# Physical Sciences Grade 11 P1 September 2018 Paper & Memo

> Grade 11 Physical Sciences Paper 1 question paper and memo from the September Exam 2018, 30 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/phy-sept-p1-qp-memo/99058038-c932-4000-ab2a-3f08d8c6b7d1
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2018
- Exam period: September Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 30

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## About this document

Physical Sciences · Grade 11 · September Exam · 2018. Question paper and memorandum, 30 pages. Read online or download the PDF.

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Physical Sciences /P1 September 2018 Common Test
Kes

INSTRUCTIONS AND INFORMATION TO CANDIDATES

1. Write your name on the ANSWER BOOK.

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

3. Questions 5.2.1 and 9.3 must be answered in the special answer sheet provided. Write
your name on this sheet and submit it together with your answer book.

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

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

6. Leave ONE line between two subsections, for example between
QUESTION 2.1 and QUESTION 2.2.

7. You may use a non-programmable calculator.
8. You may use appropriate mathematical instruments.

9. You are advised to use the attached DATA SHEET.
10. Show ALL formulae and substitutions in ALL calculations.

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

12. Give brief motivations, discussions, et cetera where required.

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3

Physical Sciences /P1 September 2018 Common Test
NSC

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, for example 1.11 D.

1.4 Which of the following pairs are both vector quantities?
A Displacement and energy
B Speed and force
Cc work and mass
D

Weight and acceleration (2)

12 Two forces A and B each of magnitude 15 N act on the same point O. If the resultant
of these forces is 15 N, what is the magnitude of the angle 0 between them?

A
—v
fe)
8
/
La
A 60°
B 120° B
Cc 90°
D 30°

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1

1.4

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Physical Sciences /P1 September 2018 Common Test

NSC

3. Three forces with magnitudes F;,F2 and Fs; act on the same point P in the same plane.

The system is in equilibrium.

v
The resultant of the Y components of the forces force is given by :

A

B
Cc
D

Fysin 0 + F2+Fs
F3-Fycos 0
F, + Fycos0
Fisin @- Fo

In which of the following cases is the resultant force acting on an object equal
to zero?

A An object is moving with constant acceleration.

B When the only force acting on the object is the gravitational force.

Cc An object undergoes equal displacements every second.

D An object is moving with increasing acceleration.

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Physical Sciences /P1 5 September 2018 Common Test
NSC
1.5 Two blocks, A and B, with masses m and 2m respectively, are in contact with

1.6

each other. The blocks are accelerated by a horizontal force F along a
frictionless, horizontal surface.

B

What is the magnitude of the force that B exerts on A?

A 2F
BOF

C 2/3

DF (2)

Which of the examples stated below best describes Weightlessness?

A An object accelerates downwards towards the Earth at an acceleration
of 9,8 mss”,

B An object moving at a constant velocity upwards away from the ground.
c An object moving at constant velocity downwards towards the ground.

D An object that is accelerated upwards away from the ground by a constant (2)
applied force.

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1.7

1.8

1.9

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Physical Sciences /P1 6 September 2018 Common Test
NSC
The diagram below represents a ray of monochromatic light passing from

air into quartz.

Normal
\

air

\
\
\
1
\
1
\
quartz '
\
'
i
1
\
\
\
\
\

Pp

Which path will the ray follow in the quartz?

iwRek
OnDOTV

The field generated around a current carrying conductor is

a magnetic field

an electric field

parallel to the conductor

greater further away from the conductor. (2)

0DOD>Y

In the diagram below, a unit positive charge q is located between two charged spheres,
A and B. Sphere A has a charge of +2q and is located x meter from charge q. Sphere B
a charge of —2q and is located ’% x meter from charge q.

+2q +q -2q

---( A} @- B }--
x xX 4

If the magnitude of the force on charge q due to sphere A is F,
what is the magnitude of the force on charge q due to sphere B?

A Ya F
B 2F
Cc AF
D 4F

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Physical Sciences /P1

1.10

7
NSC

Consider the circuit diagram below:

O-

AC)

When switch S is closed, the readings on V and V2 (respectively) will

change as follows:

2R

Vi V2
A_ | decreases increases
B_ | increases decreases
C_ | decreases decreases
D | increases remains the same

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September 2018 Common Test

(2)

[20]

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2.1

2.2

2.3

2.4

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Physical Sciences /P1 8 September 2018 Common Test
NSC

QUESTION 2

In the system below the two masses labelled m and 10 kg are connected to
each other by two light strings A and B at point O. Point O is kept stationary

by attaching it to a wall using string C.

The coefficient of static friction between the 10 kg mass and its surface is 0,51.

as]

140°

O B 10 kg

SS

Define static friction. (2)

Calculate the magnitude of the frictional force between the 10 kg mass and
its surface. (3)

Determine the mass of m. (5)

Will the force in string A increase, decrease or stay the same if string B
is detached from the 10 kg mass? (1)

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Physical Sciences /P1 9 September 2018 Common Test

NSC

QUESTION 3

3.1

3.1.3

A force of magnitude 40N is applied to a 3 kg block at an angle of 60° to
the horizontal along a rough surface (surface1) as shown below. A second
2kg block is placed on the rough upper surface (surface 2) of the first
block. The two blocks are joined by a string of negligible mass running
through a frictionless pulley. The kinetic frictional force between the 3 kg
block and surface 1 is 2 N whilst the kinetic frictional force between the
2kg block and surface 2 is 1,2 N. Let T be the tension in the strings
connecting the two blocks as the 3kg block accelerates to the

right.

Surface 2 F=40N

Surface 1

Define Newtons Second law of motion. (2)

Draw a labelled free body diagram showing all the forces acting on the

3 kg block. (5)
Calculate the Tension(T) in the string (7)
[14]

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Physical Sciences /P1 10 September 2018 Common Test
NSC
QUESTION 4
4.1 State Newton’s Law of Universal Gravitation in words. (2)

4.2 The velocity - time graph shown below describes the motion of an object
of mass m thrown upwards and then falling downwards on Planet X for 5s.
Assume that the only force acting on the object is the gravitational force of planet X.

Velocity(ms”) vs time(s)

am T t a iia T 7
Teh ay ter | i i T Eee Cee CT COP rr

i f | rt COE
ala on it L I im i

Tit

The radius of planet X is 6,22x10° m. Using the graph above and relevant formulae, (5)
calculate the mass of planet X.

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Physical Sciences /P1 an
NSC

September 2018 Common Test

4.3. Asatellite is orbiting planet X. When the satellite is on the line joining the

centres of planet X and planet Y, the net gravitational force acting on the
satellite due to planet X and planet Y is zero.

The distance between the centres of planet X and planet Y is 4x10’m

Planet Y

If the mass of planet Y is twice the mass of planet X, calculate the distance of the :
satellite from planet X. (4)

[11]

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Physical Sciences /P1 12 September 2018 Common Test
NSC

QUESTION 5

5.1 Define refraction of light. (2)

5.2. A ray of light is incident upon the surface of a semi-circular glass block as

5.2.1

5.2.2

5.3

5.3.1
5.3.2

5.3.3

shown in the sketch below.

Air Glass Air

Using the diagram sheet provided, complete the diagram

by showing the path taken by the light ray as it passes from air to glass and then to

air. Label the angle of incidence, the refracted ray, the emergent ray and draw the

normal at each surface. (4)

If the refractive index of glass is 1,50, calculate the speed of light through glass
given that the speed of light in air is 3x10° ms“
Take the refractive index of air to be 1,00. (3)

The drawing shows a ray of light travelling through a gas (n = 1.00), then through a
solid (n = 1.55), and then into a liquid (n = 1.45).
The surfaces are parallel and the angle of refraction at the first surface is 25,42°.

IQ
Gas Solid Liquid
State Snell’s law. (2)
Calculate ty ,the angle of refraction at the second surface. (4)
A
If the size of i1 is increased, will the size of angle ty increase, decrease or remain
the same? (2)
[17]

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Physical Sciences /P1 13 September 2018 Common Test
NSC

QUESTION 6

6.1 The sketch below shows a phenomenon that takes place when parallel wave fronts
approach a narrow opening in a barrier.

Direction of wave motion

! J |

6.1.1 State and define the above phenomenon. (3)

6.1.2 Sketch the wave pattern that occurs when the wave fronts pass through
the opening in a barrier. (2)

6.1.3 The wave pattern that forms is explained by Huygens principle.
State this principle. (2)

6.2 When monochromatic blue light is passed through a narrow single slit, a pattern is
observed on a screen.
Describe the diffraction pattern which is observed on the screen. (3)

[10]

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Physical Sciences /P1 14 September 2018 Common Test

NSC

QUESTION 7

fal

7.2

7.3

74

75

7.6

7.7

A small, conducting sphere, X, with a charge of +5 nC is suspended by an insulated
inelastic thread of negligible mass, which is tied to point P. Another small
conducting sphere, Y, on an insulated stand, with a charge of —3 nC, is brought
closer to X until their centres are 15 mm apart on the

same horizontal line.

Draw the resultant electric field pattern produced by spheres X and Y. (3)
State Coulomb's law in words. (2)

Calculate the magnitude of the electrostatic force that sphere Y exerts on

sphere X. (4)
Define Electric field at a point. (2)
Calculate the magnitude of the electric field 15mm away from sphere X (4)

Sphere Y is now moved closer and makes contact with sphere X
and moved back to its original position, that is 15mm from X.

Will electrons be transferred from spheres X to Y or from spheres Y to X? Give a

reason for your answer.
* (2)
Calculate the number of electrons transferred upon contact. (4)
[21]

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Physical Sciences /P1 15 September 2018 Common Test
NSC
QUESTION 8
8.1 The sketch below shows a current carrying solenoid that is wrapped around a soft
iron core.

Does A represent the North pole or the South pole of the magnetic iron core?

8.1.1 Is the magnetic field direction outside the iron core from A to B or from

Bto A?
_. ; _ (1)
8.1.2 A coil is moved into or out of a magnetic field of a bar magnet as shown below so
that the rate of change of magnetic field, AB = 0,6 Tes~'

8.2
N Ss
N iS)
The area through which the field links with the coil is given by A = 0,4 m?.
8.2.1 State Faraday’s law of electromagnetic induction in words. (2)
8.2.2 Calculate the magnitude of the average emf generated in the coil, if the coil has 8
turns. (4)
8.2.3 If the experiment is repeated with a new coil that has twice as many turns, what
would the new emf be? (2)
[10}

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Physical Sciences /P1 16
NSC

QUESTION 9

September 2018 Common Test

Learners conduct an investigation to verify Ohm’s law. They measure the current
through a resistor for different potential differences across its ends, i.e. the potential
difference was chosen as the independent variable.

The results obtained are tabulated below:

Potential Current
Difference (V) | Strength (A)
21,4 0.80
35,8 1,20
56,0 1,90
72,04 2,30
98,04 3,20
124,04 4,13

a1 State Ohm’s law in words.

g.2 The dependant variable in this investigation was the current.

9.2.1 Name one quantity that was controlled in this investigation.

9.2.2 State how the quantity in Q 9.2.1 was kept constant

9.3. Using the information in the table and the graph sheet provided, draw a graph of (5)

Current versus Potential difference.

9.4 Calculate the gradient of the graph.

9.5 What quantity does the gradient of the graph represent?

9.6 Hence determine the resistance of the resistor.

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Physical Sciences /P1 17 September 2018 Common Test

NSC
QUESTION 10

10.1 You are given four identical resistors, each of magnitude ‘R’ and a
battery of emf ‘V’ volts.

10.1.1. Sketch an appropriate circuit diagram, using all the above components,
to show how you would connect them to obtain a total resistance of R in the
circuit. (2)

10.1.2 Use relevant calculations to show that the total resistance in your circuit in
Q10.1.1 is equal to R. (2)

10.2 The circuit diagram below shows a battery of unknown EMF with negligible internal
resistance, connected as shown below.

The reading on the voltmeter is 4 V.

Calculate:
10.2.1 the effective resistance in the circuit. (3)
10.2.2 the current flowing through the 1 resistor. (3)
10.2.3 the reading on the ammeter. (4)
10.2.4 the EMF of the battery (3)

10.3 The switch S is now opened.

10.3.1 How will the energy dissipated by the 4 Q resistor change if the time it is in
operation remains constant? State whether it will increase, decrease or remain

the same. (1)
10.3.2 Briefly explain your answer using relevant formulae . (2)
[20]
TOTAL : 150

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Physical Sciences /P1 18 September 2018 Common Test
NSC

DATA FOR PHYSICAL SCIENCES GRADE 11
PAPER 1 (PHYSICS)

TABLE 1: PHYSICAL CONSTANTS

NAME SYMBOL VALUE
Acceleration due to gravity

9,8 ms?

Gravitational constant 6,67 x 10°11 N-m2-kg2

Coulomb's constant 9,0 x 109N-m2-C2

Speed of light in a vacuum

aO;xX,o!] a

3,0 x 10° m-s*

Charge on electron -1,6x107C

oO

Electron mass 9,11 x 1031 kg

Radius of earth Re 6,38 x 108m

Mass of earth

Me 5,98 x 1074 kg

TABLE 2: FORMULAE
MOTION

Vv, =v, +adt Ax = v,At+ daat?

Ve ty.
Vy =v, + 2aAx Ax = (“ga

FORCE

FE

ny

e=ma w=mg

Gm,m, f,
r2 (max)

f, = 14.N g=

WAVES, SOUND AND LIGHT

va=far T=

F= =n,N

al g
=

n, sind; =n, sin@, n==

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Physical Sciences /P1

19

September 2018 Common Test

NSC
ELECTROSTATICS
kQ,Q F
FS 2 s (k = 9,0 x 10° N-m?-C*) wart
E= LS (k = 9,0 x 10°N-m?-C-?) vet
ELECTROMAGNETISM
e--no® ®=BAcos0
At
CURRENT ELECTRICITY
j= Rar
At |
424,444, R=y+ fh +h
Rh & mht Ia His *
W=Vq WwW
“At
We=VlAt
W=PRAt Bet
> P=PR
w= wat \y?
R =

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Peo

Air

September 2018 Common Test

NSC

Glass

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Air

Physical Sciences /P1
NAME OF LEARNER :

Question 5.2
Question 9.3

L33HS sdO-NVAL

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Physical Sciences 3 September — .8 Common Test

NSC
OR Fe / 49,98 N
LI
¥
WIFg
oO tan50°= W ev OR tan 40°= 49,98 v
{o) 49,98 WwW
” WW S9.564N W =99,564N
—_ |
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a} 59,564.45 m.9,87 59,564 = m-9,8v (5)
—_ m = 6,08 kgv
E 2.4 Stay the samew” (i)
) m
. QUESTION THREE
[tag
ze} 3.1.1 Ifa non-zero NET force acts on an object, then the object accelerates in
the direction of the NET force where the acceleration of the object is
<b) directly proportional to the NET forceY and inversely proportional to the
zo} mass of the object. ¥ (2)
{o) 3.1.2
_
c F=40 Nv
S ae
ro) T/Tension ‘i
fa) ©
NB : 1 mark for all 3 vertical forces present.
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Physical Sciences 4 September 2018 Common Test

3.1.3. Take the motion of the system to the right as being positive.

NSC

Consider the 3 kg block

Fx = F-cos®
= 40 cos60°
= 20NY

Fver = ma

Fx #(T) + (fcr) + (fla) =3.a [ % anyone
20-T-2-1,2=3a

18 8-T= 38% cceccsce(1)
Consider the 2 kg block
Fuer = ma

fk2 + (— T) = 2(-a)
1,2-T=-2a

TH 2atd 2%. vee(2)

Substitute equation (2) in equation (1)

16,8-(2a+12)=3a

a=312

T =aietoy vet av
= 7,44NV

QUESTION FOUR

44

7)
[14]

Every body in the universe attracts every other body with a gravitational
force that is directly proportional to the product of their masses” and

versely propo!
centres. ¥

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Physical Sciences 7 September .. d Common Test

6.2.2 n=

ov

v

1,50 =3x108 v
v

v= 2x10 mst v

5.3.1 The ratio of the sine of the angle of incidence in one medium to the sine of
the angle of refraction in the other medium is a constant. VV ( 2 or 0)

A
5.3.2 iz = 25,420

nisin8; = nesin@2v

1,55-sin 25,42 = 1,45 sin 62¥

A
Q2=2= 27,319 v

5.3.2 Increases vv

QUESTION SIX
6.1.1 Single slit diffraction. v

It is the ability of a wave to spread out in wave fronts as they
pass through a small aperture or around a sharp edgev “ ( 2 or 0 marks)

ha i as
Ne
Ee vv
If shape is asymmetrical, minus 1 mark.

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(3)

(3)

(2)

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Physical Sciences 8 September 2018 Common Test
NSC
6.1.3 According to Huygens’ principle each point on a wave front forms (2)
secondary wavelets” that spreads out in all directions” forming the above
pattern.

6.2 A broad central blue band” is observed flanked on either side by
alternating narrower” dark and blue bands. “ (3)

[10]

QUESTION SEVEN

Fat

Marking Rubric : Electric field pattern |

Criteria Mark allocation “|
Correct direction of field lines 4 mark

Correct shape of the field 1 mark

pattern

Field lines do not touch, evenly | 1 mark
spaced, and perpendicular to
surface and touches the

surface (3)

7.2 The magnitude of the electrostatic force exerted by two point charges on
each other is directly proportional to the product of the magnitudes of the
charges” and inversely proportional to the square of the distance between
them.” (2)

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(2) / abieyo sayeBau Jo sssoxa UE SEY Ar KOLA gry
(b) /eOLXOOT =
AON OL XO07% =
cv) / eke-OLXS1)
pasa PF 6-OLXS*s0LX6 =
€'LO wo Buryew aanisog Yo GL
(Z) (0402) +, quiod yeu) ye paced oBseyo (enpisod)
wun 19d paouatadxe 80J0j aNe]SO.NDa/9 94) SI JUIO e ye Play OUNeIa aL pL
(p) AN ¢-OL X 00'L =
Lae-OLXSL)
PZ 6-OVXG'6-OLXS-501X6 =
za

AOOVs4 CL

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### Page 28

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Physical Sciences 11 September ~ 18 Common Physical Sciences 12 Sepiember 2018 Common

Test Test
NSC NSC
9.3 9.4 Gradient(1/R) = AL”
AV
Current Strength versus Potential Difference = 45-14v
140 — 40 v
+ THe =
iiabsaaaineaatata atetaed iis fatstaie = 0,031 Ov
: + + +
H +
i $ ian u e Accept any suitable set of co-ordinates that is taken from the graph. (4)
oe HH e Acceptable range (0,020 — 0.040 )
:
is jaanae + 95 inverse of resistancev/ : (1)
Ea : i :
t 9.6
3 : : : Positive marking from Q 9.4
€ : : oa
£ ; i
2 - i 4 =0,031v
: rt # H R (2)
5 : : R = 32,26 OV
° ana : + = [16]
+ QUESTION TEN
imeeaaeeea Stas aaeee Hp
TT fg i 10.1.1
+ sr anauen v
4 rH mann
asa i sean; apuasesne
0 20 40 60 30 100 120 140
Potential Difference (V)
"Marking Rubric : Graph
Criteria Mark allocation v
v
° Suitable scale used on x and y 1 mark
PON a “| re DB JR
* Xand Y axes correctly labelled 1 mark |
with correct units
* Correct plotting of points 2 marks
Best fit curve drawn joining points | 1 mark (2)

Copyright reserved Please turn over Copyright reserved Please turn over

### Page 29

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4800 UINy asealg paasasas yBAdo 49A0 WIN? BS2BId poasasas yysuAdoy
(vy) AVES
cLtee
09] + Bey = WLOLy
VL=
v 9
be ,o=
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w 4.9. 4
AEE R=
A=]
loz] v1 Seem @
QE = Wy: wy A 9= Z14Z ) = 590,
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z NOLLdO
SASBIIDOP Av ZTTOLO WosSuppew eanisod ¢7'9)
/ seseeloep -\ Aqasoum * 2A» M
“Juejsuod SUIeWS! VY pue y sous (e) a Mg =
WA) = M Yo
aa
A (8L-OL/e R=
OL BL-O/p WOs) SOSESIDOP JOVSISS! Op SU] SSOIOe SOUSISHIp |enuajod ou) v=o Zz,
PUE SESESIOUI TINO OW UT SOULTSISSL (E10) Au)) ‘pauado si ¢ yo}ims Udy,
(e) /U9 =
yO p+Ze
Sq + dy = WLOLy
seseeidap 10\s1Se1 U 7 ay} Ag pajedissip ABiaue ay) eouey
AO0Z=
“SOSESIVEP AQ ¢
/ SOSBOINOP I*9l] AQayeyAr * 2] 0 M y 2 + &
“ue}suod sulewel }V pue y souls y+ by = ay
Wal =M T TO TF ber
7 Sesesivep WJDUSTS JUoLNO OU)
Puke SOSeSIOUT INI OU} Ul SOUETSISEI [E}O} oy) ‘pauado si s yo}IMS UBUM Z°E"OL y=
SUZ YA = WLOLY
(1) 7 $9see10eg 1°01
(2) yh=dy
(e) AABLE
soMe)+g = yO ANSE Ae ea =
Ms SOXES V b b
POA + 9A = oo (— raxien + ty sey thor
t v T
SC'OLD wos Bupjew eayisog 7701
L'LOLD wo. GBuppew aaneboy
OSN OSN
SOL om
UOWWOD BLOZ Jaquuaidag bE SEQUINS |BOISAYd

UOWWOD BL0Z Jequia\das eb sa0us!sg jeashud

### Page 30

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