# Physical Sciences Grade 12 Paper 1 KZN June 2025 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the KZN June Exam 2025, 11 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-kzn/c156cdf1-1678-4be1-9650-c7487f7b8644
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: KZN June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 11

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## About this document

Physical Sciences · Grade 12 · KZN June Exam · 2025. Memorandum, 11 pages. Read online or download the PDF.

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## Document text

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### Page 1

) KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

fh PHYSICAL SCIENCES P1

: 2025 JUNE EXAMINATION

1 MARKING GUIDELINES
MARKS: 150

These marking guidelines consist of 11 pages.

Copyright reserved

Please turn over

tS

### Page 2

2025 June examination

QUESTION 1
1.1 Dvv (2)
1.2 Cvv (2)
1.3 AVY (2)
1.4 Byv (2)
1.5 Cvv (2)
1.6 Cvv (2)
4.7 AYY (2)
1.8 Dvv (2)
1.9 Cvv (2)
1.10 Byy (2)

[20]
QUESTION 2
251 Force that opposes the motion of a moving object relative to the surface. vv (2)
2:2

Accept force diagram:
N N
E F.
ij f
Ww w

Accepted labels

Fg/ Fw/ weight/gravitational forcev
F./applied forceY

(kinetic) friction/Fi/fky

Fn /Normal/Fnormat ¥

Z\—>|\n/<=

### Page 3

2.3.1

2.3.2

2.4

2025 June examination

Notes

* Mark awarded for label and arrow.

¢ Do not penalise for length of arrows since drawing is not to scale.
e Any other additional force(s): Max 3 4

e If everything correct, but no arrows: Max/Maks A

Marking criteria

If any of the underlined key words/phrases in the correct context is omitted,
deduct 1 mark

A body will remain in its state of rest or motion at constant velocity unless a
non-zero resultant/net force acts on it.v ¥

Fret = ma Y Anyone
Fcos© + mgsin © + fk = ma} i

Fcos30° - (15)(9,8)sin30° v- 0,2v v[Fsin30° + (15)(9,8)cos30°)] v = Ov
F=129,19N Vv

OPTION 1 w=mg

GMe 3800 = m(7,32) ¥
(2 m = 518,88 kg

_ (6,67*107"')((5,98«10"4)
(6,38x10°+1000«10°) w=mg

7,32 ms? = (518,88)(9,8) v

= 5085,02 N v

OPTION 2
re GmMg F

R2
3800 r= oe
* (R=1X106)2

3800 ( 6,38x10® + 1x10°)? = F(6,.38x108)2 v
F=5084,58NY

OPTION 3

Tor _ _GmMg
3800 = (R+1X10°)2

(6,67X10711)(m)}(5,98 x 1024) “
(6,38X 10° + 1X109)2

3800 =

1
| m = 518,88 kg
w.=mg ¥
= (518,88)(9,8) ”
= 5085,02 N v

(4)

(2)

(6)

(5)
[19]

### Page 4

2025 June examination

QUESTION 3
a 50 NY downwardsY
3.2.1 OPTION 1 (Accept if 0? is omitted) | OPTION 2 (Accept if 0? is omitted)
UPWARDS AS POSITIVE UPWARDS AS NEGATIVE
v2 = v2 + 2aly ¥ VP = v2 + 2aly ¥
= (0)? + (2)(-9,8)(-1,2) = (0)2 + (2)(9,8)(1,2) V
vie = 4,85 m:s! v downwardsY vi = 4,85 m:s! Y¥ downwardsY
3.2.2 OPTION 1 OPTION 1
UPWARDS AS POSITIVE UPWARDS AS NEGATIVE
POSITIVE MARKING FROM Q 3.2.1 | POSITIVE MARKING FROM Q 3.2.1
Ve = vi + aAt ¥ vr = vit alt
-4,85 = 0 + (-9,8) At ]v 4,85 = 0 + (9,8) At Jv
At=0,49s v At=0,49s v
OPTION 2 OPTION 2
UPWARDS AS POSITIVE UPWARDS AS NEGATIVE
POSITIVE MARKING FROM Q 3.2.1 | POSITIVE MARKING FROM Q 3.2.1
{7% _ (ity
7 +
a = e ¥ = = v
(1,2=(—S=)ag [1.2=(—)at)
= v_ At=0.49 sv
_ OPTION 3 OPTION 3
UPWARDS AS POSITIVE UPWARDS AS NEGATIVE
Ay = viAt+ 3a At v Ay =viAt+ a ae v
(£1,2=0+3(-9,8) Ary [1,2 = 0 +5(9,8) At] v
At=049sv¥ At=0,49s v
32.3 OPTION 1 OPTION 1

UPWARDS AS POSITIVE
v2 = vi2 + ZaAy ¥

0 =v? + (2)(-9,8)(0,8) ¥

vi = 3,96 ms! ¥

UPWARDS AS NEGATIVE
vi? = vi? + 2aly v
0 =v2 + (2)(9.8)(-0,8) ¥
vi =-3,96 ms
vi = 3,96 ms? v

OPTION 2
(U + K)top = (U + K)pbottom ~

(mgh + SMV"top = (mgh + SMV2)ootiom
[(0,5)(9,8)(0,8) + 0 = 0 +(5)(0,5) vv
v= 3,96 ms'v

(2)

(4)

(3)

(3)

### Page 5

3.2.4

2025 June examination

FretAt = m(vr- vi) ¥
Frioor + Fg = m(vr- Vi)

At = 0,098 s v

POSITIVE MARKING FROM QUESTIONS 3.2.1 & 3.2.3

50 —(0,5)(9,8)At V = 0,5[3,96 — (-4,85)] v

3.3 POSITIVE MARKING FROM Q 3.2.2

i) 0,49 0,59 time (S)
QUESTION 4
4.1 Marking criteria
If any of the underlined key words/phrases in the correct context is omitted, deduct
1 mark
In an isolated system the total linear momentum is conserved/remains constant. Vv
4.2 TO THE RIGHT AS POSITIVE

4

1,2

position (m)
eeees

0.8

Criteria

Correct shapev

Graph starts at 1,2 m att = Ov
Second maximum height at 0,8 mv
Time at which ball reaches the floorv
Time at which ball leaves the floory

dpi = Tpr

v= 6 mst ¥

MbVib + MbiVib| = (Mb + Mbi)vF } v Any one
(0,006)(200) + 0 v = (0,006 + 0,194) vi vw

TO THE RIGHT AS NEGATIVE

Dpi = dpr

vr = -6 ms?
=6 ms ¥

MbVib + MbiVibI = (Mb + Moi)vs } “Any one
(0,006)(-200) + 0 v = (0,006 + 0,194) viv

(4)

(5)
[21]

(2)

(4)

### Page 6

2025 June examination

4.3 — OPTION 1

vie = vi2 + 2aAx
02v = (6)? + 2a(5) ¥
a= -3,6 m-s?
= 3,6 ms”, to the left v

I ___ OPTION 2
' Whet = AK
fAxcos@ = Yam(vr — vi?)
f(5)(cos 180°) = %4(0,2)(02 -62) V
#=0,72N

-0,72 = 0,2av
a=-3,6
= 3.6 m-s®, to the left ¥

4.41

Ww

Accepted labels

W | Fo/ Fw/ weight/gravitational forcev

f_| (kinetic) friction/Fi/fk~”

N_| Fn/Normal/Frormai ¥

Notes

e Mark awarded for label and arrow.

¢ Do not penalise for length of arrows since drawing is not to scale.
e Any other additional force(s): Max *

e If everything correct, but no arrows: Max/Maks 74

4.4.2. Normal forcev

443K =!mvv

= 5 (60)(4)¥
= 480 Jv

4.44 Fro: = mgsin@ ~ f

(60)(9.8)(sin25°) — 180 v
68,5 Nv

ot

(4)

(3)

(1)

(3)

(2)

### Page 7

2025 June examination

4.4.5 Marking criteria
If any of the underlined key words/phrases in the correct context is omitted,
deduct 1 mark

Work done on an object by a net force is equal to the change in object's kinetic

energy. v ¥ (2)
44.6 POSITIVE MARKING FROM QUESTIONS 4.4.3 & 4.4.4
Whet = AK ¥
FnetAt = Kr - Ki
(68,5)(15)cos0°V =[ 3(60)v?—- 480 Jv
v=7,09 ms? v (4)
[25]
QUESTION 5
5.1. Rate at which work is done / energy is expended. vv (2)
52.4 p=“v
At
(0,85)(200) v = Zv
W = 5100 Jv (4)
5.2.2 U=mghv
= (90)(9,8)(3) ¥
= 2646 Jv (3)

(5)
[14]

### Page 8

2025 June examination

QUESTION 6
6.1 Doppler Effect. v The change in frequency of sound detected by a listener
because the sound source and the listener have different velocities relative to
the medium of sound propagation. vv (3)
6.2 Doppler flow meter / Doppler ultrasound machinev (1)
6.3 flL=—— -f,v
V-Vs
__ 340
1695v = Wow (1500) v
Vs = 39,12 m-s"!
OPTION 1 a OPTION 2 \
_ ak __ (39,12 + 39,12
=a Ax = (722) (1,5)
39,12=5v = 58,68m
Ax = 58,68 m ¥ (5)
4
64 £ (Hz)
Criteria
4500 e Correct shapev
; e 1,5s and 1500 Hzv
15 te

(2)
[11]

### Page 9

2025 June examination

QUESTION 7

ea Marking criteria
lf 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 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. vv

(2)
7.2
OPTION 1 OPTION 2
F = Srey Ea
e 2
(9x10°)(3x10%)(9x10°) (9x10°)(3x10°)
= v Exy= +,”
(0,15)2 (0,15)?
= 10,8N = 1,2%10° N.C"
_ (9*10°)(6x10°)(9x10%) (9«10°)(6x10°)
z= (0,1) Eyz= “one v
=46,6.N = 5,4x10° NC"
v, :
n= ,6-10,8 = tv g
Fret = 48,6~10,8 = 37, 8Ntothe right “|. 5,4x10° 1.20108
= 4,2x10° NC
Fre = QE wv
= (9x10°)(4,2x108v
= 37,8 N to the righty (5)
7.3.1 Xto YY (1)

7.3.2 QgaMit®

- (-3x10°)+(6x10°) /

2
= 1,5x10°C

n=2¥

de

_ (1,510) -(ax10*)
~ 1.61074 (4)

= 2,81x10" (electrons) ”
7.3.3 es v

(9x10°)(1,5x10°) | _ (9x10°)(9x10°) |

r2 (0,15 - r)2
r=0,04mv (4)

[16]

### Page 10

2025 June examination

QUESTION 8

8.1 Marking criteria
lf any of the underlined key words/phrases in the correct context is omitted,
deduct 1 mark
The potential difference across a conductor is directly proportional to the
current in the conductor at constant temperature. Vv

(2)

8.2.1
OPTION 1 OPTION 2
ee Rua Re
Rp R, Ro P R,+Ro
_1.1 _ 13*7 e
“a7 “1347
Rp =4,550v = 4,55 Ov (3)
8.2.2 POSITIVE MARKING FROM QUESTION 8.2.1
€=I(Rtr) v
12 = 1(4,55 + 0,3)v
|= 2,47 Av (3)
8.2.3 Vexternal = IR
= (2,47)(4,55) v
=11,24V

I through 8 Q
Vv

l=av
- 11,24 Vv

13
= 0,8646Av (4)

8.3 POSITIVE MARKING FROM QUESTION 8.2.3

Vsa=IR Isq = 2,47 — 0,8646 = 1,6054

= (0,8646)(5) v V30 = IR

= 4,323 V = (1,6054)(3) v

=4,816V

No = 4,816 % 4,323
= 0,49 Vv

(Range: 0,49 — 0,53 V)

(4)

8.4 e Total resistance increases Vv
e Total current decreases Vv
© Vinternal (lost volts) decreases V (3)

### Page 11

8.5 Cost = energy x tariff
300 “ = (P x 8 x 30) x 2,59
P = 482,625 W v

482,625
= v
no. of bulbs 80

=6v

2025 June examination

TOTAL:

(5)
[24]

150

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