# Physical Sciences Grade 11 Paper 1 November Mock Exam 2 2026 Memo

> Grade 11 Physical Sciences Paper 1 memorandum from the November Mock Exam 2 2026, 12 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/grade-11-physical-sciences-2026-november-mock-exam-2-memorandum-paper-1/dd3b8a5a-05c7-45d4-842b-cc3092489cba
- Updated: 2026-10-23
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam 2
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 12

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## About this document

Physical Sciences · Grade 11 · November Mock Exam 2 · 2026. Memorandum, 12 pages. Read online or download the PDF.

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### Page 1

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Mock
hlayiso mock exams
Practice Series:
Grade 11 Exam Preparation
WWW.HLAYISO.COM
NATIONAL
SENIOR CERTIFICATE
GRADE 11
PHYSICAL SCIENCES: PHYSICS (P1)
NOVEMBER EXAMINATION 2026 (MOCK)
MARKING GUIDELINES
MARKS: 150
These marking guidelines consist of 12 pages.

### Page 2

Physical Sciences/P1
2
NSC – Marking Guidelines
Hlayiso/Mock 2026
GENERAL MARKING GUIDELINES
1. Calculations: Marks are awarded for the correct formula, correct substitution and the
correct final answer WITH the correct unit, as indicated by the ticks (✓). A formula is not
awarded a mark if it is not followed by a substitution. If an incorrect formula is used, no
further marks are awarded for that calculation.
2. Units: The final-answer mark is only awarded if the correct unit is given. Penalise an
omitted or incorrect unit only ONCE per question. Do not penalise units in intermediate
steps.
3. Rounding: Accept final answers rounded to a minimum of TWO decimal places. Small
differences caused by rounding of intermediate values are accepted.
4. Positive marking: Where an incorrect answer to a previous question is used correctly
in a later calculation, the later calculation is marked positively.
5. Negative marking: If a candidate must choose (e.g. INCREASES/DECREASES) and
then explain, and the choice is incorrect, no marks are awarded for the explanation.
6. Definitions/laws: Where key words are underlined, deduct 1 mark for EACH key word
or phrase that is omitted or used in an incorrect context. Accept equivalent wording that
conveys the same meaning.
7. Free-body diagrams: A mark is awarded for each force with a correct arrow AND label.
Deduct 1 mark for any additional (incorrect) force. If forces do not touch the dot/object,
deduct 1 mark.
8. Sign convention: Candidates may choose either direction as positive provided it is
applied consistently.
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### Page 3

Physical Sciences/P1
3
NSC – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 1
1.1
C ✓✓
(2)
1.2
B ✓✓
(2)
1.3
D ✓✓
(2)
1.4
B ✓✓
(2)
1.5
D ✓✓
(2)
1.6
D ✓✓
(2)
1.7
A ✓✓
(2)
1.8
C ✓✓
(2)
1.9
A ✓✓
(2)
1.10
C ✓✓
(2)
[20]
Notes for markers: 1.2:
√
52 +122 = 13 N. 1.5: w ∝1
r2 , so r = 3R from the centre, i.e. a height of 2R. 1.6:
F′ = k(2Q1)(2Q2)
( 1
2r)2
= 16F. 1.7: n = 4,8×10–9
1,6×10–19 = 3×1010. 1.8: the induced pole opposes the magnet moving
away (Lenz), so X becomes a SOUTH pole that attracts the north pole. 1.9: the external resistance
decreases, so the current increases and the lost volts (Ir) increase.
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### Page 4

Physical Sciences/P1
4
NSC – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 2
2.1
The resultant (net) force acting at O is zero. ✓✓
Accept: the forces, drawn head-to-tail, form a closed vector diagram (2 marks).
(2)
2.2
w = 176,4 N
T1
T2
50◦
35◦
Marking criteria
Three force vectors drawn head-to-tail forming a CLOSED triangle with correct
directions ✓
All three forces labelled ✓
At least TWO correct angles shown (e.g. 50◦between w and T1, 35◦between w
and T2, 95◦between T1 and T2) ✓
(3)
2.3
w = mg = (18)(9,8) = 176,4 N ✓
OPTION 1 (components)
Horizontal: T1 cos40◦= T2 cos55◦✓
Vertical: T1 sin40◦+T2 sin55◦= 176,4 ✓
T2 = T1 cos40◦
cos55◦
= 1,3356T1 ✓
0,6428T1 +(0,8192)(1,3356T1) = 176,4 ✓
T1 = 101,57 N ✓
T2 = 135,65 N ✓
OPTION 2 (sine rule on the closed triangle)
T1
sin35◦= 176,4
sin95◦✓✓
T2
sin50◦= 176,4
sin95◦✓✓
T1 = 101,57 N ✓
T2 = 135,65 N ✓
The w = mg ✓line applies to BOTH options. Ranges: T1 101,5 – 101,6 N; T2 135,6 –
135,7 N.
(7)
2.4
176,4 N ✓vertically UPWARDS ✓
(Equal in magnitude and opposite in direction to the weight.) POSITIVE MARKING
FROM 2.3.
(2)
[14]
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### Page 5

Physical Sciences/P1
5
NSC – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 3
3.1
The force that opposes the motion of a moving object relative to a surface.
✓✓
(2)
3.2
N / FN ✓
w / Fg ✓
fk / f / Ff ✓
F / FA ✓
F is directed at 40◦BELOW the horizontal, to the right. Accept F drawn as a push
ending at the dot.
(4)
3.3
An object continues in its state of rest or uniform motion (moving with constant
velocity) unless it is acted upon by a net (resultant) force. ✓✓
(2)
3.4
Constant velocity, so Fnet = 0 in the horizontal and vertical directions.
N = mg+Fsin40◦✓
fk = µkN ✓
Horizontally: Fcos40◦= fk ✓
Fcos40◦= 0,35[(25)(9,8)+Fsin40◦] ✓
0,7660F–0,2250F = 85,75
F = 158,48 N ✓
Range: 158,4 – 158,6 N. If the vertical component of F is omitted from N (i.e. N = mg):
max. 3/5.
(5)
3.5
N = mg+Fsin40◦= (25)(9,8)+(158,48)sin40◦✓
N = 346,87 N ✓
Also accept N = fk
µk
= 158,48cos40◦
0,35
. POSITIVE MARKING FROM 3.4.
(2)
3.6
SMALLER THAN ✓
The vertical component of the applied force now acts UPWARDS, so the
normal force decreases (N = mg – Fsinθ < mg). Since fk = µkN, the kinetic
frictional force decreases. ✓
(2)
[17]
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### Page 6

Physical Sciences/P1
6
NSC – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 4
4.1
When a net (resultant) force acts on an object, the object will accelerate in the
direction of the net force. The acceleration is directly proportional to the net
force and inversely proportional to the mass of the object. ✓✓
(2)
4.2
N / FN ✓
w / Fg ✓
T / FT ✓
fk / f / Ff ✓
N perpendicular to the incline; T up the incline; friction down the incline (opposes
motion up the incline).
(4)
4.3
fk = µkN = µkmAgcos30◦✓
fk = (0,2)(6)(9,8)cos30◦✓
fk = 10,18 N ✓
(3)
4.4
Block A (up the incline positive):
Fnet = ma ✓
T–mAgsin30◦–fk = mAa
T–(6)(9,8)sin30◦–10,18 = 6a ✓
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Block B (downwards positive):
mBg–T = mBa
⇒
(5)(9,8)–T = 5a ✓
. . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
(1) + (2):
49–29,4–10,18 = 11a ✓
a = 0,86 m·s–2✓
Range: 0,85 – 0,86 m·s–2. POSITIVE MARKING FROM 4.3.
(5)
4.5
OPTION 1 (block B)
mBg–T = mBa:
(5)(9,8)–T = (5)(0,86) ✓✓
T = 44,70 N ✓
(44,72 N if the unrounded a = 0,856 m·s–2 is used)
OPTION 2 (block A)
T–29,4–10,18 = (6)(0,86) ✓✓
T = 44,74 N ✓
Range: 44,70 – 44,75 N. POSITIVE MARKING FROM 4.4.
(3)
4.6
SMALLER THAN ✓
Block B accelerates DOWNWARDS, so the net force on B is downwards; its
weight must be greater than the (upward) tension. ✓
(2)
[19]
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### Page 7

Physical Sciences/P1
7
NSC – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 5
5.1
T / FT ✓
w / Fg ✓
(2)
5.2
Upwards positive:
Fnet = ma
T–mg = ma ✓
T–(2500)(9,8) = (2500)(0,8) ✓
T = 26500 N
(2,65×104 N) ✓
(3)
5.3
The cage slows down while moving upwards, so its acceleration is
DOWNWARDS.
Upwards positive:
T–mg = ma
T–(2500)(9,8) = (2500)(–1,2) ✓✓
T = 21500 N
(2,15×104 N) ✓
One mark for the substitution of –1,2 m·s–2 (correct direction of a) and one for the rest
of the substitution. If +1,2 m·s–2 is used (answer 27 500 N): max. 1/3.
(3)
[8]
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### Page 8

Physical Sciences/P1
8
NSC – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 6
6.1
Every particle in the universe attracts every other particle with a force which is
directly proportional to the product of their masses and inversely proportional
to the square of the distance between their centres. ✓✓
(2)
6.2
F = GMm
r2
✓
F = (6,67×10–11)(2,4×1025)(1200)
(9,2×106)2
✓
w = 22695,65 N
(2,27×104 N) ✓
Also accept: g = GM
R2 = 18,91 m·s–2 ✓and w = mg = (1200)(18,91) ✓= 22692 N ✓.
Range: 22 690 – 22 700 N.
(3)
6.3
Weight on Earth: F = mg = (1200)(9,8) = 11760 N ✓
F = GMm
r2
:
11760 = (6,67×10–11)(2,4×1025)(1200)
r2
✓
r = 1,28×107 m ✓
h = r–R = 1,278×107 –9,2×106 = 3,58×106 m ✓
Range for h: 3,57 – 3,59 ×106 m. Accept the ratio method: r
R =
√
22695,65
11760
, r = 1,28×
107 m. POSITIVE MARKING FROM 6.2 where used.
(4)
6.4
NO ✓
According to Newton’s Third Law, the force that Kepra exerts on the lander
and the force that the lander exerts on Kepra are EQUAL in magnitude (and
opposite in direction), regardless of the masses. ✓
(2)
[11]
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### Page 9

Physical Sciences/P1
9
NSC – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 7
7.1.1
After X and Y touch:
QY = –9+0
2
= –4,5 nC ✓
After Y and Z touch:
QZ = –4,5+5
2
= +0,25 nC
(+2,5×10–10 C) ✓
(2)
7.1.2
n = ∆Q
e = (5–0,25)×10–9
1,6×10–19
✓
n = 2,97×1010 electrons ✓
Electrons moved from Y TO Z. ✓
Accept ∆Q calculated for Y: 0,25–(–4,5) = 4,75 nC. POSITIVE MARKING FROM 7.1.1.
(3)
7.2.1
The magnitude of the electrostatic force exerted by one point charge on
another point charge is directly proportional to the product of the magnitudes of
the charges and inversely proportional to the square of the distance between
them. ✓✓
(2)
7.2.2
T / FT ✓
w / Fg ✓
FE / FTS / electrostatic force ✓
Tension along the string (towards the support); electrostatic force horizontal, towards
T.
(3)
7.2.3
OPTION 1
Vertical: Tcos12◦= mg;
horizontal: Tsin12◦= FE ✓
FE = mgtan12◦= (2×10–3)(9,8)tan12◦✓
FE = 4,17×10–3 N ✓
OPTION 2
T =
mg
cos12◦= (2×10–3)(9,8)
cos12◦
= 2,004×10–2 N ✓
FE = Tsin12◦= (2,004×10–2)sin12◦✓
= 4,17×10–3 N ✓
Range: 4,16 – 4,17 ×10–3 N. If the mass is not converted to kg: max. 2/3.
(3)
7.2.4
F = kQSQT
r2
:
4,17×10–3 = (9×109)(40×10–9)QT
(0,05)2
✓
QT = 2,89×10–8 C
(28,9 nC) ✓
T is NEGATIVE (S is positive and is attracted). ✓
Range: 2,89 – 2,90 ×10–8 C. POSITIVE MARKING FROM 7.2.3.
(3)
[16]
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### Page 10

Physical Sciences/P1
10
NSC – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 8
8.1
The (electrostatic) force per unit positive charge placed at that point. ✓✓
(2)
8.2
+
–
A
B
Marking criteria
Correct shape: field lines from A to B (lines curve between the charges) ✓
Direction: arrows from A (positive) towards B (negative) ✓
Field lines do not cross and start/end on the spheres ✓
Accept: more lines ending on B than leaving A (|QB| > |QA|); do not penalise equal
densities.
(3)
8.3
E = kQ
r2 ✓
EA = (9×109)(4×10–6)
(0,15)2
✓= 1,6×106 N·C–1 towards B (away from A)
EB = (9×109)(9×10–6)
(0,15)2
✓= 3,6×106 N·C–1 towards B
Enet = 1,6×106 +3,6×106 ✓
Enet = 5,2×106 N·C–1 towards B (to the right) ✓
The final mark is awarded only if the direction is given. If the two fields are subtracted:
max. 3/5.
(5)
8.4
At P the fields of A and B are opposite in direction and equal in magnitude:
kQA
x2 =
kQB
(x+0,3)2 ✓
(9×109)(4×10–6)
x2
= (9×109)(9×10–6)
(x+0,3)2
✓
x+0,3
x
= 3
2
⇒
x = 0,6 m ✓
(3)
[13]
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### Page 11

Physical Sciences/P1
11
NSC – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 9
9.1.1
NORTH ✓
(The induced current opposes the approach of the north pole.)
(1)
9.1.2
Zero (no deflection) ✓
There is no change in the magnetic flux (linkage) through the solenoid, so no
emf/current is induced. ✓
(2)
9.2.1
The magnitude of the induced emf across the ends of a conductor is directly
proportional to the rate of change in the magnetic flux linkage with the
conductor. ✓✓
(2)
9.2.2
Φ = BAcosθ = (0,5)π(0,04)2 cos0◦✓
Φ = 2,51×10–3 Wb ✓
(2)
9.2.3
∆Φ = BA(cos60◦–cos0◦) = (0,5)π(0,04)2(0,5–1) = –1,26×10–3 Wb ✓
ε = –N∆Φ
∆t ✓
ε = –(150)(–1,26×10–3)
0,02
✓
ε = 9,42 V ✓
Range: 9,42 – 9,45 V. Ignore the sign of ∆Φ for the magnitude. If the final flux is taken
as zero (i.e. ∆Φ = BA): max. 3/4. POSITIVE MARKING FROM 9.2.2.
(4)
9.2.4
DECREASES ✓
The same change in flux linkage takes place over a longer time, so the RATE
of change of flux linkage is smaller (ε ∝1
∆t). ✓
(2)
[13]
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### Page 12

Physical Sciences/P1
12
NSC – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 10
10.1.1 The maximum energy provided (work done) by a battery per coulomb (unit) of
charge passing through it. ✓✓
Accept: the potential difference across the terminals of a battery when no current flows
(open circuit).
(2)
10.1.2 R34 = R3 +R4 = 4+8 = 12 Ω
1
Rp
= 1
6 + 1
12 ✓
Rp = 4 Ω✓
Rext = R1 +Rp = 3+4 = 7 Ω✓
Accept Rp = R2R34
R2 +R34
= (6)(12)
18
. If R3 and R4 are not added in series: max. 1/3.
(3)
10.1.3 ε = I(R+r) ✓
12 = I(7+0,5) ✓
I = 1,60 A ✓
If the internal resistance is ignored (I = 1,71 A): max. 1/3. POSITIVE MARKING FROM
10.1.2.
(3)
10.1.4 OPTION 1
V = ε –Ir = 12–(1,6)(0,5) ✓
V = 11,20 V ✓
OPTION 2
V = IRext = (1,6)(7) ✓
V = 11,20 V ✓
POSITIVE MARKING FROM 10.1.3.
(2)
10.1.5 Vp = IRp = (1,6)(4) = 6,4 V
I3 = Vp
R34
= 6,4
12 = 0,53 A ✓
P3 = I2R = (0,53)2(4) ✓
P3 = 1,14 W ✓
Range: 1,12 – 1,14 W. Also accept: V3 = I3R3 = 2,13 V and P = VI or P = V2
R . If Vp = 6,4
V is used across R3 alone: max. 1/3. POSITIVE MARKING FROM 10.1.3.
(3)
10.1.6 INCREASES ✓
When S is opened the R2 branch is removed and the external resistance
increases (from 7 Ωto 3 + 12 = 15 Ω), so the current in the battery decreases
(from 1,6 A to 0,77 A). ✓
The lost volts (Ir) across the internal resistance decrease, so the terminal
potential difference (V = ε –Ir) increases (to 11,61 V). ✓
(3)
10.2
P = V2
R = (230)2
24
= 2204,17 W
(2,20 kW) ✓
W = P∆t = (2,204 kW)(1,5 h×30) = 99,19 kW·h ✓
Cost = 99,19×R3,15 = R312,44 ✓
Range: R312,40 – R312,50. If calculated in joules: W = (2204,17)(1,5×3600×30) =
3,57×108 J = 99,19 kW·h. Full marks if correctly converted.
(3)
[19]
TOTAL:
150
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