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)
[20]

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)

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

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

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

CONSIDER MOTION UPWARD AS POSITIVE/BESKOU OPWAARTSE BEWEGING AS POSITIEF

QUESTION 4/VRAAG 4

4.1
The total linear momentum in a closed system√ remains constant./is conserved √/Die totale lineêre momentum in 'n geslote stelsel√ bly 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

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 equal √to the change in kinetic energy of the object√/Die netto arbeid wat op 'n voorwerp verrig word, is gelyk √ aan die verandering in kinetiese energie van die voorwerp√ (2)

5.3

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

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
Tcos20^o = 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

QUESTION 8/VRAAG 8

8.1

Vectors E_Q1 and E_Q2 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)

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

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

9.2
Removing the 2 Ω resistor increases the total resistance of the circuit. √ Thus the total current decreases, decreasing the V_int (V_lost). √ 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 V_int (V_verloor) verlaag. √ Dus verhoog die voltmeterlesing V. √ (3)
[15]

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

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)

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)
[12]
TOTAL/TOTAAL: 150