# Physical Sciences Grade 11 LP March Test 2026 Paper & Memo

> Grade 11 Physical Sciences question paper and memo from the LP March Test 2026, 26 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-11-physical-sciences-2026-lp-march-test-question-paper-and-memorandum/8ea3e613-33a8-4e1f-9f60-13c9e2616674
- Updated: 2026-09-28
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2026
- Exam period: LP March Test
- Curriculum: CAPS
- Pages: 26

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## About this document

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### Page 1

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PhySicalSciences/Gr 11

a
.
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1
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o
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s

¢

MARKS: 100

TIME: 2 hours

LDoE/March., 2026

“ety, LIMPOPO

es “+ PROVINCIAL GOVERNMENT

REPUBLIC OF SOUTH AFRICA

DEPARTMENT OF

EDUCATION

NATIONAL
NIOR CERTIFICATE

GRADE 11

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Physical Sciences/Gr 11 2 LDoE/March, 2026

INSTRUCTIONS AND INFORMATION

if

10.
11.
12.

Write your centre number and examination number in the appropriate spaces
on the ANSWER BOOK.

This question paper consists of SEVEN questions. Answer ALL the questions
in the ANSWER BOOK.

Start EACH question on a NEW page in the ANSWER BOOK.

Number the answers correctly according to the numbering system used in this
question paper.

Leave ONE line between two sub questions, for example between
QUESTION 2.1 and QUESTION 2.2.

You may use a non-programmable calculator.

You may use appropriate mathematical instruments.

You are advised to use the attached DATA SHEETS.
Show ALL formulae and substitutions in ALL calculations.

Round off your final n vers" 4jminimum of TWO decimal places.

Give brief motivations ) a where required.

Write neatly and legibly.

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Physical Sciences/Gr 11 3 LDoE/March. 2026
QUESTION 1: MULTIPLE-CHOICE QUESTIONS

Four options are provided as possible answers to the following questions. Each question
vas only ONE correct answer. Write only the letter (A — D) next to the question number
(1.41.10) in the ANSWER BOOK, for example 1.11 D.

1.4 The statement below refers to the vector and scalar quantities.

(i) A vector has magnitude and direction, while a scalar has magnitude
only.

(ii) | A-scalar quantity can always be added to a vector quantity.

(iii) | Force is an example of a vector quantity, while distance is an
example of a scalar quantity.

Which of the above statements is /are TRUE?
A (i) and (ii) only
B (i) and (iii) only
C (ii) and (iii) only
D_ (i) only (2)

Four forces act on a body in

jorephysics.com

and direction of the resultant force?

C 8Neast

D  8Nsouth (2)

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LDoE/March. 2026

~
If) net force F is applied on an object of mass m kg and causes an acceleration
In of am-s-. When the net force F on the same object is doubled, the resulting

I NNN cceleration, in m-s~, will be ...

¢
f
oe

| =

Nf a

(2)

ards along a frictionless slope,
own in the sketch below.

Which ONE of the following equations can be used to calculate the magnitude
of the normal force (N)?

A N=(4)(9,8)sin8
B  N=F-(4)(9,8)cos'
N =F + (4)(9,8)cos6

ie:
D  N=(4)(9,8)cos@ (2)

Sidhimorephysics.com

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Physical Sciences/Gr 11 LDoE/March, 2026

15

5
agnitude F acts on a Ree at.an.angle @ to the horizontal, as
agi e 1 DIOCK wit at rest on a flat, frictionless

DRCE and the NET
NORMAL NET HORIZONTAL FORCE
A Increases Increases
B Decreases Increases
Cc Increases Decreases
D Decreases Decreases
(2)
1.6 Three objects of mass 1 spectively, are suspended by
a light string as shown i
2kg
0,7 kg
A 15N
B 42N
C 27N
D 35N (2)

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Physical Sciences/Gr 11 6 LDoE/March. 2026

ectively exert a force F on each

What is the new force that the bodies exert on each other, in terms of F?

A 8F
B 16F
Cc 4F
D 2F (2)

1.8 The distance between two charges, Q: and Q: is r. The force exerted by the
spheres on each other is F. Both charges are now doubled without changing

the distance between them. The magnitude of the electrostatic force on Q1

will be:

A 4F

BMF

C 2F

D “AF (2)

1.9 Which one of the following diagrams indicates the direction of magnetic fields
and how it is determined with a test charge around charged sphere: correct!

(2)

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Physical Sciences/Gr 11 7 LDoE/March. 2026

1.10 A negative charge Q is placed at a distance of 5d from another charge R.

Q Pgneutteaep
e . e
#------------- caine >

RATIO OF THE CHARGES | CHARGE ONR

Q:R
A 4:9 Positive
B 3:2 Negative
Cc 5:2 Positive
D 9:4 Negative (2)

[20]

QUESTION 2 (Start on a new page)

$, as shown in the diagram below.
in rope A, is 530 N and the force
(0° with the horizontal and rope B

A heavy object is lifted using two
The two pulleys are a distance
Fs, in rope B, is 1 240 N. Rope
makes an angle of 20° with the verti

FA =530N FB = 1240N

2.1. Define the term resultant vector. (2)

2.2 Explain why the vector diagram of force Fa, force Fe and the weight will NOT
be a closed vector diagram. (2)

2.3 Calculate the:
2.3.1 Vertical component of Fa (2)

2.3.2 Horizontal component of Fa (2)
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2.4 Calculate the maximum weight that force Fa and force Fe will be able to lift

from the ground. Show ALL calculations. (4)

25 Explain why the rope and pulley system will be less effective if the distance
x between the pulleys is increased (2)
[14]

QUESTION 3 (Start on a new page)

A horizontal force, F, is applied to a 200 kg block by means of a massless, inextensible

rope, as shown in the sketch below. The block remains stationary when the rope makes

an angle of 15° with the vertical...
wig

3.1 Explain why the block is st (2)
3.2 Determine the magnitud (4)
3.3 Determine the magnitude F. (3)

[09]

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Physical Sciences/Gr 11 9 LDoE/March. 2026

QUESTION 4 (Start on a new page.)

A block Of mass 4 kg is sliding down a rough plane (AB) inclined at 30° to the
horizontal, as shown in the diagram below. The coefficient of kinetic friction

between the surface and the block is 0.2.

A

c
4.1 Define the term frictional force. (2)
4.2 Calculate the magnitude of the:
4.2.1. Kinetic frictional force between the 4 kg block and the surface (3)
4.2.2 Acceleration of block while is moving on section AB (5)
The block is NOW moving on s : n is frictionless.
4.3. Will the block in section BC” LOWER, a HIGHER or a ZERO
ACCELERATION? Give the f the answe (2)

[14]

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Physical Sciences/Gr 11 10 LDoE/March. 2026

QUESTION 5 (Start on a new page.)

4kg

respectively.
5.1 State Newton's second law of motion in words. (2)
5.2 Drawa labeled free-body diagram showing all the forces acting on the 4 kg
block (4)
5.3. Calculate the magnitude of the:
5.3.1 Acceleration of the system. (4)
5.3.2 Tension in the string connecting the two blocks. (2)

5.4

ied IS decreased to 20°. How will the
change? Write down only
SAME. (1)

The angle at which the force
frictional force experien
INCREASES, DECREA

[13]

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Physical Sciences/Gr 11 1 LDoE/March. 2026
QUESTION 6 (Start on a new page.)

A satellite of mass 1 200 kg is orbiting the Earth at a distance d from the surface of the
Earth as shown on the diagram below.

Earth

The radius of the Earth is 20 times the atstone

6.1 State Newton's Law of Universal Gravitation in words (2)
6.2 Calculate the:
6.2.1 Distance between the ce of the:Earth and the centre of satellite
(Distance d). z (2)
6.2.2 Magnitude of the fo on the satellite to keep it
in orbit (4)
6.3 Calculate the weight of the s (2)
[10]

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Physical Sciences/Gr 11 12 LDoE/March. 2026

QUESTION 7 (Start on a new page.)

Three charged spheres, A, B and C, with charges of -4 nC, +7 nC and +10 nC
respectively, are placed on rubber stands as shown below

A B c

7.1 Can metal stands be used instead of rubber stands? Write down only YES
or NO. Give a reason for the answer.

7.2 State Coulomb's law in words.

7.3 Calculate the net force acting on charge C.

7.4 Aand B are brought into n Adetinen separated.

Charge A is returned to it ile charge B is removed.

7.4.1 In which direction did |
Choose from A to B of

7.4.2 Calculate the number of ele ansferred to or from charge

AtoB.

(2)
(2)
(4)

(1)

(5)

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7.5, Acharged sphere A carries a charge of -0.65 UC. It is placed 10 cm away
from another charged sphere B, which carries a charge of +0.8 uC, along a
‘straight line in a vacuum. Point P is located 7cm to the right of

sphere A.

7.5.1 Define electric field at a point. (2)
7.5.2 Draw electric field lines on sphere A. (2)
7.5.3 Calculate the magnitude of the net electric field at point P. (5)

[23]

TOTAL: [100]

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Physical Sciences/Gr 11 14 LDoE/March. 2026

DATA FOR PHYSICAL SCIENCES GRADE 12
PAPER 1 (PHYSICS)

GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12
VRAESTEL 1 (FISIKA)

TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES

NAME/NAAM SYMBOL/S/IMBOOL VALUE/WAARDE
Acceleration due to gravity
g 9,8 m-s@
Swaartekragversnelling
Universal gravitational constant
. ote G 6,67 x 10°11 N-m2-kg?
Universele gravitasiekonstant
Speed of light in a vacuum
P pia c 3,0 x 10° m-s"
Spoed van lig in 'n vakuum
Planck's constant
h 6,63 x 10% J-s

Planck se konstante

Coulomb's constant
9,0 x 109N-m?-C2

Coulomb se konstante

Charge on electron
-1,6 x 10°C

Lading op elektron

Electron mass
9,11 x 10°" kg

Elektronmassa

Mass of Earth

M 5,98 x 10% kg
Massa van Aarde
Radius of Earth

Re 6,38 x 10 km
Straal van Aarde

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TABLE 2: FORMULAE/TABEL 2: FORMULES

MOTION/BEWEGING

vp=vitadt Ax =v At+4adt? or/of Ay = v,At+taat?

Vv =v, + 2adx orlof v,? =v, +2ady | Ax= (“ - Ja orlof Ay = (“Ja

FORCE/KRAG
Fue: = mal p=mv
C= iN f, =y,N
F,,,At=Ap

Ap= mv, - mv,

mm,

F=G ae orlof =F

ELECTROSTATICS/ELEKTROS

kQ,Q
Fe 5 2
we! pel
q q
_Q _Q
a or/ of “a

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fee, LIMPOPO

PROVINCIAL GOVERNMENT

REPUBLIC OF SOUTH AFRICA

DEPARTMENT OF EDUCATION

NATIONAL
SENIOR CERTIFICATE

MARKS/PUNTE: 100

This marking guideline consists of 11 pages/ Hierdie nasienriglyn bestaan uit 11 bladsye

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Marking Guidelines/Nasienriglyne

QUESTION 1/VRAAG 1

1.1 B vv
1.2 IB vv
13)0.C vv
14 a vv
1S Bw
1. 6a. NG
17 A vv
18 A 4
19°C vv
1.10 D vv

[20]
QUESTION 2/VRAAG 2

2.1. The sum of two or more vec is acting together on the object. ¥v/
Die som van twee ofn Saam op die voorwerp inwerk

OR/OF

as two or more vectors together.v “/

fek ee of meer vektore saam (2)
2.2. The object is lifted/mo' ere will be a resultant/net force

not equal to zero. “/ Di rp word opwaarts gelig/beweeg.

Daar sal 'n resulterende/netto krag wees wat nie gelyk is aan nul nie. (2)
2.3.1

A single vector having the
‘n Enkele vektor met diese

Fy = Fa sin 60°
= 530(sin60°) ¥
= 458,99 NY

(2)
2.3.2 POSITIVE MARKING FROM 2.2/POSITIEWE NASIEN VANAF 2.2

F x = Fa (cos60°)
= 530(cos60°) v
= 265NV

(2)

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Marking Guidelines/Nasienriglyne

2.4

Fy = Fe cos20°
ao 240)cos20°v

iv
i 1165,21N

a
y = Fe sin 70°

= (1-240)sin70°

: a

/ SIE

= 1165,21N

Maximum w = Fya + Fys
= 265 +1165,21v
= 1430,21NV

(4)

2.5 If the distance x increases, ihe vertical components of the applied forces
will decrease v and tl fill possibly not be able to pick up the
weight. “/As die af: x die vertikale komponente van die
toegepaste kragte afneem e stelsel moontlik nie die gewig
kan optel nie.

(2)
[14]

QUESTION 3/VRAAG 3

3.1 The forces acting on the block are balanced/in equilibrium. “/
Die kragte wat op die blok inwerk, is gebalanseerd/in ewewig

OR/OF

The resultant/net force is equal to zero. /Die resulterende/netto krag is
gelyk aan nul.

OR/OF

There is no resultant force acting on the box Y / Daar is geen
resulterende krag wat op die boks inwerk nie (2)

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3.2

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Marking Guidelines/Nasienriglyne

Fretiy) =O0 Vv
Ty=w=mg
= (200)(9.8) ¥
= 1960 N

OPTION 1/OPSIE 1 aa OPTION 2/OPSIE 2

Ty=T cos 6
1960 = Tcos15° ¥
«T= 2029,14N ¥

OSTIEWE NASIEN VANAF 3.2

14s v= : RIOF 2029,14cos75° =F
«F = 525,18 NV «F = 525,18 N

OR/OF

F
Tan 15° = iy

F = 1960Tan15°
= 528,18 N
OR/OF

T? = F2 + F,? (Pythag

F= Jr- EK?
g
P= 2029, 147— 19602

F?=257728.18
F=525,18N

(4)

(3)
[09]

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5
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Marking Guidelines/Nasienriglyne
QUESTION 4/VRAAG 4

4.1. Frictional force: the force that opposes the motion of an object and
acts parallel to the surface the object is in contact withY v/

Wrywingskrag: die krag wat die beweging van 'n voorwerp teenwerk en
parallel met die opperviak waarmee die voorwerp in kontak is, werk. (2)

Marking criteria/Nasien kriteria

If any of the underlined key words/phrases in the correct context is

omitted deduct/ Indien enige van die onderstreepte

sleutelwoorde/frases in die korrekte konteks weggelaat word, trek af
(2 or/of 0)

4.2
4.2.1
Fhet(y) =0 v
f = ukN Any one/enige een
= ukmg cos @
f= ! sf,
= ae
(3)
4.2.2 re

orep nn
Fnet = ma + “Se |
Fgll — fk = ma v
mg sin® — uxmgsin 8 =ma Any one/ enige een

(4)(9,8)sin30°v — (4)(9,8)sin30° v= (4)av

a= 3,20 ms? v

(5)
4.3 ZERO ACCELERATIONY/ NUL VERSNELLING

The block continues at constant velocity/no frictional force/ the net force is
equals to zero. “/Die blok beweeg voort teen konstante snelheid/geen
wrywingskrag nie/die netto krag is gelyk aan nul. (2)

[14]

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Marking Guidelines/Nasienriglyne

QUESTION 5/VRAAG 5:

5.1. Marking criteria/Nasien kriteria

If any of the underlined key words/phrases in the correct context is omitted
deduct/ Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte
konteks weggelaat word, trek af

When a resultant/net force acts on an object, the object will accelerate in the
direction of the force atan acceleration is directly proportional to_the
resultant/net force and inversely proportional to the mass of the object. vv/
Wanneer 'n resulterende/netto krag op 'n voorwerp inwerk, sal die voorwerp in die
rigting van die krag versnel teen 'n versnelling wat direk eweredig is aan die
resulterende/netto krag en omgekeerd eweredig aan die massa van die voorwerp.

(2 or 0) (2)
5.2
Nv
fe ¥

N Fn/ Normal force / Normaal krag / Fnormal / Fnormaat / v
212,18 N

fk (kinetic) friction / (kinetiese) wrywing / Fi f / 40 N v

F Applied force / Toegepaste krag / Fappiied / Ftoegepas / v
Fa

w Fg/ Fw/ mg / 245 N/ weight/ gewig / gravitational v
force / gravitasie krag

Notes/Notas: Mark is awarded for label and arrow./Punte word
toegeken vir byskrif en pyl

e Do not penalize for length of arrows/ Moenie

penaliseer vir die lengte van pyle nie

lf arrows do not touch the dot: Max 3/4/ Indien pyle nie die punt

raak nie: Maks 3/4/

e_ Any other additional force(s): Max/ % / Enige ander bykomende
krag(e): Maks/ 3/4

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Marking Guidelines/Nasienriglyne
e |f everything is correct, but no arrows: Max %/ Indien alles
korrek is, maar geen pyle nie: Maks. 3/4
e If components are drawn: Max % / Indien komponente geteken
word: Maks 3/4

(4)

5.3.1

For the 4 kg block/ Vir die 4 kg-blok

Fret = ma ‘i

T +(- fk) =ma

T-2=4av

T=4at+ 2.0.0.0... (1) Any one/enige een

For the 10 kg block/ Vir die 10 kg-blok
Fret = ma

Fx -T-fk =ma

60cos 25° ¥ —-T-3,5 =10a
54,378-T-3,5 = 10a

(4)
5.3.2, POSITIVE MARKING
T = 4(3,491) + 2v
T = 15,97 NV

SITIEWE NASIEN VANAF 5.3.1

(2)

5.4 Increases. V /Neem toe (1)
[13]

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Marking Guidelines/Nasienriglyne
QUESTION 6/VRAAG 6:

6.1 Any two objects attract each other with a force that is directly proportional to
the product of their masses and inversely proportional to the square of the
_ distance between their centresY “/ Enige twee voorwerpe trek mekaar aan
met 'n krag wat direk eweredig is aan die produk van hul massas en
omgekeerd eweredig aan die kwadraat van die afstand tussen hul
middelpunte. (2)

6.2.1 R=20d
. 7

(2)

_ (6,67 x 10711)(5,98 x 10

Fe=
(6,69 x 10°
F = 1,06x104 NY (4)
6.3 W=mgv
=1200x9,8Y Suu a

= 1,18 x 104 NV (2)
[10]

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QUESTION 7/VRAAG 7:

7.1. NO V/NEE
The spheres will lose their chargev/ Die sfere sal hul lading verloor
OR/OF

The metal stands are electrical conductors / Die metaalstaanders is
eleKtriese geleiers (2)

7.2 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 chargesv and inversely proportional to the square
of the distance between them. v / Die grootte van die elektrostatiese
krag wat deur een puntlading op 'n ander puntlading uitgeoefen word, is
direk eweredig aan die produk van die groottes van die ladings en
omgekeerd eweredig aan die kwadraat van die afstand tussen hulle.

107

(2)

7.3

OPTION 1/OPSIE 1

paki; ,
r?
— (9 x 109)( 4x 107
CA = (0,
= 4,28 x 10-°N ui

(9x 10°)(7 x 107-°)(10 x 107°)
ce = v
(0,17)?
= 2.18 x 10-5N

Fnet = Fos - Fea
= 2,18 x 1075 -v (4,28 x 10~°)
= 1,75 x 10 ©N to the right“/na regs

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Marking Guidelines/Nasienriglyne

OPTION 2/OPSIE 2

_ (9x 10°)(4x 107°) ,
~ (0,29)2
= 428,06 N.C t=

(2 736 x12 x 10° )
3,28 x 10 © N due Eastv/ reg oos

(4)
7.4.1 A tolna BY (1)
742 Q =2t&
#c5,cCOoMm
Q =15x10°C
Charge transferred/lading oorgedra = 1,5 x 10-°-v (—4 x 107°)
=5,5 x 107°C
OR/OF
= 15 x 10°°-7 x 107°) >
=-5,5 x 10-°C
n= ay
qe
_ 5,5x107?
~ 1,6 x 10719
n= 3,44 x 10?° (electrons/elektrone) V (5)

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7.5.1 The electric field at a point is the electrostatic force experienced per unit
positive charge placed at that point.vv / Die elektriese veld by 'n punt
is die elektrostatiese krag wat per eenheid positiewe lading wat by daardie
punt geplaas word, ervaar word

(2 orlof0) = (2)

152
Notes/Notas: more than one
charge or combined field_drawn
0/2 /meer as een lading of
gekombineerde veld geteken 0/2
Criteria for sketch/Kriteria vir
skets:
Correct shapev / korrekte vorm
Correct direction towards the
charge / Korrekte rigting na die
lading
/ Nota: If electric field lines
$ or touch 1/2 / As elektriese
llyne kruis of raak 1/2
(2)
753.
_ kQ
E= nm v
(9 x 10°)(0,65 x 107°)
A = YY
(0,07)?
= 1,19 x 10 ®N.C*
_ (9x 10)(0,8 x 107°) |
~ (0,03)2
= 8,0 x 10 §N.C*
Enet = Ea + Es
= (1,19 x 10 ®) + (8,0 x 10 §) v
Enet = 9,19 x 10 ®N.C1 ¥ (5)

TOTALITOTAAL: [100]

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