QUESTION 1


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

QUESTION 2

2.1 An object will remain at rest or continue moving at a constant velocity (or at constant speed in a straight line) ✓ unless acted upon by a non-zero external resultant force. ✓(2)
2.2.1

 OPTION 1   OPTION 2 
 Fy = 60.Sin30°✓
= 30 N✓
 F= 60.Cos60°✓
= 30 N✓

(2)

2.2.2

 OPTION 1  OPTION 2
 Apply positive marking from 2.2.1
fk= µk N✓
= 0,13 [(6x9,8) – 60.Sin30°] ✓
= 3,74 N (to the right) ✓/na regs
NOTE: Credit if 60.Sin30°
is
expressed as 30

Apply positive marking from 2.2.1
N = mg - Fy
N = (6x9,8) – 60.Sin30° ✓
= 58,8 - 30
= 28,8 N
fk= µk N ✓
= 0,13 x 28,8
= 3,74 N (to the right) ✓/na regs
NOTE: Credit if 28,8N is expressed
as 58,8 - 30

 (4)

2.2.3

 OPTION 1   OPTION 2 
 Fx = 60.Cos30° ✓
= 51,96 N✓
 Fx = 60.Sin60°✓
= 51,96 N✓ 

(2)

Positive marking from 2.2.2 and 2.2.3
2.3 Fnet = ma
      Fx + fk = ma
       60Cos30° + (-3,74)✓ = 6.a ✓
       a = 8,04 m.s-2 to the left ✓/na links (4)

2.4 Decrease✓
The vertical component (Fy) will increase and thus the normal force will decrease. ✓✓
OR
The force will tend to lift the object from the surface and thus decrease the friction. (3)

2.5.1 When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A. ✓✓
NOTE:Credit one mark if any of the key words is ommitted (1/2) (2)

2.5.2

TS Nov 2021 Grade 12 PP1

 ACCEPTABLE LABELS
 NOTES:
 N/FN: Normal/Normaal
Fg/w: Force due to
gravity/Weight/

F/FA/Fcar: Applied force/

f k/Ff/ f: frictional force/

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

volgende:
  • No arrows
  • There is no dot
  • Gap between the line and the dot
  • Dotted lines are used
  • A force diagram is given
  • Extra force is given

 (4)
[23]

QUESTION 3

3.1.1 Product of the mass of an object and its velocity. ✓✓(2)
3.1.2 pL= mv✓
             = 5800 x 1,5✓
             = 8700 kg.m.s-1 west✓ (3)
3.1.3 Apply positive marking from 3.1.2
∑ p before = ∑ p after
mLvL + mWvW = mC vC
mLvL + mWvW = (mL +mw) vC
5800 x 1,5 + 2500 x 0 ✓ = 8300 x vC
∴ vC = 1,05 m.s-1✓ west✓/                  (5)

3.1.4 During elastic collision, the total kinetic energy is conserved and the total linear momentum is conserved ✓✓and during inelastic collision, total kinetic energy is not conserved and the total linear momentum is conserved. ✓✓
Accept/Aanvaar : ∑ pbefore/voor = ∑ pafter/na and ∑ Ekbefore/voor = ∑ Ekafter/na
                            ∑ pbefore/voor = ∑ pafter/na and ∑ Ekbefore/voor ∑ Ekafter/na
NOTE: Do not penalise if total linear momentum is ommitted
         : Penalise one mark if the word ‘total’ is ommitted for elastic and inelastic collision (4)

3.2.1

  • If the vehicle collides or come to a standstill, the driver and passengers would contiue moving at the initial velocity. ✓
  • Safety belts will then prevent them from moving forward✓ and hurting themselves and others or even going through the windscreen.(2)

3.2.2

 OPTION  OPTION
 Let the direction towards the wall be
positive
Fnet∆t = Impulse
Impulse = ∆p
Impulse = 0 – 24300 ✓
Impulse = - 24300
Impulse = 24300 kg.ms-1✓(away
from the wall)
 Let the direction towards the wall be
negative
Fnet∆t = Impulse
Impulse = ∆p
Impulse = 0 – (- 24300) ✓
Impulse = 24300 kg.ms-1 ✓ (away
from the wall)

 (3)

Positive marking from 3.2.2
3.2.3
Fnet∆t = Impulse✓
Fnet x 1,2 = 24300✓
Fnet = 20 250 N
Force exerted by impulse/Krag deur impuls uitgeoefen = 20 250 N✓
The wall can withstand 80 000 N, so it will withstand the impact of the test. ✓/

NOTE: If Fnet is more than 80 000N then it will NOT withstand the impact of the test (resulting from positive marking). (4)
[23]

QUESTION 4

4.1.1 The product of the force applied on an object and the displacement in the direction of the force. ✓✓ (2)

4.1.2
W = FA ∆ x cosθ✓
W = 60 x 8 x cos 25°✓
W = 435,03 J✓ (3)
4.2.1 The total mechanical energy ✓of an isolated system is constant. ✓
OR
The sum of the gravitational potential energy and kinetic energy in an isolated system remains constant. ✓(2)

4.2.2
Ep = mgh ✓
Ep = 75 x 9.8 x 12✓
Ep = 8 820 J ✓
(Accept : 8,820 kJ/8,82 x 103J)      (3)

4.2.3
EK = ½ mv2
     = 0.5 x 75 x 32
      = 337,5 J✓   (3)

Positive marking from 4.2.3
4.3

MEtop = Ektop + Eptop 
11500 = 337,5 + Eptop ✓
Eptop = 11 162,5 J
Epbefore = Eptop −  Epground
                = 11 162,5 – 8820✓
                = 2342,5 J✓ (4)
[17]

QUESTION 5

5.1.1

TS Nov 2021 Grade 12 PP1 Ans 5

5.1.2

OPTION 1  OPTION 2 
ε =   Δℓ  
        L
ε =       0.5     
       3 x 103
ε = 1,666 x 10-4
NOTE: Penalise if the unit is given.
ε =   Δℓ  
        L
ε  = 5 x 10-4
           3 
ε = 1,666 x 10-4
NOTE: Penalise if the unit is given.

(3)

Positive marking from 5.1.1 and 5.1.2.

5.1.3
K = σ/ε 
K =   3.26 x 106
      1,666  x-4
K = 1,956 x 1010 Pa✓ (3)

5.2.1 Pressure at a particular point is the thrust acting on the unit area around that point. ✓✓

NOTE: Do not penalise if force is used instead of thrust (2)

5.2.2
P=F/A
P =           26           
          7,855 x 10-5
P= 3,3099 x 105 Pa ✓ (3)

5.2.3

OPTION 1 OPTION 2 
F1/ A1 = F2/A2
        26        = 1278
7855 x 10-5       A2 
A2 = 3,861 x 10-3 m2
 A2 = 3,861 x 10-3 m
P = F2/A2
33099 x 105 = 1278/A2
A2 = 3861 x 10-3m2

 


QUESTION 6
6.1 The bending of light when it passes from one medium to another (of different optical densities).✓✓  (2)
6.2 Critical angle✓/Kritieke hoek/Grenshoek (1)
6.3 90º✓ (1)
6.4 Medium 1✓/

Light ray QS bends away from the normal.✓(2)

6.5 QR✓ (1)
6.6 The light must travel from a more optically dense to a less optically dense medium.✓

The incident angle should be greater than the critical angle. ✓  (2)
[9]

QUESTION 7

7.1.1 The phenomenon whereby white light break up (spread out) into its component colours.✓✓  (2)
7.1.2

3. yellow✓/geel
6. Indigo✓ (2)

7.1.3 Refraction ✓/Refraksie/Breking (1)

7.1.4 When the wavelength increases the speed of the waves will also increase. ✓✓

Accept: Wavelength is directly proportional to the speed of the wave. (2)

7.2.1 A succession/repetition of pulses✓✓/

OR

A disturbance that transfers energy through matter or space./' (2)

7.2.2

  • can propogate in a vacuum✓/kan in 'n vakuum voortplant
  • move at the speed of light (3 x 108 m.s -1 ) ✓
    (3 x 108 m.s -1 )
  •  transfer energy
  • have a dual nature (particle and wave) nature
  • can be polarized  (any two) (2)

7.2.3

Radio-wave Micro-wave   Infrared Visible
light
Ultraviolet    X-rays Gamma
rays 


NOTE: 2 or zero(2)
[13]

QUESTION 8

8.1 Capacitance is the amount of charge a capacitor can store✓ per volt. ✓ (2)
8.2.1

C = εoA
        d
(8,85 x 10‾ ¹²)(2)
     6 x 10-3

= 2,95 x 10-9
F ✓ (3)
Positive marking from 8.2.1/Positiewe nasien vanaf 8.2.1.

8.2.2

C =Q/V
2,95 10-9 = Q/120
Q = 3,54 x 10-7 C ✓
(3)
[8]

QUESTION 9

9.1 It is the rate at which electrical energy is converted in an electric circuit. ✓✓ (2)
9.2

tsci p1 9.2
(3)

9.3

tsci p1 9.3
(5)
[10]


QUESTION 10

10.1.1 Electromagnetic induction✓(1)
10.1.2 When the speed at which the magnet is moved in and out of the coil is increased the rate of change in the magnetic flux increases/ The rate of
change of magnetic flux is directly proportional to the induced emf.✓
The induced emf/ the extent of deflection of the needle increase with the increase in change of the magnetic flux. ✓ 2)

10.1.3 Alternating current✓ (1)
10.2.1

  • The primary voltage is higher (220 V) than the secondary voltage (24 V) ✓
  • More windings on primary than on secondary coil✓
    OR
  • The secondary voltage is lower (24 V) than the primary voltage (220 V)
  • The windings on the secondary coil are fewer than that of the primary coil.
     (2)

10.2.2

OPTION 1  OPTION 2
Vs/Vp = Ns/ Np
   
24/220= 480/Np

Np = 4400 windings✓
Ratio: Vs : Vp
24 : 220✓
 ∴Ns : Np
480 : 4400 windings✓
NOTE: Maximum mark is 2/3 (for
OPTION 2)

(3)
[9]

TOTAL/TOTAAL: 150

Last modified on Wednesday, 07 December 2022 09:06