# Physical Sciences Grade 10 P1 National November 2016 Memo

> Grade 10 Physical Sciences Paper 1 memorandum from the National November 2016, 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-10-physical-sciences-p1-memo/1cc386c8-299f-447c-8a31-0357a62b792f
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 10
- Document type: Memorandum
- Year: 2016
- Exam period: National November
- Paper: 1
- Curriculum: CAPS
- Pages: 14

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## About this document

Physical Sciences · Grade 10 · National November · 2016. Memorandum, 14 pages. Read online or download the PDF.

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## Document text

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“mie basic education

wg
VESraG Department:
] th Basic Education
SZ REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE/
NASIONALE

SENIOR SERTIFIKAAT

GRADE/GRAAD 10
SITE RTT TT Tt IT TTL ee
PHYSICAL SCIENCES: PHYSICS (P1) i
FISIESE WETENSKAPPE: FISIKA (V1) "
NOVEMBER 2016 A
MEMORANDUM :
eee 2 eee ee ee ne en ne ne nn ne nn nn ne ne nn ne ne ne en ee 2”

MARKS/PUNTE: 150

This memorandum consists of 14 pages.
Hierdie memorandum bestaan uit 14 bladsye.

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Physical Sciences/P1/Fisiese Wetenskappe/V1 2 DBE/November 2016
NSC/NSS —- Grade/Graad 10 — Memorandum

QUESTION 1/VRAAG 1

14 Byy (2)
1.2 Cv (2)
1.3 Dvv (2)
14 Dvv (2)
15 Dvv (2)
16 Avv (2)
17 BYyY (2)
18  CYY (2)
1.9 DvY (2)
110 BY (2)
[20]
QUESTION 2/VRAAG 2
2.4

Mark allocation/Puntetoekenning:

csv 1 x line AB: length, arrow, label

1 x lyn AB: lengte, rigting, benoem
v¥ 1xline BC: length, arrow, label
1 x lyn BC: lengte, rigting, benoem

(2)

2.2 2m VY west/to the leftv
2mv_ wes/na linksY (2)

OR
If the learner has done 2.2 on the vector diagram above, allocate 1 mark for the

resultant from A to Cv and 1 mark for the label of the magnitude of 2m.

As leerder 2.2 op die bostaande vektordiagram aangedui het, ken 1 punt vir die
resultant van A tot C toe, en 1 punt vir die benoeming van 2m

2.3 Total distance/Totale afstand
=5+7Vv
=12mv (2)
2016 -1i- {1

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he

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Physical Sciences/P1/Fisiese Wetenskappe/V1 3 DBE/November 2016

NSC/NSS — Grade/Graad 10 — Memorandum

2.4 For the total distance, the whole path length travelled is considered.
For change in position, only the original position and final position“ of the man
are considered. (2)
Vir die totale afstand word die totale padiengte afgelé in berekening gebring,
maar slegs die begin- en eindposisie word in berekening gebring vir
verandering in posisie.
2.5 Velocity is the rate of change of displacement. ¥¥
Snetheid is die tempo waarteen verplasing (verandering in posisie) verander. (2)
ACCEPT:
Change in displacement over change in time
AANVAAR:
Verandering in verplasing gedeel deur verandering in tyd
2.6 POSITIVE MARKING FROM 2.2
POSITIEWE NASIEN VANAF 2.2
Ax
=—Vv
vet
2v
~ 20v
=0,1m-s" west/to the left”
wes/na links (4)
[14]
QUESTION 3/VRAAG 3
3.1 Acceleration is the rate of change of velocity. “vv”
Versnelling is die tempo van snelheidsverandering.
ORIOF
Acceleration is the change in velocity per unit timev”.
Versnelling is die verandering in snetheid per tydseenheid. (2)
3.2 Nov
Nee (1)
3.3 NEGATIVE MARKING FROM 3.2
NEGATIEWE MERK VANAF 3.2
Velocity to the right, acceleration to the left ¥ 2016 -11- 11
Snetheid na regs, versnelling na links.
OR/OF
Taxi slowing down so acceleration is in opposite direction” to movement.
Die taxi beweeg stadiger, dus is versnelling in die teenoorgestelde rigting van
beweging. 1)

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Physical Sciences/P1/Fisiese Wetenskappe/V1 4 DBE/November 2016
NSC/NSS — Grade/Graad 10 —- Memorandum

3.4
OPTION 1/OPS/E 1 OPTION 2/OPSIE 2
D (ve + vi)
v =a or D=vxAtv Ax= > At ¥
D 25 +25
av = Tv =25x1v¥ = y¥x1Vv
D =25mv =25mv =25mv

OPTION 3/OPSIE 3

AX = vit + Zadt?
=25x1¥ +%xOx tv
=25mv

(4)

2016 -i1- 11

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Physical Sciences/P1/Fisiese Wetenskappe/V1 5

NSCI/NSS — Grade/Graad 10 - Memorandum

DBE/November 2016

3.5 POSITIVE MARKING FROM 3.4

POSITIEWE NASIEN VANAF 3.4

distance > 40 m
nie, want die afstand is > 40 m

OR / OF

0+25
= (025) oy

2
=25mv

.. taxi will not stop at the traffic light as

. die taxi sal nie V by verkeerslig stop

.. distance available for braking time
is 40 — 25 = 15 m, therefore the taxi is
short of 10 m to stop. ”

..the taxi will not stop in time. ”

..die beskikbare afstand vir briektyd is
40-25 = 15m, dus sal die taxi10m
te min hé om fe stop.

-.dus sal die taxi nie betyds stop nie

OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
ax =(““) av Vp =v; + aAtY
2 _ VE- Vi Only one mark
~ for either
= (0425) x 2v at equation
7 _ (0 - 25) v Slegs een punt
=25m a= 2 vir die enige een
van die
= “2 lykings.
.. total distance/totale afstand = -12,5 ms —
= 25+ 25V
=50mv ve =v? + 2aAx

0 = 257 +2 x -12,5 x Ax¥

“Ax = 25m

.. total distance/fotale afstand

=25+25
= 50 mv

..taxi will not stop at the traffic light as

distance > 40 mv

.. die taxi sal nie by verkeerslig stop

nie, want die afstand is > 40 m

=25m

. total distance/fofale afstand
= 25+ 25
= 50 mv

OPTION 3/OPSIE 3
= ty
a At Only one mark for either
= (0-25) y equation
= 125 m-s? Slegs een punt vir enige een van
Ax = vit+ Yadt? die vergelykings.
=25x2 +%x-12,5x2?¥

-. taxi will not stop at the traffic light, as distance > 40 mY
.. die taxi sal nie betyds stop nie, want die afstand is > 40 m

(5)

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Physical Sciences/P1/Fisiese Wetenskappe/V1 6 DBE/November 2016
NSC/NSS — Grade/Graad 10 - Memorandum

OPTION /OPSIE 4
_(t +Mi)

Ay AtY
15 = yx aty
At =1,2sv

Actual time to reach traffic light is 1,2s
which is shorter than the 2s. Therefore

the taxi will not stop in time. ~
Ejintlike tyd of verkeerslig te bereik is

1,2s wat korter is as 2s. Dus sal die
taxi nie betyds stop nie.

3.6 x (m)

50

25

| MARKING GUIDELINES/NASIENRIGL YNE
¥ Both axes correctly labelled with unit
Beide asse korrek benoem met eenheid
¥ Straight line (t=O s and t = 1s)
Reguitlyn (t =0 s ent = 7 s)
¥~Curve shape (t= 1s andt = 3)
Kurwe (t =1 s ent = 3) A
Y values for displacement and time indicated (A)
Waardes vir verplasing en tyd aangedui [17]

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Physical Sciences/P1/Fisiese Wetenskappe/V1 7 DBE/November 2016

NSC/NSS — Grade/Graad 10 - Memorandum

QUESTION 4/VRAAG 4

4.1

4.2

4.3.1

4.3.2

441

4.4.2

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5 m-s" north “ (accept range from 4,9 to 5,1)
5 m-s" noord (aanvaar vanaf 4,9 tot 5,1)

8,3m-s' vv (accept range from 8,2 to 8,4)
8,3 ms" (aanvaar vanaf 8,2 tot 8,4)

The velocity is uniformly increasing.
Velocity increases from 5 m-s" to 10 ms” in 150s.

Pasitive acceleration.
The girl is speeding up.
The girl is uniformly accelerating

Any ONE of the options VW”
Enige EEN korrekte opsie

Snelheid neem uniform toe.

Snelheid neem van 5 m-s"' fot 10 m-s" in 150s foe.
Positiewe versnelling.

Die meisie se spoed neem toe.

Die meisie se versnelling is uniform.

Uniform/constant velocity

Zero acceleration Any ONE of the options VY”
Same speed / velocity Enige EEN korrekte opsie

Uniforme/konstante snetheid
Nil versnelling
Dieselfde spoed/ snetheid

OPTION 1/OPSIE 1 OPTION 2/OPSIE 2

Distance A to C Distance A to C

Afstand A tot C Afstand A tot C
=f{xb+%xbxhyv =lxb+lxb+%xbxhv

=2125mv =2125mv

=5 x 350 v+ %x 150 x Vv = 200 x 5 + 150 x 5V+% x 150 x 5¥

OPTION 3/OPSIE 3

Distance A to C

Afstand A tot C

={xb+%(sum of parallel sides jh”
={xb+%(som van parallele sye )hY

=5x 200 v+ %(5 + 10)(150) ¥
=2125mv

Gradient of this graph is the acceleration
ya-¥4) or Mi
XQ - X4 tr-t
_(-10)¥
~ (65-0) ¥
=-0,15 m-s*
+ a=0,15m-s® Southv/ Suid

gradient =

(2)

(2)

(2)

(2)

(4)

Ane
(4)

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Physical Sciences/P1/Fisiese Wetenskappe/V1 8 DBE/November 2016

45

4.6

NSC/NSS - Grade/Graad 10 — Memorandum

DtoE. vv
DtotE (2)

The change in speed from D to E is (-)10 m-s" v and that occurs over (65 s) a
shorter period. W

OR

From B to CG, the change in speed is 5 m-s”' over a period of 150 s.vv

OR

Gradient is the steepest

Die verandering is spoed van D tot E is (-)10m-s" »en die beweging gebeur
oor 'n korter tydperk.(65 s)

OF

Vanaf B tot C is die verandering in spoed 5 m-s" oor 'n tydperk van 150 s.
OF

Gradient is die steilste (2)
[20]

QUESTION 5/VRAAG 5

5.1

5.2

5.3

5.4

3.600
=v
14x 1000

= 50,4 km-h'Y
ORIOF

14x3,6v
= 50,4 kmh"! ¥ 2)

The energy an object possesses as a result of its motion. VV
Die energie van 'n voorwerp as gevolg van die beweging daarvan. (2)

E, = mgh¥
= 0,01 x 9,8 x Sv
= 0,49 Jv (3)

OPTION/OPSIE 1
(Ep + Ex)topibo = (Ep + Ex)bottomronder v
mgh + %4mv? = mgh + %mv*
(0,01)(9,8)(10) + OY = (0,01)(9,8)(5) +%x0,01xv ¥
v=9,89m-s? ¥
OPTION/OPSIE 2
(Ep + Etopibo = (Ep + Ex)bottom/onder v
mgh + Ymv? = mgh + Ymv2
(0,01)(9,8)(10) +0 Vv =049+%x001xv v

049% 2x00Ixv
v=9,89ms" ¥
ACCEPT/ AANVAAR: A { he.

(Ep + Ex) = (Ep + Ex)s (4)

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Physical Sciences/P1/Fisiese Wetenskappe/V1 9 DBE/November 2016
NSC/NSS — Grade/Graad 10 —- Memorandum
5.5 Equal tow’. Mechanical energy is conserved”, it is a closed system.”
Gelyk aan. Meganiese energie word behou, dit is 'n gesiote stelsel. (3)
[14]
QUESTION 6/VRAAG 6
6.1 0,4mvv (2)
6.2.1 TroughyY
Trog/buik (1)
6.2.2 CrestY
Kruin (1)
6.3 Aand CY
Aenc (1)
6.4 2uvv (2)
6.5 vafxav
0,4 =0,5 xAv
“ N= 0,8 mY (3)
6.6 2% x 0,8¥
=2mv
OR/OF
_q
vt
0,4 d v
"5
[12]

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Physical Sciences/P1/Fisiese Wetenskappe/V1 10 DBE/November 2016
NSC/NSS — Grade/Graad 10 —- Memorandum
QUESTION 7/VRAAG 7
7.1 The wavelength of notes (played). “
Die golflengte van note (gespeel) (1)
7.2 The frequency. ¥
Die frekwensie (1)
7.3
120
100
— a
2
po
og
56
s
zs
fe
Le
0 5 10 15
Wavelength (m)
Golflengte (m)
MARK ALLOCATION: | | PUNTETOEKENNING: 7 |
Y correct y-axis label and unit  korrekte benoeming en eenheid op y-as_ (4)
correct x-axis label and unit korrekte benoeming en eenheid op x-as
Y¥ all points plotted correctly “ alle punte korrek gestip en beste lyn
And best fit line drawn geteken
v shape of graph ¥~ vorm van grafiek
¥ correct scales on axes vkorrekte skaal vir asse
74 Frequency and wavelength are inversely proportional “to each other.
Frekwensie en golflengte is omgekeerd eweredig aan mekaar.
rn Me
1
fas (1)

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Physical Sciences/P1/Fisiese Wetenskappe/V1 11
NSC/NSS — Grade/Graad 10 - Memorandum

7.5

vafxav
=55x6v
= 330 m-s'!v

QUESTION 8/VRAAG 8

8.1.1
8.1.2
8.1.3
8.2.1

8.2.2

8.3.1

8.3.2

8.3.3

Cv
AY
By

Keeping food warm

Remote controls

Optical fibres v
Animals like snakes which hunt

Infrared scanners (for picking up heat)

Hou voedse!l warm
Afstandbeheerders

Optiese vesels

Sekere diere soos silange wat jag
Infrarooi skandeerders (wat hitte optel)

RADAR systems v
RADAR speed traps
Microwave ovens

Telefoon- OF satelliet- OF selfoonkonneksies
RADARsteisels

RADARspoediokvaistelsels

Mikrogolfoonde

X-ray ¥
X-strale

DBE/November 2016

2ni6 -tt 11 (10)

(3)

Any ONE v/Enige een

(1)

Telephone OR satellite OR cellphone connections

Any ONE v/Enige een

(1)

(1)

X-ray has a high frequency OR (high) penetration into soft tissues of humans, ¥

but not bones.

X-strale het 'n hoé frekwensie OF sagte weefsel van mense (hoog)indring, maar

nie been nie.

X-rays can:

e damage living tissue
e cause cancer
X-strale kan:

« weefsel beskadig

e kanker veroorsaak

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(1)

Any ONE ¥/Enige een eh

(1)

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DBE/November 2016

Physical Sciences/P1/Fisiese Weienskappe/Vi 12

NSCINSS - Grade/Graad 10 —- Memorandum

8.4 OPTION/OPSIE 1 OPTION/OPSIE 2
E=hiv f=£
a a
= 6,63 x 10% x sity =u
=6,63 x 107% J v =1x 10° Hz

E = hf v (one mark for both equations/
een punt vir albei vergelykings)

= 6,63 x 10%x 1 x 10° ¥

=6,63x 10%) v

(4)

QUESTION 9/VRAAG 9

9.1.1 A force exerted on an object without touching the object. vv

OR

A force exerted on an object that is at a distance. VW

'n Krag wat op ‘n voorwerp uitgeoefen word sonder om aan die voorwerp te

raak.
OF

'n Krag wat oor 'n afstand op ‘'n voorwerp uitgeoefen word.

9.1.2  e Gravity/Weight/Gravitational force
e Electrostatic/Coulombic force

° Gravitasiekrag/Gewig/Gravitasie
e Elektrostatiese/Coulomb-kragte

9.2 Attractive “
Aantrekkend

9.3 © North/NY
e

v

Any ONE v/Enige een

The direction of magnetic field lines is from north to southy

Noord/ N

Die rigting van magneetveldlyne is van noord na suid

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[12]

(2)

(1)

(1)

(2)
[6]

by

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Physical Sciences/P1/Fisiese Wetenskappe/V1 13 DBE/November 2016
NSC/NSS — Grade/Graad 10 —- Memorandum
QUESTION 10/VRAAG 10
10.1 BY (1)
10.2 B to AY
Btot A (1)
Qi+ Q:
10.3 Qhewinuut = = ZY
_ (43x 10% + (2x10) y
= 2
=5x10’Cv (3)
10.4 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
n=Sv n=fv
e e
— 5x 107-(-2x 10° dy = 5x 107-(+3 x 10°),
1,6 x 107° 1,6 x 10°
= 1, 56 x 10° electrons” = 1, 56x 10" electrons”
elektrone elektrone (3)
[8]
QUESTION 11/VRAAG 11
11.1.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
4 1 4 Ry xX Ro
—— 1 —_— — Ry, = v
Ry Ry * R2 v : Ry + Ro
4 4 - 8x 3,
“5 *3" 6+3
2. Ry = 20 =20.
~Riotavtotaal = 4+ 2” ©. Reotaitotaal = 4 + 2%
=6 Ov =6 a” (4)
11.1.2 [Ry : Reeves ACCEPT! AANVAAR:
20 : 40Vv ~ v2
as : : : Ran = 7%
.. potential difference is also in ratio of ly
Potensiaal verskil is ook in die verhouding 4= V2 v
2:40r1:2v 2
-. 12V +3 parts/dele = 4V V2=8VV¥
«. series = 2X 4 = BV ¥
(3)

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Physical Sciences/P1/Fisiese Wetenskappe/V1 14
NSC/NSS ~ Grade/Graad 10 - Memorandum

11.1.3

11.1.4

11.2

11.3

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DBE/November 2016

1 7
Ra yorin words:
resistance is inversely proportional to
current and

-. ratio of resistors is 6 : 3V
2:1

«ratio of current is 1:2 “”
A2: Ag

lao = : x1

ola = 067A ¥

1;
R a; in woorde:
weerstand is omgekeer eweredig aan
stroom

.. verhouding van resistors is 6 : 3¥
2:1

.. verhouding van stroom is 1: 2¥
Ae2 i A3
2
wtlao = 3 x?
wlap = 0,67 A v

ACCEPT/IAANVAAR:
v
=av¥
R
_ 12-8

6
=0,67AVY

(3)

Ai=2A
Q=lAtY
=2x120V
=240CV

Decrease ¥
Afneem

NEGATIVE MARKING FROM 11.2
NEGATIEWE MERK VANAF 11.2

(3)

(1)

e Ifthe 6 Q resistor is removed, the resistance of the whole circuit increases V
e SinceRa : ¥, if R increases, and V is constant V and I of the circuit

decreases

e Indien die 6 Q-resistor verwyder word, sal die totale weerstand van die

stroombaan verhoog.

°e R at ¥ , so indien R verhoog en V bly konstantv’, sal die stroom (I) verlaag. (3)

206 -1i- 11

[17]

<yTOTAL/TOTAAL: 150

Me

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