# Physical Sciences Grade 12 Paper 1 Limpopo Mock 2025 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Limpopo Mock Exam 2025, 18 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physics-grade-12-nsc-p1-memo-september-2025-limpopo/ee326b2b-8480-42f2-8ce0-7bcf6229bde6
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Limpopo Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 18

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## About this document

Physical Sciences · Grade 12 · Limpopo Mock Exam · 2025. Memorandum, 18 pages. Read online or download the PDF.

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### Page 1

PROVINCIAL GOVERNMENT

REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

PHYSICAL SCIENCES: PHYSICS (P1)

PREPARATORY 2025
MARKING GUIDELINE/ MERK RIGLYNE

MARKS: 150

These marking guidelines consist of 12 pages

Copyright reserved please turn over

### Page 2

Physical Sciences/P1

QUESTION/VRAAG 1

11. Dv
1.2 BYY
1.3 AVY
140 AW
1.5 Dv¥v
16 Av
1.7 Cvv
1.8 AYY
19 BYY
1.10 BYY

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LimpopoDoE/September 2025

(2 x 10 =20)

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### Page 3

Physical Sciences/P1

QUESTION/VRAAG 2

2.1

2:2

LimpopoDoE/September 2025

When a (non-zero) net force acts on an object, the object will
accelerate in the direction of the (net) force at an acceleration that is
directly proportional to the (net) force and inversely proportional to the

mass of the object.vV

Wanneer 'n (nie-nul) resulterende/netto krag op 'n voorwerp inwerk,
versnel die voorwerp in die rigting van die krag teen 'n versnelling
direk eweredig aan die (netto)krag en omgekeerd eweredig aan die

massa van die voorwerp.v ¥

FerakesY%

Fnv¥ fe

ACCEPTED SYMBOLS/AANVAARDE SIMBOLE

the body: -1 mark
-1 mark if all arrowheads are omitted
but correctly labelled.

Fn Normal force/N /Fsurtace on truck Normaalkrag/ N/Fopperviak op trok
fic Kinetic friction / friction /Frictional | Kinetiese
force wrywing/wrywing/wrywingskrag
Fg w, weight/Gravitational force/ Gewing/Gravitasiekrag/
Fearth on truck Faarde op die trok
Feraxes | Force of Brakes/ Fa/Applied force | Fremme / Krag van remme/ F7/
_. Toegepaste krag
Notes: Notas:
e Mark awarded for label! and arrow e Punt toegeken vir pyl en
e¢ Do not penalize for length of arrows pylpunt
e Any additional force/s: -1 e Addisionele krag/te:-1
e If force(s) do not make contact with e As kragte nie kontak maak

met kol nie: -1
e -7 punt as alle kragte
teenwoordig is en benoem

Copyright reserved

maar geen pylpunte nie.
(4)

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### Page 4

Physical Sciences/P1

2.3.1

2.3.2

2.4.1

2.4.2

Copyright reserved

For == mgsin® (no mark for formula)

= (40 000)(9,8)(sin 33° ¥

= 213 498,50 N ¥ (213 498,5017 N)
W = FAxcos0 v

= (10 000)(400)(cos180°) v
= -4,0 x 108 J ¥ (- 4 000 000 J)

POSTIVE MARKING FROM 2.3.1
The speed of the truck is calculated from the motion of A to B. The pre-
amble clearly states that the brakes were still functional during this motion.
Fnet = ma
Fon — Ferakes ~ fy = ma
mgsin@ — 10000 - pxmgcos® = ma
40000(9,8)(sin33°) — 10000 — (0,42)(40000)(9,8)cos33°“ = 40000aY
OR
213 498,50 — 10000 — (0,42)(40000)(9,8)cos33 Vv =40000a_ Vv
a = 1,64 m-s? (downhill)
(Range: 1,63549 — 1,64)

v? = v? + 2aAyv Carry over
= (5,4)? + 2(1,64)(400) ¥
vi = 36,62 m-s* (downhill/down the incline) Y (Range: 36,5726-36.62)
(af teen helling/afwaarts)
The product of the resultant/net force acting on an object and the time the
net force acts on the object. vv

Die produk van die resulterende/netto krag wat op 'n voorwerp inwerk en
die tyd wat die netto krag op die voorwerp inwerk VV

LimpopoDoE/September 2025

(2)

(3)

(5)

(2)

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### Page 5

Physical Sciences/P1 LimpopoDoE/September 2025

2.43

244

POSITIVE MARKING FROM QUESTION 2.4.1
The only horizontal force present, is the friction caused by the gravel.
Therefore friction is seen as the net force.

NOTE: ONLY THE IMPULSE-MOMENTUM THEORY METHOD IS
ALLOWED (4)
FnetAt = Ap ~
Fnet(2,5) ¥ = 40 000(0 - 36,62) v
Fret = 585 161,71N (range: 585 161,71 — 585 920 N) v

POSITIVE MARKING FROM QUESTIONS 2.4.1 , 2.4.3
OPTION 1

W = fAxcos68 ¥
= 585 161,71 (55) cosi80 ¥
= -3,22 x10" ¥

OPTION 2 (OPTIONS 2 and 3 Provides a different answer)
Whe = AEp + AEk¥

:. Wi=0+ % (40 000)(0)? — % (40 000)(36,62)? V
= - 2,675 x10’ N (Range: =2,675 x10’~ 2,68 x 10” N) ¥

OPTION 3

Whet = AE, ¥
= % (40 000)(0)* — % (40 000)(36,62)2V
= - 2,675 x10’ N (Range: =2,675 x10’~ 2,68 x 10” N) ¥v

(3)
[25]

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### Page 6

Physical Sciences/P1 LimpopoDoE/September 2025

3.1

3.2

3.3

3.4

QUESTION/ VRAAG 3
An object that has an initial velocity and moves under the influence of the
gravitational force only. ¥¥ (2)

‘n Voorwerp met ‘n aanvangsnelheid wat slegs onder die invloed van die
gravitasiekrag beweeg.v¥

v? = v? + 2aAy ¥
0 = vj + 2(9,8)(-7) V

v; = 11,71 m-s~t upward VY (3)

POSITIVE MARKING FROM QUESTION 3.2

Ve = Vi tat v¥

vr = (-11,71) +9,8(2,83) ¥ (3)
= 16,02 m-s downward v

POSITIVE MARKING FROM QUESTION 3.3

NOTE: Candidate may not show
calculation, but is credited if it was
calculated and substituted

75 % of 16,02= 12,015 m-s'V
Carry over mark — if initial value calculated (5)
incorrectly, candidate gets mark

Ve =Vi +adt for substituting correctly but not further.
14,938Y¥ =-12,015+9,8 -Atv

At = 2,7495 sv (Range: 2,7495 — 2,75)

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### Page 7

Physical Sciences/P1

LimpopoDoE/September 2025

3.5 POSITIVE MARKING 3.2,3.3,3.4
OPTION 1:

AB:
Ay = vjAt + SaAt? v

= (-11,71)(2,83) + % (9,8)(2,83)? ¥
= 6,10 m (6,10431 m)
BC:

Ay = v,At+ 5 adt?

= (-12,015)(2,749) + % (9,8)(2,749)? v|

OPTION 2:
AB

v? =v? + 2ady v
(16,02)? = (—11,71)* + 2(9,8)AyY
Ay = 6,0977m

BC

v? =v} + 2ady ¥

(14,93)? = (—12,015)? + 2(9,8)Ay¥
Ay = 4,00m

=4,00m
OPTION 3
AB

_ vety;

Ayae = (“*)aev
Ayan = eae) (2,83) Py,
Ay = -6,09865
BC:
Ayap = Co) At
AYan = (ees) (2,749) v
Ay = -4,015

v

(5)

f4a1

MARKING CRITERIA:

e One mark for either formula
present to calc Ay for AB or BC

e Substitution to calc AB

e Substitution to calc BC

e Addition of AB + BC

e Final answer

Total height: a = 10,11 mv

Copyright reserved

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### Page 8

Physical Sciences/P1 LimpopoDoE/September 2025

QUESTION/VRAAG 4

Afrikaans and English papers differ in mass of objects:
Afrikaans = 85 kg

English = 85 g

NOTE: Mass conversions is imperative. Penalise when conversion is

not done correctly.

4.1  Em=Ep+ Ex y Any one AFRIKAANS:
=mgh + % mv?

= (85)(9,8)(0,7) + = (85)(0)2
= (85 x 10°)(9,8)(0,7) + 5 (85 x 10°)(0)2¥ euanesemll “ienaiees

= 83,1 J
= 0,5831J ¥
(3)
4.2 POSTIVE MARKING FROM 4.1
OPTION 1 AFRIKAANS:
_nal
Emtop = Embottom Y anyone | 983,1= 0 +5 (85)v?
= Ep +%mv2 v=3.70 ms1
0,5831 = 0 +5 (85x 10°)’

v = 3,70 m-s* rightY (3,7041 m-s")

OPTION 2:

Wret = AE, + AE, ¥

0 = (Mghbottom -mghtop) + (% mv? - % mv?)

0 = [0-(85 x 10°°)(9,8)(0,7)] +[% (85 x 10°)v? -O] ¥
v= 3,70 ms right ¥

(3)

4.3. The total linear momentum of an isolated system remains constant/is
conserved. VY
Die totale linéere momentum van ‘n geisoleerde sisteem bly konstant/ bly
behoue V¥ (2)

Copyright reserved Please turn over

### Page 9

Physical Sciences/P1 LimpopoDoE/September 2025

44 POSITIVE MARKING 4.2

tpi = <Xper Yany one
mva +mve=mva + mva (4)

(85 x 10°)(3,7) ¥+0 = 0 +85 x 103 viv
vi =3,7 m-s* Right ¥
AFRIKAANS:

Zpi = =Ipr Y any one
mva +Mve=mMva + MvB

(85 )(3,7) V+0 = 0 +_(85)vi“
ve =3,7 m-s"right ¥

OPTION 2:

Apa = —Apz ¥

[0 - (85 x 10°)(3,7)] = - [(85 x 10°)(va)-0] ¥
Vb = 3,7 m.s? Right”

4.5  0,7m V Same value as A started (1)
[13]

QUESTION/VRAAG 5
5.1. The total mechanical energy (sum of gravitational potential energy and

kinetic energy) in an isolated system remains constant. Vv“

Die totale meganiese energie (som van gravitasie- potensiéle energie en

kinetiese energie) in 'n geisoleerde sisteem bly konstant. VV (2)

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### Page 10

Physical Sciences/P1 LimpopoDoE/September 2025

5.2

5.3.1

OPTION 1

Ema = Eup i y
: An
(Ep + Exa = (Ep + En ‘

(4410)(9,8)(23) + (2) (4410)(18)2V = (4410)(9,8)(37) + (-) (4410)v? V

ve =7,04 ms! ¥ (7,04273 ms") (4)
OPTION 2
Wye = AE, + AE,

0 = [(4410)(9,8)(37) — (4410)(9,8)(23)] + (5) (4410)(18)? — (6) (4410)v2|v

v=7,04m.s"

POSITIVE MARKING from 5.2
OPTION 1

Wac = AEp + OE, %

1
= (4410)(9,8)(1,5 — 37)¥ + (6) (4410) (25? — 7,042732)¥ (4)

= - 265 482,1011 J v (- 265 397,33 J)

OPTION 2
Calculate Emechanical at points B and C

Emc = Ep + Ex
Emc = =mv? +mgh v(Formula at either MC or MB)

= 5 (4410 x 252) + (4410 x9,8x1,5) Vv
= 1442 952 J

Emp = 5 (4410 x7,047) + (4410 x 9,8 x 37) ¥
=1 708 349,328 J
Wa = Emc — Emp

= 1 442 952 — 1 708 349,328
= - 265 397,328 JV

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### Page 11

Physical Sciences/P1

5.3.2 NO v

6.1

6.2

6.3

6.4

Friction is present / The external net force is not equal to zero
There is a non-conservative force acting. Vv

Neev

Wrywing (is teenwoordig)/Die eksterne netto krag is nie nul nie

Daar is ‘n nie-konservatiewe krag teenwoordig “

QUESTION/VRAAG 6

Doppler Effect
Doppler Effek V

fi, = ¢ =) fs Vv

Vivs

2428,57V = (=

) 1000 v

0~-Vs5

Vs = 200,00 m-s! v (199,9999176 m-s*)

DECREASESY

The wavelengths are longer. ~

ACCEPT: Waves are stretched out/Wave crests are
apart/wavelength increase “

a
|
VERLAAG Y

Die golflengtes is langer ¥
AANVAAR::Golwe is uitgerek/Golffronte is verder uit mekaar ¥

Ultrasound

Detection of blood clots

Detection of heart problems Any ¥
Fetal blood flow

Blood flow in the heart/organs

Copyright reserved

LimpopoDoE/September 2025

(2)
[12]

(1)

(4)

farther

(2)

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### Page 12

Physical Sciences/P1 LimpopoDoE/September 2025

Ultraklank

Bloedkionte (1)
Hart probleme

Fetale bloed vioei [8]

Bioed vioei in hart/organe

QUESTION/VRAAG 7

7.1 The electric field at a point is the electrostatic force experienced per
unit positive charge placed at that point vv

Die elektrostatiese krag wat per eenheid positiewe lading wat by daardie
punt geplaas is, ondervind word. Vv

(2)

724 8 = Fy
q
_ 3,5x 1076 Sf
6,2 x 10-9 (3)

= 564,516 N+ C~* Right

7.2.2 Take to the right as positive:

OPTION 1
_ kQe
E= 2 _Use of formula v
kQ_ kQ Substitution of any one ¥
Enet = 72 72

y — (OX10°)Qe _ (9x10°)(6,2x10-%)
(x)? (2x)?

SIMPLIFY BY MULTIPLYING WITH x* THROUGHOUT

(9x10) (6,2x109)

0 = (9x10°)Qa - ri

Qg = 155x107°C ¥

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### Page 13

Physical Sciences/P1 LimpopoDoE/September 2025

OPTION 2
kQr

E= a Use of formula ¥

(4)

Substitution of any one
Ev atp Fate ——
(9x10°)QR— (9x10°)(6,2x10~°)
~? (2x)?

Qa = 1,55x10°CV

NOTE: Consider to the left as positive also.

7.2.3 POSITIVE MARKING FROM 7.2.2
OPTION 1

¥
kQ,Q2
Fa
— (9x10°)(6,2x107°)(1,55x10"9 -v
OX =

r2
r= 0,157 —0,1572

3x = 0.1572 (Division with 3)
x=0,052m  v (0,0524 m)

OPTION 2
kQ
E = 7

— (9x10°)(1,5x10~?
7 (3x)?

X =0,0157m (4)
7.3. POSITIVE MARKING FROM 7.2.2

RtoT v
(NOTE: Wrong calculation in QUESTION 7.2.2 giving Qr > Qr: T to R)
(1)

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### Page 14

Physical Sciences/P1 LimpopoDoE/September 2025

7.4

8.1

8.2

8.3

8.4

POSITIVE MARKING FROM /.z.z
(1,55x107) + (6,2x10~°)y
me 2 (2)
Qnew = 3,875 x10-°C V (3,88 x 10°C) [16]
QUESTION/VRAAG 8
{t is the maximum energy provided by a battery per unit charge passing
through it. VW

Dit is die maksimum energie wat 'n battery lewer per eenheidslading wat

daardeur vioei. (2)
OPTION 1 OPTION 2

Vinternal = Ir ¥ £ = Vextemal + Vintemal V

12 —11,3 = 1(0,5 42 = 11,3 + (0,5) ¥ (3)
OR 1=1,40AV

0,7 = 1(0,5)”

1=14AVY

INCREASE ¥

TOENEEM (1)

S2 causes a parallel circuit:

Rtotai decreases ¥

ltotal increases since I « ; v

OR:

| is inversely proportional to R. Therefore, as Rtotai decrease, Itotal will
increase. VV

S2 veroorsaak dat die stroombaan verander in parallel skakeling:

Riotaal neem af ~
Iotaai neem toe want I a = Y
OF:
(2)
| is omgekeerd eweredig aan R. Daarom, wanneer Rootaai afneem,

Sal Itotaai toeneem. V~

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### Page 15

Physical Sciences/P1 LimpopoDoE/September 2025

85 OPTION 1

e=1(R+r) 1 1 1  ¥Formula for Rp

R, Ri Ry
e= Vexternal +Ir Pp 4 s

1 i 1

12=11,10 +1(0,5) ¥ ie Ei,

617 x 3x
I=18A Carry

over x= 8,230 ¥ (8,2267 O
mark¥
V, = IR
Pp Pp OR
11,10 = 18R,
, 1 _ Rat (R1+2+3)
Rp = 6,170 (6,1667 O) Rp ~ R4xR,
1
_ Xx +3x% 7

6,17 3x?

3x = 24,6667

X = 8,2220V
OPTION 2
Vinternat = Ir
0,9 = 1(0,5) v

= arry
y= 18 over
mark¥

e=1(R+r)
12 = 1,8(RA0,5)
R =6,1667
OPTION 3
Calculate Ra in the initial circuit,
V=IRa ¥
11,3 Y= 1(1,4) ¥
Ra = 8,070 ¥
Ro = Rav (5)

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### Page 16

Physical Sciences/P1

8.6 POSITIVIE MARKING
OPTION 1
V=IRY

LimpopoDoE/September 2025

Rs = 24,68 (8,2267 x 3)

11,10 = 1246801) ) 7

1 = 0,4498 AY

OPTION 3

V=IRVY
11,10 = (8,22) ¥

IRH = 1,3504

Trot = Tri + tra

1,8 = Ir1 + 1,3596

In = 0,45A ¥
QUESTION/VRAAG 9

OPTION 2

Ratio’s:

1.
1,8 xqVvY = 045 AV

Resistors in series are potential
dividers:

9.1. Mechanical (energy) to Electrical (energy) VY

Meganiese (energie) na Elektriese (energie)

9.2 AC/Alternating
It has slip rings ¥

WS/Wisselstroom ¥
Dit het sleep ringe V

93 ._1
~ f

me ll
50

=0,02sv

Copyright reserved

(3)
[16]

(2)

(2)

(2)

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### Page 17

Physical Sciences/P1 LimpopoDoE/September 2025

9.4.1

9.4.2

The AC potential difference which dissipates/produces the same amount

of energy as an equivalent DC potential difference. VW

Die WS _potensiaalverskil wat dieselfde hoeveelheid energie

verbruik/oordra as ‘n ekwivalente GS potensiaalverskil. VW”

(2)
Vinax
— Y
rms v2
220(V2)¥ = Vmax
Vmax = 311,13 VY
(3)

=
o 0 »
2 0,04
= ime (s)
=
NOTES:

Question does not require 2
cycles to start at the same
position as the coil in the

ACCEPT, since the instruction does not specify diagram.

That the graph must start with same orientation Period (0,02 s) ¥
As the diagram. ,
Vmax =311,13 v

CULES pga eeny Correctness: (Shape and 2
full cycles) ~
= 0
5 02 04 3)
£ Time (s) (
$
[14]

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### Page 18

Physical Sciences/P1 LimpopoDoE/September 2025

QUESTION/VRAAG 10
10.1. The minimum energy that an electron in the metal needs to be emitted from
the metal surface. VY
Die minimum energie benodig om 'n elektron uit die oppervilak van 'n
metaal vry te stel VV (2)

10.2 Photoelectric effect V and particle nature of light “
Foto-elektriese effek V en deeltjie-aard van lig V (2)

10.3 Ejight = Wo + Excmax) Y (note that subscript (max) is essential)

1
(6,63 x10-**)(1x10!7)¥ = (3,66x10-19) + (5) (9,11x10-*)v2,4.7

Vmax = 1,20 x 107 ms ¥ (12 031 243,63 m-s*) (4)
[8]

GRAND TOTAL/GROOTTOTAAL: 150

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