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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



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



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



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



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



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.


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


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



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



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


t=2s

0.4m 0.6m

Cork


t=3s

0.4m 0.6m

Cork


0.4m 0.6m

Cork


t=3.25s

0.4m 0.6m

Cork


t=3.5s

0.4m 0.6m

Cork


t=3.75s

Answer is C



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



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



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’



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



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



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



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



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



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



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



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



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.



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



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



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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