penang trial spm 2013 physics k2 skema.pdf

7/27/2019 Penang Trial SPM 2013 Physics K2 skema.pdf

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453112

PHYSICS

KERTAS 2MASA 2112 JAM

MAJLIS PENGETUA SEKOLAH MENENGAH(CAWANGAN PULAU PINANG)

MODUL LATIHAN BERFOKUS SPM 2013

FIZIR

Kertas 2

PERATURAN PEMARKA}IAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

Nota :

Skemapemarkahan

ini dalam Bahasa lnggeris sahaja

www. myschoolch ild ren . com

Skema ini mengandungi 8 halaman bercetak

http://edu.joshuatly.com/ http://fb.me/edu.joshuatly



NO. MARKING SCHEME

MARK

SUB TOTAL

1(a) 0.10c 1

4

(b) o 1

(c)

(d)

29.70C 1

Scalar quantity1

,

SECTION A

4s31(PP)

2(al Product of mass and velocity 1

5

(b) mv = 10

v=10/0.05

= 200 ms-1

t

1

1

(c)

(d)

lmpulsive force // Resultant force 1

(Show a line which is less steep)

1

3(a) sinc=11 2.42c = sin-1 2.42

= 24.4'

1

1

6

(b) . (Show 2 total intemal reflections in the diamond)

IIt

\

400/

/

o

20

I,lJ L

Ino" eoo

bo" '..50o

,, 3Oo 50q.

o

80

7

2

(c) Total intemal reflection 1

(d) Periscope // binoculars ll optical fibre 1

4s31(PP) [Lihat halaman sebelah




NO"MARKING SCHEME

MARK

SUB TOTAL

4(al3Y x2

=6V1

1

7

(b) Period=Smsx4=20ms

Frequency = 1120 ms

=50H2

1

1

(c)

(dxi)

Rectiflcation ll half wave retification 1

capasitor 1

(dxii)

t

1

4531(PP)

5(a) Pressure is the (normal) force acting per unit area.1

8

(bxi) The paper strip in Diagram 5.2 is higherthan that in Diagram 5.1

ll 5.1 < 5.21

(bxii)

(bxiii)The speed of the air in Diagram 5.2Diagram 5.1 ll 5.1 < 5.2

is higher than that in1

The air pressure above the paper is lower than that below theDaoer // vice-versa

1

(b)(iv) When the speed of air increases, the air pressure decreases //vice-versa

1

(c) t, 1

(d) Bernoulli's Principle 1

(e) Bunsen bumer ll earburelor llvacuum pump

(or any relevant application)

1

6(aXi) Potential difference across the cell 1

I

(ii) The voltrneter reading in 6.1 > 6.2 116.2 < 6.1 1

(iii)(iv)

The brightness of the bulbs in 6.1 < 6.211 6.2 > 6.1 1

The ammeter reading in 6.1 < 6.2 ll 6.2 > 6.1 1

(b) As the ammeter reading increases the voltmeter reading

decreases // The larger the ammeter reading the smaller thevoltmeter readino

1

(cXi) The voltmeter reading shows the highest reading / increases. 1

(ii) No current flowing ll apen circuit ll no load to the cellVoltmeter shows the e.m.f of the cell llno voltaoe loss due to intemal resistance.

1

1





NO.MARKING SCHEME

MARK

SUB TOTAL

7(aXi) lncreases // becomes higher 1

10

(aXii) Pressure 1

(b)

(c)

Pressure Law 1

Tz = PtJ_= 230 x 300 = 345 K ll 72 "C

Pr 200Mk 1- Conversion of unit from "C to KelvinMk 2- SubstitutionMk 3- Answer with correct unit

1

1

1

(dxi) Widersurface area ll btgger surface area t 1

(dxii) The pressure exerted on the road is reduced 1

(dxiii) Thick tyre thread 1

(dXiv) To have a better grip on the road ,l

4s31(PP)

NO.ANSWER

MARK

SUB TOTAL

8(a) lsotopes that are not stable /l emit radioactive emissions1

12

(b) D

The reading of the ratemeter is the highest1

1

(cXi)1',f n, '??an + lH" 2

(c)(ii) 100o/o + 50o/o + 25o/o -+ 12.5o/o

Sodium-24 : 3T1p = 45 hoursTrn

=15 hours

Cobalt-60 : T,tn = 15.9/3 =Radium-226 : Trn = 4860/3 =

5.3 years

1620 years

,l

1

1

1

(d) Sodium-24

Half life is short // activity decreases faster

Emits beta rays ll can penetrate ground but cannot penetrate

pipe

1

1

1





5

SECTION B

4s31(PP)

QUESTION MARKING SCHEMESUB

MARK TOTAL

I (a) The quantity of heat needed to increase the temperature of 1kg of

substance by 1oC. 1

1

(b) i) The reading of thermometer in Diagram 9.1 is greater than in

Diagram 9.2 ll 9.2 < 9.1 1

5

ii) The specific heat capacity of the metal block in Diagram 9.2 is

greater than in Diagram 9.1 ll 9.1 < 9.21

iii) The amount of heat supplied is the same1

iv) The lower the specific heat capacity, the higher the reading of the

thermometer ll vice-versa 1

v) Different metals / materials have different specific heat capacities.1

(c) The body temperature is higher than the temperature of the cooling

pad 1

4here is a net heat flow from the body to the cooling gel until

thermal equilibrium achieved.1

The body heat energy is decreased1

ln thermal equilibrium, there is no net heat flow from the body to the

cooling gel llthe temperatures are the same.1

(d)MODIFICATION EXPLANATION

H:lArarse number oron{

,

lncrease the surface area I .t

releases heat faster { I 2

10

Big fan,/

, ,T.:1 :J:'

of air can be,/,2

Cooling liquid with high Oorrry

,oint

Will not hil easily ,/vt 2

Cooling liquid with high specflc ,heat capacity

Can store more heat ll t/Sloryer increase in temperature

2

Dark radiator tank ll black,/

,colour

Absorbs heat easily ll ,/Good absorber of heat { 2

TOTAL 20

4531(PP) [Lihat halaman sebelah




MARKING SCHEME

10 (a) The maximum displacement from the equilibrium position.

10 max

Size of the tuning fork in Diagram 9.1 is greater / thicker than in

Diagram 9.2 ll 9.2 < 9.1

Amplitude of sound waves in Diagram 9.1 is greater than in

Diagram 9.2 ll 9.2 < 9.1

Frequency of sound waves are the same.

Amplitude is directly proportional to the loudness // When theamplitude increases, the loudness of the sound increases.

Amplitude is directly proportional to the size of the tuningfork llWhen the size of the tuning fork increases, the amplitudeincreases.

aeroplane is equipped with a device / microphone thatcan detect sound

the sound is sent to the computercomputer generates waves thatproduce destructive interference

1)

2)

3)4)

1+1

1+1

1+1

1+1

1+1

1+1

2) sound waves will be heardclearly

1) speakers must be placed

facing the audience

4) allow interference to

occur ll able to produce

loud sounds

3) use a minimum of 2 loud

speakers

6) interference occurs) sound output is loud

8) able to absorb the sound/ the sound waves

7) the floor is carpeted

10) reduce reflection of soundwaves

9) use wooden floor

12) reduce reflection of soundwaves

11) wall is padded //use curtains II softboard

// cushions

Any 5 characterisfics &5 anslvers

Accept any relevant answers

4531(PP)





MARKING SCHEME

11(a) Rate of change in momentum // lmpulsive force o I/(time ofimpact),Fs1ll

(i) Helmet - to protect the head by increasing the time oi irnpact(ii) Jacket - to protect the body from injuries, cuts, burn due to fall(iii)Gloves - to protect the hand from injuries/force(iv) Helmet, jacket and gloves increase the time of impact and

reduces impulsive force

1 80 x1 000

E0L6o-

50.0 ms-1

v-u=-

t50-0

=10= 5.0 ms-2

F=ma= (200x5.0)

= 1000 N

M9 - Motocycle S is chosenM10 - because it has fewer number of coils, wider tyres,

smaller mass, lowest seat height

Higher spring constant ll lower naturalewer / Lower no. of

Wide / Bigger width oftyres

Bigger surface area, bettersupport ll low pressure

acts on the tyres / more friction

Smaller mass

Lower centre of gravity /more stable / safer when

7

SECTION C

4531(PP)





MARKING SCHEME

12 (a) The ratio of potential difference to current.

Length // thickness/cross sectional area ll lype of materialb) (i)

Wire P

. Since I is constanl lsarne for wire P and wire Qo V = lR ll V is directly proportional to R //

the higher the V, the higher the R

M9 - Metal R is chosen

Ml0 - because it has low melting point, low specific heat

capacity, smal! wire diameter and high resistivity.

1+1

1+1

1+1

1+1

Can rnelt easily.

Needs a small amount of heat to

increase its temperature fi can be

heated up easily

Low specific heat

capacity

lncrease the resistance

Produces a wire of high resistance//

wire can convert electrical energy to

heat easily.

High resistivity

Correction substitutionR = V2lP

= Q4O)2

60

Correct answer with unit

= 960C)

f=PA/ =601240R=Vll=240x240 160

= 960Ct

ChooseV= 120VP = V2lR

Correction substitutionP = fi20)2

960

Correct answer with unit

= 15 W lt 15 Js-1

l=V/R= 12A196O

= 0.125 A

P=l2R= (0.125)2(960)

= 15 W tl 15 Js'1

4531(PP)

453UPP) [Lihat halaman sebelah


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