secondary school improvement programme (ssip) 2019

© Gauteng Department of Education

1

SECONDARY SCHOOL IMPROVEMENT

PROGRAMME (SSIP) 2019

GRADE 12

SUBJECT: PHYSICAL SCIENCES

TERM 01

EDUCATORS GUIDE

PAGE (1 OF 22)


2

Organic Chemistry

Question 1

Multiple-choice Questions

1.1. A

(2)

1.2. B

(2)

1.3. C

(2)

1.4. B

(2)

1.5. A

(2)

1.6. B

(2)

1.7. A

(2)

1.8. D

(2)

1.9. C

(2)

1.10. A

(2)

1.11. C

(2)

1.12. B

(2)

1.13. B

(2)

1.14. D

(2)


3

Question 2

2.1.

2.1.1. Alkynes

(1)

2.1.2. Hydroxyl group

(1)

2.1.3. C

(1)

2.1.4. 2-methylpentan-3-one

(2)

2.1.5.

C C

H

H H

H

(2)

2.1.6. 2𝐶4𝐻10 + 13𝑂2 ⟶ 8𝐶𝑂2 + 10𝐻2𝑂 𝐵𝑎𝑙.

(3)

2.2. Same molecular formula, but different positions of the

functional group.

(2)

2.3

C C O

H

H

H H

H

H

+ C C C

O

O

H

H

H H

H

H

C C O C C C

OH

H

H

H

H H

H

H

H

H

+ O

H

H

(7)

[19]

Question 3

3.1.

3.1.1. B (1)

3.1.2. E (1)

3.1.3. F (1)

3.2.

3.2.1. 2-bromo-3-chloro-4-methylpentane (3)

3.2.2. Ethene (1)


4

3.3.

3.3.1.

C C C C C C

C

CC

H

H

H H

H H

H

H

H

H

H

H

H

H

H

H

HH

H

H

(2)

3.3.2.

C C C C C

OH

H

H H

H

H

H

H

H

H

(2)

3.4.

3.4.1. Compounds with the same molecular formula

but different functional groups / different homologous series.

(2) 3.4.2. B & F (1)

[14]


5

Question 4

4.1.

4.1.1. Carboxyl group (1)

4.1.2. Ketones (1)

4.1.3. Addition (1)

4.2.

4.2.1. Ethene (1)

4.2.2. 4-methylhexan-3-one (2)

4.2.3. 4-ethyl-2,2-dimethylhexane (2)

4.3. Carbon dioxide/CO2

Water / H2O (2)

4.4.

4.4.1.

C C C C

C

O

O

H

H

H H

H

H

H

H

H

H (2)

4.4.2.

C C C C

H

H

H H H H

H

H

OR/OF

C C C C

H

H H H

H

H

H

H

(2)

4.5.

4.5.1. E (1)

4.5.2. Substitution / halogenation / bromination (1)

4.5.3

C C C C

Br BrH

H

H H H H

H

H

(2)

[8]


6

Question 5

5.1.

5.1.1. B (1)

5.1.2.

(1)

5.1.3. 𝐶𝑛𝐻2𝑛−2 (1)

5.1.4. 4-ethyl-5-methylhept-2-yne (3)

5.1.5. Butan-2one (2)

5.2.

5.2.1. Alkanes / Alkane (1)

5.2.2. Methylpropane

C C C

C

H

H

H H H

H

H

H

H

H

(4)

5.3.

5.3.1. Haloalkanes / Alkyl halides (1)

5.3.2. Substitution / halogenation / bromination (1)

[6] Question 6

6.1Temperature at which the vapour pressure of the substance

Equals atmospheric pressure. (2)

6.2 .1 Boiling point increases as the chain length / molecular mass increases.

OR Boiling point increases from methane to butane. (1)

6.2.2

Chain length increases from methane to butane.

Strength of London forces / induced dipole forces increases

from methane to butane.

More energy needed to overcome intermolecular forces in

butane than in methane. (3)


7

6.3 Between molecules of the alkanes are weak London forces or

induced dipole forces.

Between alcohol molecules are, in addition to weak London Forces or induced dipole forces, also strong hydrogen bonds. (2)

[8]

Question 7

7.1Alkanes have ONLY single bonds between C-atoms. (1)

7.2

7.2.1

OR (1)

7.2.2

C C C

C

O

H

H

H H

H

H

H

H

H

H (2)

7.3.1 What is the relationship between chain length / molecular size/

molecular structure / molecular mass / surface area and boiling point?

(2)

7.3.2

Structure:

The chain length / molecular size / molecular structure / molecular mass / surface area increases.

Intermolecular forces:

Increase in strength of intermolecular forces / induced dipole / London / dispersion / Van der Waals forces.

Energy:

More energy needed to overcome / break intermolecular forces. OR

Structure:

From propane to methane the chain length / molecular size / molecular structure / molecular mass / surface area decreases.

Intermolecular forces:

Decrease in strength of intermolecular forces / induced dipole forces /


8

London forces / dispersion forces.

Energy:

Less energy needed to overcome / break intermolecular forces. (3) 7.4

Between propane molecules are London forces/dispersion forces/

induced dipole forces.

Between propan-1-ol molecules are London forces/dispersion

forces/induced dipole forces and hydrogen bonds.

Hydrogen bonds / Forces between alcohol molecules are stronger or

need more energy than London forces / dispersion forces / induced dipole forces.

OR Between propane molecules are weak London forces / dispersion forces / induced dipole forces and between propan-1-ol molecules are strong hydrogen bonds. (3) [12]

Question 8

8.1 Alkenes / Alkene (1)

8.2

8.2.1 Addition / Hydrohalogenation / Hydrochlorination (1)

C C C

O

H

H

H

H

H

H

H

H

Propan-2-ol (3)

8.2.2 Elimination / Dehydration (1)

8.2.3 Catalyst (1)

8.3

8.3.1 Sodium hydroxide / Potassium hydroxide (1)

8.3.2 Dissolve base in ethanol / Concentrated (strong) base

Heat strongly (2)


9

8.3.3

C C C

Cl

H

H

H

H

H

H

H

+ ONa

H

C C C

H

H H H

H

H

+ ClNa + O

H

H

(5)

[15]


10

Question 9

9.1

9.1.1 Substitution / chlorination / halogenation (1)

9.1.2 Substitution / hydrolysis (1)

9.2

9.2.1 Hydrogenation (1)

9.2.2

C C C

H

H

H H H

H

+ H H C C C

H

H

H H

H

H

H

H

(3)

9.3

C C C

Cl

H

H

H

H

H

H

H

(2)

9.4

9.4.1 Esterification / Condensation (1)

9.4.2 Concentrated H2SO4 / Concentrated sulphuric acid (1)

9.4.3

C C C O C C

OH

H

H H

H H

H H

H

H

(2)

9.4.4 Propyl ethanoate (2)

9.5 Sulphuric acid / H2SO4 / Phosphoric acid / H2PO4 (1)

[15]


11

NEWTON MEMO

Question 1 Multiple choice questions

1.1. D 1.2. A 1.3. A 1.4. A 1.5. C 1.6. B 1.7. D 1.8. D 1.9. B 1.10. D

[𝟏𝟎 × 𝟐 = 𝟐𝟎]

Question 2

2.1 When a resultant / net force acts on an object, the object will accelerate in the direction of the force. This acceleration is directly proportional to the force and inversely proportional to the mass of the object. (2)

2.2 REMAINS THE SAME / BLY DIESELFDE (1)

2.3 . Accepted Labels FG Weight, gravitational force fk Friction N Normal force FT Tension

(4)

2.4 2.4.1. 𝐹𝑛𝑒𝑡 = 𝑚𝑎 = 𝐹𝑇 + 𝑓𝑘 + 𝐹𝐺‖

𝑚𝑎 = 𝐹𝑇 + 𝜇𝑘𝑚𝑔𝑐𝑜𝑠30° + 𝑚𝑔𝑠𝑖𝑛30° {𝑓𝑘 = 𝜇𝑘𝑁}

FG FG‖

FG⫠


12

{ 𝑁 = − 𝐹𝐺⫠ =𝑚𝑔𝑐𝑜𝑠30°} (6)(4) = 𝐹𝑇 − (0,2)(6)(9,8)(𝑐𝑜𝑠30°)− (6)(9,8)(𝑠𝑖𝑛30°) ∴ 𝐹𝑇 = 63,58 𝑁 (5)

2.4.2. 𝐹 + 𝑓𝑘 3𝑘𝑔 + 𝐹𝐺‖ = 𝑚𝑎

𝐹 − (0,2)(6)(9,8)𝑐𝑜𝑠30°− (0,1)(3)(9,8)𝑐𝑜𝑠30°− (3 + 6)(9,8)𝑠𝑖𝑛30° =0 𝐹 = 56,83 𝑁 (6)

2.5 DECREASES (1)

[19]

Question 3

3.1. When a resultant / net force acts on an object, the object will accelerate in the direction of the force. This acceleration is directly proportional to the forceand inversely proportional to the mass of the object. (2)

3.2. . (3)

3.3. 𝐹𝑛𝑒𝑡 = 𝑚𝑎

5kg 𝑇2 + 𝐹𝐺 + 𝑇1 = 𝑚𝑎 250 − (5)(9,8) − 𝑇1 = 5𝑎 201 − 𝑇1 = 5𝑎 𝑇1 = 201 − 5𝑎 ……..(1) 20kg 𝑇1 + 𝐹𝐺 = 𝑚𝑎 𝑇1 − [(20)(9,8)] = 20𝑎 𝑇1 = 196 + 20𝑎 ……..(2)

(1) = (2)


13

201 − 5𝑎 = 196 + 20𝑎 𝑎 = 0,2 𝑚 ∙ 𝑠−2upwards ∴ 𝑇1 = 201 − (5)(0,2) ∴ 𝑇1 = 200 𝑁 (6)

3.4. Q

(1) [12]

Question 4

4.1. When a body exerts s force on a second body, the second body exerts a force of

equal magnitude in the opposite direction on the first body. (2)

4.2. . Accepted Labels FG Weight, gravitational force FA Applied force N Normal force FT Tension f Friction

(5)

4.3.

4.3.1. 𝑓𝑘 = 𝜇𝑘𝑁


14

𝑓𝑘 = 𝜇𝑘𝑚𝑔𝑐𝑜𝑠𝜃 𝑓𝑘 = (0,29)(1)(9,8)𝑐𝑜𝑠30° 𝑓𝑘 = 2,46 𝑁 (3)

𝐹𝑛𝑒𝑡 = 𝑚𝑎 1kg 𝐹𝐴 + 𝑇 + 𝑓 + 𝐹𝐺‖ = 𝑚𝑎

40 − 𝑇 − 2,46 − (1)(9,8)𝑠𝑖𝑛30° = 1𝑎 40 − 𝑇 − 2,46 − 4,9 = 𝑎 32,64 − 𝑇 = 𝑎 ……..(1) 4kg 𝑇 + 𝐹𝐺‖ + 𝑓 = 𝑚𝑎

𝑇 − (4)(9,8)𝑠𝑖𝑛30° − 10 = 4𝑎 𝑇 − 19,6 − 10 = 4𝑎 𝑇 − 29,6 = 4𝑎 ……..(2) (1) +(2) 32,64 − 𝑇 = 𝑎 𝑇 − 29,6 = 4𝑎 3,04 = 5𝑎 𝑎 = 0,608 𝑚 ∙ 𝑠−2 ∴ 𝑇 − 29,6 = (4)(0,61) ∴ 𝑇 = 32,04 𝑁 (6)

[16]

Question 5

5.1. 5.1.1. When a body exerts s force on a second body, the second body exerts a force of

equal magnitude in the opposite direction on the first body. (2)

5.1.2. 2,5kg 𝐹𝑛𝑒𝑡 = 𝑚𝑎 = 𝐹𝑇 + 𝐹𝐺 𝐹𝑇 − (2,5)(9,8) = (2.5)(0) 𝐹𝑇 = 24,5 𝑁 (3)

5.1.3. 𝑓𝑠 = 𝜇𝑠𝑁 24,5 = 0,2𝑁


15

𝑁 = 122,5 𝑁 𝑁 = −𝐹𝐺 𝑁 = 𝑀𝑔 122,5 = 𝑀(9,8) 𝑀 = 12,5 𝑘𝑔 (5)

5.1.4. 5kg 𝑓𝑘 = 𝜋𝑘𝑁 𝑓𝑘 = (0,15)(5)(9,8) 𝑓𝑘 = 7,35 𝑁 𝐹𝑛𝑒𝑡 = 𝑚𝑎 = 𝐹𝑇 + 𝑓𝑘 5𝑎 = 𝑇 − 7,35 ……..(1) 2,5kg 𝐹𝑛𝑒𝑡 = 𝑚𝑎 = 𝐹𝐺 + 𝐹𝑇 2,5𝑎 = (2,5)(9,8) − 𝐹𝑇 2,5𝑎 = 24,5 − 𝐹𝑇 ……..(2) (1) + (2) 5𝑎 = 𝑇 − 7,35 2,5𝑎 = 24,5 − 𝐹𝑇 7,5𝑎 = 17,15 𝑎 = 2,29 𝑚 ∙ 𝑠−2 (5)

𝐹𝐺 = 𝐺𝑚1𝑚2

𝑟2

𝐹𝐺 = (6,67 × 10−11)(6,5 × 1020)(90)

(550 × 103)2

𝐹𝐺 = 12,899 𝑁 (4)

+

+


16

MOMENTUM AND IMPULSE Question 1

Multiple choice questions

1.1. D

1.2. C

1.3. B

1.4. D [4 2 = 8]

Question 2

2.1. The total linear momentum in a closed system remains constant / is

conserved.

(2)

2.2. Σ𝑝𝑖 = Σ𝑝𝑓

(𝑚1 + 𝑚2)𝑣𝑖 = 𝑚1𝑣1𝑓 + 𝑚2𝑣2𝑓 (2𝑚 + 4𝑚)(0) = (2𝑚)(2) + (4𝑚)( 𝑣2𝑓)

−4𝑚 = 4𝑚𝑣2𝑓

∴ 𝑣𝑓 = −1 𝑚 ∙ 𝑠−1

∴ 𝑣𝑓 = 1 𝑚 ∙ 𝑠−1; in the opposite direction to that of the boys

(5)

2.3. GREATER THAN .

(1)

[8]

Question 3

3.1. 𝑝 = 𝑚𝑣

𝑝 = (50)(5) 𝑝 = 250 𝑘𝑔 ∙ 𝑚 ∙ 𝑠−1 , (downwards) (3)

3.2. The product of the net force and thee time interval (during which the force acts)

.

(2)


17

3.3. ∆𝑝 = 𝐹𝑛𝑒𝑡∆𝑡

0 − 250 = 𝐹𝑛𝑒𝑡(0,2) 𝐹𝑛𝑒𝑡 = −1 250 𝑁 𝐹𝑛𝑒𝑡 = 1 250 𝑁 (3)

3.4. GREATER THAN

(1)

3.5. For the same momentum change , the stopping time (contact time) will be

smaller (less) . ∴ the (upward) force exerted (on her) is greater . (3)

[12]

Question 4

4.1. Σ𝑝𝑖 = Σ𝑝𝑓

(𝑚1 + 𝑚2)𝑣𝑖 = 𝑚1𝑣1𝑓 + 𝑚2𝑣2𝑓 (3 + 0,02)(0) = (3)(−1,4) + (0,02)𝑣2𝑓

𝑣2𝑓 = 210 𝑚 ∙ 𝑠−1

(4) 4.2. 𝑣𝑓

2 = 𝑣𝑖2 + 2𝑎∆𝑥

(0) = 2102 + (2)(𝑎)(0,4) 𝑎 = −55 125 𝑚 ∙ 𝑠−2 𝐹𝑛𝑒𝑡 = 𝑚𝑎 𝐹𝑛𝑒𝑡 = (0,02)(−55 125) 𝐹𝑛𝑒𝑡 = −1 102,5 𝑁 ∴ 𝐹𝑛𝑒𝑡 = 1 102,5 𝑁 (5)

4.3. THE SAME .

(1)

[10]

Question 5 5.1. The total linear momentum in a closed system remains constant/is

conserved.

(2)

5.2.

5.2.1. 𝑝𝑖 = 𝑝𝑓

𝑚1𝑣𝑖1 + 𝑚2𝑣𝑖2 = 𝑚1𝑣𝑓1 + 𝑚2𝑣𝑓2

+


18

(𝑚1 + 𝑚2)𝑣𝑖 = 𝑚1𝑣𝑓1 + 𝑚2𝑣𝑓2

0 = (0,4)𝑣𝑓1 + (0,6)(4)

𝑣𝑓1 = −6 𝑚 ∙ 𝑠−1

= 6 𝑚 ∙ 𝑠−1 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 (4)

5.2.2. ∆𝑝 = 𝐹𝑛𝑒𝑡∆𝑡

(0,6)(4 − 0) = 𝐹𝑛𝑒𝑡(0,3) 𝐹𝑛𝑒𝑡 = 8 𝑁 (4)

5.3. NO

(1)


19

SESSION 06

Doppler effect MEMO

Question 1 1.1. A 1.2. C 1.3. A 1.4. A 1.5. C

Question 2

2.1. Smaller than (1)

2.2. Doppler effect (1)

2.3. 𝑣 = 𝑓𝜆 345 = 𝑓0.55 𝑓 = 627,27 𝐻𝑧

𝑓𝐿 = 𝑣 ± 𝑣𝐿

𝑣 ± 𝑣𝑠 𝑓𝑠

𝑓𝐿 = 345± 0

345−33,33 (627,27)

𝑓𝐿 = 694,35 𝐻𝑧 (7)

2.4. Decrease (1) [10]

Question 3

3.1.1. An (apparent) change in observed/detected frequency (pitch), (wavelength) as a result of the relative motion between a source and an observer (listener). (2)

3.1.2. Towards

Observed/detected frequency is greater than the actual frequency. (2)

3.1.3. 𝑓𝐿 = 𝑣 ± 𝑣𝐿


1200 = 343 ± 0

343− 𝑣𝑠 1130

𝑣𝑠 = 20,01 𝑚 ∙ 𝑠−1 (5)

3.2. The star is approaching the earth.

OR


20

The earth and the star are approaching (moving towards) each other. The spectral lines in diagram 2 are shifted towards the blue end / blue shifted. (2)

[11]

Question 4

4.1.1. 𝑣 = 𝑓𝜆 340 = 520 𝜆 𝜆 = 0,65𝑚 (2)

4.1.2. 𝑓𝐿 = 𝑣 ± 𝑣𝐿


𝑓𝐿 = 340± 0

340− 15 (520)

𝑓𝐿 = 544 𝐻𝑧

𝑣 = 𝑓𝜆 340 = 544𝜆 𝜆 = 0,63 𝑚 (6)

4.2. The wavelength in QUESTION 4.1.2 is shorter because the waves are compressed as they approach the observer. (2)

4.3. The red shift occurs when the spectrum of a distant star moving away from the earth is shifted toward the red end of the spectrum. (2)

[12]

Question 5 5.1.1. Frequency (of sound detected by the listener (observer)) (1)

5.1.2. The apparent change in frequency or pitch of sound (detected (by a listener) because the sound source and the listener have different velocities relative to the medium of sound propagation. (2)

5.1.3. Away Detected frequency of source decreases (2)

5.1.4. OPTION 1 EXPERIMENT 2



874 = 𝑣 ± 0

𝑣+10(900)

𝑣 = 336.15 𝑚 ∙ 𝑠−1 (323.33 – 336.15) (5)


21

EXPERIMENT 3



850 = 𝑣 ± 0

𝑣+20(900)

𝑣 = 340 𝑚 ∙ 𝑠−1 (313.33 – 340) (5) EXPERIMENT 4



827 = 𝑣 ± 0

𝑣+30(900)

𝑣 = 339,86 𝑚 ∙ 𝑠−1 (339,86 – 345) (5) OPTION 2 EXPERIMENT 2 AND 3



874(𝑣+10)

𝑣 =

850(𝑣+20)

𝑣 both frequencies

874𝑉 + 8740 = 850𝑉 + 1700 𝑣 = 344,17 𝑚 ∙ 𝑠−1

EXPERIMENT 2 AND 4



874(𝑣+10)

𝑣 =

827(𝑣+30)


874𝑉 + 8740 = 827𝑉 + 24810 𝑣 = 341,91 𝑚 ∙ 𝑠−1 EXPERIMENT 3 AND 4



850(𝑣+20)

𝑣 =

827(𝑣+30)


850𝑉 + 1700 = 827𝑉 + 24810 𝑣 = 339,57𝑚 ∙ 𝑠−1

5.2. Away from the Earth (1)

[11] Question 6

6.1. 𝑣 = 𝑓𝜆

𝑣 = (222 × 103)(1,5 × 10−3) 𝑣 = 333𝑚 ∙ 𝑠−1

(3)

6.2. 6.2.1. Towards the bat (1)


22

6.2.2. POSITIVE MARKING FROM QUESTION 6.1



230,3 = 333± 0

333− 𝑣𝑠 (222)

𝑣𝑠 = 12 𝑚 ∙ 𝑠−1 (6)

[10]

secondary school improvement programme (ssip) 2019

Documents