MAMELODI EAST CLUSTER CONTROL TEST 1 2016

MAMELODI EAST CLUSTER CONTROL TEST 1 2016

GRADE 11 PHYSICAL SCIENCES

MARKING MEMORANDUM Please read the marking guidelines on pages 33 ? 37 of the Physical Sciences Examination Guidelines, Senior Certificate, Grade 12, 2015.

Page 1 of 6

QUESTION 1

1.1

B

(2)

1.2

C

(2)

1.3

C

(2)

1.4

D

(2)

[8]

QUESTION 2

2.1 They have both magnitude and direction.

(2)

2.2 = + 45? = 3 + 4 cos 45? = 5,828 , right

= - 45? = 5 - 4 sin 45? = 2,172 , up

Fnet

F Y

F X

Fnet Fx2 Fy2

5,8282 2,1722

6,22N

tan =

tan = 2,172

5,828

= 20,44?

Resultant force = 6,22 N in direction: 20,44o OR Bearing

69,56o, N69,56oE or 20,44o North of East

(7)

Page 2 of 6

2.2 SCALE DRAWING:

3 N

5 N

45O 4 N

70o 20O

Resultant = 6 N

CRITERIA Each angle correctly measured Correct tail to head drawing of PQR Resultant both magnitude and direction correct starting from origin to head of vector R

TOTAL

MARK 1 ? 2 1 ? 3 1 ? 2

7 MARKS

2.3 The resultant of all forces acting at point S is zero

(2)

2.4

55o

90o

35o

(3)

Correct shape of triangle

Forces labelled and point correct directions

Labelled angles are all correct

2.5

sin

35?

=

sin 35? = 60

= 104,61

ACCEPT: Sine rule, cosine rule, component

tan 35? =

tan

35

=

60

= 85,69

method and any other trigonometric (4) method

[18] Page 3 of 6

QUESTION 3

3.1 The force that opposes the motion of an object and which acts parallel to the surface.

3.2

fs

FN

w

(2)

Other labels to be accepted: Normal/ N (3) Friction/ f Weight/ gravity/ Fg

3.3

3.3.1

= mg sin

(3)

= 95,0 ? 9.8 ? sin 23.2?

= 366.76 N

3.3.2 N = mg cos N = 95.0 ? 9.8 ? cos 23.2? N = 855.72 N

= N

366,76 = x 855,72

= 0,43

(5)

3.4 Less than

(1)

[14]

QUESTION 4

4.1 When a resultant (net) force acts on an object, the object will

accelerate in the direction of the force. This acceleration is

(2)

directly proportional to the force and inversely proportional to

the mass of the object.

4.2 Force of the block on the table and the force of the table on

(2)

the block.

OR

Force of the block on the string and the force of the string on

the block.

Page 4 of 6

4.3

4.3.1

6 kg:

N

f

T

w

= + (-) = - 11.76 = 6 ... ... ... ... (1)

2 kg:

T

F w

= + + (-) = (2 ? 9.8) + 2 - = 2

21,6 - = 2 ... ... ... ... (2) (5)

(1) + (2): 9,84 = 8 = 1,23 -2

4.3.2 T ? 11,76 = 6(1,23)

(2)

T = 19,14 N

4.4 Increases

(1)

4.5

=

12

2

=(6,67?10(-51,51)?(160,55?)21020)(90)

= 12,90

OR

= 21 =(6,67?1(50,-51?11)0(65,)52?1020) = 0,143 -2

= = 90 ? 0,143 = 12,89

(4) [16]

Page 5 of 6

QUESTION 5 5.1 B 5.2 A

[4]

QUESTION 6

6.1

Molecule Lewis structure

Shape of the Polarity of Polarity of

molecule

the bonds the molecule

CO2 (g)

Linear

Polar

Non polar

H2O (g)

Angular/bent Polar

Polar

(10)

(1) 6.2.1 Dative covalent bond

6.2.2

(4)

[15]

Page 6 of 6

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