TECHNICAL SCIENCES PAPER 1
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
SEPTEMBER 2018

QUESTION 1:
(MULTIPLE-CHOICE QUESTIONS)
1.1 C ✓✓ (2)
1.2 B✓✓ (2)
1.3 B ✓✓ (2)
1.4 C ✓✓ (2)
1.5 B ✓✓(2)
1.6 C ✓✓ (2)
1.7 D✓✓ (2)
1.8 B ✓✓(2)
1.9 A ✓✓(2)
1.10 D ✓✓ (2)
[20]

QUESTION 2
2.1 When a non-zero resultant force acts on an object, the object will accelerate in the direction of the resultant force. This acceleration is directly proportional to the force and inversely proportional to the mass of the object. ✓✓ (2)
2.2 Tension (T) ✓
Applied force (F) ✓
Force of friction ✓ (Any TWO) (2)
2.3 

(5)
2.4 On block A
F_net = ma ✓
Fcosθ – T – f_kA = ma
(500)(cos30º) – T – 8.8 ✓ = (30)(a)✓
424.21 – T = 30a ………(1)

On block B
Fnet = ma
T- fkB = 20a ✓
T- 39.4 = 20a
T = 39,4 + 20a …………(2)

Subs (2) in (1)
424,21 – (39,4 + 20a) = 30a ✓
a = 7,70 m.s^-2 ✓ (6)
2.5

(3)
2.6 Newton’s Third Law ✓
When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A. ✓✓ (3)
[21]

QUESTION 3
3.1 It is the product of mass and velocity. ✓✓ (2)
3.2 The total linear momentum in an isolated system remains constant. ✓✓ (2)
3.3 OPTION 1
East is taken as positive
Σpi = Σpf ✓
mCvCi + mTvTi = mCvCf + mTvTf
(800)(33,33) ✓+ (2500)(- 19,44) ✓ = (800+2500) (vf) ✓✓
vf = - 6,65 m.s^-1
vf = 6,65 m.s^-1 due west.✓ (5)

OPTION 2
East is taken as positive.
Σpi = Σpf ✓✓
mCvCi + mTvTi = mCvCf + mTvTf
(800)(120) ✓+ (2500)(-70) ✓ = (800+2500) (vf) ✓✓
vf = - 23,94 km.h^-1
vf = 23,94 km.h^-1 due west. ✓
(6,65 m.s^-1)
3.4 Consider the formula: Fnet = Δp 
                                                     Δt
The change in momentum (Δp) remains constant during a collision. ✓
If the contact time (Δt) is increased, the net force decreases.✓
The airbag increases the contact time, thereby decreasing the force acting on the driver ✓ and thereby reducing the extent of injuries. (3)
[12]

QUESTION 4
4.1 Work is the product of the force applied on an object and the displacement in the direction of force. ✓✓ (2)
4.2 W = FΔxcosθ ✓
W = (70)(5)(cos20º) ✓
W = 328.89 J ✓ (3)
4.3 Greater than ✓ (1)
[6]

QUESTION 5
5.1 Rate at which energy is transferred/Rate at which work is done. ✓✓(2)
5.2 P = Fv ✓
P = (100)(5) ✓
P= 500 W ✓ (3)
5.3 At point A ✓✓ (2)
5.4 (Ep + Ek)A = (Ep + Ek)B ✓
(mgh + ½mv²)A = (mgh + ½mv²)B
(10)(9,8)(10) + 0 ✓ = 0 + ½(10)(v²) ✓
v = 14 m.s^-1. ✓ (4)
5.5 Law of conservation of mechanical energy. ✓
The total mechanical energy of an isolated system remains constant. ✓✓✓(3)
[14]

QUESTION 6
6.1 6.1.1 1,2 cm ✓ (1)
6.1.2 Force applied for the length of 1,7 cm = 4,2 N ✓✓
F= mg✓
4,2 = m(9,8) ✓
m = 0,43 kg ✓(4)
6.1.3 Strain is defined as the ratio of change in dimension to the original dimension. ✓✓ (2)
6.1.4
ε=Δl
    L
ε= 0,02
     0.15
= 0,13 ✓ (4)
6.2 6.2.1 Hooke’s law states that within the limit of elasticity, stress is directly proportional to the strain. ✓✓ (2)
6.2.2
K = σ 
      ε
2×10¹¹ =     σ     
           3,6×10^-4
σ =7,2 x 10⁷ Pa✓ (3)
[16]

QUESTION 7
7.1 7.1.1 Pressure at a particular point is the thrust acting on the unit area around that point. ✓✓ (2)
7.1.2 Less than.✓
Pressure increases with depth/ Druk neem toe met diepte. ✓(3)
7.1.3 Pressure of water at point 2 is higher than pressure of water at point 1. ✓✓(2)
7.2 7.2.1 Pascal’s law states that in a continuous liquid at equilibrium, the pressure applied at any point is transmitted equally to the other parts of the liquid. ✓✓ (2)
7.2.2
F₁ = F₂
A₁    A₂

   F    = (1 200)(9,8)
0,001         0,06
F = 196 N ✓ (4)
[13]

QUESTION 8
8.1 A capacitor is a device for storing electrical charge. ✓✓(2)
8.2 The high voltage across plates can cause electric shock or even death when the capacitor discharges. ✓✓(2)
8.3
8.3.1
C =   ε₀A  
         d
C = (8,85×10^-12)(2×10^-2)
               0,00003
C = 5,9 x 10^-11 F✓ (3)
8.3.2 N.B. Positive marking from 8.3.1
Q = CV ✓
Q = (5,9 x 10^-11)(6) ✓
Q = 3,54 x 10^-10 C ✓ (3)
[10]

QUESTION 9
9.1
9.1.1 5 ✓ (1)
9.1.2 Electrons or negative charges ✓ (1)
9.2
(2)
9.3
9.3.1 0,5 V ✓✓ (2)
9.3.2 Decreases ✓✓ (2)
[8]

QUESTION 10
10.1 Energy transferred to the kettle per second is 1500 J when operating at 240 V. ✓✓(2)
10.2
P =  V²
       R
1 500 = (240)²
                R
R = 38,4 Ω  (4)
10.3 W = Pt✓
= (1500)(3)(60) ✓✓
= 270x10³ J ✓ (4)
[10]

QUESTION 11
11.1 11.1.1 If transformer defined as a device used to step up or step down the voltage. ✓✓ (2)
11.1.2
V_S = N_S
V_p      N_p
 12  = 300
240     N_p
N_p = 6 000 ✓ (3)
11.1.3 Step down ✓ (1)
11.2 11.2.1 If the magnetic field linked with the coil changes, the emf induced is directly proportional to the rate of change of magnetic flux. ✓✓(2)
11.2.2
ε = - N ΔΦ 
            Δt
ε = - 11  2,7- 5,34
                  12
ε = 2,42 V ✓ (3)
[11]

QUESTION 12
12.1 AC generator ✓ (1)
12.2 Mechanical energy to electrical energy. ✓(1)
12.3
12.3.1 Slip rings ✓ (1)
12.3.2 Brushes ✓ (1)
12.4
(3)
Axes ✓
One complete cycle ✓
Shape ✓
12.5 Electroplating plants. ✓
Locomotives.✓
Ships ✓ (2)
(Any TWO)
[9]

TOTAL: 150