1 and falls back to the ground. At the same time, object B is projected vertically downwards at a velocity of 8 m·s-1. Ignore the effects of air friction. balcony B 1,5 m 8m A ground 3.1 Define the term projectile. (2) 3.2 Calculate the: 3.2.1 Velocity of object A at the bottom edge of the window when moving upwards (4) 3.2.2 Velocity with which object A is projected from the ground (3) 3.2.3 Height above the ground at which object A and B will pass each other (6) 3.3 Draw a position-time graph (not to scale) for the entire motion of objects A and B on the same set of axes from the moment they are projected. Label the graphs and clearly indicate the following:  Height where the objects pass each other  Time the objects pass each other as t (4) [19] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 10 com September 2024 QUESTION 4 (Start on a new page.) A toy trolley of mass 95 g is moving east at a speed of 2,45 m·s-1. It collides with a stationary wooden block of mass 120 g, which is suspended with a rope, as shown in the diagram below. The impact causes the block to move to a maximum height of 0,15 m from its original position, while the trolley continues moving eastwards. Ignore the effects of friction and air friction. 2,45 m·s-1 0,15 m 95 g 120 g 4.1 State the principle of conservation of linear momentum in words. (2) 4.2 Draw a force diagram for the block at a height of 0,15 m above the ground. (2) 4.3 Calculate the speed of the trolley immediately after the collision. (5) 4.4 The contact time between the trolley and the block is 0,08 s. Calculate the magnitude of the force applied on the trolley by the block. (3) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 11 com September 2024 QUESTION 5 (Start on a new page.) A skier of mass 100 kg starts from rest at point B, which is 30 m below point A, and moves down a frictionless curve to point C. The skier passes point C and moves up a rough plane inclined at 30° to the horizontal. Ignore the effects of air friction. A ● 30 m 50 m B 30o ● C 5.1 Write down the energy conversion of the skier from point B to C. (1) 5.2 Define the term isolated system. (2) 5.3 Use ENERGY PRINCIPLES to calculate the speed of the skier at point C. (4) 5.4 The skier experiences a frictional force of 200 N on the inclined plane. 5.4.1 Calculate the maximum height reached by the skier from point C. (6) 5.4.2 If the angle is reduced to 20°, will the maximum height be GREATER THAN, SMALLER THAN or EQUAL TO the maximum height calculated in QUESTION 5.4.1? Give a reason for the answer. (2) [15] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 12 com September 2024 QUESTION 6 (Start on a new page.) A man is sitting in a train which is travelling at a constant velocity. The siren of the train emits sound waves of frequency 2 000 Hz. A detector placed on the railway line records a frequency of 2 450 Hz. The train takes 0,5 s to reach the detector. Assume the speed of sound in air is 342 m·s-1. 6.1 Define the phenomenon described above. (2) 6.2 Name any medical instrument that makes use of the phenomenon in QUESTION 6.1. (1) 6.3 Calculate the distance between the train and the detector. (5) 6.4 Draw a frequency versus time graph (not to scale) of the frequency heard by the man sitting in the train as it moves towards and away from the detector. Indicate the values of frequency and time on the graph. (2) 6.5 What is the significance of red shift? (1) [11] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 13 com September 2024 QUESTION 7 (Start on a new page.) Three charged spheres, A, B and C, with charges of -6 nC, +9 nC and +12 nC respectively, are placed on rubber stands as shown below. A B C -6 nC +9 nC +12 nC 10 cm 15 cm 7.1 Can metal stands be used instead of rubber stands? Write down only YES or NO. Give a reason for the answer. (2) 7.2 State Coulomb's law in words. (2) 7.3 Calculate the net force acting on charge C. (5) 7.4 Charge A and B are brought into contact and then separated. Charge A is returned to its original position, while charge B is removed. 7.4.1 In which direction did electrons flow? Choose from A to B or B to A. (2) 7.4.2 Calculate the number of electrons transferred to or from charge A. (4) 7.4.3 Point X is placed between charge A and C. Calculate the distance between charge A and point X where the net electric field is equal to zero. (4) [19] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 14 com September 2024 QUESTION 8 (Start on a new page.) A battery of emf 12 V and an internal resistance of 0,8 Ω is connected to three resistors of resistance R Ω, 6 Ω and 4 Ω. A high resistance voltmeter and an ammeter of negligible resistance are also connected in the circuit, as shown below. The reading on the ammeter is 0,5 A. V R r 0,5 A A 6Ω 4Ω 8.1 Define the term internal resistance. (2) 8.2 Calculate the: 8.2.1 Reading on the voltmeter (3) 8.2.2 Resistance of resistor R (6) 8.2.3 Energy dissipated by resistor R in 3 minutes (4) 8.3 How would the reading on the voltmeter change if resistor R is removed? Write down INCREASES, DECREASES or REMAINS THE SAME. Explain the answer. (3) [17] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 15 com September 2024 QUESTION 9 (Start on a new page.) A simplified diagram of an electric machine is shown below. 9.1 Is this electric machine a generator or an electric motor? Give a reason for the answer. (2) 9.2 Write down the names of the components labelled A, B, C and D. (1) 9.3 Give a reason why carbon is preferred over iron for the brushes in electric machines. (1) 9.4 In which direction is the current flowing? Choose from A to B or B to A. (2) 9.5 The graph below of voltage vs time was obtained from an alternating current generator. 9.5.1 Calculate the frequency of this voltage. (2) 9.5.2 Calculate the average power generated if the generator produces a maximum current of 2 A. (5) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 16 com September 2024 QUESTION 10 (Start on a new page.) 10.1 The table below shows different metals with their respective threshold frequencies. Metal Threshold frequency (Hz) A 2,06 × 1014 B 1,18 × 1014 C 0,06 × 1014 D 8,52 × 1014 Light of frequency 3,15 × 1014 Hz is radiated on each metal. 10.1.1 Define the term threshold frequency. (2) 10.1.2 In which metal(s), A, B, C or D, will the photoelectric effect not take place? Give a reason for the answer. (2) 10.1.3 Calculate the highest maximum speed of the ejected electrons. (5) 10.1.4 If the intensity of light radiated is decreased, what will happen to the speed of the electrons calculated in QUESTION 10.1.3? Choose from INCREASES, DECREASES or REMAINS THE SAME. (1) 10.2 Explain the formation of the atomic emission spectrum. (2) [12] TOTAL: 150 Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 1 com September 2024 DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 1 (PHYSICS) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12 VRAESTEL 1 (FISIKA) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE Acceleration due to gravity g 9,8 m·s-2 Swaartekragversnelling Universal gravitational constant G 6,67 x 10-11 N·m2·kg-2 Universele gravitasiekonstante Radius of the Earth RE 6,38 x 106 m Radius van die Aarde Mass of the Earth M 5,98 x 1024 kg Massa van die Aarde Speed of light in a vacuum c 3,0 x 108 m·s-1 Spoed van lig in 'n vakuum Planck's constant h 6,63 x 10-34 J·s Planck se konstante Coulomb's constant k 9,0 x 109 N·m2·C-2 Coulomb 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 Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 2 com September 2024 TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING v f  vi  a t x  v i t  21 at 2 or y  v i t  21 at 2  v  vi   v  vi  v f 2  vi2  2ax or v f  v i  2ay 2 2 x   f  t or y   f  t  2   2  FORCE/KRAG Fnet  ma p  mv fs max = μsN fk = μkN Fnet t  p w  mg p  mv f  mv i m1m 2 m1m 2 M M FG or/of F=G g =G or/of g =G 2 d2 r2 d2 r WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING W  Fx cos  U mgh or/of EP  mgh Wnet  K or/of Wnet  Ek 1 1 K  mv 2 or/of Ek  mv 2 2 2 K  K f  K i or/of Ek  Ekf  Eki W Wnc  K  U or Wnc  Ek  Ep P t Pave = Fvave / Pgemid = Fvgemid WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG 1 vf T f v  vL v  vL c fL  fs fL  fb E  hf or E h v  vs v  vb  E = Wo + Ek(max) or/of E = W o + K max where/waar 1 1 E  hf and/en W0  hf 0 and/en E k (max) = mv max 2 2 or/of K max = mv max 2 2 Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 3 com September 2024 ELECTROSTATICS/ELEKTROSTATIKA kQ 1Q 2 kQ F E r2 r2 W F V E q q Q or/of Q n= n e qe ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE R V emf ( ε )= I(R + r) I R s  R1  R 2  ... 1 1 1 q I t    ... R p R1 R 2 W = Vq W P Δt W = VI  t P = VI W = I2R  t P = I 2R V 2 Δt W= V2 R P R ALTERNATING CURRENT/WISSELSTROOM Pav e  Vrms I rms / Pgemid  Vwgk I wgk I max I maks I rms  / I wgk  2 2 Pav e  Irms 2 R / Pgemid  I 2wgk R Vmax Vmaks Vrms  / Vwgk  2 Vrms 2 Vwgk 2 2 Pav e  / Pgemid  R R Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com ————————————————————— DEPARTMENT OF EDUCATION ————————————————————— DEPARTEMENT VAN ONDERWYS ————————————————————— LEFAPHA LA THUTO ————————————————————— ISEBE LEZEMFUNDO ————————————————————— PROVINCIAL PREPARATORY EXAMINATION/ PROVINSIALE VOORBEREIDENDE EKSAMEN GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS FISIESE WETENSKAPPE: FISIKA PAPER/VRAESTEL 1 SEPTEMBER 2024 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 17 pages./ Hierdie nasienriglyne bestaan uit 17 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 2 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 D  (2) 1.2 B  (2) 1.3 B  (2) 1.4 C  (2) 1.5 C  (2) 1.6 D  (2) 1.7 B  (2) 1.8 A  (2) 1.9 A  (2) 1.10 C  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 3 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted, deduct one mark./Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. An object will remain in its state of rest or motion at constant velocity unless a (non-zero) resultant force/net force acts on it.  'n Voorwerp sal in sy toestand van rus of beweging teen konstante snelheid bly, tensy 'n (nie-nul) resulterende krag/netto krag daarop inwerk.  (2) 2.2 N FA N f OR T f T FA w w Force diagram 0/2./Kragte diagram 0/2. Accepted labels/Aanvaarde benoemings w Fw /Fg/mg/88,2 N/Weight/gravitational force Fw /Fg/mg/88,2 N/Gewig/gravitasiekrag T FT /Tension/Spanning Fapplied F /Applied force/FA F /Toegepaste krag/FA f Ff / fk /frictional force/kinetic frictional force Ff / fk /wrywingskrag/kinetiese wrywingskrag N FN/Normal/Normaal Notes/Aantekeninge • Mark is awarded for label and arrow./Punt word toegeken vir byskrif en pyltjie. • Do not penalise for length of arrows/Moenie vir die lengte van die pyltjies penaliseer nie. • If arrows do not touch the dot/Indien pyle nie die kolletjie raak nie: Max/Maks 4 5 • Any other additional force(s)/Enige ander addisionele krag(te) : Max/Maks: 4 5 • If everything correct, but no arrows/Indien alles korrek, maar geen pyltjies Max/Maks: 4 5 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 4 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne 2.3 EAST AS POSITIVE EAST AS NEGATIVE OOS AS POSITIEF OOS AS NEGATIEF 4 kg mass/4 kg massa 4 kg mass/4 kg massa Fnet = ma Fnet = ma T – w = ma  (Any one) T – w = ma  (Any one) T – mg = ma T – mg = ma T – (4)(9,8) = (4)(0)  T – (4)(9,8) = (4)(0)  T = 39,2 N T = 39,2 N 9 kg mass/9 kg massa 9 kg mass/9 kg massa Fnet = ma Fnet = ma Fx + T + f = ma Fx + T + f = ma -Fcos75o + 39,2 + 2,5 = (9)(0)  Fcos75o - 39,2 - 2,5 = (9)(0)  F = 161,12 N  F = 161,12 N  OR/OF OR/OF -Fsin15o + 39,2 + 2,5 = (9)(0)  Fsin15o - 39,2 - 2,5 = (9)(0)  F = 161,12 N  F = 161,12 N  (5) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 5 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1 Marking criteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutel- woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. A projectile is an object which has been given an initial velocity and then it moves under the influence of gravitational force/gravity/weight only. 'n Projektiel is 'n voorwerp wat 'n beginsnelheid snelheid gegee is en dan beweeg dit slegs onder die invloed van gravitasiekrag/swaartekrag/ gewig.  (2) 3.2.1 UPWARDS AS POSITIVE DOWNWARDS AS POSITIVE OPWAARTS POSITIEF AFWAARTS POSITIEF ∆y = vi∆t + ½ a∆t2✓ ∆y = vi∆t + ½ a∆t2✓ 1,5 ✓= vi (0,8) + ½ (-9,8)(0,8)2✓ -1,5 ✓= -vi(0,8) + ½ (9,8)(0,8)2✓ vi = 5,80m.s-1upwards /opwaarts✓ vi = 5,80m.s-1upwards /opwaarts✓ (4) 3.2.2 POSITIVE MARKING FROM QUESTION 3.2.1 POSITIEWE NASIEN VANAF VRAAG 3.2.1 OPTION 1/OPSIE 1 UPWARDS AS POSITIVE DOWNWARDS AS POSITIVE OPWAARTS POSITIEF AFWAARTS POSITIEF vf2 = vi2 + 2aΔy✓ vf2 = vi2 + 2aΔy ✓ (5,80)2 = ( vi)2+ 2 (-9,8)(8)✓ -(5,80)2 = -(vi)2 + 2 (9,8)8✓ vi = 13,80m.s upwards/opwaarts✓ vi = 13,80m.s-1upwards/opwaarts✓ -1 OPTION 2/OPSIE 2 ½ m v2i + mghi = ½ m v2f + mghf ✓ ½ mvi 2 + 0 = ½ m(5,80)2 + m(9,8)(8)✓ vi =13,80 m.s-1upwards/opwaarts  (3) 3.2.3 POSITIVE MARKING FROM QUESTION 3.2.2 POSITIEWE NASIEN VANAF VRAAG 3.2.2 vf2 = vi2 + 2aΔy 1,42 = 13,82 + 2 (-9,8)y✓ y = 9,62 m OR/OF ½ m v2i + mghi = ½ m v2f + mghf ½ m(13,8 )2 + 0 = ½ m(1,4)2 + m(9,8)(y) ✓ y = 9,62 m ∆y = vi∆t + ½ a∆t2✓ ∆y = 13,8∆t + ½ (-9,8) ∆t2…………….(1) 9,62 - ∆y = 8∆t + 4,9∆t2…………….(2) ∆t = 0,44 s ∆y = (13,8)(0,44) – (4,9)(0.44)2✓ ∆y = 5,12 m ✓ (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 6 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne 3.3 POSITIVE MARKING FROM QUESTION 3.2.3 POSITIEWE NASIEN VANAF VRAAG 3.2.3 Position/Posisie (m) Marking guidelines Nasienriglyne B • Shape of A/Vorm van A A • Shape of B/Vorm van B 5,21 • Position of t and 5,21/ Posisie van t en 5,21 t time(s)/tyd(s) (4) [19] QUESTION 4/VRAAG 4 4.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark/Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. In an isolated/closed system the total (linear) momentum is conserved/remains constant.  In 'n geïsoleerde/geslote sisteem bly die totale (lineêre) momentum behoue/konstant.  (2) 4.2 T W If there is free-body diagram 0/2./As ñ vrye liggaam diagram get eken word 0/2 Accepted labels/Aanvaarde benoemings w Fw /Fg /mg /gravitational force Fw /Fg /mg/gravitasiekrag T FT/Tension FT/spanning Notes/Aantekeninge • Mark awarded for label and arrow./Punt word toegeken vir byskrif en pyltjie. • Do not penalise for length of arrows./Moenie vir die lengte van die pyltjies penaliseer nie. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 7 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne • Any other additional force(s)/Enige ander addisionele krag(te): Max/Maks ½ • If everything is correct, but no arrows /Indien alles korrek, maar geen pyltjies Max/Maks ½ • If force(s) do not make contact with the dot:/Indien krag(te) nie die kolletjie raak nie Max/Maks ½ Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 8 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne 4.3 vf2 = vi2 + 2ay 02 = vi2 + 2(-9,8)(0,15)  vi = 1,71 m·s-1 OR/OF (Ek + Ep)i = (Ek + Ep)f ½ m v2i + mghi = ½ m v2f + mghf (½)(0,12)(vi 2) +0 = 0 + (0,12)(9,8)(0,15)✓ vi = 1,71 m.s-1 OR/OF Wnet = ½ m v2f - ½ m v2i (0,12)(9,8)(0,15)(cos180) = 0 – (½ )(0,12)( vi 2) ✓ vi = 1,71 m.s-1 OR/OF Wnc = mghf - mghi + ½ m v2f - ½ m v2i 0 = (0,12)(9,8)(0,15) – 0 + 0 - (½ )(0,12)(vi 2) ✓ vi = 1,71 m.s-1 OPTION 1/OPSIE 1 EAST/RIGHT AS POSITIVE/OOS/REGS AS POSITIEF Σpi = Σpf ✓Any one/Enige een m1vi+ m2vi = m1vf + m2vf (0,095)(2,45) + (0,12)(0) ✓ = (0,095)(vf) +(0,12)(1,71)✓ vf = 0,29 m∙s-1✓ WEST/LEFT AS POSITIVE/WES/LINKS AS POSITIEF Σpi = Σpf ✓Any one /Enige een m1vi+ m2vi = (m1+ m2)vf (0,095)(-2,45) + (0,12)(0) ✓ = (0,095)(vf) + (0,12)(-1,71) ✓ vf = - 0,29 m∙s-1 vf = 0,29 m∙s-1✓ Range (0,28 m∙s-1 - 0,29 m∙s-1) OPTION 2/OPSIE 2 EAST/RIGHT AS POSITIVE/OOS/REGS AS POSITIEF ∆px = -∆py ✓Any one/Enige een m(vxf – vxi) = - m(vyf – vyi) 0,095(vf – 2,45) ✓ = -0,12(1,71– 0) ✓ vf = 0,29 m∙s-1✓ WEST/LEFT AS POSITIVE/WES/LINKS AS POSITIEF ∆px = -∆py m(vxf – vxi) = - m(vyf – vyi) ✓Any one/Enige een 0,095(vf + 2,45) ✓= -0,12(-1,71 – 0) ✓ vf = -0,29 m∙s-1 vf = 0,29 m∙s-1✓ (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

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Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 10 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne 4.4 POSITIVE MARKING FROM QUESTION 4.3 POSITIEWE NASIEN VANAF VRAAG 4.3 OPTION 1/OPSIE 1 EAST AS POSITIVE/OOS AS POSITIEF Fnet ∆t = ∆p ✓Any one/Enige een Fnet ∆t = m(vf – vi) /  Fnet (0,08) = 0,095(0,29 – 2,45) ✓ OR/OF Fnet (0,08) = 0,012(1,71 – 0) Fnet = -2,565 N Fnet = 2,57N Fnet = 2,57 N✓ WEST AS POSITIVE/WES AS POSITIEF Fnet ∆t = ∆p ✓Any one/Enige een Fnet ∆t = m(vf – vi)  Fnet (0,08) = 0,095(-0,29 + 2,45)✓OR/OF Fnet (0,08) = 0,12(-1,71 – 0) Fnet = 2,57 N✓ Fnet = 2,57 Fnet = 2,57 N✓ OPTION 2/OPSIE 2 EAST AS POSITIVE WEST AS POSITIVE OOS AS POSITIEF WES AS POSITIEF vf = vi + a∆t vf = vi + a∆t 0,29 = 2,45 + a(0,08) ✓ -0,29 = -2,45 + a(0,08) ✓ a = -27 m∙s -2 a = 27 m∙s-2 Fnet = ma✓ Fnet = ma✓ = 0,095 (-27) = 0,095 (27) = - 2,565 N = 2,57 N✓ = 2,57 N (3) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 11 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 Potential energy to kinetic energy/Potensiële energie na kinetiese energie✓ (1) 5.2 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark/Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. A system in which the net external forces acting on the system is zero.✓✓ 'n Sisteem waarin die netto eksterne kragte wat op die sisteem inwerk nul is (2) 5.3 OPTION 1/OPSIE 1 ½ m v2i + mghi = ½ m v2f + mghf✓ (100)(9,8)(20) +0 ✓ = 0 + (½ )(100)( vf2 ) ✓ vf = 19,80 m.s-1✓ OPTION 2/OPSIE 2 Wnet = ½ m v2f - ½ m v2i✓ (100)(20)(9,8)(cos0)✓ = (½ )(100)( vf 2) ✓ - 0 vf = 19,80 m.s-1✓ OPTION 3/OPSIE 3 Wnc = mghf - mghi + ½ m v2f - ½ m v2i✓ 0✓ = 0 – (100)(9,8)(20) + ½ (100) v2f - 0✓ vf = 19,80 m.s-1✓ (4) 5.4.1 POSITIVE MARKING FROM QUESTION 5.3 POSITIEWE NASIEN VANAF VRAAG 5.3 OPTION 1/OPSIE 1 Wnet = ΔEk ✓ Wnet = (½ )(100)(0)2 - (½ )(100)(19,80)2✓ Wnet = -19 602 J Wnet = WG + Wf -19 602✓ = (100)(9,8)(sin30o)(x)(cos180o) + (200)(x)(cos180o)✓ x = 28,41 m h = (28,41)(sin30o)✓ h = 14,21 m✓ OPTION 2/OPSIE 2 Wnc = ΔEp + ΔEk ✓Any one/Enige een Wnc = mghf - mghi + ½ m v2f - ½ m v2i /  h  (200)   ✓cos180o✓ = (100)(9,8)(h) – 0✓ + 0 – ½(100)(19,80)2✓  sin 30  o h = 14,20 m✓ Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 12 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3 Fnet = ma ✓ -(100)(9,8)(sin30o)✓ – 200 = 100a✓ a = -6,9 m.s-2 v f 2 = v i2 + 2aΔy 02= 19,802 + 2 (-6,9)(x) ✓ x = 28,41 m h = (28,41)(sin30o)✓ h = 14,21 m✓ (6) 5.4.2 Smaller than/kleiner as ✓ The net force acting on the object will decrease./Die netto krag wat op die voorwerp inwerk sal verminder ✓ (2) [15] QUESTION 6/VRAAG 6 6.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The 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.  Die 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. OR/OF An (apparent) change in observed/detected frequency (pitch), as a result of the relative motion between a source and an observer (listener). 'n Skynbare verandering in waargenome frekwensie (toonhoogte), as gevolg van die relatiewe beweging tussen die bron en 'n waarnemer (luisteraar). (2) 6.2 Doppler ultrasound machine or Doppler flow meter/Doppler- ultraklankmasjien of Doppler-vloeimeter (Any one/Enige een). ✓ (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 13 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne 6.3 v  vL v fL = fs OR/OF fL = fs ✓ v  vs v − vs 342 2 450  = (2 000 )  342 − v s vs = 62,82 m·s-1 x  v + vf  Δv = ∆x =  i t t  2  x 62,82 =   62,82 + 62,82  0,5 Δx =  (0,5)  2  Δx = 31,41 m  Δx = 31,41 m (5) 6.4 Marking guidelines/Nasienriglyne • Shape of graph/Vorm van grafiek • 2000 Hz and/en 0,5 s  f(Hz) 2000 0,5 t(s) (2) 6.5 Shows that the Universe is expanding/Toon dat die heelal besig is om uit te brei ✓ (1) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 14 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1 NO/NEE✓ The spheres will lose their charge./The metal stands are electrical conductors./Die sfere sal hul lading verloor./Die metaalstaanders is elektriese geleiers. (2) 7.2 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The (magnitude of the) electrostatic force exerted by one point charge on another point charge is directly proportional to the product of the (magnitudes of the) charges and inversely proportional to the square of the distance between them./Die (grootte van die) elektrostatiese krag wat een puntlading op 'n ander puntlading uitgeoefen, is direk eweredig aan die produk van die (groottes van die) ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle (2) If mass is used 0/2./Indien massa gebruik word in plaas van lading 0/2 7.3 OPTION 1/OPSIE 1 kQ 1Q 2 F= ✓ r2 (9  10 9 )(12  10 −9 )(6  10 −9 ) FCA = 2  = 1,04 x 10-5 N (0,25) (9  10 9 )(12  10 −9 )(9  10 −9 ) FCB = 2 = 4,32 x 10-5 N (0,15) Fnet =4,32×10 N - (1,04×10-5 N )  -5 = 3,28×10-5 N due East/ Right/reg/ Oos ✓ OPTION 2/OPSIE 2 kQ E= 2  r (9  10 9 )(6  10 −9 ) ECA =  = 864 N·C-1 (0,25) 2 (9  10 9 )(9  10 −9 ) ECB =  = 3 600 N·C-1 (0,15) 2 Enet =3 600 - 864 = 2 736 N.C-1due East F = EQ ✓ = (2 736 × 12 × 10-9)✓ = 3,28×10-5 N due East/Right/reg/Oos✓ (5) 7.4.1 A to B /A na B (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

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Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 16 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne 7.4.2 Q1 + Q 2 Q= 2 ( −6  10 −9 ) + (9  10 −9 ) =  2 Q = 1,5 × 10-9 C Charge transferred/Lading oorgedra = 1,5 × 10-9 C – (-6 × 10-9 C ) = 7,5×10-9 C OR/OF = (1,5 × 10-9 C – (9 × 10-9 C ) = -7,5×10-9 C Q n=  qe 7,5  10 −9 n=  1,6  10 −19 n = 4,69 × 1010 (electrons/elektrone)  (4) 7.4.3 OPTION 1/OPSIE 1 kQ E= 2  r (9  10 9 )(1,5  10 −9 ) (9  10 9 )(12  10 −9 ) =  (b) 2 (0,25 − b) 2 b = 0,07 m  OPTION 2/OPSIE 2 kQ 1Q 2 F= ✓ r2 (9  10 9 )(1,5  10 −9 )Q (9  10 9 )(12  10 −9 )Q  =  (b) 2 (0,25 − b) 2 b = 0,07 m  (4) [19] QUESTION 8/VRAAG 8 8.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The resistance within a battery that causes a drop in the potential difference of the battery when there is a current in the circuit. Die weerstand binne 'n battery wat 'n daling in die potensiaalverskil van die battery veroorsaak wanneer daar 'n stroom in die stroombaan is. Accept/Aan vaar Opposition to the flow of current/charges inside the battery due to its chemicals./Werk teen die vloei van lading/stroom bihne die battery as gevolg van chemikaliee (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 17 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne 8.2.1 V = IR = (0,5)(6 + 4) =5V (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 18 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne 8.2.2 POSITIVE MARKING FROM QUESTION 8.2.1 POSITIEWE NASIEN VANAF VRAAG 8.2.1. OPTION 1/OPSIE 1  = I(R + r) ✓ 12✓ = 5 + 0,8I✓ I = 8,75 A V = IR✓ 5 = 8,25R✓ R = 0,61 Ω✓ OPTION 2/OPSIE 2 V = IR  10R  5  = I  ………………….(1)  10 + R   10R  12  = I  + 0,8  …..……..(2)  10 + R  R = 0.61 Ω  (6) 8.2.3 POSITIVE MARKING FROM QUESTION 8.2.1 AND 8.2.2 POSITIEWE NASIEN VANAF VRAAG 8.2.1 EN 8.2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 W = I Rt  2 W = VIt  V 2 t = (5)(8,25)(180)  W =  = (8,25)2(0,61)(180) R = 7 473,26 J = 7 425 J  (5) 2 (180 ) =  0,61 = 7 377,05 J  (3) 8.3 Increases/Toeneem Total external resistance will increase and total current will decrease,  emf remains constant and internal voltage decreases. Totale eksterne weerstand sal toeneem en totale stroom sal afneem, emk bly konstant en interne spanning neem af. (3) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 19 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1 Generator/Generator✓ , It converts mechanical energy to electrical energy./Dit skakel meganiese energie om na elektriese energie. Accept/Aan vaar There is no energy source./Daar is geen energie bron. (2) 9.2 Coil/Spoel/Conductor/Geleier/Armature/roterende spoel✓ (1) 9.3 Good conductor of electricity/Goeie geleier van elektrisiteit/Does not rust/Roes nie ✓(Any one/Enige een) (1) 9.4 A to/na B✓✓ (2) 9.5.1 1 f= T 1 =  0,02 = 50 Hz  (2) 9.5.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Vmax Vmax Vrms = Vrms = 2 2 250 250 =  =  2 2 = 176,78… V = 176,78… V Imax Imax Irms = Irms = 2 2 I I I rms = max I rms = max 2 2 2 2 =  =  2 2 = 1,41A = 1,41A V R = rms Pave = VrmsIrms Irms = (176,78)(1,41) 176,78 = 249,26 W = 1,41 = 125,38 Ω 2 Vrms Pave =  R (176,78) 2 =  125,38 = 249,25 W  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 20 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3 V Vrms = max 2 250 =  2 = 176,78… V I Irms = max 2 I I rms = max 2 2 =  2 = 1,41A V R = rms Irms 176,78 = 1,41 = 125,38 Ω 2 Pave = I rms R = (1,41)2(125,38)  = 249,27 W  (5) [13] Note/Let wel: Range (249,25 – 250 W) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Downloaded Physical Sciences f 12 rom St anmorephysics. P1/Grade 21 com September 2024 Fisiese Wetenskappe/Graad 12 Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1.1 Marking criteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark /Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The minimum frequency needed to emit an electron  from (the surface of) a metal.  Die minimum frekwensie wat nodig is om 'n elektron vanaf (die oppervlak van) 'n metaal uit te straal. NOTE: If reference to work function 0/2 LET WEL: Indien verwysing na werkfunksie 0/2 (2) 10.1.2 D Frequency of light is less than the threshold frequency /f < f 0 / Energy of light is less than the work function of the metal /E < W0  Frekwensie van lig is minder as die drumpelfrekwensie /f < f0 / Energie van lig is minder as die werkfunksie van die metaal /E < W0  (2) 10.1.3 E = W o + E k(max) E = W o + K max (any one) (6,63×10-34)(3,15×1014)= (6,63×10-34)(0,06×1014) + ½(9,11×10-31)v2 v = 6,71×105 m·s-1 (5) 10.1.4 Remains the same /Bly dieselfde  (1) 10.2 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark/Indien enige van die onderstreepte sleutel woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. It is formed when certain frequencies of electromagnetic radiation are emitted due to an atom making a transition from a higher energy state to a lower energy state./Dit word gevorm wanneer sekere frekwensies van elektromagnetiese straling uitgestraal word as gevolg van 'n atoom wat 'n oorgang maak van 'n hoër energietoestand na 'n laer energietoestand. (2) [12] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief