When a (non-zero) 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 (non-zero) 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 or 0) 
ACCEPT
Acceleration is directly proportional to the net force and inversely proportional  to the mass of the object. 

NOTE

• If any of the underlined key words in the correct context is omitted deduct 1  mark. 

Notes

• Mark is awarded for label and arrow
• Do not penalise for length of arrows 
• If T is not shown but T|| and T⊥ are shown, give 1 mark for both
• If force(s) do not make contact with body : Max/Maks: 3/4
• Deduct 1 mark for any additional force  

Accept the following symbols 

N 

FN; Normal;Normal force ✔

f

Ff / fk / frictional force ✔

w

Fg; mg; Weight;F_{Earth on block;}Fw  ;Gravitational force ✔

T 

Tension ; FT /FA, F /16,96 N ✔

The 2/8 kg block /system is accelerating ✔
OR
The acceleration is not zero / a ≠ 0 (m·s-2) / a = 1,32 m.s-2✔
OR
Velocity is /increasing/changing/not constant ✔
OR
F_net is not equal to zero  ✔
OR
The acceleration is changing  ✔
Accept

• An unbalanced force is acting on it  ✔ 

For 2 kg 
F_net = ma  {1 mark for any }    F_net = ma  
mg -T= ma                              mg + T = ma  ✔ 
(2)(-9,8) + T = 2(-1,32) ✔   (2)(9,8) – T = 2(1,32) ✔ 
T = 16,96 N ✔                       T = 16,96 N ✔ 

POSITIVE MARKING FROM 2.3.2

OR

• ANY ONE
  Normal force changes/decreases ✔
• The angle (between string and horizontal) changes/increases. 
• The vertical component of the tension changes/increases 

Yes✔
The frictional force (coefficient of friction) depends on the nature of the surfaces  in contact. ✔ 
ACCEPT
The nature of the surface changes / µk changes  

Downwards ✔ 
The only force acting on the object is the gravitational force/weight which acts  downwards.✔
ACCEPT

• The only force acting is gravitational/weight.✔
  OR
• Gravitational force/weight acts downwards.✔
  OR
• The ball is in free-fall ✔✔
  OR
• (Gravitational) acceleration is downwards✔✔

OPTION 1
Upward positive
vf = vi + aΔt ✔ 
0 = 7,5 + (-9,8)Δt ✔ 
Δt = 0,77 s ✔ 

Downward positive
vf = vi + aΔt ✔ 
0 = -7,5 + (9,8)Δt ✔
Δt = 0,77 s ✔

OPTION 2
Upward positive 
At highest point vf is zero 
v_f² = v_i² + 2aΔy
0 = (7,5)² + (2)(-9,8)Δy
Δy = 2,87 (2,869)m

Δt = 0,77 s✔ 

OPTION 2
Downward positive 
At highest point vf is zero 
v_f² = v_i² + 2aΔy
0 = (-7,5)² + (2)(9,8)Δy
Δy = 2,87 (-2,869)m

Δt = 0,77 s✔ 

OPTION 3
Upward positive 
F_netΔt = m(vf – vi) ✔
mgΔt = m(vf – vi) 
(-9,8)Δt = 0 – 7,5 ✔
∴ Δt = 0,76531 s (0,77 s) 

OPTION 3
Downward positive 
FnetΔt = m(vf – vi) ✔
mgΔt = m(vf – vi) 
(9,8)Δt = 0 –(- 7,5) ✔
∴ Δt = 0,76531 s (0,77 s)✔

OPTION 4
Upward positive 
(Top to Bottom ) 
vf = vi + aΔt ✔
-7,5 = 0 + (-9,8)Δt ✔
∴ Δt = 0,76531 s (0,77 s) ✔

OPTION 4
Downward positive 
(Top to Bottom ) 
vf = vi + Δt ✔
7,5 = 0 + (9,8)Δt✔
∴ Δt = 0,76531 s (0,77 s) ✔

OPTION 5
Upward positive 
(Top to Bottom) 
vf² = vi² + 2aΔy 
(7,5)² = (0)² + 2(-9,8)Δy 
Δy = -2,87 m 
Δy = viΔt + ½ aΔt² ✔
-2,87 = (0)Δt + ½ (-9,8)(Δt)² 
Δt = 0,765 s ✔

OPTION 5
Downward positive 
(Top to Bottom ) 
vf² = vi² + 2aΔy 
(7,5)2 = (0)² + 2(9,8)Δy 
Δy = 2,87 m Δy = viΔt + ½ aΔt² ✔
2,87 = (0)Δt + ½ (9,8)Δt² ✔
Δt = 0,765 s ✔

NOTES for marking QUESTION 3.3 

Formula mark

✔

Substitution mark

✔✔

Mark for height/distance  

✔

Mark for comparison

✔

Mark for conclusion

✔

OPTION 1
Upward positive
At highest point vf is zero
v_f² = v_i² + 2aΔy
0✔ = (7,5)2 + (2)(-9,8)Δy ✔ 
Δy = 2,87 (2,869) m✔ 
This is higher than height needed to reach point T (2,1 m)✔therefore the ball will pass point T.✔ 

Downward positive
At highest point vf is zero
v_f² = v_i² + 2aΔy
0✔ = (-7,5)2 + (2)(9,8)Δy✔ 
Δy = - 2,87 (-2,869) m✔ 
This is higher than height needed to reach point T (2,1 m)✔therefore the ball will pass the target. ✔ 

OPTION 2
(POSITIVE MARKING FROM 3.2) 
Upward positive
Δy = viΔt + ½ aΔt²✔ 
Δy = (7,5)(0,77) ✔ + ½ (-9,8)(0,77)²✔  
Δy = 2,87 m (2,86 m)✔ 
This is higher than height needed to reach point T (2,1 m)✔therefore the ball will pass point T. ✔ 
Downward positive
Δy = viΔt + ½ aΔt²✔ 
Δy = (-7,5)(0,77) ✔ + ½ (9,8)(0,77)²✔  
Δy = -2,87 m (2,869 m)✔ 
This is higher than the height needed to reach point T (2,1 m)✔therefore the  ball will pass point T.

OPTION 3
(E_mech)_Top = (E_mech)_Ground
1 mark for any 
(E_P +E_K)Top = (E_P +E_K)_Bottom
(mgh + ½ mv2)_Top = (mgh + ½ mv2)_Bottom 
(9,8)(h) + 0 ✔= 0 + (½ )(7,5)2 ✔ 
h = 2,87 m (2,869 m)✔ 
This is higher than height needed to pass the target (2,1 m)✔therefore the ball will pass the target.✔ 

OPTION 4
W_net = ∆E_K 
mgΔxcosθ = ½ mvf²- ½ mvi² ✔ 
(9,8)Δxcos180º✔ = 0 - ½(7,5)2✔ 
Δx = 2,87 m (2,869 m)✔ 
This is higher than point height needed to pass point T (2,1 m)✔therefore the  ball will pass point T. ✔ 

OPTION 5
Upward positive
If the highest point is yf then Δy = (y_f – y_1,6) At highest point vf is zero
v_f² = v_i² + 2aΔy
0✔ = [(7,5)2 + (2)(-9,8)(yf – 1,6)]✔ 
yf = 4,47 (4,469) m✔ 

Yes ✔ ✔ 
OR
This point (4,47m) is higher than point T ✔ ✔  (or even the required height of  2,1 m) therefore the ball will pass point T. 

Downward positive
If the highest point is yf then Δy = (yf – y1,6)
At highest point vf is zero
v_f² = v_i² + 2aΔy
0✔ = [(-7,5)2 + (2)(9,8){yf – (-1,6)}] ✔ 
yf = -4,47 (-4,469) m✔ 
height is  is 4,47 m. 
This point (4,47 m) is higher than point T✔✔ (or even the required height of  2,1 m) therefore the ball will pass point T. 

OPTION 6 (POSITIVE MARKING FROM 3.2) 
Upward positive
If the highest point is yf then Δy = (yf – y1,6) At highest point vf is zero
Δy = viΔt + ½ aΔt²✔ 
(y_f – 1,6) = (7,5)(0,77) ✔ + ½ (-9,8)(0,77)²✔  
y_f = 4,47 m (4,469 m)✔ 
This point (4,47m) is higher than point T ✔✔ (or even the required height  of 2,1 m) therefore the ball will pass point T. 

Downward positive
If the highest point is yf then Δy = (yf – y1,6) At highest point vf is zero 
Δy = viΔt + ½ aΔt²✔ 
{y_f – (-1,6)} = (-7,5)(0,765) ✔ + ½ (9,8)(0,765)²✔  
y_f = -4,47 m (-4,469 m)✔ 
This point (4,47m) is higher than point T✔✔ (or even the required height of  (2,1m) therefore the ball will pass point T. 

OPTION 7 (POSITIVE  MARKING FROM 3.2) 
Upward positive

This is higher than height needed to  pass the target (2,1 m)✔therefore the  ball will pass the target.✔ 

OPTION 7(POSITIVE  MARKING FROM 3.2 
Downward positive

Height  is 2,89m 
This is higher than height needed to  pass the target (2,1 m)✔therefore the  ball will pass the target.✔ 

OPTION 8 
Upward positive
At highest point vf is zero
v_f² = v_i² + 2aΔy
0✔ = v_i² – (2)(9,8)(2,1) ✔ 
v_i = 6,42 m∙s^-1 ✔ 

• This is the actual velocity needed to reach the target.
• The given velocity is greater than the actual velocity needed. ✔
• The ball will pass the target. ✔ 

Downward positive
At highest point vf is zero 
v_f² = v_i² + 2aΔy
0✔ = v_i² + (2)(9,8)(-2,1) ✔ 
v_i = 6,42 m∙s^-1 ✔ 

• This is the actual velocity needed to pass the target.
• The given velocity is greater than the actual velocity needed. ✔
• The ball will reach the target. ✔ 

OPTION 9
W_nc = ΔE_p + ΔE_k ✔ 
0 = mgh_f – mgh_i + ½ mv_f² – ½ mv_i² 
0 ✔ = (9,8)hf – (9,8)(1,6) + ½ (0) 2 – ½ (7,5)2✔ 
 0 = (9,8)hf – 43,805 
∴ hf = 4,47 m ✔ 
∴ The ball will pass point T ✔ ✔ 

OPTION 10
POSITIVE MARKING FROM 3.2
Upward positive
Δt(max. height/maks. hoogte) = 0,77 s 
Δy = viΔt + ½ aΔt² ✔ 
2,1✔  = (7,5)Δt + ½ (-9,8)Δt² ✔ 
∴ Δt = 0,36 s ✔ 
∴ Δt (max height, 0,77 s) > Δt (to pass point T 0,36 s) ✔ 
∴ The ball passed point T ✔ 

Downward positive
Δt (max height) = 0,77 s 
Δy = viΔt + ½ a Δt² ✔ 

2,1✔ = (7,5)Δt + ½ (-9,8)Δt² ✔ 
∴ Δt = 0,36 s✔ 
∴ Δt (max height, 0,77 s) > Δt (to reach point T, 0,36 s) ✔ 
∴ The ball passed point T ✔ 

OPTION 11
Upward positive
∆y = vi∆t + ½ a∆t²✔ 
(3,7 – 1,6)✔  = 7,5 ∆t + ½ (- 9,8) ∆t² ✔ 
∆t = 0,375 s ✔ 
The time to pass point T is less than time to reach maximum height ✔ . Ball  will pass point T.✔ 

Downward positive
∆y = vi∆t + ½ a∆t²✔ 
(3,7 – 1,6)✔  = -7,5 ∆t + ½ (9,8) ∆t² ✔ 
∆t = 0,375 s ✔ 
The time to reach point T is less than time to reach maximum height ✔ . Ball  will pass point T.✔ 

OPTION 12
Upward positive
v_f² = v_i² + 2aΔy ✔ 
v_f² = (7,5)² ✔ + 2(-9,8)(2,1) ✔ 
v_f = 3,88 m·s^-1 ✔ 
Velocity at T is 3,88 m·s-1therefore the ball still moving towards its maximum height ✔ ✔ 

Downward positive
v_f² = v_i² + 2aΔy ✔ 
v_f² = (-7,5)² ✔ + 2(9,8)(-2,1) ✔ 
v_f = -3,88 m·s^-1 ✔ 

Velocity at T is -3,88 m·s-1therefore the ball is still moving towards its  maximum height ✔✔

Notes

Initial velocity and time for final velocity shown  

✔

Correct straight line (including orientation) drawn  

✔

POSITIVE MARKING FROM 4.4 /POSITIEWE NASIEN VANAF 4.4 32 kg⋅m⋅s-1/ N·s  to the right/opposite direction /na regs /teenoorgestelde  rigting 

NOTE

• If any of the underlined key words/phrases in the correct context is omitted  deduct 1 mark. If the word work is omitted 0 marks  .

OPTION 1
P =  W✔ 
       Δt
= 4,8×10⁶✔ 
      (90) 
 = 53 333,33 W  
 = 5,33 x 104 W (53,33 kW) ✔ 

OPTION 2

 W_F = FΔxcosθ 
4,80 x 10⁶ = F(1 125)cos0º
 F = 4 266,667 N 
P_ave = F_vave 
 = (4 266,667)(12,5) 
 = 53 333, 33 W 

NOTE

If any of the underlined key words/phrases in the correct context is omitted  deduct 1 mark. .

OPTION 1
W_net = ΔK✔ 
W_w + W_f + WF = ½ mv_f²- ½ mv_i² 
mgΔxcosθ + W_f + W_F = ½ mvf2- ½ mv_i² 
(1 500)(9,8)200cos180º✔ + W_f + 4,8 x 10⁶ ✔= ½ (1 500)(25² – 0) ✔
-2 940 000 + W_f +4,8 x 10⁶ = 468 750 
W_f = -1 391 250 J 
 = -1,39 x 10⁶ J✔ 

OR

W_net = ΔK✔ 
W_w + W_f + W_F = ½ mvf²- ½ mvi² 

-ΔEp + W_f + W_F = ½ mvf²- ½ mvi² 
-(1 500)(9,8)(200 – 0)✔ + W_f + 4,8 x 10⁶ ✔= ½ (1 500)(25² – 0) ✔
-2 940 000 + W_f +4,8 x 106 = 468 750 
Wf = -1 391 250 J 
 = -1,39 x 106 J ✔ 

NOTE

• 0 can be omitted in above substitutions. 

OPTION 2
1 mark for any of  these
W_nc = ΔK + ΔU 
W_nc = ½ mvf²- ½ mvi² + mgh_f - mgh_i 
 = ½ m (v_f²- v_i²) + mg(h_f - hi) 
W_nc = ½ mv_f² + mgh_f - ½ mv_i²- mgh_i 
 W_f + W_F = ½ mv_f²- ½ mv_i² + mgh_f - mgh_i 
W_f + 4,8 x 106 ✔= [½ (1 500)(25)² + -0] ✔+[(1 500)(9,8)(200) - 0]✔ 
W_f = - 1,39 x 106 J (-1,40 x 106J )✔ 

OR

1 mark for any of  these
W_nc = ΔK + ΔU 
Wnc = ½ mvf²- ½ mvi² + mghf - mghi 
 = ½ m (vf²- vi²) + mg(hf - hi) 
Wnc = ½ mvf² + mghf - ½ mvi²- mghi 
Wf + 4,8 x 106 ✔= [½ (1500)(25)² + (1500)(9,8)(200)] - [0 + 0] 
Wf = - 4,8 x 106 + 3,4 x 106 
 = - 1,39 x 106 J (-1,40 x 106J )✔ 

ACCEPT THE FOLLOWING FOR: 3 /5
POSITIVE MARKING FROM 5.3
v_f = v_i + aΔt 
25 = 0 + a(90) 
a = 0,277…m\s-2 
F_net = ma  
 =(1 500)(0,2777…) = 416,66… N 
F + (w||) + (-fk) = 416,666… 
4 266,6667 – 1 500(9,8)sinθ - fk = 416,666… 
fk = 1 236,6667 N 
W_f = fkΔxcosθ 
 = (1 236,6667)(1 125)(cos180°)
 = - 1 391 250 J 

NOTE

If any of the underlined key words/phrases in the correct context is omitted  deduct 1 mark. 

v = fλ ✔ 
340 = f (0,34) ✔ 
 f = 1 000 Hz ✔ 

POSITIVE MARKING FROM 6.3

Accept:

Q_net  = Q₁ + Q₂ + Q₃ 
                     3 
-3 x 10^-9 = 15 × 10^-9 +  Q  + 2x10^-9 
                                  3 
Q = + 4 x 10^-9 C✔ 
NOTE

• for addition of the three correct charges 
• correct answer 

NOTE

• If any of the underlined key words/phrases in the correct context is omitted  deduct 1 mark. If masses used (0/2) 

POSITIVE MARKING FROM 7.4 
OPTION 1
E = F
       q
= 8,15 × 10^-6
    3 ×  10^-9
 = 2,72 x 103 N.C^-1 ✔ 

OPTION 2
E_S = KQ
         r²
= (9×10⁹)(3×10^-9) 
         (0,1)²
 = 2 700 N.C^-1 

E_T = KQ
         r²
= (9×10⁹)(3×10^-9) 
         (0,3)²
 = 300 N.C^-1 

E_net = √ (E_S² + E_T²)
= √  (2700)² +(30)²
 = 2 716,62 N.C^-1  

NOTE
Mark Allocation

• Correct formula
• both subsitutions
• correct answer

If calculation done in 7.4 award full marks for answer written here.

NOTE
If any of the underlined key words/phrases in the correct context is omitted  deduct 1 mark. 

OPTION 1
V_lost = I r 
 = (2) (0,5) 
 = 1 V 
V_ext  = Emf – V_lost 
 = (12 - 1) 
 = 11 V

OPTION 2
ε = I(R + r) 
12 = V_ext +(2)(0,5) 
V_ext = 11 V

OPTION 3
ε = I(R + r)
12 = 2(R + 0,5)
R = 5,5 Ω 
V = IR  
 = 2(5,5)
 = 11 V 

POSITIVE MARKING FROM 8.2.1
OPTION 1
R = V 
       I
 =  11 
      2
 = 5,5 Ω

OPTION 2
0,5:R 
 1:11 
R = 5,5 Ω 

OPTION 3
 1  =  11 
0,5    R 
R = 5,5 Ω 

OPTION 4
V_total = IR_total 
12 = (2)R_total 
R_total = 6 Ω
R = 6 – 0,5 
 = 5,5 Ω 

OPTION 5 
ε = I(R + r) 
12 = 2(R + 0,5) 
R = 5,5 Ω 

ε = slope (gradient) of the graph✔

• Accept any correct values from the graph 

ε = 7,5 -(-3)
       1,5 -0 
 = 7 V  ✔

OR

POSITIVE MARKING FROM 9.2.1 
R = ε- r 
       I

• Accept any correct values on the line from the graph 

7,5 = 1,5ε -3
ε= 7 V  ✔

OR

ε = I(R + r) 
 = 0,5(11 + 3) 
ε= 7 V  ✔

OPTION 3
P_ave =  V_rmsI_rms 
1600 = (240,416) I_rms 
I_rms = 6,66 A 
R = V_rms
       I_rms 
= 240,416 
       6,66 
 = 36,1 Ω (36,09 Ω )

(Do not penalise if rms is omitted V) in R = V_rms
                                                                     I_rms  

OPTION 4
P_ave  =  V_maxI_max
                    2 
1600 = 340I_max 
              2 
I_max = 9,412 A 
R = V_rms
       I_rms 

=   340    
    9,412 
 = 36, 12 Ω 

(Do not penalise if rms is omitted V) in R = V_rms
                                                                     I_rms 

Work function of a metal is the minimum energy needed to eject an electron  from the metal surface ✔✔

NOTE
If any of the underlined key words/phrases in the correct context is omitted  deduct 1 mark. 

Potassium / Kalium / K ✔
f_o for potassium is greater than fo for caesium 
OR
Work function is directly proportional to threshold frequency ✔
ACCEPT
W_o = hf_o

W_o α f_o

OPTION 1
c = fλ 
3 x 10⁸ = f(5,5 x 10^-7) 
f = 5,45 x 10¹⁴ Hz 
f^uv < f_{o of K(potassium)} 
∴Ammeter in circuit B will not show a reading 

OPTION 2
E = hc =  (6,63 × 10^-34)(3 × 10⁸) 
       λ                5,5 × 10^-7

 = 3,6164 x 10-19 J 
W_o = hf_o = (6,63 x 10^-34)(5,55 x 10¹⁴) = 3,68 x 10^-19 J 
W_o > E or hf_o > hf 
∴The ammeter will not register a current 
Mark allocation

• both correct formulae: E = hc and W_o = hf_o 
                                              λ
• both substitutions:  (6,63 × 10^-34)(3 × 10⁸) 
                                            5,5 × 10^-7
   (6,63 x 10^-34)(5,55 x 10¹⁴)
• correct conclusion

OPTION 3
c = f_oλ_o
3 x 10⁸ = (5,55 x 10¹⁴)λ 
λ_o = 5,41 x 10^-7 m 
λ_o (threshold wavelength) < λ (incident wavelength) 
∴ the ammeter will not register a current  

OPTION 1

NOTE: If EK of the incorrect   photocell is calculated, candidate   forfeit the mark for the final  answer. 
(6,63 × 10^-34 )(3 x 10⁸) = (6,63 x 10^-34)(5,07 x 10¹⁴) + E_k(max) 
      5,5  ×10⁷
 E_K = 2,55 x 10^-20 J  (Range: 2,52 x 10^-20 – 2,6 x 10^-20 J) 

OPTION 2
POSITIVE MARKING FROM 11.3

NOTE: If EK of the incorrect  photocell is calculated, candidate  forfeit the mark for the final  answer. 
(6,63 x 10^-34)(5,45 x 10¹⁴) = (6,63 x 10^-34)(5,07 x 10¹⁴) + E_k(max) 
E_K = 2,52 x 10^-20 J  (Range: 2,52 x 10^-20 – 2,6 x 10^-20 J)