NATIONAL SENIOR CERTIFICATE GRADE 12

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1 NATIONAL SENIOR CERTIFICATE GRADE 1 PHYSICAL SCIENCES: PHYSICS (P1) NOVEMBER 015 MARKS: 150 TIME: 3 hours This question paper consists of 17 pages, 3 data sheets and 1 answer sheet. Copyright reserved Please turn over

2 Physical Sciences/P1 DBE/November 015 NSC INSTRUCTIONS AND INFORMATION 1. Write your centre number and examination number in the appropriate spaces on the ANSWER BOOK and ANSWER SHEET This question paper consists of 11 questions. Answer QUESTION 11. on the attached ANSWER SHEET. Answer ALL the other questions in the ANSWER BOOK. Start EACH 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.1 and QUESTION.. You may use a non-programmable calculator. You may use appropriate mathematical instruments. You are advised to use the attached DATA SHEETS. Show ALL formulae and substitutions in ALL calculations. Round off your final numerical answers to a minimum of TWO decimal places. Give brief motivations, discussions et cetera where required. Write neatly and legibly. Copyright reserved Please turn over

3 Physical Sciences/P1 3 DBE/November 015 NSC QUESTION 1: MULTIPLE-CHOICE QUESTIONS Four 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 number ( ) in the ANSWER BOOK, for example 1.11 E. 1.1 Two forces, F 1 and F, are applied on a crate lying on a frictionless, horizontal surface, as shown in the diagram below. The magnitude of force F 1 is greater than that of force F. F 1 F W E The crate will N S A B C D accelerate towards the east. accelerate towards the west. move at a constant speed towards the east. move at a constant speed towards the west. () 1. A person stands on a bathroom scale that is calibrated in newton, in a stationary elevator. The reading on the bathroom scale is W. The elevator now moves with a constant upward acceleration of 4 1 g, where g is the gravitational acceleration. What will the reading on the bathroom scale be now? A B C D W 5 4 W W W () Copyright reserved Please turn over

4 Physical Sciences/P1 4 DBE/November 015 NSC 1.3 Which ONE of the graphs below correctly represents the relationship between the kinetic energy (K) of a free-falling object and its speed (v)? A B K (J) K (J) C v (m s -1 ) D v (m s -1 ) K (J) K (J) v (m s -1 ) v (m s -1 ) () 1.4 The simplified diagram below shows a rocket that has been fired horizontally, accelerating to the west. N Rocket W E exhaust gases S Which ONE of the statements below best explains why the rocket accelerates? A B C D The speed of the exhaust gases is smaller than the speed of the rocket. The pressure of the atmosphere at the back of the rocket is less than at the front. The air outside the rocket exerts a greater force on the back of the rocket than at the front. The rocket pushes the exhaust gases to the east and the exhaust gases push the rocket to the west. () Copyright reserved Please turn over

5 Physical Sciences/P1 5 DBE/November 015 NSC 1.5 The graph below represents the relationship between the work done on an object and the time taken for this work to be done. Work done (J) The gradient of the graph represents the Time (s) A B C D power. momentum. kinetic energy. potential energy. () 1.6 A line emission spectrum is formed when an excited atom moves from a... A B C D higher to a lower energy level and releases energy. higher to a lower energy level and absorbs energy. lower to a higher energy level and releases energy. lower to a higher energy level and absorbs energy. () 1.7 Two charged spheres of magnitudes Q and Q respectively are placed a distance r apart on insulating stands. If the sphere of charge Q experiences a force F to the east, then the sphere of charge Q will experience a force A B C D F to the west. F to the east. F to the west. F to the east. () Copyright reserved Please turn over

6 Physical Sciences/P1 6 DBE/November 015 NSC 1.8 The four resistors P, Q, R and T in the circuit below are identical. The cell has an emf ε and negligible internal resistance. The switch is initially CLOSED. P S ε Q R T Switch S is now OPENED. Which ONE of the following combinations of changes will occur in P, R and T? CURRENT IN P CURRENT IN R CURRENT IN T A Decreases Remains the same Decreases B Increases Remains the same Increases C Increases Increases Increases D Decreases Increases Decreases () 1.9 A DC current passes through a rectangular wire loop OPQR placed between two pole pieces of a magnet, as shown below. Which TWO segments of the loop will experience an electromagnetic force when the loop is in the position above? A B C D OP and PQ QR and RO OP and QR RO and OP () Copyright reserved Please turn over

7 Physical Sciences/P1 7 DBE/November 015 NSC 1.10 When light of a certain wavelength is incident on a metal surface, no electrons are ejected. Which ONE of the following changes may result in electrons being ejected from the metal surface? A B C D Increase the intensity of the light. Use light with a much shorter wavelength. Use metal with a larger work function. Increase the surface area of the metal. () [0] Copyright reserved Please turn over

8 Physical Sciences/P1 8 DBE/November 015 NSC QUESTION (Start on a new page.).1 Two blocks of mass M kg and,5 kg respectively are connected by a light, inextensible string. The string runs over a light, frictionless pulley, as shown in the diagram below. The blocks are stationary. table M kg,5 kg.1.1 State Newton's THIRD law in words. ().1. Calculate the tension in the string. (3) The coefficient of static friction (μ s ) between the unknown mass M and the surface of the table is 0,..1.3 Calculate the minimum value of M that will prevent the blocks from moving. (5) The block of unknown mass M is now replaced with a block of mass 5 kg. The,5 kg block now accelerates downwards. The coefficient of kinetic friction (µ k ) between the 5 kg block and the surface of the table is 0, Calculate the magnitude of the acceleration of the 5 kg block. (5). A small hypothetical planet X has a mass of 6,5 x 10 0 kg and a radius of 550 km. Calculate the gravitational force (weight) that planet X exerts on a 90 kg rock on this planet's surface. (4) [19] Copyright reserved Please turn over

9 Physical Sciences/P1 9 DBE/November 015 NSC QUESTION 3 (Start on a new page.) Ball A is projected vertically upwards at a velocity of 16 m s -1 from the ground. Ignore the effects of air resistance. Use the ground as zero reference. 3.1 Calculate the time taken by ball A to return to the ground. (4) 3. Sketch a velocity-time graph for ball A. Show the following on the graph: (a) Initial velocity of ball A (b) Time taken to reach the highest point of the motion (c) Time taken to return to the ground (3) ONE SECOND after ball A is projected upwards, a second ball, B, is thrown vertically downwards at a velocity of 9 m s -1 from a balcony 30 m above the ground. Refer to the diagram below. B 9 m s m 16 m s -1 A ground 3.3 Calculate how high above the ground ball A will be at the instant the two balls pass each other. (6) [13] Copyright reserved Please turn over

10 Physical Sciences/P1 10 DBE/November 015 NSC QUESTION 4 (Start on a new page.) A bullet of mass 0 g is fired from a stationary rifle of mass 3 kg. Assume that the bullet moves horizontally. Immediately after firing, the rifle recoils (moves back) with a velocity of 1,4 m s Calculate the speed at which the bullet leaves the rifle. (4) The bullet strikes a stationary 5 kg wooden block fixed to a flat, horizontal table. The bullet is brought to rest after travelling a distance of 0,4 m into the block. Refer to the diagram below. before after 0 g 5 kg 4. Calculate the magnitude of the average force exerted by the block on the bullet. (5) 4.3 How does the magnitude of the force calculated in QUESTION 4. compare to the magnitude of the force exerted by the bullet on the block? Write down only LARGER THAN, SMALLER THAN or THE SAME. (1) [10] Copyright reserved Please turn over

11 Physical Sciences/P1 11 DBE/November 015 NSC QUESTION 5 (Start on a new page.) The track for a motorbike race consists of a straight, horizontal section that is 800 m long. A participant, such as the one in the picture above, rides at a certain average speed and completes the 800 m course in 75 s. To maintain this speed, a constant driving force of 40 N acts on the motorbike. 5.1 Calculate the average power developed by the motorbike for this motion. (3) Another person practises on the same motorbike on a track with an incline. Starting from rest, the person rides a distance of 450 m up the incline which has a vertical height of 5 m, as shown below. 5 m 450 m The total frictional force acting on the motorbike is 94 N. The combined mass of rider and motorbike is 300 kg. The average driving force on the motorbike as it moves up the incline is 350 N. Consider the motorbike and rider as a single system. 5. Draw a labelled free-body diagram for the motorbike-rider system on the incline. (4) 5.3 State the WORK-ENERGY theorem in words. () 5.4 Use energy principles to calculate the speed of the motorbike at the end of the 450 m ride. (6) [15] Copyright reserved Please turn over

12 Physical Sciences/P1 1 DBE/November 015 NSC QUESTION 6 (Start on a new page.) 6.1 The data below was obtained during an investigation into the relationship between the different velocities of a moving sound source and the frequencies detected by a stationary listener for each velocity. The effect of wind was ignored in this investigation. Experiment number Velocity of the sound source (m s -1 ) Frequency (Hz) of the sound detected by the stationary listener Write down the dependent variable for this investigation. (1) 6.1. State the Doppler effect in words. () Was the sound source moving TOWARDS or AWAY FROM the listener? Give a reason for the answer. () Use the information in the table to calculate the speed of sound during the investigation. (5) 6. The spectral lines of a distant star are shifted towards the longer wavelengths of light. Is the star moving TOWARDS or AWAY FROM the Earth? (1) [11] Copyright reserved Please turn over

13 Physical Sciences/P1 13 DBE/November 015 NSC QUESTION 7 (Start on a new page.) A very small graphite-coated sphere P is rubbed with a cloth. It is found that the sphere acquires a charge of + 0,5 µc. 7.1 Calculate the number of electrons removed from sphere P during the charging process. (3) Now the charged sphere P is suspended from a light, inextensible string. Another sphere, R, with a charge of 0,9 µc, on an insulated stand, is brought close to sphere P. As a result sphere P moves to a position where it is 0 cm from sphere R, as shown below. The system is in equilibrium and the angle between the string and the vertical is 7 o. 7 o P 0 cm R 7. Draw a labelled free-body diagram showing ALL the forces acting on sphere P. (3) 7.3 State Coulomb's law in words. () 7.4 Calculate the magnitude of the tension in the string. (5) [13] Copyright reserved Please turn over

14 Physical Sciences/P1 14 DBE/November 015 NSC QUESTION 8 (Start on a new page.) Two charged particles, Q 1 and Q, are placed 0,4 m apart along a straight line. The charge on Q 1 is + x 10-5 C, and the charge on Q is 8 x 10-6 C. Point X is 0,5 m east of Q 1, as shown in the diagram below. Q 1 0,5 m X Q 0,4 m W N S E Calculate the: 8.1 Net electric field at point X due to the two charges (6) 8. Electrostatic force that a x 10-9 C charge will experience at point X (4) The x 10-9 C charge is replaced with a charge of 4 x 10-9 C at point X. 8.3 Without any further calculation, determine the magnitude of the force that the 4 x 10-9 C charge will experience at point X. (1) [11] Copyright reserved Please turn over

15 Physical Sciences/P1 15 DBE/November 015 NSC QUESTION 9 (Start on a new page.) A battery with an internal resistance of 1 Ω and an unknown emf (ε) is connected in a circuit, as shown below. A high-resistance voltmeter (V) is connected across the battery. A 1 and A represent ammeters of negligible resistance. 8 Ω A 1 0 Ω 16 Ω R A S ε r = 1 Ω V With switch S closed, the current passing through the 8 Ω resistor is 0,5 A. 9.1 State Ohm's law in words. () 9. Calculate the reading on ammeter A 1. (4) 9.3 If device R delivers power of 1 W, calculate the reading on ammeter A. (5) 9.4 Calculate the reading on the voltmeter when switch S is open. (3) [14] Copyright reserved Please turn over

16 Physical Sciences/P1 16 DBE/November 015 NSC QUESTION 10 (Start on a new page.) 10.1 A teacher demonstrates how current can be obtained using a bar magnet, a coil and a galvanometer. The teacher moves the bar magnet up and down, as shown by the arrow in the diagram below. galvanometer coil Briefly describe how the magnet must be moved in order to obtain a LARGE deflection on the galvanometer. () The two devices, A and B, below operate on the principle described in QUESTION above. output output Write down the name of the principle. (1) Write down the name of part X in device A. (1) 10. A 0 V, AC voltage is supplied from a wall socket to an electric kettle of resistance 40,33 Ω. Wall sockets provide rms voltages and currents. Calculate the: Electrical energy consumed by the kettle per second (4) 10.. Maximum (peak) current through the kettle (3) [11] Copyright reserved Please turn over

17 Physical Sciences/P1 17 DBE/November 015 NSC QUESTION 11 (Start on a new page.) In an experiment to demonstrate the photoelectric effect, light of different wavelengths was shone onto a metal surface of a photoelectric cell. The maximum kinetic energy of the emitted electrons was determined for the various wavelengths and recorded in the table below. INVERSE OF WAVELENGTH 1 ( 10 6 m -1 ) λ MAXIMUM KINETIC ENERGY E k(max) ( J) 5,00 6,60 3,30 3,30,50 1,70,00 0, What is meant by the term photoelectric effect? () 11. Draw a graph of E k(max) (y-axis) versus λ 1 (x-axis) ON THE ATTACHED ANSWER SHEET. (3) 11.3 USE THE GRAPH to determine: The threshold frequency of the metal in the photoelectric cell (4) Planck's constant (4) [13] TOTAL: 150 Copyright reserved

18 Physical Sciences/P1 DBE/November 015 NSC DATA FOR PHYSICAL SCIENCES GRADE 1 PAPER 1 (PHYSICS) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 1 VRAESTEL 1 (FISIKA) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE Acceleration due to gravity Swaartekragversnelling g 9,8 m s - Universal gravitational constant Universele gravitasiekonstante G 6,67 x N m kg Radius of the Earth Radius van die Aarde R E 6,38 x 10 6 m Mass of the Earth Massa van die Aarde M E 5,98 x 10 4 kg Speed of light in a vacuum Spoed van lig in 'n vakuum c 3,0 x 10 8 m s -1 Planck's constant Planck se konstante h 6,63 x J s Coulomb's constant Coulomb se konstante k 9,0 x N m C Charge on electron Lading op elektron e 1,6 x C Electron mass Elektronmassa m e 9,11 x kg Copyright reserved

19 Physical Sciences/P1 DBE/November 015 NSC TABLE : FORMULAE/TABEL : FORMULES MOTION/BEWEGING 1 1 vf = vi + a t Δx = v iδt + a t or/of Δy = v iδt + a t v i + v f v i + v f v f = v i + a x or/of v f = v i + a y Δx = Δt or/of Δy = Δt FORCE/KRAG F net = ma p = mv max fs = μsn Fnet t = p p = mv - mv f i fk = µ kn w = mg m m d m m r 1 1 F =G or/of F =G M M g = G or/of g = G d r WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING W = FΔx cos θ U = mgh or/of E P = mgh 1 K = mv or/of W nc E = k 1 mv = K + U or/of Wnc = Ek + Ep W net = K or/of Wnet = Ek K = K f K i or/of Ek = Ekf Eki W P = t P ave = Fv ave / P gemid = Fv gemid WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG v = f λ 1 T = f v ± v L L L = fs fl fb v ± v s v ± v b f v ± v = E = hf or /of E = W o + E k(max) or/of E = W o + Kmax where/waar 1 E = hf and/en W 0 = hf0 and/en E k(max) = mv max or/of E = hc λ 1 K = max mv max Copyright reserved

20 Physical Sciences/P1 DBE/November 015 NSC ELECTROSTATICS/ELEKTROSTATIKA kq Q 1 F= E = r kq r W V = q F E = q Q n = e or/of n = Q q e ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE R = V I R = R1 + R = + R R R s + p W = Vq 1 W = VI t W = I R t W = V Δt R emf ( ε ) = I(R + r) emk ( ε ) = I(R + r) q = I t P = W Δt P = VI P = I R V P = R ALTERNATING CURRENT/WISSELSTROOM Imax I rms = / I = wgk I maks V V = max rms / V V = wgk maks P P = I / Pgemiddeld = V wgk Iwgk ave V rms rms = R / P = R ave I rms Vrms Pave = / R P gemiddeld I wgk gemiddeld = V wgk R Copyright reserved

21 Physical Sciences/P1 DBE/November 015 NSC CENTRE NUMBER: EXAMINATION NUMBER: ANSWER SHEET QUESTION 11. Hand in this ANSWER SHEET with your ANSWER BOOK. Graph of E k(max) versus λ Ek(max) x (J) (m -1 ) λ -4 Copyright reserved

22 NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 1 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 015 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 8 pages. Hierdie memorandum bestaan uit 8 bladsye. Please turn over/blaai om asseblief

23 Physical Sciences P1/Fisiese Wetenskappe V1 DBE/November 015 QUESTION 1/VRAAG B () 1. D () 1.3 C () 1.4 D () 1.5 A () 1.6 A () 1.7 A () 1.8 D () 1.9 C () 1.10 B () [0] Please turn over/blaai om asseblief

24 Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/November 015 QUESTION /VRAAG.1.1 When body A exerts a force on body B, body B exerts a force of equal magnitude in the opposite direction on body A. Wanneer liggaam A 'n krag uitoefen op liggaam B, oefen liggaam B 'n krag van gelyke grootte in die teenoorgestelde rigting op liggaam A uit. OR/OF If body A exerts a force on body B, then body B exerts an equal and opposite force on body A Indien liggaam A 'n krag uitoefen op liggaam B, dan sal liggaam B 'n gelyke maar teenoorgestelde krag op liggaam A uitoefen ().1. For,5 kg block/vir,5 kg blok T = mg T = (,5)(9,8) = 4,5 N OR/OF F net = ma T mg = (,5)(0) T (,5)(9,8) = 0 T = 4,5 N OR/OF F net = ma mg - T = (,5)(0) (,5)(9,8) - T = 0 T = 4,5 N (3).1.3 POSITIVE MARKING FROM.1. POSITIEWE NASIEN VANAF.1. For mass M/Vir mass M f s = μ s N 4, 5 N = = 1,5 N 0, N = Mg = 1,5 N M(9,8) = 1,5 N M = 1,5 kg.1.4 For the 5 kg block/vir die 5 kg blok: f k = μ k N f k = (0,15)(5)(9,8) = 7,35 N F net = ma T f k = ma T 7,35 = 5a For the,5 kg block/vir die,5 kg blok w T = ma (,5)(9,8) T =,5 a OR/OF μ s N = μ s Mg 4,5= (0,)M(9,8) M = 1,5 kg (5) 17,15 = 7,5 a a =,9 m s - (5) Please turn over/blaai om asseblief

25 Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 015. m1m F=G r (6,67 10 )(6,5 x 10 )(90) F = 3 ( ) = 1,90 N (1,899 N) OR/OF Gm g = g = r ( 6,67 10 )( 6,5 10 ) 3 ( ) = 0,143...m s - w = mg = (90)(0,143..) = 1,89 N(downwards/afwaarts) (4) (4) [19] Please turn over/blaai om asseblief

26 Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November 015 QUESTION 3/VRAAG OPTION 1/OPSIE 1 Upwards positive/opwaarts positief: v f = v i + a t -16 = 16 9,8( t) t = 3,7s OPTION /OPSIE Upwards positive/opwaarts positief: v f = v i + a t To the top/by bopunt: 0 = 16 9,8( t) t = 1,63s Total time/totale tyd = 1,63 x = 3,6(7) s 3.1 OPTION 3/OPSIE 3 Upwards positive/opwaarts positief: y = v i t + ½ a t 0 = 16 t + ½ (-9,8) t t(16-4,9 t) = 0 t = 0 or/of 3,7 s Time taken/tyd geneem = 3,7 s (accept/aanvaar 3,6 s) OPTION 4/OPSIE 4 Upwards positive/opwaarts positief: vf = vi + aδy At highest point/by hoogste punt 0 = 16 + (-9,8) y y = 13,06 m y = v i t + ½a t 13,06 = 16 t 4,9 t t =1,6 or 1,65 Total time/totale tyd = (1,6/1,65)x = 3,4 sor/of 3,3 s Downwards positive/afwaarts positief: v f = v i + a t 16 = ,8( t) t = 3,7s (4) Downwards positive/afwaarts positief: v f = v i + a t To the top/by bopunt: 0 = ,8( t) t = 1,63s Total time/totale tyd = 1,63 x = 3,6(7) s (4) Downwards positive/afwaarts positief: y = v i t + ½ a t 0 = -16 t + ½ (9,8) t t(-16 +4,9 t) = 0 t = 0 or/of 3,7 s Time taken/tyd geneem = 3,7 s (accept/aanvaar 3,6 s) (4) Downwards positive/afwaarts positief: vf = vi + aδy At highest point/by hoogste punt 0 = (-16) + (9,8) y y = 13,06 m y = v i t + ½a t 13,06 = -16 t + 4,9 t t =1,6 or 1,65 Total time/totale tyd = (1,6/1,65) x = 3,4 sor/of 3,3 s (4) Please turn over/blaai om asseblief

27 Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November 015 OPTION 5/OPSIE 5 Upwards positive/opwaarts positief: vf = vi + aδy At highest point/by hoogste punt 0 = 16 + (-9,8) y y = 13,06 m v f + v i Δy = t ,06 = t Δt = 1,63 s Total time/totale tyd = 3,6 s 3.1 OPTION 6 /OPSIE 6 Upwards positive/opwaarts positief: F net t = p mg t = m (v f -v i ) -9,8 t = (0-16) t = 1,63 s Total time/totale tyd = (1,63)() = 3,6 s OPTION 7 /OPSIE 7 Upwards positive/opwaarts positief: F net t = p mg t = m (v f -v i ) -9,8 t = [-16 (+16)] t = 3,6 s Total time/totale tyd = 3,6 s Downwards positive/afwaarts positief: vf = vi + aδy At highest point/by hoogste punt 0 = (-16) + (9,8) y y = -13,06 m v f + v i Δy = t ,06 = t Δt = 1,63 s Total time/totale tyd = 3,6 s (4) Downwards positive/afwaarts positief: F net t = p mg t = m (v f -v i ) 9,8 t = {0 (-16)} t = 1,63 s Total time/totale tyd = (1,63)() = 3,6 s (4) Downwards positive/afwaarts positief: F net t = p mg t = m (v f -v i ) 9,8 t = [16 (-16)] t = 3,6 s Total time/totale tyd = 3,6 s (4) Please turn over/blaai om asseblief

28 Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November POSITIVE MARKING FROM 3.1./POSITIEWE NASIEN VANAF 3.1 Upwards positive/opwaarts positief: Velocity/snelheid (m s -1 ) ,63 3,6 t(s) -16 (3) POSITIVE MARKING FROM 3../POSITIEWE NASIEN VANAF 3. Downwards positive/afwaarts positief: Velocity/snelheid (m s -1 ) ,63 3,6 t(s) -16 Criteria for graph/kriteria vir grafiek Correct shape for line extending beyond t = 1,63 s. Korrekte vorm vir lyn verleng verby t = 1,63 s Initial velocity correctly indicated as shown. Beginsnelheid korrek aangedui soos getoon. Time to reach maximum height and time to return to the ground correctly shown. Tyd om maksimum hoogte te bereik en om na die grond terug te keer. Marks/Punte (3) Please turn over/blaai om asseblief

29 Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/November OPTION 1 / OPSIE 1 Upwards positive/opwaarts positief: Take y A as height of ball A from the ground. (no penalising)/neem y A as hoogte van bal A vanaf die grond. (geen penalisering) y A = v i t + ½ a t y A - 0 = 16 t + ½(-9,8) t =16 t 4,9 t Take y B as height of ball B from the ground./neem y B as hoogte van bal B vanaf die grond. y B = v i t + ½ a t y B 30 = (v i t + ½ a t ) y B = 30 - [ -9( t -1) + ½(-9,8)( t 1) = 34,1 +0,8 t - 4,9 t y A = y B 16 t 4,9 t = 34,1 + 0,8 t - 4,9 t 15, t = 34,1 t =,4 s y A = 16 (,4) - 4,9(,4) = 11,5 m (6) Downwards positive/afwaarts positief: Take y A as height of ball A from the ground.(no penalising)/neem y A as hoogte van bal A vanaf die grond. (geen penalisering) y A = v i t + ½ a t y A - 0 = -16 t + ½(9,8) t = -16 t + 4,9 t Take y B as height of ball B from the ground/neem as hoogte van bal B vanaf die grond.. y B = v i t + ½ a t y B 30 = (v i t + ½ a t ) y B = 30 [ 9( t -1) + ½(9,8)( t 1) = 34,1 + 0,8 t - 4,9 t y A = y B 16 t 4,9 t = 34,1+ 0,8 t 4,9 t 15, t = 34,1 t =,4 s y A = (- 16 (,4) + 4,9(,4) ) = 11,5 m (6) Please turn over/blaai om asseblief

30 Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/November OPTION /OPSIE Upwards positive/opwaarts positief: Δy A = v i t + ½ a t = 16 t + ½(-9,8) t =16 t 4,9 t Distance travelled by ball A = y A = 16 t 4,9 t Δy B = v i t + ½ a t = -9( t -1) + ½(-9,8)( t 1) = 0,8 t 4,9 t + 4,1 Distance travelled by ball B = y B = 0,8 t 4,9 t + 4,1 y A +(-y B ) = t 4,9 t (0,8 t 4,9 t + 4,1) = 30 15, t = 34,1 t =,4 s y A = v i t + ½ a t y A = 16 (,4) - 4,9(,4) = 11,5 m (6) 3.3 Downwards positive/afwaarts positief: y A = v i t + ½ a t = -16 t + ½(9,8) t = -16 t + 4,9 t y B = v i t + ½ a t = 9( t -1) + ½(9,8)( t 1) = - 0,8 t + 4,9 t - 4,1 (-y A ) + y B = 30 -(-16 t + 4,9 t ) - 0,8 t + 4,9 t - 4,1 = 30 15, t = 34,1 t =,4 s y A = v i t + ½ a t y A = -16 (,4)+ 4,9(,4) = - 11,5 m Height of ball A/Hoogte van bal A = 11,5 m (6) Please turn over/blaai om asseblief

31 Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/November 015 OPTION 3/OPSIE 3 Upwards positive/opwaarts positief: v f = v i + a t After 1 s, speed of ball A/Spoed van bal A na 1 s v f = 16 + (-9,8)(1) = 6, m s -1 Distance travelled by ball A in 1 s/afstand deur bal A afgelê in 1 s y A = v i t + ½ a t = (16)(1) + ½(-9,8)1 = 11,1 m For ball A, after 1 s/vir bal A na 1 s y A = 6, t 4,9 t For ball/vir bal B, y B = v i t + ½ a t = -9 t + ½(-9,8) t y A + (-y B ) = (30-11,1) = 18,9 6, t 4,9 t [ -9 t + ½(-9,8) t ] = 18,9 15, t = 18,9 t = 1,4 s The balls meet after/die balle ontmoet na (1,4 +1) =,4 s y A = [6, (1,4) 4,9 (1,4) ] = 0,154 m Meeting point/ontmoetingspunt = (11,1 + 0,154) = 11,5 m OR/OF Δy = (-9)(1,4) + ½ (-9,8)(1,4) = -18,69 m Meeting point/ontmoetingspunt = (30-18,69) = 11,31 m (6) Please turn over/blaai om asseblief

32 Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/November 015 Downwards positive/afwaarts positief: v f = v i + a t After 1 s, speed of ball A/Spoed van bal A na 1 s v f = (9,8)(1) = -6, ms -1 Distance travelled by ball A in 1 s/afstand deur bal A afgelê in 1 s y A = v i t + ½ a t = (-16)(1) + ½(9,8)(1) = - 11,1 m For ball A, after 1 s/vir bal A na 1 s y A = - 6, t + 4,9 t For ball/vir bal B y B = v i t + ½ a t = 9 t + ½(9,8) t - y A + y B = 18,9 6, t - 4,9 t + [ 9 t + ½(9,8) t ] = 18,9 15, t = 18,9 t = 1,4 s The balls meet after/die balle ontmoet na (1,4 +1) =,4 s y A = -6, (1,4) + 4,9 (1,4) = - 0,154 m Meeting point/ontmoetingspunt = (-11,1-0,154) = 11,5 m OR/OF Δy = (9)(1,4) + ½ (9,8)(1,4) = 18,69 m Meeting point/ontmoetingspunt = (30-18,69) = 11,31 m (6) [13] Please turn over/blaai om asseblief

33 Physical Sciences P1/Fisiese Wetenskappe V1 1 DBE/November 015 QUESTION 4/VRAAG OPTION 1/OPSIE 1 Take motion to the right as positive/neem beweging na regs as positief. Σp i = Σp f (m 1 + m )v i = m 1 v f1 + m v f (m 1 +m )v i = m 1 v f1 + m v f (3 + 0,0)(0) = (3)(-1,4) + (0,0) v f v f = 10 m s -1 OR/OF Take motion to the left as positive/neem beweging na links as positief. Σp i =Σ p f (m 1 + m )v i = m 1 v f1 + m v f (m 1 +m )v i = m 1 v f1 + m v f (3 + 0,0)(0) = (3)(1,4) + (0,0) v f v f = -10 m s -1 Speed/Spoed = 10 m s -1 (4) OPTION /OPSIE Take motion to the right as positive/neem beweging na regs as positief. Δp bullet = -Δp block m(v f v i ) = - m(v f v i ) (0,0)(v f 0) = -(3)(-1,4 0) v i = 10 m s -1 Any one/enige een Any one/enige een OR/OF Take motion to the left as positive/neem beweging na links as positief Δp bullet = -Δp block m(v f v i ) = - m(v f v i ) (0,0)(v f 0) = -(3)(1,4 0) v i = -10 m s -1 Speed/Spoed = 10 m s -1 (4) Please turn over/blaai om asseblief

34 Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/November OPTION 1/OPSIE 1 v f = v i + a x 0 = 10 + a(0,4) a = m s - F net = ma = (0,0)(-55 15) = -1 10,5 N Magnitude of force = 1 10,5 N Grootte van krag = 1 10,5 N OPTION /OPSIE vi + v f x = t ,4 = t t = 0,004 s (0,00381s) F net t = p = m v (0,0)(0-10) F net = ( 0,004) = N Magnitude of force = N Grootte van krag = N (Accept/Aanvaar: 110,5 N) (5) OPTION 3/OPSIE 3 v f = v i + a x 0 = 10 + a(0,4) a = m s - v f = v i + a t 0 = 10 (55 15) t t = 0,004 s (0,00381 s) F net t = p = m v (0,0)(0-10) F net = ( 0,004) = N Magnitude of force = N (Accept/Aanvaar: 1 10,5 N) Grootte van krag = N (5) OPTION 4/OPSIE 4 W net = K F net xcosθ = K = ½ m(v i - v f) Any one/enige een F net (0,4) cos180 o = ½ (0,0)(0-10 ) F net = 1 10,5 N OR/OF W nc = ΔE p + ΔE k Any one/enige een F net Δxcosθ = 0 + ½ m(v i - v f) F net (0,4) cos180 o = ½ (0,0)(0-10 ) F net = 1 10,5 N (5) 4.3 The same as/equal Dieselfde as/gelyk (1) [10] Please turn over/blaai om asseblief

35 Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/November 015 QUESTION 5/VRAAG OPTION 1/OPSIE v ave = = 10,67m s P ave = Fv ave P ave = (40)(10,67) = 560,8 W (,56 kw) OPTION 3/OPSIE 3 W P = t F x cos θ = t o (40)(800)cos0 = 75 = 560 W OPTION /OPSIE 800 v ave = = 10,67m s Distance covered in 1s = 10,67m W(Work done in 1 s) = FΔxcosθ = (40)(10,67)(1) = 560,8 J s -1 P ave = 560,8 W (,56 kw) OPTION 4/OPSIE 4 W P = t F x cos θ = t o (40)(10,67)cos0 = 1 = 560 W (3) 5. F D /F A F N /N F f /f w/f g w f N F D Accepted labels/aanvaarde benoemings F g / F w / weight / mg / gravitational force/ 940 N F g / F w / gewig / mg / gravitasiekrag F friction / F f / friction /94 N /f k F wrywing / F w / wrywing/94 N /f k F N / F normal / normal force F N / F normaal / normaalkrag F Applied/toegepas /350 N/Average driving force F driving/dryfkrag /350/Gemiddelde aandrywingskrag (4) 5.3 The net/total work done on an object is equal to the change in the object's kinetic energy Die netto/totale arbeid verrig op 'n voorwerp is geyk aan die verandering in die voorwerp se kinetiese energie. OR/OF The work done on an object by a resultant/net force is equal to the change in the object's kinetic energy. Die arbeid verrig op 'n voorwerp deur 'n resulterende krag is gelyk aan die verandering in die voorwerp se kinetiese energie. () Please turn over/blaai om asseblief

36 Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/November OPTION 1/OPSIE 1 W nc = U + K W f + W D = U + K f xcos θ + W D = mg(h f - h i ) + ½ (m)(v f v i ) (f xcosθ + F D x cos θ = mg(h f - h i ) + ½ m(v f v i ) (94)(450)(cos180 o ) + (350)(450)cos0 o = (300)(9,8)(5-0) + ½(300)(v f 0) v f = 8,37 m s -1 OPTION /OPSIE W net = K W net = W D + W g + W f + W N = (F D xcos θ) + (mg sinα) xcosθ) + (f xcosθ) + 0 (6) 5 W net = [350(450)](cos0)+ (300)(9,8) (450)(cos180) (450)(cos180 o ) = = J OR/OF W net = K α = sin = ½ (300)(v f 0) 450 v f = 8,37 m s -1 = 0,64 o (6) OPTION 3/OPSIE 3 W net = W D + W g + W f + W N = (F D xcos θ) + mg xcosθ) + f xcosθ + 0 W net = (350)(450)(cos0 o )+(300)(9,8)(450)cos(90 + 0,64) +94(450)(cos180 o ) = , = ,51 J OR/OF W net = K α = sin ,51 = ½ (300)(v f 0) 450 v f = 8,34 m s -1 = 0,64 o (6) OPTION 4/OPSIE 4 F net = F D + (-mgsinα) +(-f) = 350 +[-(300)(9,8)sin0,64 o ] + (- 94) = 3,16 N W net = F net x cos θ = (3,16)(450) cos0 o = 10 4 J OR/OF F net = (300)(9,8)sin0,64 o - 94 = 350-3,84-94 = 3,16 N W net = K 10 4 = ½ (300)(v f 0) v f = 8,34 m s -1 (6) [15] 5 Please turn over/blaai om asseblief

37 Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/November 015 QUESTION 6/VRAAG Frequency (of sound detected by the listener (observer)) Frekwensie van klank deur luisteraar (waarnemer) waargeneem (1) 6.1. The apparent change in frequency or pitch of 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 deur 'n luisteraar waargeneem omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium van klankvoortplanting het. () Away/Weg van Detected frequency of source decreases Waargenome frekwensie van bron neem af () OPTION 1/OPSIE 1 EXPERIMENT/EKSPERIMENT v ± v L v fl = fs OR/OF fl = fs v ± v s v + v s v 874 = ( 900) v + 10 v = 336,15 m s -1 (Accept/Aanvaar : 336,15 m s -1 33,33 m s -1 ) (5) EXPERIMENT/EKSPERIMENT 3 v ± v L v fl = fs OR/OF fl = fs v ± v s v + v s v 850 = ( 900) v + 0 v = 340 m s -1 (Accept/Aanvaar : 313,33 m s m s -1 ) (5) EXPERIMENT 4/EKSPERIMENT 4 v ± v L v fl = fs OR/OF fl = fs v ± v s v + v s v 87 = ( 900) v + 30 v = 339,86 m s -1 (Accept/Aanvaar : 339,86 m s m s -1 ) (5) Please turn over/blaai om asseblief

38 Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/November 015 OPTION /OPSIE v ± vl v fl = fs OR/OF fl = fs v ± v v + v s s Experiment/Eksperiment and/en (v + 10) 850(v + 0) = v v 874v = 850v both frequencies / beide frekwensies v = 344,17 m s -1 Experiment/Eksperiment and/en (v + 10) 87(v + 30) = v v 874v = 87v both frequencies / beide frekwensies v = 341,91 m s -1 Experiment/Eksperiment 3 and/en (v + 0) 87(v + 30) = v v 850v = 87v both frequencies / beide frekwensies v = 339,57 m s -1 (5) 6. Away from the Earth/Weg vanaf die aarde (1) [11] Please turn over/blaai om asseblief

39 Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/November 015 QUESTION 7/VRAAG Q n = e -6 0,5 10 n = -19 1,6 10 n = 3,13 x 10 1 electrons/elektrone (3) 7. T/F T P F E w/ F g Accepted labels/aanvaarde benoemings F w g / F w / weight / mg / gravitational force F g / F w / gewig / mg / gravitasiekrag F T T / tension F T / spanning F E Electrostatic force/f C / Coulombic force/f Q /F RP/PR Elektrostiesekrag / Coulombkrag / F Q /F RP/PR (3) 7.3 The magnitude of the electrostatic force exerted by one point charge (Q 1 ) on another point charge (Q ) is directly proportional to the product of the (magnitudes of the) charges and inversely proportional to the square of the distance (r) between them. Die grootte van die elektrostatiese krag wat deur een puntlading (Q 1 ) op 'n ander puntlading (Q )uitgeoefen word, is direk eweredig aan die produk van die (groottes van die) ladings en omgekeerd eweredig aam die kwadraat van die afstand (r) tussen hulle. () Please turn over/blaai om asseblief

40 Physical Sciences P1/Fisiese Wetenskappe V1 19 DBE/November OPTION 1/OPSIE 1 Q1Q F E = k r Tsinθ/(Tcosθ) = F E T sin7 o /(Tcos83 o ) = 9-6 ( 9 10 )(0,5 10 )(0,9 ( 0, ) T = 0,83 N (Accept/Aanvaar 0,8 N) (5) 10-6 ) OPTION /OPSIE kq1q F E = r 9-6 (9 10 )(0,5 10 )(0,9 10 FE = (0,) = 0,101 N o Tx 0,101 tan 7 = = TY TY T Y = 0,83 N -6 ) T = TX + TY = (0,101) + (0,83) = 0,83N (5) OPTION 3/OPSIE kq1q (9 10 )(0,5 10 )(0,9 10 ) F = = = 0,101 N r (0,) FE T = o o sin7 sin90 0,101 T = o o sin7 sin90 T = 0,83 N (5) [13] Please turn over/blaai om asseblief

41 Physical Sciences P1/Fisiese Wetenskappe V1 0 DBE/November 015 QUESTION 8/VRAAG E X = E + E (-8) kq kq = + 8 correct equation /korrekte vergelyking r r (9 10 )( 10 ) (9 10 )(8 10 ) = + (0,5) (0,15) =,88 x , x 10 6 = 6,08 x 10 6 N C -1 to the east/na oos OR/OF Q E = k r 9-5 (9 10 )( 10 ) E = (0,5) =,88x 10 6 NC -1 to the east/na oos 9-6 (9 10 )( 8 10 ) E -8 = (0,15) = 3, x 10 6 N C -1 to the east/na oos E X = E + E (-8) = (,88 x , x 10 6 ) = 6,08 x 10 6 N C -1 to the east/na oos (6) 8. OPTION 1/OPSIE 1 F E = QE = (- x 10-9 ) (6,08 x 10 6 ) = -1,16 x 10-3 N = 1, x 10 - N to the west/na wes (4) OPTION /OPSIE F (-)Q1 = qe () = ( x 10-9 ) (,88x 10 6 ) = 5,76 x 10-3 N to the west/na wes F (-)Q = qe (8) = ( x 10-9 )(3, x 10 6 ) = 6,4 x 10-3 N to the west/na wes F net = 5,76 x ,4 x 10-3 = 1, x 10 - N to the west/na wes (4) Please turn over/blaai om asseblief

42 Physical Sciences P1/Fisiese Wetenskappe V1 1 DBE/November 015 OPTION 3/OPSIE 3 Q1Q F = k r ( 9 10 )( 10 )( 10 ) F (-) = (0,5) = 5,76 x 10-3 N to the west/na wes ( 9 10 )( 10 )(8 10 ) F (-)(-8) = (0,15) = 6,4 x10-3 N to the west/na wes F net = (5,76 x ,4 x 10-3 ) = 1, x 10 - N to the west/na wes (4) 8.3,44 x 10 - N (1) [11] QUESTION 9/VRAAG The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. (provided temperature and all other physical conditions are constant) Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur (mits temperatuur en alle fisiese toestande konstant bly) OR/OF. The current in a conductor is directly proportional to the potential difference across the conductor, provided temperature and all other physical conditions are constant Die stroom in 'n geleier is direk eweredig aan die potensiaalverskil oor n geleier by konstante temperatuur mits temperatuur en alle fisiese toestande konstant bly () 9. OPTION 1/OPSIE 1 V = IR V 8 = (0,5)(8) = 4 V V 8 = V 16 V 16 = 4 V I 16 = R V = 16 4 = 0,5 A I tot// = A 1 = (0,5 + 0,5) = 0,75 A OPTION /OPSIE V = IR V 8 = (0,5)(8) = 4 V 1 R = = R1 R R = 5,33 Ω 4 I tot// = 5,33 A 1 = 0,75 A (4) Please turn over/blaai om asseblief

43 Physical Sciences P1/Fisiese Wetenskappe V1 DBE/November 015 OPTION 3/OPSIE 3 I 1 R 1 = I R (0,5)(8) = I 16 (16) I 16 = (8)(0,5) = 0,5 A 16 I tot// = A 1 =(0,5 + 0,5) = 0,75 A 9.3 OPTION 1/OPSIE 1 V = IR V 0Ω =( 0,75) (0) = 15 V V //tot = (15 + 4) = 19 V V R = 19 V OPTION 4/OPSIE 4 R 8Ω = R 16Ω I R16 = ½ I R8 I R16 = ½ (0,5) = 0,5 A A 1 = (0,5 + 0,5) = 0,75 A (4) P = VI 1 = (19)I I R = A = 0,63 A (5) OPTION /OPSIE = + = + R R1 R 8 16 R // = 5,33 Ω R // + R 0 = (5,33 + 0) = 5,33Ω V //tot = I(R // + R 0 ) = (0,75)(5,33) = 19 V OR/OF R1R R = R + R 1 = 8 16 = 5,33 Ω P = VI 1 = I(19) I R = A = 0,63 A (5) OPTION 3/OPSIE 3 V = IR V 0Ω =( 0,75) (0) = 15 V V //tot = (15 + 4) = 19 V V R = 19 V P = R V (19) 1 = R R = 30,08 Ω P = I R 1 = I (30,08) I = 0,63 A (5) Please turn over/blaai om asseblief

44 Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/November OPTION 1/OPSIE 1 ( ε ) = I(R + r) = V terminal + V int = 19 + (0,75 + 0,63)(1) = 0,38 V OPTION /OPSIE V int = Ir = (0,75 + 0,63)(1) = 1,38 V ε = V terminal + V int = ,38 = 0,38 V (3) OPTION 3/OPSIE 3 V 19 R = = = 30,16 Ω I 0, = + = + R p = 13,77 Ω Rp R1 R 30,16 5,33 I tot = 0,63 + 0,75 = 1,38 A ε = I(R + r) = (1,38)(13,77 + 1) = 0,38 V [14] Please turn over/blaai om asseblief

45 Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 015 QUESTION 10/VRAAG Move the bar magnet very quickly up and down inside the coil Beweeg die staafmagneet baie vinnig op en af binne in die spoel. () Electromagnetic induction/elektromagnetiese induksie (1) Commutator/kommutator / split rings/spleetringe (1) OPTION 1/OPSIE 1 Vrms Paverage = R 0 = 40,33 = 1 00,10 W (J s -1 ) OPTION /OPSIE Vrms I = rms R 0 = 40,33 = 5,45 A Paverage = I rms R = (5,45 )(40,33) = 1 197,9 W OR/OF 1 00,10 W Vrms W = t R 0 = (1) 40,33 = 100,10 J I rms Vrms = R 0 = 40,33 = 5,45 A W = I rmsr t = (5,45 )(40,33)(1) = 1 197,9 J OR/OF 1 00,10 J (4) (4) OPTION 3/OPSIE 3 Vrms I = rms R 0 = 40,33 = 5,45 A I rms Vrms = R 0 = 40,33 = 5,45 A P = Irms = (0 )(5,45) = W or/of 1 00,10 W average V rms W = VrmsI rms t = (0 )(5,45)(1) = J or/of 1 00,10 J (4) Please turn over/blaai om asseblief

46 Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November OPTION 1/OPSIE 1 Vmax Vrms = V 0 = max V max = 311,13 V I max V = R max 331,13 = 40,33 = 7,71 A OR/OF Vmax Imax P ave = 311,13)I 100,1 = I max = 7,71 A ( max OPTION /OPSIE Paverage = V rms Irms 100,1 = (0)I rms I rms = 5,455 A I max = (5,455) = 7,71 A (7,715 A) OPTION 3/OPSIE 3 P average = I rms R 100,1 = I rms(40,33) I rms = 5,455 A (3) (3) I max = I rms = (5,455) = 7,71 A (3) OPTION 4/OPSIE 4 V rms = I rms R 0 = I rms (40,33) I rms = 5,455 A I max = I rms = (5,455) = 7,71 A (3) [11] Please turn over/blaai om asseblief

47 Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November 015 QUESTION 11/VRAAG It is the process whereby electrons are ejected from a metal surface when light (of suitable frequency) is incident on it. Dit is die proses waartydens elektrone vanaf n metaaloppervlak vrygestel word wanneer van geskikte frekwensie daarop inval () Graph of/grafiek van E k(max/maks) vs λ 1 4 Ek(max) x (10-19 J) λ (m ) -4 Please turn over/blaai om asseblief

48 Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November OPTION 1/OPSIE = 1,6 10 m -1 λ 1 f o = c λ = (3 x 10 8 )(1,6 x 10 6 ) = 4,8 x Hz (Accept/Aanvaar 4,8 x Hz to/tot 5,1 x Hz) (4) OPTION /OPSIE By extrapolation: y-intercept = -W o /Deur ekstrapolasie : y-afsnit = -W o W o = hf o 3, x = (6,63 x )f o f o = 4,8 x Hz (Accept/Aanvaar 4,8 x Hz to/tot 4,83 x Hz) (4) OPTION 3/OPSIE 3 (Points from the graph/ Punte vanaf grafiek) E = W o + E k(max) hc = hf0 + Ek(max) λ ( 6,63 10 )(3 10 )(1,6 10 ) = (6,63 10 )f0 + 0 f o = 4,8 x Hz OR/OF (6,63 10 )(3 10 )(5 10 ) = (6,63 10 )f0 + 6,6 10 f o = 4,9 x Hz OR/OF (6,63 10 )(3 10 )(3,3 10 ) = (6,63 10 )f0 + 3,3 10 f o = 4,8 x Hz OR/OF (6,63 10 )(3 10 )(,5 10 ) = (6,63 10 )f0 + 1,7 10 f o = 4,94 x Hz OR/OF (6,63 10 )(3 10 )(, 10 ) = (6,63 10 )f0 + 0,7 10 f o = 5,54 x Hz (4) Please turn over/blaai om asseblief

49 Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/November OPTION 1/OPSIE 1 hc = Gradient/ Helling y = x -19 6,6 10 = 6 (5-1, 6) 10 = 1,941 x10-5 (J m) gradient / helling h = c 1, h = = 6,47 x10-34 J s OPTION /OPSIE W 0 = y intercept/afsnit = 3, x10-19 J Accept /Aanvaar 3, x10-19 J to/tot 3,4 x10-19 J) W o = hf o 3, x = h(4,8 x ) h = 6,66 x J s Accept /Aanvaar 6,66 x10-34 J s to/tot 7,08 x10-34 J s) (4) OPTION 3/OPSIE 3 (Points from the graph (Punte vanaf grafiek) hc = W0 + K max = 3, ,6 10 λ -19 9, h = = 6, J s 8 6 (3 10 )(5 10 ) OR/OF hc = W0 + K max = 3, ,3 10 λ -19 6, h = = 6, J s 8 6 (3 10 )(3,3 10 ) OR/OF hc = W0 + K max = 3, ,7 10 λ -19 4, h = = 6, 7 10 J s 8 6 (3 10 )(,5 10 ) OR/OF hc = W0 + K max = 3, ,7 10 λ -19 3, h = = 6, 5 10 J s 8 6 (3 10 )( 10 ) OPTION 4/OPSIE 4 hc 1 Wo = or / of Wo = hc λo λo 3, x = h(3 x 10 8 )(1,6 x 10 6 ) h = 6,66 x J s (4) [13] TOTAL/TOTAAL: 150

50 NATIONAL SENIOR CERTIFICATE GRADE 1 PHYSICAL SCIENCES: CHEMISTRY (P) NOVEMBER 015 MARKS: 150 TIME: 3 hours This question paper consists of 15 pages and 4 data sheets. Copyright reserved Please turn over

51 Physical Sciences/P DBE/November 015 NSC INSTRUCTIONS AND INFORMATION Write your centre number and examination number in the appropriate spaces on the ANSWER BOOK. This question paper consists of TEN questions. Answer ALL the questions in the ANSWER BOOK. Start EACH 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.1 and QUESTION.. You may use a non-programmable calculator. You may use appropriate mathematical instruments. You are advised to use the attached DATA SHEETS. Show ALL formulae and substitutions in ALL calculations. Round off your final numerical answers to a minimum of TWO decimal places. Give brief motivations, discussions et cetera where required. Write neatly and legibly. Copyright reserved Please turn over

52 Physical Sciences/P 3 DBE/November 015 NSC QUESTION 1: MULTIPLE-CHOICE QUESTIONS Four 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 number ( ) in the ANSWER BOOK, for example 1.11 E. 1.1 Which ONE of the following pairs of reactants is used in a reaction during the contact process? A B C D N (g) and H (g) SO (g) and O (g) NH 3 (g) and O (g) H SO 4 (l) and NH 3 (g) () 1. The rate of a chemical reaction is most correctly defined as the... A B C D time taken for a reaction to occur. speed at which a reaction takes place. change in the amount of reactants or products. change in the concentration of reactants or products per unit time. () 1.3 Consider the reaction represented by the balanced equation below. H 3 PO 4 (aq) + H O(l) H 3 O + (aq) + H PO (aq) Which ONE of the following is a conjugate acid-base pair? 4 A B C D H 3 O + (aq) and H O(l) H 3 PO 4 (aq) and H O(l) H 3 PO 4 (aq) and H 3 O + (aq) H 3 O + (aq) and H PO 4 (aq) () 1.4 Which ONE of the following compounds has dipole-dipole forces between its molecules? A B C D Ethanal Ethane Ethene Ethyne () Copyright reserved Please turn over

53 Physical Sciences/P 4 DBE/November 015 NSC 1.5 The energy changes represented by P, Q and R on the potential energy graph below take place during a reversible chemical reaction. Potential energy (kj mol -1 ) P Q R Course of reaction Which ONE of the following changes will decrease both P and R, but leave Q unchanged? A B C D A decrease in volume The addition of a catalyst A decrease in temperature A decrease in concentration () 1.6 Which ONE of the following is a product formed during the hydrolysis of bromoethane? A B C D Water Ethene Ethanol Bromine () 1.7 Which ONE of the following is the EMPIRICAL FORMULA of 1,-dichloroethane? A B C D CHCl CH Cl CHCl C H 4 Cl () Copyright reserved Please turn over

54 Physical Sciences/P 5 DBE/November 015 NSC 1.8 The reaction represented by the balanced equation below reaches equilibrium in a closed container. Cl (g) + H O(l) Cl (aq) + ClO (aq) + H + (aq) Which ONE of the following reagents will favour the forward reaction when added? A B C D Hydrogen Sodium chloride Hydrogen chloride Sodium hydroxide () 1.9 The following half-reactions take place in a galvanic cell: Co 3+ + e - Co + Al e - Al Which ONE of the following is the cell notation for this cell? A B C D Al Al 3+ Co 3+, Co + Al Al 3+ Co 3+, Co + Pt Al Al 3+ Co +, Co 3+ Pt Pt Co +, Co 3+ Al 3+ Al () 1.10 Chlorine gas (Cl ) is bubbled through a potassium iodide solution (KI). The reducing agent in this reaction is: A B C D Potassium ions Chlorine gas Iodide ions Chloride ions () [0] Copyright reserved Please turn over

55 Physical Sciences/P 6 DBE/November 015 NSC QUESTION (Start on a new page.) The letters A to D in the table below represent four organic compounds. H CH 3 H H H H O H A H C C C C C C H B H C C C C H CH 3 H CH CH 3 H H H H C CH 3 CH CHO D Butane Use the information in the table to answer the questions that follow..1 Write down the:.1.1 Letter that represents a ketone (1).1. Structural formula of the functional group of compound C (1).1.3 General formula of the homologous series to which compound A belongs (1).1.4 IUPAC name of compound A (3).1.5 IUPAC name of compound B (). Compound D is a gas used in cigarette lighters...1 To which homologous series does compound D belong? (1).. Write down the STRUCTURAL FORMULA and IUPAC NAME of a structural isomer of compound D. (4)..3 Is the isomer in QUESTION.. a CHAIN, POSITIONAL or FUNCTIONAL isomer? (1).3 Compound D reacts with bromine (Br ) to form -bromobutane. Write down the name of the:.3.1 Homologous series to which -bromobutane belongs (1).3. Type of reaction that takes place (1) [16] Copyright reserved Please turn over

56 Physical Sciences/P 7 DBE/November 015 NSC QUESTION 3 (Start on a new page.) 3.1 The flow diagram below shows two organic reactions. The letter P represents an organic compound. Alcohol + Reaction 1 P Ethyl propanoate Reaction Alkene Use the information in the flow diagram to answer the questions that follow. Write down the: Type of reaction of which Reaction 1 is an example (1) 3.1. STRUCTURAL FORMULA of the functional group of ethyl propanoate (1) IUPAC name of compound P (1) Reaction takes place in the presence of an acid catalyst and heat. Write down the: Type of reaction of which Reaction is an example (1) NAME or FORMULA of the acid catalyst (1) STRUCTURAL FORMULA of the alkene () 3. The condensed formula of a polymer is shown below. H C H H C H n Write down the: 3..1 STRUCTURAL FORMULA of the monomer that is used to prepare the above polymer () 3.. Type of polymerisation reaction (ADDITION or CONDENSATION) that is used to prepare this polymer (1) [10] Copyright reserved Please turn over

57 Physical Sciences/P 8 DBE/November 015 NSC QUESTION 4 (Start on a new page.) Four compounds of comparable molecular mass are used to investigate the effect of functional groups on vapour pressure. The results obtained are shown in the table below. COMPOUND VAPOUR PRESSURE (kpa at 0 C) A Butane 04 B Propan--one 4,6 C Propan-1-ol D Ethanoic acid 1,6 4.1 Define the term functional group of an organic compound. () 4. Which ONE of the compounds (A, B, C or D) in the table has the: 4..1 Highest boiling point (Refer to the vapour pressures in the table to give a reason for the answer.) () 4.. Weakest intermolecular forces (1) 4.3 Refer to the type of intermolecular forces to explain the difference between the vapour pressure of compound A and compound B. (3) 4.4 The vapour pressures of compounds C and D are much lower than those of compounds A and B. Name the type of intermolecular force in A and B that is responsible for this difference. (1) 4.5 Briefly explain the difference in vapour pressure between compound C and compound D. () 4.6 During a combustion reaction in a closed container of adjustable volume, 8 cm 3 of compound A (butane) reacts in excess oxygen according to the following balanced equation: C 4 H 10 (g) + 13O (g) 8CO (g) + 10H O(g) If the initial volume of the oxygen in the container was 60 cm 3, calculate the TOTAL volume of the gases that are present in the container at the end of the reaction. All the gases in the container are at the same temperature and pressure. (5) [16] Copyright reserved Please turn over

58 Physical Sciences/P 9 DBE/November 015 NSC QUESTION 5 (Start on a new page.) Dilute acids, indicated in the table below, react with EXCESS zinc in each of the three experiments to produce hydrogen gas. The zinc is completely covered with the acid in each experiment. EXPERIMENT DILUTE ACID cm 3 of 0,1 mol dm -3 H SO 4 50 cm 3 of 0, mol dm -3 H SO cm 3 of 0,1 mol dm -3 HCl The volume of hydrogen gas produced is measured in each experiment. 5.1 Name TWO essential apparatuses needed to determine the rate of hydrogen production. () The graph below was obtained for Experiment 1. Volume (cm 3 ) Experiment 1 0 t 1 t t 3 Time (s) Use this graph and answer the questions that follow. 5. At which time (t 1, t or t 3 ) is the: 5..1 Reaction rate the highest (1) 5.. Mass of zinc present in the flask the smallest (1) 5.3 In which time interval, between t 1 and t OR between t and t 3, does the largest volume of hydrogen gas form per second? (1) 5.4 Redraw the graph for Experiment 1 in the ANSWER BOOK. On the same set of axes, sketch the graphs that will be obtained for Experiments and 3. Clearly label the three graphs as EXPERIMENT 1, EXPERIMENT and EXPERIMENT 3. (4) Copyright reserved Please turn over

59 Physical Sciences/P 10 DBE/November 015 NSC 5.5 The initial mass of zinc used in each experiment is 0,8 g. The balanced equation for the reaction in Experiment 3 is: Zn(s) + HCl(aq) ZnCl (aq) + H (g) Calculate the mass of zinc present in the flask after completion of the reaction in Experiment 3. (5) 5.5. How will the mass of zinc present in the flask after completion of the reaction in Experiment compare to the answer to QUESTION 5.5.1? Write down only LARGER THAN, SMALLER THAN or EQUAL TO. (1) [15] QUESTION 6 (Start on a new page.) An unknown gas, X (g), is sealed in a container and allowed to form X 3 (g) at 300 C. The reaction reaches equilibrium according to the following balanced equation: 3X (g) X 3 (g) 6.1 How will the rate of formation of X 3 (g) compare to the rate of formation of X (g) at equilibrium? Write down only HIGHER THAN, LOWER THAN or EQUAL TO. (1) The reaction mixture is analysed at regular time intervals. The results obtained are shown in the table below. TIME (s) [ X ] (mol dm -3 ) 0 0,4 0 0, 0,10 4 0,08 0,13 6 0,06 0,6 8 0,06 0,6 10 0,06 0,6 [ X 3 ] (mol dm -3 ) 6. Calculate the equilibrium constant, K c, for this reaction at 300 C. (4) 6.3 More X 3 (g) is now added to the container How will this change affect the amount of X (g)? Write down INCREASES, DECREASES or REMAINS THE SAME. Use Le Chatelier's principle to explain the answer to QUESTION (1) () Copyright reserved Please turn over

60 Physical Sciences/P 11 DBE/November 015 NSC The curves on the set of axes below (not drawn to scale) was obtained from the results in the table on page 10. Concentration (mol dm -3 ) t 1 Time (s) 6.4 How does the rate of the forward reaction compare to that of the reverse reaction at t 1? Write down only HIGHER THAN, LOWER THAN or EQUAL TO. (1) The reaction is now repeated at a temperature of 400 C. The curves indicated by the dotted lines below were obtained at this temperature. Concentration (mol dm -3 ) 300 C 400 C Time (s) 6.5 Is the forward reaction EXOTHERMIC or ENDOTHERMIC? Fully explain how you arrived at the answer. (4) The Maxwell-Boltzmann distribution curve below represents the number of particles against kinetic energy at 300 C. Number of particles 300 C Kinetic energy 6.6 Redraw this curve in the ANSWER BOOK. On the same set of axes, sketch the curve that will be obtained at 400 C. Clearly label the curves as 300 C and 400 C respectively. () [15] Copyright reserved Please turn over

61 Physical Sciences/P 1 DBE/November 015 NSC QUESTION 7 (Start on a new page.) 7.1 Ammonium chloride crystals, NH 4 Cl(s), dissolve in water to form ammonium and chloride ions. The ammonium ions react with water according to the balanced equation below: + NH 4 (aq) + H O(l) NH 3 (aq) + H 3 O + (aq) Write down the name of the process described by the underlined sentence. (1) 7.1. Is ammonium chloride ACIDIC or BASIC in aqueous solution? Give a reason for the answer. () 7. A certain fertiliser consists of 9% ammonium chloride. A sample of mass x g of this fertiliser is dissolved in 100 cm 3 of a 0,10 mol dm -3 sodium hydroxide solution, NaOH(aq). The NaOH is in excess. The balanced equation for the reaction is: NH 4 Cl(s) + NaOH(aq) NH 3 (g) + H O(l) + NaCl(aq) 7..1 Calculate the number of moles of sodium hydroxide in which the sample is dissolved. (3) During a titration, 5 cm 3 of the excess sodium hydroxide solution is titrated with a 0,11 mol dm -3 hydrochloric acid solution, HCl(aq). At the endpoint it is found that 14,55 cm 3 of the hydrochloric acid was used to neutralise the sodium hydroxide solution according to the following balanced equation: HCl(aq) + NaOH(aq) NaCl(aq) + H O(l) 7.. Calculate the mass x (in grams) of the fertiliser sample used. (8) 7.3 Calculate the ph of a 0,5 moldm -3 sodium hydroxide solution at 5 C. (4) [18] Copyright reserved Please turn over

62 Physical Sciences/P 13 DBE/November 015 NSC QUESTION 8 (Start on a new page.) Learners are given the following two unknown half-cells: Half-cell 1: Q + (aq) Q(s) Half-cell : Pt R (g) R - (aq) During an investigation to identify the two half-cells, the learners connect each half-cell alternately to a Cd + (aq) Cd(s) half-cell under standard conditions. For each combination of two half-cells, they write down the net cell reaction and measure the cell potential. The results obtained for the two half-cell combinations are given in the table below. COMBINATION NET CELL REACTION CELL POTENTIAL I Q + (aq) + Cd(s) Cd + (aq) + Q(s) 0,13 V II R (g) + Cd(s) Cd + (aq) + R - (aq) 1,76 V 8.1 Write down THREE conditions needed for these cells to function as standard cells. (3) 8. For Combination I, identify: 8..1 The anode of the cell (1) 8.. Q by using a calculation (5) 8.3 For Combination II, write down the: Oxidation half-reaction () 8.3. NAME or FORMULA of the metal used in the cathode compartment (1) 8.4 Arrange the following species in order of INCREASING oxidising ability: Q + ; R ; Cd + Explain fully how you arrived at the answer. A calculation is NOT required. (4) [16] Copyright reserved Please turn over

63 Physical Sciences/P 14 DBE/November 015 NSC QUESTION 9 (Start on a new page.) The simplified diagram below represents an electrochemical cell used for the purification of copper. DC source Electrode A Electrolyte Electrode B (impure copper) 9.1 Define the term electrolysis. () 9. Give a reason why a direct-current (DC) source is used in this experiment. (1) 9.3 Write down the half-reaction which takes place at electrode A. () 9.4 Due to small amounts of zinc impurities in the impure copper, the electrolyte becomes contaminated with Zn + ions. Refer to the attached Table of Standard Reduction Potentials to explain why the Zn + ions will not influence the purity of the copper obtained during this process. (3) 9.5 After the purification of the impure copper was completed, it was found that,85 x 10 - moles of copper were formed. The initial mass of electrode B was,0 g. Calculate the percentage of copper that was initially present in electrode B. (4) [1] Copyright reserved Please turn over

64 Physical Sciences/P 15 DBE/November 015 NSC QUESTION 10 (Start on a new page.) Ammonia is an important fertiliser. Large amounts are prepared from hydrogen and nitrogen in industry For the industrial preparation of ammonia, write down: The name of the process used (1) A balanced equation for the reaction that occurs (3) The source of nitrogen (1) 10. The yield of ammonia changes with temperature and pressure during its industrial preparation. The graphs below show how the percentage of ammonia in the reaction mixture that leaves the reaction vessel varies under different conditions. GRAPHS OF THE PERCENTAGE OF AMMONIA IN THE REACTION MIXTURE VERSUS PRESSURE Percentage 300 C 400 C 500 C Pressure (atmosphere) Use the appropriate graph to estimate the percentage of ammonia present in the reaction mixture at 40 atmosphere and 400 C. (1) 10.. State TWO advantages of using high pressure in the preparation of ammonia. () The advantage of using a low temperature is the large percentage of ammonia formed. What is the disadvantage of using a low temperature? (1) 10.3 Ammonia is also used in the preparation of other fertilisers such as ammonium nitrate. Calculate the mass of nitrogen in a 50 kg bag of pure ammonium nitrate fertiliser. (3) [1] TOTAL: 150 Copyright reserved