# Physical Sciences Grade 12 Paper 1 KZN March 2018 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the KZN March 2018, 16 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/phy-march-p1-qp-memo-2018-gr12/cd865a81-c29e-486e-90c1-96bdb6f08f03
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: KZN March
- Paper: 1
- Curriculum: CAPS
- Pages: 16

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## About this document

Physical Sciences · Grade 12 · KZN March · 2018. Memorandum, 16 pages. Read online or download the PDF.

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## Document text

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Education

KwaZulu-Natal Department of Education
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE
GRADE 12

" PHYSICAL SCIENCES: PHYSICS (P1) *
: COMMON TEST ,
4 MARCH 2018 :
MARKS: 50
TIME : 1 hour

This question paper consists of 6 pages and a 1-page data sheet.

Copyright Reserved Please Turn Over

———— <<

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Physical Sciences P1

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March 2018 Common Test
NSC-GRADE 12

INSTRUCTIONS AND INFORMATION TO CANDIDATES

10.

11.

12.

Write your name on the ANSWER BOOK.

This question paper consists of FOUR 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 subsections, 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 SHEET.
Show ALL formulae and substitutions in ALL calculations.

Round off your final numerical answers to a minimum of TWO decimal places.

Give brief motivations, discussions, et cetera where required.

Write neatly and legibly.

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3

March 2018 Common Test
NSC-GRADE 12

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 — 1.4)
in the ANSWER BOOK, for example 1.5 D.

1.1

M2

1.3

A heavy crate is at rest on a rough horizontal surface. Three different forces, each
applied to the crate, attempt to move the crate as shown in the diagram.

F2
Fa a a
CRATE }#—— > CRATE CRATE

I Il Il
In each case, the crate is just at the point of moving.

The maximum force of static friction between the crate and surface is ...

A Greatest in situation (1)

B Greatest in situation (II)

Cc Greatest in situation (III)

D Equal in all three situations. (2)

A satellite orbits Earth at a point x, where the gravitational force is a
quarter (1) of the gravitational force it experiences on the surface of the Earth.

x

©

Satellite

Earth

If the radius of the Earth is R, then the height of the satellite ABOVE THE
SURFACE of Earth will be ...

A 4R

B 2R

Cc R

D eR (2)
The net force acting on an object is a measure of the of the object.

A Change in kinetic energy

B Rate of change of kinetic energy

Cc Rate of change of momentum

D Change in momentum (2)

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NSC-GRADE 12

Physical Sciences P1

1.4 Two balls P and Q are thrown into the air simultaneously from the same height
above the ground. P is thrown vertically upwards and Q vertically downwards

Mo 8 Common Test

with the same initial speed. Neglect air resistance. Take downwards as the positive

direction.

Which of the following is TRUE for P and Q when they hit the ground?

VELOCITY ACCELERATION

A Both have the same velocity Both have the same acceleration

B P has a greater velocity than Q | P has a negative acceleration; Q has
a positive acceleration

Cc P has a greater velocity than Q | Both have the same acceleration

D Both have the same velocity P has a negative acceleration; Q has
a positive acceleration.

QUESTION 2

A train carriage A of mass 3000 kg is moving horizontally to the right at 6 m.s~". It
collides with another train carriage B of mass 4200 kg that is initially at rest, as
shown in the diagram below.

6m.s*
—>

A

The graph below shows the variation with time t¢ of the velocities of the two train
carriages before, during and after the collision.

6

\ train|cafriggd BI

° \

v(m.s*)

aj t (Ss)

“4
o

train Fayriage ry

2.1 Define momentum.

22 Determine, by means of appropriate calculations, whether the total momentum
of the system is conserved or not after the carriages collided.

2.3. Calculate the average net force exerted by train carriage A on B.

(2)
[8]

(2)

(6)

(5)
[13]

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NSC-GRADE 12

QUESTION 3

March 2018 Common Test

A hot-air balloon is rising vertically upwards with a constant speed of 2.5 m.s“'(Figure 1). When
the balloon is at a certain height (h) above the ground, an object is accidentally dropped from
the balloon. The object reaches the ground 4 s later.

Figure 2 shows the velocity time sketch graph (where down is positive) for the vertical motion
of the object from the moment it is dropped until it hit the ground.

Neglect air resistance. (g = 9,8 m.s*)

3.1

3.2

3.3
3.4

3.5

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fas ms v(m.s")
R ay RRS TS AES St Bh ee sere eee 1
ay”
Py
ground
Figure 1 Figure 2

Define a projectile.
What is the magnitude and direction of the velocity of the object at point P.

What physical quantity does the gradient of the graph correspond to?

What is the significance of point Q on the graph with respect to the motion of
the object?

USING THE GRAPH ONLY (No equations of Motion), calculate:
3.5.1 The velocity of the object when it hits the ground.

3.5.2 The height, h, from which the object was dropped.

MA

(4)
[16]

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Physical Sciences P1 March 2018 Common Test

NSC-GRADE 12
QUESTION 4

Block A, that is at rest on a horizontal rough surface, is used as an anchor to hold block B, of
mass of 60 kg, in the air at a certain height above the ground. The two blocks are connected
with rope S that makes an angle of 25° with the vertical. Block B is suspended from the ceiling
by cable R that makes an angle of 8 with the ceiling.

60kgl =B

Block A experiences a frictional force of magnitude 150 N. The entire system is
in equilibrium . Neglect air resistance.

44 Define frictional force. (2)

4.2 What is the magnitude of the resultant force acting on block B? (1)
43 Drawa labelled free-body diagram indicating all the forces acting on

block B. (3)

4.4 Calculate the vertical component of the force in cable R. (5)

4.5 Calculate the angle, B, between cable R and the ceiling. (2)

[13]

TOTAL: [50]

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Physical Sciences P1 March 2018 Common Test

NSC-GRADE 12
DATA FOR PHYSICAL SCIENCES (PHYSICS) GRADE 12

GEGEWENS VIR FISIESE WETENSKAPPE (FISIKA) GRAAD 12

TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE
KONSTANTES

NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE
Acceleration due to gravity e2
Swaartekragversnelling g opts
Universal gravitational constant le 6,67 x 1071 N.m2.kg2

TABLE 2: FORMULAE/TABEL 2: FORMULES

MOTION/BEWEGING

vp =Vitaat Ax = v,At+ 4aAt® orf/of Ay =v,At+ $adt?
5G va +V Vv, +V;
vp? =vi2 + 2adx orfof v,? =v, +2ady_ | AX = (aes) At or/ofAy = (SS Ja
12
K=E, =. mv
2
FORCE/KRAG
Fret = Ma p=my
Fie Mt =Ap=mv , my; Fy =mg
Fe Sm,m,
r
£0 = y,N f= LN

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UIT

EASL4A4TRIITOQVVOO00L1

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education

) Department:
Education

PROVINCE OF KWAZULU-NATAL

: PHYSICAL SCIENCES P14 ‘ |
MEMORANDUM

: COMMON TEST |
: MARCH 2018 ; |

NATIONAL
SENIOR CERTIFICATE

N.B. This memorandum consists of 4 pages including this page.

Copyright Reserved Please Turn Over
bh

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Physical Sciences 2 March 2018 Common Test
NSC — Memorandum (GRD 12)

PHYSICAL SCIENCES PAPER ONE

QUESTION 1
11. Cv (2)
12  Cvv (2)
13. Cvv (2)
14 Avv (2)
[8]
QUESTION 2
24 The product of mass and velocity of the object. vv (2)
2.2 total Poefore = Mavat Mave — —
= (3000)(6) + (4200)(0) v
= 18000 kg.m.s" ,to the right ~ anyone vy
total pafter = Ma vat Meve
=(3000)(-1) + (4200)(5) ~
= 18000 kg.m.s" ,to the right”
Total Pvetore = total pater
Momentum is conserved v (6)
[ m(v,- v,) / __m(v,- vj)
2.3 Fret = aA “ OR Fret ~ At
_ 4200v(6 -0) ¥ _ 3000¥(-1 - 6)”
~ av - 24
= 10500N tothe right ¥ = -10500 N (to the right)
Fg = 10500 N to the righty (5)
- ~ [13]
QUESTION 3
3.1 An object upon which the only force acting is the force of gravity. v ¥ (2)
3.2. 2.5m.s! Vupwards ¥ (2)
3.3 (Gravitational) acceleration “ (1)
3.4. Maximum height reached by the object. v ” (2)

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Physical Sciences =] March 2018 Common Test
NSC — Memorandum (GRD 12)

3.5 If equations of motion are used for the 3.5.1 and 3.5.2 subtract 2 marks for each

question. de V4 wef bab” va
3.5.1 gradient = —— _ + )(« )
At 2-2 the) y
98 = oe = LET toa NK
v Vv, +25 . (> |
98 =—R8 numerator and denominator ‘
‘ 4-0 v
Vz = 36,70 m.s'v (downwards) ¥
Note: If point R and Q are used the answer is 36,75 m.s! (5)
Positive marking from 3.5.1 Ue
3.5.2 Height = Area ofA, +AreaofA, ¥ BN " Ye wat
= (A) x (0,25) x (2.5) + (4) x (4 - 0,25) x (36,7) ¥¥ Af wot sale Ke
= 68,69 m% a > ste) + 4/9 (6 ]

iv

OR GPs botiy a (»/

Height = Area of trapezium of the v — t graph. ¥

= %4(2,5 + 36,7)(3,5) vv
= 68,60mY” (4)

[16]

QUESTION 4

4.1 The force that opposes the motion of an object Vand acts parallel with the

surface. v (2)
42 ONY (Accept 0 / Zero) (1)
4.3
F,
R Fs
w
[ Notes : Accepted Labels Mark |
w weight /Fe/ Fy 4
FR Tension force in cable / Tz iv
Fs Tension force in rope / Ts v
| Any additional force: deduct 1 mark (maximum %)
Lines must touch object otherwise (maximum % )
Subtract one mark if arrows are not shown (maximum % )

(3)

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Physical Sciences 4
NSC — Memorandum (GRD 12)

com

March 2018 Common Test

44

2Fx=0 tan 25°=

Fs.sin 25° - 150 = 0 W

Fs= 354.93 N Anyone FE =
OR SV _

ik-o ———= Then, 2Fy =

Fry + Fsy=mg Fey =

Fry + (354.93).cos 25° Y= (60)(9.8) ” =

Fry = 266.32 N ¥ a

Positive marking from 4.4
45

266,32
tan B= 59
B = 60,612 v

OR

W = 588N
S = 354,93 N
R=?

R? = W? + S?— 2WSCos 25°
R? = 588? + 354,932 — 2(588)(354,93)Cos 25° ¥
R = 305,66

305,66Sin (90° - B) = 354,93Sin25°
B= 60,61°V

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150 v
sw
= 321,68
Ov
60 x 9,8 Y- 321,68”
266,32N v
nas (.))
(2)
[13]

TOTAL MARKS: 50

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