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Suggested solution for
2009 CE Physics Paper 2 This suggested solution is not represented the official solution or answer from the HKEAA.
物理學人 PHYSICS CE 2009 Solutions
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Suggested answer for 2009 CE physics paper II
1.C 2. A 3.C 4.A 5.A 6.B 7.A 8.C 9.C 10.B
11. D 12. A 13. D 14. C 15. A 16. C 17. A 18. B 19. C 20. B
21. D 22. D 23. D 24. B 25. B 26. D 27. B 28. C 29. C 30. D
31. C 32. B 33. A 34. B 35. A 36. C 37. B 38. D 39. D 40. D
41. A 42. C 43. D 44.A 45 B
物理學人 PHYSICS CE 2009 Solutions
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Q1.
1 150 100050 13.89
3600
displacement travelledvelocity=
time taken
mkmh ms
s
1
50035.997 36
13.89
displacement travelledtime taken=
velocity
ms s
ms
So, the answer is C
Q2.
Statement (1) Displacement = final position – initial position.
Since the final position and initial position of P1, P2 and P3 are equal, so, statement (1) is
correct.
Statement (2) Distance: P3 > P1 > P1 so, statement (2) is incorrect.
Statement (3) total distance travelled
Average speed=time taken
Since statement (2) is incorrect statement (3) is also incorrect
Answer is A, statement (1) only
物理學人 PHYSICS CE 2009 Solutions
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Q3
u
10 m
Fresistive =3N
v=0
0.5kg 0.5kgFresistive =3N
u
10 m
Fresistive =3N
v=0
0.5kg 0.5kgFresistive =3N
Fnet
a
Let right hand side be positive:
2
3 0.5
36
0.5
netF ma
a
a ms
2 2 2by v u as
2 2
2
1
0 2( 6)( 10)
120
120 10.95
u
u
u ms
So, the answer is C
物理學人 PHYSICS CE 2009 Solutions
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Q4.
Rough friction exists
a=2ms-2
Fexternal =5N1kg friction
u=0
Fexternal =2N1kg friction
At rest, u=0 a=0 Fnet = 0
So, the answer is A
Q.5
v
0
P
Q
uP
uQ
Reaction
time
t
uP ≠ uQ statement (1) is incorrect
For P and Q, their reaction time is equal statement (2) is correct
Area of v-t graph = displacement travelled by P or Q respectively.
Since Area of P > Area of Q statement (3) is incorrect
So, the answer is A
物理學人 PHYSICS CE 2009 Solutions
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Q.6
Initial at rest u = o
Smooth no friction
For the average power: P =W.D.byF
time taken
P =Fs
t=
F(ut +12
at2)
t=
F(12
at2)
t=
Fa
2t
Remark: need not to consider the instantaneous power. i.e. derivatives of energy
Constant F constant Fnet constant a
P t
Answer is B
Q.7
F1
F
F2
1 2F F F
So, answer is A
Q.8
Weight on the Earth’s surface = 120N = mgearth = m(10) m =12 kg
Mass is constant in everywhere.
Weight at the Moon’s surface = mgmoon = mgearth/6 = 120/6 = 20N
So , the answer is C
物理學人 PHYSICS CE 2009 Solutions
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Q.9
Internal energy = K.E. + P.E.
Average K.E depends on Temperature of the object in absolute scale
P.E. depends on the state and mass of the object
Statement (1) same temperature same average K.E. ΔK.E. = 0
ΔInternal energy = ΔK.E. + ΔP.E.
Different masses but same state different P.E.
Different Internal energy
So statement (1) is incorrect
Statement (2) Hot’s temperature > Cold’s temperature
Hot’s average K.E. > Cold’s average K.E.
ΔInternal energy = ΔK.E. + ΔP.E.
Assume same state (solid = block) same P.E. ΔP.E. = 0
ΔInternal energyhot >ΔInternal energycold
So statement (2) is correct
Statement (3) same temperature at 0ºC same average K.E. ΔK.E. = 0
ΔInternal energy = ΔK.E. + ΔP.E.
Different states P.E.water > P.E.ice since P.E.liquid > P.E.solid
ΔInternal energywater >ΔInternal energyice
So statement (3) is correct
So, the answer is C statement (2) and (3) are correct
Q10
0ºC ice absorb heat 0ºC water
20 ºC soft drink release heat 0ºC soft drink
Energy gained by ice = Energy loss by the soft drink
E=mwaterl = msoft drinkcΔT
mwater = msoft drinkcΔT/ l = 0.3*5300* (20-0) / 3.34x105
mwater = 0.1 kg
Answer is B
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Q.11
Same heater Same power of heater Same P
Same substance Same specific heat capacity same c
For the slope of the curve:
ΔTemperature ≠0ΔK.E. average ≠0 use specific heat capacity
Since E = mc∆T = Pt ∆T
t=
P
mc= slope of the graph
Mass , m ↑ slope ↓
Also, for the time, as mass , m ↑, mc∆T = Pt, the time for heating up to the same temperature is
longer
Answer is D
Q12
lExperimental =Energy supplied
mass
(I) Mass of the water decreases, lExperimental increases
(II) Mass of the water increases, lExperimental decreases
Answer is A
Q13
Erect object and the image are in same side concave lens
Lens moves further away from the paper paper as the object is also away from the lens
relatively object moves to infinity, images tends to form at the focus
Magnification = v/u, as u increases and v decreases, then the magnification decreases.
The size of the image decreases also
Answer is D
物理學人 PHYSICS CE 2009 Solutions
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Q.14
From the graph 2λ=0.4 m λ=0.2m
By v = fλ 0.2 = f(0.2)
f=1Hz T=1s T/4 = 0.25s
0.4m 0.6m
Cork
Direction of wave propagation
t=0s
t=1s
0.4m 0.6m
Cork
Direction of wave propagation
t=2s
0.4m 0.6m
Cork
Direction of wave propagation
t=3s
0.4m 0.6m
Cork
Direction of wave propagation
0.4m 0.6m
Cork
Direction of wave propagation
t=3.25s
0.4m 0.6m
Cork
Direction of wave propagation
t=3.5s
0.4m 0.6m
Cork
Direction of wave propagation
t=3.75s
Answer is C
物理學人 PHYSICS CE 2009 Solutions
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Q.15
ηglass =sini
sinr
For the incident ray X and Y, ix > iy
For the refracted ray X and Y, rx = ry
For angle is less than 90º sin θ increases with θ
As a result,
ηglass ,x =sinix
sinrx
>siniy
sinry
= ηglass ,y
For EM wave, speed of coloured light is constant in vacuum.
Answer is A
Q16
For EM wave in vacuum, speed of light is constant, c = 3x108 ms-1
By v = fλ=c = constant
The speed of wave just depends on the medium, it’s independent of the frequency of the
wave
The frequency of wave just depends on the frequency of the source.
Statement (1) is correct v is independent of frequency statement (1) is correct
Statement (2) frequency is proportional to wavelength statement (2) is incorrect
Statement (3) is correct v is independent of wavelength statement (3) is correct
So, the Answer is C
物理學人 PHYSICS CE 2009 Solutions
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Q17.
C = 2f C = 2ff f
L
f f
L
X
X
image
image
f' f' C'C'
L
f f
L
X image
f' f' C'C'
Only, Statement (1) and (2) will place the object with C and focus, then the image tends to
infinites. Answer A is correct
Q.18
Identical bulbs same R
In parallel Same potential difference Same R Same i pass through R
ix = 0.4A iy = 0.4A
Total current pass through the circuit = 0.8A
Px=IxVx = 0.4A× 220V
1000= 0.088kW
Ex = Pxt = 0.088kW * 5 h = 0.44kWh
Ex = Ey
Total energy consumed by X and Y = Ex + Ey = 2* 0.44kWh = 0.88kWh
Answer is B
物理學人 PHYSICS CE 2009 Solutions
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Q19
Identical resistors same R
In series same current Different potential difference.
S is open Voltmeter is in series with the RLeft
112 12
112
1
0
12
voltmeter
voltmeter
P.D.
as ideal voltmeter
P.D.
V
VV
V V
V
V
V
RV V
R RR R
R R
VR
R
RR
R
V
S is closed Voltmeter is in parallel with the RRight
'
1
0
' 112 12
' 2
6
voltmeter R'
as ideal voltmeter
R'
P.D. P.D.
V
V V
VV
VV
V
V
R R
R R R RR
R R RR R
RR
RR
R
R
RV V
R R
V
Answer is C
V
12V
RRV
V
12V
R
RV
R
12V
R R’
物理學人 PHYSICS CE 2009 Solutions
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Q.20
Rating voltage for “24W, 6Ω” within 24W P ≤V2
R V ≤ PR = 24 ∗ 6 = 12V
P ≤ i2R i ≤ P
R =
24
6= 2A
Rating voltage for “9W, 4Ω” within 9W P ≤V2
R V ≤ PR = 9 ∗ 4 = 6V
P = i2R i ≤ P
R =
9
4= 1.5A
Consider the supply voltage, for “9W, 4Ω”, the voltage supplied to it cannot be more than 6V.
Otherwise, it cannot work at within its rated power.
Assume both light bulbs working at their maximum safe current when they are connected in
parallel.
Total current drawn from 2+1.5A, equivalent resistance in the parallel circuit is 2.4Ω the
common supply voltage = 2.4* 3.5 = 8.4V > 6V, so “9W, 4Ω” cannot work at its rated power.
Contradiction!
For the maximum common supply voltage would be 6V, then “9W, 4Ω” is working at its rated
power while “24W, 6Ω” is working WITHIN its rated power. So it’s acceptable to simply connect
them in parallel.
R=2.4Ω
6V
i=2.5A
6V
6Ω
4Ω
For I total =2.5A, both light bulb will not be burnt out.
4Ω resistance will have 6
4+6× 2.5A = 1.5A
6Ω resistance will have 4
4+6× 2.5A = 1.0A < maximum safety current
物理學人 PHYSICS CE 2009 Solutions
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Q21
obviously, answer is D
Q22
Current into paper
Current in an external magnetic field Magnetic force produced use Left hand rule
Fmagnetic
i
Magnetic fieldN
S
So, statement (1) is correct
Fmagnetic
i
Magnetic fieldN
S
So, statement (2) is correct
sin
Reference: Magnetic force of current carried wire in an extenal magnetic field
is the angle between the wire and the magnetic field, L is the length of wire within t
magnetic ext
magnetic ext
F B
F ILB
he magnetic field
So, Statement (3) is correct
As a result, the answer is D
Q23.
Fuse and switch must be connected to live. Answer is D
Q24
This is an example of nuclear fusion. Answer is B
物理學人 PHYSICS CE 2009 Solutions
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Q25
Alpha decay of X: 4 4
2 2
A A
Z ZX Y
Alpha decay of X: 4 4 0
2 1 1
A A
Z ZY Z
X and Z have different atomic number different number of proton Different element
Statement (1) is incorrect
No. of neutron = Mass number – atomic number
= (No. of proton + No. of neutron) – No. of proton
For X: No. of neutron of X = A – Z
For Z: No. of neutron of Z = A-4 – (Z-1) = A-Z -3
Difference of no. of neutron of X and Z = (A-Z)- (A-Z-3) = 3 not 2 Statement (2) is
incorrect
No. of proton = Atomic number
No. of proton of Y = Z-2
No. of proton of Z = Z-1
No. of proton of Z – No. of proton of Y = Z-1 –(Z-2) = 1 Statement (3) is correct
As a result, answer B is correct
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Q26
Given
1
2
15hours
1000counts per mintueoC
After 15 hours, 15 528counts per mintuehrsC
o oC A bg 12
1
2
C A bg
Equation 1. 1000o oC A bg
Equation 2. 12
1
2
528C A bg
Equation 3. 1
22
oAA
Sub equation 3 into equation 2Equation 4: 12
2 2*528 1056oC A bg
Equation 4 minus Equation 1 bg = 56 counts per min answer is D
物理學人 PHYSICS CE 2009 Solutions
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Q27
A necessary condition of a statement must be satisfied for the statement to be true
Statement P is a necessary condition of a statement Q if Q implies P
即 Q 發生,P 才會發生。
但 P 發生,Q 不一定會發生。
Only statement (2) is necessary condition for the CHAIN reaction
Q28
Total area in magnitude of v-t graph = distance travelled from P to Q
v/ms-1
3
13
21 1.6A1
A2 A30.3
Total distance travelled from P to Q = A1+A2 =3× 0.3
2+
1.6−0.3 × 13
2=0.45+8.45=8.9m
Height of the platform above water surface = A2-A1 = 8.45 – 0.45 = 8m
Answer is A
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Q29
5 kg
S1
S2
30º
S1
S2
30º
5 kg
50N
T
T
S1sin30º
S1cos30º
Consider the above free body diagrams:
T= 50N
Equation 1: T = S1Sin30º=50
Equation 2: S2=S1Cos30º
Equation 2/ equation 1 S2
50=
1
tan 30°= 1.73205
S2= 86.6 N The answer is C
物理學人 PHYSICS CE 2009 Solutions
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Q30
Smooth No friction no external force before and after collision. No change in the
momentum Answer A is incorrect
Rebound after collision change direction +ve direction to –ve direction Answer C
and B are incorrect
During the collision Reaction force
between the wall and the block
Action – reaction pair Net force of the
system is still zero
Considering the wall and the block as a
system, its total momentum is
conserved.
Magnitude:
, , , ,
, ,
, ,
0
block initial wall initial block final wall final
block final wall final
block final wall final
P P P P
P P P
P P P P
Since Pwall, final is non-zero
Direction:
Rebound Opposite to its initial direction -ve direction after collision
Answer is D
Q31
Uniform speed u=v a = 0 Fnet = 0 K.E. = constant
W.D. by the pulling force = P.E. gain + W.D. against resistive force
= mgh + Fresistive * s = 1500*10*100*sin30º + 200*100
= 770kJ
Answer is C
P
+ve
Force acting on
the block by the
wall
+ve
Force acting on
the wall by the
block
P
+ve
Before
collision
During
collision
After
collision
物理學人 PHYSICS CE 2009 Solutions
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Q32
Algebraic solution:
Assume downward as positive
By 2 2 2v u as
u=0, a=g
2 0 2v gs
2
1 22
hv g while 2
2 2v gh
2
2
2
1
22
22
v gh
hvg
So 2 12v v Statement (1) is incorrect
By v u at
u=0, a=g
1 1v gt and 2 2v gt
2 2
1 1
2t v
t v Statement (2) is incorrect
K.E. at 2
2 2
2 2 1 1
1 1 12 2
2 2 2t mv m v mv
=2 K.E. at 1t
Statement (3) is correct
As a result Answer is B
Q33
Statement (1) true
Statement (2) Same temperature Same average K.E. True
Statement (3) P.E. depends on the state and mass of object.
Heavier one having more P.E.
Statement (3) is incorrect
Answer is A
Graphical method:
a/ms-2
t/s0
v/ms-1
t/s0
s/m
t/s0
h/2
h
t1 t2
v1
v2
物理學人 PHYSICS CE 2009 Solutions
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Q34
For X,Y and Z, their incident angles are the same.
For Z, its incident angle equals to the medium’s critical angle.
For X, its incident angle is larger than the medium’s critical angle it occurs total internal
reflection.
For Y, it still occurs refraction, its incident angle is small than the medium’s critical angle
CX>CZ> CY
Sin CX> Sin CZ> Sin CY
X> Z > Y
Answer is B
Q35
Same loudspeakers Same sources Same frequency + in phase Coherent source
By v f 330
0.5660
vm
f
The path difference at Y = 0λ Constructive interference
The path difference at X = 4 – 2 = 2 =4 * 0.5 = 4λ = nλ\
n=4 = integer Constructive interference at X
As a result, the answer is A
Q36
Displacement/ m
Distance/ m0b jf
+
d
h
c
+
Wave direction
The orange coloured line is the waveform just after at the instant shown in Figure (B)
Particle c will be upward, i.e. towards to the positive direction, moves to right
Particle f will be downward, i.e. towards to the negative direction moves to left
As a result, answer is C
Conditions for total internal reflection:
1. Traveling from optically denser to optically less dense medium
2. Incident angle is larger than the critical angle.
物理學人 PHYSICS CE 2009 Solutions
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Q37
Same source same frequency
Difference medium string and air different wave speed different wavelength
So, answer is B
Q38
Coherent source Constructive interference = nλ
The path difference at P = 4.5 λ – 3.5λ = 1λ n=1 Constructive interference at P
Since the position P is at trough, the total displacement after the constructive interference -A +
-A = -2A = constant at the instant occurring constructive interference.
Answer is D
Q39
Opposite the change at Q, so the dotted magnet is having North pole towards Q
P is south pole then
Answer is D
Q40
Obviously, the answer is D
物理學人 PHYSICS CE 2009 Solutions
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Q 41
S1 and S2 are closed and S3 is open
R
R
S1 and S2 are open and S3 is closed 2 2 2
2
2total
V V VP
RR R
R R
2
2 2
,
122 2
2 4 4total new
V
V V PP
R R R
So, the answer is A
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Q42
1st statement at the highest point, the object needs to change its direction from positive to
negative or negative to positive. As a result, it must has zero velocity at its highest point.
1st statement is true
2nd statement a = g = 10ms-2 = constant throughout the motion
2nd statement is false
Answer is C
Q43
1st statement frequency of Ultrasonic waves is larger than that of audible sound waves. For
speed of waveform is constant, then the wavelength of Ultrasonic waves is smaller than that of
audible sound waves.
The degree of diffraction for the same slit width of Ultrasonic waves is smaller than that of
audible sound waves 1st statement is false
2nd statement it is true
Answer is D
Q44
1st statement True.
2nd statement True. It’s also the explanation of 1st statement
Answer is A
Q45
1st statement True beta particles of course are emitted from beta decay
2nd statement True it is the nature of beta particles but it is not the explanation of 1st
statement
Answer is B
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