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NATIONAL SENIOR CERTIFICATE
NASIONALE SENIOR SERTIFIKAAT
GRADE/GRAAD 11
NOVEMBER 2013
PHYSICAL SCIENCES P1/
FISIESE WETENSKAPPE V1
MEMORANDUM
MARKS/PUNTE: 150
This memorandum consists of 8 pages./
Hierdie memorandum bestaan uit 8 bladsye.
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Physical Sciences Grade 11 P1 2013 Memo
Grade 11 Physical Sciences Paper 1 memorandum from the Eastern Cape November 2013, 8 pages. Read online or download the free PDF.
Physical Sciences · Grade 11 · Eastern Cape November · 2013. Memorandum, 8 pages. Read online or download the PDF.
- Subject
- Physical Sciences
- Grade
- Grade 11
- Document type
- Memorandum
- Year
- 2013
- Exam period
- Eastern Cape November
- Paper
- 1
- Curriculum
- CAPS
- Pages
- 8
- File size
- 912.7 KB
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This exam
Eastern Cape November 2013
Questions in this paper
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2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (NOVEMBER 2013)
QUESTION/VRAAG 1
1.1 Normal force/Normaalkrag (1)
1.2 Refractive index/Brekingsindeks (1)
1.3 Critical angle/Kritiesehoek (grenshoek) (1)
1.4 Electric field (strength)/Elektriese veld (sterkte) (1)
1.5 Energy (Work) done/Energie (Arbeid) verrig (1)
[5]
QUESTION/VRAAG 2:
2.1 C (2)
2.2 D (2)
2.3 C (2)
2.4 C (2)
2.5 B (2)
2.6 A (2)
2.7 C (2)
2.8 C (2)
2.9 A (2)
2.10 B (2)
[20]
TOTAL SECTION/TOTAAL AFDELING A: 25
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(NOVEMBER 2013) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 3
QUESTION/VRAAG 3
3.1 3.1.1 F = 60 mm √
= 55° √
X = 34 mm √ (accept/aanvaar 33 – 35)
F = 60 mm 102 N √ (accept/aanvaar 99 N – 105 N)
left/links
Y = 49 mm (accept/aanvaar 48 – 50) √
55º 147 N √ (accept/aanvaar 144 N – 150 N)
down/af √
All 3 arrows correct/Al 3 pyle korrek √ (7)
If a calculation was done instead of a construction – max 4 out of 7
Indien ʼn berekening gedoen is i.p.v. ʼn konstruksie – maks. 4 uit 7
Right angled triangle with information shown/Reghoekige driehoek met
inligting getoon √
All 3 arrows correct/Al 3 pyle korrek √
X = 180 cos 55º = 103,24 N √
Y = 180 sin 55 º = 147,45 N √
Positive marking from Q3.1.1/Positiewe nasien vanaf Vr3.1.1
3.1.2 w = mg = 30(9,8) √ = 294 N √
normal force/normaalkrag = 294 + 147 = 441 N √ (3)
3.2 A single vector √ with the same effect as a number of vectors acting on an object. √
ʼn Enkel vektor √ wat dieselfde effek het as ʼn aantal vektore wat saam op ʼn
voorwerp inwerk. √ (2)
3.3 3.3.1 one mark for correct vector diagram/een punt vir korrekte vektordiagram
R=√ = 25 N √√ (3)
3.3.2 tan θ =
θ = 36,87° √
thus/dus 6,87° √
(clockwise from the positive y-axis/
kloksgewys vanaf die y-as)
OR/OF (2)
3.4 When three or more vectors drawn head to tail form a closed figure, √ their
resultant is zero or they are in equilibrium/balanced. √
Wanneer drie of meer vektore kop-stert geteken word en ʼn geslote figuur vorm, √
is hul resultant nul of is hulle in ewewig/gebalanseerd √ (2)
[19]
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4 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (NOVEMBER 2013)
QUESTION/VRAAG 4
4.1
FN – normal force/normaalkrag
w – weight/gewig
fk – friction/wrywing
(3)
4.2 4.2.1 The kinetic frictional force is the only force acting on the sled in x direction
and opposes the motion of the sled so the net force is given by/Die
kinetiese wrywingskrag is die enigste krag op die skild in x-rigting en
staan die beweging van die slee teen sodat die krag gegee word deur:
Fnet = ma √ OR/OF fk = μk N √ = 0,06 m (9,8) √ = 0,588 m N
µkmg √ = ma Fnet = ma √ 0,588 m = m a
µkg = a a = 0,588 m.s-2 √ or/of 0,59 m.s-2
-2
6 x10 x 9,8 = a √
a = 0,588 m.s-2 √ or/of 0,59 m.s-2 (4)
4.2.2 POSITIVE MARKING FROM QUESTION 4.2.1
POSITIEWE NASIEN VAN VRAAG 4.2.1
vf2 = vi2 + 2a∆x √
02 = 62 + 2 (-0,588) ∆x √ (the sled decelerates/die slee vertraag)
∆x = 30,61 m √ (if/as a = 0,59 m.s-2 then/dan ∆x = 30,51 m) (3)
[10]
QUESTION/VRAAG 5
5.1 Bulie’s applied force/toegepaste krag/250 N √
friction/wrywing/500N √
Douglie’s applies force/toegepaste krag/400 N √
(3)
5.2 Fnet = ma = 400 + 250 + (-500) √ = 2000 a √
150 = 2000 x a
a = 0,075 m.s-2 √ to the right/na regs √ (4)
5.3 Fg//= Fg sin θ = mg sin θ √
= 2000 (9,8) sin 5˚
=1 708,25 N √ (2)
5.4 accelerate down the incline/versnel teen die steilte af √ (1)
5.5 Inertia is the tendency of an object to resist change √
Traagheid is ʼn voorwerp se vermoë om verandering teen te staan √ (1)
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(NOVEMBER 2013) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 5
5.6 A person will keep on moving forward in a straight line at constant velocity unless
acted on by a resultant force √ The seatbelt acts as a net force √ which hold you
safely in the seat.
ʼn Liggaam sal aanhou beweeg in ʼn reguit lyn teen ʼn konstante snelheid tensy ʼn
resultante krag daarop inwerk. √ Die veiligheidsgordel is die net krag √ wat jou
veilig in die sitplek hou.
OR/OF
In an accident seatbelts hold you safely in place because of Newton’s First law of
motion and inertia. √ When a car suddenly slows down the seatbelt acts as net
force √ which helps the passengers from being hurt.
In ʼn ongeluk hou die veiligheidsgordel jou veilig in plek a.g.v. Newton se eerste
wet/traagheid. √ Wanneer ʼn motor skielik stadiger ry, tree die veiligheidsgordel
as die netto krag op √ wat help dat die passasier nie seerkry nie. (2)
5.7 Newton’s third law of motion states that when object A exerts a force on object B,
object B simultaneously exerts an oppositely directed force of equal magnitude
on object B. √√
Volgens Newton se derde bewegingswet: as voorwerp A ʼn krag op voorwerp B
uitoefen, oefen voorwerp B terselfde tyd ʼn teenoorgestelde krag van dieselfde
grootte op A uit. √√ (2)
5.8 force of book on table –
force of table on book √
krag van boek op tafel –
krag van tafel op boek
force of floor on table - force of
table on floor √
krag van vloer op tafel – krag
van tafel op vloer
force of earth on table - force of table on earth √
krag van aarde op tafel – krag van tafel op aarde
(3)
[18]
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6 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (NOVEMBER 2013)
QUESTION/VRAAG 6
6.1 F sun earth = G Msun Mearth √
d2
= 6,67 x 10-11 √ (1,99 x 1030)(5,98 x 1024) √
(2 x 1011)2 + (4 x 108)2 √
= 1,98 x1022 N √ (5)
6.2 gmoon = GMmoon √ OR/OF F =mg=GmM
2
d d2
= 6,67 x10-11 (7,35 x 1022) √
(1,6 x 106)2 √
= 1,92 m.s-2 √ (4)
6.3 W = mg √ = 50 x 10-3 x 9,8 √ = 0,49 N √ (3)
[12]
QUESTION/VRAAG 7
7.1 7.1.1 n1sinӨ1 = n2sinӨ2√
1,00 x sin 47°√ = 1,33 sinӨ2 √
sinӨ2 = 0,55 Ө2
= 33,36° √ (4)
7.1.2 n1sinӨ1 = n2sinӨ2
1,33 sin 47° √ = 1,00 sinӨ2 √
Ө2 = 76,58°√ (3)
7.2 n= √
1,33 = 3 x 108 √
v
v = 2,26 x 108 m.s-1 √ (3)
7.3 n1sinӨ1 = n2sinӨ2
1,33 sinӨ1 √ = 1,00 sin 90° √ Ө1 = 48,75° (2)
7.4 7.4.1 7.4.2
(2) (2)
7.5 (any of the following/enige van die volgende)
-telecommunications/telekommunikasie √
-video communications/videokommunikasie
-computer-data communications/rekenaardatakommunikasie
-medicine - endoscope)/medisyne - endoskoop (1)
[17]
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(NOVEMBER 2013) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 7
QUESTION/VRAAG 8
8.1 Every point in a wave front acts as the source of secondary wavelets √ that
spread out in all directions with the same speed as a wave. √
Elke punt op die golffront dien as ʼn bron van sekondêre golwe √ wat uitsprei in
alle rigtings met dieselfde spoed as die golf. √ (2)
8.2 A – Central bright broad band/breë sentrale ligte band √
B – dark band/donker bande √ (2)
8.3 A√ (1)
8.4 8.4.1 broader/breër √ (1)
8.4.2 broader/breër √ (1)
8.5 diff α wavelength/diff α golflengte √√ (2)
[9]
QUESTION/VRAAG 9
9.1 9.2 F12 = k q1q2 √
r2
= 9 x 109 (4 x 10-6)(6 x 10-6) √
(0,15)2 √
= 9,6 N √
(3) (4)
9.3 F13 = 9 x 109 (4 x 10-6)(5x10-6) √
(0,1)2 √
= 18 N √
Fnet = √ √
= 20,4 N √
tan θ = 9,6 √
18
θ = 28,07° √ (anticlockwise from positive x-axis)
(antikloksgewys vanaf die positiewe x-as)
OR/OF = 61,93° (clockwise from the positive y-axis)
(kloksgewys vanaf die positiewe y-as) (7)
[14]
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8 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (NOVEMBER 2013)
QUESTION/VRAAG 10
10.1 10.1.1
Current direction: up out of page/
Stroomrigtin: uit papier
circular field/sirkelvormige veld √
anticlock wise/antikloksgewys √
(2)
10.1.2 shape/vorm √
position of north and
direction N to S/
posisie van noord
en rigting N na S√
(2)
10.2 10.2.1 Φ = BAcos θ √ = 0,72 (0,0176) cos 0° √ = 0,013 Wb √ (0,012672 Wb) (3)
POSITIVE MARKING FROM Q10.2.1/POSITIEWE NASIEN VANAF VR10.2.1
10.2.2 ε = -N∆Φ √ = -450 (0 – 0,013) √ = 26,59 V √
∆t 0,22 (3)
[10]
QUESTION/VRAAG 11
11.1 11.1.1 3V √ (1)
11.1.2 I = V/R √ = 3/5 √ = 0,6 A √ (3)
11.1.3 I = V/R = 3 √ /7 √ = 0,43 A
V2 = IR √ = 0,43 x 4 √ = 1,72 V √ (accept/aanvaar 1,71 V) (5)
11.2 11.2.1 P = V2 √ OR/OF P = VI
R 2600= 220 I √
both equations/
2600 √= 2202 √ I = 11,82 A
beide vergelykings
R R = V/I
√
R = 18,62 Ω √ = 220/11,82 √
= 18,62 Ω √ (4)
11.2.2 Cost = (2,6 x 3,5) √ x 1,04 √ = R9,46 √
OR/OF E = Pt = 2,6 kW x 3,5 h √ = 9,1 kWh
Cost = 9,1 kWh x R1,04 √ = R9,46 √ (3)
[16]
TOTAL SECTION/TOTAAL AFDELING B: 125
GRAND TOTAL/GROOTTOTAAL: 150
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