PHYSICAL SCIENCES: PHYSICS (PAPER 1)
GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
MEMORANDUM
MAY/JUNE 2019

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

QUESTION 2
2.1.1

Notes

• Mark awarded for label and arrow 
• Do not penalise for length of arrows since drawing is not to scale.
• Any other additional force(s) Max ³/₄
• If force(s) do not make contact with body: Max: ³/₄
• Deduct 1 mark for an arrow/arrows omitted (4)

2.1.2 It is moving at constant speed in a straight line/, the acceleration is zero/ the net force (resultant) acting on it is zero/it is moving at constant velocity (1)
2.1.3 Fnet = ma
Fnet = 0
Fx= f
Fx – f = 0
Fcos 40º- f = 0
90 cos 40º – f = 0✓
f = 68,94 N✓

OR
Fnet = ma
Fnet = 0
Fx= f
Fx – f = 0
Fcos 320º- f = 0
90 cos 320º – f = 0✓
f = 68,94 N✓
✓any one

Fnet = ma
Fnet = 0
Fx = f
Fx – f = 0
Fsin 50º – f = 0
90 sin 50º – f = 0✓
f = 68,94 N✓
✓any one

NOTE:
1 mark for formula/equation, 1 mark substitution with zero, 1 mark answer.
2.1.4 POSITIVE MARKING FROM 2.1.3
OPTION 1
vf = vi + aΔt
2 = 0 ✓+ a(3) ✓
a = 0,67 m∙s^-2

F_net = ma✓
F cos 40º ✓– 68,94 ✓ = 15 (0,67)
F = 103,11N✓ (103,05 N – 103,11N)

Fnet = ma✓
Fcos 320º – f = 15(0,67)
F cos 320º ✓– 68,94✓ = 15(0,67)
F = 103,11N✓

POSITIVE MARKING FROM 2.1.3
OPTION 2
Fnet.Δt = Δp✓
Fcos 40º ✓- (68,94) ✓(3) ✓= 15(2 – 0) ✓
F = 103,11 N✓

POSITIVE MARKING FROM 2.1.3 
OPTION 3
Fnet = ma
Fx – fk = ma
Fx – 68,94✓ = 15(2 - 0)
                               3
Fx = 78,94 N
tanθ = Fy
           F_x
tan40º=   Fy   
             78.94
Fy = 66,24 N
F² = F²_x + F²_y
F² = (78,94)² + (66,24)²✓
F = 103,05 N✓

POSITIVE MARKING FROM 2.1.3
OPTION 4
Δx =v1 + vf=Δt
            2
=(2 + 0)
      2
Δx = 3m
W_net = ΔK
W_F + W_f = ΔK
FΔx cos θ + fΔx cos θ = ΔK
F(3)cos40º ✓+ 68,94(3)cos180º ✓ = ½ (15)(2²) – ½ (15)(0)²
F = 103.06 N✓ (6)
2.2 OPTION 1
G = m₁m₂
           r²
20= (6.67 x 10^-11)m _planet(10)  
                              (6 x 10⁵)²
m_planet = 1,08 x 10²² kg✓

OPTION 2
w = mg
20 = (10)(g)✓
g = 2 m·s^-2
g = GM 
       R²
2 =(6.67 x 10^-11)M
           (6 x 10⁵)²
M = 1,08 x 10²² kg✓ (4)
[18]

QUESTION 3
3.1 Motion of an object under the influence of gravity/gravitational force (weight) only✓✓
OR
Motion in which the only force acting on the object is gravity/weight. ✓✓
ACCEPT
Vertical motion in which friction/air resistance is absent. ✓✓
Motion in air with an acceleration of 9,8 m∙s^-2. ✓✓ (2)
NOTE: 2 OR ZERO
3.2.1 OPTION 1
Upwards positive:
vf = vi +aΔt ✓
0 = vi+(-9,8) (1,53) ✓
∴ vi = 14,99 m∙s^-1 (15 m∙s^-1) ✓
Downwards positive
vf = vi +aΔt ✓
0 = vi+(9,8) (1,53) ✓
∴ vi = -14,99 m∙s^-1
vi = 14,99m∙s^-1 (15 m∙s^-1) ✓

OPTION 2
Fnet = ma
= 9,8 (m)
Fnet Δt = mΔv✓
(9,8)(m)(1,53) = (m)(vf – 0) ✓
vf = 14,99 m∙s^-1 (15 m∙s^-1) ✓

OPTION 3
Upwards positive
Δy = viΔt + ½aΔt² ✓
0 = vi(3,06) + ½ (- 9,8)(3,06)²✓
vi = 14,99 m∙s^-1✓(15 m∙s^-1)
Downwards positive
Δy = viΔt + ½aΔt² ✓
0 = vi(3,06) + ½ (9,8)(3,06)²✓
vi = 14,99 m∙s^-1✓ (15 m∙s^-1)
NOTE: initial and final velocities can be swopped if starting from top, as long as sign of g is changed accordingly.(3)
3.2.2 OPTION 1
POSITIVE MARKING FROM 3.2.1
Upwards positive
Δy = viΔt + ½aΔt² ✓
= 14,99 (1,53) + ½ (- 9,8)(1,53)²✓
= 11,47 m✓(11,46-11,48)
Maximum height is 11,47 m
Downwards positive
Δy = viΔt + ½aΔt² ✓
= -14,99 (1,53) + ½ (9,8)(1,53)² ✓
= -11,47 m (11,46-11,48)
Maximum height is 11,47 m✓

OPTION 2
POSITIVE MARKING FROM 3.2.1
Upwards positive:
vf2 = vi² + 2aΔy✓
0 = (14,99)2 + 2(-9,8)(Δy)✓
Δy= 11,47 m∙✓(11,46-11,48)
Maximum height reached is 11,47 m
Downwards positive:
vf2 = vi² + 2aΔy✓
0 = (-14,99)² + 2(9,8)(Δy)✓
Δy= -11,47 m∙(11,46-11,48)
Maximum height reached is 11,47 m✓

OPTION 3
POSITIVE MARKING FROM 3.2.1
Upwards positive:

Δy = 11,47 m✓
Maximum height reached is 11,47 m
Downwards positive:

Δy = -11,47 m (11,46-11,48)
Maximum height reached is 11,47 m✓

OPTION 4
POSITIVE MARKING FROM 3.2.1
ΔE = ΔK + ΔU
½ mvi² + mghi = ½ mvf² + mghf
½ (14,99⁴)² + (9,8)(0) = 0 + 9,8 hf ✓
hf = 11,47 m✓ (11,46-11,48)
Maximum height reached is 11,47 m
OR
ΔK = - ΔU✓
½ m(vf² – vi²) = - mg(hf – hi)
½ (0 - 14,99²) = - 9,8( hf - 0) ✓
hf = 11,47 m (11,46-11,48)
Maximum height reached is 11,47 m✓ (3)
3.3 OPTION 1
POSITIVE MARKING FROM 3.2.1
Upwards positive:
Δy = viΔt + ½aΔt² ✓
= (14,99) (4) + ½ (- 9,8)(4)²✓
= -18,4 m
Position is 18,4 m downwards (below the edge of the roof) ✓
Downwards positive
Δy = viΔt + ½aΔt² ✓
= (-14,99) (4) + ½ (9,8)(4)²✓
= 18,4 m
Position is 18,4 m downwards (below the edge of the roof) ✓
1 mark for any✓

OPTION 2
POSITIVE MARKING FROM 3.2.1
Upwards positive
vf = vi +aΔt
= (14,99) + (-9,8) (4)
= - 24,2 m∙s^-1
vf² = vi² + 2aΔy✓
(-24,2)² = (14,99)² + 2(-9,8)(Δy)✓
Δy= - 18,4 m∙
Ball is 18,4 m downwards (below the edge of the roof)
Downwards positive
vf = vi +aΔt
= (-14,99) + (9,8) (4)
= 24,2 m∙s^-1
vf2 = vi² + 2aΔy✓
(24,2)² = (-14,99)2 + 2(9,8)(Δy)✓
Δy= 18,4 m∙
Ball is 18,4 m downwards (below the edge of the roof)

OPTION 3
POSITIVE MARKING FROM 3.2.1
Upwards positive/Opwaarts positief:

Δy =- 18,4 m

vf = vi +aΔt
= (-14,99) + (9,8) (4)
= 24,2 m∙s^-1
Ball is 18,4 m downwards (below the edge of the roof) ✓
Downwards positive

Δy = 18,4 m
Ball is 18,4 m downwards (below the edge of the roof) ✓

vf = vi +aΔt
= (14,99) + (-9,8) (4)
= - 24,2 m∙s^-1

OPTION 4
Total time to return to starting point/totale tyd terug na beginpunt
= 2(1,53) = 3,06 s
∴ time from reference point to ground/ tyd vanaf verwysingspunt tot by grond = (4 – 3,06) = 0,94 s
Δy = viΔt + ½ (g)Δt² ✓
= (14,99)(0,94) + ½(9,8)(0,94)² ✓
= 18,43 m ✓downwards (below the edge of the roof). (3)
3.4 No✓
The motion of the ball is only dependent on its initial velocity✓✓/the initial velocity depends on the time taken to reach maximum height.
ACCEPT for 1 mark
The ball will still be in the air.✓
OR
The ball is still falling.✓
OR
The ball would not have reached the ground.✓
OR
The motion of the ball is independent of the height of the building. ✓
NOTE: If learners gave separate answers for 3.2 and 3.3, mark them together. Thus, if one answer is correct and the other incorrect 0/3 (3)
[14]

QUESTION 4
4.1 The total (linear) momentum in a isolated/closed system remains constant./ is conserved ✓✓
OR
In an isolated/closed system the total momentum before a collision is equal to the total momentum after the collision. ✓✓
NOTE:
-1 for each key word/phrase omitted.
Take the whole statement in context  (2)
4.2 OPTION 1
Σpi = Σpf
m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f
m₁v_1i + m₂v_2i = (m₁ +m₂)_vf
{0,45(9) + 0,20(0)}✓ = (0,45 + 0,20)v✓
v = 6,23 m⋅s^-1✓
1 mark for any
OR
Δpball = - Δpcont✓
0,45(v – 9) ✓ = - 0,2(v – 0) ✓
v = 6,23 m⋅s^-1✓
If – sign omitted from formula 0/4

y6yOPTION 2
Σpi = Σpf
pf Total = pi Total
(Thus change in total momentum = 0
0 ✓= (0,65vf) – (9)(0,45)✓
vf = 6,23 m∙s^-1✓ (4)
1 mark for any✓
4.3 POSITIVE MARKING FROM 4.2
K = ½ mv²✓ (or EK = ½ mv²)
Total kinetic energy before collision:
½ (0,45)(9)2 + 0✓= 18,225J
Total kinetic energy after collision:
½ (0,45 + 0,20)(6,23)²✓ = 12,614J
ΣK_before ≠ ΣK_after
Collision is inelastic.
If start with ΣEKi =ΣEKf 4/5 max
No calculation: 0
Do not accept a conclusion of inelastic collision based on any other calculation (such as that of momentum or mechanical energy). (5)
[11]

NOTE: maximum ¾ if friction and tension are not on a straight line
NOTE: maximum ¾ if N and w⊥ are not on a straight line

QUESTION 5
5.1 Tension/Spanning✓ (1)
5.2 There is friction/ tension in the system ✓
OR
Friction/tension is a non-conservative force ✓
OR
The system is not isolated because there is friction/tension 
OR
The internal energy increases because of friction ✓
OR
The applied force is non-conservative✓
OR
It is not an isolated system✓ (1)
5.3

ACCEPT

Notes

• Mark awarded for label and arrow 
• Do not penalise for length of arrows since drawing is not to scale.
• Any other additional force(s) Max ³/₄
• If force(s) do not make contact with body: Max: ³/₄

Deduct 1 mark for an arrow/arrows omitted

(4)
5.4 W = FΔxcosθ✓
Wf = [178,22(4)cos180º] ✓
= - 712,88 J✓ (3)
5.5 OPTION 1
POSITIVE MARKING FROM QUESTIONS 5.4 
Wnet = ΔEK
Wf + Wg + W_T = ΔK
Wf + mgsinθΔxcosθ + W_T = ΔK
-712,88 + (70)(9,8) (sin 30º)(4) cos 180º ✓+ (700 x 4 x cos 0º)✓ = ½ 70(vf² – 0)✓
vf = 4,52 m⋅s^-1✓
NOTE: Wg can be obtained using any of the following formulae:
W_gravity = mgΔxcosθ
= (70)(9,8)(4).(cos 120º)
∴-712,88 + (70)(9,8)(4)cos 120º) ✓+ (700 x 4 x cos 0º)✓ = ½ 70(vf² – 0)✓
vf = 4,52 m⋅s^-1✓
Wgravity= -Δ mgh = - mg(hf – h₀)
= mgΔycosθ
= ((70)(9,8) 4(sin 30º).cos 180º
Wgravity = mgsinθΔxcosθ
= (70)(9,8) (sin 30º)(4).cos 180º
1 mark for any one✓

OPTION 2
POSITIVE MARKING FROM 5.4 
Wnc = ΔE_K + ΔE_p ✓
WT + Wf = ΔEK + ΔEp
(700)(4) cos 0º)✓ + (-712,88) = [(70)(9,8) 4(sin 30º). - 0 ]✓+ ½ 70(vf² – 0)✓
vf = 4,52 m⋅s^-1✓

OPTION 3
Fnet = FT – [mgsinθ + f_k]
= 700 – [(70 x 9,8 sin 30º) + 178,22] ✓
= 178,78 N
W_net = ΔEK✓
F_net . Δxcosθ =ΔE_K
(178,78)(4)cos0 º ✓=.½ 70(vf² – 0)✓
vf = 4,52 m⋅s^-1✓ (5)
5.5 WHERE EQUATIONS OF MOTION ARE USED:
F_net = ma
FT – [mgsinθ + fk] = ma
700 – [(70 x 9,8 sin 30º) + 178,22] ✓= 70a
a = 2,554 ms^-2
vf² = vi² + 2aΔx
= 0 + 2(2,554)(4)
vf = 4,52 m⋅s^-1
5.6 POSITIVE MARKING FROM 5.4
2(-712,88) = -1425,76 J✓
OR
Double the answer (in question 5.4). ✓ (1)
[15]

QUESTION 6
6.1.1 Δx = viΔt + ½ aΔt²
300 = vi (10) ✓
vi = 30 m∙s^-1✓
NOTE:
AcceptΔx = viΔt (2)
v = d  =  300  = 30m-s^-1
      t        10
6.1.2 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 sound propagation. ✓✓
OR
An (apparent) change in observed/detected frequency (pitch), (wavelength) as a result of the relative motion between a source and an observer (listener). ✓✓
NOTE:
-1 for each key word/phrase omitted.(2)
6.1.3 Car/source (just) passes observer ✓✓
Accept:
Car moves away from observer ✓✓
No relative motion between car and observer ✓✓
Car and observer at the same place/position✓✓ (2)
6.1.4 POSITIVE MARKING FROM 6.1.1
fL = v ± vLf_s    OR fL=    v    f_s
       v ± v_s                    v - v_s
932=   340   
       340 - 30
fs = 849,76 Hz ✓ (4)
Notes:

• Any other Doppler formula, e.g.
  fL=   v   f_s 
        v - v_s

Marking rule 1.5: No penalisation if zero substitutions are omitted.
6.2 Doppler / Blood flow meter
Measuring the heartbeat of a foetus
Radar
Sonar
Used to determine whether stars are receding or approaching earth(2)
[12]

QUESTION 7
7.1 The electric field at a point is the electrostatic force experienced per unit positive charge placed at that point. ✓✓
NOTE:
-1 for each key word/phrase omitted. If definition of electric field: 0/2 (2)
7.2 q₂ is positive ✓
The electric field due to q₁ points to the right because q₁ is negative. ✓ Since the net field is zero, field due to q₂ must point to the left away from q₂, ✓ hence q₂ is positive.
OR
q₂ is positive ✓
Since Enet is zero,E₁ and E₂ are in opposite directions✓ therefore q₁ and q₂ are oppositely charged. ✓ (3)
7.3 E =k Q 
               r²
Enet = 0
∴k q₁ =kq₂  OR q₁ = q₂
   r₁²     r₂²        r₁²     r₂²
(9 x 10⁹)(3 x 10^-9) = (9 x 10⁹)q₂
         (0.1)²                    (0.1)⁴
q₂ = + 4,8 x 10^-8 C✓ (4)
1 mark for formula
1 mark for equating the two fields
1 mark for both substitutions
1 mark for answer
7.4 The electrostatic force (of attraction/repulsion) between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. ✓✓
NOTE:
-1 for each key word/phrase omitted.If masses used instead of charges 0 (2)
7.5 POSITIVE MARKING FROM 7.3
E =k Q₁Q₂
          r²
F = (9 x 10⁹)(3 x 10^-9)(4.8 x 10^-8)
                         (0.3)²
= 1,44 x 10^-5 N✓ (3)
7.6 POSITIVE MARKING FROM 7.2 AND 7.3
YES✓
Both charges are equal and positive ✓
Accept calculation which shows charges the same and positive
If the answer is YES, mark according to the memo, if NO check 7.2 first for sign of charge. If stated NEGATIVE at 7.2, then answer is:
No✓, the direction is incorrect. ✓ (2)
[16]

QUESTION 8
8.1.1 The rate at which (electrical) energy is converted (to other forms) (in a circuit)
The rate at which energy is used/Energy used per second
The rate at which work is done ✓✓
(2 or zero) (2)
8.1.2

(3)
8.1.3 POSITIVE MARKING FROM 8.1.2
OPTION 1
 1  =  1  + 1 
 R_//    R₁   R₂
= 1  + 1 
  24   24
R_// = 12 Ω
R_ext = (R_s + R_//)
R_ext = (24 +12) ✓
= 36 Ω
V = IR
OR
ε = I(R + r)
12 = I(36 + 2) ✓
I = 0,32 A✓(0,316 A)

R_tot= (R_s +  R₁R₂ )
                   R₁R₂
R_tot= (24 + (24)(24) )
                      48
= 36 Ω

POSITIVE MARKING FROM 8.1.2
OPTION 2
 1  =  1  + 1 
 R_//    R₁   R₂
= 1  + 1 
  24   24
R_// = 12 Ω
Rext = (R_s + R_//)
Rext = (24 +12) ✓
= 36 Ω

P = I²R =  V ²
                 R
I²(36 + 2) = (12)²
                    38
I = 0,32 A✓(0,316) ✓ (5)

8.1.4 POSITIVE MARKING FROM 8.1.3
OPTION 1
V = IR
V = I(RA+r)
= 0,316(26) ✓
= 8,216 V (8,32 V)
V// = (12 – 8,216) ✓
= 3,784 V(3,68 V)
∴VC = 3,78 V (3,68 V)✓

POSITIVE MARKING FROM 8.1.3
OPTION 2
V = IR
For the parallel portion (or from 8.1.3):
 1  =  1  + 1  = R =  R₁R₂ 
 R     R₁   R₂         (R₁R₂)
R = (24)(24) 
          48
= 12 Ω
V// = VC ✓
V = IR_//
= (0,316)(12) ✓
= 3.79 V (3,84 V)✓ (3)

POSITIVE MARKING FROM 8.1.3
OPTION 3
I_A = I_B + I_C
= 2 I_B
0,316 = 2I_B✓
I_B = 0,158 A
V = 0,158 (24) ✓
= 3,79 V✓ (3)

8.1.5 OPTION 1
The power rating (output voltage) of the bulb is 6 W, 12 V.
P = V²
      R
[For a given resistance, power is directly proportional to V²] ✓
Since the potential difference across light bulb C is less than the operating voltage,✓ the output/power will be less,✓

OPTION 2
P = V²
      R
The potential difference across light bulb C is less than the operating voltage. ✓ Thus for the same resistance, ✓ brightness decreases.

OPTION 3
P =I²R
For a given resistance✓, power is directly proportional to I² Since current decreases✓, brightness decreases.]

OPTION 4
P =I²R
In the circuit, the total current in light bulb C is less than the optimum current required (0,5 A). ✓ Thus for the same resistance, ✓ the power will be less✓ hence brightness will decrease.

OPTION 5
P = IV✓ [Power is directly proportional/equal to product of V and I. ✓
Since current decreases✓, brightnes decreases
OR
The voltage across light bulb C, as well as the current in the bulb are all less✓ than the optimum values✓ hence power is less✓ and brightness is less. (3)
NOTE: No mark if only equation is given.
8.2.1 The total current passes through resistor A. ✓ For the parallel portion, the current branches, therefore only a portion of the total current passes through resistor C.✓ (2)
ACCEPT for 1 mark: Resistor C is connected parallel to resistors B and D together. Current is dividing ✓at the junction.
8.2.2 The current in B is equal✓ to the current in A. The circuit becomes a series circuit.✓(2)
[21]

QUESTION 9
9.1 Slip rings✓(1) 
9.2 B ✓ (1)
9.3 2 V_rms = V_max
                      √2
= 312 
    √2
= 220,62 V ✓ (3)
9.4.1 POSITIVE MARKING FROM 9.3
OPTION 1
P_aver= V²rms 
              R
=(220,62)²
      40
= 1216,83 W ✓

POSITIVE MARKING FROM 9.3
OPTION 2
I_rms = Vrms 
            R
=(220,62)²
      40
= 5,515
P_ave = I²rmsR
= (5,515)²(40) ✓
= 1216,61 W ✓
OR
P_ave = V_rmsI_rms
= (220,62)(5,515) ✓
= 1216,72 W✓
✓for any

OPTION 3
I_max = V max
              R
= 312 
    40
= 7,80 A
P_ave =I_max V_max
                  2
= (7.8)(312) 
          2
= 1216,80 W (3)
9.4.2 OPTION 1
I_max = V_max
            R
= 40312
= 312 
    40
= 7,8 A

POSITIVE MARKING FROM 9.3 AND 9.4.1
OPTION 2
P_ave = V_rmsI_rms
1 216,83 = 220,62 Irms ✓
Irms = 5,515 A
Irms = I_max
           √2
5,515 = I_max
             √2
I_max = 7,8 A✓

OPTION 3
P_ave = I²_rmsR
1 216,83 = I²_rms(40)✓
I_rms = I_max
           √2
5,515 = I_max
             √2
I_max = 7,8 A✓ (4)
[12]

QUESTION 10
10.1 The minimum frequency of light needed to eject electrons from a metal (surface)✓✓
NOTE:
-1 for each key word/phrase omitted.(2)
10.2 OPTION 1
E= h c 
        λ
= (6.63 x 10^-34)(3 x 10⁸)
              5×10^-7
= 3,98 x 10^-19 J✓

OPTION 2
c = fλ
3 x 10⁸ = f(5 x 10^-7)
f = 6 x 10¹⁴ Hz
E = hf
= (6,63 x 10^-34)(6 x 10¹⁴) ✓
= 3,98 x 10^-19 J✓ (3)
NOTE: do not penalise if v is used in place of c.
10.3 OPTION 1
POSITIVE MARKING FROM QUESTION 10.2
E = W₀ + Ek_max
hf = W₀ + ½mv²max
h c = W₀ + EK(max)
   λ 
h c = hf₀ + EK(max)
   λ
3,98 x 10^-19 = (6,63 x 10^-34)(5,55 x10¹⁴) + EK(max)✓
EK(max) = 3,0 x10^-20 J✓
EK(max) > 0 ✓
(The electrons emitted from the metal plate have kinetic energy to move between the plates, hence the ammeter registers a reading.
1 mark any one
✓Both equations

OPTION 2
POSITIVE MARKING FROM QUESTION 10.2
W_o = hf_o ✓
= (6,63 x 10^-34)(5,55 x10¹⁴) ✓
= 3,68 x 10^-19 J
Ephoton > W_o ✓
(The energy of the incident photon is greater than the work function of potassium. From the equation hf = W₀ + EKmax, the ejected photoelectrons will move between the plates,✓ hence the ammeter registers a reading.

OPTION 3
c = fλ ✓
3 x 10⁸ = f(5 x 10^-7) ✓
f = 6 x 10¹⁴ Hz
f > f₀ ✓
The frequency of the incident photon is higher than the threshold frequency.
From the equation hf = hf₀ + EK(max), the ejected photoelectrons will be able to move between the plates ✓ (for the given frequency), hence the ammeter registers a reading. (4)
10.4 The increase in intensity increases the number of photons per second.✓
Since each photon releases one electron✓ the number of ejected electrons per second increases.✓
ACCEPT: Flow of electrons per unit time increases ✓ (1 mark)
This causes the current /ammeter reading to increase. (3)
[12]
TOTAL: 150