# Physical Sciences Grade 11 Paper 1 KZN November 2024 Memo

> Grade 11 Physical Sciences Paper 1 memorandum from the KZN November Exam 2024, 25 pages. Read online or download the free PDF.

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- Updated: 2026-09-28
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Memorandum
- Year: 2024
- Exam period: KZN November Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 25

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### Page 1

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

150

TIME : 3hours

This question paper consists of 12 pages and 3 data sheets.

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Physical Soweshaaded from Stanmexapdaysics.com November 2024

INSTRUCTIONS AND INFORMATION

il Write your NAME in the appropriate spaces on the ANSWER BOOK.

2 This question paper consists of 8 questions. Answer ALL the questions in the
ANSWER BOOK.

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

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

question paper.

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

6. You may use a non-programmable calculator.

te You may use appropriate mathematical instruments.

8. Show ALL formulae and substitutions in ALL calculations.

9. Round off your final numerical answers to a minimum of TWO decimal places.
10. Give brief motivations, discussions et cetera where required.

Tt You are advised to use the attached DATA SHEETS.

12. 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. Which ONE of the following pairs can be classified as scalar quantities?

A Mass and inertia

B Inertia and weight
Cc Weight and frictional force
D __ Frictional force and mass (2)

1.2. The vertical component of a force acting on an object moving horizontally...

A accelerates the object horizontally.
B accelerates the object vertically.
c has a tendency to move the object vertically.

D has a tendency to move thesobjectyh (2)
1.3 Acrate is moved, fr ; horizontal surface by a force F.
Which O} f the fo! ents about the kinetic frictional force
acting on t
The kinetic onal force is ...
A greater than the maximum static frictional force and less than F.
B greater than the maximum static frictional force and greater than F.
Cc less than the maximum static frictional force and greater than.F..
D less than the maximum static frictional force and less than F. (2)
1.4 Consider the following statements related to Newton's third law:

When a force acts on an object,

(i) the reaction to this force acts on the same object.

(ii) the reaction to this force acts on a different object.

(iii) the time for which the force acts is greater than the time

for which the reaction to this force acts.
(iv) _ the force and the reaction to this force are in opposite
directions.
Which of the above statements is/are true?

A (i) and (iv) only.
B (ii) and (iv) only.
Cc (i), (iii) and (iv) only.
D (ii), (iii) and (iv) only. (2)

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Physical

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ciences P1 NSC Grade 11 November 2024

The acceleration versus resultant force graphs for the motion of objects X and
Y are shown below. Which ONE of the following statements is true?

A Object X has a greater mass than object Y.

B Object Y has a greater mass than object X.

Cc If both objects experience the same acceleration, the resultant force on
object X is greater than that on object Y.

D For the same resultant force, object X has a smaller acceleration than
object Y.

1.6 The magnitude of the gravitational acceleration on Earth is g.
What will the value of the gravitational acceleration be on planet X, which has
the SAME mass as Earth, but HALF the radius of Earth?

A “wg
B 4g
Cc ’Ag
D 29

1.7 Asmall test charge, q, is placed exactly halfway between two fixed identical
charged spheres, X and Y, each carrying a charge of +Q, as shown in the
diagram below.

xX ¥'
> O<

eo ; *@

+Q +Q
The net electrostatic force experienced by the test charge will be...
A towards X.
B towards Y.
Cc zero.
D greater than zero.

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(2)

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Physical Sciences P1 November 2024

1.8 The minimum resistance that can be obtained by connecting two 4 Q
resistors in a circuit is ...

A 8a
B 40
Cc 3Q
D 20 (2)

1.9 The two resistors in CIRCUIT 1 below are identical. They are connected to a
cell with emf V and negligible internal resistance.
The power dissipated by each resistor is P.

i
|!

CIRCUIT 1

The same resistors are now GéAhed parallel, as shown in CIRCUIT 2.

What is the power dissipated by each resistor in CIRCUIT 2?

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

1.10 Abar magnet is moved into and out of a solenoid. Which ONE of the following
changes will result in a LARGER induced emf in the solenoid?

A Using a weaker magnet.

B Using a solenoid with fewer turns.

[e} Using a solenoid with a smaller diameter.

D Moving the magnet faster into and out of the solenoid. (2)
[20]

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Physical Sciences P1 NSC Grade 11 November 2024

QUESTION 2 : (Start on a new page)

Three co-planar forces, F1, F2 and Fs of magnitude 6 N, 3 N and 4 N respectively
act on an object. The sketch below, not drawn to scale, shows the directions of the

forces.

2.1
2:2

Define resultant force. (2)

Calculate the:

2.2.1 Magnitude of the resultant force acting on the object (8)
2.2.2 Direction of the resultant force (3)
[13]

QUESTION 3: (Start on a new page)

3.1
3.2

3.3

Define inertia. (2)

A passenger is seated in a fast-moving taxi when the taxi driver suddenly
slams on the brakes. Use your knowledge of Newton's laws to explain why
the passenger moves forward when the brakes are applied. (3)

Explain why wearing a seat belt is a safety measure for the occupants of a
vehicle. (2)

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3.4. Acrate of mass 80 kg lies on a rough plane inclined at 25° to the horizontal.
The crate begins to slip when the angle of inclination exceeds 25°. Refer to
the diagram below.

3.4.1 Define the term static friction. (2)

3.4.2 Draw a free-body diagram showing all the forces acting on the crate. (3)

3.4.3 Define the term normal force. (2)
3.4.4 Calculate the magnitude of the normal force. (3)
3.4.5 Calculate the magnitude of the maximum static frictional force acting

on the crate. (3)
3.4.6 Calculate the coefficient of static friction between the crate and the

plane. (3)
3.4.7 _ If the mass of the crate is increased, will the crate begin to slide?

Choose from YES or NO.

Explain the answer. (3)

3.4.8 If the angle of inclination is changed from 25° to 20°, how would this
change affect the static frictional force acting on the crate? Choose
from INCREASES, DECREASES, or REMAINS THE SAME.
Explain the answer. (3)

[29]

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Physical

QUESTION 4: (Start on a new page)

Two blocks of masses 1 kg and 3 kg are connected by a light, inextensible string.
The blocks are initially at rest on a smooth horizontal surface. A force F of
magnitude 9 N is applied to the 3 kg block, as shown in the diagram below.

4.1 State Newton's Se in in words. (2)

4.2 Drawa free-body di showing all the forces acting on the 3 kg block. (4)
4.3 By applying news eenane aw to each block separately, calculate the:
4.3.1 Magnitude of the acceleration of the blocks (5)
4.3.2 Tension in the string (2)

4.4 The 1 kg block is replaced by a block with a bigger mass, while the force
applied on the 3 kg block remains unchanged. State whether the tension
calculated in Question 4.3.2 will INCREASE, DECREASE or STAY THE

SAME. Give a reason for the answer. (3)
QUESTION 5: (Start on a new page) =
5.1. State Newton's Law of universal gravitation in words. (2)
5.2 Differentiate between the MASS and the WEIGHT of an object. (2)

5.3. A70kg mountain climber reaches the top of Mount Everest, 8,85 km above
sea level. Take the surface of Earth to be sea level.

At the top of Mount Everest, determine the:

5.3.1 Acceleration due to gravity (4)

5.3.2 Mountain climber’s weight (3)
5.4 Alearner makes the following statement:

“An astronaut orbiting Earth experiences weightlessness because he or she
is beyond the pull of Earth’s gravity’.

5.4.1. Define weightlessness. (2)
5.4.2 Is the statement correct? Choose from YES or NO.

Explain the answer. (3)

[16]

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QUESTION 6: (Start on a new page)

The diagram below shows two identical spheres, A and B on insulated stands,

carrying charges of -4yC and +4C respectively. The distance between the centres
of the spheres is 0,40 m.

0,40 m

-4uC (8 )+4uc

6.1. State Coulomb's Law in words. (2)
6.2 Calculate the electrostatic force that sphere A exerts on sphere B. (4)
6.3 Draw the electric field pattern due to the two spheres. (3)

sphere, sphere C, carrying a charge of
ining the centres of the spheres, as

6.4. Define electric field at a point. (2)
6.5 _ If the net electric field at point P is zero, calculate the distance between point

P and the centre of sphere A. (6)

[17]

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QUESTION 7: (Start on a new page)

The battery inthe circuit diagram below has an internal resistance r. The ammeters
and connecting wires have negligible resistance.

©

Si

oly

Both switches S: and Sz are initially closed. The reading on voltmeter V2 is
18 V and the power dissipated by resistor R is 12 W.

7.1 Calculate the:

7.1.1 Resistance of resistor R (3)
7.1.2 Reading on the ammeter (5)
7.1.3 Potential difference across the 10 Q resistor (3)

7.2 When switch Si is opened the reading on voltmeter Vi changes to
41,9 V. Calculate the internal resistance of the battery. (4)

7.3 Switch S2 is now opened while S: is closed. Will the reading on the ammeter

INCREASE, DECREASE or REMAIN THE SAME?
Give a reason for the answer. (2)

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Physical Sciences P1 NSC Grade 11 November 2024

7.4 Alearner sets up the circuit shown below to investigate Ohm's Law for each
of two resistors X and Y. Ignore the internal resistance of the battery.

12V

Si
@) We

XorY

fw)
ay,
The following graphs were obtained from the results of the investigation.

12

10
= s
3)
8
=e6
$
4
2
01 02 03 04 05 06 07 08 0,9 rh
Current(A) Yr
7.4.1. State Ohm’s Law in words. (2)
7.4.2 Use the graph to determine the resistance of resistor Y. (4)

7.4.3 Resistors X and Y are connected in series to a battery in a circuit.
Which resistor, X or ¥ will be more suitable to boil the same mass of
water at a faster rate in an electric kettle?

Give a reason for the answer. (3)
[26]

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Physical Sciences P1

QUESTION 8: (Start on a new page)

In the diagram below the north pole of a bar magnet is being pushed into a coil.
The emf induced in the coil is recorded by a galvanometer.

8.1 When the bar magnet moves in and out o}
magnet exert a force on each other | S
ELECTROSTATIC or MAGNETIC? (1)

8.2 State the direction of the induced current through the galvanometer.
Choose A to B or B to A (2)

8.3. Write down the reading on the galvanometer when the bar magnet is at rest
inside the coil. (1)

8.4 Acoilis made up of 100 turns of insulated copper wire, each having a circular
area of 4,8 x 10% m2. The coil is placed in a uniform magnetic fi eld of
3,9 x 10 T in such a way that the angle between the magnetic field and the
normal to the circular plane of the coil is 30°. The coil is then rotated so that

the angle changes to 70° in 0,2 s.

8.4.1 State Faraday's law of electromagnetic induction in words. (2)
8.4.2 Calculate the magnitude of the emf induced in the coil. (4)
8.4.3 Calculate the current induced in the coil if it has an effective
resistance of 2 Q. (3)
[13]
TOTAL: 150

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DATA FOR PHYSICAL SCIENCES GRADE 11

PAPER 1 (PHYSICS)

GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 11

VRAESTEL 1 (FISIKA)

TABLE 1: PHYSICAL CONSTANTS / TABEL 1: FISIESE KONSTANTES

Radius van Aarde

NAME / NAAM SYMBOL / SIMBOOL VALUE / WAARDE
Acceleration due to gravity 9,8 m-s?
Swaartekragversnelling
Universal gravitational constant fey 6,67 x 10° N-m?-kg?
Universele gravitasiekonstante
Speed of light in a vacuum 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 k 9,0 x 109N-m2-C2
Coulomb se konstante
Charge on electron e -1,6 x 10°79C
Lading op electron
Electron mass Me 9,11 x 10°31 kg
Elektronmassa
Mass of Earth M 5,98 « 104 kg
Massa van Aarde
Radius of Earth Re 6,38 x 10®m

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TABLE 2: FORMULAE / TABEL 2: FORMULES

MOTION / BEWEGING

VE =v +aAt

Ax =v, At + tadt? or/of Ay = v,At + taat?

v2 =Vv\2-+2aAxorlofv,” =v,’ + 2aAy

Ax = (4 Jae orlof Ay = (“sts Jas

FORCE / KRAG
Fret =a
fs(™* = sN fk = UkN
w=mg
F= Gm, M2 x Gm
2 ee

ELECTROSTATICS / ELEKTROSTATIKA

kQ4Q2 kQ

F=—t2 Ex

r2 r@

fet e-F

q q
n=2 ORIOF n=2
e Ge

ELECTRIC CIRCUITS / ELEKTRIESE STROOMBANE

R Vv emf (€) =1(R+r)
“aT

R, =R, +R, +

a q-tet

R, R, R,

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W=Vq
pW
W=VIAt At
‘ P=VI
W=FRAt P=?R
_ V2At ”
We R R
ELECTRODYNAMICS
A@
e=-N>-
@ = BACos 0

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NA) KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

GRADE 11

MARKS: 150

These marking guidelines consist of 10 pages.

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Physica] Sciences P1 2 . November 2024 Term 4 Test
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QUESTION 1
1.1. Avv
12 Cvv'
1.3. Dvv
14 BYY
15 ByYv
16 ByYy
17 Cv¥¥
18 Dvv
19 BYv
1.10 Dvv
QUESTION 2
2:4 Marking criteria

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

Resultant force is the vector sum of two or more forces acting togetherY
OR
That single force having the same effectv as two or more forces acting together. v

2.2.1

(2)
(2)
(2)
(2)
(2)
(2)
(2)
(2)
(2)
(2)
[20]

(2)

Vector Resultant

x-component

5,83 N v (to the right)

y-component

3,17 N v¥ (Upwards)

NOTE: If the 0 N is implied for the x—cot

y—component of F2, award the

marks indicated by the rin

R2= Rx? + Ry a

R? = 5,832 + 3,172¥
R =6,64NV

2.2.2 POSITIVE MARKING FROM Q 2.2.1

Ry
NOTE: Award the formula mark H
to the substitution if formula is R
not given. 1
1
8
R Rx
=

tan 8= ad

237

tan 0= SB

6 = 28,53° North of East ¥ OR 6 = 61,47° East of North / Bearing.

(8)

(3)
[13]

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Physica] Sciences P1 November 2024 Term 4 Test
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QUESTION 3
3.1 Inertia is the resistance of an object to any changes in its state of motion. vv (2OR0). (2)
3.2 e Passenger and taxi move with the same velocity before brakes are applied. v
e When brakes are applied, the net force acts on the taxiv (and not the passenger).
«Due to inertia the passenger continues to move forward.” (3)

3.3. The occupant moving forward exerts a force on the seat belt. By Newton’s third law, the
seat belt exerts an equal force in the opposite direction” to prevent the occupant from
moving forward.” (2)

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

Static frictional force is the force that opposes the tendency of motion of a stationary
object relative to a surface. Vv (2)
3.4.2

W | Fg/Fw/Fearth on P/weight /mg /784 N /gravitational force

f | Fr/fs/(static) friction
N | Fn/Normal /Fnormal

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 2/5
e If everything correct, but no arrows: Max 2/5
e If force(s) do not make contact with the dot: Max 2/5 (3)

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Q3.4.7
Correct answer is :
NO.✔Both the parallel component of the gravitational force and the maximum static frictional force
increase✔ by the same magnitude.✔

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Physical Sciences P1

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QUESTION 4

November 2024 Term 4 Test

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

41

4.2

When a resultant/net for

the force at ania

object will accelerate in the direction of
fo the force” and inversely proportional

(2)

Accepted labels
W | Fg/Fw/Fearth on P/weight /mg /88,2 N /gravitational force
T | Fr/Fstring /Ft/tension.
F | Fapplied/ Fapp / 9 N
N | Fn/Normal /Frormat
Notes:
Mark awarded for label and arrow.
e Do not penalise for length of arrows since drawing is not to scale.
e Any other additional force(s): Max 3 A
e If everything correct, but no arrows: Max x
e If force(s) do not make contact with the dot: Max %

(4)

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Physical Sciences P1 November 2024 Term 4 Test
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4.3.1. MARKING CRITERIA
e Fhet = ma (for either block)¥
e Substitution of T and mass for 1 kg blockY
e Substitution on the left-hand side for the 3 kg block.”
e Substitute 3 kg on the right-hand side for the 3 kg block.
e Answer. ¥

For the 1 kg block
Fret = ma ¥
T = 1av -----------------------------------eqn. 1

For the 3 kg block
Fret = ma
Fx-T=ma
(9 cos25° - T)v= BAW aaccnccsesicceasances eqn. 2
Therefore a = 2,04 m:s?v

4.3.2 POSITIVE MARKING FROM Q 4.3

OPTION 1: OPTION 2:

T=1a Fapp- T=ma

T = 1(2,04) (9 cos25° - T) = 3 (2,04)v
T = 2,04 NVY T = 2,04 NY

44  IncreaseY
The increased mass of the smaller block increases the load, “ requiring a bigger force
to pull it.v OR
T = Fcos25 — 3a. Y Acceleration decreases, VY hence T increases.

QUESTION 5

Marking criteria

niverse attracts every other particle with a gravitational force that is
| to the product of their masses and inversely proportional to the

f matter contained in the body. Weight is the force with which the
y-v

NOTE: If candidate states: Mass is a scalar quantity, while weight it is a vector quantity,
award 1 mark only.

(2)

(3)
[16]

(2)

(2)

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Physical Sciences P1 November 2024 Term 4 Test

5.3.1

5.3.2

5.4.1

5.4.2

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g> Slt
_ 6a x10" x 5,98 x10%v
(6,38%10°+ 8850)”
= 9,77 m:s?v downwards / towards the Earth’s centre
OPTION 1 OPTION 2
POSITIVE MARKING FROM Q 5.3.1 Fg= = Smim
amg sereio" ga UL
Se ee Fo= (6,38%10° 8850)" "
= Ze |, 38x +
acini w = 684,04 Nv
The sensation experienced when all contact forces are removed / in free fall. vv

(2 OR 0).

No.¥ The gravitational force exerted by Earth still exists in orbit. ~ Astronauts experience
weightlessness because the only force acting on them is the gravitational force. ¥

QUESTION 6
6.1 Marking criteria

6.2

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If 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 (Q1) on another
point charge (Qz2) is directly proportional to the product of the magnitudes of the charges
and inversely proportional to the square of the distance (r) between them. Vv

Fe=
910° x4 x10%x 4 x10%)  y
(0,4)?
= 0,9 NY towards A/ to the leftY

k Q4Q)
2 v

NOTE: Direction must be marked independent of answer.

(4)

(3)

(2)

(3)
[16]

(2)

(4)

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Physica] Sciences P1 November 2024 Term 4 Test
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6.3 . 3)

Marking Criteria:

e Correct direction of field — from positive to negative charge. v

e Field lines not touching each other and starting from / touching the
surfaces of the spheres. W

e Correct shape. ¥

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

The electric field at a point is the electrostatic force experienced per unit positive charge

placed at that point. vv (2)
6.5 | Marking Criteria: Ea + Ec =0 Any one v
Ea = Ec

e Correct vector equation equating Ea and Es.v kKQa , _ _kQp

¢ Correct formula to calculate electric field. v 2” 4-12

° Substitution to calculate Ea. ¥ 4x 10° 9x10 8v

¢ Substitution to calculate Es. vv 2 ys oa ney

e Final answer (r = 0,16 m). “ (0,4 - 1)

r=0,16 mv
(6)

[17]

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Physical Sciences P1

November 2024 Term 4 Test

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QUESTION 7
711 5_¥
p= tv
_ 18?
12=2v
R= 2704
7.1.2 | OPTION 1: OPTION 2:
POSITIVE MARKING FROM Q7.1.1: lIr= vy
1 if 1 =
—=—+—v¥ OR =Ri*Roy P =Viv OR
Rp Ry Re Ro Re Rp 12 = 18 IRV = By
| 1 1 27
Bae = 2M2, Ir =0,67A
Reesgiq0 l= Bv
V2 18
lp = Rp” lyo= aa”
18. ly2=1,5A
331 ing= +15 =2,17AV
= 2,17 AV -. ammeter reading is 2,17 A Arinety teading “0.87 + 15° 2.178

7.1.3 POSITIVE MARKING FROM Q7.1.2:
V=IRVY
V =2,17 x 10¥
V=21,7 V¥
7.2 | OPTION 1: OPTION 2:
POSITIVE MARKING FROM Q7.1.2 & Q7.1.3 POSITIVE MARKING FROM Q7.1.2
~ Viost ="44,9 = (21,7 +18) V E=\(R+nv
haul 220, 41,9Y = 2,17((8,31+10) +r) ¥
Most = Irv aa r= 1,00 QV
2,20 = 2,17rv OR €= Vext +Ir ¥
Wy =40inv 41,9Y = (21,7418) + 2,171
7.3. Decrease.v Rrot increases. V
7.4.1 | Marking criteria
If 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
va a NOTES:
“-& Gradient = rid Any ordered pair, correctly read from the graph
= 120) "= 43339 can be used. ; a
(0,9-0) W y The fourth mark is for clearly indicating that the
Therefore, the resistance of Y is 13,33 Qv |_aradient is the resistance.
7.4.3 Y. ¥ The higher the resistance the higher the power output / P « Rv (I is constant) thus

more energy dissipated in the resistor per unit time / per second / in a given time. ¥

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(3)

(4)

(2)

(4)

(3)

[26]

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Physical Sciences P1 November 2024 Term 4 Test

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QUESTION 8

8.1 Magnetic force. v (1)
8.2 AtoBYY (2)
8.3 Zero. (1)
8.4.1 | Marking criteria

8.4.2

8.4.3

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

The magnitude of the induced emf across the ends of a conductor is directly proportional

to the rate of change in the magnetic flux linked with the conductor. vv (2)
A
7 ne
Ve -D 39°)
g= sa a ny one ¥
(BACos 70 - BACos30)
= -NQ
At
(3,9x10“x4,8 x 10% Cos 70¥ - 3,9x10% x 4,8x10% Cos30v)
€ =-100
0,2
& =4,90 x 105 vv (4)
POSITIVE MARKING FROM Q8.4.2
e€=IR/V=IRY
4,90 x 105 = (2) v
| =2,45x 10° Av (3)

[13]
TOTAL: 150

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