# Physical Sciences Grade 12 Paper 1 KZN Mock 2025 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the KZN Mock Exam 2025, 12 pages. Read online or download the free PDF.

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- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: KZN Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 12

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Physical Sciences · Grade 12 · KZN Mock Exam · 2025. Memorandum, 12 pages. Read online or download the PDF.

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### Page 1

} KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

4 PHYSICAL SCIENCES P11 (PHYSICS)
C PREPARATORY EXAMINATION
: MARKING GUIDELINES
4 SEPTEMBER 2025
Seenee rere reese settee eee eens
MARKS: 150

These marking guidelines consist of 9 pages.

Copyright reserved

Please turn over

### Page 2

Physical Sciences P1 September 2025 Preparatory Examination

QUESTION 1

1.1 Dyvv
h Dyv
1.3 Byy
1.4 Dvv
1.5 AVY
1.6 Cvv
LF Byv
1.8 AVY
1.9 AVY
1.10 Cvv
QUESTION 2
2

22

2.3.1

T Accept force diagram:
-F+——_@——”
-«—_ "|

! Accepted labels
F | Fafapplied force/50 NY
f_| (kinetic) friction/Fi/fky”
_T | Fr/Tension/Ftension ~
Notes
e Mark awarded for label and arrow.
e Do not penalise for length of arrows since drawing is not to scale.
e Any other additional force(s): Max 7%
e If everything correct, but no arrows: Max %

When a net force is applied on an object, the object will accelerate in the
direction of the force and the acceleration is directly proportional to the forceY
and inversely proportional to the mass of the object.v

Accept Newton’s second law in terms of momentum

fk = UkKNY
= 0,3{(6)(9,8) +(3)(9,.8)]”

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Physical Sciences P1

2.3.2

2.4.1

24.2

= 26,46 Nv’

September 2025 Preparatory Examination

| POSITIVE MARKING FROM Q 2.3.1

LEFT AS NEGATIVE

Block X

Fret = ma v
-—F+T+f=ma

-50+ T+ 2646 V= Gave (1)

Block Y
Fnet = ma

Og Solving equation (1) and (2):
-50 +3a + 26,46 = -6av

a= 2,62 ms? v

POSITIVE MARKING FROM Q 2.3.1
LEFT AS POSITIVE

Block X

Fret = ma v
F-T-f=ma

50 - T - 26,46 V= 6a... A)

Block Y
Fret = ma
Ts 88 4 nono (2)

ee Solving equation (1) and (2):
+50 -3a - 26,46 = +6av

a=2,62 ms? v

Each body in the universe attracts every other body with a force that is directly
proportional to the product of their masses “and inversely proportional to the

square of the distance between their centres.”

If charges are mentioned ;

OPTION 1
Foplanet = YF earth

G(2Me)(m) /_ >G(Me)(m) ey
x2 ~ R2
x = 2RV

OPTION 2
Qplanet M = YeQeEarth v
1
G(2Me) ,_ 37 S(Me) /
x2 Re
x = 2RY

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### Page 4

Physical Sciences P1

September 2025 Preparatory Examination

QUESTION 3
3.1 0,6 mv
3.2.1. [DOWNWARD AS POSITIVE DOWNWARD AS NEGATIVE

3.2.2

3.2.3

3.3

vie = v,2 + 2aAyv

= (1,5)? + 2(9,8)(2) v

w=6,44m-s' v

vi? = vi2 + 2aAy
= (-1,5)2 + 2(-9,8)(-2) V

w=6,44 mst v

OPTION 1
POSITIVE MARKING FROM Q 3.2.1
DOWNWARD AS POSITIVE
vi=vit alt ¥
6,44 = 1,5 + (9,8)At v
At = 0,50 s¥

OPTION 1
POSITIVE MARKING FROM Q 3.2.1
DOWNARD AS NEGATIVE
ve = vi t+ alt Vv
-6,44 = -1,5 + (-9,8)At Vv
At = 0,50 sv

OPTION 2
POSITIVE MARKING FROM Q 3.2.1
DOWNWARD AS POSITIVE

OPTION 2
POSITIVE MARKING FROM Q 3.2.1
DOWNWARD AS NEGATIVE

- (it™ _ (ity
+6, -1,5 + (-6,44
[2=(-S)vae p= (Nv at
At = 0,50 At=0,50s Vv
OPTION 3 OPTION 3

DOWNWARD AS POSITIVE
Ay = viAt+ adv

[2=1,5 At +3(9,8))vAt
At= 0,50 s ¥

DOWNWARD AS NEGATIVE
Ay = v,At+ sadtv

[-2=-1,5 At +3(-9,8))VAt
At=0,50s¥_

OPTION 1
DOWNWARD AS POSITIVE
ve = vi2 + 2aly ¥
02V = vi + 2(9,8)(-0,8) ”
Vi = 3,96 m:s*, upwards Vv

OPTION 1
DOWNWARD AS NEGATIVE
vie =v? + 2ahy ¥
0?v = vit + 2(-9,8)(0,8) “
Vi = 3,96 ms‘, upwards Vv

POSITIVE MARKING FROM QUESTION 3.2.1 & QUESTION 3.2.3

Fret At = m(vr— vi) ¥
Fnet(0,2) ¥ = 0,5(3,96 — (-6,44)) ¥
Fret = 26 N, upwards Vv

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### Page 5

Physical Sciences P1 September 2025 Preparatory Examination

3.4

[ POSITIVE MARKING FROM QUESTION 3.1 TO QUESTION 3.3

Criteria
« First part of starts at v = 1,5 m-s’ and ends at 6,44 m-s! Vv

« Second part starts at v = 3,96 m:s* until v= 0 m-s? ¥
« Shape (parallel lines) v

« Both axes labelled v

DOWNWARD AS POSITIVE

»

6,44

b
£
>

1,5

t(s)

-3,96+---------------

DOWNWARD AS NEGATIVE

a

3,96
‘e
—
> 15 t(s)
-6,44
(4)
[19]

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### Page 6

Physical Sciences P1

QUESTION 4
41 Easiv

4.2 | Newton’s Third Law (of Motion) “
(Mark independently)

September 2025 Preparatory Examination

When object A exerts a force on object B, object B simultaneously exerts an
oppositely directed force of equal magnitude on object A. Vv v

4.3
OPTION 1 OPTION 1
TO THE LEFT AS POSITIVE TO THE LEFT AS NEGATIVE
2pi = Lpr v 2pi = Xpr } v
mtvit = Mive + Mave2 anyone mitvit = Maver + Mavee Any one
Ov = 6Ovn + (4)(3) ¥ Ov = 60Ovi + (4)(-3) ¥
vi = -0,20 vr = 0,20 ms! ¥
* Vwe= 0,20 mst v
OPTION 2
Apboy =- APobject
mb(Vor —Vbi) = = Mo(Vor= Voi) PY Any one
60(Vvor - 0)” = -4(3-0)”
Vor = -0,20
<. Vor = 0,20 mst
4.4.1 Increases Vv
4.4.2 Increases Vv

4.4.3 ApobjectincreasesY, which causes Appboy to increasev

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### Page 7

Physical Sciences P1 September 2025 Preparatory Examination

QUESTION 5
5.1
N Accept force diagram:
- N
f
f
w Ww

Accepted labels

Fg/ Fw/ weight/588 N/gravitational forcev”

F./applied force’.

(kinetic) friction/Fefky”

Fn /Normal/Fnormat “

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 a

Z\>| n=

, : 3
e If everything correct, but no arrows: Max/Maks % (4)

5.2 Work done on an object by a constant force F is F Ax cos 6 , where F is the
magnitude of the force, Ax the magnitude of the displacement and 0 the angle
between the force and the displacement. v

OR

The work done on an object is the product of the force and the displacement of
the object in the direction of the displacement. vv (2)

5.3 [OPTION1

Whet = AK

We + Wr = Yem(ve? — vi?) } v Any one

(80)(2)cos© + (8)(2)cos180° = [¥4(7)(1,52 - 02)]
© = 81,42 °V

OPTION 2

We = AK + AU

We + Wr = ¥em(ve2 — v2) + mg(hr— hi) a aiden
(80)(2)cos© + (8)(2)cos180° = [¥4(7)(1,52 - 02) + 0] ¥

8 = 81,42 °v (4)
5.4‘ Increasesv Vv (2)
[12]

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### Page 8

Physical Sciences P1 September 2025 Preparatory Examination

QUESTION 6
6.1 Frequency” (1)
6.2 The sound source (bat) and the listener (bird watcher) have different velocities (2)

relative to the medium of sound propagation. vv
6.3 0,4 mv (1)
6.4.1 v=fxdAv

340 =f x 0.4v
f = 850 Hzv (3)

6.4.2 POSITIVE MARKING FROM QUESTION 6.4.1
_ Viv,
i= viv, ‘

y Y Any one
fi=—— -f,
V—Vs,

_ 340
850v = = v(875) ¥

Vs = 10 m-s"' (away from the batwatcher) v (5)
[12]

QUESTION 7

7.1 Q= aren
2

(9x10*)+(-Sx10°) y @uC)+C5HC)

5 Accept Q= :
= 2x108Cv = 2uC Vv (2)

had

Criteria for sketch Marks
Correct direction of field lines v
Shape of electric field v
Field lines not crossing each other / v
Field lines start from sphere /
No field lines inside the spheres

(3)

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### Page 9

Physical Sciences P1 September 2025 Preparatory Examination

7.3 Fr = S12 v a (9x10°)(2x10°)(1x10) |
(9x10°)(2x10°)(1«10°) = (0,2)?
= 5 v = 0,45 N to the left
(0,1)
= 1,8 N to the right

Fret = 1,8 + (-0,45) ¥
= 1,35.N to the right v (5)

7.4 Electrostatic force experienced per unit positive charge placed (at that point). vv (2)

7.5 POSITIVE MARKING FROM QUESTION 7.3
OPTION 1

= 1,35x10° N-C"' to the right ¥ (3)

OPTION 2
kQ
E=>

 _(9.0x10")(2 x 10°)
(0,1)°
= 1,80 x 10° N:C* to the right “

nl _ 6.0% 10° 2x 10°)
7 (0,2)°
= 4,50 x 105 N-C" to the left v

Enet = 1,80 x 10®N— 4,50 x 108
= 1,35x10° N-C"' to the right (3)

[15]

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### Page 10

Physical Sciences P1

September 2025 Preparatory Examination

QUESTION 8
8.1 12V¥
B21 Vy,
R
Sy
8
= 0,75A V
8.2.2
OPTION 1 OPTION 2
| 1 4 Rix Ro
— eee ae = v
Rp R,; Ra” Rp Ry+ Ro
-',1, _14«12 )
14 12 ~ 44412
Rp =6,46Q v =6,450V

8.2.3. POSITIVE MARKING FROM QUESTION 8.1 & 8.2.2

€=(Rtr) v

12v = 0,75(6,46 + 18 + r)
r= 1,54 QV

8.2.4 OPTION 1
POSITIVE MARKING FROM
QUESTION 8.2.1 & 8.2.2

OPTION 2
POSITIVE MARKING FROM
QUESTION 8.2.1

Vi =IR
= (0.75)(6,46) ¥
=4,845V

ly20 = Ms wi
_ 4,845,
~~ 42—
= 0,404 Av

Viost = Ir
= (0,75)(1,54)
= 1,155 V

Vian = 12 — 1,155 — 6Y
= 4,845 V

haa = RY

4,845
liq = |

=0,404A Vv

8.3 Cost = P x At x tariff
5

=3,5v x 2v x 265v

60
=R7,73v

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### Page 11

Physical Sciences P1

QUESTION 9
9.1 Split ring (commutator) ¥ Accept: Commutator
9.2 Anticlockwise vv
9.3 Electrical to mechanicalv
9.4.1 Vmax
Vime = v
rms V2
_ 311
v2
= 220 Vv

September 2025 Preparatory Examination

9.4.2 POSITIVE MARKING FROM QUESTION 9.4.1

p.. = Vins
ave ~ “R

2
_ 220 J

30

= 1 613, 33 W
OPTION 1 OPTION 2

Pave = ImsVims ¥ Pave = l?rmsR V
1 613,33 = Irs (220) v 1 613,33 = I?rms(30) ¥
Ims = 7,33 A ¥ Ims = 7,33 Av
OPTION 3
Reavis v
ne
v v
30= a
lms = 7,33 AY

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### Page 12

Physical Sciences P1 September 2025 Preparatory Examination

QUESTION 10
10.1 Minimum frequency of light needed to emit electrons from a certain metal

surface. vv (2)
10.2 4 vv

7,2x 10° Hz (2)

10.3. OPTION4
POSITIVE MARKING FROM QUESTION 10.2
Eximax)= hf - hfo ~
X = (6,63x 10) 1x 10°) v- (6,63x 10) 7,2x 10") v
= 1,86x 10? Jv
(4)
OPTION 2

- v
Gradient < Ek(max)2 - Ek(max)1 r

ff,
X-0
10 x 10" — 7,2 x 10"
= 719 vw
x = 1,86x 10% J (4)

v
6,63 x 10% =

10.4 Remains the samev

Intensity does not affect kinetic energy of the photonsY
OR
Only frequency affects Ex(max)

The frequency remains the same. (2)

[10]
TOTAL: 150

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