s Kema Pahang

7/30/2019 s Kema Pahang

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Answers For Physics Paper 1 Trial 2010

1. C 6. A 11.B 16. A 21. C 26. C 31. D 36. A 41. C 46. A

2. B 7. B 12.D 17.B 22. C 27, D 32. C 37. C 42. A 47. A

3. B 8. A 13.B 18 A 23. A 28. D 33. B 38. C 43. C 48. B

4. D 9. C 14. A 19 A 24.C 29. C 34. B 39. C 44. A 49. B

5. D 10.B 15. C 20. D 25.C 30. B 35. B 40. C 45. D 50. B


2/15

Pahang SPM Trial Exam 2010

Marking Scheme Physics Paper 2

Section A

Question 1 Answer Note

(a) 1 0.2 s With correct unit(b) 1 - 0.4 s With correct unit

(c) 2 50.0s (0.4 s) =50.4 s With correct unit

Total : 4 marks


(a) 1 e.m.f induced/ current induced

(b) 1 North

(c)(i)

(ii)

(iii)

1

1

1

Total: 5 marks


(a) 1 Force that oppose acted force forward

(b)(i) 1 W= mgh = 430 N With correct unit

(b)(i) 1

1

Resultant force, F =W sin 20 147.1 N

= 147.1 N 147.1 N

= 0 N With correct unit

(c) 1

1

Stationary

Because balance force/ Force in equilibrium/

acted force = frictional force

Total: 6 marks

Galvanometer

S

CompassKompas


3/15


(a)(i) 1 Thermionic emission

(a)(ii) 1 To accelerate electrons

(b) 1 Kinetic energy heat energy + light energy

(c)(i) 1 Alternating current/ AC

(c)(ii) 1

1Period, T = 3 0.02s = 0.06 s

F=1/T = 1/0.06s = 16.67 Hz

With correct unit

(d) 1

Total: 7 marks

Question 5 Answer Note(a) 1 Atmospheric pressure/ Air pressure

(b)(i) 1 The volume of air trapped in the beaker diagram 5.1 is larger Quantities must be

same as stated in

questions.(b)(ii) 1 The pressure of air trapped in the beaker diagram 5.2 is bigger

(c)(i) 1 The higher the pressure, the lower the volume of air trapped

(c)(ii) 1 Boyles law

(d) 1

1

1

Density of air > water

Water exerts upthrust

Upthrust > Weight of beaker + air trapped

Total : 8 marks

0.01

s / div

4

V / div


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(a) 1 Sources that have the same frequency and in phase.

(b)(i) 1 The wavelength of water wave in Diagram 6.1 is shorter

(b)(ii) 1 The distance between two consecutive nodal lines,x in Diagram6.1 is shorter

(c)(i) 1 The shorter the wavelength of water wave, the shorter the distance

between two consecutive nadal lines,x.

(c)(ii) 1 Interference

(d)(i) 2

(d)(ii) 1 A: Bright

B: Dark/Dim

Total : 8


(a) 1 Elastic potential energy

(b)(i) 1 X= 4.0 cm With correct unit(b)(ii) 1

1

F = kx

k= 0.9 N cm-1 @ 90.0 N m

With correct unit-1

(b)(iii)

1

1

cmx

x

x

F

x

F

56.5

5

4

6.3

2

2

2

2

1

1

=

==

=

l = 12 5.56 = 6.44 cm With correct unit

(c)(i) 1 Compression of the spring is directly proportional to the load in the

balance/ The higher the weight of load, the higher the compression(c)(ii) 1 The spring will not return to its original shape// spoil

(d) 2 Use a larger diameter of spring wire

Add more springs in parallel

Use stiffer spring/ any two

Total: 10 marks

A B


5/15


(a)(i) 1

1

15 hours

Short time taken to decay/Can decay faster and they leave

harmless daughter nuclei

(a)(ii) 1

1

Gamma ray

Strong penetration power/ Can penetrate the soil

(a)(iii) 1 Liquid

Easier to dissolve in water/ Produce more even radiations.

(b)(i) 1

1

8/8 4/8 2/8 1/8

T1/2 T1/2 T1/2

t = n (T1/2

With correct unit

)= 3 (28 years) = 84 years

* 2nd method:

Undecay value =

n

2

1Original value

8

8

2

1

8

1

=

n

2n = 8 n = 3

t = 3 28 years = 84 years

(c) 1 Thickness control/Examine contamination in canned food/

Medical screening/treatment// smoke detector/ sterilizing/

(d)(i) 1

1

Mass defect = (2.014012u + 3.016029u) (4.0022603u +

1.008665u)

= 0.018863 u = 0.018863 1.66 10-27 kg

= 3.13 10-29

With correct unitkg

(d)(ii)

1

E= mc2

= 3.13 10-29

kg (3.0 108ms-1)

2

= 2.82 10-12

With correct unitJ

Total: 12 marks


6/15

Section B


(a)(i) 1 Refraction

(a)(ii) 1 Ratio of sin i / sin r // The ratio of speed of light in vacuumrelative to that speed through a medium

(b) 1

1

1

1

1

Refractive index of the glass is higher.

The density of glass is higher

The angle of refraction of light ray in glass is shorter

The higher the density of medium, the smaller the angle of

refraction of light.

The higher the density of medium, the higher the refractive

index.

(c)

- He must shoot the target at the lower position of the image.

1- m

Light refractedaway from normal

1 m

Extrapolation to

show position of

the observing

image 1 m

(d)(i)

Eyepiece

Total internal

reflection

4545

45 45

Objective lens

Total internal

reflection


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Distribution of marks:

1 mark - Labeled 90prism

1 mark - Arrangement of prisms --- facing each other

1 mark - Location of objective lens

1 mark - Location of the eyepiece lens

1 mark - Light ray with 2 times total internal reflection at the 1 stprism

1 mark - Light ray with 2 times total internal reflection at the 2nd

prism

(d)(ii)

2

2

Modifications Reasons

Objective lens with largerdiameter.

More light passes throughthe lens

Eyepiece lens with higherpower // Thicker eyepiece

lens

Act as a strong magnifyingglass

Total : 20 marks

Question 10 Answer Note(a) 1 A wave in which the vibration of particles in the medium is

parallel to the direction of propagation of the wave

(b)(i) 1 The amplitude in Diagram 10.2 is higher

(b)(ii) 1 The peak value, a2 in Diagram 10.2 is higher

(b)(iii) 1 The higher the amplitude of vibration of tuning forks, the higher

the peak value

(b)(iv) 1 The higher the peak value, the louder the sound

(b)(v) 1 The higher the amplitude, the louder the sound

(c) 1

11

1

1

- Use ultrasound

- Ultrasound is transmitted to the sea bed- a receiver will then detect the reflected the reflected pulses

- the time taken by the pulse to travel to the seabed and return tothe receiver being recorded, t

- the depth of the sea can be calculated using the formula,2

vtd =

Max 4 marks


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(d)

1

1

1

1

1

1

1

1

11

1

1

Suggestions Reason

Loudspeakers are

positioned at quite a

distance away.

So that the distance between

consecutive constructive /

destructive interference is

smaller.

The two main

loudspeakers are not

positioned opposite to

each other

To prevent multiple

reflections

Fix soft boards/

wooden/ materials

which are sound

absorbers

Reflection effects can be

reduced

Use thick carpet/

Wooden floor/ Rubber

floor

To prevent echo

Assemble a high power

speaker system

To produce a high amplitude

of sound wave

Assemble the speaker at

a highplace

Wide coverage // the wave is

not blocked

Max 10 marks

Total : 20


9/15

Section C


(a)(i) 1 Archimedes principle states that the buoyant force on an

object immersed in a fluid is equal to the weight of fluid

displaced by the object.(a)(ii) 1

1

1

1

- Volume of air displaced equal to volume of a balloon

- Density of air decreased as a altitude increase

- Weight of displaced air become smaller

- At certain height weight of displaced air equal to weight of theballoon

(b)

1

1

11

1

11

1

1

1

Characteristics Explaination

Large ballon To produce bigger buoyant/ upthrust // Increase the

volume of the air displaced

Use 2 burners // Many

burners

To produce bigger flame //

heat up the gas in theballoon faster

Synthetic nylon Light-weight/ strong /air-

proof material

High temperature of the air

in the balloonReduce the density /weight

of the air in the balloon

Hot air balloon Q is chosen Large balloon, use 2

burners / many burners,use synthetic nylon and

has high temperature of the

air in the balloon.

(c)(i) 1

1

Mass = density x volume

Mass = 0.169 kg m-3 x 1.2 m3

With correct unit

= 0.20 kg

(c)(ii)1

1

1

Calculate mass of displaced air correctlym = 1.3 kg m-3 x 1.2 m3

With correct unit

=1.56kg

Calculate weight of displaced air correctly and state that

bouyant force equal to weight of displaced airWeight of displaced air = bouyant force

= mg = 1.56 x 10= 15.6N

Total : 20 marks


10/15


(a) 1 Afuse is a very thin wire, which either melts or vaporizes when

too much current flows through it

(b) 1

1

1

1

- A parallel circuit can run several devices using the full voltage

of the supply.

- If one device fails, the others will continue running normally

- If the device shorts, the other devices will receive no voltage,

preventing overload damage.- A failure of one component does not lead to the failure of the

other components.

- More components may be added in parallel without the need

for more voltage.

- Each electrical appliance in the circuit has it own switch.

Max 4 marks

(c)(i) 1 - The electrical appliance use 240 V of voltage to generates 500

W of power.

(c)(ii) 1

1

Current = Power/Voltage

Current = 500/240 = 2.08 A

With correct unit

(c)(iii) 1

1

Efficiency = Output Power x 100 %

Input Power

Output Power = 85

With correct unit

x 500100

Output power = 425 W

(d)

Characteristics Explanation

Thin fuse wire Less space needed/ to carry a

limited electrical current/ less

mass hence low heat

capacity/ shorter time to heat

up to melting point and blow.

Ceramic cartridge Can withstand highertemperature because sparkscreated by high voltage,240V can be huge/

Fuse rating is 13 A Maximum rating must be

higher than normal current.

Low melting point For fast blow/ Melting faster

when excessive currentflow/ Easy to cut the current

flow.

R is chosen because Because it has thin fuse wire,

ceramic cartridge, fuse ratingis 13 A and low melting

point.

Total : 20 marks
http://www.gcse.com/glos.htm#voltage*http://www.gcse.com/glos.htm#voltage*


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12/15

1

MARKING SCHEME PHYSICS PAPER 3 (TRIAL 2010)

No Answer Marks

1 a) (i) Distance between 2 coherent sources of sound waves, a 1

(ii) Distance between two consecutive instances of loud sound, x 1

(iii) Distance between loudspeakers and where soun is

detected/frequency of sound waves/ wavelength of sound waves 1b) (i)

(ii)

a = 1.0m x = 15.2cm

a = 1.5m x = 10.3 cm

a = 2.0m x = 7.3 cm

a = 2.5 m x = 5.9 cm

a = 3.0 m x = 4.7 cm

a

1/ m x / cm

-1x / m

1.00

0.67

0.500.40

0.33

15.2

10.3

7.35.9

4.7

3.04

2.06

1.461.18

0.94

Tabulate data

1. Shows a table which havea

1and x

2. State the correct unit (a

1: m-1

3. All values x are correctand x : m)

4. Values ofa

1are consistent to 2 decimal point

5. Values of x are consistent to 2 decimal point

All correct 2M

4 correct 1M

3 correct 0M

( accept : 0.1 cm)

1

1

1

1

1 (7)

c)Draw graph x against

a

1

1. The responding variable, x at y axis2. The manipulated variable,

a

1at x axis

3. Sates the unit of variable correctly4. Both axis with the even and uniform scale5. 5 points correctly plotted6. A smooth best fit straight line7. Minimum size (50% of graph paper)

No of ticks Score

7 5

5-6 4

3-4 3

2 2

1 1 5

d) x is inversely propotional to a OR x is directly proportional to 1/a 1

TOTAL 16


13/15

2

No Answer Marks

2 a) Extrapolation on graph to cross y-axis where h=0m

Correct answer with unit1.00 x 10

5Nm

-2

1

1 (2)

b) (i) Large triangle ( 4 x 3 larger square)on graph

Correct substitution (refer to triangle drawn) exp:

07.0

)100.1()106.1(55

=k

Correct answer with correct unitk = 5999.3 Nm

-3

(ii) Method with correct substitution refer to answer in (i)

0.12 (5.999.3)

Correct answer

= 719.92 kgm-3

1

1

1 (3)

1

1 (2)

c) Line on graph at h = 0.5 m

Correct substitution OR final answer with unit (accept the valuehalf of small square )

P = 1.043 x 105 + 1.0 x 105 Nm-2 OR 2.043 x 105 Nm

1

1 (2)-2

d) (i) Increase

(ii) The higher the density, the higher the pressure and produce

the higher gradient of graph P - h

1

1 (2)

e) Describe the method to avoid parallax errorThe eye at the same level of meniscus of water 1

TOTAL 12

3. a) Correct inference refer to actual situation and correct directionThe distance between paper to the lens / focal length is depends on

the thickness of lens. 1

b) Correct hypothesis with correct direction (refer to variables that

choose for the experiment)The higher the thickness of lens, the longer its focal length 1

c) i) Aim of the experimentTo Investigate the relationship between the focal length and

the thickness of lens.

ii) Variables (MV and RV should can be measured)

MV : The thickness of lens

RV : The focal lengthFV : Refractive index of lens / the diameter of lens / type of

lens

iii) List of apparatus (ruler and vernier caliper/micrometer

screw gauge must be in the list)

5 convex lenses of different thickness but samediameter, light box/candle with flame, low voltage

power supply, screen, plasticine, ruler and

micrometer screw gauge/ vernier caliper

1

1

1

1


14/15

3

iv) Arrangement of apparatus (Should be relevant and

functional)

to a.c

screen lens with holder ray box (far from thelens as a distance object)

v) Procedures

1. The thickness of lens is measured using the vernier

caliper/micrometer screw gauge and record.

2. The lens is adjusted until the clear sharp image offilament is obtain on the screen.The distance between

centre of lens to the screen is measured as the focal

length.

3. The experiment is repeated use another four differentthickness of lenses. (accept without values because the

thickness is unknown and measured before the

experiment)

NOTES : ACCEPT OTHERS EXPERIMENT THAT RELATE

TO THE SITUATION SUCH AS

1. relationship between u and v where f is measure usethe formula

fvu

111=+

2. use distance object (outside object) to determine thefocal length of lens

vi) Show the table

Thickness / cm Focal length / cm

vii) State the graph should be drawn or draft the graph

Draw a graph, the focal length against the thickness of lens

OR focal length / cm

Thickness / cm

1

1

1

1

1

1

Total Marks 12

4. a) Correct inference refer to actual situation and correct directionThe brightness of lamp depend on the speed of wheel. 1

b) Correct hypothesis with correct direction (refer to variables that

choose for the experiment)The higher the speed of magnet, the higher the current induced 1

c) i) Aim of the experiment

To Investigate the relationship between the high of

magnet and the current induced. 1


15/15

4

ii) Variables (MV and RV should can be measured)

MV : Speed (of the magnet)

RV : current (induced)

FV : Numbers of turns of the coil / the diameter of coil / the

thickness of wire

iii) List of apparatus (ruler and galvanometer must be in the

list)coil/selonoid made of copper wire, ruler , galvanometer and

connecting wire

iv) Arrangement of apparatus (Should be relevant and

functional)

v)

Procedures1. The apparatus is set up as the diagram shown.

2. The strong magnet is released at 10 cm of height.

3. The division of the galvanometers pointer deflect isrecorded.

4. The experiment is repeated at the height of magnet is

15cm, 20 cm, 25 cm and 30 cm

vi) Show the table

height / cm division

vii) State the graph should be drawn or draft the graph

Draw a graph, the division against the height of magnet

released

OR division

height / cm

1

1

1

1

1

1

1

1

1

Total Marks 12

h

G

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