# Physical Sciences Grade 11 2023 QP & Memo

> Grade 11 Physical Sciences question paper and memo from the KZN Local March Test 2023, 13 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/amajuba-test-physical-sciences-february-2023-qp-and-memo/90b573af-fd86-45f4-b30b-05580b90b198
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2023
- Exam period: KZN Local March Test
- Curriculum: CAPS
- Pages: 13

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## About this document

Physical Sciences · Grade 11 · KZN Local March Test · 2023. Question paper and memorandum, 13 pages. Read online or download the PDF.

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

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

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education

Department:
Education
PROVINCE OF KWAZULU-NATAL

AMAJUBA DISTRICT

MARKS: 50

TIME: 1 hour

This question paper consists of 7 pages and 1 data sheet.

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INSTRUCTIONS AND INFORMATION

Write your NAME and SURNAME in the appropriate spaces on the ANSWER BOOK
provided.

Answer Att the questions.

You may use a non-programmable calculator.

You may use appropriate mathematical instruments.

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

Data sheets and a periodic table are attached for your use.

Give brief motivations, discussions, et cetera where required.

PAPER 1: PHYSICS

QUESTION 1: Multiple choice questions

Four options are provided as possible answers to the following questions. Each question has
only ONE correct answer. Choose the correct answer and write the letter (A — D) next to the
question number (1.1 — 1.4)

1.1

1.2

The frictional force acting on a sliding object is ...

A dependent of the apparent area of contact.

B__ proportional to the normal force.

C dependent of the velocity of object.

D__ independent of the type of surface. (2)

When a spaceship moves at constant velocity, it means that the resultant
force acting on the body is zero. This phenomenon is best ined by

Newton’s First Law.
Newton’s Second Law.

Newton's Third Law.

0 QO DBD >

Newton's Universal Gravitational Law. (2)

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1.3

00ND >

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A learner pulls a block ata CONSTANT SPEED over a rough horizontal
surface with a force F. The force diagram below shows all the forces
acting on the block.

Soh

Y

Z

>

Which ONE of the following relationships between the magnitudes of the
forces F, X, Y and Z is true?

>

>

nma.mnn 7

Y andX =
Y andX <
Y andX =

Z
Z
Z

Y andX < Z

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(2)
[6]

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QUESTION 2

The diagram below shows three forces P, Q and R of 5 N, 3 N and 4 N respectively
acting on an object in the same Cartesian plane.

(2)
2.1. Give areason why the three forces are classified as vectors.
2.2 Determine the magnitude and direction of the resultant force, either by
CALCULATION or by ACCURATE CONSTRUCTION AND MEASUREMENT.
Use scale 10 mm: 1N
(7)

A box of weight (w) 60 N hangs on a ceiling as shown in the sketch below. A horizontal force F
acts horizontally to the right through knot S. The knot S is in equilibrium.

2.3 Explain what is meant by the knot S is in equilibrium. (2)
2.4 Draw the triangle of the three forces T, F and w. Clearly label th s and all (3)
the angles.
2.5 Calculate the magnitudes of the force F and the tension T. (4)
[18]

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QUESTION 3

A crate of mass 95 kg crate lies on a plank inclined at 23,2°. At this angle the crate is just
about to move down the incline. Refer to the diagram below.

3.1 Define the term frictional force. (2)

3.2. Sketch a free body diagram showing the force(s) acting on the crate at its current (3)
position.

3.3. CALCULATE:

3.3.1 The magnitude of the static frictional force (3)

3..3.2 The coefficient of static friction between the plank and the block. (5)
3.4 The plank is now tilted at an angle of 20.0°. State whether the static friction force

will be LESS THAN; EQUAL TO; OR GREATER THAN Question 3.3.1 above.

(1)
[14]

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QUESTION 4

A 6 kg block on a horizontal rough surface is joined to a 2 kg block by a light, inelastic string
running over a frictionless pulley. The frictional force between the 6 kg block and the table is
11,76 N. Ado ards force F of 2 N is applied to the 2 kg block as indicated in the diagram
below.

6kg
l
2kg
F=2N

4.1 State Newton’s Second Law of motion in words. (2)
4.2 Identify ONE action-reaction force pair acting on the 6 kg block. (2)
4.3, CALCULATE:

4.3.1 The magnitude of the acceleration of the 6 kg block. (5)

4.3.2 The magnitude of the tension (T) in the string connecting the two blocks (2)

4.4 The rough surface is replaced by a smooth frictionless surface. How will this
change affect the answer in QUESTION 4.4.1? Write only INCREASES,
DECREASES or REMAINS THE SAME. (1)

[12]

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DATA FOR PHYSICAL SCIENCES

TABLE 1: PHYSICAL CONSTANTS

=
av SYMBOL VALUE

Acceleratioiralg to gravity 8 9,8 m-s™

Universal gravitational constant G 6,67 x 10" N-m™-kg~

Electron mass Me 9,11 x 10°" kg

Mass of Earth Me 5,98 x 10” kg

Radius of Earth Re 6,38 x 10° m

TABLE 2: FORMULAE

MOTION

Vp =v tat Ax =v At + dat? or/of Ay = v,At+ dadt?

v, +Vv v, +Vv
V, =v, + 2aAxor/ofv, =v, +2ady | Ax= (Sea or/of Ay = fem

FORCE
Fret = Ma w=mg
fs¢maxy = HsN fx = KN
F-S oR FS <S OR gS
(aTaYay

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PROVINCE OF KWAZULU-NATAL

AMAJUBA DISTRICT

MEMORANDUM

This question paper consists of 7 pages and 1 data

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QUESTION 1

11 BV )

1.2 Cv him (2)

1.3 c rng (2)
jaca

— [6]
QUESTION 2
21 They have both magnitude and direction. ¥W (2)

2.2

F, = Q+Rcos 45°
=3+4cos45°V
= 5,828 N, right

F, = P —Rsin 45°
=5-—4sin45°V
= 2,172 N, up

F, Fy

net

Fy

F,, =F? +F;
= 5,828" +2172 v
=6,22N
tang =
ang = &
2,172
tan == vv
6 = 20,44°

Resultant force = 6,22 NV in direction: 20,44° YOR Bearing
69,56°, N69,56°E or 20,44° North of East

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2.2
SCALE DRAWING:

3N

Resultant = 6 N

CRITERIA MARK
Each angle correctly measured 1x2
Correct tail to head drawing of PQR 1x3
Resultant both magnitude and direction correct starting from origin to head of 1x2
vector R
TOTAL | 7 MARKS
2.3 The resultant of all forces Y acting at point S is zero ¥ (2)
24
55°
w
90° 35° (3)
F
e Correct shape of triangle VW
e Forces labelled and point correct directions ¥
e Labelled angles are all correct V
25
sin35° == ACCEPT:
oy Sine rule,
sin35° = | cosine rule,
T = 104,61 NY component
method and
tan 35° =t any other
60 trigonometric
tan35 =>¥ method
F = 85,69 NY
[18]

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QUESTION 3
3.1 | The force that opposes the motion of an object Wand which
acts par, ‘o the surface. ¥ (2)
Fu
3.2 Other labels to
be accepted:
Normal/ N
(3) | Friction/ f
Weight/ gravity/
Fy
wv
3.3
3.3.1
f; = mg sin 0Y (3)
= 95,0 x 9.8 x sin 23.2°%
= 366.76NY
3.3.2 N = mgcos OY
N = 95.0 x 9.8 x cos 23.2°V
N = 855.72N
fo = Us NV¥
366,76 = ux 855,72¥
Ms = 0,43 (5)
3.4 | Less than ¥ (1)
[14]
QUESTION 4
4.1 | When a resultant (net) force acts on an object, the object will
(2)

accelerate in the direction of the force. This acceleration is

directly proportional to the forceY and inversely proportional to
the mass of the object. ¥

is

4.2

Force of the block on the table “and the force of the table on
the block. ¥

OR

Force of the block on the string Y and the force of the string on
the block. ¥

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4.3

= mav
+(-f) =ma
T = 11.76 = 60M over os oe (1)

2 kg: T
F
Ww
Fret = ma

w+F+(-T)=ma
(2x 9.8) +2-—T = 2av

21,6 — TH 20 ore vee vve os (2)
(5)
(1) + (2): 9,84 = 8av¥
a=1,23m:s?v¥
43.2 T—11,76=6(1,23)V Q)
T=19,14 NY
4.4 | Increases V (1)
[12]

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