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

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

QUESTION 2
2.1

Accepted Labels 
Fg     W/FW/Weight/gravitational force
N       FN/ Fnormal / normal force 
T       FT / Tension
FA     Applied force/ Fapplied/ 24 N (4)
2.2 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)
2.3 For object A
Fnet = ma ✓
F + (− T) = (5)(a) ✓
24 – T = 5a ……………(1)
For object B
F_net = ma
T = 3a … (2) ✓
Substitute (T) in (1)
24 – 3a = 5a
a = 24 = 3 m.s^-2 ✓
      8 
Accept (3 marks)
F_net = ma ✓
F_app = (m_A + m_B)a
24 = ( 5+3) (a) ✓
a = 3 m.s^-2 (4)
2.4 POSITIVE MARKING FROM 2.3
Option 1
From eqn (2)
T= 3a ✓
T = (3) (3) ✓
T = 9 N ✓✓

Option 2
From equation (1)
24 – T = 5a ✓
24 – T = (5)(3) ✓
T = 9 N ✓
(3)
2.5
2.5.1 Decreases✓ (1)
2.5.2 Increases ✓ (1)
[15]

QUESTION 3
3.1
3.1.1 Q. ✓
Gradient of graph Q is greater than that of graph P ✓or Graph Q is steeper than graph P. ✓ (2)
3.1.2 Gradient of graph P:
m_P = 𝛥𝑦
         𝛥𝑥
= 0,4 −0 (Accept any two sets of correct values)
   12−0 ✓
= 0,033 ✓
1𝑚 = 0,033 ✓ ∴ m = 30 kg ✓
(4)
3.2
3.2.1 Force exerted by John on Tom:
F_{netJ on T} = ma ✓
= (45)(0,2) ✓
= 9 N ✓
(3)
3.2.2 Positive marking from 3.2.1
F_{net T on J} = 9 N right ✓
(1)
3.2.3 Newton’s third law.✓
When object A exerts a force on object B, object B simultaneously exerts an opposite directed force of equal magnitude on object A.✓✓ (3)
[13]

QUESTION 4
4.1 Impulse is defined as the product of net force acting on the object and time during which the force is in action Any one Impulse is defined as the change in momentum. ✓✓ (2)
4.2 Impulse = F_netΔt ✓
Impuls = (7)(0,005) ✓
= 0,035 N.m ✓
(3)
4.3 Positive marking from 4.2
Impulse = Δp
= mv_f −mv_i ✓ Any one
0,035 ✓ = 0,005( vf – 0) ✓
vf = 7 m.s^-2 ✓

Option 2
Fnet = ma ✓
7 = (5x10^-3)(a) ✓
a = 1400 m.s^-2
vf = v_i + aΔt
vf = 0 + (1400)(0,005) ✓
vf = 7 m.s^-1 ✓ (4)
[9]

QUESTION 5
5.1 The total linear momentum in an isolated (closed) system remains constant. ✓✓ (2)
5.2 Σpi = Σpf ✓
m_bv_bi + m_bcv_bci = m_bv_bf + m_bcv_bcf
(0,023)(230) + (2)(0) ✓ = (0,023)(170)+ (2)(v_bcf) ✓
vf = 0,69 m.s^-1 in the direction of motion of the bullet ✓
(4)
5.3 Kinetic energy is not conserved as the collision is inelastic. ✓✓ (2)
[8]

QUESTION 6
6.1
6.1.1 The total mechanical energy of an isolated system remains constant. ✓✓ (2)
6.1.2 E_PA = mgh ✓
E_PA = (30)(9,8)(40) ✓
= 11760 J ✓
(3)
6.1.3 (E_p + E_k)A = (E_p + E_k)C ✓
( mgh + ½mv² )A = ( mgh + ½mv² )C
11760 + 0 ✓ = 0 + ½(30)(v²) ✓
v = 28 m.s^-1. ✓
(4)
6.1.4 Positive marking from 6.1.3
Option 1
a = vf−vi
        Δt ✓
a = 0−28
      10 ✓
a = − 78,4 m.s^-2
F_net = ma ✓
F_net = (30)(−78,4)
F_net = − 2352 N ✓

Related Items

Option 2
v_f = v_i + aΔt ✓
(0)² = (28)² + (a)(10) ✓
a = −78,4 m.s^-2
F_net = ma ✓
F_net = (30)(−78,4)
= −2352 N ✓
(4)
6.2.1 Rate at which work is done. 
Rate of transfer of energy. (Any one)
6.2.2 P = Fv ✓
= (10000)(55) ✓
= 550000 W ✓
P = 550000 = 737,27 hp / (pk) ✓✓
         746 
(4)
6.2.3 Ff = 3181 N ✓
Opposite to the direction of motion of the locomotive. (Any one)
Locomotive is moving with constant velocity. ✓ (2)
[21]

QUESTION 7
7.1 Work is defined as the product of the force applied on an object and the displacement in the direction of the force.✓✓ (2)
7.2 fk = μkN =( 0,1)(100)(9,8) = 147 N
F_net = ma ✓
F + f_k = ma
F + (−147) ✓= (100)(0,5) ✓
F = 197 N ✓
(6)
Positive marking from 7.2
7.3 W = FΔcosθ 
W= (197)(5)(cos 0⁰) ✓
W= 985 J ✓
(3)
[11]

QUESTION 8
8.1.150 mm ✓ (1)
8.1.2

Axes ✓
Any two points plotted correctly ✓✓
Correct shape✓ (4)
8.1.3 The maximum force that can be applied to a body so that the body regains its original form completely on removal of the force. (2)
8.1.4 3,7 N ✓✓Accept 3,8 N (2)
8.2
8.2.1 Hooke’s law states that within the limit of elasticity, stress is directly proportional to the strain.✓✓ (2)
8.2.2 Stress: σ = 𝐹 
                           𝐴 ✓
F = mg = (1)(9,8) = 9,8 N ✓
A = πr² = π(0,00018)² = 1,02 x 10^-7 m² ✓
σ =    9,8       ✓ = 9,61x10⁷ N.m² ✓
    1,02𝑥10−7 
(5)
8.2.3
ε = 𝛥𝑙
      𝐿 ✓
ε = 0,0018 ✓= 4,5x10-4 ✓
          4 
(3)
8.2.4 K = 𝜎 
               𝜀 ✓
K = 9,61𝑋10⁷  ✓= 2,14 x10¹¹ Pa ✓
       4,5𝑋10⁻⁴ 
(3)
[22]

QUESTION 9
9.1
9.1.1 Viscosity is the property of the fluid to oppose relative motion between the two adjacent layers.✓✓ (2)
9.1.2 Viscosity decreases with increase in temperature.✓✓ (2)
9.1.3 This means that the oil acts as an SAE 20 when it is cold, for easy cold starting ✓ and it acts like an SAE 50 oil (which is thicker than SAE 20) when it is hot. ✓ (2)
9.2
9.2.1 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)
9.2
9.2.2  F₁ = F₂ 
          A₁    A₂ ✓
1176 =  F 
0,03    0,5 ✓
F = 19600 N ✓
(3)
9.2.3 F = mg
19600 = m(9,8) ✓
m = 2000 kg ✓
(2)
9.2.4 W = FΔxcosθ ✓
W = (19600) (1,2) cos0 ✓
= 23520 J ✓
(3)
9.2.5 Hydraulic braking system (Any two)
Hydraulic jack
Dentist chair ✓✓
(2)
[18]

QUESTION 10
10.1 A semiconductor is a material which has electrical conductivity between that of a conductor and an insulator.✓✓ (2)
10.2.1 A pure semiconductor is called an intrinsic semiconductor. ✓✓(2)
10.2.2 Doping is a process of adding impurities to a pure semiconductor.✓✓(2)
10.2.3 n-type ✓(1)
10.2.4 Electron ✓ (1)
10.3.1
(2)
10.3.2

• the free electrons in the n-type material will be repelled by the negative terminal voltage supply.
• the electrons will move and break through the junctions and start flowing towards the positive terminal and it seems as if the holes are moving towards negative terminal.
• the junction region becomes smaller and the break over voltage is overcome and the diode will begin to conduct.
  (3)

[13]
TOTAL: 150