# Physical Sciences Grade 10 P1 2016 Memo

> Grade 10 Physical Sciences Paper 1 memorandum from the Mpumalanga National Exemplar Exam (NSC) 2016, 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-10-physical-sciences-p1-memo/384b2b6b-8ff7-4602-bce3-3669ee045d2e
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 10
- Document type: Memorandum
- Year: 2016
- Exam period: Mpumalanga National Exemplar Exam (NSC)
- Paper: 1
- Curriculum: CAPS
- Pages: 14

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## About this document

Physical Sciences · Grade 10 · Mpumalanga National Exemplar Exam (NSC) · 2016. Memorandum, 14 pages. Read online or download the PDF.

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## Document text

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### Page 1

basic education

Department:
Basic Education
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE/
NASIONALE
SENIOR SERTIFIKAAT

GRADE/GRAAD 10

PHYSICAL SCIENCES: PHYSICS (P1)
FISIESE WETENSKAPPE: FISIKA (V1)

NOVEMBER 2016
MEMORANDUM

MARKS/PUNTE: 150

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

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### Page 2

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

QUESTION 1/VRAAG 1

14 Byv (2)
1.2 Cv (2)
13 Dyvv (2)
1.4 Dvv (2)
15 Dvv (2)
1.6 AvY (2)
1.7 BYY (2)
1.8 Cvy (2)
1.9 DvY (2)
110 BYv (2)
[20]

QUESTION 2/VRAAG 2

2.1

Mark allocation/Puntetoekenning:

Cc  —® Iv 1 xiine ab: length, arrow, label

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

### Page 3

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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### Page 4

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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### Page 5

Physical Sciences/P1/Fisiese Wetenskappe/V1 5
NSCI/NSS — Grade/Graad 10 - Memorandum

3.5

POSITIVE MARKING FROM 3.4
POSITIEWE NASIEN VANAF 3.4

DBE/November 2016

.. taxi will not stop at the traffic light as
distance > 40 m

. die taxi sal nie V by verkeerslig stop
nie, want die afstand is > 40 m

OR / OF

- oe ov
=25mv

.. 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

O = 257 +2 x -12,5 x AxY
“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

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?¥

=25m

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

-. 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

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(5)

### Page 6

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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### Page 7

Physical Sciences/P1/Fisiese Wetenskappe/V1 7
NSC/NSS — Grade/Graad 10 - Memorandum

QUESTION 4/VRAAG 4

4.1

4.2

4.3.1

4.3.2

441

4.4.2

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"
8,3 ms" (aanvaar vanaf 8,2 tot 8,4)

The velocity is uniformly increasing.
Pasitive acceleration.

The girl is speeding up.
The girl is uniformly accelerating

Snelheid neem uniform toe.

Positiewe versnelling.
Die meisie se spoed neem toe.
Die meisie se versnelling is uniform.

Uniform/constant velocity
Zero acceleration
Same speed / velocity

Uniforme/konstante snetheid
Nil versnelling
Dieselfde spoed/ snetheid

¥¥ (accept range from 8,2 to 8,4)

DBE/November 2016

(2)

(2)

Velocity increases from 5 m-s" to 10 ms” in 150s.

Any ONE of the options VW”
Enige EEN korrekte opsie

Snelheid neem van 5m-s" tot 10 m-s" in 150s toe.

(2)

Any ONE of the options v “
Enige EEN korrekte opsie

(2)

OPTION 1/OPSIE 1

OPTION 2/OPSIE 2

Distance A to C
Afstand A tot C
={xb+%xbxhv

= 5x 350 w+ “4x 150 x 5”
=2125mv

Distance A to C

Afstand A tot C
=lxb+lxb+%xbxhv

= 200 x5 + 150 x 5v+% x 150 x 5¥
=2125mv

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
y2-Y4 VE Vi
XQ - X4 tr-t
_(-10)¥
~ (65-0) ¥
=-0,15 m-s*
+ a=0,15m-s® Southv/ Suid

gradient = Yor

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(4)

Ane
(4)

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### Page 8

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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### Page 9

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 v=ftxay
0.4=0,5xAv
“= 0,8 mY (3)
6.6 2% x 0,8¥
=2mv
ORIOF
=
vt
oaetv
"5
[12]

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### Page 10

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
=P
ze
po
og
5 5
s
zs
es
Le
0 5 10 15
Wavelength (m)
Golflengte (m)
MARK ALLOCATION: | | PUNTETOEKENNING: |
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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### Page 11

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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### Page 12

Physical Sciences/P1/Fisiese Weienskappe/Vi 12

DBE/November 2016

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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### Page 13

Physical Sciences/P1/Fisiese Wetenskappe/V1 13
NSC/NSS - Grade/Graad 10 —- Memorandum

QUESTION 10/VRAAG 10

10.1

10.2

10.3

10.4

11.1.1

11.1.2

DBE/November 2016

By (1)
Bto Av
Btot A (1)
Qi+ Q:
Qhew/nuut = 5 ay
_ (43x 10% + (2x10) y
= 3
=5x107CV (3)
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
OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
1 1 1 Ry x Ro
—— 1 —_— — Ry, = v
Ry Ra * Ro ” me Rt + Ro
4 4 = 8x 3,
“5*3” 6+3
2. Ry = 20 =22
--Riotavtotaal = 4 + 2 ©. Riotaitoteal = 4 + 2¥”
=6 OY =6 OY (4)
Ry : Reeries ACCEPT/ AANVAAR:
20 : 40v _Y2
an . . . Ran = 7%
.. potential difference is also in ratio of ly
Potensiaal verskil is ook in die verhouding 4= V2 v
2: 407 1:24 2
-. 12V +3 parts/dele = 4V V2=8VV¥
-. Veeries = 2x 4 = BV 4
(3)

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### Page 14

Physical Sciences/P1/Fisiese Wetenskappe/V1

14

NSC/NSS ~ Grade/Graad 10 - Memorandum

DBE/November 2016

11.1.3. | Ro orin words: ACCEPT/IAANVAAR:
I . Vv
resistance is inversely proportional to =R v
current and _ 12-8
.. ratio of resistors ia =0,67AY
«ratio of current is 1:2 “”
Az: Ag
lao = : x1
wb2=067A v

1,
R azin woorde:

weerstand is omgekeer eweredig aan

2:1

Ae2 i A3

stroom

.. verhouding van resistors is 6 : 3¥
.. verhouding van stroom is 1: 2¥
wtlao = z x?

wlap = 0, 67A v

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

11.2 Decrease ¥
Afneem

11.3 NEGATIVE MARKING FROM 11.2
NEGATIEWE MERK VANAF 11.2
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.

(3)

(3)

(1)

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

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206 -1i- 11

[17]

<yTOTAL/TOTAAL: 150

Me

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