NATIONAL SENIOR CERTIFICATE GRADE 11

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1 NATIONAL SENIOR CERTIFICATE GRADE 11 PHYSICAL SCIENCES: PHYSICS (P1) NOVEMBER 014 MARKS: 150 TIME: 3 hours This question paper consists of 16 pages, 1 graph sheet and data sheets. *IPHSCE1* Copyright reserved Please turn over

2 Physical Sciences/P1 DBE/November 014 CAPS Grade 11 INSTRUCTIONS AND INFORMATION Write your name in the appropriate space on the ANSWER BOOK. This question paper consists of ELEVEN questions. Answer ALL the questions in the ANSWER BOOK except QUESTION 6. which has to be answered on the graph paper attached to this question paper. Write your name in the appropriate space on the graph paper. 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 014 CAPS Grade 11 QUESTION 1: MULTIPLE-CHOICE QUESTIONS Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A D) next to the question number ( ) in the ANSWER BOOK. 1.1 Two vectors, P and Q, act simultaneously at point O as shown in the diagram below. The magnitude of Q is greater than the magnitude of P. y Q O x P Which ONE of the following could represent the resultant R of the two vectors? R y y A O x B R O x y y R C O R x D O x () Copyright reserved Please turn over

4 Physical Sciences/P1 4 DBE/November 014 CAPS Grade Forces P, Q, R and S all have the same magnitude. The forces act at the same point in the directions shown in the diagram. y Q R 45 o 60 o 45 o 30 o S P x Which ONE of the following combinations CORRECTLY shows the vectors having the greatest magnitude for the x-component and for the y-component? x-component y-component A Vector P Vector R B Vector P Vector Q C Vector R Vector Q D Vector R Vector S () 1.3 If the resultant (net) force acting on an object is zero, the object A B C D slows down. accelerates uniformly. changes its direction of motion. continues moving with constant velocity. () Copyright reserved Please turn over

5 Physical Sciences/P1 5 DBE/November 014 CAPS Grade A graph of the gravitational force versus the mass of an object is shown below. Gravitational force 0 Mass of object Which ONE of the following CORRECTLY represents the slope of the graph? A B C D Velocity of the object Weight of the object Acceleration due to gravity (g) Universal gravitation constant (G) () 1.5 A light wave travels obliquely from air into a glass block and its speed changes. Which ONE of the combinations below CORRECTLY describes the changes in the FREQUENCY of the wave and REFRACTIVE INDEX of the block compared to that of air? FREQUENCY REFRACTIVE INDEX A Remains the same Increases B Remains the same Decreases C Increases Decreases D Decreases Increases () 1.6 Sound waves bend readily around buildings whereas light waves only bend very slightly around buildings. Which ONE of the following statements BEST explains this observation? A B C D Sound waves have much longer wavelengths than light waves. Sound waves have much shorter wavelengths than light waves. Sound waves have higher frequencies compared to light waves. Sound waves have greater amplitudes compared to light waves. () Copyright reserved Please turn over

6 Physical Sciences/P1 6 DBE/November 014 CAPS Grade The electrostatic force between two charged spheres, a distance r apart, is F. When the charge on each sphere is doubled and the distance between the spheres is also doubled, the force between the spheres will now be A B C D ½ F F F 4F () 1.8 The electrostatic force F between two charged particles is positive. Which ONE of the following is CORRECT? A B C D The magnitudes of the two charges are equal. One charge is positive while the other is negative. The electrostatic force between the charges is attractive. The electrostatic force between the charges is repulsive. () 1.9 A conducting wire, XY, moves between two magnets as shown below. Which ONE of the following actions can lead to an increased induced current in wire XY? Move the wire A B C D quickly and parallel to the magnetic field. slowly and parallel to the magnetic field. quickly and perpendicular to the magnetic field. slowly and perpendicular to the magnetic field. () Copyright reserved Please turn over

7 Physical Sciences/P1 7 DBE/November 014 CAPS Grade A learner wants to measure the current in and the potential difference across a resistor R in a circuit. In which ONE of the following circuits will the learner be able to take these readings? A R A V B A V R A C R V D R V A () [0] Copyright reserved Please turn over

8 Physical Sciences/P1 8 DBE/November 014 CAPS Grade 11 QUESTION (Start on a new page.) The diagram below shows a rope and pulley arrangement of a device being used to lift an 800 N object. Assume that the ropes are light and inextensible and also that the pulley is light and frictionless. P T T o 10 o 800 kg Determine the:.1 Magnitudes of the tensions T 1 and T (7). Magnitude and direction of the reaction force at pulley P (4) [11] Copyright reserved Please turn over

9 Physical Sciences/P1 9 DBE/November 014 CAPS Grade 11 QUESTION 3 (Start on a new page.) A block Q of mass 70 kg is at rest on a table. It is connected to block P by means of two light inextensible strings knotted at S. A third string is arranged in such a way that the string connecting block Q is horizontal as shown in the diagram below. The coefficient of static friction between block Q and the surface of the table is 0,5. The knot S is in equilibrium. T 1 Q T S 35 o P The tension in the string connecting block Q is T and that for the string that pulls at 35 o is T 1 as shown in the diagram. 3.1 Define the term static frictional force in words. () 3. Explain what is meant by the knot S is in equilibrium. () 3.3 Draw a labelled free-body diagram to show all the forces acting on: The knot at S 3.3. Block Q (3) (4) 3.4 Calculate the maximum weight of block P for which block Q will just begin to slip. (7) [18] Copyright reserved Please turn over

10 Physical Sciences/P1 10 DBE/November 014 CAPS Grade 11 QUESTION 4 (Start on a new page.) A block of mass 8 kg resting on a rough horizontal table is connected by a light inextensible string which passes over a light frictionless pulley to another block of mass 5 kg. The 5 kg block hangs vertically as shown in the diagram below. A 15 N force is applied to the 8 kg block at an angle of 30 o to the horizontal, causing the block to slide to the left. 15 N 30 o 8 kg 5 kg The coefficient of kinetic friction between the 8 kg block and the surface of the table is 0,5. Ignore the effects of air friction. 4.1 Draw a free-body diagram showing ALL the forces acting on the 8 kg block. (5) 4. Write down Newton's second law of motion in words. () Calculate the magnitude of the: 4.3 Normal force acting on the 8 kg block (3) 4.4 Tension in the string connecting the two blocks (6) [16] QUESTION 5 (Start on a new page.) 5.1 Write down Newton's law of universal gravitation in words. () An object weighing 140 N on the surface of the earth is moved to a position which is 6,7 x 10 6 m above the surface of the earth. 5. Calculate the percentage by which its weight will change. (8) [10] Copyright reserved Please turn over

11 Physical Sciences/P1 11 DBE/November 014 CAPS Grade 11 QUESTION 6 (Start on a new page.) Learners investigate how the path of a light ray incident on an air-glass boundary changes as it enters the glass medium. Their results are shown in the table below. angle i o angle r o sin i sin r ,59 0, ,43 0, ,707 0, ,819 0, ,866 0, ,940 0, For this investigation, write down the: Dependent variable (1) 6.1. Independent variable (1) Constant (control) variable (1) 6. Draw an appropriate graph of the data in the table and use it to obtain the refractive index of the glass material. USE THE GRAPH PAPER ATTACHED TO YOUR QUESTION PAPER TO ANSWER THIS QUESTION. (8) 6.3 Use the result in QUESTION 6. to calculate the speed of light through the glass material. (3) [14] Copyright reserved Please turn over

12 Physical Sciences/P1 1 DBE/November 014 CAPS Grade 11 QUESTION 7 (Start on a new page.) In the diagram below (not to scale), a ray of light, PO, is travelling from flint glass towards the boundary with crown glass. The angle of incidence of ray PO at the boundary between the two surfaces ( PON / ) is 1 o. Crown glass (n = 1,5) N O Q P 1 o N / Flint glass (n = 1,66) 7.1 Write down Snell's law in words. () The refractive indices of crown glass and flint glass are 1,5 and 1,66 respectively as shown in the diagram above. 7. Calculate the critical angle for the boundary between the two glass materials. (3) Copy the diagram into your ANSWER BOOK. 7.3 On your diagram draw a ray to show what happens to light ray PO at the boundary between the two glass surfaces. Label the ray OX. (1) 7.4 Ray QO is incident at the boundary at 40 o. Draw a ray to show what happens to light ray QO at the boundary between the two glass surfaces. Label the ray OY. Include the angle N / OY on your drawing. () 7.5 How does the speed of light in the crown glass compare to that in the flint glass? Write down only GREATER THAN, LESS THAN or EQUAL TO. (1) [9] Copyright reserved Please turn over

13 Physical Sciences/P1 13 DBE/November 014 CAPS Grade 11 QUESTION 8 (Start on a new page.) Diffraction provides evidence that light can behave as a wave. 8.1 Define the term diffraction in words. () In the diagram below a plain wave front of light of wavelength 6 x 10-7 m approaches a narrow opening. Diffraction effects are observed on a screen placed some distance from the slit as shown in the diagram below. slit screen 8. Describe the pattern observed on the screen. () 8.3 Two important principles explain the diffraction pattern. Write down the NAME of each of these principles. () 8.4 The width of the slit (opening) is increased slightly. Describe how this change will affect the: Diffraction pattern observed (1) 8.4. Brightness of the diffraction pattern observed (1) 8.5 The width of the slit is kept constant but light of wavelength 4 x 10-7 m is now used. Describe how this change will affect the diffraction pattern obtained. (1) [9] Copyright reserved Please turn over

14 Physical Sciences/P1 14 DBE/November 014 CAPS Grade 11 QUESTION 9 (Start on a new page.) Two point charges of + μc and +3 μc are placed a distance of 0, m apart. P is a point on the line joining the two charges, a distance of x m from the 3 μc charge such that the net electric field at point P is zero. x + μc +3 μc P 0, m 9.1 Define the term electric field at a point in words. () 9. Calculate the distance x. (7) A -4 μc charge is now placed a distance of 0,1 m from the +3 μc charge as shown in the sketch below. 0, m + μc +3 μc 0,1 m -4 μc 9.3 Calculate the magnitude of the electrostatic force experienced by the +3 μc charge due to the presence of the other two charges. (5) [14] Copyright reserved Please turn over

15 Physical Sciences/P1 15 DBE/November 014 CAPS Grade 11 QUESTION 10 (Start on a new page.) In the diagram below a bar magnet is being pushed into a coil. The current induced in the coil is in the direction indicated Write down the polarity (north pole or south pole) of the end of the coil facing the bar magnet, as the bar magnet approaches the coil. () 10. Which end of the bar magnet is approaching the coil? Write down only NORTH POLE or SOUTH POLE (1) 10.3 Write down what will be observed on the galvanometer if the bar magnet is held stationary inside the coil. Give a reason for the answer. () Faraday's law of electromagnetic induction plays a very important role in the generation of electricity Write down Faraday's law of electromagnetic induction in words. () A coil of 100 turns, each of area 4,8 x 10-4 m, is made from insulated copper wire. The coil is placed in a uniform magnetic field of 4 x 10-4 T in such a way that the angle between the magnetic field and the normal to the plane of the coil is 30. The coil is then rotated so that the angle changes to 70 in a time interval of 0, s. Calculate the: 10.5 Magnitude of the emf induced in the coil (5) 10.6 Current induced in the coil if it has an effective resistance of Ω (3) [15] Copyright reserved Please turn over

16 Physical Sciences/P1 16 DBE/November 014 CAPS Grade 11 QUESTION 11 (Start on a new page.) In the circuit below the internal resistance of the 6 V battery is negligible. The resistance of the connecting wires is negligible. When switch S is closed, the current in the 6 Ω resistor is 0,6 A. 6 V A S 4 Ω X 6 Ω V 1 V 11.1 State Ohm's law in words. () Calculate the: 11. Current passing through the 4 Ω resistor (4) 11.3 Total current in the circuit () 11.4 Resistance X (3) The 4 Ω resistor gets hotter than the 6 Ω resistor after a while Explain this observation. (3) [14] GRAND TOTAL: 150 Copyright reserved

17 Physical Sciences/P1 DBE/November 014 CAPS Grade 11 NAME OF THE LEARNER: GRAPH PAPER FOR QUESTION 6. 1,0 0,8 0,6 0,4 0, 0 0, 0,4 0,6 0,8 1,0 Copyright reserved

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19 Physical Sciences/P1 DBE/November 014 CAPS Grade 11 DATA FOR PHYSICAL SCIENCES GRADE 11 PAPER 1 (PHYSICS) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 11 VRAESTEL 1 (FISIKA) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE Acceleration due to gravity Swaartekragversnelling Gravitational constant Swaartekragkonstante Radius of Earth Straal van Aarde Coulomb's constant Coulomb se konstante Speed of light in a vacuum Spoed van lig in 'n vakuum Charge on electron Lading op elektron Electron mass Elektronmassa Mass of the earth Massa van die Aarde g 9,8 m s - G 6,67 x N m kg - R E 6,38 x 10 6 m K 9,0 x N m C c 3,0 x 10 8 m s -1 e m e M -1,6 x C 9,11 x kg 5,98 x 10 4 kg TABLE : FORMULAE/TABEL : FORMULES MOTION/BEWEGING v v f f = v i + a t x = v t + i 1 a t v f + v i = v i + a x x = t FORCE/KRAG F net = ma w = mg Gm1m F= r µ = f k k N µ = s f s(max) N Copyright reserved

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21 Physical Sciences/P1 DBE/November 014 CAPS Grade 11 WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG v = f λ n i sin θ i = n r sin θ r 1 T = f c n = v ELECTROSTATICS/ELEKTROSTATIKA kq Q 1 F = (k = 9,0 x 10 9 N m C - ) r kq E = (k = 9,0 x 10 9 N m C - ) r F E = q W V = Q ELECTROMAGNETISM/ELEKTROMAGNETISME Φ ε = N Φ = BA cos θ t CURRENT ELECTRICITY/STROOMELEKTRISITEIT I = Q t W = VI t W= I R t W= V Δt R V R = I = R= r1 + r + r R r1 r r3 W = Vq W P = Δt P = VI P = I R V P = R Copyright reserved

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25 NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) EXEMPLAR/MODEL 014 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 15 pages. Hierdie memorandum bestaan uit 15 bladsye. Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

26 Physical Sciences P1/Fisiese Wetenskappe V1 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum QUESTION 1/VRAAG D () 1. B () 1.3 D () 1.4 C () 1.5 A () 1.6 A () 1.7 B () 1.8 D () 1.9 C () 1.10 D () [0] QUESTION /VRAAG.1 CHECK ANSWER BY CONSTRUCTION AND MEASUREMENT VERGELYK ANTWOORD DEUR KONSTRUKSIE EN METING 40 o T 800 N 700 N 50 N T T 1 60 o 800 N 700 N 40 o 60 o 50 N T 1 NOTES/AANTEKENINGE Weight (w) measured to accuracy of ±5 N / Gewig (w) gemeet tot akkuraatheid van ±5 N Anything beyond ± 10 N should attract 1 mark only Enigiets bo ±10 N kan slegs 1 punt kry Tension force T 1 measured to accuracy of ±10 N Spanning T 1 korrek gemeet tot 'n akkuraatheid van ±10 N Tension force T measured to accuracy of ±10 N Spanning T korrek gemeet tot 'n akkuraatheid van ±10 N Angle of 60 o and 40 o accurately obtained (may not be indicated, but must be measured to ascertain). Hoek van 60 o en 40 o akkuraat verkry (mag nie aangedui word nie, maar moet gemeet word om te bepaal). Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

27 Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum Penalise 1 mark if distance (cm) instead of forces indicated on diagram (if final answer is correctly written). Penaliseer 1 punt indien afstand (cm) in plaas van kragte op diagram aangedui word (indien finale antwoord korrek geskryf is). Penalise 1 mark if no scale is indicated. Penaliseer 1 punt indien geen skaal aangedui is nie If no answer is given but sketch correctly shown in cm, award only 3 marks. Indien geen antwoord gegee is nie, maar die skets is korrek in cm getoon, ken slegs 3 punte toe..1 BY CALCULATION / DEUR BEREKENING OPTION 1/OPSIE 1 T 1 sin 30 + T sin 50 = 800.(1) T 1 cos 30 = T cos 50..() T T = 1 cos 30 cos 50 (T1 cos 30)(sin 50) + T1 sin30 = 800 cos 50 T T 1 50 o 30 o 1,53 T 1 = 800 T 1 = 5,19 N T = (5,19)(cos30) cos 50 = 703,54 N OPTION /OPSIE T 800 = sin 60 sin 80 T = ( 800 )( sin 60 ) sin 80 T = 703,51N T = sin 40 sin 80 T 1 = ( 800 )( sin 40 ) sin 80 = 5,16 N 800 N NOTE/LET WEL: Do not penalise if sketch is not shown. Moenie penaliseer indien skets nie getoon is nie NOTE/LET WEL: Do not penalise if sketch is not shown Moenie penaliseer indien skets nie getoon is 40 O 60 O (7) Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

28 Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum. BY CONSTRUCTION AND MEASUREMENT: DEUR KONSTRUKSIE EN METING POSITIVE MARKING FROM QUESTION.1/ POSITIEWE NASIEN VANAF VRAAG.1 703,5 N P 703,5 N 5 o 175 N Parallelogram drawn with adjacent sides equal to T Parallelogram getrek met aangrensende kante gelyk aan T. Angle of 50 o or 5 o accurately obtained May not be indicated, but must be measured to ascertain. Hoek van 50 o of 5 o akkuraat verkry Mag nie aangedui word nie, maar moet gemeet word om te bepaal. Reaction force recorded as 1 75 N ±10 N Reaksiekrag aangeteken as 1 75 N ±10 N Penalise 1 mark if distance (cm) instead of forces indicated on diagram (if final answer is correct). Penaliseer 1 punt indien afstand (cm) in plaas van kragte aangedui word op diagram (indien finale antwoord korrek is). If no answer is given but sketch shown in cm, award only 1 mark. Indien geen antwoord gegee word nie, maar skets getoon in cm, ken slegs 1 punt toe.. BY CALCULATION: (POSITIVE MARKING FROM QUESTION.1) DEUR BEREKENING (POSITIEWE NASIEN VANAF VRAAG.1) OPTION 1/OPSIE 1 P 703,54 703,,54 x = sin 5 sin ,54 5 o ( 703,54 x = )( sin130 ) sin5 reaction force / reaksiekrag = 175,5 N at 5 o below the horizontal/ onder die horisontaal (or/of 335 o ) OPTION /OPSIE P 703,54 5 o PX sin 65 = 703,54 703,54 5 o X PX = 637,6 Reaction force/reaksiekrag = PX = 175,5 N at 5 o horizontal/onder die horisontaal( or/of 335 o ) OPTION 3/OPSIE 3 c = a + b abcos C 5 o x 5 o below the Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

29 Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum PR = 703, ,54 (703,54)(703,54) cos 130 PR = 175,5 N at 5 o below the horizontal /onder die horisontaal (or/of 335 o ) P 703,54 5 o 703,54 5 o 5 o 703,54 R (4) [11] QUESTION 3/VRAAG The force that opposes the tendency of motion of a stationary object relative to a surface. / Die krag wat die neiging tot beweging van 'n stilstaande liggaam relatief tot 'n oppervlak teenwerk. OR/OF The force of friction developed between two surfaces that are at rest. / Die krag of wrywing wat ontwikkel word tussen twee oppervlakke wat in rus is. () 3. The resultant of all forces acting at point S is zero / Die resultaat van al die kragte wat op punt S inwerk, is nul. Net force at S equals zero / Netto krag by S is gelyk aan nul. There is no acceleration / Daar is geen versnelling nie () (3) N T w/f g T 1 Do not penalise if angle is notshown./moenie penaliseer indien die hoek nie aangedui is nie. Deduct maximum 1 mark if arrows do not touch the dot. Trek n maksimum van 1 punt af as pyltjies nie die kolletjie raak nie 3.3. f (4) Deduct maximum 1 mark if arrows do T not touch the dot. Trek n maksimum van 1 punt af as pyltjie nie die kolletjie raak nie. w/f g Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

30 Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum 3.4 For the string/ Vir die toutjie -T + T 1 cos 35 o = 0 T 1 cos 35 o = T.(1) T 1 sin35 o = w P () For the block/vir die blok T - f s = 0 T = f s = μ s N T = 0,5 (70)(9,8) (3) T 1 = ( )( )( ) 0,5 70 9,8 o cos 35 from (1)/ vanaf (1) From () and (3)/Vanaf () en (3) 0,5(70)(9,8) sin 35 o = w P 10,09 N (7) [18] QUESTION 4/VRAAG 4 N/F 4.1 N (5) 15 N/F A f w/f g /mg T NOTE/LET WEL: 1 mark for each force correctly shown emanating from the dot. 1 punt vir elke krag korrek aangetoon wat uit die kolletjie voortspruit. 4. When a net force is applied to an object of mass m, it accelerates in the direction of the force at an acceleration directly proportional to the force and inversely proportional to the mass of the object. Wanneer 'n netto krag op 'n liggaaam met massa m toegepas word, versnel dit in die rigting van die krag teen 'n versnelling wat direk eweredig is aan die krag en omgekeerd eweredig is aan die massa van die liggaam. OR/OF When a net force acts on an object of mass m, the acceleration that results is directly proportional to the net force, has a magnitude that is inversely proportional to the mass and a direction that is the same as that of the net force. Wanneer 'n netto krag op 'n liggaam met massa m inwerk, is die gevolglike versnelling direk eweredig aan die netto krag, het 'n grootte wat omgekeerd eweredig is aan die massa en 'n rigting wat diedelfde is as dié van die netto krag. () 4.3 N = w F A sinθ = 8(9,8) 15 sin30 = 70,9 N (3) Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

31 Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum 4.4 POSITIVE MARKING FROM QUESTION 4.3 POSITIEWE NASIEN VANAF VRAAG 4.3 For the 8 kg block/vir die 8 kg-blok 15 cos 30 o T f k = ma 15 cos 30 o T μ k N = 8a 15(0,866) T (0,5)(70,9) = 8a -4,735 T = 8a.(1) For the 5 kg block/vir die 5 kg-blok T w = ma T 5 (9,8) = 5a () From (1) and ()/ Vanaf (1) en () -53,735 = 13 a a = -4,133 m s - from/vanaf (1) -4,735 T = 8(-4,133) T = 8,3(9) N OR/OF From /vanaf () T 5(9,8) = 5(-4,133) T = 8,33(5)N (6) [16] QUESTION 5/VRAAG Every body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. Elke liggaam in die heelal trek 'n elke ander liggaam aan met 'n krag wat direk eweredig is aan die produk van hul massas en omgekeerd eweredig is aan die kwadraat van die afstand tussen hul middelpunte. () 5. w = mg m = 14,9 (14,86 kg) m1m MEm F = G OR/OF F = G r R = 6, ( 5,98 10 )( 14, 86) 6 ( ) [ 6, ,7 E 10 = 33,31 N Change /Verandering = (140 33,31) = 106,69 N 106,699 % change/verandering = = 76,1 % ] Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

32 Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum OPTION B/OPSIE B w = mg m = 14,9 (14,86 kg) MEm w = mg = G RE g / ME = G R E 4 ( 5,98 10 ) 6 ( ) -11 = 6,67 10 [ 6, ,7 g / =,331 m s - 10 (New weight / Nuwe gewig) w / = mg / = 14,86 x,331 ] = 33,301 N Change/Verandering = (140 33,30) = 106,699 N 106,699 % change/verandering = = 76,1 % (8) [10] QUESTION 6/VRAAG r/ sin r (1) 6.1. i/sin i (1) The type of block used/temperature of the surroundings/source of light/surface on which block is placed Die tipe blok gebruik/temperatuur van die omgewing/bron van die lig/ oppervlak waarop blok geplaas word. (1) Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

33 Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum 6. Graph of sin i versus sin r/grafiek van sin i teenoor sin r 1,0 0,8 Sin i o 0,6 0,4 0, 0 0, 0,4 0,6 0,8 sin r o Δ sin i Slope of graph/helling van grafiek = n = Δsin r (0,9-0) = (0,6-0) = 1,5 40 RUBRIC FOR MARKING GRAPH / RUBRIEK VIR NASIEN VAN GRAFIEK Axes correctly chosen and labelled / Asse korrek gekies en benoem Graph has a descriptive title / Grafiek het 'n beskrywende titel Correctly plotted points (minimum of 4 points) / Punte korrek geteken (minimum van 4 punte) Best line of fit / Beste lyn van passing Deduct a maximum of 1 mark if more than 3 points are incorrectly plotted Trek 'n maksimum van 1 punt af indien meer as 3 punte verkeerd getrek is (8) Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

34 Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum 6.3 n = v c v = 1,5 = x 10 8 m s -1 (3) [14] QUESTION 7/VRAAG The index of refraction of the incident medium multiplied by the sine of the incident angle is equal to the index of refraction of the refracting medium multiplied by the sine of the refracted angle. Die brekingsindeks van die invallende medium vermenigvuldig met die sinus van die invalshoek is gelyk aan die brekingsindeks van die refraktiewe medium vermenigvuldig met die sinus van die gebreekte hoek. NOTE/LET WEL: Only 1 mark for Slegs 1 punt vir sin i = a constant. sin r = 'n konstante OR/OF When light passes from one medium into another, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. As lig van een medium na 'n ander beweeg, is die verhouding van die sinus van die invalshoek tot die sinus van die brekingshoek 'n konstante. () 7. Sin c = n 1 1 = 1,66 c = 37,04 o (3) 7.3 &7.4 Crown glass/kroonglas (1) () N X O 40 Q P N / Y Flint glass/flintglas Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

35 Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum 7.5 Greater than/groter as (1) [9] QUESTION 8/VRAAG The bending of a wave as it passes around the edges of an object. Die buiging van 'n golf soos dit om die kante van 'n voorwerp beweeg. OR/OF The bending of a wave around an obstacle or the corners of an narrow opening. Die buiging van 'n golf om 'n versperring of deur die hoeke van 'n nou spleet/opening. OR/OF The ability of a wave to spread out in wave fronts as they pass through a small aperture or around a sharp edge. Die vermoë van 'n golf om in golffronte uit te sprei soos hulle deur 'n klein opening of om 'n skerp kant beweeg. () 8. A broad central bright band with alternating bright and dark band (of decreasing intensity) on either side of it. / 'n Breë, sentrale helder band met afwisselende helder en donker bande (van afnemende intensiteit) aan weerskante. () 8.3 Huygen's principle/ Huygen se beginsel Principle of superposition / Beginsel van superposisie () Patterns become narrower / Patrone word nouer. (1) 8.4. Brightness is unchanged/remains the same./ Helderheid is onveranderderd / bly dieselfe (1) 8.5 The patterns become narrower / Patrone word nouer. (1) [9] QUESTION 9/VRAAG Electric field at a point is defined as the force acting per unit charge. Elektriese veld by 'n punt word gedefinieer as die krag wat inwerk per eenheidslading. OR/OF It is the force experienced by a unit positive charge placed at that point. Dit is die krag wat deur 'n eenheid positiewe lading geplaas by daardie punt ondervind word. () Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

36 Physical Sciences P1/Fisiese Wetenskappe V1 1 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum 9. E net = 0 OR/OF E 1 + E = 0 kq1 KQ + = 0 r1 r 9-6 (9 10 )( 10 ) ( (0, - x) 3 = (0, - x) x 9 )(3 10 x - 6 ) = 0 Taking square root/neem vierkantswortel 1,414 1,73 = (0, - x) x x = 0,11 m (7) 9.3 kq1q F= r Force experienced by the +3 μc charge due to the + μc charge = F 3, Krag ondervind deur die +3 μc -ading as gevolg van die + μc-lading = F 3, ( 10 )(3 10 ) 9 10 F 3, = (0,) = 1,35 N to the right (east) / na regs (oos) Force experienced by the +3 μc charge due to the presence of the - 4 μc charge = F 3,-4 Krag ondervind deur die +3 μc-lading as gevolg van die 4 μc-lading = F 3, (4 10 )(3 10 ) 9 10 F 3,-4 = (0,1) = 10,8 N downwards (southwards)/afwaarts (suid) F = net = F (10,8) 1 + F + (1,35) = 10,88 N (5) [14] F 3,-4 F 3, F net Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

37 Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum QUESTION 10/VRAAG North Pole/Noordpool () 10. North Pole/Noordpool (1) 10.3 There will be no reading (deflection) Daar sal geen lesing (afwyking) waargeneem word nie An emf is induced only when the magnetic (flux) links with the coil. This is achieved when either the magnet (producing the field) or coil is moving. 'n Emk word slegs geïnduseer as die magnetiese vloedlyne met die spoel koppel. Dit word bereik wanneer óf die magneet (wat die veld verskaf) óf die spoel beweeg. ACCEPT/AANVAAR Either the coil or magnet must be moving to induce an emf. Óf die spoel óf die magneet moet beweeg om 'n emk te induseer. () 10.4 The magnitude of the induced emf (in a conductor) is equal to the rate of change of magnetic flux linkage. Die grootte van die geïnduseerde emk (in 'n geleier) is gelyk aan die tempo van verandering van magnetiese vloedkoppeling. OR/OF The emf induced in a conducting loop is equal to the negative of the rate at which the magnetic flux through the loop is changing with time Die geïnduseerde emk in 'n geleidende lus is gelyk aan die negatiewe van die tempo waarteen die magnetiese vloedlyne deur die lus verander oor tyd. ACCEPT/AANVAAR: The emf induced in a conductor is proportional to the rate at which magnetic field lines are cut by a conductor. Die geïnduseerde emk in 'n geleier is eweredig aan die tempo waarteen die magneetveldlyne deur 'n geleier gesny word. () 10.5 Φ ε = N t OR Φ ε = - N ( - Φ ) o o (BA cos70 - BAcos 30 ) = - N Δt Δt -4-4 o -4-4 o [(4 10 )(4,8 10 )cos70 - (4 10 )(4,8 10 )cos 30 ) = , OR/OF -4-4 o o [(4 10 )(4,8 10 )](cos70 - cos 30 ) , ε = 5,03 x 10-5 V (5) 10.6 ε = IR -5 5,03 10 I = =,5 x 10-5 A (3) [15] Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

38 Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum QUESTION 11/VRAAG The potential difference across a conductor is directly proportional to the current in the conductor at constant tempearature. Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur. OR/OF Provided temperature and other physical conditions are constant, the potential difference across a conductor is directly proportional to the current. Mits die temperatuur en ander fisiese toestande konstant is, is die potentsiaalverskil oor 'n geleier direk eweredig aan die stroom. () 11. OPTION 1/OPSIE 1 V 1 = IR 6Ω = 0,6 (6) = 3,6 V 3, 6 I 4Ω = 4 I 4Ω = 0,9 A OPTION /OPSIE V = IR (0,6)(6) = I 4Ω (4) (0,6)(6) I 4Ω = 4 = 0,9 A (4) 11.3 POSITIVE MARKING FROM QUESTION POSITIEWE NASIEN VANAF VRAAG I tot = I 6Ω + I 4Ω = (0,6 + 0,9) I tot = 1, 5 A () 11.4 POSITIVE MARKING FROM QUESTION AND QUESTION 11.. POSITIEWE NASIEN VANAF VRAAG EN VRAAG 11.. V X = V tot V 1 =(6-3,6) =,4 V V = IR,4 X = 1,5 = 1,6 Ω (3) Copyright reserved/kopiereg voorbehou Please turn over/blaai om asseblief

39 Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/November 014 CAPS/KABV Grade 11/Graad 11 Memorandum 11.5 Energy/Energie W = I R t For the same time interval I R t will be greater for the 4Ω resistor than for the 6Ω resistor. Vir dieselfde tydinterval sal I RΔt groter wees vir die 4 Ω-resistor as vir die 6 Ω-resistor. OR/OF Energy/Energie W = V R Δt For the same potential difference and time resistance than for the larger resistance. Vir dieselfde potensiaalverksil en tyd is as vir die groter weerstand. V R Δt V R Δt is greater for the smaller groter vir die kleiner weerstand (3) [14] TOTAL/TOTAAL: 150 Copyright reserved/kopiereg voorbehou