PHYSICAL SCIENCES PAPER 1
GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
MEMORANDUM
MAY/JUNE 2021

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

QUESTION 2
2.1 Marking criteria
If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase.
The perpendicular force exerted by a surface on an object in contact with the surface. (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 all forces are correctly drawn and labelled, but no arrows, deduct 1 mark.(5)

2.3 For the 20 kg:
Fnet = ma
T – f – F_Ax = ma
T – 5 – 35 cos40° = 0 
T = 31,81 N
For m:
Fnet = ma
mg – T = ma
m(9,8) – 31,81 = 0
m = 3,25 kg 

Marking criteria

• Formula for 20 kg or m kg = ma 
• Substitution of zero into either formula
• All substitutions into Fnet for 20 kg as shown
• Substitution of value of T in eqn for m
• Final answer: 3,25 kg (5)

2.4.1 Decreases (1)
2.4.2 POSITIVE MARKING FROM QUESTION 2.3
Moving to the right
Velocity decreases
Accelerates/Net force to left
OR
As the tension force decreases, the net force/acceleration acts in the opposite direction of motion /to the left.
Moving to the left
Velocity increases
Accelerates/Net force to left(3)
[16]

QUESTION 3
3.1 (Motion of an object) under the influence of gravity (weight) only. (2 or 0)
OR
(Motion in which) the only force acting on the object is gravity (weight). (2)
3.2.1 Δt = 0,67 – 0,64 = 0,03 s (2)
3.2.2
OPTION 1
Δt =(1,90-0,67)
             2
= 0,62 s (0,615 s)

OPTION 2
Δx = viΔt + ½ aΔt²
(-1,85) = 0 + ½ (-9,8)Δt² 
Δt = 0,61 s (0,6145 s) (2)

OPTION 3
Δt =(1,90 - 0,67)
               2
= 1,285 s
Δt = 1,285 – 0,67 
= 0,62 s (0,615 s)

OPTION 4
vf² = vi² + 2aΔx
0 = vi² + 2(-9,8)(1,85)
vi = 6,02 m·s^-1
vf = vi + aΔt
0 = 6,02 + (-9,8)Δt
Δt = 0,61 s

3.2.3 POSITIVE MARKING FROM QUESTION 3.2.2
Marking Criteria

• Any appropriate formula
• Correct substitution
• Final answer: 5,94 to 6,08 m·s^-1

OPTION 1
Upwards positive
vf = vi + aΔt 
0 = vi + (-9,8)(0,62) 
vi = 6,08 m·s^-1 (6,076 m·s^-1)
Downwards positive
vf = vi + aΔt
0 = vi + (9,8)(0,62)
vi = -6,08
6,08 m·s^-1 (6,076 m·s^-1) 

OPTION 2
Upwards positive
Δy = viΔt + ½ aΔt²
1,85 = vi (0,62) + ½ (-9,8) (0,62)²
vi = 6,02 m·s-1 (6,022 m·s^-1)
Downwards positive
Δy = viΔt + ½ aΔt²
1,85 = vi (0,62) + ½ (9,8) (0,62)²
vi = -6,02
vi = 6,02 m·s-1 (6,022 m·s-1)

OPTION 3
Motion from top to bottom
Downwards positive
vf² = vi² + 2aΔy
vf² = 0 + 2(9,8)(1,85)
vf = 6,02 m·s^-1
initial velocity =6,02 m·s^-1
Upwards positive
vf² = vi² + 2aΔy
vf² = 0 + 2(-9,8)(-1,85)
vf = 6,02 m·s^-1
initial velocity =6,02 m·s^-1

Motion from bottom to top
Downwards positive
vf² = vi² + 2aΔy
0² = vi² = 0 + 2(9,8)(1,85)
vf = 6,02 m·s^-1
Upwards positive
vf² = vi² + 2aΔy
0² = vi² = 0 + 2(-9,8)(-1,85)
vf = 6,02 m·s^-1

OPTION 4
Upwards positive/Opwaarts positief
Δy = viΔt + ½ aΔt²
0 = vi(1,23) + ½ (-9,8)(1,23)²
vi = 6,03 m·s^-1
Downwards positive
Δy = viΔt + ½ aΔt²
0 = vi(1,23) + ½ (9,8)(1,23)²
vi = - 6,03 m·s^-1
speed = 6,03 m·s^-1

OPTION 5
Δy = (vf + vi) Δt
             2
1,85 =(0 + vi) (0.62)
              2
vi = 5,97 m·^s-1

OPTION 6
FnetΔt = mΔv
FnetΔt = m(vf – vi)
m(9,8)(0,62) = m(0 – vi)
vi = 6,08 m·s^-1

OPTION 7
(Ep + Ek)floor = (Ep + Ek)top
(mgh + ½ mv²)floor = (mgh + ½ mv²)top
0 + ½ v² = (9,8)(1,85) + 0
v = 6,02 m·s^-1 (3)

3.2.4 OPTION 1, 2, 3, 4: Marking criteria
Calculate initial velocity:

• Appropriate formula
• Substitution

Calculate Δt:

• Appropriate formula
• Substitution
• 1,97 s + Δt
• Fin answer: 2,95 – 2,97 s

Calculate initial velocity:
OPTION 1
Downwards positive
vf² = vi² + 2aΔy
0 = vi² + 2(9,8)(-1,2)
vi = - 4,85 m·s^-1
Upwards positive
vf² = vi² + 2aΔy
0 = vi² + 2(-9,8)(1,2)
vi = - 4,85 m·s^-1 

OPTION 2
(Emech)top = (Emech)bot
(Ep +Ek)top = (Ep +Ek)Bot
(Any one)
(mgh + ½mv²)top = (mgh + ½mv²)Bot
(9,8)(1,2) + 0 = 0 + (½)v²
vi = 4,85 m·s^-1 upwards

OPTION 3
Wnc = ΔEp + ΔEk
0 = (0 – mgh) + ½m(vf² - vi² )
0 = -(9,8)(1,2) + ½vi²
= 4,85 m·s^-1 upwards

OPTION 4
Wnet = ΔEk
wΔxcos180° = ½mm(vf² - vi² )
(Any one)
(9,8)(1,2)cos180° = ½vi²
= - 4,85 m·s-1

OPTION 5
Downwards positive
Δy = viΔt + ½ aΔt²
1,2 = 0 + ½(9,8) Δt²
Δt = 0,49 s
t = 1,97 + 2(0,49) 
= 2,96 s
Upwards positive
Δy = viΔt + ½ aΔt²
-1,2=  0 + ½(-9,8)Δt²
Δt = 0,49 s
t = 1,97 + 2(0,49)
= 2,96 s

Calculate time Δt
Upwards positive
Δy = viΔt + ½ aΔt²
1,2 = (4,85)Δt + ½(-9,8)Δt²
Δt = 0,4898 s / 0,5 
t = 1,97 + 2(0,4898)
= 2,95 s / 2,97 s 

OR
Δy = viΔt + ½ aΔt²
0 = (4,85)Δt + ½(-9,8)Δt²
Δt = 0,9898 s (or Δt = 0)
t = 1,97 + 0,9898 = 2,96 s
Downwards positive
Δy = viΔt + ½aΔt²
1,2 = (-4,85)Δt + ½(9,8)Δt²
Δt = 0,4898 s / 0,5 s
t = 1,97 + 2(0,4898) 
= 2,95 s / 2,97 s 

OR
Δy = viΔt + ½aΔt²
0 = (4,85)Δt + ½(9,8)Δt²
Δt = 0,9898 s (or Δt = 0)
t = 1,97 + 0,9898 = 2,96 s

OR
vf = vi + aΔt
-4,85 = 4,85 + (-9,8)Δt
Δt = 0,9898 s
Δt = 1,97 + 0,9898= 2,96 s

OR
Upwards positive
vf = vi + aΔt
0 = 4,85 + (-9,8)Δt
Δt = 0,4949 s
Δt = 1,97 + (2)(0,4949) 
= 2,96 s

OR
Δy =(vi + vf)Δt
            2
1.2 =(0 + 4.85) Δt  Δt = 0,4948s
             2
Δt_total = 2(0,4948) = 0,99 s
Δt = 1,97 + 0,99 = 2,96 s

OPTION 5: Marking criteria

• Formula
• Substitution Δy = 1,2 
• Substitution 0 + ½(9,8) Δt²
• 1,97 s + 
• 2 Δt 
• Final answer: 2,95 - 2,97 s (6)

[15]

QUESTION 4
4.1 (Linear) momentum (of an object) is the product of mass and velocity.
(2 or 0) (2)

4.2.1
OPTION 1
East as positive
Σpi = Σpf
m_pv_pi + m_Qv_Qi = m_pv_pf + m_Qv_Qf
(Any one)
(0,16)(10) + (0,2)(-15) = (0,16)(-5) +(0,2)v_Qf 
v_Qf = -3 m∙s^-1
v_Qf = 3 m∙s^-1 west

OPTION 2
West as positive
Σpi = Σpf
m_pv_pi + m_Qv_Qi = m_pv_pf + m_Qv_Qf
(Any one)
(0,16)(-10) + (0,2)(15) = (0,16)(5) +(0,2)Q_Nf 
v_Qf = 3 m∙s^-1 west

OPTION 3
Δpp = -Δp_Q 
(0,16)(-5 – 10) = -(0,2)(v – (-15))
v = -3 m·s^-1
= 3 m·s^-1 west (5)

4.2.2 For ball P:
West as negative
Impulse = Δp
FnetΔt = Δp
Δp = m(vPf – vPi)
= 0,16(-5 – 10)
= - 2,4
2,4 N∙s (2,4 kg∙m∙s^-1)

OR
West as positive
Impulse = Δp
FnetΔt = Δp
= m(vPf - vPi)
= 0,16(5 - (-10)) 
= 2,4 N·s 

POSITIVE MARKING FROM QUESTION 4.2.1 (3)
For ball:
West as negative
Impulse = Δp
FnetΔt = Δp
= m(vQf – vQi)
= 0,2[-3 – (-15)] 
= 2,4 N∙s (2,4 kg∙m∙s^-1)

OR
West as positive
Impulse = Δp
FnetΔt = Δp
= m(vQf - vQi)
= 0,16(3 – (15)) 
= - 2,4 N·s
2,4 N∙s (2,4 kg∙m∙s^-1)
[10]

QUESTION 5
5.1 Marking criteria
If any of the underlined key words/phrases in the correct context are omitted:

• 1 mark per word/phrase. However, IF: The word “work” is omitted 0 marks

A force is non-conservative if the work it does on an object (which is moving between two points) depends on the path taken.

OR
A force is non-conservative if the work it does on an object depends on the path taken.

OR
A force is non-conservative if the work it does in moving an object around a closed path is non-zero.(2)

5.2 K = ½ mv² /Ek = ½ mv²
ΔK = Kf - Ki
K = ½mvf² - ½mvi²
= ½m(vf²- vi²)
=½(200)(2² - 4²)
K = - 1 200 J (3)

5.3 POSITIVE MARKING FROM QUESTION 5.2.
Marking criteria/Nasienriglyne

• Appropriate formula
• Substitution into appropriate formula together with -3,40 × 10³
• Final answer: 8,88 m

OPTION 1
Wnc = K + U
Wnc = ½ mvf² - ½ mvi² + mghf - mghi
= ½ m (vf²- vi²) + mg(hf - hi)
-3,40 × 10³ = -1 200 + 200(9,8)(hf - 10)
h = 8,88 m (8,87765 m)

OPTION 2
E(mech/meg)A + Wf = E(mech)_B
(Ep +Ek)A + Wf = (Ep +Ek)_B
(mgh + ½mv²) A + Wf = (mgh + ½mv²)_B
200(9,8)(10) + ½(200)(4²) - 3,40 × 10³ = 200(9,8)(h) + ½(200)(2)²
h = 8,88 m (8,87755)

OPTION 3
Wnet = ΔK
Wf + Ww = ½mvf² - ½mvi²
Wf – ΔEp = ½mvf² - ½mvi²
Wf - mg(hf - hi) = ½m(vf²- vi²)
-3,40 × 10³ - 200(9,8)(h-10)= -1 200 
h = 8,88 m  (8,87755 m) (4)

5.4 OPTION 1 AND 2: Marking criteria

• Appropriate formula
• Work done by friction
• Substitution of (200)(9,8)(13,12)
• Appropriate formula
• Substitution into power formula
• Final answer: 1 814,35 W

OPTION 1
Wnc = ΔK +ΔU
Wengine + Wf = ½mvf² - ½ mvi² + mghf - mghi
= ½m(vf²- vi²) + mg(hf - hi)
W_engine + (50)(15)(2)cos180° = 0 + 200(9,8)(22 – 8,88)
W_engine = 27 215,20 J
P_engine = W_engine
                   Δt
=27 215,20
       15
= 1 814,35 W

OPTION 2
Wnet = ΔK
WN + W_engine + Ww + Wf = 0
WN + W_engine - ΔEp + Wf = 0
0 + W_engine - (200)(9,8)(13,12) + (50)(2)(15)cos180° = 0
W_engine = 27 215,20 J

OR
Wnet = ΔK
WN + W_engine + Ww|| + Wf = 0
WN + W_engine +mgsinθΔxcos180^o + W_f = 0
0 + W_engine - (200)(9,8)(13,12)(-1) + (50)(2)(15)cos180°  = 0
                                         Δx
W_engine = 27 215,20 J
P_engine = W_engine
                   Δt
= 27 215,20
        15
= 1 814,35 W

OPTION 3: Marking criteria

• Appropriate formula
• Substitution of - 50
• Substitution of (-200)(9,8)(0,4373) or (-200)(9,8)(0,44)
• Appropriate formula
• Substitution into P_ave = Fv_ave
• Final answer: 1 814,35 W - 1 824,8 W

OPTION 3
F_net = ma
F_engine + F_friction + Fg// = 0
F_engine + (-50) + (-200)(9,8)(0,4373) = 0
F_engine = 906,52 N (906,52 – 912,4)
P_ave = Fvave
P_ave = (908,52)(2)
= 1 813,04 W (1 824,8 W)

OR
W = F_engineΔxcosθ
= (906,52)(30)cos0o
= 27 195,6 J (27 372 W)
P = W = 27195,6 = 1 813,04 W(1 824,8 W)
      Δt       15
(5)
[14]

QUESTION 6
6.1 Marking criteria
If any of the underlined key words/phrases in the correct context are omitted:
- 1 mark per word/phrase.
The change in frequency(or pitch) (of the sound) detected by a listener because the source and the listener have different velocities relative to the medium of propagation.
OR
An (apparent) change in (observed/detected) frequency (pitch), as a result of the relative motion between a source and an observer (listener). (2)

6.2 Towards (1)

6.3
f_L = v ± v_L f_s   OR  f_L =    v     f_s   OR f_L =    v      f_s
       v ± v_s                    v - v_s                     v + v_s
3148 = 340 + 0                                  2073 = 340 - 0
             340 v_s                                                         340 v_s
3148(340 - v_s) = 2073(340 + v_s)
    340 + 0                340 - 0
v_s = 70 m·s^-1 (69,95 - 70,16 m·s^-1) (6)

6.4 POSITIVE MARKING FROM QUESTION 6.3
OPTION 1
Δt =Δx 
       v
Δt = 350 
        70
Δt = 5s

OPTION 2
Δx = viΔt + ½ aΔt²
350 = 70Δt + 0
Δt = 5 s 

OPTION 3
Δx = (vi + vf)Δt
             2
350 =(70 + 70) Δt
               2
Δt = 5 s (2)
[11]

QUESTION 7
7.1
n = Q 
      e
=  ( - )4 x 10 ^-6
    ( - )1,6 10^-19
= 2,5 x 10¹³ (3)

7.2 Electrostatic force on B due to A:
F_AB = kQ₁ Q₂
             r²
=[9 x 10⁹(4 x 10^-6)(3 x 10^-6)
                 0,2
= 2,7 N(3)

7.3 Marking criteria
If any of the underlined key words/phrases in the correct context are omitted:
- 1 mark per word/phrase.
Electric field is a region (in space) where (in which) an (electric) charge experiences a (electric) force. (2)
Ignore negative signs

7.4 Marking criteria

• Appropriate formula
• Correct substitution for A and B
• Subtraction of electric fields
• Final answer: 2,3 x10⁶ N∙C^-1

OPTION 1
Electric field at M due to: -4 x10^-6 C
E_AM =k Q 
             r²
= 9 x 10⁹(4 x10^-6)
                (0,3)²
= 4,0 x 10⁵ N∙C^-1 (to left)
Electric field at M due to: +3 x 10^-6 C,
E_BM =k Q 
             r²
=9 x 10⁹(3 x10^-6)
                (0,1)²
= 2,7 x106 N∙C^-1 (to right)
Net electric field at M
E_net = E_BM + E_AM
= 4,0 x105 – 2,7 x10⁶ 
= 2,3 x106 N∙C^-1
(right)

OR
Net electric field at M
E_net = E_BM + E_AM
= -4,0 x10⁵ + 2,7 x10⁶ 
= - 2,3 x10⁶ N∙C^-1
= 2,3 x10⁶ N∙C^-1 (right)
(5)

OPTION 2
F_AM = kQ₁ Q₂ = (9 x 10⁹)(4 x 10^-6)Q = 4 x 10⁵Q N
              r²                          (0,3)²
F_BM =  kQ₁ Q₂ = (9 x 10⁹)(4 x 10^-6)Q = 2.7 x 10⁶Q N
                r²                          (0,1)²
F_net = 2,7 x 10⁶Q + (-4 x 10⁵Q) = 2,3 x 10⁶Q
E = F  = 2,3 x 10⁶Q = 2,3 x 10⁶ N·C^-1 (right)
      q             Q 
7.5 Positive (1)

7.6 POSITIVE MARKING FROM 7.2
Marking criteria

• Correct substitution into Pythagoras’s equation
• Correct substitution into Coulomb’s Law
• Correct answer(3)

(Fnet)² = (FAD)² + (FAB)²
(7,69)² = (FAD)² + (2,7)² 
F_AD = 7,2 N

F_AD = kQ₁ Q₂
             r²
7,2 = (9 x10⁹)(4 x10^-6)Q
                (0,15)²
= 9 x10⁹ (4 x10^-6)Q
                  0,15²
= 1,6 x 10⁶ Q
F_net = √F_AB² + F_AD² OR F_net = F_AB² + F_AD²
7,69 = √2,7² + (1,6x10⁶Q)²
Q = 4,50 x10^-6 C
[17]

QUESTION 8
8.1 Marking criteria
If any of the underlined key words/phrases in the correct context are omitted:
- 1 mark per word/phrase.
(Maximum) energy provided (work done) by a battery per coulomb/unit charge passing through it.
Work done by the battery to move a unit coulomb of charge across the circuit. (2)

8.2 Energy (per coulomb of charge) is converted to heat in the battery due to the internal resistance.(2)

8.3.1
I = V 
     R
I = 1.5 
     0.5
= 3 A (3)

8.3.2 OPTION 1
  1   =  1   +   1  
 R_P     R₁      R₂
  1   =   1   +  1  
 R_P     25     15 
 R_P = 9,375 Ω
R_ext = 9,375 + 4 = 13,38
(13,375)

OPTION 2
 R_P =   R₁ R₂ 
          R₁ + R₂
R_P = (25)(15) 
          25 +15
Rp = 9,375 Ω
Rext = 9,375 + 4= 13,38Ω
(13,375Ω)
(4)

8.3.3 POSITIVE MARKING FROM QUESTIONS 8.3.1 AND 8.3.2. (3)
OPTION 1
Ɛ = I(R + r) 
= 3(13,38 + 0,5) 
= 41,64 V (Range: 41,625 – 41,64)

OPTION 2
Ɛ = Vext + Vint 
= (3)(13,38) + 1,5 
= 41,64 V (Range: 41,625 – 41,64)

8.4 Yes.
For the same voltage/potential difference,a larger current will flow through a smaller resistor (I = V )
                                                                                                                                                         R

OR
I α ¹/_R , V = constant
I is inversely proportional to R and V is constant.

OR
V|| = IR
= (3)(9,38)
= 28,14 V
I_R2 = V  = 28,14  = 1.13A
         R       25
I_R3 =  V  = 28,14  = 1.88A
          R       25

OR
V is the same
I_15Ω = 25  I
          40
I_25Ω = 15  I
          40
(3)

8.5 Remains the same(1)
[18]

QUESTION 9
9.1.1 (DC) motor/(GS-)motor (1)

9.1.2 POSITIVE MARKING FROM QUESTION 9.1.1
Electrical to mechanical /kinetic (energy) (2 or 0)(2)
9.1.3 Split ring/commutator(1)
9.1.4 Anticlockwise(2)
9.2.1 (The rms voltage/value of AC is) the AC voltage/potential difference which dissipates the same amount of energy/heat/power as an equivalent DC voltage/potential difference. (2 or 0)

ACCEPT
The rms voltage/value of AC is the DC potential difference which dissipates the same amount of energy/heat/power as AC. (2)

9.2.2 Marking criteria

• Appropriate formula for P_ave
• Substitution to calculate R
• Final answer: 242 Ω

OPTION 1
P ave = V_rms²  
                R
200= 220²
          R
R = 242 Ω

OPTION 2
P ave = V_rmsI_rms
200 = Irms (220)
I_rms = 0,909 A (0,91)
R = V_rms or R = V 
       I rms           I
R = 200 =
     0,909 
R = 242 Ω (241.76Ω)

OPTION 3
P ave = V_rmsI_rms
200 = Irms (220)
I_rms = 0,909 A (0,91)

P_ave = I_rms² 
R= 242 Ω
(241,76 Ω) (3)

9.2.3 Marking criteria for options 1,2 and 3

• Appropriate formula to calculate P or Irms 
• Substitution
• Formula for P or W containing Δt
• Substitution
• Final answer: 55 785,12 J 

POSITIVE MARKING FROM QUESTION 9.2.2.
OPTION 1
Marking criteria

• Appropriate formula for W containing V
• Substitution
• Final answer: 55 785,12 J (5)

W = V² Δt
         R
= (150²) (10 x 60)
          242
= 55 785,12 J 

OPTION 2
P_ave = V²rms 
              R
= 150²
   242
P_av = 92,975 W
P = W 
      Δt
92,975 =     W      
               (10)(60)
W = 55 785,12 J
(55785,12 – 55896 J)

OPTION 3
P_ave = V²rms 
              R
242 =  150 
          I rms
I rms = 0,620 A
W = I²RΔt
= (0,62)²(242)(10)(60)
= 55 785,12 J
(55785,12 – 55896 J)

OR
w = VIΔt
= (150)(0,62)(600)
= 55 800 J

OPTION 5
P ave = V²rms =150² = 92,975 W
                R        242
Pave = Irms²R
92,975 = Irms² (242)
Irms = 0,6198 A
W = I2RΔt 
= (0,6198)2(242)(10)(60) 
= 55 778,88 J 
[16]

QUESTION 10
10.1 Photoelectric effect(1)
10.2 Work function (of potassium)(1)
10.3 Potassium
It has the lowest work function / threshold frequency / highest threshold wavelength. (2)
10.4 Marking criteria/Nasienriglyne
If any of the underlined key words/phrases in the correct context are omitted:
- 1 mark per word/phrase.
The work function of a metal is the minimum energy that an electron (in the metal) needs to be emitted/ejected from the metal / surface. (2)
10.5.1 W_o= hf_o
= (6,63 x 10^-34)(1,75 x 10¹⁵)
= 1,160 x 10^-18 J
OR
E = Wo + Ek(max)
hf = Wo + Ek(max)
(6,63 x 10^-34)(1,75 x 10¹⁵) = W_o + 0
W_o = 1,160 x 10^-18 J (3)

10.5.2 POSITIVE MARKING FROM QUESTION 10.5.1.
E = W_o + Ek(max)
hf = hf_o + ½mv²
(6,63 x 10-34)f = 1,160 x 10^-18 + (9,11 x 10^-31) (5,60 x 10⁵)²
f = 1,97 x 10¹⁵ Hz (4)
[13]
TOTAL: 150