# Physical Sciences Grade 12 P1 NSC November 2026 Memo (Eng)

> Grade 12 Physical Sciences Paper 1 memorandum from the NSC November Exam 2026 (English), 21 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-physical-sciences-2026-nsc-november-exam-unofficial-memorandum-paper-1/3d3a4a7b-4da0-4918-b291-d3737a972511
- Updated: 2026-10-30
- Subject: Physical Sciences
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Exam
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 21

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## About this document

Physical Sciences · Grade 12 · NSC November Exam · 2026. Unofficial memorandum, Paper 1, 21 pages. Unofficial marking guidelines with full worked answers by Hlayiso for the Grade 12 Physical Sciences Paper 1 NSC November 2026 exam, plus how our prediction compared. Not the official DBE memorandum. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

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### Page 1

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Unofficial memo
hlayiso mock exams
Practice Series:
Grade 12 NSC Preparation
WWW.HLAYISO.COM
NATIONAL
SENIOR CERTIFICATE
GRADE 12
PHYSICAL SCIENCES: PHYSICS (P1)
NOVEMBER 2026 — UNOFFICIAL MARKING GUIDELINES
MARKS: 150
These marking guidelines consist of 21 pages.

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Physical Sciences/P1
2
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
IMPORTANT: These are unofficial marking guidelines prepared by Hlayiso for revision. They
are not the DBE’s official memorandum; the official marking guidelines may differ in mark
allocation and accepted alternatives.
GENERAL MARKING GUIDELINES
• Calculations: a tick (✓) shows one mark. Marks are given for the formula, the substitution and
the final answer WITH the unit. A formula on its own (no substitution) earns no mark; after a
wrong formula, no further marks are given for that calculation.
• Units: penalise a missing or wrong unit in a final answer once per question only. Accept answers
rounded to a minimum of TWO decimal places; ranges are given where rounding affects the
answer.
• Positive marking: an incorrect answer carried into a later calculation is marked positively.
• Negative marking: if the choice in a “choose and explain” item (e.g. GREATER THAN/SMALLER
THAN) is wrong, no marks are given for the explanation.
• Definitions and laws: deduct 1 mark for each underlined key word/phrase that is omitted or
used in the wrong context. “(2 or 0)” means no part marks.
• Diagrams and graphs: marked against the “Marking criteria” box. A force earns its mark only
with BOTH an arrow and a label.
• Sign convention: candidates may take either direction as positive if they apply it consistently.
Both versions are shown where the signs change.
READING OF THE QUESTION PAPER
• Drawn up from photographs of pages 2–18 of the question paper and the three data sheets (g =
9,8 m·s–2, k = 9,0×109 N·m2·C–2, h = 6,63×10–34 J·s, c = 3,0×108 m·s–1, me = 9,11×10–31
kg).
• Answers that depend on our reading of a diagram, or on how the DBE has marked a similar item
before, carry a short Hlayiso note. Hlayiso notes are for learners and teachers, not for marking.
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Physical Sciences/P1
3
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
QUESTION 1
1.1
B ✓✓
(2)
1.2
C ✓✓
(2)
1.3
D ✓✓
(2)
1.4
C ✓✓
(2)
1.5
B ✓✓
(2)
1.6
C ✓✓
(2)
1.7
B ✓✓
(2)
1.8
A ✓✓
(2)
1.9
D ✓✓
(2)
1.10
A ✓✓
(2)
[20]
Hlayiso notes (why each answer is correct):
1.1: Newton’s Second Law: a ∝Fnet (constant mass) and a ∝1
m (constant net force).
1.2: F = GMm
r2
with equal masses, so FA
FB = (2x)2
x2
= 4, i.e. 4 : 1.
1.3: When it is released, the stone has the balloon’s upward velocity, so (i) is false. It first rises and slows down
to zero at its highest point, so (ii) is false, and its maximum height is above 50 m, so (iii) is true.
1.4: A constant force gives a constant rate of change of momentum (a straight line). The ball reverses direction,
so its momentum changes sign during the contact time. Only graph C shows both.
1.5: Mechanical energy is conserved from B to C: Ek,C = Ek,B +mgh = E+mgh.
1.6: Graph A is steeper than B (fL > fs), so the source approaches and fL =
v
v–vs fs, with gradient
v
v–vs > 1. Graph
C (gradient < 1) is for a receding source, with gradient
v
v+vs .
1.7: E = kQ
r2 does not depend on the test charge q, so the field stays E (only the force on q doubles).
1.8: The battery has no internal resistance, so the total resistance doubles and the current halves. P = I2R, so
P1 = ( I
2)2R = 1
4P.
1.9: The current in AB (next to X) flows upwards from A to B. With X = North, the field points to the right, and the
motor rule (⃗F = I⃗L ×⃗B) gives a force INTO the page. A motor converts electrical energy to mechanical energy.
With X = South the force would be out of the page, which no option pairs with “electrical to mechanical”. Only D
is consistent.
1.10: One photon interacts with one electron, so photon energies do not add up. Each photon has only 1
2W0 < W0,
so no electron is ejected, however many photons arrive (A). B, C and D all assume an electron is ejected.
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Physical Sciences/P1
4
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
QUESTION 2
2.1
N ✓
w ✓
T ✓
F ✓
15◦
Accepted labels
w Fg / Fw / weight / mg / 19,6 N / gravitational
force / FEarth on B
N FN / normal (force) / Fnormal
T FT / tension / Fstring
F FA / 20 N / applied force
Notes
• Mark awarded for label AND arrow. Do not penalise for the length of the arrows (not
drawn to scale).
• F must act at an angle above the horizontal (to the right); T acts to the left (towards
block A).
• Block B is on the frictionless section RS. Any additional force (e.g. friction): Max 3
4.
• If everything is correct but there are no arrows: Max 3
4.
• If force(s) do not make contact with the dot: Max 3
4.
• Accept a force diagram (forces drawn on a block) instead of a dot.
(4)
2.2
If any of the underlined key words/phrases in the correct context is omitted, deduct 1
mark.
The force that opposes the motion of an object and which acts parallel to the
surface (with which the object is in contact). ✓✓
Accept: the force that opposes the motion of a moving object relative to a
surface (kinetic frictional force).
(2)
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Physical Sciences/P1
5
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
2.3
Marking criteria
hlayiso.com
• Formula Fnet = ma (any one) ✓
• Correct equation for block A: T–11 = 5a ✓
• Correct equation for block B (20cos15◦– T = 2a) OR for the system
(20cos15◦–11 = 7a) ✓
• Final answer: T = 16,94 N (range: 16,94 – 16,95 N) ✓
• Conclusion: T > 16 N, therefore the string breaks ✓
Direction of motion (to the right) as positive.
OPTION 1 (EACH BLOCK SEPARATELY)
Block A:
Fnet = ma ✓
T+(–11) = 5a ✓. . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Block B:
Fcos15◦+(–T) = 2a
20cos15◦–T = 2a ✓. . . . . . . . . . . . . . . . (2)
(1) + (2):
19,32–11 = 7a
⇒
a = 1,19 m·s–2
T = 11+5(1,19) = 16,94 N ✓
16,94 N > 16 N, therefore the string will break. ✓
OPTION 2 (SYSTEM, THEN ONE BLOCK)
Fnet = ma ✓
System:
20cos15◦–11 = (5+2)a ✓
⇒
a = 1,19 m·s–2
Block A:
T–11 = 5(1,19) ✓
OR
Block B:
20cos15◦–T = 2(1,19) ✓
T = 16,94 N ✓
> 16 N, so the string breaks. ✓
OPTION 3 (ASSUME THE STRING HOLDS AT 16 N)
Fnet = ma ✓
Block A:
16–11 = 5aA ✓
aA = 1,00 m·s–2
Block B:
20cos15◦–16 = 2aB ✓
aB = 1,66 m·s–2✓
B would accelerate faster than A, which a taut inextensible string cannot allow,
so the tension must be greater than 16 N: the string breaks. ✓
NOTE: If 20 N is used as the horizontal force (no cos15◦component): Max 3
5.
(5)
2.4
Block A: continues to move to the right with a decreasing velocity (it slows
down; friction is the only horizontal force on it). ✓
Block B: moves to the right with an increasing velocity (it accelerates; Fcos15◦
is now unopposed). ✓
Accept: A decelerates; B accelerates (at a greater rate than before).
(2)
[13]
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Physical Sciences/P1
6
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
QUESTION 3
3.1
An object upon which the only force acting is the force of gravity. ✓✓(2 or 0)
Accept: an object that experiences only the gravitational force (weight).
Hlayiso note: DBE Nov 2025 gave 0/2 when a learner defined a projectile instead of free fall.
Expect the same strictness the other way round: the definition must describe an object, not a
motion.
(2)
3.2
Equal in magnitude (same speed) ✓but opposite in direction (upwards at x,
downwards at t). ✓
(2)
3.3.1
UPWARDS AS POSITIVE
v2
f = v2
i +2a∆y ✓
02 = (20)2 +2(–9,8)∆y ✓
∆y = 20,41 m ✓
DOWNWARDS AS POSITIVE
v2
f = v2
i +2a∆y ✓
02 = (–20)2 +2(9,8)∆y ✓
∆y = –20,41 m; height = 20,41 m ✓
OPTION 2
vf = vi + a∆t: 0 = 20 + (–9,8)∆t, ∆t = 2,04 s;
∆y = vi∆t + 1
2a∆t2 ✓= (20)(2,04) +
1
2(–9,8)(2,04)2 ✓= 20,41 m ✓
Range: 20,40 – 20,41 m. NOTE: If energy principles are used: 1
3 (final answer only).
(3)
3.3.2
Marking criteria
hlayiso.com
• Time from h to the maximum height: 1
2(1,5) = 0,75 s ✓
• Formula ✓
• Correct substitution ✓
• Subtraction from the maximum height / correct method to reach h ✓
• Final answer: 17,65 m (range: 17,64 – 17,65 m) ✓
POSITIVE MARKING FROM QUESTION 3.3.1
The motion is symmetrical: the ball spends 1
2(1,5) = 0,75 s rising from h to the
top and 0,75 s falling back to h. ✓
OPTION 1 (FROM THE TOP DOWN TO h)
UPWARDS POSITIVE:
∆y = vi∆t + 1
2a∆t2 ✓
= (0)(0,75) + 1
2(–9,8)(0,75)2 ✓
= –2,76 m
h = 20,41–2,76 ✓
h = 17,65 m ✓
(DOWNWARDS POSITIVE: ∆y = 0 + 1
2(9,8)(0,75)2 = 2,76 m; h = 20,41 – 2,76 = 17,65
m.)
OPTION 2 (VELOCITY AT h)
vf = vi +a∆t = 0+(9,8)(0,75) = 7,35 m·s–1 (speed at h) ✓
v2
f = v2
i +2a∆y ✓
(7,35)2 = (20)2 +2(–9,8)∆y ✓
h = 17,65 m ✓
OPTION 3 (TIME x)
x = 2,04–0,75 = 1,29 s ✓
∆y = vi∆t+ 1
2a∆t2 ✓
= (20)(1,29)+ 1
2(–9,8)(1,29)2 ✓
h = 17,65 m ✓
(5)
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Physical Sciences/P1
7
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
3.3.3
POSITIVE MARKING FROM QUESTIONS 3.3.1 AND 3.3.2
OPTION 1
vf = vi +a∆t ✓
0 = 20+(–9,8)∆t ✓
∆ttop = 2,04 s
t = 2,04+0,75 = 2,79 s ✓
OPTION 2
∆y = vi∆t+ 1
2a∆t2 ✓
17,65 = 20∆t+ 1
2(–9,8)∆t2 ✓
∆t = 1,29 s or 2,79 s
t = 2,79 s (the second time the ball is at h) ✓
OPTION 3
t = x+1,5 s = 1,29+1,5 ✓✓= 2,79 s ✓
(x from equations of motion)
Range: 2,79 – 2,80 s.
(3)
3.4
POSITIVE MARKING FROM QUESTIONS 3.3.1 – 3.3.3
Marking criteria
hlayiso.com
• Graph A: straight line starting at v = 20 m·s–1 at t = 0 that crosses the time
axis (at 2,04 s) ✓
• Graph B starts at v = 0 at t = 2,79 s (the time ball B was dropped) ✓
• Graphs A and B are parallel (same gradient) ✓
• Both graphs end with a negative velocity; B ends after A with a smaller final
speed (A: –20 m·s–1 at 4,08 s; B: –18,60 m·s–1 at 4,69 s) ✓
NB: If the graphs are not labelled A and B, deduct 1 mark. Downwards as positive:
mirror image (accept).
UPWARDS AS POSITIVE
t (s)
Velocity (m·s–1)
0
A
B
20
2,04
2,79
–20
4,08 4,69
–18,60
DOWNWARDS AS POSITIVE
t (s)
Velocity (m·s–1)
0
A
B
–20
2,04
2,79
20
4,08 4,69
18,60
Hlayiso note: The end points are computed from the data: A lands after 2(2,04) = 4,08 s at 20
m·s–1; B falls 17,65 m from rest in 1,90 s, so it lands at 2,79 + 1,90 = 4,69 s at 18,60 m·s–1.
These values are not required on the graph.
(4)
[19]
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Physical Sciences/P1
8
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
QUESTION 4
4.1
If any of the underlined key words/phrases in the correct context is omitted, deduct 1
mark.
The total (linear) momentum in an isolated system remains constant (is
conserved). ✓✓
Accept for 1 mark: in an isolated system the total momentum before a collision
(explosion) is equal to the total momentum after it.
IF: closed system: –1 mark.
(2)
4.2
RIGHT AS POSITIVE. From the graph: after separation vP = –3 m·s–1 and
vQ = 4 m·s–1.
Σpi = Σpf
OR
(mP +mQ)vi = mPvfP +mQvfQ ✓
(any one)
(1,5+2)vi ✓= (1,5)(–3)+(2)(4) ✓
vi = 1,00 m·s–1✓
LEFT AS POSITIVE: (1,5 + 2)vi = (1,5)(3) + (2)(–4), so vi = –1 m·s–1, magnitude 1,00
m·s–1.
NOTE: 1 mark for the left-hand side, 1 mark for the right-hand side. Accept pbefore =
pafter with the same substitutions.
(4)
4.3
GREATER THAN ✓
The spring exerts forces of equal magnitude on P and Q (Newton’s Third Law),
and P has the smaller mass, so a = F
m is greater for P. ✓
OR: in the same time interval P’s velocity changes by 4 m·s–1 (1 to –3) and Q’s by
only 3 m·s–1 (1 to 4), so the graph for P is steeper.
(2)
4.4
POSITIVE MARKING FROM QUESTION 4.2
OPTION 1 (TROLLEY Q)
Fnet∆t = ∆p = m(vf –vi) ✓
(any one)
(10)∆t = (2)(4–1) ✓
∆t = 0,6 s
∴
t = 3+0,6 = 3,60 s ✓
OPTION 2 (TROLLEY P)
Fnet∆t = m(vf –vi) ✓
(–10)∆t = (1,5)(–3–1) ✓
∆t = 0,6 s; t = 3,60 s ✓
OPTION 3
a = Fnet
m = 10
2 = 5 m·s–2;
vf = vi +a∆t ✓: 4 = 1+5∆t ✓
∆t = 0,6 s; t = 3,60 s ✓
NOTE: If only ∆t = 0,6 s is given (not added to 3 s): Max 2
3.
(3)
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Physical Sciences/P1
9
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
QUESTION 5
5.1
If any of the underlined key words/phrases in the correct context is omitted, deduct 1
mark.
A force is non-conservative if the work done by the force on an object (moving
between two points) 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. ✓✓
IF the word “work” is omitted: 0 marks.
(2)
5.2.1
Marking criteria
hlayiso.com
• Formula Wnc = ∆Ek +∆Ep OR Wnet = ∆Ek ✓
• Correct substitution for the work done by friction: 55∆xcos180◦✓
• Correct substitution of ∆Ek and ∆Ep (or Wg) ✓
• Final answer: 9,09 m ✓
The block moves down: ∆Ep = –2500 J and ∆Ek = +2000 J.
OPTION 1
Wnc = ∆Ek +∆Ep ✓
fk∆xcosθ = ∆Ek +∆Ep
(55)∆xcos180◦✓= 2000+(–2500) ✓
∆x = 9,09 m ✓
OPTION 2
Wnet = ∆Ek ✓
Wg +Wf = ∆Ek
(Wg = –∆Ep = 2500 J)
2500 ✓+(55)∆xcos180◦✓= 2000
∆x = 9,09 m ✓
(4)
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Physical Sciences/P1
10
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
5.2.2
POSITIVE MARKING FROM QUESTION 5.2.1
Marking criteria
hlayiso.com
• Formula ∆Ep = mg∆h / Wg = mg∆xcosθ ✓
• Correct substitution of 2 500 J and (40)(9,8) ✓
• Correct trigonometric ratio (sinθ = h
∆x) / use of the distance from 5.2.1 ✓
• Final answer: θ = 44,55◦(range: 44,55◦– 44,58◦) ✓
OPTION 1
∆Ep = mg∆h ✓
–2500 = (40)(9,8)(0–h) ✓
h = 6,38 m
sinθ = 6,38
9,09 ✓
θ = 44,55◦✓
OPTION 2
Wg = –∆Ep:
mg∆xcos(90◦–θ) = 2500 ✓
(40)(9,8)(9,09)cos(90◦– θ) = 2500 ✓✓
θ = 44,56◦✓
(accept sinθ for
cos(90◦–θ))
OPTION 3
Wnet = ∆Ek ✓
[(40)(9,8)sinθ –55](9,09) ✓✓= 2000
θ = 44,55◦✓
(4)
5.3
SMALLER THAN ✓
The acceleration down the slope (a = gsinθ – µkgcosθ) does not depend on
the mass, so both blocks reach Y with the same speed. Wnet = ∆Ek = 1
2mv2
f is
therefore smaller for the smaller mass. ✓
OR: Fnet = ma with the same a, so the net force on the 20 kg block is half as large
over the same displacement. Accept: net work is directly proportional to the mass.
Hlayiso note: For the 20 kg block, Wnet = 1000 J (half of 2 000 J).
(2)
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Physical Sciences/P1
11
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
QUESTION 6
6.1.1
If any of the underlined key words/phrases in the correct context is omitted, deduct 1
mark.
The change in frequency (or pitch) of the sound detected by a listener because
the sound source and the listener have different velocities relative to the
medium of sound propagation. ✓✓
OR: An (apparent) change in (observed/detected) frequency (pitch) as a result
of the relative motion between a sound source and an observer (listener). ✓✓
(2)
6.1.2
In front
(towards the
listener)
Behind
direction of motion of the ambulance
Marking criteria
hlayiso.com
• Wavefronts closer together in front of the ambulance (in the direction of
motion, towards the listener) ✓
• Wavefronts further apart behind the ambulance ✓
The dot (ambulance) must be off-centre, displaced towards the front. Concentric
circles: 0/2.
(2)
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Physical Sciences/P1
12
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
6.1.3
Marking criteria
hlayiso.com
• Formula ✓
• Police car:
340
340–(v+5)fP ✓
• Ambulance:
340
340–vfA ✓
• Equating the two detected frequencies ✓
• Using fP
fA = 0,984 (or fP = 0,984fA) ✓
• Final answer: 27,50 m·s–1✓
Both vehicles move TOWARDS the stationary listener (vL = 0; subtract vs in
the denominator).
fL = v±vL
v±vs
fs
OR
fL =
v
v–vs
fs ✓
Police car:
fL =
340
340–(v+5)fP ✓
Ambulance:
fL =
340
340–vfA ✓
Equal detected frequencies:
340
335–vfP =
340
340–vfA ✓
fP
fA
= 335–v
340–v = 0,984 ✓
335–v = 334,56–0,984v
⇒
0,016v = 0,44
v = 27,50 m·s–1✓
NOTE: If (v+vs) is used in the denominators (sources moving away): Max 3
6 (formula,
equating, ratio).
(6)
6.2.1
Absorption (line) spectra ✓
(1)
6.2.2
Red shift ✓
Hlayiso note: In the photo every line of Spectrum 2 sits further to the right (the red,
longer-wavelength end) than the matching line of Spectrum 1: at about 25 %, 72 % and 85 %
of the width, against 18 %, 65 % and 79 %.
(1)
6.2.3
Away from (the Earth) ✓
(1)
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Physical Sciences/P1
13
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
QUESTION 7
7.1
If any of the underlined key words/phrases in the correct context is omitted, deduct 1
mark.
The magnitude of the electrostatic force exerted by one point charge (Q1)
on another point charge (Q2) is directly proportional to the product of the
(magnitudes of the) charges and inversely proportional to the square of the
distance (r) between them. ✓✓
(2)
7.2
w / Fg
FE
T
5◦
90◦
85◦
Marking criteria
hlayiso.com
• w (Fg) vertical and FE horizontal
(away from A, to the right), correctly
labelled ✓
• T
completes
a
closed
triangle
(arrows head to tail) ✓
• Any
TWO
correct
angles:
5◦
(between T and w), 90◦(between w
and FE), 85◦(between FE and T) ✓
Not drawn to scale. Accepted labels: T / FT
/ tension; w / Fg / mg / weight; FE / FAB /
electrostatic (Coulomb) force.
(3)
7.3
EQUAL TO 5◦✓
The electrostatic forces that A and B exert on each other are equal in
magnitude (Newton’s Third Law) AND the spheres have equal masses (equal
weights), so their force triangles are identical. ✓
(2)
7.4
If any of the underlined key words/phrases in the correct context is omitted, deduct 1
mark.
The (electrostatic) force experienced per unit positive charge placed at that
point. ✓✓
NOTE (1 mark only): an electric field is a region of space in which an electric charge
experiences a force.
(2)
7.5
RIGHT ✓
Hlayiso note: Both spheres are positive and X lies to the right of both, so both fields at X point
away from the charges, to the right.
(1)
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Physical Sciences/P1
14
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
7.6
Marking criteria
hlayiso.com
• Formula E = kQ
r2 ✓
• Substitution for EB (distance 2r) ✓
• Substitution for EA (distance 3r) ✓
• Adding: Enet = EA +EB ✓
• Equating to 1,05×104 N·C–1✓
• Final answer: r = 0,05 m ✓
RIGHT AS POSITIVE
E = kQ
r2 ✓
EB = (9×109)(9×10–9)
(2r)2
✓= 20,25
r2
to the right
EA = (9×109)(6×10–9)
(3r)2
✓= 6
r2 to the right
Enet = EA +EB ✓
1,05×104 = 6
r2 + 20,25
r2
= 26,25
r2
✓
r2 = 2,5×10–3
r = 0,05 m ✓
NOTE: If EA and EB are subtracted: Max 4
6. Wrong distances (e.g. r and 2r): Max 4
6.
(6)
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Physical Sciences/P1
15
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
QUESTION 8
8.1.1
OPTION 1
1
Rp
= 1
R1
+ 1
R2
:
1
0,72 = 1
R + 1
R ✓
RL1 = 1,44 Ω✓
OPTION 2
Rp = R1R2
R1 +R2
:
0,72 = R2
2R ✓
RL1 = 1,44 Ω✓
Accept: identical bulbs in parallel, so R = 2Rp = 2(0,72) ✓= 1,44 Ω✓.
(2)
8.1.2
POSITIVE MARKING FROM QUESTION 8.1.1
OPTION 1
P = I2R ✓
25 = I2(1,44) ✓
I = 4,17 A ✓
OPTION 2
P = V2
R :
25 = V2
1,44,
V = 6,00 V
P = VI ✓
25 = 6I ✓
I = 4,17 A ✓
Range: 4,16 – 4,17 A.
(3)
8.1.3
POSITIVE MARKING FROM QUESTIONS 8.1.1 AND 8.1.2
Marking criteria
hlayiso.com
• Total current: I = 2(4,17) = 8,33 A ✓
• Formula ε = I(R+r) OR ε = Vext +Ir ✓
• Correct substitution of 7 V and the total current ✓
• Correct addition of resistances (0,72+R+0,02) OR of potential differences ✓
• Final answer: 0,10 Ω(range: 0,099 – 0,10 Ω) ✓
OPTION 1
Itotal = 2(4,17) = 8,33 A ✓
ε = I(R+r) ✓
7 = 8,33 ✓(0,72+Rrh +0,02) ✓
Rrh = 0,10 Ω✓
OPTION 2
Itotal = 8,33 A ✓;
V∥= IR = (4,17)(1,44) = 6,00 V;
Vlost = Ir = (8,33)(0,02) =
0,17 V
ε = V∥+Vrh +Ir ✓:
7 = 6,00+Vrh +0,17 ✓✓
Vrh = 0,83 V
Rrh = V
I = 0,83
8,33 = 0,10 Ω✓
(5)
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Physical Sciences/P1
16
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
8.1.4
GREATER THAN ✓
•
The resistance of the L2 branch increases, so the resistance of the parallel
combination increases. ✓
•
The total (external) resistance of the circuit increases. ✓
•
The total current (through the rheostat and battery) decreases. ✓
•
The potential difference across the rheostat (IRrh) and the lost volts (Ir)
decrease. ✓
•
The emf is constant, so the potential difference across the parallel part
(across L1) increases, and with RL1 constant the current through L1
(ammeter) increases. ✓
(6)
8.1.5
BRIGHTER THAN ✓
(1)
8.2.1
Higher than (battery B) ✓
Hlayiso note: Battery A has the larger internal resistance, so more of its energy is converted
to heat inside it (DBE Nov 2020 8.2: “energy is converted to heat in the battery due to the
internal resistance”). Strictly, the heat per second in a battery is I2r =
ε2r
(R+r)2 , which grows with
r only while r is smaller than the heater’s resistance R. That is the normal case for a heater,
so “higher” is the expected answer.
(1)
8.2.2
Lower than (heater H2) ✓
The internal resistance of battery A is greater, so the total resistance of circuit
A is greater and the current through H1 is smaller. ✓
The heaters have the same resistance, so less electrical energy is converted
to heat in H1 per second (P = I2R is smaller). ✓
OR: more volts are lost inside battery A (Ir larger), so the potential difference across
H1 is smaller and P = V2
R is smaller.
NOTE: No marks for a calculation (“WITHOUT A CALCULATION”).
(3)
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Physical Sciences/P1
17
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
QUESTION 9
9.1
Electromagnetic induction ✓
Accept: Faraday’s Law (of electromagnetic induction).
(1)
9.2
Vrms = Vmax
√
2
✓
= 180
√
2
✓
= 127,28 V ✓
(3)
9.3
POSITIVE MARKING FROM QUESTION 9.2
NO ✓
OPTION 1 (CURRENT)
The speakers are in series: Rtotal = 8+8 = 16 Ω✓
Irms = Vrms
R
✓= 127,28
16
✓= 7,95 A
7,95 A < 11 A: the current through each speaker is less than the 11 A needed,
so neither speaker gives its maximum sound output. ✓
OPTION 2 (POWER)
Needed: Pave = I2rmsR = (11)2(8) = 968 W per speaker (1 936 W for both) ✓
Available: Pave = V2rms
R
= (127,28)2
16
✓✓= 1012,50 W in total (506,25 W per
speaker)
1012,50 W < 1936 W (506,25 W < 968 W per speaker), so the power is not
enough. ✓
OPTION 3 (POTENTIAL DIFFERENCE)
Needed: Vrms = IrmsR = (11)(16) ✓✓= 176 V ✓;
127,28 V < 176 V, so not
enough. ✓
NOTE: The 11 A rating of an AC device is an rms value. Comparing the PEAK current
(180
16 = 11,25 A) with 11 A and answering YES is incorrect: Max 1
5 (total resistance).
(5)
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Physical Sciences/P1
18
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
9.4
Time (s)
Induced voltage (V)
0
360
180
–180
–360
0,5
1
new graph (2 Hz)
original (1 Hz)
Marking criteria
hlayiso.com
• Two complete cycles in 1 s (period 0,5 s) ✓
• Maximum induced voltage of 360 V (and –360 V) indicated ✓
• Correct shape: sine graph starting at 0 and going positive first, like the original
✓
Hlayiso note: The emf is proportional to the speed of rotation (εmax = NBAω), so doubling the
speed doubles both the frequency (2 Hz) and the peak voltage (2×180 = 360 V).
(3)
[12]
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Physical Sciences/P1
19
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
QUESTION 10
10.1
If any of the underlined key words/phrases in the correct context is omitted, deduct 1
mark.
The process whereby electrons are ejected from a metal surface when light
(of suitable frequency) is incident on that surface. ✓✓
(2)
10.2
Light has a particle nature (light consists of photons/quanta). ✓
Accept: light energy is quantised.
(1)
10.3
Marking criteria
hlayiso.com
• Dependent (current) and independent (intensity of the light) variables
correctly identified ✓
• A question about the relationship between the independent and dependent
variables ✓
Example: What is the relationship between the intensity of the incident light
and the current in the circuit (photocurrent)?
(2)
10.4.1 REMAINS THE SAME ✓
Accept: DECREASES (fewer photons per second at the same intensity)
Hlayiso note: This is the answer the DBE is most likely to accept. DBE treats the photocurrent
as depending on the intensity of the light only (Feb/March 2016 11.5: “the photocurrent
is directly proportional to the intensity”). It has twice rejected “the current increases when
the frequency increases” (Nov 2014 1.10 statement (iii); Nov 2020 1.10 option C). Strict
physics gives a different answer: at the SAME intensity (energy per unit area per second),
higher-frequency photons each carry more energy, so FEWER photons and photoelectrons
arrive per second and the gradient DECREASES. We did not find a DBE memo that settles
this exact item, so check the official memo, which may also accept DECREASES.
(1)
10.4.2 REMAINS THE SAME ✓
The number of photoelectrons ejected per second (the current) depends
only on the number of incident photons per second (the intensity), which
is unchanged; each photon ejects one electron. A lower work function only
increases the (maximum) kinetic energy of the ejected electrons. ✓
(2)
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Physical Sciences/P1
20
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
10.5
Marking criteria
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• Formula E = W0 +Ek(max) (any form) ✓
• Substitution of hc
λ : (6,63×10–34)(3×108)
390×10–9
✓
• Substitution of W0 = 3,78×10–19 J ✓
• Substitution into 1
2mv2max with m = 9,11×10–31 kg ✓
• Final answer: 5,38×105 m·s–1 (range: 5,38 – 5,39×105 m·s–1) ✓
OPTION 1
E = hc
λ = (6,63×10–34)(3×108)
390×10–9
✓= 5,10×10–19 J
E = W0 + 1
2mv2max ✓
5,10×10–19 = 3,78×10–19 ✓+1
2(9,11×10–31)v2max ✓
vmax = 5,38×105 m·s–1✓
OPTION 2
hc
λ = W0 + 1
2mv2max ✓
(6,63×10–34)(3×108)
390×10–9
✓= 3,78×10–19 ✓+1
2(9,11×10–31)v2max ✓
vmax = 5,38×105 m·s–1✓
(Ek(max) = 1,32×10–19 J.)
(5)
[13]
TOTAL:
150
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Physical Sciences/P1
21
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
HOW DID OUR PREDICTION DO?
The real November 2026 P1 compared with the Hlayiso 2026 Physical Sciences P1 Prediction
Q
Topic (marks)
Real paper asked
Our closest prediction item(s)
Verdict Covered
1
MCQ (20)
N2 proportionalities; gravitation ratio; stone
from rising balloon; p–t graph; E + mgh;
Doppler fL–fs graph; E vs test charge; series
power; motor; photon < W0
Gravitation ratio, identical-resistor
power, projectile symmetry, impulse
graph, N2/N3, energy graph. No
Doppler graph, field or motor MCQ
/
10/20
2
Newton (13)
FBD with F at 15◦(4); define friction (2);
prove string breaks (5); motion after break
(2)
Angled force on a horizontal surface,
FBD, a for EACH block separately.
Defined N2L, not friction

9/13
3
Projectiles (19)
Define projectile (2); v at x vs t (2); max
height from 20 m·s–1 (3); h from “1,5 s
above h” (5); t (3); v–t graphs, two balls (4)
Free fall definition; same 20 m·s–1
launch and 20,41 m; time to top;
two-ball v–t graphs. No “time above a
level” item

14/19
4
Momentum (11)
Conservation principle (2); vi from a v–t
graph (4); a of P vs Q (2); F∆t gives t (3)
Principle; collision read from a v–t
graph; Newton III force; impulse with
contact time

11/11
5
Work–energy
(12)
Non-conservative force (2); distance from
Wnc (4); θ (4); Wnet vs mass (2)
Same definition (even-year call); crate
down a rough incline with Wnc

10/12
6
Doppler, spectra
(13)
Doppler (2); wavefront drawing (2); two
approaching sources, speed from a ratio
(6); absorption, red shift, away (3)
Definition; moving listener (two
equations, eliminate fs); wavefront
explanation, no drawing; red-shift
MCQ; absorption spectrum
/
9/13
7
Electrostatics
(16)
Coulomb (2); vector triangle (3); θ vs 5◦(2);
define E at a point (2); direction (1);
unknown r from net E (6)
Coulomb’s law (our strongest call); net
E at a point. We drilled 2D forces at
90◦, not an equilibrium triangle
/
9/16
8
Circuits (21)
R from parallel (2); I from 25 W (3); rheostat
from emf (5); ammeter change, explain (6);
brightness (1); battery and heater
temperatures (4)
Rated bulb, branch currents,
ε = I(R+r), “switch opened, explain”
chain, heat in the battery

21/21
9
AC (12)
Principle (1); Vrms (3); speakers, enough
power? (5); sketch at double speed (3)
rms and average-power calculation;
emf–time sketch. No “name the
principle”

11/12
10
Photoelectric
(13)
Define effect (2); particle nature (1);
investigative question (2); gradient changes
(3); vmax (5)
vmax from λ (even-year call); “one
photon, one electron” reasoning. We
defined the work function instead
/
7/13
Marks of the real paper covered by our prediction
111/150
A learner who had practised our prediction paper would have met about 111 of the 150 marks (74%)
as familiar question types. The ten-question structure and topic order held, Q4 and Q8 were covered
almost item for item, and Q3 used our 20 m·s–1 launch speed. Most losses were definitions, drawings
(vector triangle, wavefronts), Q1 and Q10.
What we got right
• Even-year calls that landed: Coulomb’s law,
non-conservative force, vmax of photoelectrons,
and the “due” gravitation-ratio MCQ.
• An applied force at an angle on a horizontal
surface, solved block by block (Q2).
• Two-ball vertical projectile with v–t graphs;
conservation of momentum read from a v–t graph.
• Rough incline with Wnc; heat in a battery; the
6-mark “resistance up, current down, lost volts
down” chain.
• Red shift and an absorption spectrum; an AC
sketch and rms/power work.
What we missed
• No pulley and no moving listener: the paper used a
string-breaking test and two sources approaching
one listener.
• An equilibrium vector triangle (Q7.2), not the 2D
forces at right angles we drilled.
• Definitions we did not set: frictional force, projectile,
electric field at a point, photoelectric effect.
• The investigative question and the “gradient of I vs
intensity” items in Q10.
• MCQs on a Doppler fL–fs graph, a motor and E vs
test charge. Our emf, rms and work-function
definitions were not asked.
Rating: 7/10. The structure, topic order and most calculation types were called correctly, and
Q4, Q5 and Q8 were very close to the real paper (111/150 covered). Several definitions, the
vector triangle, the wavefront drawing and the Q10 concept items were missed.
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