Monday, 21 February 2022 07:00

## TECHNICAL SCIENCES PAPER 1 GRADE 12 MEMORANDUM - NSC EXAMS PAST PAPERS AND MEMOS MAY/JUNE 2021

TECHNICAL SCIENCES PAPER 1
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
MEMORANDUM
MAY/JUNE 2021

QUESTION 1
1.1 B (2)
1.2 C  (2)
1.3 A  (2)
1.4 D  (2)
1.5 C  (2)
1.6 B  (2)
1.7 C  (2)
1.8 A (2)
1.9 B(2)
1.10 D  (2)
[20]

QUESTION 2
2.1 An object will remain at rest, or continue moving at a constant velocity, unless acted upon by a/an unbalanced/resultant/net force.(2)
2.2 Tension is the force acting in a string or rope.(2)
2.3 Increases(2)
2.4.1 Moves forward.(2)
2.4.2 According to Newton's first law of motion, the book will continue moving forward at a constant velocity (of 30 m s-1)until a net force acts on it.
OR
According to inertia, the book will continue moving forwards at a constant velocity (of 30 m s-1) as it resists a change in its state of motion. (2)
[10]

QUESTION 3
3.1 Normal force is the perpendicular force exerted by a surface on an object that lies /rests on that surface.(2)
3.2 OPTION 1

ACCEPTABLE LABELS:

N/FN: Normal
Fg/w: Force due to gravity/Weight
Ff/fk/f: 5,82 N/friction
FY/FV: Vertical component of FA
FX/FH: Horizontal component of FA
FA: Applied force/50 N
T: Tension/force in the string

OPTION 2

NOTES:

One mark for each force represented by an arrow with a correct label.
Penalise (once) for each of the following:

• No arrows
• There is no dot
• Gap between the line and the dot
• Dotted lines are used
• Additional force is included
• A force diagram is given(5)

3.3 When a net force is applied to an object of mass, m, it accelerates the object in the direction of the net force.  The acceleration is directly proportional to the net force and inversely proportional to the mass. (2)

3.4.1 OPTION 1
N = W - FV
= mg – FASin28º
= 25 x 9,8 – 50Sin28º
= 221,53 N
fk = μkN
5,82 = μk221,53
μk = 0,026

OPTION 2
fk = μkN
= μk(W - FV)
= μk(mg – FASin θ )
= μk(mg – FASin28º)
{5,82 = μk(25 x 9,8 – 50Sin28º)}
μk = 0,026
NOTE:

• 1 mark for fk = μkN
• 1 mark for expanding N to W - FV OR (mg – FASin28)
• 1 mark for correct substitution: 25 x 9,8 – 50Sin28º
• 1 mark for the whole substitution: 5,82 = μk(25 x 9,8 – 50Sin28º)
• 1 mark for the correct answer(5)

3.4.2 OPTION 1
Let east be positive
For block A
Fnet = ma
T + fk + Fx = ma
T + fk + FAcos28 = ma
T – 5,82 – 50cos28 = 25a
T – 49,97 = 25a
a =T - 49,97  ………….. (1)
25
For block B
Fnet = ma
FB + fk + T = ma
350 – 8,35 – T = 45a
341,65 – T = 45a ……… (2)
Substitute a in (2):
341,65 – T = 45(T - 49,97)
25
T = 154,14 N

OPTION 2
Let east be positive
For block A
Fnet = ma
T + fk + Fx = ma
T + fk + FAcos28º = ma
T – 5,82 – 50cos28º = 25a
T – 49,97 = 25a ………… (1)
For block B
Fnet = ma
FB + fk + T = ma
350 – 8,35 – T = 45a ..….. (2)
Add equations (1) and (2)
291,68 = 70a
a = 4,167 m.s-2
T – 49,97 = 25(4,167)
T = 154,15 N (5)
[19]

QUESTION 4
4.1 Momentum is the product of an object's mass and its velocity.(2)

4.2 The total linear momentum of an isolated system remains constant (is conserved)(2)

4.3 Zero
OR
0 N (2)

4.4 OPTION 1
Σpi = Σpf
m1vi1 + m2vi2 = m1vf1+ m2vf2
(1 120)(25) + (m2)(6,25)= (1 120)(7,45) + (m2)(8,45)
(1 120)(25) - (1 120)(7,45) = (m2)(8,45) - (m2)(6,25)
19 656 = 2,2 m2
m2 = 8 934,55 kg
mass of the construction vehicle
= 8 934,55 – 100
= 8 834,55 kg

OPTION 2
Σpi = Σpf
m1vi1 + m2vi2 = m1vf1+ m2vf2
(1 120)(25) + (mcv+100) (6,25) = (1 120)(7,45) + (mcv+100)(8,45)
19 436 = 2,2 mcv
mass of the construction vehicle (mcv)= 8 834,55 kg (5)

4.5 Positive marking from 4.4
ΣEki  = ½m1v21i +  ½m2v22i
= ½(1 120)(25)2 +  ½(8 934,55)(6,25)2
= 524 502,93 J
Σkf E =  ½m1v2if +  ½m2v22f
= ½(1120)(7,45) +  ½(8 934,55 )(8,45)
= 350 056,0 J
ΣEki ≠ ΣEkf
⇒collision was inelastic. (5)
[16]

QUESTION 5
5.1 Impulse is the product of the net/resultant force acting on an object and the time the net/resulatant force acts (on the object).(2)
5.2 OPTION 1
Let left to be positive
Fnet Δt = Δp
Fnet Δt = f i m(v -v )
Fnet(1,28) = 950(- 1,24 – 6)
Fnet = - 5 373,44 N
= 5 373,44 N backwards/to the right

OPTION 2
Let left to be negative
Fnet Δt = Δp
Fnet Δt = f i m(v -v )
Fnet(1,28) = 950{1,24 – (- 6)}
Fnet = 5 373,44 N backwards/to theright (4)
[6]

QUESTION 6
6.1 Work is the product of the force applied on an object and the displacement in the direction of the force. (2)
6.2 OPTION 1
WFg = FgΔ xCosθ
= mgΔ xCosθ
= mgΔ xCos180º
= 0,145 x 9,8 x 0,5 x Cos180º
= 0,145 x 9,8 x 0,5 x -1
= - 0,71 J
WFg = 0,71 J against the motion of the phone

OPTION 2
Work done by Fg (WFg) = - ΔEp
WFg = - mg(hf - hi )
= - (0,145)(9,8)(0,5 - 0)
WFg = - 0,71 J
WFg = 0,71 J against the motion of the phone

OPTION 3
Work done by Fg (WFg) = - ΔEp
ΔEp = mg(hf - hi
ΔEp = (0,145)(9,8)(0,5 - 0)
ΔEp = 0,71 J
WFg = - 0,71 J
ΔWFg = 0,71 J against the motion of the phone
NOTE:
Penalise if the first statement is missing (3)

6.3 Power is the rate at which work is done.
OR
Power is the rate at which energy is expended/transferred.
OR
Power is the rate at which energy changes/transferred.(2)

6.4 OPTION 1
W = FΔxCosθ
= 4,9x0,5xCos 0°
= 2,45 J
P =
Δt
P =2,45
4
P = 0,613 W

OPTION 2
P =
Δt
P =FΔxCosθ
Δt

P = 0,613 W

OPTION 3
Assume that the object moves at constant velocity.
Pave = Fvave
= 4,9(0,5/4)
= 0,613 W
If the assumption is not indicated:3/4marks (4)

### Related Items

6.5 The total mechanical energy in an isolated system remains constant /is conserved.
OR
The sum of gravitational potential energy and kinetic energy in an isolated system remains constant/ is conserved.(2)

6.6 ME at top = ME at bottom
(Ep + Ek) at top = (Ep + Ek) at bottom
(mgh + ½mv2) at top = (mgh + ½mv2) at bottom
(0,145 x 9,8 x 0,5 + ½ x 0,145 x 02) = (0,145 x 9,8 x 0 + ½ x 0,145 x v2)
0,145 x 9,8 x 0,5 + 0 = 0 + ½ x 0,145 x v2
v = 3,13 m. s-1
NOTE: Do not penalize if zero substitution is not shown.(4)
[17]

QUESTION 7
7.1.1 OPTION 1
A= πd 2
4
A= π x (0,08)2
4
A = 5,027 x 10-3m2
σ =F/A
σ =  30 x 10 3
5,027 x 10-3
σ = 5 967 774,02 Pa
Accept:
5 967 774,02 – 6 000 000,00 as a range

OPTION 2
A = πr2
A = π(0,04)2
A = 0,005027/ 0,005 m2
A = 5,027 x 10-3m2
σ =F/A
σ =  30 x 10 3
5,027 x 10-3
σ = 5 967 774,02 Pa
Accept:
5 967 774,02 – 6 000 000,00 as a range

OPTION 3

σ = 5 967 774,02 Pa
Accept:
5 967 774,02 – 6 000 000,00 as a range

OPTION 4

σ = 5 967 774,02 Pa
Accept :
5 967 774,02 – 6 000 000,00 as a range (4)

7.1.2 OPTION 1
ε =Δl/L
ε =  0,4
3000
ε = 1,33 x 10-4 / 0,000133

OPTION 2
ε = Δl/L
ε =0,0004
3
ε = 1,33 x 10-4 / 0,000133
Note: Penalise if unit is included (3)

7.2.1 Perfect plastic body(1)
7.2.2 Perfect plastic body (1)
7.2.3 Perfect elastic body(1)
7.2.4 Perfect plastic body (1)
7.3 As the temperature of the liquid increases its viscosity decreases.
OR
As the temperature of the liquid decreases its viscosity increases. (2)
7.4 Pascal's law states that in a continuous liquid at equilibrium, the pressure applied at any point is transmitted equally to other parts of the liquid.(2)
7.5 ANY TWO

• Hydraulic Car lifts
• Hydraulic Jacks
• Hydraulic brakes
• Dentist chairs
• Forklifts/ Hysers
• Hydraulic press (2)

7.6 OPTION 1
A= πd 2
4
A =π x (0,12)2
4
A = 1,13 x 10-2 m2
F1 = F2
A1    A2
=   2 x 10 3=      F2
2,827x10-3  1,13 x10-2
F2 = 7 994,34 N
Accept:
7 994,34 – 8 001,23 as a range

OPTION 2
A = πr2
A = π(0,06)2
A = 1,13 x 10-2 m2
F1 = F2
A1    A2
=   2 x 10 3=      F2
2,827x10-3  1,13 x10-2
F2 = 7 994,34 N
Accept:
7 994,34 – 8 001,23 as a range

OPTION 3

F2 = 7 994,34 N
Accept :
7 994,34 – 8 001,23 as a range

OPTION 4

F2 = 7 994,34 N
Accept :
7 994,34 – 8 001,23 as a range (5)
[22]

QUESTION 8
8.1 The outer electrons (valence electrons) in silicon are involved in perfect covalent bonds while metals have free (outer) electrons to conduct electricity (2)

8.2 Doping.
It (doping) is the process of adding impurities to intrinsic semiconductors.(3)

8.3. A p-n junction diode is a combination of p-type semiconductor material with n-type semiconductor material( to achieve its practical utility/ to make current to flow in one direction).
NOTE: 2 marks or nothing.(2)

8.4.1
C = Q/V
4 x 10-6 =2/V
V = 500 000 V(3)

8.4.2
C = εοA
d
4 x 10-6 =  8,85 x10-12
0,0008
A = 361,58 m2 (4)
8.4.3 Capacitance is directly proportional to the area of the plates.
Accept:
The capacitance will increase with an increase in the area of the plates.
OR
The capacitance will decrease with an decrease in the area of the plates.(1)
8.5.1

NOTE: Penalise once if the resistors and the cell are not connected in parallel (3)
8.5.2 OPTION 1
P = V2/R
P =2
5
P= 16,2 W
P = V2/R
P =2
12
P= 6,75 W

OPTION 2
I = V/R
9/5
= 1,8 A
P = VI
= 9 x 1,8
= 16,2 W
I = V/R
9/12
= 0,75 A
P = VI
= 9 x 0,75
= 6,75 W

OPTION 3
I = V/R
9/5
= 1,8 A
P = I2R
= 1,82 x 5
= 16,2 W
I = V/R
9/12
= 0,75 A
P = I2R
= 0,752 x 12
= 6,75 W (5)
[23]

QUESTION 9
9.1 Magnetic flux density is the number of field lines (perpendicular) to a given surface
OR
Magnetic flux density is the number of (perpendicular) field lines per unit area. (2)

9.2 OPTION 1
A = l x b
A= 0,09 x 0,15
A= 0,0135 m2
Φ= BA
9 = 0,0135 B
B = 666,67 T

OPTION 2
Φ= BA
Φ = B(lb)
9 = {B(0,09 x 0,15)
B = 666,67 T (4)
[6]

QUESTION 10
10.1 A transformer is a device that is used to step up or step down the voltage. (2)
10.2 Vs  = Ns
Vp     Np
Vs 1500
120     80
=Vs = 2 250 V (3)
10.3.1 Figure 1 – DC generator (1)
10.3.2 Figure 2 – AC generator(1)
10.3.3 A – Commutator
Accept: Split ring (1)
10.3.4 B - Slip ring (1)
10.3.5 Positive marking from question 10.3.3
Commutator maintains electrical contact between the load and the (rotating) coil in a DC generator.
OR
Commutator helps to maintain polarity on a brush (as the shaft rotates through the magnetic field).
OR
Commutator ensures that current flows in one direction.(2)
[11]
TOTAL: 150

Last modified on Friday, 25 February 2022 07:43