Monday, 07 March 2022 09:33

## PHYSICAL SCIENCES PAPER 1 GRADE 12 MEMORANDUM - NSC EXAMS PAST PAPERS AND MEMOS NOVEMBER 2020

PHYSICAL SCIENCES: PHYSICS (PAPER 1)
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
NOVEMBER 2020

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

Accepted symbols
N - FN/Normal/Normal force/173,5N /Normaal
f -  Ff /fk /frictional force/kinetic frictional force/5 N
w - Fg /mg/Weight/FEarth on block/Fw/Gravitational force/196 N
T - Tension/FT
Fapplied  - F/Applied force/35 N/ FA

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 – FAx = 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 20 kg of m kg / Fnet = 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/finale antwoord: 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
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 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/Afwaarts positief
vf = vi + aΔt ✓
0 = vi + (9,8)(0,62) ✓
vi = -6,08
6,08 m·s-1 (6,076 m·s-1 )✓

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

OPTION 4
Upwards positive
Δy = viΔt + ½ aΔt
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
(mgh + ½mv2)top = (mgh + ½mv2)Bot
(9,8)(1,2) + 0 = 0 + (½)v2 ✓
vi = 4,85 m·s-1

OPTION 3
Wnc = ΔEp + ΔEk
0 = (0 – mgh) + ½m(v2f - v2i)
0 = -(9,8)(1,2) + ½vi2
vi = 4,85 m·s-1 upwards

OPTION 4
Wnet = ΔEk
wΔxcos180° = ½m(v2f - v2i)
(9,8)(1,2)cos180° = ½vi2
vi = - 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

Calculate time Δt
Upwards positive
Δy = viΔt + ½ aΔt2✓
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 + 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
mpvpi + mQvQi = mpvpf + mQvQf
(0,16)(10) + (0,2)(-15) ✓= (0,16)(-5) +(0,2)vQf ✓
vQf = -3 m∙s-1
vQf = 3 m∙s-1 ✓ west ✓

OPTION 2
West as positive
Σpi = Σpf
mpvpi + mQvQi = mpvpf + mQvQf
(0,16)(-10) + (0,2)(15) ✓= (0,16)(5) +(0,2)QNf ✓
vQf = 3 m∙s-1 ✓ west ✓

OPTION 3
Δpp = -ΔpQ✓
(0,16)(-5 – 10) ✓= -(0,2)(v – (-15))✓
v = -3 m·s-1
= 3 m·s-1 ✓west ✓✓✓ (5)
Any one

4.2.2 For ball:
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 /Wes as positief
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)
Any one

5.3 POSITIVE MARKING FROM QUESTION 5.2.
Marking criteria

• 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 = ✓K
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 = ✓K + ✓U
Wengine + Wf = ½mvf2 - ½ mvi2 + mghf - mghi
= ½m(vf2- vi2) + mg(hf - hi)
Wengine+ (50)(15)(2)cos180° = 0 + 200(9,8)(22 – 8,88)
Wengine = 27 215,20 J
Pengine =Wengine
t
=27 215,20
15
= 1 814,35 W
Any one

OPTION 2
Wnet = K
WN + Wengine + Ww + Wf = 0
WN + Wengine - ΔEp + Wf = 0
0 + Wengine - (200)(9,8)✓(13,12) + (50)(2)(15)cos180° ✓✓= 0
Wengine= 27 215,20 J
OR
Wnet = ✓K✓✓
WN + Wengine + Ww|| + Wf = 0
WN + Wengine +mgsinθΔxcos180o + Wf = 0
0 + Wengine - (200)(9,8)(13,12)Δx(-1)+ (50)(2)(15)cos180° = 0
Δx
Wengine= 27 215,20 J
Pengine = Wengine
Δt
=25 215,20
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 Pave = Fvave
• Final answer: 1 814,35 W - 1 824,8 W

OPTION 3
Fnet = ma
Fengine + Ffriction + Fg// = 0
Fengine + (-50) + (-200)(9,8)(0,4373) = 0
Fengine = 906,52 N (906,52 – 912,4)
Pave = Fvave
Pave = (908,52)(2)
= 1 813,04 W (1 824,8 W)

sin = h/Δt
=13.12
2(15)
= 0.4373

OR
W = FengineΔxcosθ
= (906,52)(30)cos0o
= 27 195,6 J (27 372 W)
P = W27195,6 = 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 fL = v ± vL fs  OR fL =     v       OR fL =    v     fs
v ± vS                      v - vs                 v + vs
3148 = 340 + 0 fs                           2073 = 340 - 0 fs
340 - vs                                          340 + vs
3148(340 - vs)2073(340 + vs
340 + 0                   340 - 0
vs = 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 =5 s ✓

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 =
e
=  (-)4 x 10 -6
(-)1,6 x 10-19
= 2,5 x 1013 (3)

7.2 Electrostatic force on B due to A:
FAB = kQ1Q2
r2
= [9 x 109(4 x 10-6)(3 x 10-6)]
0.22
= 2,7 N (3)
Ignore negative signs

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)

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
EAM = k
r2
=9 x 109(4 x 10-6)
(0.3)2
= 4,0 x 105 N∙C-1 (to left )
Electric field at M due to : +3 x 10-6 C,
EBM = k
r2
9 x 109(3 x10-6)
(0,1)2
= 2,7 x106 N∙C-1 (to right)

Net electric field at M
Enet = EBM + EAM
= 4,0 x105 – 2,7 x 106
= 2,3 x106 N∙C-1 (right)
OR
Net electric field at M
Enet = EBM + EAM
= -4,0 x 105 + 2,7 x 106
= - 2,3 x 106 N∙C-1
= 2,3 x 106 N∙C-1 (right) (5)

OPTION 2
FAM = kQ1Q (9 x 109)(4 x 10-6)Q = 4 x 105Q N
r2                  (0,3)2
FBM = kQ1Q (9 x 109)(4 x 10-6)Q = = 2,7 x 106Q N
r2                  (0,1)2
Fnet = 2,7 x 106Q + (-4 x 105Q) = 2,3 x 106Q
E = = 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
FAD = 7,2 N
r2
7,2 = (9 x109)(4 x10-6)Q
(0,15)2
QD = 4,5 x 10-6 C

OR
r2
= 9 x109(4 x10-6)Q
0,152
= 1,6 x 106Q
Fnet = √FAB+ FAD2 OR Fnet2 = FAB2 + FAD2
7,69 =√2,72 + (1,6x106Q)2
Q = 4,50 x10-6
[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
=
RP  R1    R2
=
RP  25    15
Rp = 9,375 Ω
Rext = 9,375 + 4 = 13,38 Ω
(13,375Ω)

OPTION 2
R= R1 + R2
R1 + R2
R=(25)(15)
25 + 15
Rp = 9,375 Ω
Rext = 9,375 + 4 = 13,38 Ω
(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 αI/R , V = constant
I is inversely proportional to R and V is constant.
OR
V|| = IR
= (3)(9,38)
= 28,14 V
IR2 =28,14 = 1,13 A✓
R       25
IR3 =28,14 = 1,88 A✓
R       15
OR
V is the same
I15Ω25/40 I
I25Ω15/40 I (3)

8.5 Remains the same✓ (1)
[18]

QUESTION 9
9.1.1 (DC) 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 Pave
• Substitution to calculate R ✓
• Final answer: 242 Ω

OPTION 1
Pave = V2rms
R
200= 2202
R
R = 242 Ω 

OPTION 2
Pave = Vrms Irms
200 = Irms (220)
Irms = 0,909 A(0,91)
R = Vrms or R =
Irms             I
200 =   220
0.909
R = 242 Ω (241,76 Ω)

OPTION 3
Pave = Vrms Irms
200 = Irms =  0,909 A (0,91)
Pave = Irms2R
200 = (0,909)2R ✓
R = 242 Ω
(241,52 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 ✓t
• 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
242
= 55 785.12J

OPTION 2
Pave = V2rms
R
Pav =  92,975 W
P = W/Δt
92,975 =    W
(10)(60)
W = 55 785,12 J ✓
(55785,12 – 55896 J)

OPTION 3
R = V/I
242 = 150
Irms
Irms = 0,620 A
Pave = Irms Vrms
= (0,62)(150)
= 92,97 W (93 W)
P =
Δt
92,975 =    W
(10)(60)
W = 55 785,12 J ✓
(55785,12 – 55896 J)

OPTION 4
R = V/I
242 = 150
Irms
Irms = 0,620 A
W = I2RΔt
= (0,62)2(242)(10)(60) ✓
= 55 814,88 J ✓
(55785,12 – 55896 J)
OR
W = VIΔt
= (150)(0,62)(600)
= 55 800 J

OPTION 5
Pave = V2rms = 1502 = 92.975W
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 Wo= hfo
= (6,63 x 10-34)(1,75 x 1015) = Wo + 0
= 1,160 x 10-18 J ✓
OR
E = Wo + Ek(max)
hf = Wo + Ek(max)
(6,63 x 10-34)(1,75 x 1015) = Wo + 0
Wo = 1,160 x 10-18 J  (3)

10.5.2 POSITIVE MARKING FROM QUESTION 10.5.1.
E = Wo + Ek(max)
hf = hfo + ½mv2max
(6,63 x 10-34)f  = 1,160 x 10-18 + ½(9,11 x 10-31) (5,60 x 105)2
f = 1,97 x 1015 Hz (4)
[13]
TOTAL: 150

Last modified on Tuesday, 15 March 2022 06:41