# Physical Sciences Grade 11 Limpopo March Test 2023 QP & Memo

> Grade 11 Physical Sciences question paper and memo from the Limpopo March Test 2023, 21 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/limpopo-physical-sciences-grade-11-march-2023-qp-and-memo/002a8b49-f43d-4b5d-b974-bfb9bf1a1810
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2023
- Exam period: Limpopo March Test
- Curriculum: CAPS
- Pages: 21

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## About this document

Physical Sciences · Grade 11 · Limpopo March Test · 2023. Question paper and memorandum, 21 pages. Read online or download the PDF.

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## Document text

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7 CIE

PROVINCIAL GOVERNMENT
REPUSLUC OF SOUTH AFRICA

DEPARTMENT OF

EDUCATION

MOGALAKWENA DISTRICT

PHYSICAL SCIENCES

NATIONAL SENIOR CERTIFICATE

TERM 1 PRE-CONTROL TEST
07 MARCH 2023
GRADE 11

MARKS : 100 DURATION: 2 hours

STARTING TIME: 08HO00

This question paper consists of 13 pages including this one

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INSTRUCTIONS

1. This question paper consists of 13 pages including the cover page
2. Answer all the questions in the answer book

3. You are advised to use the attached DATA SHEETS.

4. Round off your final answer to a minimum of TWO decimal places
5. Show all your calculations including formulae where applicable.

6. Candidates may use non-programmable calculators.

7. Write neatly and legibly.

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QUESTION 1: MULTIPLE-CHOICE QUESTIONS

Four options are provided as possible answers to the following questions.
Each question has only ONE correct answer. Write only the letter (A-D) next
to the question number (1.1 - 1.10) in the ANSWER BOOK

1.1 A boy applies a force of 400 N to push a crate as shown below. The
frictional force acting on the crate is 300 N.

400 N
i—>

—
300 N

The magnitude of the force exerted by the crate on the boy is...

A ON

B 50N

c 300 N

D 400 N (2)

1.2 The following graph represents the relationship between acceleration
and net force F for two objects, X and Y, moving on the same
frictionless surface.

& | Object X

Object Y

Which one of the following statements is TRUE?

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A X has a greater mass than Y.
B X has a smaller acceleration than Y for equal forces.
Cc X experiences a greater force than Y for equal accelerations.
D The gradient of graph X represents the inverse of the mass of (2)

1.3

1.4

1.5

1.6

X.

Which ONE of the following physical quantities is a measure of the
inertia of a body?

A Mass

B Energy

Cc Velocity

D Acceleration (2)

Acar moving forward collides with a tree and comes to rest very quickly.
The passenger who was not wearing a seat belt continued moving
forward and collided with a car’s windscreen. Which law explains this
scenario?

A Newton’s first law of motion

B Newton’s second law of motion

Cc Newton's third law of motion

D Newton's law on universal gravitation (2)

A laptop rests on a table. According to Newton's 3rd law, what is the
reaction force to the weight of the laptop?

A The upward force of the table on the laptop.

B The upward force of the laptop on the earth.

Cc The downward force of the Earth on the laptop

D The normal force on the laptop. (2)

A lady pushes against a heavy crate on a horizontal surface. One
action-reaction force pair applicable to this situation is indicated in the
diagram below.

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Ferate on lady F lady on crate

ty

Which one of the following statements is correct for this situation?

A The crate accelerates because the magnitude of the force of the
lady on the crate is greater than the magnitude of the force of
the crate on the lady.

B The crate remains stationary because the force of the lady on
the crate is equal in magnitude to the force of the crate on the
lady.

Cc The crate remains stationary because the force of the lady on
the crate is equal in magnitude to the force of friction on the
crate.

D The crate moves at constant velocity because the force of the (2)
lady on the crate is equal in magnitude to the force of the crate
on the lady.

1.7. The gravitational force between two uniform spheres will change to one
half of the original force if the ...

A Mass of each sphere is halved.

B Distance between them is halved.
Cc Distance between them is doubled.
D

Mass of one of the spheres is halved. (2)

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1.8 Two small objects with masses M and m exert a gravitational force of
F on each other when separated by a distance d.

What gravitational force would be exerted by the objects with masses
3M and 2m at a separation distance of 6d?

A -36F

B 6F

Cc OF

D F (2)
6

1.9 Three point charges of magnitudes -3 uC, +3 uC and +2 UC are placed
at the three corners of a right angled triangle as indicated in the

diagram.
-3 uC e
42 uC e e t3uC

Which one of the following best represents the direction of the resultant
force exerted on the +2 uC charge by the other two charges?

D. Sa (2)

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1.10 Which one of the following sketch graphs best represents the
relationship between the electric field strength E and the distance d
from a given charge Q?

> > >» | > (2)

[20]
QUESTION 2
Two forces, of magnitude 50 N and 80 N, act at a point on a Cartesian plane
in the directions shown in the sketch below.
x 50|N
IW
30° INS 80N
2.1 Give the correct term for the following description: A single vector (1)
having the same effect as two or more vectors together
2.2 Calculate the:
2.2.1 Magnitude of the vertical component of the 50 N force (2)
2.2.2 Magnitude of the horizontal resultant (net) force (3)
2.2.3 Magnitude of the resultant (net) force (2)
2.2.4 Direction of the resultant (net) force (2)
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2.3. An athlete at point A runs 100 m north, then 70 m east, then 100 m
south and then 70 m south.

2.3.1 Draw a neat labelled vector diagram in the Cartesian place (2)
acting from the origin using the tail-to-head method

2.3.2 Explain why the vector diagram from Question 2.3.1 is referred (1)
to as a closed vector diagram

[13]
QUESTION 3
Grade 11 learners investigate the relationship between the net force and
acceleration by pulling a trolley across a surface which is slightly inclined to
compensate for friction. The trolley is connected to different masses by a
string of negligible mass. The string passes over a frictionless pulley. Refer to
the diagram below.
f oe trolley
mass
Ticker-tape attached to the trolley passes through the ticker-timer. The
acceleration of the trolley is determined by analysing the ticker-tape. The
results of the net force produced by the different masses and the acceleration
of the trolley were recorded in the table below.
Net Force (N) Acceleration (m.s)
0,3 0,41
0,6 0,81
0,9 1,22
1,2 1,63
3.1 Write down a hypothesis for this experiment (2)
3.2 — Identify the:
3.2.1 Independent variable (1)
3.2.2 Dependent variable (1)
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3.2.3 Law being investigated (1)
3.3. Use the graph paper on the ANSWER SHEET and draw a graph of (4)
the acceleration versus net force.
3.4 Calculate the gradient of the graph (3)
3.5 Use the gradient of the graph calculated in Question 3.4 to determine (2)
the mass of the trolley.
3.6 Use the information gained from your calculation and investigation to (2)
explain why overloading the car is dangerous.
[16]
QUESTION 4
A moving sled pulled by dogs begins to climb a slope which is inclined at 20°
to the horizontal. The total mass of the sled and the two passengers is 150
kg. The coefficient of kinetic friction for the sled is 0,145. The dogs exert a
combined force of 600 N on the sled.
an i
4.1 Drawa labelled free-body diagram (4)
4.2 Calculate the kinetic frictional force acting on the sled (3)
4.3. Calculate the acceleration of the sled. (6)
4.4 — If the direction of the acceleration is down the slope, what can you (1)
say about the motion of the sled?
4.5 Consider a car engine which hangs from a chain. A mechanic pulls the
engine sideways by applying a force of 100 N to the right. The engine
is held in such a position that the chain makes an angle of 30° to the
vertical
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Calculate the tension T in the chain. (3)

[17]

QUESTION 5

©

5.1

5.2

5.3

5.4

State Newton’s law of universal gravitation in words. (3)
Distinguish between the mass and the weight of an object (2)

Mars has a mass of 6.4 x 1023 kg and its moon Phobos has amass of 9,6 = (5)
x 10% kg. If the magnitude of the gravitational force between the two bodies
is 4,6 x 107° N, calculate how far apart Mars and Phobos are.

Calculate the weight of an astronaut whose mass is 72kg on the (3)
Phobos if the radius of Phobos is 7,4 x 106 m.

[13]

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QUESTION 6

Two small conducting spheres which carry charges of +8 nC and +5 nC are
placed a distance of 40 mm apart as shown in the diagram below.

+8 nC ia +5 nC
.
« 2
40 mm
6.1 State Coulomb’s Law in words (2)

6.2 Calculate the magnitude of the electrostatic force that one charge (4)
exerts on the other.

6.3 Define an electric field. (2)

6.4 Draw the electric field that would surround these two small conducting (2)
spheres. Show the conducting spheres as small circles.

6.5 Along the line joining the two small spheres, there is a point P, which (8)
is 10 mm from the +8 nC charge. Show that the magnitude of the
resultant or net electric field strength at P is 6,7 x 105 N.C7.

6.6 Anelectron is placed at point P. Calculate the force experienced by (3)
the electron due to the electric field.

[21]

GRAND TOTAL: 100 MARKS

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Term 1 Control Test

DATA FOR PHYSICAL SCIENCES GRADE 11

PAPER 1 (PHYSICS)

TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES

NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE
Sas! ven cle Inarde Re 6,38 x 10° m
Nason ven ‘io Aarde Me 5,98 x 10° kg

TABLE 2: FORMULAE/TABEL 2: FORMULES

MOTION/BEWEGING
Vv, =V, +aAt AX = vjAt+4aat?
V2 =v2 +2aax ax=[ "ae
FORCE/KRAG
Fee =Ma w=mg
Gm,m Fo smanm
F=- 7 2 b= ( A laks )
= fi
My = N
ELECTROSTATICS/ELEKTROSTATIKA
F
F= Kai, (k=9,0x 10°N-m?-C*) | E “5
r
g-kO (k= 9,0 x 10°Nm?-c) | n=2
r €

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ANSWER SHEET
HAND IN THIS ANSWER SHEET TOGETHER WITH THE ANSWER BOOK.
NAME: CLASS:
Question 3.3 Graph of acceleration versus net
EEE
a ro Toor
(m.s%)
Hi
oem BA)
Hee oH
HH ASSSRGGen Gene a a
Lt t
oe BETEEETEOEETUGSEETE . ro
I
Fret (N)

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m Stanmorephysics.com

gece LIMPOPO
PROVINCIAL GOVERNMENT

REPUSLUC OF SOUTH AFRICA

CEPARTMENT OF

EDUCATION

MOGALAKWENA DISTRICT

PHYSICAL SCIENCES

NATIONAL SENIOR CERTIFICATE

TERM 1 CONTROL TEST
07 MARCH 2023
GRADE 11

MEMORANDUM

MARKS : 100 DURATION: 2 hours

STARTING TIME: 08H00

This question paper consists of 8 pages including this one

Page 1 of 8
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QUESTION 1
1.4 Dvv
1.2 Dvv
1.3 AVY
1.4 AVY
1.5 BYY
1.6 BYY
1.7 D¥vv
1.8 Dvv
1.9 Dvv
1.10 BYY
QUESTION 2
2.1 Resultant (net) vector ¥
2.2

2.21 F =Fsin@
=50sin 30° v OR 50cos 60°
=25NV

2.2.2 F,=Fcos 8
=50cos 30° OR 50sin60°

=43.30 N to the left ¥

Substitution marks awarded within the
question even if calculations for F, and
Fy are wrong

Horizontal resultant
Frx=(-43.30)+(80) ¥
=36.70 N to the right ¥

Mogalakwena District

[20]

(1)

(2)

(3)

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2.2.3

2.2.4

2.3
2.3.1

2.3.2

Positive marking from Questions 2.2.1 and 2.2.2 (2)

Vertical resultant
Fry= 25 N upwards

Resultant Force
FRo= Frx + Fry”
FR’ = (36.70)*+ (25)° ¥

Fr=44.41N¥
Positive marking from Question 2.2.1 and 2.2.2 (2)
tane=*Z v
25
0=55.74° v
OR
inQ= 20
sinO= a
8=55.74°
OR Also accept direction as:
cos@=—28 @ = 90 — 55.74°
aa @ = 34.26°
0@=55.74°
. (2)
70m ¥v Correct diagram
,
100m 100m
70m
The first vector starts at the origin and the last vector ends at (1)

the origin. (The resultant would have a magnitude of zero since
itis drawn from the tail of the first vector to the head of the final

vector) ¥
[13]

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QUESTION 3

3.1

3.2

3.3

3.4

Marking criteria
e The dependent and independent variables are stated
correctly.

e State the relationship between dependent and independent
variable

The (net) force is directly proportional to acceleration if the mass of
the trolley is kept constant VW

3.2.1 Net force V
3.2.2 Acceleration ¥
3.2.3 Newton's second law of motion ¥

The relationship between acceleration and net force

ra

Pe
00

PP
oy

s P
or N

Soo
roy

Acceleration (m.s?)

o
nS

os
Nu

°

Fret (N)

Marking criteria for the graph

Axes with correct/ appropriate scale v

3 or more coordinates correctly plotted vv

Drawing a line of best fit v

1,63-0 v

Gradient= 72-07

= 1,36 ¥

(2)

(1)
(1)
(1)

(4)
(3)

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OR

— 1,63-1,22 _

Gradient 7205 ~ 1,36

OR

1,22—0,81 _

Gradient= Fo-96 —

1,36

3.5 POSITIVE MARKING FROM QUESTION 3.4.

a 1
Gradient= m

a
FNet
1
1,36=-—v
m

m=0.74kg ¥

3.6 For aconstant applied force, an increase in mass decreases the
negative acceleration. This cause the stopping time to increase and
therefore the vehicle has a greater stopping distance. VW

QUESTION 4

4.1 Normal force N

Applied force F = 600 N
Friction f:

Weight w

Accept the following symbols
Nv | Fy/ Normal force

Ff¥ | Kinetic friction force/ f,/ Fl F,

Fg¥ | W /mg

FY | Fa/Force Applied/600N

NB: Do not penalise for the length of the arrows

(2)

(2)

[16]

(4)

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4.2 {= uN (3)
= umgcosé ¥
= 0,145 (150)(9,8) cos 20° ¥

= 200,30 N ¥

4.3. Wy,=mgsiné (6)

=(150)(9,8) sin 20°
=502,77 N down the slope ¥

Fret=ma

(Fapp)+(-Wi)+CH=ma v

(600)+(-502,77)+(-200,30)”=150a ¥

-103.07=150a

a=-0,69 m.s?¥

a=0,69 m.s* down the slope ¥

4.4 It means that the sled will slow down. ¥ (1)
45
(3)
30° T
w
90° ~=60°
F=100N
«ano 100
sin 30 => v
100
~ sin 30°
T=200NV¥
[17]
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QUESTION 5

5.1 Every object in the universe attracts every other object in the universe (3)
with a force that is directly proportional to the product of their masses
vand inversely proportional to the square of distanceY between their

centres, ¥
5.2. The mass of an object is the amount of matter found in an object. ¥ (2)
The weight of an object is the force with which the centre of a planet attracts
an object. ~
5.3 F= Gmime v (5)
r
46x10 oy = (6:87 10"1 106.4% 1079)(9,6x10')w

r2

r=9,44x10°m ¥

5.4 — Smimg 3
Fe Stim (3)

_ (6,67x107"')(9,6x10'°)(72)¥

2
(7,4*10°)

=8,42x10 NV

[13]

QUESTION 6

6.1 The force between two charges is directly proportional to the product (2)
of the charges and inversely proportional to the distance between the
charges squared¥¥

6.2 =p Q1Q2 4
Fok ov (4)
-9 -9
8x10 5x10™ |v
“en 9 SNe"
(40x10) v
=2,25 x104N ¥
6.3 Aregion in space in which an electric charge V will experience a (2)
force ¥
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6.4 (2)

Y direction v shape

6.5 Eenc = kS Esne = kS (8)
I r
— (9%10°)(8x10%v) _ (9*10°)(5%10) v
re i, er i,
(10x10 v)* (30x10)? v
= 720000 N.C? ¥ =50000N.C? ¥

Enet = 720 000-50 000 Y (towards the right — positive direction)

670 000 NC? ¥
6.6 F = QE = (1,6x10"!°) x (670 000) ¥ (3)
=1,07x 10" NV towards the +8nC charge ¥
[21]
GRAND TOTAL: 100 MARKS

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