Physics – Marking Scheme – Form 4 Secondary – Track 3 – 2016 Page 1 of 4

DIRECTORATE FOR QUALITY AND STANDARDS IN EDUCATION

Department of Curriculum Management

Educational Assessment Unit

Annual Examinations for Secondary Schools 2016

FORM 4 PHYSICS MARKING SCHEME

SECTION A 40 MARKS

Question Answer Mark Additional Guidelines

1 a

Correct labelling of: a compression and a rarefaction.

1 1

1 b

one complete wavelength marked on the diagram with the symbol 𝜆

1

1 c

5 1

1 d v =

s

t=

16.5

0.05= 330 m/s 2

Deduct 1 mark for missing or incorrect unit.

1 e

𝜆 =16.5

5= 3.30 m

𝑣 = 𝑓𝜆 → 𝑓 =𝑣

𝜆=

330

3.30= 100 Hz

1

1

2 a

correct drawing of ray that bends away from the normal as it exits the water; correct drawing of normal;

1 1

Formation of image is not expected.

2 b increases; virtual;

1 1

2 c

Light changes speed as it emerges from the water into air.

2

Accept also ‘Ray is travelling from a dense to a less dense medium’.

2 d

𝑛 =𝑟𝑒𝑎𝑙 𝑑𝑒𝑝𝑡ℎ

𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑑𝑒𝑝𝑡ℎ

→ 𝑟𝑒𝑎𝑙 𝑑𝑒𝑝𝑡ℎ = 𝑛 × 𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑑𝑒𝑝𝑡ℎ → 𝑟𝑒𝑎𝑙 𝑑𝑒𝑝𝑡ℎ = 1.33 × 1.5 = 1.995 m

1 1

3 a No; microwaves are not in the visible part of the electromagnetic spectrum

1

1

3 b All electromagnetic waves do not require a medium to travel

1 Accept any other correct answer.

3 c 18.0 m/s 1

3 d

1 s → 18.0 m 3600 s → 64800 m 1 hr → 64.8 km Yes as he is moving at 64.8 km/hr

1

1

Accept also ‘yes, he is moving with 18m/s, which is more than the speed limit of 16.67m/s’

3 e

𝑐 = 𝑓𝜆 10700MHz = 10700 × 106 Hz

𝜆 =300000000

10700000000= 0.028 m

1

1

Track 3

Page 2 of 4 Physics – Marking Scheme – Form 4 Secondary – Track 3 – 2016

4 a

𝑎 = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡

𝑎 =Δ𝑦

Δ𝑥=

𝑣 − 𝑢

𝑡=

2 − 12

5 − 0= −2 m/s2

𝑑𝑒𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 2 m/s2

2

4 b 𝑠 = 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑔𝑟𝑎𝑝ℎ

𝑠 = 2 × 10 +1

2× 10 × 5 = 45 m

2

4 c

𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 =𝑠

𝑡=

45

10= 4.5 m/s

1

1

4 d

𝑠 =(𝑢 + 𝑣)𝑡

2

200 = (2 + 18

2) 𝑡

200 = 10𝑡 𝑡 = 20 s

1

1

5 a

2

5 b

𝑉 = 𝐼𝑅

𝐼 =𝑉

𝑅=

1.5

15= 0.10 𝐴

2

Deduct 1 mark for missing or incorrect unit.

5 c

1

𝑅𝑡𝑜𝑡𝑎𝑙=

1

𝑅1+

1

𝑅2

1

𝑅𝑡𝑜𝑡𝑎𝑙=

1

15+

1

30=

1

10

𝑅𝑡𝑜𝑡𝑎𝑙 = 10 Ω

1

1

5 d stay the same 1

5 e

due to the heat generated in the filament, its resistance increases with increasing voltage across the lamp

1

Accept ‘Because I and V are not directly proportional’. Do not accept: ‘As it gets hot’.

SECTION B 45 MARKS

Question Answer Mark Additional Guidelines

6 a

ammeter; series

1 1

6 b variable resistor or rheostat 1

A

V

Physics – Marking Scheme – Form 4 Secondary – Track 3 – 2016 Page 3 of 4

6 c tight connections 1 Accept any other correct precaution.

6 d

Set up apparatus with the rheostat (set at maximum setting) connected in series; Record and tabulate the voltage and current readings as the resistance of the rheostat is changed. Plot a graph of I against V

1

1 1

6 e

Graph: has correctly labelled and positioned

axes; is drawn over at least 75% of graph

paper; has the correct title; is correctly plotted; points are joined by a smooth curve

5

6 f 0.9 ± 0.025 A 1

6 g No; the IV graph is not a straight line passing through the origin

1 Do not award mark if the correct reasoning is missing.

6 h the resistance of the thermistor decreases with increasing voltage

1

7 a i

Newton’s first law of motion; An object at rest remains at rest while a moving object continues to move with constant velocity unless an external resultant force acts on it.

1

1

7 a ii Crumple zones or Air bags

1 Accept any other correct safety measure. Do not accept ‘seat belt’.

7 a iii 𝑎 =𝑣 − 𝑢

𝑡=

0 − 15

1.2= −12.5 m/s2 2

Deduct 1 mark for missing or incorrect unit.

7 a iv 𝐹 = 𝑚𝑎 = 900 × −12.5 𝐹 = −11250 N 𝐹𝑜𝑟𝑐𝑒 𝑜𝑓 𝐼𝑚𝑝𝑎𝑐𝑡 = 11250 N

2

7 a v 𝑇ℎ𝑖𝑛𝑘𝑖𝑛𝑔 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 15 × 0.5 = 7.5 m 𝐵𝑟𝑎𝑘𝑖𝑛𝑔 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 100 − 7.5 = 92.5 m

1 1

7 a vi 𝑣2 = 𝑢2 + 2𝑎𝑠 0 = 152 + 2𝑎 × 92.5 𝑎 = −1.22 m/s2

1 1

7 b i 𝑀𝑜𝑚𝑡𝑜𝑡𝑎𝑙 = (10000 × 4) + (30000 × 0) Deduct 1 mark for missing or

0

0.5

1

1.5

2

2.5

3

0 2 4 6 8

Cu

rren

t I /

A

Voltage V / V

Graph of I /A vs V /V

Page 4 of 4 Physics – Marking Scheme – Form 4 Secondary – Track 3 – 2016

𝑀𝑜𝑚𝑡𝑜𝑡𝑎𝑙 = 40000 kg m/s 2 incorrect unit.

7 b ii

𝑀𝑜𝑚𝑎𝑓𝑡𝑒𝑟 = 40000 kgm

s

𝑀𝑜𝑚𝑎𝑓𝑡𝑒𝑟 = (40000 × 𝑣)

40000 = 40000 × 𝑣 𝑣 = 1 m/s

1

1

8 a

Convex lens is positioned to face a distant object. Convex lens is moved to produce a sharp image on the screen; The distance between the lens and the screen is measured. This distance is roughly equal to the focal length of the lens.

1

1

1

8 b i

2

8 b ii 4 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 → 8 cm

1

8 b iii real; inverted

1 1

Accept any other correct property of the image.

8 b iv 𝑀 =𝐻𝐼

𝐻𝑂=

4

8= 0.5 2

8 c i

2 Award 1 mark for each complete ray and its respective label.

8 c ii Dispersion 1 8 c iii Rainbow 1 8 c iv Red 1

Please Note: When marking questions that involve calculations, apply the ‘follow

through’ rule. This means that if a student gives a wrong value for part (a) of a

question and then uses the value of (a) in the subsequent calculations, marks should be

deducted for part (a) only. The subsequent parts should be given full marks if these are

correct.

Object

Lens

Screen

White LightGlass Prism

Screen

F

Red

Blue