Physical Sciences Grade 12 P1 KwaZulu Natal Mock 2021 Memo

e av +2; RE 7 age Bo Sh ate yeRe ge ee ) RWAZULU-NATAL PROVINCE é EDUCATION " REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE PHYSICAL SCIENCES P1 (PHYSICS) PREPARATORY EXAMINATIONS SEPTEMBER 2021 150 TIME : 3hours This question paper consists of 17 pages and 3 data sheets.

Downloaded from hlayiso. om NSC INSTRUCTIONS AND INFORMATION TO CANDIDATES 1. Write your name on the ANSWER BOOK. 2 This question paper consists of TEN 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 sub questions, for example between QUESTION 2.1 and QUESTION 2.2. You may use a non-programmable calculator. You may use appropriate mathematical instruments. You are advised to use the attached DATA SHEET. oo oO Show ALL formulae and substitutions in ALL calculations. 10. Round off your final numerical answers to a minimum of TWO decimal places. 11. Give brief motivations, discussions, et cetera where required. 12. Write neatly and legibly. Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 3 Preparatory Examintion September 2021 NSC QUESTION 1: MULTIPLE CHOICE Various options are provided as possible answers to the following questions. Choose the answer and write only the letter (A-D) next to the question number (1.1-1.10) in the ANSWER BOOK, for example 1.11 D. 11 The free-body diagram below shows the relative magnitudes and directions of all the forces acting on an object. Normal force Frictional force Renee 1 E Gravitational force s The object is... A accelerating eastward. B accelerating westward. Cc at rest. D moving with constant velocity. (2) 1.2 Acceleration due to gravity on the surface of the Earth is g. What will be the acceleration due to gravity on the surface of another planet of the SAME mass as that of the Earth, but HALF the radius of the Earth?: A rg B g Cc 2g D 4g (2) 1.3. From the top of a building, ball X is thrown vertically upwards and another ball Y, is thrown vertically downwards. Both balls are thrown at the same speed. Ignore the effects of friction. Which ball(s) will have a greater speed when it hits the ground? A xX B Y Cc X and Y would have the same speed D Depends on the mass of the ball. (2) Copyright reserved Please turn over

ee Downloaded from hlayiso.com NSC 1.4. In movies, Superman hovers in mid - air, grabs a villain (bad guy), and throws him forward. Superman, however, remains stationary. This is not possible in real life, because it goes against one of the following laws or principles: A B Cc D The principle of conservation of energy Newton’s first law of motion Newton's Universal law of Gravitation The principle of conservation of linear momentum (2) 1.5 A force, F, acts in three different directions on a box moving with a velocity, v, as shown below. F li Y, F ‘, |, A B c Select the order in which the work done by the force F from minimum to maximum A A; B,C B C;B;A Cc B,C; A D CAB (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 5 Preparatory Examintion September 2021 NSC 1.6 A police car, with its siren on, is moving away at constant speed from a 1.7 stationary observer. The siren emits a sound of constant frequency Which of the following characteristics, associated with the sound emitted by the siren is/are CORRECT, as perceived by the observer? (i) The speed remains the same. (ii) The frequency increases. (iii) The wavelength increases. (iv) The frequency decreases A (iii) only B (i), (iii) and (iv) c (i) and (iii) only D (i) and (ii) only (2) A point charge exerts a force of magnitude F on another point charge. One of the charges is changed to one third and the distance between the charges is doubled. What is the new electrostatic force in terms of F? a 3 BF 12 C8 4 DF 6 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com NSC 1.8 The sketch below, show a battery of emf &, connected in a circuit containing three resistors. Ignore the internal resistance of the battery. GS) | LEY) | A: WV} 3) Which ONE of the following combinations best describes the readings of ammeter Ag3 and voltmeter V2? Reading on A3 Reading on V2 A ly-lb e-V B l-l2 E+V, Cc +h o+Vi D +h e-V; 1.9 White light is passed through a cold gas, then through a prism as shown in the sketch. . Screen Cold Gas Prism White light What type of spectrum is observed on the screen? A Line absorption spectrum B Line emission spectrum Cc Continuous absorption spectrum D Continuous emission spectrum (2) | Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 7 Preparatory Examintion September 2021 NSC 1.10 Which ONE of the following provides evidence that light behaves as a particle? A Light can be diffracted. B Cc D Light can be refracted. Light ejects electrons from a metal surface. The speed of light increases when it travels in a vacuum. (2) [20] Copyright reserved Please turn over

pe eee Downloaded from hlayiso.com NSC QUESTION 2 Two crates of masses 4 kg and 6 kg joined by a light inextensible string, are placed on a rough inclined plane. A force F pulls on the 6 kg crate such that the systems accelerates up the plane at 2,5 ms”. The frictional forces exerted by the surface is 1 N for 4 kg crate and 1,5 N for the 6 kg crate. 2.1 2.2 2.3 2.4 Copyright reserved 6k g 4k 30° State Newton’s Second Law of Motion in words. (2) Draw a labelled free body diagram for the 6 kg crate. (5) Calculate the magnitude of the force F, exerted on the 6 kg crate. (5) The string joining the crates now snaps, whilst the same force F acts on the 6 kg crate. How will this change affect the following: 2.4.1 The acceleration of the 6 kg crate. (Choose from INCREASES, DECREASES or REMAIN THE SAME) (1) 2.4.2 The motion of the of the 4 kg block. (3) [16] Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 9 Preparatory Examintion September 2021 NSC QUESTION 3 A hot — air balloon is rising at a constant speed of 4,9 m’s’. At a height of 80 m above the ground, a ball is released. Ignore the effects of friction. M GROUND 3.1 Write down the acceleration of the ball: 3.1.1 Before it was released (1) 3.1.2 After it is released (2) 3.2 Calculate the: 3.2.1 Position of the ball 0,5 s after it was released relative to the ground (4) 3.2.2 Time taken by the ball to reach the ground (4) 3.3 Sketch a velocity — time graph for the motion of the ball from the instant it is released until it reaches the ground. Indicate the following in your graph: e Initial velocity of the ball e The time for the entire motion of the ball (3) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com ee eee NSC QUESTION 4 A car of mass 1200 kg travelling east with a speed of 25 m's”, collides with the van of mass 1800 kg. The van of mass 1800 kg is stationary on a frictionless track. The two vehicles move together as a unit after the collision. N A we rE vy Sy} 41 State the principle of conservation of momentum in words (2) 4.2 Calculate the speed of the vehicles after the collision. (4) 4.3 Determine by means of a relevant calculation whether the collision was ELASTIC or INELASTIC. (5) After the collision the combination of the vehicles move into a real road and stop, covering a distance of 20 m. 4.4 Calculate the magnitude of the frictional force. (5) [16] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 11 Preparatory Examintion September 2021 NSC QUESTION 5 A lift arrangement shown below, comprises an electric motor, a cage and its counterweight. The counterweight moves vertically downwards as the cage moves upwards. The cage and counterweight move at the same constant speed. OC Electric motor Cage Counterweight | The cage, carrying passengers, moves vertically upwards at a constant speed, covering 60 m in 2 minutes. The counterweight has a mass of 900 kg. The total mass of the cage and passengers is 1 300 kg. The electric motor provides the power needed to operate the lift system. Ignore the effects of friction. 5.1 Calculate the work done by the: 5.1.1 Gravitational force on the cage (3) 5.1.2 Counterweight (2) 5.2 Calculate the average power required by the motor to operate the lift arrangement in 2 minutes as explained above. Assume that there are no energy losses due to heat and sound. (5) [10] Copyright reserved Please turn over

Py ere Seen NSC QUESTION 6 A siren is attached to a sea shore that emits sound waves of frequency 755 Hz. A speed boat is travelling with a constant speed away from the shore. 6.1 State Doppler Effect in words . The speed boat travels 45 min 3s. 6.2 6.3 6.4 Calculate the frequency of the sound waves emitted by the siren as heard by a person on the boat. Take the speed of sound in air as 340 ms”. Write down any two uses of the Doppler Effect. Spectral lines of star X at an observatory are observed to be red shifted. 6.4.1. Explain the term red shifted in terms of wavelength. 6.4.2 Will the frequency of the light observed from the star INCREASE, DECREASE or REMAIN THE SAME? Copyright reserved Downloaded from hlayiso.com (2) (5) (1) [12] Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 13 Preparatory Examintion September 2021 NSC QUESTION 7 7.41 Three metal spheres X, Y, and Z carrying charges are placed on the vertices of a right angled triangle as shown below. -2,10 x 10°C 1,5x10°C 3x10%m 1,8x 10°C 7.1.1 State Coulomb’s Law in words (2) 7.1.2 Calculate the magnitude of the net electrostatic force on sphere Z. (8) 7.2 Two identical spheres, A and B, carrying charges +5 uC and -10 uC respectively, are separated by a distance of 30 mm. P is a point located at a distance of 15 mm from sphere B as shown below. : 30 mm *p : The spheres A and B are allowed to touch each other and then return to their original positions. 15mm 7.2.1 Define electric field at a point in words (2) 7.2.2 Draw an electric field pattern between spheres A and B, after touching. (3) 7.2.3 Calculate the magnitude of the net electric field at point P (6) [21] Copyright reserved Please turn over

Downloaded from hlayiso.com riysical OGiIenvesir | 14 ritpdl awry CxaiNuUl Oepleniver 2uzZ | NSC QUESTION 8 A battery with an internal resistance of 0,10 Q is connected to the circuit as shown in the sketch below. The three light bulbs in the circuit, Ri, Re and Rs, are all identical, with a power rating of 40 W each. Ignore the resistance of the wires and that of the ammeter. 8.1 Define emf in words (2) Switch $ is closed. Now bulb R3 works at optimum condition and the ammeter reads 2A. 8.2 Calculate the: 8.2.1 TOTAL EXTERNAL RESISTANCE of the circuit (5) 8.2.2 Emf of the battery (3) 8.2.3. The energy dissipated by the bulb labelled R3in 20 minutes (3) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 15 Preparatory Examintion September 2021 NSC QUESTION 9 9.1 The sketch below represents a simplified version of a generator used to light up a light bulb. The resistance of the light bulb is 807 Q Brushes 9.1.1 Name the type of generator in the sketch. Choose from AC or DC (1) 9.1.2 Name the part labelled P in the sketch. (1) 9.1.3 Write down the principle on which the generator operates. (1) 9.2 The graph of output emf versus time obtained when the above generator was in use is shown below. 811,13 p-prynnrnnmno : E ts mf (V 9.2.1 Define the term root mean square of an AC voltage. (2) 9.2.2 Calculate the power dissipated through the light bulb connected to the generator above. (5) ' Copyright reserved Please turn over

Downloaded from hlayiso.com PiysiGal OUieHces/r 1 10 Preparatory Examintion september 2021 NSC | Some changes were made to the above generator, a new machine, shown below, was obtained. 9.3.1 Draw a sketch graph of potential difference versus time obtained from this machine. (2) 9.3.2 Write down the function of the part labelled Q in the new machine. (2) [14] QUESTION 10 The simplified diagram below illustrates how a cathode emits electrons when light shines on it. Incident Light o> Cathode Potential —__—* Difference * (a) The incident light has an energy of 2,12 x 10° J and the cathode has a threshold frequency of 2,21 x 10° Hz. 10.1 Define threshold frequency. (2) The incident light releases 2,01 x 10° photo-electrons per second from the cathode. 10.2 Calculate the current flowing through the ammeter. (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 17 Preparatory Examintion September 2021 NSC 10.3. The brightness of the incident light is now decreased. What effect will this change have on the current in QUESTION 10.2? Choose from INCREASES, DECREASES or REMAIN THE SAME Give a reason for the answer. (2) 10.4 Calculate the maximum kinetic energy of the photo - electrons. (4) The cathode is replaced by another one with a higher threshold frequency and the same light was used. 10.5 How does this change affect the kinetic energy of the photoelectrons released? Choose from INCREASES, DECREASES or REMAIN THE SAME Give a reason for your answer (2) [14] TOTAL: 150 Copyright reserved Please turn over

Downloaded from hlayiso.com Pnysical Sciences/r' 1 NSC Preparatory Examintion September 2021 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 Swaartekragversnelling 9,8 m-s@ Radius van Aarde Universal gravitational constant G 6,67 x 10" N-m?-kg* Universele gravitasiekonstante Speed of light in a vacuum c 3,0 x 10° m-s™ Spoed van lig in 'n vakuum Planck's constant h 6,63 x 10 Js Planck se konstante Coulomb's constant k 9,0 x 10°N-m?-C* Coulomb se konstante Charge on electron e -1,6x 10°C Lading op electron Electron mass Me 9,11 x 10°" kg Elektronmassa Mass of Earth M 5,98 x 10% kg Massa van Aarde Radius of Earth Re 6,38 x 10°m Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 Preparatory Examintion September 2021 NSC TABLE 2: FORMULAE / TABEL 2: FORMULES MOTION / BEWEGING Vp =Vi taAt Ax =v, At + dadt? or/of Ay = v,At + faAt? v, +¥ vit v, =v? +2adxor/ofv,’ =v," +2ady Ax = (“eta or/of Ay = (“eta FORCE / KRAG Fie = Ma p=mv fsqmax) =u N fk = UN FierAt =Ap w=mg Ap = mv, -my, F= Gmym 2 _GM ar oe WORK, ENERGY AND POWER / ARBEID, ENERGIE EN DRYWING W=FAxcos 0 U=mgh or/ofE, =mgh K= 3 mv? or/of E, = 3 mv? Whet = AK tei Whet = AE AK = Kr-K, or/of AE, = Exg¢ - Exi Ww Whe = AK + AU orlof Whe = AE + AE, ar Pay = F-Vay / Pgem = F-Vgem WAVES, SOUND AND LIGHT / GOLWE, KLANK EN LIG 1 v=fr f= f viv Cc f= Lf = =h- L viv, E=nhf or/of E a E = Wo + Eximax) Or/of E = Wo + Kimax) “’*here/waar 3 E=hf and/en Wo=hf. and/en Eximax) = 3 ™Vinax orlof Kimax) = a mVAax ELECTROSTATICS / ELEKTROSTATIKA Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 20 Preparatory Examintion September 2021 NSC k r r Ww F vo E=* q q n=2 OR/IOF n=2 e Ge ELECTRIC CIRCUITS / ELEKTRIESE STROOMBANE Vv emf (€) =1(R +r) R=— I emk (¢) = (R +n) R, =R,+R2+ 14 q=1At —=—+—+.. R, R, R, W=Vq ph W=VIAt At P=VI =/2 W=IRAt P=[’R v2 V7 At = We R R ALTERNATING CURRENT / WISSELSTROOM I Pave = Vans Las / Pyemiadeta= Vngk ! wok Tome Tye = v2 v2 Pive =TinsR / Premiddeld =TegR V, V, Vv. = max i Vv. : — maks v2 v2 ™ v2 “ v2 Pave AR i Premiddela a Copyright reserved

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES P1 (PHYSICS) PREPARATORY EXAMINATIONS MARKING GUIDELINE SEPTEMBER 2021 MARKS : 150 TIME: 3 hours This marking guideline consists of 15 pages.

Downloaded from hlayiso.com Physical Sciences/P1 2 Preparatory Examination September 2021 NSC QUESTION 1: MULTIPLE CHOICE 1.1 A (2) 1.2 D (2) 1.3 C (2) 1.4 D (2) 1.5 D (2) 1.6 B (2) 1.7 B (2) 1.8 A (2) 1.9 A (2) 1.10 C (2) [20] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 3 Preparatory Examination September 2021 NSC QUESTION 2 2.1 When a resultant/net force acts on an object, the object will accelerate in the direction of the force at an acceleration directly proportional to the force and inversely proportional to the mass of the object. OR 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 force)(2 or 0). (2) 2.2 ACCEPT Accept the following symbols F FA/Fapp/FApplied N FN/Normal/Normal force fk Kinetic friction force/f/Ff/fr T Tension force/FT Fg W/58,8N Notes • Mark is awarded for label and arrow. • Do not penalise for length of arrows. • Deduct 1 mark for any additional force. • If force(s) do not make contact with body/dot : Max:4/5 • If arrows missing but labels are there: Max:4/5 (5) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 4 Preparatory Examination September 2021 NSC 2.3 For the 4 kg crate Fnet = ma T – f – Fg// = ma T – 1 – (9,8)(4)sin30o = 4(2,5) T = 30,6 N upwards  For the 6 kg crate  Any of the two Fnet = ma F – f – T – Fg// = ma F–1,5– 30,6 – (9,8)(6)sin 30o = 6(2,5) F = 76,5 N  (5) 2.4.1 INCREASES (1) 2.4.2 The 4 kg block will move upwards/forward (for a brief moment), stop and then slide down the plane / backward. (3) [16] QUESTION 3 3.1.1 Zero (1) 3.1.2 9,8 m.s-2 downwards  (2) 3.2.1 OPTION 1 UPWARD POSITIVE UPWARD NEGATIVE Δy= viΔt + ½aΔt2 Δy= viΔt + ½aΔt2 Δy= (4,9)( 0,5) + ½(-9,8)(0,5)2 Δy= (-4,9)( 0,5) + ½(9,8)(0,5)2 Δy = 1,225 m Δy = -1,225 m Height after 0,5 s = 80 + 1,225 Height after 0,5 s = 80 + 1,225 = 81,23 m = 81,23 m ∴The ball is 81,23 m above the ∴The ball is 81,23m above the ground ground Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 5 Preparatory Examination September 2021 NSC OPTION 2 UPWARD POSITIVE UPWARD NEGATIVE vf = vi + aΔt vf = vi + aΔt vf = 4,9 + (–9,8) (0,5) vf = -4,9 + (9,8) (0,5) vf = 0 m.s-1 vf = 0 m.s-1 vf2 = vi2 + 2aΔy vf2 = vi2 + 2aΔy 0 = 4,92 + 2(–9,8)Δy 0 = - 4,92 + 2(9,8)Δy Δy = 1,225 m Δy = -1,225 m Height after 0,5 s = 80 + 1,225 Height after 0,5 s = 80 + 1,225 = 81,23 m = 81,23 m ∴The ball is 81,23 m above the ground ∴The ball is 81,23 m above the ground OPTION 3 UPWARD POSITIVE UPWARD NEGATIVE vf = vi + aΔt vf = vi + aΔt vf = 4,9 + (–9,8) (0,5) vf = 4,9 + (–9,8) (0,5) vf = 0 m.s-1 vf = 0 m.s-1 v f + vi v f + vi Δy = Δt  Δy = Δt  2 2 0 + 4,9 0 - 4,9 [ Δy = 0,5 ] [ Δy = 0,5 ] 2 2 Δy = 1,225 m Δy = -1,225 m Height after 0,5 s = 80 + 1,225 Height after 0,5 s = 80 + 1,225 = 81,23m = 81,23 m ∴The ball is 81,23 m above the ∴The ball is 81,23 m above the ground ground Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 6 Preparatory Examination September 2021 NSC OPTION 4 ½mvi2 + mghi = ½mvf2 + mghf  ½m(4,9)2 + m(9,8)(80)  = ½m(0)2 + m(9,8)hf  hf = 81,225 m  OPTION 5 ½mvi2 + mghi = ½mvf2 + mghf  ½m(4,9)2 + m(9,8)(0) = ½m(0)2 + m(9,8)hf  h = 1,225 m Height after 0,5 s = 80 + 1,225 = 81,23 m ∴The ball is 81,23 m above the ground (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 7 Preparatory Examination September 2021 NSC 3.2.2 OPTION 1 UPWARDS POSITIVE UPWARDS NEGATIVE Δy= viΔt + ½aΔt2 Δy= viΔt + ½aΔt2 -80 = (4,9)Δt + ½(-9,8)Δt2 80 = (-4,9)Δt + ½(9,8)Δt2 Δt = 4,57 s Δt = 4,57 s OPTION 2 UPWARDS POSITIVE UPWARDS AS NEGATIVE vf2 = vi2 + 2a∆y vf2 = vi2 + 2a∆y vf2 = (4,9)2 + 2(-9,8)(-80)  vf2 = (-4,9)2 + 2(9,8)(80)  vf = -39,9 m.s-1 vf = 39,9 m.s-1 vf = vi + a∆t vf = vi + a∆t -39,9 = 4,9 + (-9.8)∆t 39,9 = -4,9 + (9.8)∆t ∆t = 4,57 s ∆t = 4,57 s  OPTION 3: POSITIVE MARKING FROM QUESTION 3.2.1 Considering ball from the maximum height UPWARDS POSITIVE UPWARDS AS NEGATIVE Δy= viΔt + ½aΔt2 Δy= viΔt + ½aΔt2 -81,23 = (0)Δt + ½(-9,8)Δt2 81,23 = (0)Δt + ½(9,8)Δt2 Δt = 4,07 s Δt = 4,07 s Time to reach ground = 0,5 + 4,07 Time to reach ground = 0,5 + 4,07 = 4,57 s = 4,57 s Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 8 Preparatory Examination September 2021 NSC 3.3 POSITIVE MARKING FROM Q3.2.2 UPWARD POSITIVE 4,9 4,57 0 Velocity (m.s-1) Time (s) OR UPWARD NEGATIVE Velocity (m.s-1) 0 4,57 Time (s) -4,9 CRITERIA FOR MARKING OF GRAPH Correct shape  Indication of initial velocity  Indication of the time for the entire motion  (3) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 9 Preparatory Examination September 2021 NSC QUESTION 4 4.1 The total linear momentum in an isolated system is conserved (accept “closed” instead of “isolated”) OR In an isolated system, the total linear momentum before a collision is equal to the total linear momentum after the collision. (2) 4.2 ptotal (before) = ptotal (after) Σpi =Σpf  m1vi1 + m2vi2 = (m1 + m2)v 1200 (25) + 1800 (0) = (1200 + 1800)v  v = 10 m.s-1. ∴ Speed of the cars after is 10 m.s-1 (4) 4.3 ΣKf = ½ (m1 + m2)v2 ΣKi = ½ m1 vi12 + ½m2 vi22 = ½ (1200 + 1800) 10 2 = ½ (1200 )(25)2 + ½(1800)(0)2 = 150 000 J = 375 000 J ΣKf ≠ ΣKi ∴Collision is Inelastic  (5) NOTE: If it is assumed that ΣKf = ΣKi at the outset, Max:2/5 4.4 POSITIVE MARKING FROM QUESTION 4.2 OPTION 1 OPTION 2 vf2 = vi2 + 2a∆y v f + vi Δy = Δt  2 02 = (10)2 + 2a(20)   0  10  20    t   2  a = - 2,5 m.s-2 Δt = 4 s Fnet = ma vf = vi + aΔt Ff = ma 0 = 10 + a(4) Ff = 3000(-2,5)  a = - 2,5 m.s-2 Ff = -7500 N ∴ Magnitude of frictional force is 7500 N (5) [16] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 Preparatory Examination September 2021 NSC OPTION 3 OPTION 4 v f + vi Wnet = ΔEk Δy = Δt   2 f Δx cos θ = ½mvf2 - ½mvi2  0  10  f (20) cos 180⁰  = ½(3000)(0)2 - ½(3000)(10)2 20    t  ∴ Δt = 4 s  2  f = 7500 N  p Fnet = f = t m( v f v i ) f =  t ( 0 10 ) 3000  4 = -7500 N ∴ Magnitude of f is 7500 N (4) [16] QUESTION 5 5.1.1 OPTION 1 W = FΔxcosθ Wgravity = mgΔycosθ = (1 300)(9,8)(60)cos180o = - 764 400 J (-7,64 x105 J) OPTION 2 W = - ΔEp = -(1300)(9,8)(60 -0)  = - 764 400 J (-7,64 x105 J) -1 mark if either negative is omitted (3) 5.1.2 Wcounterweight = mgΔycosθ = (900)(9,8)(60)cos0o  = 529 200 J (5,29 x 105 J)  (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 11 Preparatory Examination September 2021 NSC 5.2 OPTION 1 OPTION 2 POSITIVE MARKING FROM 5.1.1 Fmotor = Tright - Tleft AND 5.1.2 = 1300(9,8) - 900(9,8) Wnet = ΔEK = 3920 N Wgravity + Wcountweight + Wmotor = 0  Wmotor = Fmotor Δy cos0  Wmotor = - (W gravity + W countweight) = (3920)(60) cos0  = 235 200 J -764 400 + 529 200 + Wmotor = 0 ∴Wmotor = 235 200 J W Pave motor =  Δt 235200 Pave motor =  120 = 1960 W  (5) [10] QUESTION 6 6.1 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.  (2) 6.2 d 45 v= v=  ∴vL = 15 m.s-1 Δt 3 v ± vL v vL fL = fS fL  fs v ± vS v 340 - 15  fL = ( )755  340 + 0 fL = 721,69 Hz (5) 6.3 Any two  Ultrasound waves (to measure the heartbeat of a foetus in the womb).  Doppler flowmeter (to measure the rate of blood flow)  Traffic management systems, (especially speed control)  Radar, (allowing for the tracking of weather systems)  Astronomy, (where the application of the red-shift and blue-shift of light from the stars has revolutionised our understanding of the universe) (2) 6.4.1 The spectral lines (light) from the star are shifted towards longer wavelengths.  (2) 6.4.2 Decrease (1) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 12 Preparatory Examination September 2021 NSC QUESTION 7 7.1.1 The magnitude of the electrostatic force exerted by one point charge (Q 1) on another point charge (Q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them (2) NB DO NOT MARK QUESTION 7.1.2. MAX MARK FOR QUESTION 7 WILL BE 13. CONVERT THIS MARK TO 21 USING THE CONVERSION TABLE PROVIDED AT THE END OF THESE GUIDELINES 7.1.2 q1q F=k 2 2 r -6 1,5×10 - 6×1,8×10 FYZ = 9×109 ( ) (3 x10 - 2 )2 FYZ = 27 N downwards. -6 2,1×10 - 6×1,8×10 FXZ = 9×109 ( ) 2 (6x10 - 2 ) FAC = 9,45 N at 60o to the vertical But FXZ has two perpendicular components i.e. FXZx and FXZy FYZx = FYZSin60o Vector diagram FYZx = 9,45Sin60o FACx = 8,184 N to the left FXZ FYZy = FYZCos60o (FXZ)y 60o FYZy = 9,45Cos60o FACy = 4,725 N upwards (FXZ)x Fnet x = 8,814 N to the left Fnet y = 27 – 4,725 = 22,275 N downwards FYZ (Fnet)2 = (Fnet x)2 + (Fnet y)2 (Fnet)2 = (8,814)2 + (22,275)2 Fnet = (8,814 2 + 22,275 2 ) Fnet = 23,73 N (8) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 13 Preparatory Examination September 2021 NSC 7.2.1 The electric field at a point is the (electrostatic) force experienced per unit positive charge placed at that point.  (2) 7.2.2 A B CRITERIA FOR MARKING THE ABOVE ELECTRIC FIELD PATTERN Correct direction of field lines  Shape of the electric field lines (At least 4 lines on each sphere)  No field lines crossing each other/No field lines inside the spheres  (3) 7.2.3 Q1+Q2 Q= 2 5 10  6 (-10 10  6 ) Q=  2 = - 2,5 x 10-6 C (-2,5 μC) Q E=k 2  r 2,5×10 - 6 EAP = 9×109 ( ) 0,045 2 EQP = 1,11x 107N.C-1 to the left 2,5×10-6 9 EBP = 9×10 ( ) 0,015 2 EBP = 1,00x 108 N.C-1 to the left Enet = EAP + EBP Enet = 1,11 x 107 +1,00 x 108 Enet = 1,11 x 108 N.C-1 (6) [21] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 Preparatory Examination September 2021 NSC QUESTION 8 8.1 (Maximum) energy provided (work done) by a battery per coulomb/unit charge passing through it  (2) 8.2.1 OPTION 1 OPTION 2 P = I2R Ps = V3I 40 = 22R3 40 = V3(2) V3= 20 V R3 = 10 Ω v3 20 ∴ R3= =  = 10 Ω I 2 1 1 1 1 1 𝑅1 𝑅2 = + 𝑅𝑃 = ( + )−1 𝑅𝑃 = 𝑅𝑃 𝑅1 𝑅2 𝑅1 𝑅2 𝑅1 + 𝑅2 1 1 1 1 1 −1 (10)(10) 𝑅𝑃 =( 10 + 10 ) 𝑅𝑃 = (10 + 10)  𝑅𝑃 =  10+10 RP = 5 Ω RP = 5 Ω RP = 5 Ω REXT = RP + Rs REXT = RP + Rs REXT = RP + Rs REXT = 5 + 10 REXT = 5 + 10 REXT = 5 + 10 REXT = 15 Ω REXT = 15 Ω REXT = 15 Ω (5) 8.2.2 POSITIVE MARKING FROM QUESTION 8.2.1 Ɛ = I(R + r)  Ɛ = 2(15 + 0,1)  Ɛ = 30,2 V  (3) OPTION 1 OPTION 2 (Positive marking from Q8.2.1) 8.2.3 W = P x t W = I2Rt W = 40 x 20 x 60 W = (2)2(10)(20 x 60) W = 48000 J W = 48000 J (3) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 15 Preparatory Examination September 2021 NSC QUESTION 9 9.1.1 AC (generator)  (1) 9.1.2 Slip rings (1) 9.1.3 (Faraday’s Law of) Electromagnetic induction (1) 9.2.1 The rms value of AC is the DC potential difference which dissipates the (2) same amount of energy as AC.  (2 OR 0) 9.2.2 OPTION 1 OPTION 2 V Max VMax Imax   VRMS =  R 2 311,13 311,13 VRMS =    2 807 VRMS = 220,00 V = 0,3855 A 2 VRMS Pave = R  Imax V max Pave   2 2 311,13  0,3855 Pave = 220  Pave =  807 2 Pave = 59,98 W  Pave = 59,97 W  (5) 9.3.1 Voltage/V OR time/s CRITERIA FOR MARKING THE ABOVE GRAPH Correct shape  At least one complete cycle  (2) 9.3.2 (Commutator) allows the induced current to flow in the same direction / in one direction in the external circuit (2) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 Preparatory Examination September 2021 NSC QUESTION 10 10.1 Threshold frequency, fo, is the minimum frequency of light needed to emit electrons from a certain metal surface.  (2 or 0) (2) 10.2 Q = nqe Q= 2,01 x 109 X 1,6 x 10-19 Q = 3,22 x 10-10 C Q I=  Δt 3,22×10 -10 I=  1 I = 3,22 x 10-10 A (4) 10.3 Decreases When the intensity of the light is decreased, the number of photons (2) per second will decrease 10.4 E = W o + Kmax  E = hfo + Kmax 2,12x10-18  =(6,63x10-34)(2,21x1015) + Kmax Kmax = 6,55 x 10-19 J (4) 10.5 Decreases - More energy is used to release the electrons.  OR Work function is greater.  (2) [14] TOTAL: 150 Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 17 Preparatory Examination September 2021 NSC CONVERSION OF MARKS FOR QUESTION 7 MARK CONVERTED OBTAINED MARK OUT OUT OF 13 OF 21 0 0 1 2 2 3 3 5 4 6 5 8 6 10 7 11 8 13 9 15 10 16 11 18 12 19 13 21 Copyright reserved Please turn over