GAUTENG DEPARTMENT OF EDUCATION

GAUTENGSE DEPARTEMENT VAN ONDERWYS

PROVINCIAL EXAMINATION
PROVINSIALE EKSAMEN
JUNE/JUNIE 2017
GRADE 11

PHYSICAL SCIENCES /
FISIESE WETENSKAPPE
Paper / Vraestel 1

Physics / Fisika

MEMORANDUM

14 pages / bladsye
 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

GAUTENG DEPARTMENT OF EDUCATION

PROVINCIAL EXAMINATION

PHYSICAL SCIENCES:Physics /
FISIESE WETENSKAPPE: Fisika

MEMORANDUM

QUESTION 1 / VRAAG 1

1.1 B ✓✓ (2)

1.2 A ✓✓ (2)

1.3 D ✓✓ (2)

1.4 C ✓✓ (2)

1.5 A ✓✓ (2)

1.6 B ✓✓ (2)

1.7 D ✓✓ (2)

1.8 C ✓✓ (2)

1.9 A ✓✓ (2)

1.10 C ✓✓ (2)
[20]

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 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

QUESTION 2 / VRAAG 2

2.1 The resultant or net force is the vector sum of all the forces acting on that
object. ✓✓
Die resulterende of netto krag is die vektor som van al die kragte wat op daardie
voorwerp inwerk.✓✓ (2)

2.2.1 Y – direction: = 9 - Fy ✓
Y – rigting: = 9 – ( 6 x sin 55°)✓
= 9 - 4,91
= 4,085... N upwards / opwaarts / 0 °✓
X – direction: = -2 + 8 + (6 x cos 55°) ✓
X – rigting: = 9,4416... N to the right / regs / 90° ✓

Fnet 2 = Fy2 +Fx 2✓

= (4,085)...2 + (9,4416)...2
Fnet = 10,29 N ✓ (7)

2.2.2 OPTION 1/ OPSIE 1 OPTION 2/ OPSIE 2 OPTION 3/ OPSIE 3

𝑜 𝑜 𝑎
sin  = ℎ
✓ OR / OF tan  =
𝑎
OR / OF cos  = ℎ
4,09 4,09 9,44
=
10,29
✓ = 9,44
= 10,29

 = 23,42° ✓ = 23,43° = 23,45°

(3)
[12]

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 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

QUESTION 3 / VRAAG 3

3.1 An object is in equilibruim if it remains in a state of rest or motion at constant

velocity with a zero resultant / net force acting on it.✓ ✓
ʼn Voorwerp is in ewewig as dit in rus is of met ʼn konstante snelheid beweeg met
geen resulterende / netto krag wat daarop inwerk nie.✓✓ (2)

3.2
✓ Fg / w (lower case) and direction of arrow ANY
Fg / w (klein letter), rigting en pylpunt CORRECT
✓ TA and direction correct TRIANGLE
TA en rigting korrek
✓ TB and direction correct
TB en rigting korrek
ENIGE
✓ an angle shown / ‘n hoek KORREKTE
aangetoon DRIEHOEK
(4)

3.3 3.3.1 OPTION 1/ OPSIE 1

𝑎
cos θ = ℎ
cos 26° = Fg/ TA ✓
TA = (1620 x 9,8) / cos 26° ✓
= 17663,67 N✓

OPTION 2/ OPSIE 2
𝑜
sin θ = ℎ
sin 64° = Fg/ TA ✓
TA = (1620 x 9,8) ÷ sin 64° ✓
= 17663,67 N✓ (3)

3.3.2 OPTION 1/ OPSIE 1 OPTION 3 / OPSIE 3

𝑜
tan θ = 𝑎 𝑎
cos Ɵ = ℎ
tan 26° = TB / Fg ✓ cos 64° = TB / TA ✓
TB = tan 26° x (1620 x 9,8) ✓ TB = (17663.67) x cos 64° ✓
= 7743,24 N✓ = 7743,24 N✓

OPTION 2/ OPSIE 2 OPTION 4 / OPSIE 4

𝑜 𝑜
sin θ = tan Ɵ = 𝑎
ℎ
sin 26° = TB / TA ✓ tan 64° = Fg / TB ✓
TB = (17663,67 )sin 26° ✓ TB = (1620 x 9,8) / tan 64° ✓ (3)
= 7743,24 N✓ = 7743,24 N✓ (3)
[12]

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 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

QUESTION 4 / VRAAG 4

4.1 4.1.1 Acceleration / Versnelling✓

4.1.2 Mass of trolley / massa van trollie ✓

4.1.3 Force applied / toegepaste krag✓ (3)

4.2 Criteria for investigative question /Kriteria vir ondersoekende Mark /

vraag: Punt
The dependent and independent variables are stated.
✓
Die afhanklike en onafhanklike veranderlikes is genoem.
Ask a question about the relationship between dependent and
independent variables.
✓
Vraag moet gestel word oor die verwantskap tussen die afhanklike en
onafhanklike veranderlikes.

OR / OF

What effect does a change in mass have on the acceleration of the trolley? /
Watter effek het ʼn verandering in massa op die versnelling van die trollie?

Independent variable / Onafhanklike veranderlike:

Mass of trolley / Massa van trollie

Dependent variable / Afhanklike veranderlike:

Acceleration of trolley / Versnelling van trollie.

Example / Voorbeeld:

What is the relationship between the acceleration of the trolley and the mass of
the trolley while the applied force is kept the same?/ Wat is die verwantskap
tussen die versnelling van die trollie en die massa van die trollie terwyl die
toegepaste krag konstant gehou word?
Notes/Aantekeninge:
A question that does not contain a relationship: Max 1/2
’n Stelling wat geen verwantskap bevat nie: Maks. 1/2
A question that starts with WILL, DO, DOES, SHALL. No MARKS
ʼn Vraag wat begin met SAL/ WIL. Geen PUNTE (2)

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 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

4.3 See next page/ Sien volgende bladsy

4.4 Acceleration is inversely proportional to mass✓ if force is kept constant.✓

Refers to the dependent, independent and controlled variables.
Die versnelling is omgekeerd eweredig aan die massa ✓van die voorwerp as die
krag konstant gehou word. ✓ Moet na die afhanklike, onafhanklike en kontrole
verwys word (2)

4.5 Friction increases with an increase in mass as fk = μkN. ✓ The incline was kept
constant,  Ff  as mass  ,  a  ✓
Wrywing verhoog met ʼn toename in massa omdat fk = μkN. ✓ Die helling is
konstant gehou,  Ff  soos massa  ,  a  ✓ (2)

4.6 For a constant applied force, an increase in mass decreases the negative
acceleration. (a  1/m)✓/ Vehicle is slow to respond / Causes the stopping time to
increase ✓ and therefore the vehicle has a greater stopping distance.
As die toegepaste krag konstant gehou word is versnelling omgekeerd eweredig
aan die massa (a  1/m)✓/ Voertuig is stadig om te reageer/ veroorsaak
verhoogde stoptyd en die voertuig benodig dus ʼn groter afstand om te stop.✓ (2)

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 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

QUESTION 4.3/ VRAAG 4.3

4
Acceleration/
Versnelling
(m.s-2)

4 Points plotted correct ✓✓

4 Punte korrek geplot.
Close fit line drawn ✓
Naaste pas lyn getrek.
2 Line extrapolated to Y-axis✓
Lyn verleng tot by Y -as (4)

0
0,2 0,4 0,6 0,8 1,0
__1__
Mass/Massa (per trolley
units/ in trollie eenhede)
[15]

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 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

QUESTION 5 / VRAAG 5
5.1 The force exerted by a surface on an object✓ in contact with it and it always
acts perpendicular to the surface. ✓
Die krag wat deur die oppervlak waarop die voorwerp rus ✓ uitgeoefen word
loodreg vanaf die oppervlak.✓ (2)

5.2 Criteria for Free body diagram / Marks/

Kriteria vir vryliggaam : Punte
T – upwards /opwaarts ✓
N -  up from plane /  opwaarts ✓
✓
Fg – down towards centre of earth/ ✓
afwaarts na middel van aarde

fk - // downwards / //afwaarts ✓

- 1 for any extra forces /

-1 vir enige ekstra kragte. (max ¾)
(4)

OR is the force that opposes the tendency of motion of a stationary object

relative to a surface /
OF is die krag wat die geneigdheid tot beweging van ʼn statiese objek opponeer
relatief tot ʼn oppervlak

5.3 Static friction is the frictional force when an object is standing still on a
surface✓✓, while kinetic friction is the frictional force when an object moves /
slides over a surface✓✓ OR is the force that opposes the motion of a moving
object relative to a surface
Statiese wrywing is die wrywingskrag van ʼn voorwerp wat op ʼn oppervlak
stilstaan✓✓ terwyl kinetiewe wrywing ʼn wrywingskrag is wanneer ʼn voorwerp
oor ʼn oppervlak beweeg / gly ✓✓ (4)

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 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

5.4
5.4.1 OPTION 1 / OPSIE 1
N = Fg  = Fg x cos 30°✓
= (5 x 9,8) x cos 30°
= 42,44 N  up slope /  opwaarts ✓
𝑓𝑘 f k = µx N
µx = ✓
N f k = 0,18.42,44✓✓
0,18 = __fk___
42,44 f k = 7,64N
= 7,64 N // down plane / // afwaarts ✓
OPTION 2 / OPSIE 2
𝑓𝑘
µx = ✓
N
0,18 = __fk___
(5 x 9,8)✓ x cos 30°✓
= 7,64 N // down plane / // afwaarts ✓ (4)

5.4.2 Fnet on 7kg = m x a = Fx -T – fk✓ F net on 5kg = m x a = - Fg// + Tx - fk ✓

7a = (68 x cos 12°) - T - 6 5a = (- 5 x 9,8 x sin 30°) + T - 7,64
T = - 7a + 60,51……(1) ✓ T = 5a + 32,14 ……. (2) ✓
(1) + (2) - 7a + 60,51 = 5a + 32,14 ✓
12a = 28,37
a = 2,36 m.s-2 ✓ to the left ✓ / na links (7)

5.4.3 OPTION 1 / OPSIE 1 OPTION 2 / OPSIE 2

T = - 7a + 60,51……(1) ✓ T = 5a + 32,14 ……. (2) ✓
= - 7 (2,36) ✓+ 60,51 = 5 (2,36)✓ + 32,14
= 43,99 N ✓ = 43,94 N ✓ (3)
[24]

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 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

QUESTION 6 / VRAAG 6
6.1 Newton’s Law of Universal Gravitation / Newton se Universele Gravitasie Wet✓
Every particle in the universe exerts a force of gravitational attraction on every
other particle. The force between the two particles is directly proportional to the
product of their masses ✓and inversely proportional to the square of the
distance between them. ✓
Elke voorwerp in die heelal trek elke ander voorwerp aan met ʼn krag wat direk
eweredig is aan die produk van die massas van die voorwerpe ✓ en omqekeerd
ewerediq is aan die kwadraat van die afstand tussen die massa-middelpunte van
die twee voorwerpe.✓ (3)

6.2 EQUAL TO / GELYK AAN✓✓ (2)

6.3 The mass of the Earth is so much greater than the mass of the plane.✓✓
Aangesien die massa van die aarde soveel groter is as die massa van die
vliegtuig.✓✓
(2)

6.4 FEP = GmEmP ✓ ✓

= (6,67 x 10-11 )( 5,98 x 1024 ✓ )(30000)✓
r2
(3500 + 6,38 x 106) 2✓
= (6,67 x 10-11 )( 5,98 x 1024 ✓ )(3 x 104)✓
(3,5 x103 + 6,38 x 106) 2✓
= 293 650,37 N  downwards / afwaarts ✓
OR / OF 29,37 x 106 N
2,94 x 1075 N  downwards / afwaarts 2,94 x 107 N ✓

2,94 x 105 N (5)

6.5 Fn = GmBmC
r2

= 1x2x1 ✓
( ⅓)2 ✓

Fn = 18F ✓ (3)
[15]

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 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

QUESTION 7 / VRAAG 7

7.1 Diffraction / Diffraksie ✓✓ (2)

7.2 Huygen’s principle / Huygen se beginsel ✓✓

Every point on a wavefront is a source of secondary wavelets. These wavelets
spread out in the forward direction, at the same speed as the source wave.✓✓
Elke punt op ʼn golffront reageer as die bron van sekondêre golfies wat in alle
rigtings met dieselfde spoed as die golf uitsprei. ✓✓ (4)

7.3 A central bright band of coloured light ✓with alternating dark and light bands✓ to
the sides getting dimmer.✓
ʼn Helder breë sentrale kleurband ✓afgewissel deur swart bande ✓en helder
bande, wat dowwer word na die kante. ✓ (3)

7.4 Destructive interference / Destruktiewe interferensie✓✓ (2)

7.5 7.5.1 INCREASE / VERMEERDER ✓

The angle of interference is inversely proportional to the width of the
slit. ✓✓
Die diffraksiehoek is omgekeerd eweredig aan die wydte van die
spleet.✓✓ (3)

7.5.2 DECREASE / VERMINDER ✓

The angle of interference is directly proportional to the wavelength. ✓✓
Die diffraksiehoek is direk eweredig aan die golflengte.✓✓ (3)
[17]

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 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

QUESTION 8 / VRAAG 8

8.1 8.1.1 Refraction / Refraksie ✓✓ (2)

8.1.2 When light moves from a more dense to a less dense✓ substance, it will
break (bend) away ✓ from the normal and because light travels in a
straight line it appears as if the pencil is broken.✓
Lig wat van ʼn opties meer digte medium na opties minder digte medium
beweeg breek weg ✓ van die normale en aangesien lig in ʼn reguit lyn
beweeg lyk dit asof die potlood gebuig is.✓ (3)

8.2.1 Normal line / Normaal lyn✓✓ (2)

8.2.2 ni sin θi = nr sin θr ✓

1 x sin 39°✓ = 1,33 ✓sin r
r = 28,24 °✓ (4)

8.2.3 Criteria for RAY diagram / Kriteria vir : Marks/ Punte

At D, break towards normal/ ✓
By D, breek na die normaal toe
Exiting the water: bend away from the normal /
Wanneer uit water beweeg, breek weg van die ✓
normaal
CD and exiting ray are parallel to each other./ ✓
CD en uitgaande straal is parallel aan mekaar.

Angles shown/ Hoeke aangetoon ✓

- 1 for any extra forces /

- 1 vir enige ekstra kragte (max / maks ¾)

(4)

8.3 DECREASES / VERLAAG. ✓✓ (2)

8.4 The patricles of the water✓ are much closer together and thus it will slow down
the light ray✓/ die deeltjies van die water is baie meer kompak✓ en sal dus die
ligstraal se spoed verlaag. ✓ (2)
[19]

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 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

QUESTION 9 / VRAAG 9
9.1 Light must travel from an optically more dense medium to an optically less
dense medium.✓✓
Die invalstraal moet vanaf ʼn opties meer digte medium na ʼn opties minder digte
medium beweeg.

The angle of incidence must be bigger than the critical angle of that medium.✓✓
Die invalshoek moet groter as die grenshoek van die betrokke medium wees. (4)

9.2 Is that angle of incidence that provides an angle of refraction of 90°.✓✓

Die invalshoek wat ʼn brekingshoek van 90° tot gevolg het. (2)

9.3 9.3.1 ni sin θi = nr sin θr ✓

1,45 x sin θi ✓ = 1,40 ✓sin 90°
i = 74,91 °✓ (4)

9.3.2 _c_
n = V ✓
3 x 108✓
1,45 ✓ = V

V = 2,07 x 108 m.s-1 ✓ (206 896 551,70 // 206,9 x 106) (4)

9.4 High speed data transfer / hoë oordragspoed van data✓

No resale value  doesn’t get stolen / lae herverkoop waarde, geen diefstal ✓
Not as easy to tap / bigger confidentiality / Moeliker om data te steel
(Any two applicable reasons / Enige twee gepaste redes) (2)
[16]

TOTAL/ TOTAAL: 150

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 PHYSICAL SCIENCES: Physics /
MEMORANDUM FISIESE WETENSKAPPE: Fisika
GRADE 11

Taxonomy Grid
Recall Comprehension Analysis Evaluation
Q no: Mark Q no: Mark Q no: Mark Q no: Mark

1.1 2 1.5 2 1.7 2 1.10 2

1.2 2 1.6 2 1.8 2 4.5 2
1.3 2 2.2 9 1.9 2 4.6 2
1.4 2 3.2 4 4.1 3 5.4.2 8
2.1 2 3.3 3 4.4 2
3.1 2 3.4 3 5.4.3 3
5.1 2 4.2 2 6.3 2
6.1 2 4.3 4 6.4 5
7.1 2 5.2 4 6.5 3
8.1 2 5.3 2 7.4 3
8.2 2 5.4.1 4 7.5 8
6.2 2 8.5 3
7.2 2
7.3 2
8.2 3
8.3 4
8.4 4

22 56 38 14
Total mark 15% 22.5 35% 52,5 40% 60 10% 15
Total %/100% P1&2: 15% P1:35%/P2:40% P1:40%/P2:35% P1&2: 10%

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