PHYSICAL SCIENCES: PHYSICS (PAPER 1)
GRADE 12
NOVEMBER 2019
MEMORANDUM
NATIONAL SENIOR CERTIFICATE

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

QUESTION 2
NOTE: -1 mark for each key word/phrase omitted in the correct context (2)
2.1 When a resultant/net force acts on an object, the object will accelerate in the direction of the force with an acceleration that is 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 

Notes

• Mark is awarded for label and arrow. 
• Do not penalise for length of arrows.
• Deduct 1 mark for any additional force. 
• If force(s) do not make contact with body/dot: Max: 43
• If arrows missing/Indien pyltjies uitgelaat is: Max./Maks: 43 (4)

2.3

(8)
2.4 Greater than 
Fnet increases. 

ACCEPT
There is no friction.
OR
The surface is smooth (2)
[16]

QUESTION 3
3.1 (Motion during which) the only force acting is the force of gravity. ✓✓ (2 or 0) (2)
3.2 Marking criteria:

• Any appropriate formula for Δy
• Whole substitution to calculate 5,1 m
• 40 + answer from calculation
• Final answer: 45,10 m ✓(Accept/aanvaar 45,1 m)

OPTION 1
UPWARDS AS POSITIVE:
vf² = vi² + 2aΔy ✓
0 = (10)² + (2)(- 9,8)Δy ✓
Δy = 5,10 m (5,102 m)
Height = 40 + 5,10 ✓
= 45,10 m ✓

DOWNWARDS AS POSITIVE:
vf² = vi² + 2aΔy✓
0 = (-10)2 + (2)( 9,8)Δy ✓
Δy = - 5,10 m (5,102)
Height = 40 + 5,10 ✓
= 45,10 m ✓

OPTION 2
UPWARDS AS POSITIVE:
vf = vi + aΔt
0 = (10) + (- 9,8)Δt
Δt = 1,02 s
Δy = viΔt + ½ aΔt^2✓
= (10)(1,02) + ½ (-9,8)(1,02)^2✓
= 5,10 m

OR
Δy=(v₁  + v_f)Δt
           2
=(10 + 0)(1,02)
       2
= 5.10m
Height = 40 + 5,10 ✓
= 45,10 m ✓

DOWNWARDS AS POSITIVE:
vf = vi + aΔt
0 = (-10) + ( 9,8)Δt
Δt = 1,02 s
Δy = viΔt + ½ aΔt² ✓
= (-10)(1,02) + ½ (9,8)(1,02)² ✓
= 5,10 m
Height = 40 + 5,10 ✓
= 45,10 m ✓
Accept swopping of vi and vf

OPTION 3
UPWARDS AS POSITIVE:
Δy = viΔt + ½ aΔt²
0 = (10) Δt + ½(-9,8)Δt²
Δt = 2,04 s
Δy = viΔt + ½ aΔt^2✓
= (10)(1,02) +½ (-9,8)(1,02)^2✓
= 5,10 m
Height = 40 + ✓5,10
= 45,10 m ✓

DOWNWARDS AS POSITIVE:
Δy = viΔt + ½ aΔt²
0 = (-10) Δt + ½(9,8)Δt²
Δt = 2,04 s
Δy = viΔt + ½ aΔt^2✓
= (-10)(1,02) + ½ (9,8)(1,02)^2✓
= - 5,10 m
Height = 40 + 5,10 ✓
= 45,10 m ✓

OPTION 4
UPWARDS AS POSITIVE:
Δy = viΔt + ½ aΔt²
0 = (10)Δt + ½(-9,8)Δt²
Δt = 2,04s
Δy=(v₁  + v_f)Δt
           2
=(10 + 0)(1,02)
       2
= 5.10m
Height = 40 + 5,10 ✓
= 45,10 m ✓

DOWNWARDS AS POSITIVE:
Δy = viΔt + ½ aΔt²
0 = (-10) Δt + ½(9,8)Δt²
Δt = 2,04 s
Δy=(v₁  + v_f)Δt
           2
=(10 + 0)(1,02)
       2
= 5.10m
Height = 40 + 5,10 ✓
= 45,10 m ✓
Accept swopping of vi and vf

OPTION 5
E_(mech)roof = E_(mech)top
(Ep + Ek)_roof = (Ep + Ek)top
(mgh + ½ mv²)roof/dak = (mgh + ½ mv²)top
[m(9,8)(0) + ½m (10)² = m(9,8) (h)+ 0)] ✓
h = 5,10 m
Height = 40 + 5,10 ✓
= 45,10 m✓

OPTION 6
Wnet = ΔEk ✓
wΔxcosθ = ½ mvf² – ½ mvi²
(m)(9,8)Δxcos180º= 0 – ½ m(10)^2✓
Δx = 5,10 m
Height = 40 + 5,10 ✓
= 45,10 m ✓

OPTION 7
Wnc = ΔEp + ΔEk✓
0 = m(9,8)(hf – 0) + ½ m( 0 – 10²) ✓
hf = 5,10 m
Height = 40 + 5,10✓
= 45,10 m✓

OPTION 8
Marking criteria:

• Appropriate formula
• Substitution left
• Substitution right
• Final answer: 45,10 m✓
  Any one

E_(mech)roof = E_(mech)top
(Ep + Ek)_roof = (Ep + Ek)top
(mgh + ½ mv²)roof/dak = (mgh + ½ mv²)top
[m(9,8)(0) + ½m (10)² = m(9,8) (h)+ 0)] ✓
h = 45,10 m (4)
3.3 9,8 m·s^-2 ✓downwards✓ (2)
3.4 Marking criteria

• Calculation/use of 10,26 m.
• Appropriate formula to calculate Δt
• Substitution for stone A ✓
• Substitution for stone B ✓
• Calculating time difference between two stones. ✓
• Final answer: 1,34 (s) ✓

OPTION 1
UPWARDS AS POSITIVE:
Displacement from roof to meeting point = -40 + 29,74 = - 10,26 m
Stone A
ΔyA = viΔt + ½aΔt² ✓
-10,26 ✓= 10Δt + ½(-9,8)Δt² ✓
Δt = 2,79 s
vf² = vi² + 2aΔy
= 0² + 2(-9,8)(-10,26)
vf = -14,18 m·s^-1
vf = vi + aΔt
-14,18 = 0 + (-9,8)Δt✓
Δt = 1,45 s
x = 2,79 – 1,45 ✓
= 1,34 (s) ✓

Stone B
ΔyB = viΔt + ½aΔt²
-10,26 = 0 + ½(-9,8)Δt² ✓
Δt = 1,45 s (1,447 s)
x = 2,79 – 1,45✓= 1,34 (s) ✓
OR
[-10,26 = 0(2,79 – x) + ½ (-9,8)(2,79 – x)^2✓]✓
x = 1,34 (s)✓

DOWNWARDS AS POSITIVE:
Displacement from roof to meeting point= 40 - 29,74 = 10,26 m
Stone A
Δy_A = viΔt + ½aΔt² ✓
10,26 = -10Δt + ½(9,8)Δt² ✓
Δt = 2,79 s
vf² = vi² + 2aΔy
= 0² + 2(-9,8)(-10,26)
vf = -14,18 m·s^-1
vf = vi + aΔt
-14,18 = 0 + (-9,8)Δt✓
Δt = 1,45 s
x = 2,79 – 1,45 ✓
= 1,34 (s)✓

Stone B
ΔyB = viΔt + ½aΔt²
10,26 = 0 + ½(9,8)Δt² ✓
Δt = 1,45 s (1,447 s)
x = 2,79 – 1,45 ✓ = 1,34 (s) ✓
OR
[-10,26 = 0(2,79 – x) + ½ (-9,8)(2,79 – x)^2✓]✓
x = 1,34 (s) ✓

OPTION 2
UPWARDS AS POSITIVE:
Displacement from roof to meeting point = - 40 + 29,74 = - 10,26 m
Displacement of stone A from max height to meeting point = -15,36 m

Stone A
vf = vi + aΔt
0 = 10 +(-9,8)Δt
Δt = 1,02 s
ΔyA = viΔt + ½aΔt² ✓
-15,36 = 0 + ½(-9,8)Δt² ✓
Δt = 1,77 s
Δt_tot = 1,77 + 1,02 = 2,79s

Stone B
ΔyB = viΔt + ½aΔt²
-10,26 ✓= 0 + ½(-9,8)Δt² ✓
Δt = 1,45 s (1,447 s)
x = 2,79 – 1,45 ✓ = 1,34 (s) ✓

vf² = vi² + 2aΔy
= 0² + 2(-9,8)(-10,26)
vf = -14,18 m·s^-1
vf = vi + aΔt
-14,18 = 0 + (-9,8)Δt✓
Δt = 1,45 s
x = 2,79 – 1,45 ✓
= 1,34 (s) ✓

DOWNWARDS AS POSITIVE:
Displacement from roof to meeting point = 40 - 29,74 = 10,26 m ✓
Displacement of ball A from max height to meeting point = 15,36 m

Stone A
vf = vi + aΔt
0 = -10 + (9,8)Δt
Δt = 1,02 s
ΔyA = viΔt + ½aΔt² ✓
15,36 = 0 + ½(9,8)Δt² ✓
Δt = 1,77 s
Δttot = 1,77 + 1,02 = 2,79s

Stone B
ΔyB = viΔt + ½aΔt²
10,26 = 0 + ½(9,8)Δt² ✓
Δt = 1,45 s (1,447 s)
x = 2,79 – 1,45 ✓ = 1,34 (s) ✓

vf² = vi² + 2aΔy
= 0² + 2(-9,8)(-10,26)
vf = -14,18 m·s^-1
vf = vi + aΔt
-14,18 = 0 + (-9,8)Δt✓
Δt = 1,45 s
x = 2,79 – 1,45 ✓
= 1,34 (s) ✓

OPTION 3
UPWARDS AS POSITIVE:
Displacement of stones A and B from roof to meeting point = - 40 + 29,74 = - 10,26 m

Stone A
vf = vi + aΔt
0 = 10 + (-9,8)Δt
Δt = 1,02 s
ΔyA = viΔt + ½aΔt^2✓
- 10,26 = -10 + ½(-9,8) Δt² ✓
Δt = 0,75 s
Δt_tot = 1,02 + 1,02 + 0,75 = 2,79 s

Stone B
ΔyB = viΔt + ½aΔt²
-10,26 = 0 + ½(-9,8)Δt² ✓
Δt = 1,45 s (1,447 s)
x = 2,79 – 1,45 = 1,34 (s) ✓

vf² = vi² + 2aΔy
= 0² + 2(-9,8)(-10,26)
vf = -14,18 m·s^-1
vf = vi + aΔt
-14,18 = 0 + (-9,8)Δt✓
Δt = 1,45 s
x = 2,79 – 1,45 ✓
= 1,34 (s)✓

DOWNWARDS AS POSITIVE:
Displacement of stones A and B from roof to meeting point = 40 - 29,74 = 10,26 m

Stone A
vf = vi + aΔt
0 = -10 + (9,8)Δt
Δt = 1,02 s
ΔyA = viΔt + ½aΔt² ✓
10,26 ✓= 10 + ½ (9,8)Δt² ✓
Δt = 0,75s
Δt_tot = 1,02 + 1,02 + 0,75 = 2,79 s

Stone B
ΔyB = viΔt + ½aΔt²
10,26 = 0 + ½(9,8)Δt² ✓
Δt = 1,45 s (1,447 s)
x = 2,79 – 1,45 ✓= 1,34 (s) ✓

vf² = vi² + 2aΔy
= 0² + 2(-9,8)(-10,26)
vf = -14,18 m·s^-1
vf = vi + aΔt
-14,18 = 0 + (-9,8)Δt✓
Δt = 1,45 s
x = 2,79 – 1,45✓
= 1,34 (s)✓

OPTION 4
UPWARDS AS POSITIVE:
Displacement from roof to meeting point = - 40 + 29,74 = - 10,26 m

Stone A
ΔyA = viΔt + ½aΔt²
- 5,10 = 0 + ½(-9,8)Δt²
Δt = 1,02 s
ΔyA = viΔt + ½aΔt2 ✓
- 10,26 = -10 + ½(-9,8)Δt² ✓
Δt = 0,75s
Δt_tot = 1,02 + 1,02 + 0,75 = 2,79 s

Stone B
ΔyB = viΔt + ½aΔt²
-10,26 = 0 + ½(-9,8)Δt² ✓
Δt = 1,45 s (1,447 s)
x = 2,79 – 1,45 ✓= 1,34 (s) ✓

vf² = vi² + 2aΔy
= 0² + 2(-9,8)(-10,26)
vf = -14,18 m·s^-1
vf = vi + aΔt
-14,18 = 0 + (-9,8)Δt✓
Δt = 1,45 s
x = 2,79 – 1,45 ✓
= 1,34 (s)✓

DOWNWARDS AS POSITIVE:
Displacement from roof to meeting point = 40 - 29,74 = 10,26 m

Stone A
ΔyA = viΔt + ½aΔt²
5,10 = 0 + ½(9,8)Δt²
Δt = 1,02 s
ΔyA = viΔt + ½aΔt² ✓
10,26 = 10 + ½(9,8)Δt² ✓
Δt = 0,75s
Δt_tot = 1,02 + 1,02 + 0,75 = 2,79 s

Stone B
ΔyB = viΔt + ½aΔt²
10,26 = 0 + ½(9,8)Δt² ✓
Δt = 1,45 s (1,447 s)
x = 2,79 – 1,45 = 1,34 (s)✓

vf² = vi² + 2aΔy
= 0² + 2(-9,8)(-10,26)
vf = -14,18 m·s^-1
vf = vi + aΔt
-14,18 = 0 + (-9,8)Δt✓
Δt = 1,45 s
x = 2,79 – 1,45 ✓
= 1,34 (s) ✓

OPTION 5
UPWARDS AS POSITIVE:
Displacement from roof to meeting point = - 40 + 29,74 = - 10,26 m
Displacement of stone A from max height to meeting point = -15,36 m

Stone A
vf² = vi² + 2aΔy
vf² = (0)2 + (2)(- 9,8)(-15,36)
vf = - 17,35 m·s^-1
vf = vi + aΔt
-17,35 = 0 + (-9,8)Δt ✓
Δt = 1,77 s
Δt_tot = 1,02 + 1,77 = 2,79 (s)

Stone B
ΔyB = viΔt + ½aΔt²
-10,26 = 0 + ½(-9,8)Δt² ✓
Δt = 1,45 s (1,447 s)
x = 2,79 – 1,45  = 1,34 (s) ✓

vf² = vi² + 2aΔy
= 0² + 2(-9,8)(-10,26)
vf = -14,18 m·s^-1
vf = vi + aΔt
-14,18 = 0 + (-9,8)Δt✓
Δt = 1,45 s
x = 2,79 – 1,45 ✓
= 1,34 (s)✓

DOWNWARDS AS POSITIVE:
Displacement from roof to meeting point = 40 - 29,74 = 10,26 m
Displacement of stone A from max height to meeting point = 15,36 m

Stone A
vf² = vi² + 2aΔy
vf² = (0)2 + (2)(9,8)(15,36)
vf = - 17,35 m·s^-1
vf = vi + aΔt
17,35 = 0 + (9,8)Δt ✓
Δt = 1,77 s
Δt_tot = 1,02 + 1,77 = 2,79 (s)

Stone B
ΔyB = viΔt + ½aΔt²
10,26 = 0 + ½(9,8)Δt² ✓
Δt = 1,45 s (1,447 s)
x = 2,79 – 1,45 ✓ = 1,34 (s) ✓

vf² = vi² + 2aΔy
= 02 + 2(9,8)(10,26)
vf = 14,18 m·s^-1
vf = vi + aΔt
14,18 = 0 + (9,8)Δt✓
Δt = 1,45 s
x = 2,79 – 1,45 ✓
= 1,34 (s) ✓

OPTION 6
UPWARDS AS POSITIVE:
Displacement from roof to meeting point/Verplasing vanaf dak tot by ontmoetingspunt = - 40 + 29,74 = - 10,26 m
Stone/Klip A
vf² = vi2 + 2aΔy
vf² = (-10)2 + (2)(- 9,8)(-10,26)
vf = - 17,35 m·s^-1
vf = vi + aΔt
-17,35 = -10 + (-9,8)Δt ✓
Δt = 0,75 s
Ball A: Δt = 1,02 + 1,02 + 0,75 = 2,79 (s)

Stone B
ΔyB = viΔt + ½ aΔt²
-10,26 ✓= 0 + ½ (-9,8)Δt² ✓
Δt = 1,45 s (1,447 s)
x = 2,79 – 1,45 = 1,34 (s)✓
vf² = vi² + 2aΔy
= 02 + 2(-9,8)(-10,26)
vf = -14,18 m·s^-1
vf = vi + aΔt
-14,18 = 0 + (-9,8)Δt✓
Δt = 1,45 s
x = 2,79 – 1,45 ✓
= 1,34 (s) ✓

DOWNWARDS AS POSITIVE:
Displacement from roof to meeting point = 40 - 29,74 = 10,26 m
Stone/Klip A
vf² = vi2 + 2aΔy
vf² = [(10)2 + (2)(9,8)(10,26)
vf = 17,35 m·s^-1
vf = vi + aΔt
17,35 = 10 + (9,8)Δt ✓
Δt = 0,75 s
Ball A: Δt = 1,02 + 1,02 + 0,75 = 2,79 (s) (6)

Stone B
ΔyB = viΔt + ½aΔt²
10,26 = 0 + ½(9,8)Δt^2✓
Δt = 1,45 s (1,447 s)
x = 2,79 – 1,45 ✓ = 1,34 (s) ✓
vf² = vi² + 2aΔy
= 02 + 2(9,8)(10,26)
vf = 14,18 m·s^-1
vf = vi + aΔt
14,18 = 0 + (9,8)Δt✓
Δt = 1,45 s
x = 2,79 – 1,45 ✓
= 1,34 (s) ✓
3.5.2 a ✓ (1)
3.5.3 f ✓ (1)
3.5.4 c ✓ (1)
[18]
3.5.1 d ✓ Accept / Aanvaar ( 0 – e; 0 – d; d – e) (1)

QUESTION 4
4.1 NOTE: -1 mark for each key word/phrase omitted in the correct context.
Isolated system is a system on which the resultant/net external force is zero.✓✓
OR
Isolated system is one that has no net / external force acting on it.(2)
4.2.1 p = mv ✓
24 = m (480) ✓
m = 0,05 kg ✓ (3)
4.2.2 Marking criteria

• Appropriate formula including Fnet or Wnet.
• Substitutions ✓✓
• Final answer: 2 000 N ✓
• Correct direction: west or left ✓

POSITIVE MARKING FROM QUESTION 4.2.1
OPTION 1
F_netΔt = Δp
F_netΔt = (p_bullet)f - (p_bullet)i
F_netΔt = (mv_bullet)f - (mv_bullet)i
F_net(0,01) ✓ = (0,05)(80) – 24 ✓ or/of (0,05)(80) – (0,05)(480)
F_net = -2 000 N
F_net= 2 000 N ✓ west ✓

OPTION 2
vf = vi + aΔt
80 = 480 + a(0,01) ✓
a = - 40 000 m∙s^-2
F_net= ma ✓
= (0,05)(-40 000) ✓
= - 2 000 N
F_net= 2 000 N ✓ west ✓
✓Any one
Note: p and v must have the same sign

OPTION 3
Δx =(v₁  + v_f) Δt
             2
=480 + 80(0.01)
        2
= 2,80 m
vf² = vi² + 2aΔx
(80)2 = (480)2 + 2a(2,80) ✓
a = - 40 000 m∙s^-2
= ma ✓
= (0,05)(-40 000) ✓
= - 2 000 N
Fnet = 2 000 N ✓ west ✓

W_net = ΔK ✓
F_net Δxcosθ = ½mvf² - ½mvi²
F_net (2,80)cos0° ✓= ½(0,05)(80² - 480²) ✓
F_net = -2 000 N
F_net = 2 000 N ✓ west ✓

OR
F_net (2,80)cos180° ✓ = ½(0,05)(80² - 480²) ✓
F_net = 2 000 N ✓ west ✓ (5)
[10]

QUESTION 5
5.1 Note: -1 mark for each key word/phrase omitted in the correct context.
IF: The word "work" is omitted - 0 marks.
A conservative force is a force for which the work done (in moving an object between two points) is independent of the path taken. ✓✓
OR
A conservative force is a force for which the work done in moving an object in a closed path is zero.(2)
5.2 Gravitational (force) ✓
ACCEPT: Gravitation /Gravity/Weight (1)
5.3 No ✓
There is friction/non-conservative force (doing work)/It is not isolated system.✓
OR
The net work done by the non-conservative forces is not zero (2)
5.4 OPTION 1
EP = mgh ✓
= (1,8)(9,8) (1,5) ✓
= 26,46 J ✓

OPTION 2
Ww = -ΔEp ✓
(1,8)(9,8)(h – 0)cos180o = -(EpA – Ep(ground))
(1,8)(9,8)(1,5)(-1) = -EpA ✓
Ep = 26,46 J✓

OR
W = FΔxcosθ
= mgΔhcosθ
= (1,8)(9,8) (1,5)cos0o ✓
= 26,46 J✓ (3)
5.5 POSITIVE MARKING FROM QUESTION 5.4 
OPTION 1
Wnc = ΔK + ΔU
Wf = ½m(vf² – vi²) + mg(hf – hi)
= ½(1,8)(42 – 0,952) ✓+ (0 – 26,46) ✓
= -12,87 J ✓

OPTION 2
W_net = ΔK
Wf + Wg = ½mvf² - ½mvi²
Wf + mgh = ½m(vf²- vi²)
Wf + mgh = ½mvf2 - ½mvi²
Wf + 26,46✓= ½(1,8)[(4)2 - (0,95)²] ✓
Wf = -12,87 J (- 12,872 J)✓

OPTION 3
E_(mech)A = E_(mech)B - Wf
(Ep + Ek)A = (Ep + Ek) B - Wf
(mgh + ½ mv2) A = (mgh + ½ mv²) B - Wf
26,46 + ½(1,8)(0,95²) ✓ = 0 + ½(1,8)(4²) - Wf ✓
Wf = -12,87 J ✓ (4)
5.6 W_net = 0 (J ) / zero✓(1)
[13]

QUESTION 6
6.1 Doppler effect ✓(1)
6.2 (Q): (records sounds with) longer period/ longer time per wave / lower frequency.
OR
P: (records sounds with) shorter period/ shorter time per wave / higher frequency. ✓
ACCEPT
(Q): longer wavelength. /P: shorter wavelength. (1)
6.3 OPTION 1
f = 1  ✓=    1     ✓ = 5,88 x 10² = 588,24 Hz ✓
    T       17 x 10^-6

OPTION 2
speed =distance
                time
340 =   distance   
          25.5 x 10^-6
Distance = 0,867 m
Distance = 1 ½ λ
∴λ = 0,578 m
v = fλ ✓
340 = f(0,578) ✓
f = 588,24 Hz ✓

OPTION 3
v = λ 
      T
340 =      λ      
         17 × 10^-4
∴ λ = 0,578 m
v = fλ ✓
340 = f(0,578) ✓
f = 588,24 Hz ✓ (3)
6.4 POSITIVE MARKING FROM QUESTIONS 6.2 AND 6.3.
Do not penalise if 10-4 is again omitted.

OPTION 1
f =     1     = 5,56 x 10² = 555,56 Hz
    18 x 10^-4
f_L  = v ± v_Lf_s  OR f_L =   v    f_s
        v ± v_s                 v + v_s
555,56 =   340    588.24
              340 + v
v = 20 m∙s^-1 ✓
Range 19,57 – 20,09 m·s^-1

OPTION 2
f =     1      
    18 x 10^-4
f_L  = v ± v_Lf_s  OR  f_L =   v     f_s
        v ± v_s                   v + v_s
      1       =     340           1    
18 x 10^-4     340 + v  17 x 10^-4
v = 20 m∙s^-1 ✓
Range 19,57 – 20,09 m·s^-1 (6)
[11]

QUESTION 7
7.1.1 Positive ✓ (1)
7.1.2 Marking criteria:

• Appropriate formula 
• Whole substitution
• Final answer: 2,26 x 10^-6 C ✓

OPTION 1
F=kQ₁Q₂ 
       r² 
3.05 = (9 x 10⁹)(6 x 10^-6)Q
                     0.2²
Q = 2,26 x 10^-6 C ✓
(2,259 x 10^-6 C)

OPTION 2
E = kQ
       r²
=  (9 x 10⁹)(6 x 10^-6)
               0.2²
= 1,35 x 10⁶ N·C^-1
F = Eq✓
3,05 = (1,35 x 10⁶)q✓
q = 2,26 x 10^-6 N✓ (3)
7.1.3

Notes

• Mark is awarded for label and arrow. 
• Do not penalise for length of arrows.
• Deduct 1 mark for any additional force 
• If force(s) do not make contact with dot: Max 32
• If arrows missing: Max 32 (3)

7.1.4 F_net = 0
F_E = Tsin10°
F_E = Tcos80°
[3,05 = Tsin10°✓]✓ [IF Tcos10º = 3,05 (¹/₃)]
OR
[3,05 = Tcos80°✓]✓ [IF Tsin80º = 3,05 (¹/₃)]
T = 17,56 N  (17,564 N) (3)
7.2.1 Marking criteria

• 1 mark for each key word/phrase omitted in the correct context.

The electric field at a point is the (electrostatic) force experienced per unit positive charge placed at that point. ✓✓
[IF the word “unit” or phrase “positive charge” is omitted in this definition: -1 for each
OR
The electric field at a point is the (electrostatic) force experienced by a UNIT positive charge placed at that point. ✓✓
[If “UNIT” is omitted in this definition, then 0 marks. (2)
7.2.2 OPTION 1
Electric field at M due to A (+2 x10^-5 C):
E_A = kQ 
         r²
= 9 x 10⁹(2 x 10^-6)
                  (0.2)²      
= 4,5 x106 N∙C^-1
Electric field at M due to B (-4 x10^-5 C):
E_B = kQ     or     q_B =2q_A
         r²
= 9 x 10⁹(4 x 10^-6)    E_B=2E_A
                  (0.2)²
= 9 x10⁶ N∙C^-1 = 9 x 10⁶ N∙C^-1
E_net at M = E_A + E_B
= (4,5 x 10⁶ + 9 x 10⁶) ✓
= 1,35 x 107 N∙C^-1 ✓ to the right/towards B/away from A

OPTION 2
Net electrostatic force at M
F_net = kQ₁Q₂  +  kQ₁Q₂ 
               r²                r²
= (9 x 10⁹)(2 x 10^-5)q  + (9 x 10⁹)(4 x 10^-5)q  (any one/ enige een)
               (0.2)²                         (0.2)² 
= 4,5 x 10⁶q +✓ 9 x 10⁶q
= 1,35 x 10⁷q N
F_net = E_netq ✓
1,35 x 10⁷q ✓= E_netq
E_net = 1,35 x 107 N.C^-1 ✓to the right ✓/towards B  (6)
[18]

QUESTION 8
8.1 (Maximum) energy provided (work done) ✓ by a battery per coulomb / unit charge passing through it. ✓
ACCEPT:
The reading on a voltmeter connected across a battery when there is no current/ in an open circuit.✓✓ (2)
8.2 13 V ✓ (1)
8.3.1 R = V 
                I
5.6 =10.5
           I
I = 1,88 A ✓ (1,875 A)
Marking criteria:

• Appropriate formula ✓
• Whole substitution✓
• Final answer: 1,88 A ✓ (3)

8.3.2 POSITIVE MARKING FROM QUESTION 8.3.1.
OPTION 1
P = VI ✓
= (10,5)(1,88) ✓
= 19,74 W ✓ (19,688 W)

OPTION 2
P = I²R ✓
= (1,88)2(5,6) ✓
= 19,79 W ✓ (19,688 W)

OPTION 3
P =  V²
       R
=10.5² 
   5.6
= 19,69 W ✓ (19,688 W) (3)
8.3.3 POSITIVE MARKING FROM QUESTIONS 8.2 AND 8.3.1.
OPTION 1
Ɛ = I(R + r) ✓
13 = 1,88 (5,6 + r) ✓
r = 1,31 Ω  (1,31 – 1,33 Ω)

OPTION 2
r = V_internal✓
            I
= 2.5  
  1.88
= 1,33 Ω ✓ ( 1,31 – 1,33 Ω)

OPTION 3
Ɛ = V_ext + V_int
13 = 10,5 + V_int
V_int = 2,5 V
V_int = Ir ✓
2,5 = (1,88)r✓
r = 1,31 Ω ✓ (1,31 – 1,33 Ω) (3)
8.4.1 Decreases ✓
Vinternal resistance ✓
Vinterne weerstand (2)
8.4.2 Marking criteria

• Formula Ɛ = I(R + r) ✓
• Correct substitution into Ɛ = I(R + r) ✓
• Substitution of values into Rp formula ✓
• Halving value of R2X
• Final answer: 1,49 Ω ✓ Range: 1,46 Ω – 1,49 Ω

POSITIVE MARKING FROM QUESTIONS 8.2 AND 8.3.3
OPTION 1
Ɛ = I(R + r) ✓
13 = 4 (R_ext + 1,31) ✓
R_ext = 1,94 Ω (1,92 Ω)
  1   =  1   +  1  
 R_P     R₁      R₂
  1   =  1   +  1  
1.94   5.6    R₂
R₂ = 2.97Ω (2,92 Ω)
X= ½(2.97)
49,1 = Ω ✓ (1,46 – 1,49 Ω)

OPTION 2
Ɛ = I(R + r) ✓
13 = 4 (R_ext + 1,31) ✓
R_ext = 1,94 Ω (1,92 Ω)
R_P =  R₁ R₂  
         R₁ + R₂ 
1.94 =  5.6 R₂ 
           5.6 + R₂
R₂ = 2.97Ω (2,92 Ω)
X= ½(2.97)
49,1 = Ω ✓ (1,46 – 1,49 Ω)

OPTION 3
Ɛ = I(R + r) ✓
13 = 4 (R_ext + 1,31) ✓
R_ext = 1,94 Ω (1,92 Ω)
  1   =  1   +  1  
 R_P     R₁      R₂
  1   =  1   +  1  
1.94   5.6    2X
X 49,1 = Ω ✓ (1,46 – 1,49 Ω)

OPTION 4
Ɛ = I(R + r) ✓
13 = 4 (R_ext + 1,31) ✓
R_ext = 1,94 Ω (1,92 Ω)
R_P =  R₁ R₂  
         R₁ + R₂ 
1.94 =  5.6(2X)  
           5.6 + 2X
X 49,1 = Ω ✓

OPTION 5
Ɛ = I(R + r) ✓
Vext = 13 – (4)(1,31) ✓
= 7,76 V
Vp = IR_5,6
7,76 = I(5,6)
I_5,6Ω = 1,37 A
IT = I_2X + I_5,6
4 = I_2x + 1,37
I_2X = 2,63 A
V = IR_2X
[7,76 = (2,63)2X✓]✓
X = 1,46 Ω

OPTION 6
Ɛ = I(R + r) 
V_ext = 13 – (4)(1,31) ✓
V_ext = 7,76 V
I_5,6Ω = 7.76 = 1,39 A
            5.6
I_2x = 4 – 1,39 = 2,61 A
V_2x = I_2XR_2X
[7,76 = (2,61)2X✓]✓
2X = 2,97 Ω
X = 1,49 Ω✓
V_X = I_XR_X
3,88=(2,61)RX✓
R_X = 1,49 Ω✓

OPTION 7
Ɛ = I(R + r)✓
Vext = 13 – (4)(1,31)✓
= 7,76 V
V_ext = IR_ext
7,76 = (4)( 1  + 1 )^-1
                2X   5.6
X = 1,48 Ω ✓ (5)
[19]

QUESTION 9
9.1
9.1.1 DC ✓ (1)
9.1.2 NOTE: -1 mark for each key word/phrase omitted in correct context.
Emf is induced as a result of change of magnetic flux (linked) with the coil. ✓✓(2)
9.1.3 POSITIVE MARKING FROM QUESTION 9.1.1

9.2.1 The AC potential difference/voltage which dissipates the same amount of energy ✓ as DC potential difference. ✓
OR
(The rms value of AC is) the DC potential difference/voltage which dissipates the same amount of energy ✓ as AC potential difference/voltage. ✓(2)
9.2.2

(5)
[12]

QUESTION 10
10.1 Note: -1 mark for each key word/phrase omitted in correct context.
The process whereby electrons are ejected from a metal / surface when light (of suitable frequency) is incident on that surface. ✓✓ (2)
10.2 7,48 x 10^-19 (J) ✓
E = W_o + E_k(max) (= Wo + ½mv²_max) ✓
When E_k(max) = 0 / v = 0 / v² = 0 / E = W_o / W_o is the y-intercept ✓ (3)
10.3 Mass (of photo-electron) ✓
ACCEPT :½m (1)
10.4 OPTION 1
Gradient = ½m
11.98 x 10^-19 - 7.48 x 10^-19 = ½(9,11 x 10-31) ✓
             X - 0
X = 0,9879 ✓ (0,99 or 0,988)
ACCEPT
X = 0,9879 x 10¹² (m²∙s^-2)

POSITIVE MARKING FROM 10.2
OPTION 2
E = Wo + E_k(max)
E = Wo + ½mv²_(max)
11,98 x 10^-19 ✓ = 7,48 x 10^-19 ✓+ ½(9,11 x 10^-31) v^2✓ [or ½(9,11 x 10^-31)X]
4,5 x 10^-19 = 4,56 x 10^-31v²
v² = 0,9868 x 10¹²
X/v² = 0,9868 ✓ (0,99)
Range (0,9868 – 0,9879 / 9,87 x 10¹¹ – 9,88 x 10¹¹)

ACCEPT:
X = 0,9868 x 10¹² (m²∙s^-2) / 9,868 x 10¹¹ (m²∙s^-2) (5)
10.5.1 Remains the same ✓ (1)
10.5.2 Increases ✓ (1)
[13]
TOTAL: 150