NATIONAL SENIOR CERTIFICATE GRADE 11€¦ · ... PHYSICS (P1) IPHSCE1 Physical Sciences/P1 2...

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This question paper consists of 16 pages, 1 graph sheet and 2 data sheets.

GRADE 11

NATIONAL SENIOR CERTIFICATE

PHYSICAL SCIENCES: PHYSICS (P1)

NOVEMBER 2014

*IPHSCE1*

Physical Sciences/P1 2 DBE/November 2014 CAPS – Grade 11


INSTRUCTIONS AND INFORMATION 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12

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 6.2 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 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 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.



Q

P

x

y

O

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 (1.1–1.10) 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.

Which ONE of the following could represent the resultant R of the two

vectors?

A

B

(2)

C

D

R

y

x

R

y

x

R

y

x

R

y

x

O O

O O



x

y

P 30o

60o

45o

45o

Q

R S

1.2 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.

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 (2) 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.

(2)



Gra

vita

tiona

l for

ce

Mass of object 0

1.4 A graph of the gravitational force versus the mass of an object is shown

below.

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)

(2) 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 (2) 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.

(2)



1.7 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

2F 4F

(2) 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.

(2) 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.

(2)



V

A

R

V

A

R

V R

A

V

R

A

1.10 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

B

C

D

(2) [20]



QUESTION 2 (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.

Determine the: 2.1 Magnitudes of the tensions T1 and T2 (7) 2.2 Magnitude and direction of the reaction force at pulley P (4)

[11]

T1

T2

800 kg

140o 120o

P



S 35o

P

Q

T1

T2

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 0,25. The knot S is in equilibrium.

The tension in the string connecting block Q is T2 and that for the string that pulls at 35o is T1 as shown in the diagram.

3.1 Define the term static frictional force in words. (2) 3.2 Explain what is meant by the knot S is in equilibrium. (2) 3.3 Draw a labelled free-body diagram to show all the forces acting on: 3.3.1 The knot at S (3) 3.3.2 Block Q (4) 3.4 Calculate the maximum weight of block P for which block Q will just begin to

slip.

(7) [18]



5 kg

8 kg

30o

15 N

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 30o to the horizontal, causing the block to slide to the left.

The coefficient of kinetic friction between the 8 kg block and the surface of the table is 0,25. Ignore the effects of air friction.

4.1 Draw a free-body diagram showing ALL the forces acting on the 8 kg block. (5) 4.2 Write down Newton's second law of motion in words. (2) 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. (2) An object weighing 140 N on the surface of the earth is moved to a position which is 6,7 x 106 m above the surface of the earth.

5.2 Calculate the percentage by which its weight will change. (8)

[10]



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 io angle ro sin i sin r

15 10 0,259 0,174 25 16 0,423 0,276 45 28 0,707 0,469 55 33 0,819 0,545 60 35 0,866 0,574 70 39 0,940 0,629

6.1 For this investigation, write down the: 6.1.1 Dependent variable (1) 6.1.2 Independent variable (1) 6.1.3 Constant (control) variable (1) 6.2 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 6.2 to calculate the speed of light through the

glass material.

(3) [14]



O

N

P

Crown glass (n = 1,52)

Flint glass (n = 1,66) Q

N/

21o

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 21o.

7.1 Write down Snell's law in words. (2) The refractive indices of crown glass and flint glass are 1,52 and 1,66 respectively as shown in the diagram above.

7.2 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 40o.

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.

(2)

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]



screen slit

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. (2) 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.

8.2 Describe the pattern observed on the screen. (2) 8.3 Two important principles explain the diffraction pattern.

Write down the NAME of each of these principles.

(2) 8.4 The width of the slit (opening) is increased slightly. Describe how this change

will affect the:

8.4.1 Diffraction pattern observed (1) 8.4.2 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]



+2 μC +3 μC

0,2 m

-4 μC

0,1 m

+2 μC +3 μC

0,2 m P

• x

QUESTION 9 (Start on a new page.) Two point charges of +2 μC and +3 μC are placed a distance of 0,2 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.

9.1 Define the term electric field at a point in words. (2) 9.2 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.

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]



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.

10.1 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.

(2) 10.2 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.

(2) Faraday's law of electromagnetic induction plays a very important role in the generation of electricity.

10.4 Write down Faraday's law of electromagnetic induction in words. (2) A coil of 100 turns, each of area 4,8 x 10-4 m2, 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 0,2 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 2 Ω (3)

[15]


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A

4 Ω

6 Ω

X

V2 V1

6 V

S

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.

11.1 State Ohm's law in words. (2) Calculate the: 11.2 Current passing through the 4 Ω resistor (4) 11.3 Total current in the circuit (2) 11.4 Resistance X (3) The 4 Ω resistor gets hotter than the 6 Ω resistor after a while. 11.5 Explain this observation. (3)

[14] GRAND TOTAL: 150

Physical Sciences/P1 DBE/November 2014 CAPS – Grade 11

Copyright reserved

NAME OF THE LEARNER: GRAPH PAPER FOR QUESTION 6.2

0,2

0,4

0,6

1,0

0 0,2 0,4 0,6 0,8

0,8

1,0


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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

g 9,8 m·s-2

Gravitational constant Swaartekragkonstante

G 6,67 x 10-11 N⋅m2⋅kg-2

Radius of Earth Straal van Aarde

RE 6,38 x 106 m

Coulomb's constant Coulomb se konstante

K 9,0 x 109 N⋅m2·C-2

Speed of light in a vacuum Spoed van lig in 'n vakuum

c 3,0 x 108 m·s-1

Charge on electron Lading op elektron

e -1,6 x 10-19 C

Electron mass Elektronmassa

me 9,11 x 10-31 kg

Mass of the earth Massa van die Aarde

M 5,98 x 1024 kg

TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING

tavv if ∆+= 221

i tatvx ∆+∆=∆

xa2vv 2i

2f ∆+= t

2vvx if ∆

+=∆

FORCE/KRAG

ma=Fnet w = mg

221

rmGmF=

Nf (max)s

s =µ

Nfk

k =µ


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WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG

λ= fv f1T =

rrii sinnsinn θ=θ vcn =

ELECTROSTATICS/ELEKTROSTATIKA

221

rQkQ

=F (k = 9,0 x 109 N⋅m2·C-2) qF

=E

2rkQE= (k = 9,0 x 109 N⋅m2·C-2)

QW

=V

ELECTROMAGNETISM/ELEKTROMAGNETISME

tN

∆∆Φ

−=ε θ=Φ cosBA

CURRENT ELECTRICITY/STROOMELEKTRISITEIT

tQI∆

= IVR=

...r1

r1

r1

R1

321

+++= ...rrrR 321 +++=

W = Vq W = VI∆ t W= I2R∆ t

W= RΔtV2

ΔtWP=

P = VI P = I2R

RVP

2

=

Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

MARKS/PUNTE: 150

This memorandum consists of 15 pages. Hierdie memorandum bestaan uit 15 bladsye.

PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1)

EXEMPLAR/MODEL 2014

MEMORANDUM

NATIONAL SENIOR CERTIFICATE/

NASIONALE SENIOR SERTIFIKAAT

GRADE/GRAAD 11

Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/November 2014 CAPS/KABV – Grade 11/Graad 11 – Memorandum


QUESTION 1/VRAAG 1 1.1 D (2) 1.2 B (2) 1.3 D (2) 1.4 C (2) 1.5 A (2) 1.6 A (2) 1.7 B (2) 1.8 D (2) 1.9 C (2) 1.10 D (2)

[20] QUESTION 2/VRAAG 2 2.1 CHECK ANSWER BY CONSTRUCTION AND MEASUREMENT

VERGELYK ANTWOORD DEUR KONSTRUKSIE EN METING

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 T1 measured to accuracy of ±10 N Spanning T1 korrek gemeet tot 'n akkuraatheid van ±10 N

Tension force T2 measured to accuracy of ±10 N Spanning T2 korrek gemeet tot 'n akkuraatheid van ±10 N

Angle of 60o and 40o accurately obtained (may not be indicated, but must be measured to ascertain). Hoek van 60o en 40o akkuraat verkry (mag nie aangedui word nie, maar moet gemeet word om te bepaal).

60o

40o 800 N 700 N

520 N

T2

T1

60o

40o

800 N

700 N

520 N

T2

T1



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.

2.1 BY CALCULATION / DEUR BEREKENING OPTION 1/OPSIE 1

T1 sin 30 + T2 sin 50 9= 8009……….(1) T1 cos 30 = T2 cos 509……………..(2)

T2 = 50 cos30 cosT1

∴ 800 = 30sinT + 50 cos

50) 30)(sin cosT(1

1 1,532 T1 = 800 T1 = 522,19 N9

T2 = 50 cos

)30)(cos19,522(

= 703,54 N9

(7)

OPTION 2/OPSIE 2

80 sin800

= 60 sin

T2

T2 = ( )( )

80 sin60 sin8009

T2 = 703,51N9

80 sin800

= 40 sin

T1

T1 = ( )( )

80 sin40 sin8009

= 522,16 N9

50o 30o

T2 T1

800 N 9

9

9 9

40O

60O

9

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



P

703,5 N

703,5 N

25o 1275 N

2.2 BY CONSTRUCTION AND MEASUREMENT: DEUR KONSTRUKSIE EN

METING POSITIVE MARKING FROM QUESTION 2.1/ POSITIEWE NASIEN VANAF VRAAG 2.1

Parallelogram drawn with adjacent sides equal to T2 Parallelogram getrek met aangrensende kante gelyk aan T2.

Angle of 50o or 25o accurately obtained May not be indicated, but must be measured to ascertain. Hoek van 50o of 25o akkuraat verkry Mag nie aangedui word nie, maar moet gemeet word om te bepaal.

Reaction force recorded as 1 275 N ±10 N Reaksiekrag aangeteken as 1 275 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.

2.2 BY CALCULATION: (POSITIVE MARKING FROM QUESTION 2.1)

DEUR BEREKENING (POSITIEWE NASIEN VANAF VRAAG 2.1) OPTION 1/OPSIE 1

130 sinx

= 25 sin54,,703

x = ( )( )

25sin130sin54,703

reaction force / reaksiekrag = 1275,25 N9 at 25o below the horizontal/ onder die horisontaal 9(or/of 335o)

OPTION 2/OPSIE 2

703,54PX

= 65 sin 9

PX = 637,62 Reaction force/Reaksiekrag = 2PX9 = 1275,25 N9 at 25o below the horizontal/onder die horisontaal9( or/of 335o) OPTION 3/OPSIE 3 c2 = a2 + b2 – 2abcos C9

P 703,54

703,54

25o

25o

x

9 9

P 703,54

703,54

25o

25o

X

25o



T1

w/Fg

∴ PR2 = 703,542 + 703,542 – 2(703,54)(703,54) cos 1309 PR = 1275,25 N9 at 25o below the horizontal /onder die horisontaal9 (or/of 335o)

(4) [11]

QUESTION 3/VRAAG 3 3.1 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.

(2) 3.2 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

(2) 3.3.1 (3) 3.3.2 (4)

w/Fg

T2 f

N

T2

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

Deduct maximum 1 mark if arrows do not touch the dot. Trek ‘n maksimum van 1 punt af as pyltjie nie die kolletjie raak nie.

P 703,54

703,54

25o

25o

R 25o

703,54



w/Fg/mg

T

N/FN

15 N/FA f

3.4 For the string/ Vir die toutjie

-T2 + T1 cos 35o = 0 ∴ T1 cos 35o = T2………….(1) 9 T1 sin35o = wP ……………(2) 9 For the block/Vir die blok T2 - fs = 0 T2 = fs = μsN 9 T2 = 0,25 (70)(9,8) ………(3) 9

T1 = ( )( )( )

o35 cos8,97025,0

from (1)/ vanaf (1)

From (2) and (3)/Vanaf (2) en (3) 0,25(70)(9,8) sin 35o 9= wP 120,09 N9

(7) [18]

QUESTION 4/VRAAG 4 4.1 (5) 4.2 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.99 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. 99 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.

(2) 4.3 N = w – FA sinθ9

= 8(9,8) – 15 sin309 = 70,9 N9

(3)

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.



4.4 POSITIVE MARKING FROM QUESTION 4.3

POSITIEWE NASIEN VANAF VRAAG 4.3 For the 8 kg block/Vir die 8 kg-blok 15 cos 30o – T –fk = ma9 15 cos 30o –T – μkN = 8a 15(0,866) – T –(0,25)(70,9) = 8a9 -4,735 – T = 8a…………….(1) For the 5 kg block/Vir die 5 kg-blok T – w = ma9 T – 5 (9,8) = 5a9……………(2) From (1) and (2)/ Vanaf (1) en (2) -53,735 = 13 a a = -4,133 m∙s-2 from/vanaf (1) -4,735 – T = 8(-4,133) 9 T = 28,32(9) N9 OR/OF From /vanaf (2) T – 5(9,8) = 5(-4,133) 9 T = 28,33(5)N9

(6) [16]

QUESTION 5/VRAAG 5 5.1 Every body in the universe attracts every other body with a force that is

directly proportional to the product of their masses9 and inversely proportional to the square of the distance between their centres. 9 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.

(2) 5.2 w = mg9

m = 14,29 (14,286 kg) 9

F = 221

rmm

G OR/OF F = 2E

E

RmM

G 9

= ( )( )( ) 26

2411-

]10 ×6,7 + 38 ,6[286 ,1410 × ,985

× 10 67 ,6

= 33,31 N Change /Verandering = (140 – 33,31) = 106,69 N

% change/verandering = 100×140

699,106

= 76,21 %9

9

9

9



OPTION B/OPSIE B w = mg9 m = 14,29 (14,286 kg) 9

w = mg = 2E

E

RmM

G

g/ = 2E

E

RM

G 9

= ( )

( ) 26

2411-

]10×6,7 + 38 ,6[ 10 × ,985

10× 6,67

g/ = 2,331 m∙s-2 (New weight / Nuwe gewig) w/ = mg/ = 14,286 x 2,331 = 33,301 N Change/Verandering = (140 – 33,30) = 106,699 N

% change/verandering = 100×140

699,106

= 76,21 %9

(8) [10]

QUESTION 6/VRAAG 6 6.1.1 r/ sin r (1) 6.1.2 i/sin i (1) 6.1.3 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)

9 9

9



6.2

Slope of graph/Helling van grafiek = n = r sinΔi sinΔ

=0) - (0,60) - (0,9

= 1,5

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)

9

9

Sin

io

0,2

0,4

0,6

1,0

sin ro

0 0,2 0,4 0,6 0,8

Graph of sin i versus sin r/Grafiek van sin i teenoor sin r

0,8

40



O

N

P

Crown glass/Kroonglas

Flint glass/Flintglas

X

Q Y

40

N/

6.3

n = vc

v =1,5

10 × 3 8

= 2 x 108 m∙s-1

(3) [14]

QUESTION 7/VRAAG 7 7.1 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. 99 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. 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.

NOTE/LET WEL: Only 1 mark for Slegs 1 punt vir

r sini sin = a constant.

= 'n konstante

(2) 7.2

Sin c = n19

=1,66

19

c = 37,04o9

(3) 7.3 &7.4

(1) (2)

9



7.5 Greater than/Groter as9 (1)

[9] QUESTION 8/VRAAG 8 8.1 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.

(2) 8.2 A broad central bright band 9with alternating bright and dark band (of

decreasing intensity) on either side of it. 9 / 'n Breë, sentrale helder band met afwisselende helder en donker bande (van afnemende intensiteit) aan weerskante.

(2) 8.3 Huygen's principle/ Huygen se beginsel9

Principle of superposition / Beginsel van superposisie9

(2) 8.4.1 Patterns become narrower / Patrone word nouer.9 (1) 8.4.2 Brightness is unchanged/remains the same./ Helderheid is onveranderderd /

bly dieselfe9

(1) 8.5 The patterns become narrower / Patrone word nouer.9 (1)

[9] QUESTION 9/VRAAG 9 9.1 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.

(2)



9.2 Enet = 0

OR/OF E1+ E2 = 0

0=rKQ

+r

kQ2

22

21

1

0=x

)10×3)(10×(9 -

x)- (0,2)10×)(210×(9

2

6 -9

2

-69

22 x3

= x)- (0,2

2

Taking square root/Neem vierkantswortel

22 x732,1

= x)- (0,2

1,414

x = 0,11 m

(7) 9.3

221

rQkQF=

Force experienced by the +3 μC charge due to the + 2 μC charge = F3,2 Krag ondervind deur die +3 μC -ading as gevolg van die + 2 μC-lading = F3,2

F3,2 = 2

-6-69

)2,0()10× 3)(10 ×(2

10×9

= 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 = F3,-4 Krag ondervind deur die +3 μC-lading as gevolg van die – 24 μC-lading = F3,-4

F3,-4 = 2

-6-69

)1,0()10× 3)(10 ×(4

10×9

= 10,8 N downwards (southwards)/afwaarts (suid)

22

12

net F+F=F 22 )35,1(+)8,10(=

= 10,88 N

(5) [14]

F3,2

F3,-4 Fnet



QUESTION 10/VRAAG109 10.1 North Pole/Noordpool 99 (2) 10.2 North Pole/Noordpool 9 (1) 10.3 There will be no reading (deflection) 9

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. 9 '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.

(2) 10.4 The magnitude of the induced emf (in a conductor) is equal to the rate of

change of magnetic flux linkage. 99 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 time99 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. 99 Die geïnduseerde emk in 'n geleier is eweredig aan die tempo waarteen die magneetveldlyne deur 'n geleier gesny word.

(2) 10.5

tN

∆∆Φ

−=ε

OR

ε = ( )

tΔΦ -Φ

N- 3070 = tΔ

)30 BAcos - 70cosBA( N-

oo

= 2,0

)30 )cos10× )(4,810× (4 - 70cos)10 ×8,4)(10 ×4[( 100-

o-4-4o-4-4

OR/OF

2,0)30 cos - 70)](cos10 ×8,4)(10 ×4[(

100-oo-4-4

ε = 5,03 x 10-5 V9

(5) 10.6 ε = IR9

I = 2

10 × 03,5 -5

9

= 2,52 x 10-5 A 9

(3) [15]

9

9 9

9

9 9

9



QUESTION 11/VRAAG 11 11.1 The potential difference across a conductor is directly proportional to the

current in the conductor 9at constant tempearature.9 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 constant9, the potential difference across a conductor is directly proportional to the current9. Mits die temperatuur en ander fisiese toestande konstant is, is die potentsiaalverskil oor 'n geleier direk eweredig aan die stroom.

(2) 11.2 OPTION 1/OPSIE 1

V1 = IR6Ω = 0,6 (6) = 3,6 V

I4Ω =46,3

∴ I4Ω = 0,9 A OPTION 2/OPSIE 2 V = IR (0,6)(6) = I4Ω(4)

I4Ω = 4

)6)(6,0(

= 0,9 A

(4)

11.3 POSITIVE MARKING FROM QUESTION 11.2.1

POSITIEWE NASIEN VANAF VRAAG 11.2.1 Itot = I6Ω + I4Ω = (0,6 + 0,9) Itot = 1, 5 A

(2) 11.4 POSITIVE MARKING FROM QUESTION 11.2.1 AND QUESTION 11.2.2

POSITIEWE NASIEN VANAF VRAAG 11.2.1 EN VRAAG 11.2.2 VX = Vtot – V1 =(6 -3,6) =2,4 V V = IR

X = 5,14,2

= 1,6 Ω

(3)


Copyright reserved/Kopiereg voorbehou

11.5 Energy/Energie W = I2 R∆t

For the same time interval I2 R∆t will be greater for the 4Ω resistor than for the 6Ω resistor. Vir dieselfde tydinterval sal I2RΔt groter wees vir die 4 Ω-resistor as vir die 6 Ω-resistor.

OR/OF

Energy/Energie W = tΔRV 2

For the same potential difference and time tΔRV 2

is greater for the smaller

resistance than for the larger resistance.

Vir dieselfde potensiaalverksil en tyd is tΔRV 2

groter vir die kleiner weerstand

as vir die groter weerstand.

(3)

[14]

TOTAL/TOTAAL: 150

NATIONAL SENIOR CERTIFICATE GRADE 11€¦ · ... PHYSICS (P1) IPHSCE1 Physical Sciences/P1 2...

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NATIONAL SENIOR CERTIFICATE GRADE 11€¦ · ... PHYSICS (P1) IPHSCE1 Physical Sciences/P1 2...

Transcript of NATIONAL SENIOR CERTIFICATE GRADE 11€¦ · ... PHYSICS (P1) IPHSCE1 Physical Sciences/P1 2...