PHYSICAL SCIENCES: PHYSICS P1 Past Paper FEBRUARY/MARCH 2016 Memo/Memorandum - GRADE 12 NATIONAL SENIOR CERTIFICATE

QUESTION 1

1.1 B √√ (2)
1.2 D √√ (2)
1.3 B √√ (2)
1.4 D √√ (2)
1.5 A √√ (2)
1.6 C√√ (2)
1.7 B√√ (2)
1.8 B √√ (2)
1.9 C √√ (2)
1.10 A √√ (2)
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QUESTION 2

2.1 For the 5 kg mass/Vir die 5 kg massa:
2.1.1
T – f = ma
T - μk(mg) = ma√
T – (0,4)(5)(9,8)√ = 5a√….…..(1)

NOTE/LET WEL:
1 mark for any of the 2 formulae
1 punt vir enige van die 2 formules

For the 20 kg mass/Vir die 20 kg massa

mg – T = ma
20(9,8) – T = 20a√……….(2)

176,4 = 25a (1) + (2)
∴a = 7,06 (7,056) m∙s-2√ (5)

ACCEPT/AANVAAR (4 marks/4 punte)

Fnet = ma
Mg – f = (M + m) a√
[20(9,8) – (0,4)(5)(9,8)]√ = 25a√
∴a = 7,06 m∙s-2√ (4)

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2.1.3 6 m√ (1)

2.2
2.2.1
Each body in the universe attracts every other body with a force that is directly proportional to the product of their masses √and inversely proportional to the square of the distance between their centres.
Elke liggaam in die heelal trek elke ander liggaam aan met 'n krag wat direk eweredig is aan die produk van hul massas √ en omgekeerd eweredig is aan die kwadraat van die afstand tussen hul middelpunte. √ (2)

2.2.2
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3.2
POSITIVE MARKING FROM 3.1.2; 3.1.3/POSITIEWE NASIEN VANAF 3.1.2; 3.1.3
CONSIDER MOTION DOWNWARD AS POSITIVE/BESKOU BEWEGING AFWAARTS AS POSITIEF

 

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CONSIDER MOTION UPWARD AS POSITIVE/BESKOU OPWAARTSE BEWEGING AS POSITIEF

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QUESTION 4/VRAAG 4

4.1
The total linear momentum in a closed systemremains constant./is conserved √/Die totale lineêre momentum in 'n geslote stelselbly konstant/bly behoue. √

OR/OF
In a closed/isolated system, the total momentum before a collision is equal to the total momentum after the collision./In 'n geslote/geïsoleerde stelsel is die totale momentum voor 'n botsing gelyk aan die totale momentum na die botsing. (2)

4.2
4.2.1

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4.3

QUESTION 5/VRAAG 5
5.1
It is a ratio of two forces √ (hence units cancel out)./Dit is 'n verhouding van twee kragte √ (dus word eenhede uitgekanseleer) (1)

5.2
The net work done on an object is equalto the change in kinetic energy of the object√/Die netto arbeid wat op 'n voorwerp verrig word, is gelykaan die verandering in kinetiese energie van die voorwerp√ (2)

5.3

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QUESTION 6/VRAAG 6
6.1
v = fλ √
= (222 x 103)(1,5 x 10-3)√
= 333 m.s-1 √ (3)

 

6.2

6.2.1
Towards the bat/Na die vlermuis toe √ (1)

6.2.2
POSITIVE MARKING FROM QUESTION 6.1/POSITIEWE NASIEN VANAF VRAAG 6.1

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QUESTION 7/VRAAG 7

7.1
The magnitude of the charges are equal√/ The balls repel each other with the same/identical force or force of equal magnitude√/Die grootte van die ladings is gelyk√/Die balle stoot mekaar af met dieselfde/identiese kragte of krag van dieselfde grootte. √ (1)

7.2
The electrostatic force of attraction between two point charges is directly proportional to the product of the charges √and inversely proportional to the square of the distance between them. √/Die elektrostatiese aantrekkingskrag tussen twee puntladings is direk eweredig aan die produk van die ladings√ en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle.√ (2)

7.3
7.3.1
Tcos20o = w√
= mg
= (0,1)(9,8)√ = 0,98 N
∴T = 1,04 N√ (3)

7.3.2
POSITIVE MARKING FROM 7.3/POSITIEWE NASIEN VANAF 7.3

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QUESTION 8/VRAAG 8

8.1

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Vectors EQ1 and EQ2 in the same direction√√/Vektore EQ1 en EQ2 in dieselfde rigting√√

Correct drawing of vectors EQ1 and EQ2√√/Korrekte tekening van vektore EQ1 en EQ2√√

The fields due to the two charges add up because they come from the same direction. Hence the field cannot be zero./Die velde as gevolg van die twee ladings word bymekaar getel omdat hulle uit dieselfde rigting inwerk. Die veld kan dus nie nul wees nie. (4)

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QUESTION 9/VRAAG 9

9.1
9.1.1
V = IR√
= (0,2)(4+8)√
= 2,4 V√ (3)

9.1.2
POSITIVE MARKING FROM QUESTION 9.1.1/POSITIEWE NASIEN VANAF VRAAG 9.1.1

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9.1.3
POSITIVE MARKING FROM QUESTION 9.1.2/POSITIEWE NASIEN VANAF VRAAG 9.1.2

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9.2
Removing the 2 Ω resistor increases the total resistance of the circuit. √ Thus the total current decreases, decreasing the Vint (Vlost). √ Therefore the voltmeter reading increases. V√/Wanneer die 2 Ω-resistor verwyder word, verhoog dit die totale weerstand van die kring. √ Dus verklein die totale stroom, wat die Vint (Vverloor) verlaag. √ Dus verhoog die voltmeterlesing V. √ (3)
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QUESTION 10/VRAAG 10
10.1
10.1.1 North pole/Noordpool√ (1)

10.1.2 Q to P√ (1)

10.2
10.2.1

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QUESTION 11/VRAAG 11
11.1
It is the minimum energy that an electron in the metal needs to be emitted from the metal surface. √/Dit is die minimum energie wat 'n elektron in die metaal benodig om elektrone uit die metaaloppervlak vry te stel. √ (2)

11.2 Frequency/Intensity √/Frekwensie/Intensiteit (1)

11.3
The minimum frequency required to remove an electron from the surface of the metal√/Die minimum frekwensie benodig om 'n elektron vanaf die oppervlak van die metaal te verwyder√ (2)

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11.5
The photocurrent is directly proportional to the intensity of the incident light.√√/Die fotostroom is direk eweredig aan die intensiteit van die invallende lig. √√ (2)
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TOTAL/TOTAAL: 150