# Physical Sciences Grade 11 2023 QP & Memo

> Grade 11 Physical Sciences question paper and memo from the Limpopo Local March Test 2023, 15 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/limpopo-east-physical-sciences-grade-11-march-2023-qp-and-memo/2992dac7-015d-4763-a657-a6516dde6d0d
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2023
- Exam period: Limpopo Local March Test
- Curriculum: CAPS
- Pages: 15

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## About this document

Physical Sciences · Grade 11 · Limpopo Local March Test · 2023. Question paper and memorandum, 15 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

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To
MOPANI EAST DISTRICT

NATIONAL
SENIOR CERTIFICATE

MARKS : 100
TIME : 2 Hours

This question paper consists of 9 pages including this cover page.

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INSTRUCTIONS AND INFORMATIO

10.

Write your NAME in the appropriate space on the ANSWER BOOK.
This question paper consists of 4 questions. Answer ALL the questions.

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.

Show ALL formulae and substitutions in ALL calculations.

Round off your FINAL numerical answers to TWO decimal places.
Diagrams are NOT necessarily drawn to scale.

Write neatly and legibly.

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QUESTION 1

Four options are provided as possible answers to the following questions. Each question
has only one correct answer. Write down ONLY the letter (A-D) next to the question
number.

1.1

1.2

1.3

1.4

1.5

UNWU>Y GVNN>

QUOUN>Y

when?

leone

Choose-the set of physical quantities which are vectors only.

work, energy, weight, distance

force, mass, time, power

force, displacement, velocity, acceleration

force,.distance, speed, acceleration (2)

When resolving the vector into horizontal component, which one of the
following is relevant?

Ry = Rsiné@ for the resultant component

Ry = Rcosé6 for the resultant component

Rx = Rcosé@ for the resultant component

Rx = Rsin@ for the resultant component (2)
Which of the following forces is equal but in opposite direction to the
normal?

Fgt to the surface

Fg||to the surface

Applied force

Frictional force (2)
Frictional force is:

inverse to normal

inverse to applied force

proportional to applied force

proportional to the normal force (2)
A negative charge of 1uC which is free to move is placed a distance 2r
from a positive charge of 4 uC as shown below:

Which one of the following statements regarding the -1uC charge, when

it is at distance r is correct:

The electrostatic force experienced by the -1yuC charge will:

remain the same

be halved

be doubled

increase four times (2)
[10]

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2.1
2.2
2.2.1
2.2.2
2.3

2.3.1
2.3.2

Fi = 8N

450\ Fe= 6N

>
oe 7

Three forces Fi, F2 and Fs act at a common point O as shown in the
diagram above.
The magnitude of vector Fi = 8 N, F2=6 Nand F3=5N.

A

Define the term resultant of a vector. (2)
Use calculations to determine the following:

The vertical and the horizontal components of vector F1. (4)
The resultant force acting at point O. (9)

The structure below shows a system that is in an equilibrium state. The
strings with forces T1, T2 and Tz are joined at point O. The weight of the
block suspended from T3 is 2 OOON.

70°
2
Ta oO
T3
2 000N
Explain the meaning of a closed vector diagram. (2)

Draw a labelled closed vector diagram of forces showing all the forces (2)
acting on the point. Indicate at least one angle in your diagram.
[19]

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crate of mass 2kg Is pu e right across a rough horizontal
surface by a constant force F. This constant force F is applied at an angle
of 20° to the horizontal as shown in the diagram below.

F

20°
6 kg
3.1 State Newton’s first law of motion. (2)
3.2 Draw a free body diagram, showing al the forces acting on the crate. (4)
3.3 A constant frictional force of 3N acts between the surface and the crate.

The coefficient of kinetic friction between the crate and the surface is
0.12. Calculate the magnitude of:

3.3.1 Normal force acting on the crate. (3)
3.3.2 force F. (4)
3.3.3 acceleration of the crate. (4)
3.4 A 5 kg block, resting on a rough horizontal table is connected by a light

inextensible string passing over a frictionless pulley, is connected to
another block of mass 2 kg hanging vertically as shown in the diagram
below:

A force of 60 N is applied to the 5 kg block at an angle of 10° with the
horizontal, thus making the block to accelerate to the left.

60N
if

2kg

The coefficient of kinetic friction between the 5 kg block and the surface
of the table is 0.5. Ignore the effects of air friction.

3.4.1 Draw the free body diagram, showing ALL the forces acting on the 5 kg (4)
block.

3.4.2 State Newton’s Second Law of motion in words. (2)
3.4.3. Calculate the vertical component of the 60 N force. (3)
3.4.4 Calculate the horizontal component of the 60 N force. (3)
3.4.5 Calculate the normal force acting on the 2 kg block. (2)
3.4.6 Calculate the tension force connecting the two blocks. (5)
3.5 A massive rock from outer space is moving towards the Earth.

3.5.1 State Newton’s Law of Universal Gravitation in words. (2)
3.5.2 How does the magnitude of the gravitational force exerted by the Earth (2)

on the rock changes as the distance between the rock and the Earth
becomes smaller?
Choose from INCREASES, DECREASES or REMANS THE SAME.
Give a reason for your answer.

[40]

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Q .
A pouiMaded From Stanmorephysics.CQM a1. 0, are at a distance

4.1.1
4.1.2

4.1.3

4.1.4

4.2.

4.2.1
4.2.2

d apart and are suspended from an insulated rigid wooden bar as shown
in the diagram below. The charges on the spheres are -4uC and +8uC
respectively.

Pp Q

Sphere Q experiences an electrostatic force.

Define electric field.

In which direction will sphere Q move? Write down only TO THE LEFT
or TO THE RIGHT.

The spheres are now made to touch and then separated. Were electrons
REMOVED FROM or TRANSFERRED TO P?

Give a reason for your answer.

Calculate the distance d between the two spheres if the force
experienced by P is 0.8 N.

Charges of +4uC and -3uC are placed at a distance of 0.2 m apart ona
straight line as shown below. Point P is located 0.1m to the right of the -
3yC charge.

CO 0.2m >( \e 0.1m ,p

+4uC -3yuC

Calculate the:

net electric field at point P

mass of a +6uC charge placed at P that will experience an acceleration
of magnitude 5 x10* m.s?.

Page 6 of 9

(2)
(1)

(2)

(4)

(5)
(3)

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4. D ownlogde small fF Om teh Sfannorgph Baal 694 carrying a charge of +7

4.3.1
4.3.2

4.3.3

4.3.4

x10° C, is suspended from a horizontal surface by a string of negligible
mass. A second sphere B, carrying a charge of -8 x10° C, on an isolated
stand, attracts sphere A, so that the string forms an angle of 20° with the
vertical. The horizontal distance between the centres of the two is 3cm.
Refer to the diagram below:

‘0°

C) 3.cm -8x10°C

State Coulomb’s law in words. (2)
Draw a vector diagram of the force acting on sphere A and indicate at (4)
least ONE angle.

Calculate the magnitude of the electrostatic force that sphere B exerts (4)
on sphere A.

Calculate the magnitude of the tension force in the string. (4)

[31]
TOTAL= 100 MARKS

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Downloaded from StannrekeplsiésiGsin@es crave 11
PAPER 1 (PHYSICS)

GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 11
VRAESTEL 1 (FISIKA)

TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES

NAME/VAAM SYMBOL/SIMBOOL | _ VALUE/WAARDE
Sree btraghonstante G 6,67 x 10°" Nm?kg?
ee Re 6,38 x 10° m
radi, op olekon e -1,6x 10°C
Eeetron mass Me 9,11 x 10°" kg
Massa van die Aarde M 5,98 x 10" kg
FORCE/KRAG
F..=ma w=mg
p-Smim My = fem
r N
f,
B= N

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Downdess| GO UND ANTSINeRIICOLWE YKLANKCEMLIG

v=fa T= 7
nj sin®; =n, sin @, n=<
ELECTROSTATICS/ELEKTROSTATIKA
pO (k = 9,0 x 10°N-m?-C) Es
= KO (k= 9,0x 10°N.m2-C%) | n= e

ELECTROMAGNETISM/ELEKTROMAGNETISME

e--na®
At

®=BAcos@

ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE

\-2 r-Y
At |
44,444, R=,+ fh ++
Ron h fs _ aes
WwW
W=Vq p-—
At
We=VIAt
P=VI
W=PRAt
P=PR
12
we V*At ye
R p=
R

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DEPARTMENT OF EDUCATION
MOPANI EAST DISTRICT

NATIONAL

SENIOR CERTIFICATE

GRADE 11

MARKS : 100

This memorandum consists of pages including this cover page.

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2 Balitaded from Stanmorephysics.com

Cw
Add
Dv
Dw

1.2
1.3
14
1.5

QUESTION 2

2.1

together. /V

Option 1

Fix = Ficos30° ¥
= 8cos30°
=6.93N/

2.2.1

Fiy = Fisin30° V
= 8sin30°
=40N/

Option 2

Fix = Fisin60° J
= 8sin60°
=6.93N/

Fiy = Ficos60° ¥
= 8cos60°
=4.0N/
Horizontal vectors
Fix = -6.93 N
Fox = +6 N
Fsx = Fscos60° or
= 5cos60° J
=2.5N

2.2.2

Frx = -6.93 +6-2.5
= -3.43N
= 3.43 N to the left /

Vertical forces

Fiy = Fasin30°
= 8sin30° /
=4.0N

Fay =ON

Fay = Fssin60° or
= 5sin60° J
=4.33N

Fry = 4-4.33
= -0.33 N = 0,33 N downwards J

Frx

A single vector having the same effect as two or more vectorsacting

(2)

(2)

(2)

(2)
[10]

(2)
(4)

(9)

Fx = Fasin30°
= 5sin30° J
=2.5N

Fay = Fscos30°
= 5cos30° JV
=4.33N

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Fr

Fr = \/ (Frx)? + (Fry)? Jv
=\/ (3.43)? + (0.33)?

=3.45N J

tan 6 = 0.33 JV
3.43
6 =5.50°

So Fr = 3.45 N at the bearing of 264.50° J (270° — 5,50°)

2.3.1 A closed vector diagram is a set of vectors drawn on the cartesian plane
using the tail-to-head method and that has a resultant with a magnitude
of zero, Jv

2.3.2. th

Ts , VJ

[19]
QUESTION 3
3.1. An object will continue in its atate of rest / uniform motion unless a
resultant force acts on it. / /
3.2. Ny Nn
' Fy

J ~ OR JF Fx v

Fg / wi Fe/wi

3.3.1 Fk = ykN J
3 = (0.12)(N) J
N=15N/
3.3.2 Fnet = ma /
N+Fy-W=0
15 + Fsin20°/ —(6)(9.8) / =0
F=128.06N/
3.3.3 Fnet = ma /
Fcos20° - Fy = ma
128.06 cos20° /- 3 = 6aJ/

(2)

(4)

(3)

(4)

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3.4.1

3.4.2

3.4.3

3.4.4

3.4.5

3.4.6

3.5.1

Fa

>TOR Fx« mT

Fe/W Fg/W

Marking guidelines

W/ Weight/ Fg /

T/ Tension /

Fa/ Applied force / Fx J

N/ Normal force /

When force is applied on an object it produces acceleration in the (2)

direction of the force and this acceleration is directly proportional to the

applied force and inversely proportional to the mass of the object. / /

Fx = Fcos10° J (3)
= 60cos10° /
=59.1N/

Fy = Fcos10° V (3)
= 60sin10° V
=104N/

Fy +N=WJ/ (2)

10.4 + N =5(9.8)

N=38.6N/

Fnet = ma J (5)

For the 5kg block

Fx-T=ma

59.1-T=5a/

T = 59.1 58 .oeeecceeeeees (1)

For the 2kg block

Fnet = ma

T-W=ma

T — (2)(9,8) = 2a/

T= 19.6 + 28 oc ceeeeeeeee (2)

Equating (1) and (2)

59.1-5a=19.6+2a

a=5.64m,s?/

substituting a = 5.64 m,s° in either (1) or (2) gives:

T=30.88N/

Each body in the universe attracts every other body with a force that is (2)
directly proportional to the product of their masses and inversely
proportional to the square of the distance between their centres. //

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r2

QUESTION 4

4.1.1. Electric field is a region in space where wherein an electric charge

experiences a force. v /
4.1.2 To the left /
4.1.3 Removed /
The charge on P is positive /
414 F=kOiQ2 J
d2

0,8/ =(9 x109) x( 2 x10°) x (2 x10°9) /
d2

d=0,21m/
4.2.1 E=kO J
2

r

E4= (9 x109) x (4.x10°) JV
(0.3)

= 4x10° N,C* to the left

Es = (9 x109) x (3.x10°°) /
(0.1)2
= 2.7x10® N,C* to the left

Ep = Ea + E3
= (0.4 — 2.9) x10 N.C /
= 2,3 x 10° N.C to the left /
4.2.2 F=maJ
QE =ma
m = (6 x10°)(2,3 x10°)_J/
(5x10?)
= 2.76 x10? kg V

(2)

(2)

(1)
(2)

(4)

(5)

4.3.1 The magnitude of the electrostatic force exerted by two charges oneach (2)

other is directly proportional to the product of the charges and inversely

proportional to the square of the distance between them.

>

Fg T

Marking guidelines

4.3.2

Fe

W / Weight/ Fg /
T/ Tension J
FE/ Electrostatic force /
Angle indicated ¥
4.3.3 F = kQ1Q2V/
r2

(4)

(4)

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_ -9 -9 .
Downloade ; ephysics.com

= 5,60 x107N J
4.3.4 Fg = mg (4)
= (2,0 x10%)(9.8) ¥
= 1.96 x10° NV

T? = Fg2+ Fe?
= (1.96 x10°)2 + (5,6 x10%)2/
= 2.04 x10? NJ
[31]
TOTAL = 100 MARKS

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