stpm trials 2009 physics answer scheme (terengganu)w

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JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR



JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR

JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR



JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR





JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR


JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR

MARK SCHEME

PAPER 1 AND 2

PHYSICS

960

JABATAN PELAJARAN

TERENGGANU

960/1/2 TRIAL

2009

more examination papers at :

www.myschoolchildren.com


2/18

2

ANSWER PAPER 1

Question Answer Explanation

1 AStrain= 0/ le is dimensionless

[stress] = F/A= [pressure]

2 B

During the time from 0 to T, the body slides along the horizontal surface withzero resultant vertical force since its weight is balanced by the reaction forcefrom the surface. After time T, the resultant force acting on the ball is its ownweight W under free fall situation.

3 B

m

F

mm

Fa

32=

+=

To block X alone, )3

)(2(m

FmmaFF Y ==

Hence, YF = force exerted by block Y on block X= 3

2F

4 B

Kinetic energy of sphere =2

2

1mv = constant since v is constant.

Potential energy of sphere = mgvdt

dxmg =

v x

5 Br

vw= is constant because v and r are constant. Hence angular acceleration = 0

6 C

The force constant of the elastic string is140 Nm . Hence, the force exerted by

the string on the mass to keep it in circular motion is

NkxF 8)50.070.0(40 === If v is the linear velocity of the mass, it is related to this centripetal force by

r

vmF

2

=

7.0)05.0(8

2v

= 1

1158.10

== msv




3/18

3

7 C

T

F

W

If T is the tension in the light thread, then

FT =sin

WT =cos W

F=tan

8 A

22

2

1

2

1Imvmgh += rwv=

2

22

+= rvImvmgh

=21 v

rm

+

2

2

4

05.0

100.62.0)2.0)(10)(2.0(2 v

+=

135.1 = msv

9 D

From

r

MmGUEP ==

PE is the gravitational potential energy of a body of mass mat distance rfrom

the centre of a planet of massM

Force of attraction =2r

MmG

dr

dU+=

Thus, the gradient at any point on the gravitational potential energy curverepresent the force pulling the body towards the planet.

10 C

22

1

RW

R

GMmW =

Gravitational force on the satellite when at the highest R/50

50

RR + from the centre of the earth = W

RR

R2

2

50

+




4/18

4

= W

R

R2

2

2

50

11

+

= 0.96 W

11 A 2222222 2)2(2

1

2

1fmafmamaE ===

12 D

13 C

The frequencyis determined by the source and not the medium through which it

flows. In a denser medium, the velocity will decrease and as a result thewavelength will decrease.

14 CV = f

(100 x 106) = 3 x 108 = 3 m

l = / 2 = 2/3= 1.5 m

15 B

16 B

Work = area under the graph

=310)]59)(8060(

2

1)560(

2

1[ ++

= [(150) + (704) x310 ]

= 430 x310

= 0.43 J

17 A

k =e

F

E = eA

Fl

k =l

EA

18 D

19 C = 2

1

V

VW Vp

l

Dipole aerial




5/18

5

= 2

1

V

VV

V

nRT )( nRTpV=

= nRT1

2lnV

V

=4

6ln)400)(31.8(3 = 4043.30 J = 4kJ

20 D

Since V = 0,W = p V = 0,

Q = U + W

U = Q = mcvV= 5.0 x 10-3x 2.4 x 102x (600 300) J

= 360 J

21 A

pV= k

p = kV

ln p = ln k ln V

= - lnV + ln k

(gradient = - )

22 C

23 A

24 BVqw =

since 00 == wV

25 B

Initial energy

( ) ( )J

CQ

CQ

02.0103

10300

104

10200

2

1

21

21

6

26

6

26

2

2

2

1

2

1

=

+

=

+

Equivalent capacitance

( ) ( ) FCCC

666

21

107103104 =+=

+=

Final energy

JC

Q018.0

107

10500

2

1

2

16

62

=

=

Energy lost,

J3

100.2018.002.0 =




6/18

6

26 B

The resistance of a copper wire remains unchanged at constant temperature and

is given byA

lR =

Potential difference, V

( ) venvAeAl

IRV ln =

==

Since I is directly proportional to drift velocity of electrons in the wire, thusdirectly proportional to potential difference.

27 AWhen temperature increases, electrons are freed from covalent bonds. Hence the

number of electron-holes increases.

28 B

Kirchhoffs first law is related to current and depends on the conservation ofcharge

Kirchhoffs second law is related to electric potentials and depends on theconservation of energy.

29 D( )( )

A

t

QI

ItQ

4

15

19

102.3

100.1

106.12

=

==

=

30 C

31 A

32 B

Induced e.m.f is given by BAN= where

N number of turnsA areaB flux density

angular velocity

33 A

34 BP is inductor, wLXL =

Q is resistor

35 A

36D

37B r = 20cm, f = -10 cm, u = +5 cm




7/18

7

cmv

v

vuf

3.3

5

1

10

11

111

=

=

+=

38 A

39C From the formula of fringe separation

a

Dy

= , Dy

40

C

From Einsteins equation,

cf

WhfmvK

=

== 2maxmax2

1

Iffincreases or decreases,Kmaxor vmaxincreases

41

D

mf

c

h

Kcf

hfKhf

oo

o

o

7

15

34

19

9

8max

max

1002.8

10374.01063.6

106.10.2

10350

100.3

==

=

==

+=

42

C

43

D

mE

hc

hcE

JEEE

6

19

834

1919

2214

1046.11036.1

100.31063.6

1036.1)106.1(1

1

4

1)6.13(

=

==

=

=

==

44

A

The K characteristic line is produced by electronic transition in the target atom.The intensity is determined by the number of electrons hitting the target and thenumber of collisions determines the number of electronic transition.

45 D The charge of an electron can still be determined if its mass is not given




8/18

8

46 BM.massandqcharge

ondependnotdoesitbecauseconstantB

Ev,

1

1

alwaysspeed

qEqvB

=

=

47 B

hours

e

eNN t

100693.0

2ln2lnlifehalf

0693.0

204ln

4010 )1030(0

===

=

=

=

=

48 C

0 min 4.0 min 8.0 min 12.0 min

16 mg 8 mg 4 mg+ 8 mg6 mg

49 DNuclear fission :

nHeHH 103

2

2

1

2

1 ++

50 BA strong nuclear force is experienced by nucleons such as protons and neutrons

in the nucleus and are of very short range such as 1.7 x 10 -15m




9/18

9

Paper 2

Question Answer Mark

Structure 1

(a)

Total change in potential energy of water in the buckets in one revolution of the wheel

= hmg8 = )26.1(81.9408

= J4100.1

1

1

(b)

Average input power to wheel = number of revolutions made per unit time X change inpotential energy of the water in the buckets per revolution of the wheel

=4100.1

60

6

= W3100.1

=1kW

1

1

(c)A large number of a small buckets is preferred because the rotation of the wheel would

be smoother than the case would be when a smaller number of large buckets is used. 1

Structure 2

2(a) resonance 1

2(b) From graph, fresonance= 12.5Hz = 2f= 2 (12.5)

=78.5 rads-1

1

1

2(c)

1

2(d) Microwave oven/radio signal receiver

1

Structure 3

(a) dx

dkA

dt

dQ =

1

1

masswith card

mass without

card




10/18

10

= (380)(2.5 x10-4)21020

)20130(

= 52.25 W

1

(b)

The temperature at 15.0 cm = ( )1020

20130)(1015

2

2

= 82.5 oC

The temperature 15.0 cm from the hot end = 130oC 82.5

oC

= 47.5oC

1

1

Structure 4

(a)velocity is a vector quantity. The velocities cancel out each other in any directions,since the number of free electrons is very large. So mean random velocity is zero.

1

(b-i)

Power,P

W

IVP

2.1

)0.5)(24.0(

=

==

1

1

(b-ii)Power,

22

IPRIP =

Since, nevI= , therefore 2vP

Hence, when power increases drift velocity increases.

11

Structure 5

(a) Adder operational amplifier 1

(b)

V

VR

RVR

RV

f

i

f

60.11

2.010

3305.0

33

330

22

10

=

+

=

+

=

1+1

1

(c) Output voltage will become V9 only as saturation occurs. 1

Structure 6

(a)( )

=

21

111

1

rrn

f

wheref = focal length,

n = refractive index of the material of the lens,r1= radius of curvature of the front surface receiving the incoming rays,

r2= radius of curvature of the hind surface where the outgoing rays emerges.

[Rubric: Formula 1 mark; Defining symbols 1 mark]

1

1

(b)In water applying the formula

=

211

2 1111

rrn

n

f

mf

cmf

f

25.1

125

20

1

60

11

33.1

65.11

=

=

+

+

=

1(formula)

1(sign of

r)

1(answer)




11/18

11

Structure 7

(a) Using Einsteins equation, maximum kinetic energy is

15

31

19

max

19

199

834

2max

1023.6

1011.9

)1077.1(2

1077.1

)106.1(30.210365

)1000.3)(1063.6(

2

1

=

=

=

=

=

sm

v

J

Whc

mv

1

1

1

(b)Momentum of the photoelectrons,

hmv=

m

mv

h

9

531

34

1017.1

)1023.6)(1011.9(

1063.6

=

=

=

1

1

Structure 8

8

min4.240.0284

2ln2lnTlifeHalf

0.0284

182810-3

5ln

e

e

12000

20000:

)........(12000).......(20000

dt

dN

21

28-

10-

)28(

0

)10(

0

0

0

===

=

=+=

=

==

=

=

ii

i

iieNieN

eNrateCount

eNN

t

t

11

1

1

1




12/18

12

Essay

Essay 9

(a) The moment of a force, F is the force multiplied by the perpendicular distance, d from

the point about which the moment is being measured example the axis of rotation

d

F

1

1

(b) For a body to be in a equilibrium, there must be no resultant force and no resultanttorque

1

(c)(i)

S

F

BF 025

Weight of section S, W= N5100.3

All

correct2

or

Two

correct

1

(c)(ii) Resolving forces vertically

=0

25sinSF W for equilibrium

NW

Fs5

0

5

01010.7

25sin

100.3

25sin=

==

Resolving forces horizontally,050 25cos1010.725cos == SB FF

= N51043.6

1

1

(d)(i)

F/N

e/cm7 13

Draw

graph

correct

1

Area

shaded

1

(d)(ii)(a)Loss of elastic potential energy of the spring =

2

1

2

22

1

2

1kxkx

1

Axis of rotation




13/18

13

= )07.013.0(20002

1 22

= 12 J

1

(d)(ii)(b) Kinetic energy of the disc = Energy lost by the spring

1221 2 =mv

080.0

212=v

Initial speed =13.17 ms

1

1

(d)(iii)(c) Gain in gravitational potential energy of the disc = lose in kinetic energy

2

2

1mvmgh =

81.9080.0

12

=h

=15.3 m

1

1

Essay 10

(a)

Progressive wave Stationary wave

1. Wave profile moves

2. Adjacent particles of a

medium vibrate in a differentphases

3. The amplitude in constant for

all particles of the medium

- Wave profiles does not move

- Particles between two adjacent

nodes of a medium vibrate in thesame phase

- Particles between two adjacent

nodes vibrate with different

amplitudes

1

1

1

(b)

Thick curtains will absorb the sound.

This will reduce echo and interference of sound waves in the hall and hence theaudience will be able to hear a performance clearly.

1

1

c(i)

Conditions:- the frequency of the two sound must be almost the same

- the amplitudes from the two sound sources must be the same or almost the

same- waves from the two sound sources must be propagated in the same direction

1

1

or 1

c(ii)Frequency of beats = 894 890 = 2 HzFrequency of resultant wave = (894 + 890) 2 = 892 Hz

1

1

c(iii)Intensity level = log10

10

0

I

I= 10 log10

12

9

100.1

103.6

= 38.0 dB

1

1

c(iv)

=

2

1

21 log10I

I 1




14/18

14

=

2

1log100.980.38I

I

6log2

1 =

I

I

6

2

1 10=

I

I

1

d

=

=

4.47343

343418'

svv

vff

= 485 Hz

1

1

Essay 11

(a) Hookes Law : States that the extension is directly proportional to the stress

(force) applied in an object,if the elastic limit is not exceeded.

2

(b) Elastic Deformation Plastic Deformation

Wire can return to original shape &

size when the stress has beenremoved

Wire does not return to its original

shape and size when the stress hasbeen removed.

(Permanent deformation occurs)

3

(c) (i) Brittle

1

(ii) from graph : m10x1.0x -3= for N250F= ( )( )

( ) ( )Pa10x1.41

10x0.110x75.0

0.1250

Ax

FLE

11

323===

3

(iii) Energy = area under the curve

J26=

3

(iv) at P :( )

Pa10x1.9810x0.75

350

A

Fstress 8

23-===

3




15/18

15

Essay 12

(a)

When charge, qis moving, there will be a current,Iwill produce. By using Fleming lefthand rule for charge, q moving in a uniform magnetic field, B a force, F will be exerted

on the charge, q.

B, I and F must perpendicular to each others. Thus q will move in a circular path.

An expression for the force,Fexerted on the charged particle is given by,).(sinBqvF=

where

F is the force exerted on the chargeB is the magnetic fieldq for chargev for velocity

angle between v and B

1

1

1

1

(b-i)

Centripetal force = Magnetic force

m

Bew

m

Berrwv

m

Berv

Bevr

mv

=

==

=

=2

1

1

(b-ii)

mBe

Tw == 2

The period,

s

meBe

m

BBe

mT

9

112

1078.1

1076.1

1

100.2

2

1222

=

=

=

==

1

1

(b-iii)

From equationBe

mT

2= , the period is independent of the velocity and thus independent

of the kinetic energy of the electron.

So the period is still same// equal to s91078.1

1

1

(c-i)




16/18

16

2

(c-ii)

Electrostatic force = magnetic force

16

4

105.7

008.0

106

=

==

=

ms

B

Ev

BeveE

1

1

1

Essay 1313(a) - An electromagnetic radiation is given out when an electron makes a transition fromone state of higher energy level to another of lower energy level.

- The energy of a photon of the electromagnetic radiation is given by E = hf

1

1

13(b)(i) - The electron in the ground state gains energy that is exactly equal to the energy

difference between the initial energy level of the electron and the final energy level.- Electron move up to the higher energy level than the ground state 1

1

13(b)(ii) Energy absorbed

eV

EE

8.12

)1

6.13(

4

6.1322

14

=

=

=

1

1

13(b)(iii) Photon energy

m

hcE

8

83419

1071.9

1000.31063.6)106.1(8.12

=

=

=

1

1

1

13(c)(i) Continuous X-rays:- When an electron strikes a metal, it can lose any portion of its energy. This energyloss of the incident electron is converted into energy of a X-ray photon. Hence, theenergies of X-ray photons are different. Since the wavelength of a photon is inversely

proportional to its energy, the wavelengths of the photons emitted are different.- If all the energy of the incident electron is lost as energy of a photon, X-ray photonwith the minimum wavelength is produced.

Line X-rays:

- The incident electrons may penetrate deep into the inner-most shell of the targetatoms, causing the electron from the inner K or L shells to be excited to higher energylevels.- When an electron from a higher energy level falls to fill up these vacancies, the

1

1

1

FE

FB




17/18

17

difference in energy is emitted as energy of line X-rays

1

13c(ii) When all the energy of the incident electron is converted into energy of an X-rayphoton,

m

hceV

10min

min

834319

min

1049.2

)1000.3()1063.6()100.5()106.1(

=

=

=

1

1

Essay 14

14 a

(i)

(ii)

(iii)

Nuclides which have the same number of proton but different number of neutrons

The disintegration of a heavy nucleus to lighter nuclei with the release of a lot of

energy

The combination of lighter nuclei at very a high temperature to produce a heavynucleus with release a lot of energy

1

1

1

14b

(i)

(ii)

Mass number, AAtomic number , Z

Mass- energy

HCHeB 1113

6

4

2

10

5 ++

HeLi 427

3

1

1 2H +

HeLinB 427

3

1

0

10

5 ++

1

1

1

1

1

1

14c

(i)

(ii)

Energy released E

J

mcE

13-

827-

2

107.44103.00101.6627.97693]-[27.98191

=

=

=

The energy of -ray photon ,

hcE=

hcE=

1

1

1

1




18/18

18

J

Total

J

E

13

1313-

13-

13

834

1059.4

1085.2107.44

productsdecayofenergykinetic

102.85

1099.6

)1000.3)(1063.6(

==

=

=

1

1

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