Physical Science
Controlled Test 1 - 2023
Gauteng Department of Education
Johannesburg North District (D10)
GRADE 12
PHYSICAL SCIENCES
CONTROLLED TEST 1
17 MARCH 2023
MARKING GUIDELINES
MARKS:
100
TIME:
2 Hours
Physical Sciences
Controlled Test 1 - 2023
2
SECTION A: PHYSICS
QUESTION 1
1.1
A

(2)
1.2
B

(2)
1.3
C

(2)
[6]
QUESTION 2
2.1
When a resultant/net force acts on an object, the object will accelerate in
the direction of the force at an acceleration directly proportional to the
forceand inversely proportional to the mass of the object
OR
The resultant/net force acting on an object is equal to the rate of change
of momentum of the object (in the direction of the resultant/net force.) 
(2)
2.2
Accepted Labels
w
Fg/Fw/w/weight/19,6 N/gravitational force
T
FT/Tension/
fk
f / friction /
F
FA/ Fapplied / 20 N
N
Normal force / FN
Notes
• Mark is awarded for label and arrow.
• Do not penalise for length of arrows.
• Deduct 1 mark for any additional force.
• If all forces are correctly drawn and labelled, but no arrows, deduct
1 mark.
• Accept FA represented as its horizontal and vertical components
2.3.1
fk = µkN
(5)

fk = µk(w – Fsinθ)
fk = (0,2)(19,6 – 20sin20°)
fk = 2,55 N 
(4)
Physical Sciences
Controlled Test 1 - 2023
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2.3.2
POSITIVE MARKING FROM 2.3.1
2 kg block
Fnet = ma 
T – fk - Fx = ma
T – 2,55 – 20cos20°= 8
T = 29,34 N 
X kg block
Fnet = ma
w – T = ma
(m)(9,8) – 29,34 = m(4) 
5,8m = 29,34
m = 5,06 (5,059) kg 
(5)
[16]
QUESTION 3
3.1
Motion of an object upon which the only force acting is the force of gravity
(gravitational force). (2 or 0)
(2)
3.2
9,8 m.s-2 downwards (No marks if no/incorrect units or direction)
(1)
3.3.1
UPWARDS AS POSITIVE
UPWARDS AS NEGATIVE
vf2 = vi2 + 2aΔy 
vf2 = vi2 + 2aΔy 
vf2 = 02 + 2(-9,8)(-8) 
vf2 = 02 + 2(9,8)(8) 
vf = -12,52 m.s-1 
vf = 12,52 m.s-1 

3.3.2
∴ vf = 12,52 m.s-1 
∴ vf = 12,52 m.s-1 
UPWARDS AS POSITIVE
UPWARDS AS NEGATIVE
vf = vi + aΔt
vf = vi + aΔt
-12,52 = 0 + (-9,8)(Δt) 
12,52 = 0 + (9,8)(Δt) 
Δt = 1,2775..s (1,28)
Δt = 1,2775..s (1,28)
Δt = 1,2775 – 0,6 = 0,6775.. (0,68)
Δt = 1,2775 – 0,6 = 0,6775.. (0,68)
Δy = viΔt + ½ aΔt2 
Δy = viΔt + ½ aΔt2 
-8= vi(0,677..)+ ½(-9,8)(0,677..)2 
8= vi(0,677..)+ ½(9,8)(0,677..)2 
vi = -8,49 m.s-1
∴ vi = 8,49 m.s-1 
3.4
(3)
vi = 8,49 m.s-1
(8,43 – 8,49)
∴ vi = 8,49 m.s-1
 (8,43 – 8,49)
OPTION 1
UPWARDS AS POSITIVE
UPWARDS AS NEGATIVE
vf2 = vi2 + 2aΔy 
vf2 = vi2 + 2aΔy 
(4)

(0)2 = vi2 + 2(-9,8)(6,5) 
(0)2 = vi2 + 2(9,8)(-6,5) 
vf = 11,29 m.s-1
vf = -11,29 m.s-1
vf = 11,29 m.s-1 upwards 
vf = 11,29 m.s-1 upwards 
(3)
Physical Sciences
Controlled Test 1 - 2023
4
OPTION 2
Emech(i) = Emech(f)
(mghi + ½mvi2) = (mghf + ½mvf2)
m(-9,8)(0) + ½mvi2

= m(-9,8)(6,5) + ½m(0)2 
(3)
vi = 11,29 m.s-1 
OPTION 3
Can accept options using work calculations such as: Wnc = ΔEK + ΔEP
or Wnet = ΔEK (However very unlikely any learners would use these options
as Work, Energy and Power is only taught in Term 2)
3.5
POSITIVE MARKING FROM 3.3.1 and 3.4
Marking criteria
Correct shape
Similar gradients
Velocity values (12,52 and 11,29)
3.4
UPWARDS AS POSITIVE



UPWARDS AS NEGATIVE
(3)
Physical Sciences
Controlled Test 1 - 2023
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QUESTION 4
4.1
The total linear momentum in an isolated (closed) system remains
constant (is conserved) (2 or 0)
4.2
Δp = m(vf – vi) 
= 500(-1 – 5) 
= - 3 000 kg∙m∙s-1
= 3 000 kg∙m∙s-1 West (-1 if no direction)
4.3
(2)
(3)
POSITIVE MARKING FROM 4.2
FnetΔt = m(vf – vi) = Δp 
Fnet(1,4) = - 3 000 
Fnet = - 2 142, 86 N
(3)
Magnitude = 2 142,86 N 
4.4
EAST AS POSITIVE
Σpi = Σpf

1 mark for any

(500)(+5) + m2(0)  = (500)(-1) + 
m2(3) 
m1v1i + m2v2i = m1v1f + m2v2f
2 500 = -500 + 3m2
3 000 = 3m2
m2 = 1000 kg 
WEST AS POSITIVE
Σpi = Σpf

1 mark for any

(500)(-5) + m2(0)  = (500)(1) + m
2(-3) 
m1v1i + m2v2i = m1v1f + m2v2f
-2 500 = 500 - 3m2
-3 000 = -3m2
m2 = 1000 kg 
(4)
[12]
TOTAL SECTION A: [50 MARKS]
Physical Sciences
Controlled Test 1 - 2023
6
SECTION B: CHEMISTRY
QUESTION 1
1.1
B

(2)
1.2
C

(2)
1.3
B

(2)
[6]
QUESTION 2
2.1
A series of organic compounds that can be described by the same general
formula and functional group 
OR
2.2
A series of organic compounds in which one member differs from the next by
a CH2 group
(2)
CnH2n 
(1)

2.3
Q2.3 Marking criteria
• 1 mark for correct functional group
• 1 mark for whole structure correct

(2)
2.4.1
Alkyl halides (Haloalkanes) 
(1)
2.4.2

Q2.4.1 Marking criteria
• 1 mark for main chain with 6 C’s
• 1 mark for methyl and bromo
substituent on correct positions.
• If condensed or semi-structural
Max: 1 mark
(2)
Q2.5 IUPAC Naming marking criteria
• -1 mark for every mistake including numbering, sequence,
hyphens and commas
2.5.1
4-ethylhex-2-yne
(2)
2.5.2
4-fluro-3,3-dimethylhexane
(2)
2.6
M(C3H6O) = 58 g∙mol-1
116 ÷ 58 = 2
Hence molecular formula of ester is C6H12O2 
Molecular formula of carboxylic acid is C2H4O2 
(if final answer only - award full marks)
(3)
[15]
Physical Sciences
Controlled Test 1 - 2023
7
QUESTION 3
3.1.1
3.1.2
3.1.3
The temperature  at which the vapour pressure of a liquid is equal to the
atmospheric (external) pressure
(2)
Strength (type) of intermolecular force / Functional group/ Homologous
series
(1)
•
•
•
Butan-1-ol (C) has Hydrogen bonds 
Butan-2-one (B) has dipole-dipole forces (in addition to London
forces)
Hydrogen bonds are stronger  than the dipole-dipole forces in B
OR
Dipole-dipole forces are weaker than the hydrogen bonds in C
•
More energy is required to overcome the intermolecular (hydrogen
bonds) in Butan-1-ol (C)
OR
Less energy is required to overcome the intermolecular/dipole-dipole
forces in Butan-2-one (B)
(4)
3.1.4
Higher than
(1)
3.2.1
p1 = 100 (kPa) / 101,3 (kPa) / 1 atmosphere 
(1)
3.2.2
Gas . A is above / has past its boiling point 
(2)
3.2.3
Lower than 
(1)
3.2.4
Compound C only reached its boiling point at 118 °C where its vapour
pressure will equal 101,3 kPa. 
(2)
[14]
Physical Sciences
Controlled Test 1 - 2023
8
QUESTION 4
4.1.1
Elimination / Dehydrogenation 
(1)
4.1.2
Substitution
(1)
4.1.3
Addition / Hydrohalogenation
(1)
4.2.1
Hydration 
(1)
4.2.2
Butan-2-ol (if butanol or butan-1-ol : only 1 mark)
(2)
4.2.3
Sulphuric acid (H2SO4) or phosphoric acid (H3PO4) 
(1)
4.3.1
Hydrogen bromide 
4.3.2
CH3CH=CHCH3  + HBr  → CH3CHBrCH2CH3 
(do NOT accept HBr – name only)
(1)
OR
CH3CH=CHCH3 + HBr → CH3CH2CHBrCH3
(double bond in but-2-ene can be omitted)
If structural or molecular formula max: 2/4
4.4
(4)
2C4H10  + 13O2 → 8CO2 + 10H2O (products)Correct balancing 
(3)
[15]
TOTAL SECTION B: [50 MARKS]
[TOTAL = 100 MARKS]
Physical Sciences
Controlled Test 1 - 2023
9
COGNITIVE LEVELS FOR PHYSICAL SCIENCE
GRADE 12 – CONTROLLED TEST
2023
COGNITIVE LEVELS
1
QUESTION
2
3
4
Recall
Comprehension
Analysis
Application
Evaluation
Synthesis
(15 %)
(40 %)
(35 %)
(10 %)
MARKS
SECTION A: PHYSICS
1.1
1.2
1.3
2.1
2.2
2.3.1
2.3.2
3.1
3.2
3.3.1
3.3.2
3.4
3.5
4.1
4.2
4.3
4.4
2
2
2
2
5
4
5
2
1
3
4
3
3
2
3
3
4
2
2
2
2
5
4
5
2
1
3
4
3
3
2
3
3
4
SECTION B: CHEMISTRY
1.1
1.2
1.3
2.1
2.2
2.3
2.4.1
2.4.2
2.5.1
2.5.2
2.6
3.1.1
3.1.2
3.1.3
3.1.4
3.2.1
3.2.2
3.2.3
3.2.4
4.1.1
4.1.2
4.1.3
4.2.1
4.2.2
4.2.3
4.3.1
4.3.2
4.4
2
2
2
2
1
2
1
2
2
2
3
2
1
4
1
1
2
1
2
1
1
1
1
2
1
1
4
3
2
2
2
2
1
2
1
2
2
2
3
2
1
4
1
1
2
1
2
1
1
1
1
2
1
1
4
3
TOTAL
100
13
40
35
12
%
100%
13%
40%
35%
12%