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NATIONAL
SENIOR CERTIFICATE
GRADE/GRAAD 11
NOVEMBER 2015
PHYSICAL SCIENCES P1/
FISIESE WETENSKAPPE V1
MEMORANDUM
MARKS/PUNTE: 150
This memorandum consists of 10 pages./
Hierdie memorandum bestaan uit 10 bladsye.
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Physical Sciences Grade 11 P1 2015 Memo
Grade 11 Physical Sciences Paper 1 memorandum from the Eastern Cape November 2015, 10 pages. Read online or download the free PDF.
Physical Sciences · Grade 11 · Eastern Cape November · 2015. Memorandum, 10 pages. Read online or download the PDF.
- Subject
- Physical Sciences
- Grade
- Grade 11
- Document type
- Memorandum
- Year
- 2015
- Exam period
- Eastern Cape November
- Paper
- 1
- Curriculum
- CAPS
- Pages
- 10
- File size
- 913.2 KB
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This exam
Eastern Cape November 2015
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2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2015)
QUESTION/VRAAG 1
1.1 A
1.2 A
1.3 A
1.4 D
1.5 B
1.6 B
1.7 D
1.8 D
1.9 A
1.10 C
[20]
QUESTION/VRAAG 2
2.1 The resultant of the forces is zero.
Die resultant van die kragte is nul. (1)
2.2
T 20° 45° FL/FB
Fg
ACCEPTED LABELS/AANVAARDE BYSKRIFTE
Fw/W/Weight/mg/gravitational force/force due to gravity
Fg
Fw/W/gewig/mg/gravitasiekrag
FT/Tension/Tension in rope
T
FT/Spanning/Spanning in tou
FL Flegs/Force of legs on wall
FB Fbene/Krag van bene op muur (3)
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(EC/NOVEMBER 2015) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 3
2.3 Fg = mg = (50)(9,8) = 490 N
SCALE/SKAAL 10 mm = 50 N
Fg = 490 N = 98 mm (Accept/Aanvaar 97 – 99mm)
FL 45°
T Fg = 9,8mm
20°
T = 76,5 × 5 = 382,5 N (Accept/Aanvaar 76,4 – 76,6 mm)
FL = 37,0 × 5 = 185 N (Accept/Aanvaar 184,5 – 186,5 mm) (8)
NOTES/AANTEKENINGE
Mark awarded for correct label and direction./
Punt toegeken vir korrekte benoeming en pyltjie.
ANY TWO angles indicated./
ENIGE TWEE hoeke aangedui.
OR/OF
CALCULATIONS instead of CONSTRUCTION/ max/maks
BEREKENINGE in plaas van KONSTRUKSIES
Fg = FT FT = 490 sin 45° = 382,50 N
sin 115° sin 45° sin 115°
Fr = Fg FR = 490 sin 20° = 184,91 N
sin 20° sin 115° sin 115° [12]
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4 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2015)
QUESTION/VRAAG 3
3.1 fs = Fg‖ = mg sin = (60)(9,8) sin 30° = 294 N
Any one/Enige een (3)
3.2 s = tan θ = tan 30° = 0,58
OR/OF
fs = μs FN = μs Fg = μs mg cos
294 = μs (60)(9,8) cos 30°
μs = 0,58 (3)
3.3 3.3.1 REMAINS THE SAME/BLY DIESELFDE (1)
3.3.2 DECREASES/NEEM AF
Θ
Fg will decrease, therefore FN will decrease/fk FN
Fg sal afneem, dus sal FN afneem /fk FN (3)
[10]
QUESTION/VRAAG 4
4.1 When a resultant/net force acts on an object, the object accelerates in the
direction of the force. This acceleration is directly proportional to the force
and inversely proportional to the mass of the object.
Indien ʼn resulterende/netto krag op ʼn voorwerp inwerk, sal die voorwerp in
die rigting van die krag versnel. Hierdie versnelling is direk eweredig aan die
(resulterende/netto) krag en omgekeerd eweredig aan die massa van die
voorwerp.
OR/OF
The net force acting on an object is equal to the rate of change of
momentum of the object (in the direction of the force). (2 or 0)
Die netto krag wat op ʼn voorwerp inwerk is gelyk aan die tempo van
verandering in momentum van die voorwerp (in die rigting van die krag).
(2 of 0) (2)
4.2 N/FN
f/Ff T
w/Fg/mg (4)
NOTES/AANTEKENINGE
Mark awarded for label and arrow./Punt toegeken vir benoeming en
pyltjie.
Any other additional force(s)/Enige ander addisionele krag(te).
Max/Maks ¾
If force(s) do not make contact with body/Indien krag(te) nie met die
voorwerp kontak maak nie. Max/Maks ¾
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(EC/NOVEMBER 2015) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 5
4.3 4.3.1 OPTION/OPSIE 1
For the 6kg block./Vir 6kg blokkie:
Fnet = ma
T + (-ff ) = ma
T – 11,76 = 6a ...........(1)
For the 2kg block./Vir 2kg blokkie:
For either/vir of/6a or/of 2a
Fnet = ma
2 + w + (-T) = ma
2 + (2 × 9,8) – T = 2a
T = 21,6 – 2a .......(2)
Substitute (2) into (1)/Stel (2) in (1):
21,6 – 2a – 11,76 = 6a
a = 1,23 m.s-2
OPTION/OPSIE 2:
Fnet = ma
2 + w +(-Ff ) = ma
2 + (2 × 9,8) -Ff = 8a
2 + (2 × 9,8) – 11,76 = 8a
a = 1,23 m.s-2 (5)
4.3.2 POSITIVE MARKING from QUESTION 4.3.1
POSITIEWE NASIEN van VRAAG 4.3.1
T = 21,6 – 2a
= 21,6 – 2(1,23)
= 19,14 N
OR/OF
T – 11,76 = 6a
T = 11,76 + 6(1,23)
= 19,14 N (2)
4.4 INCREASES/TOENEEM (1)
[14]
QUESTION/VRAAG 5
5.1 BACKWARDS/AGTERTOE (1)
5.2 An object continues in a state of rest or uniform velocity, unless it is acted
upon by an external unbalanced force (resultant force).
ʼn Voorwerp sal in ʼn toestand van rus bly of teen ʼn konstante snelheid bly
voortbeweeg, tensy ʼn eksterne ongebalanseerde (resulterende krag)
daarop inwerk.
OR/OF
The tendency of an object to maintain its state of rest or constant motion in
a straight line.
Die weerstand van ʼn voorwerp wat enige verandering in sy toestand rus
of beweging teenstaan. (2)
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6 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2015)
5.3 5.3.1 EQUALS/GELYK (1)
5.3.2 Kay will have a greater acceleration, because her mass is
smaller.
Kay sal ʼn groter versnelling het, want haar massa is kleiner. (2)
[6]
QUESTION/VRAAG 6
6.1 Action: Force of seat on astronaut (upwards)
Reaction: Force of astronaut on seat (downwards)
Aksie: Krag van sitplek op ruimtevaarder (opwaarts)
Reaksie: Krag van ruimtevaarder op sitplek (afwaarts) (2)
6.2 FRES = m ∆v = (2,6 × 106)(160 – 0) = 1,04 × 106 N upwards/opwaarts
∆t 400
Mark direction independently/Merk rigting onafhanklik (5)
QUESTION/VRAAG 7
7.1 F = G m1.m2
r2
= (6,67 × 10-11)(7,35 × 1022)(6 × 1024)
(3 × 109)2
18
= 1,99 × 10 N (4)
7.2 NEWTON’s LAW OF GRAVITATION/NEWTON se GRAVITASIEWET. (1)
7.3 The distance between the earth and the moon changes due to the elliptical
orbit.
Die afstand tussen die aarde en die maan verander as gevolg van elliptiese
omwenteling.
OR/OF
F α 1/r2 (1)
7.4 FEM(new) = GME1/4 MM = FEM (1/4)(1/4) = 1/16 FEM
(2r)2
OR/OF
FEM(new) = 1/16 (1,99 × 1018)
OR/OF
FEM(new) = 1,24 × 1017N (1,24375 × 1017 N) (3)
[9]
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(EC/NOVEMBER 2015) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 7
QUESTION/VRAAG 8
8.1 Angle of incidence/ i/i/sin i
Invalshoek/ i/i/sin i (1)
8.2 Type of block used/Temperature of the surroundings/Source of light/Surface
on which block is placed. .
Tipe blok gebruik/Temperatuur van die omgewing/Bron van die lig/Oppervlak
waarop blok geplaas word. (1)
8.3 Snell’s law has been verified:
* The graph is a straight line through the origin, because sin i sin r.
* Gradient = sin i = constant
sin r
Snell se wet is bewys:
Die grafiek vorm ʼn reguitlyn deur die oorsprong, omdat sin i sin r.
Gradiënt = sin i = konstante
sin r (2)
8.4 Gradient/Gradiënt = n = Δ sin i = 0,77 – 0 = 1,51
Δ sin r 0,51 – 0 (4)
[8]
QUESTION/VRAAG 9
9.1 n sin i = n sin r
2,42 sin i = 1 sin 90°
-1
i = sin 2,42
= 24,2° (4)
9.2
No arrows: max
Geen pyltjies: maks
(2)
9.3 Total internal reflection /Totale interne weerkaatsing (1)
9.4 Light moves from medium of higher optical density to medium of lower
optical density.
Angle of incidence must be larger than critical angle.
Lig beweeg van medium van hoër optiese digtheid na medium met laer
optiese digtheid.
Invalshoek moet groter wees as kritiese hoek. (2)
9.5 Endoscope/Endoskoop (1)
[10]
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8 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2015)
QUESTION/VRAAG 10
10.1 Light consisting of only one wavelength.
Lig wat uit net een golflengte bestaan. (1)
10.2 B
A
Broader Central band
Breër sentraleband
Alternative light (bright) and dark bands.
Afwisselende helder (lig) en donker bande
A: central band and bright (light) bands = CONSTUCTIVE INTERFERENCES
sentraleband en helder (lig)bande = KONSTRUKTIEWE INTERFERENSIE/
STEURINGE
B: dark bands = DESTRUCTIVE INTERFERENCES
donkerbande = DESTRUKTIEWE INTERFERENSIE/STEURINGE (4)
10.3 10.3.1 INCREASES/NEEM TOE (1)
10.3.2 INCREASES/NEEM TOE
Θ
The narrower the opening, the greater the degree of
diffraction.
Hoe smaller die opening, hoe groter die mate van diffraksie.
OR/OF
Degree of diffraction 1
Width of opening
Mate van diffraksie 1
Wydte van opening (2)
[8]
QUESTION/VRAAG 11
11.1 Force experienced per unit charge at a point in the field.
Krag ondervind per eenheidslading by ʼn punt in die veld.
OR/OF
It is the force experienced by a unit positive charge placed at that point.
Dit is die krag wat deur ʼn eenheids positiewe lading geplaas by daardie
punt ondervind word. (2)
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(EC/NOVEMBER 2015) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 9
11.2
CRITERIA FOR SKETCH/KRITERIA VIR SKETS MARKS/PUNTE
Correct shape / Korrekte vorm
Direction from positive to negative
Rigting van positief na negatief
Fieldlines start on sphere and do not cross
Veldlyne begin op elke sfeer en kruis nie (3)
11.3 Ep = kQ = (9 × 109)(4 × 10-9) = 3 600 N.C-1 to the right/na regs
r2 (0,10)2
Any ONE
Enige EEN
EQ = kQ = (9 × 109)(2 × 10-9) = 45 000 N.C-1 to the left/na links
r2 (0,02)2
Enet = 3 600 – 45 000 = 41 400 N.C-1 to the left/na links (7)
11.4 POSITIVE MARKING FROM QUESTION 11.3
POSITIEWE NASIEN VAN VRAAG 11.3
E = F
Q
41 400 = F
1,6 × 10-19
F = 6,62 × 10-15 N (3)
[15]
QUESTION/VRAAG 12
12.1 12.1.1 Φ = BA cos θ
= (0,6) r2 cos θ
= (0,6)( × 0,062 ) cos 0°
= (0,6)( 0,01 )
= 6,79 × 10-3 Wb (4)
12.1.2 POSITIVE MARKING FROM QUESTION 12.1.1
POSITIEWE NASIEN VAN VRAAG 12.1.1
= -N Δ Φ
Δt
= (-1)(0 - 6,79 × 10-3)
0,04
= 0,17 V (4)
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10 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2015)
1.2.2 12.2.1 DECREASES/NEEM AF (1)
12.2.2 INCREASES/NEEM TOE (1)
[10]
QUESTION/VRAAG 13
13.1 13.1.1 P = I 2R
0,5 = I2 8
I = 0,25 A
1 =1 + 1 = 1 + 1 = 3 Rp = 4 Ω
Rp r1 r2 6 (6 + 6) 12
RT = 8 + 4 = 12 Ω
VT = I RT = (0,25) (12) = 3V (8)
13.1.2 Vp = I R = (0,25)(4) = 1V
I = Vp = 1 = 0,17 A
R 6 (4)
13.2 13.2.1 P = V2
R
750 = 2402
R
R = 76,80 Ω
OR/OF
P = VI
750 = 240 I
I = 3,13 A
R = V
I
= 240
3,13
= 76,80 Ω (4)
13.2.2 E = Px t = 750 × 6 = 4 500 Wh = 4,5 kWh
Cost/Koste = 4,5 × 1,14 = R 5,13
OR/OF
Cost/Koste = (0,75)(6) (1,14) = R5,13 (3)
[19]
TOTAL/TOTAAL: 150
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