# Physical Sciences Grade 12 Paper 1 Limpopo June 2025 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Limpopo June Exam 2025, 15 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-sciences-physics-grade-12-nsc-memo-p1-may-june-2025-limpopo/82f7a5dc-e27f-400b-84ae-bcf3217097d6
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Limpopo June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 15

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## About this document

Physical Sciences · Grade 12 · Limpopo June Exam · 2025. Memorandum, 15 pages. Read online or download the PDF.

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## Document text

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### Page 1

PROVINCIAL GOVERNMENT

REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

Pe

4

f
UY
in
a
a
i
it
i
it
s

J

PHYSICAL SCIENCES: PHYSICS (P1)
JUNE 2025

FINAL AND AMENDE MARKING GUIDELINE/

FINALE EN AANGEPASTE MERK RIGLYNE

These marking guidelines consist of 9 pages

POPPER UPECTORTETORCTEOTTLECTITTTT TTT kd

MARKS: 150

### Page 2

Limpopo DoE/June 2025

Copyright reserved ; please turn over
QUESTION 1
11. A¥¥
12 Cvv
13. A¥vv
14 Dv¥v
15 BYY
16 BYY
17 Cv¥v
18 Bvv
19 Avv
1.10 Dvv
(2 x 10 =20)
QUESTION/VRAAG 2
2.1 Abody will remain in its state of rest or motion at constant velocity unless a

2:2

non-zero resultant/net force acts on it. ¥W

‘n Voorwerp sal in ‘n toestand van rus of konstante bewegin bly tot ‘n nie-sero
resultante krag daarop inwerk. VV (2)

Fa

<

fx

Copyright reserved Please turn over

### Page 3

Limpopo DoE/June 2025

ACCEPTED SYMBOLS/AANVAARDE SIMBOLE

e Mark awarded for label and arrow

¢ Do not penalize for length of arrows

e Any additional force/s: -1

e If force(s) do not make contact with
the body: -1 mark

e -1 mark if all arrowheads are omitted
but correctly labelled.

FN | Normal force/N /Fesurface on box Normaalkrag/ N/Fopperviak op krat
fk : Kinetic friction / friction /Frictional | Kinetiese

| force wrywing/wrywing/wrywingskrag
Fg "w, weight/Gravitational force/ Gewing/Gravitasiekrag/

;_ Fearth on box Faarde op die voorwerp
Fa/F Ap F7/ FToegepaste

t ‘2c ars
Notes: Notas:

e Punt toegeken vir pyl en pylpunt

e Moenie lengte van pyle penaliseer
nie

e Addisionele krag/te:-1

e As kragte nie kontak maak met kol
nie: -1

e -1 punt as alle kragte teenwoordig

(4) is en benoem maar geen pylpunte
nie.
2.3.1 OPTION 1
NOTES:
Free = 0 7]
Formula for N1/N2
Fret = ma ¥ Any one/ Enige een Substitution of zero

Fax — f,, — mgsiné = 0

3f,cos28°Y — f, — (48 + 4,3)(9,8)sin15° Vv = 0 ¥

f, = 80,45 NV

Substitution for Fax
Substitution for Fgu
Answer = 80,45 N

{This value is checked and correct)

OPTION 2
fy = Np ¥

= ux(mgcos@; — Fr sin @2)

= 0,22v (48 +4,3)(9,8)cos15 v — 3fsin28v’)

NOTES:

Formula for fx

Substitution of 0,22
Substitution for Fax
Substitution for Fgu

= 83,15N v

Copyright reserved

Answer = 83,15 N

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### Page 4

Limpopo DoE/June 2025

2.3.2 POSITIVE MARKING FROM QUESTION 2.3.1

Tension in the rope is the applied force/ Spanning in die tou is die toegepaste krag

Fy, = 3f, ¥ (Ratio)
Fa = 3¥(80,45)¥
Fa= 241,35 NY

OR

Range: 241,35 — 249,46 N
OPTION 2

Fret = Fy -— Fey —-i¥ OR
0 = Facos 28 — 52,3(9,8)sin 15“ — 80,45 ¥

Fa = 241,36 NY

OPTION 3
Fx2=Fy2+ Fy? v
= (3fc0s28)? + (3fsin28)?_¥
= (3(80,45)cos28)? + (3(80,45)sin28)? v
= (213,21)? + (113,36)
Fa=241,47N Vv

F, = 3f, ¥ (Ratio)
Fa = 3¥(83,15)¥
Fa =249,4557

Fa= 249,46 NV

Faae= Feo Foy — fy ¥

0 = Facos 28 — 52,3(9,8)sin 15” — 83,15 ¥
Fa = 244,41 NY

(4)

2.3.3 (POSITIVE MARKING FROM QUESTION 2.3.1 and 2.3.2

OPTION 1
fi, = Ni
80,45 = N (0,22) ¥
N = 365,68 NV
377,95 N

OR

Copyright reserved

fy = Ni
83,15 = N (0,22) v
N = 377,95NY

Please turn over

### Page 5

OPTION 2:
Fret = 0

N = Fg. - Fy v Any One
= mg cos81 — Fa sin@2

= (48 + 4,3)(9,8)(cos15) — 241,35 sin 28 Vv

= 381,76866
= 381,77NV

Limpopo DoE/June 2025

(3)

2.4 Rate at which work is done/ Rate at which energy is expended/transferred VV

Tempo waarteen arbeid verrig word/ Tempo waarteen energie oorgedra word. v~~~

(2 OR 0) (2)

2.5 (POSITIVE MARKING FROM QUESTION 2.3.2) check alternative answers

from 2.3.2
Ww = FA x cos@ calculate Fx
241,35(5)(cos28°) ¥ OR Wex = Fx Axcos@
= 1065,497 J
Range: 1065,497 - 1101,43 J = (241,35c0s28)(5)(cos0) ¥
= 1065,497 J
w Carry over mark. Correctly
P= t v v substituting the value calculated for
work in the power formula. But then
1065,497 ‘ P
= ——— the final (wrong) answer is not
aN awarded a mark.
= 332,9678
= 332,97 W¥ Range: 332,97- 344,20 W (5)

Copyright reserved

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### Page 6

Limpopo DoE/June 2025

OPTION 3:

Speed = — = — = 1,5625 ms*
at 3,2v Carry over mark. Correctly

substituting speed in power formula

Pave =FaxVave ¥ (Fa fropé 2.3.2 Range: 241,35 — 249,49 N
= (241,35c0s 28) ¥ (1,5625) ¥

= 332,97 W v (Range: 332,97 - 344,20W)

QUESTION 3

| In an isolated system, the total linear momentum is conserved/ remains
constant. “~~
In ‘n geisoleerde sisteem sai die totale lineére momentum behoue
bly/constant bly. WW (2 or 0) (2)

B22) LP = Lpr Y Anyone/ Enige een
my, = ny + Mp
(90000 + mpr)4500Y = (90000)(11000)¥ + m,-(—3045)¥
Mer = 77 534,79 kg¥
OPTION 2

Apy = —Appr ¥
m(vp — Yj) = —M(vy — Vj)
9000 (11000V — 4500v) = -m(4500- (-3045) Vv
m=7753,48kg  ¥ (5)

3.3 POSITIVE MARKING FROM 3.2

figs = mv? ise vAny one/ Enige een

= + (90000 + 77534,79)(4500)? A

= 1,696 x 1014]

1 2 1 2
Exe = zw + gm

= = (90000)(11000)? +5 (77534,79)(3045)°¥
=5,8 x 1077]

Bay # Exe
Decoupling is not elastic“ / Ontkoppeling is onelasties (5)

Copyright reserved Please turn over

### Page 7

Limpopo DoE/June 2025

QUESTION 4
44 3 m:s1 downwardsv /afwaarts (1)
42 N
IF
§ 2
vp = vj + 2ady 1 2
= (-3)? + 2(-9,8)(-18) Ayah + pat
, = —19,02 m-s7
ve anil -18 V = (-3)t +% (-9,8)t2
ve= vyitadtv t=1,63s”
19,02V = 3+ 9,8Atv eons
At = 1,63 s¥ (4)

4.3.1

For the follov.:ng, options, vandidate calculated final velocity first:
vp = vp + 20d

= (+37 +26
vi = -19,02104 =?
Frec(tA) = mvp — v,)~ Ay = (2) ty
rag. At = (vr -vi)
—19,02104+(-3
-9,8 (At) = (-19,02104 — (-3)) ¥ -18 = eae yat
ee At = 1,63 sv

Calculate the distance that the crane moves in 1 second at constant velocity:
Bereken die afstand wat die hyskraan in 1 sekonde beweeg teen konstante
snelheid:

Ay = viAt+Fa(At)? ¥
= 3(1) V +0 (non-essential zero may be omitted)

Ay = 3m
distance vitve
=———__ ¥ Ay = ( ) tv
sbaed time y 2
i = i =—3-3
Distance = speed x time _ ( - ) ()v
= 3(1)¥
=3m = 3m
C was thrown from: 18 — 3 = 15 mv (3)

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Limpopo DoE/June 2025

8
4.3.2 POSITIVE MARKING 4.2 and 4.3.1
IF
OPTION 1:

Ay = v,At+ 5 a(At)? v
~15 = v;(0,63) +5 (9,8) (0,63)?
vy; = —20,72 ms"

vy, = 20,72 m.s‘! Ydownwards ¥ Range: 20,519 — 20,72

OPTION 2:
Use of energy principles:

Emech top = Emech bottom
Ep + Extop = Ep + Exbottom Any one ¥

mgh een? = 0=mv?
2 2

(9,8)(15) + Ye v2 = % (26,74)°v
vit = 20,519 m.s! ¥ downward ¥

(4)
4.3.3. POSITIVE MARKING FROM QUESTION 4.2

Crane moves at constant velocity for 1,63s./ Hyskraan beweeg teen
konstante snelheid vir 1,63s.

Ay = vAt¥ Ay = (=) tv
= (3)(1,63) ¥ OR eer
= 4,89 m = (=) (1,63) v
= 4,89 |
Height = 18 =“ 4,89 (Mark is for subtracting the answer from 18)
= 13,114mv

(3)

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### Page 9

Limpopo DoE/June 2025

9
44 POSITIVE MARKING FROM 4.2, 4.3.1, 4.3.2, 4.3.3
18 4
13,11
=
c
co]
£
Time {s) 1 163
NOTES: :
NOTAS:

Starting height (AB) ¥
Drop height of C ¥

Throw time of C ¥

Fina position of A“

B &C hit ground together v
Final time for atl graphs “

Aanvanklike hoogte (AB) “
Gooi/val hoogte van C ~

Gooi tyd van C ~

Finale posisie van A ~

B & C tref grond op dieselfde tyd “
Finale tyd vir al drie grafieke ~ (6)

QUESTION 5

5.1 A system on which the external net force is zero. VW
‘n Sisteem waarop die eksterne netto krag nulis VV

(2 or 0) (2)
5.2 * Any additional forces/Addisionele kragte: -1
| * Allow w/ weight /Aanvaar gewig
Fov (1)
5.3 Mechanical Energy gottom = Mechanical Energyrop
Meganiese energieonder = Meganiese energiebo
ty Ex) bottom = (Ep 5 ee Any one ¥
(mgh += mv?) = (mgh +smv*)
2 bottom 2 top
Wie — Alig + ABR NOTES:
Wet = BE, Y Formula
0 +5 (S2)(v*)” = (52)(9,8)(0,47) ¥+ 0 V Subs Ep
vy, =3,04m-s tv v Subs Ek
v Answer
(4)

Copyright reserved

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### Page 10

5.4 Ap = mv; — mv;
Ap = m(vr— vi)
= 52(3,04 — 0)¥
= 158,08 kg: m-s~! V upwards¥ /opwaarts

10

iY

5.5 POSITIVE MARKING FROM 5.4
158.08 N-s upwardsY /Opwaarts

5.6  EreAt= Ap |

OR/OF
Ap
Fret = At
OR/OF
Fnet & x

AND: momentum remains constant “

OR:

Any onev”

Limpopo DoE/June 2025

(4)

(1)

If time increase and force decreaseY,, the momentum will remain the same”

As tyd toeneem en krag afneem V vir dieselfde verandering in momentum, V

Copyright reserved

(2)
[14]

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### Page 11

Limpopo DoE/June 2025
11
QUESTION 6

6.1 A force for which work done in moving an object between two points
depends on the path taken. vw
‘n krag waarvoor die arbeid verrig word om die voorwerp tussen twee punte
beweeg afhang van die pad van beweging.V¥ (2 or 0) (2)

6.2 W, = fAx cosO¥
= (7,5)(5)cos180°Y

= -37,5]v
(3)
6.3. POSITIVE MARKING FROM 6.2 NOTE:
Wac — AE, + AE, -
Wnet = AE v Subst Ws
Weg Cd We = AEx v Subst AEp
v Subst AEx

v Final answer

1
Wrc = (mghs — mgh,) + 2 m(v7 = vi)
-37,5Y = (35)(9,8)(20 x 10% - 2,5) ¥ + % (35)(v? — 0,37) ¥
Vp = 656m:s tv
NOTE: If equations of motion is used: 0/5

(5)

[10]
QUESTION 7

7.1 Doppler effect ¥
Change in the frequency/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. vv /

Doppler effek

Verandering in frekwensie/toonhoogte van die waargenome klank agv

relatiewe beweging tussen die bron en luisteraar (3)
7.2.1. Speed (of car)/Motion (of car) /Spoed (van die motor) Vv (1)
7.2.2 (observed) frequency” / (waargenome)frekwensie (1)
oe BY (1)

74 The detected frequency is higher “/Die waargenome frekwensie is hoér (1)

Copyright reserved Please turn over

### Page 12

Limpopo DoE/June 2025

75 cee

Vv =)
= ae v
fi ( V+Vs

V;=55,24m:-s"1¥

(4)
[11]

QUESTION 8

8.1 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:¥v/
Twee ladings sal ‘n elektrostatiese krag uitoefen deur een puntlading op ‘n
ander is direk eweredig aan die produk van die ladings en omgekeerd

eweredig aan die kwadraat van die afstand tussen hulle. (2)
8.2 Fe/Fc/Faons ¥ (Coulombic force)
Fov (2)
8.3 Fe a v MARKING CRITERIA:
re
2 ,
(20 x 10-3)(9,85 2 — -v ¢ Coulomb’s Law
(25xa0%) ¢ Correct substitution into
Fg and Fe

Q=1,17 x 10-7C¥ :
e Equating Fg and Fe

e Final answer
(sign not considered)

(4)

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### Page 13

Limpopo DoE/June 2025

13
8.4
\ * All lines curve away (shape) /alle lyne buig
weg (vorm)
. SAIL i * All arrows point towards the charge (direction
; _ 3 ‘ of the field lines)V/Alle pyle wys na die
i Ye lading(rigtin van veldlyne)
a4 * No lines touching/crossing and all make
contact with the chargesY /Geen lyne
—\ Loa raak/kruis of maak kontak met die ladings nie
a (3)
85 pK < v
_ (9 x 10°)(1,17 x 107-7)¥
- (25 x 10-3)2v
= 1,68 x 10°N:Cl¥ (4)
QUESTION 9
9.1 The potential difference across a conductor is directly proportional to the
current in the conductor at constant temperature. VW (2 or 0)
Die potensiaalverskil oor die punte van ‘n geleier is direk eweredig aan die
stroom in die geleier mits temperatuur constant bly (2)

9.2.1 Vinternal = Ir 7

L7=1(0,7)¥
I = 2,43 AY (3)

9.2.2 (POSITIVE MARKING FROM QUESTION 9.2.1)
Ra:
V = IRY

5 = 2,43 RY
Rz = 2,060

Ri =2RoV¥
:.Ri=4,120 Vv (4)

9.2.3 (POSITIVE MARKING FROM QUESTIONS 9.2.1 AND 9.2.2)
OPTION 1

e=I(R+rn)¥
= (2,43)(0,7 + 2,06+ 4,12) Vv
€ = 16,72 WW

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### Page 14

9.3.1

14
OPTION 2

lar = Ire
Var _ V2

Ry Re

4% S8y

2k, Rz

Vi=10V
€=1,7+10+5V

=16,7VV

(POSITIVE MARKING FROM 9.2)

Rp = 1,811

Rr = Rp +R,

Ry = 1,811 + 4,12V

Ry = 5,930.
e=I(R+nVv

16,72 = 1(5,9 0.7)”
1=2,52AV

Copyright reserved

Limpopo DoE/June 2025

(3)

MARKING CRITERIA

e Correct substitution into
R,/Korrekte invervanging

e Addition of Rp and R1/Som van
Rp en Ri

e EMF formula/ EMK formule

e Substituting into EMF formula
Invervanging in EMK formule

e Final answer/Finale antwoord

(5)

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### Page 15

9.3.2

15

Limpopo DoE/June 2025

(POSITIVE MARKING FROM QUESTION 9.3.1)

MARKING CRITERIA
V over Ra: ¢ Substituting into Ohm's
V=IR law/Invervanging in Ohm se
_ wet
= 2,52(1.811) ¥
=456 e Power formula/Drywing formule
e Substituting into power
P= ve Y formula//nvervanging in
gece drywing
P= ; ne v e Final answer/Finale antwoord
P = 10,09 WY
OPTION 2
lin Re
1=2,52Y x
17,02
1=2,221A
R=lRY
= (2,221)?(2,52) ¥
= 10,16 WY
OPTION 3
lin Re
15
1=2,52Y x
17,02
1=2,221A
P=VIV¥
= (4,56)(2,221) ¥
= 10,127WV¥

RANGE: 10,09 — 10,16 W

(4)
[21]

GRAND TOTAL/GROOTTOTAAL: 150

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