stpm trials 2009 physics answer scheme (terengganu)w
TRANSCRIPT
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
1/18
1
JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR
JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR
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 JABATANPELAJAR
JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR
JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR
JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR
JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR
JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR
JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR
JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR
JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJARJABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU 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
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
+
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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 =
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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)
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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)
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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
more examination papers at :
www.myschoolchildren.com
-
8/12/2019 STPM Trials 2009 Physics Answer Scheme (Terengganu)w
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