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2014 HKDSE

Physics Paper 1A

Suggested Solution

Prepared by C.M. Wong

@Tutor360HK <HKCEE & HKALE A in Physics>

http://www.tutor360hk.com/



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2014 HKDSE Physics Paper 1A Suggested Solution

1. D 2. A 3. C 4. A 5. B

6. D 7. C 8. C 9. B 10. B

11. B 12. B 13. B 14. A 15. C

16. A 17. B 18. C 19. A 20. C

21. D 22. D 23. B 24. C 25. D

26. A 27. D 28. D 29. B 30. B

31. D 32. A 33. C

2014 HKDSE Physics Paper 1A Question Analysis

Number of MCQ Percentage

I. Heat & Gas 2 6.1%

II. Force & Motion 10 30.3%

III. Wave Motion 7 21.2%

IV. Electricity & Magnetism 11 33.3%

V. Radioactivity & Nuclear Energy 3 9.1%

33 100%

Heat & GasForce &Motion

Wave MotionElectricity &Magnetism

Radioactivity& Nuclear

Energy

Paper IA 6.1% 30.3% 21.2% 33.3% 9.1%

0.0%

10.0%

20.0%

30.0%

40.0%

Topic Covered in Physics Paper IA




3

Q1. D 1) Temp. difference ∝ heat transfer (from high T to low T)

∵ T of surrounding > T of ice-cream

∴ Heat gain by ice-cream from surrounding

2) Vacuum flask (Y) ↓ rate of heat gain from surrounding by

conduction & convection

3) Rate of heat gain = 𝐸

𝑡 =

𝑚𝑙

𝑡 (identical ice-cream ⇒ same m)

∴ Rate of heat gain ∝ 1

𝑡

Rate of heat gain in Y < Rate of heat gain in X

∴ ty > tx

Q2. A 1) Same electric heater ⇒ same P

2) By 𝑃𝑡 = 𝑚𝑐 △ 𝑇 (for changing T)

𝑃(2𝑥60) = 𝑚(800)(80 − 20)

P = 400m

3) By 𝑃𝑡 = 𝑚𝑙𝑓 (for changing state)

400𝑚(8 − 2)(60) = 𝑚𝑙𝑓

lf = 144 kJkg-1

Remarks: CANNOT use the change in T in LIQUID state to calculate

∵ Question only give specific heat capacity in SOLID state

Q3. C 1) Uniform density ⇒ Center of mass at the middle

2) Rod suspended at Q ⇒ Take moment at Q

Clockwise moment = Anti-clockwise moment

MQR

(3) = MPQ

(2)

𝑀𝑃𝑄𝑀𝑄𝑅

=3

2 ⇒

𝑀𝑃𝑄𝑀𝑄𝑅

= 3 : 2




4

Q4. A

1) Two strings (non-identical) ⇒ Two tensions (T1 for left, T2 for right)

2) At equilibrium ⇒ no Fnet

Considering 30N objects, Considering 20N objects,

↑ F = ↓ F ↑ F = ↓ F

T1 = 30N T2 = 20N

3) Considering object W,

↑ F = ↓ F ← F = → F

30sinθ + 20sinϕ = W 30cosθ = 20cosϕ

4) ∵ θ & ϕ < 90o

⇒ sinθ & sinϕ < 1

∴ W < 30+20 ⇒ W < 50N

Q5. B 1) Consider the 1st phase,

By 𝑠1 = 𝑢𝑡1 +1

2𝑎(𝑡1)

2

36 = 4𝑢 +1

2𝑎(4)2 ⇒ 36 = 4u + 8a … (1)

2) Consider the whole motion,

By 𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

72 = 6𝑢 +1

2𝑎(6)2 ⇒ 72 = 6u + 18a … (2)

3) By solving (1) & (2)

𝑎 = 3𝑚𝑠-2




5

Q6. D 1) Identical blocks ⇒ same mass

From rest ⇒ u = 0

2) By law of conservation of energy ⇒ Loss in G.P.E. = Gain in K.E.

𝑚𝑔𝐻 − 0 =1

2𝑚𝑣2 − 0

𝑣 = √2𝑔𝐻 ⇒ 𝑣 ∝ √𝐻

∵ same 𝐻 ⇒ ∴ 𝑣1 = 𝑣2

3) By Newton’s 2nd

law,

𝐹𝑛𝑒𝑡 = 𝑚𝑎

𝑚𝑔 sin 𝜃 = 𝑚𝑎

𝑎 = 𝑔 sin 𝜃 ⇒ 𝑎 ∝ sin θ ∝ θ

∵ θ1 > θ2 ⇒ ∴ 𝑎1 > 𝑎2

4) By 𝑎 =𝑣−𝑢

𝑡 ⇒ 𝑎 ∝

1

𝑡 ( ∵ 𝑣 & 𝑢 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒)

𝑎 ∝ 1

𝑡

∴ 𝑡1 < 𝑡2

Q7. C Take right as +ve ⇒ Q can only move to right after collision

( ∵ 𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄 𝑖𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑠𝑖𝑑𝑒)

By Law of conservation of momentum

𝑚𝑝𝑢𝑃 + 𝑚𝑄𝑢𝑄 = 𝑚𝑝𝑣𝑃 + 𝑚𝑄𝑣𝑄

1) (2)(+6) + (1)(0) = (2)( 𝑣𝑃) + (1)(+2)

𝑣𝑃 = +5𝑚𝑠−1 > 𝑣𝑄 ( +2 𝑚𝑠−1)

It is not possible as Q cannot exceed P

2) (2)(+6) + (1)(0) = (2)( 𝑣𝑃) + (1)(+4)

𝑣𝑃 = +4𝑚𝑠−1 = 𝑣𝑄

It is possible as Q move with P together

3) (2)(+6) + (1)(0) = (2)( 𝑣𝑃) + (1)(+6)

𝑣𝑃 = +3𝑚𝑠−1 < 𝑣𝑄 (+6 𝑚𝑠−1)

It is possible. Q move faster than P




6

Q8. C Consider whole system

By Newton’s 2nd

law ⇒ 𝐹𝑛𝑒𝑡 = 𝑚𝑎

5𝑔 − 3𝑔 = 8𝑎

𝑎 =𝑔

4𝑚𝑠−2

Q9. B By law of conservation of energy ⇒ Loss in G.P.E. = Gain in K.E.

𝑚𝑔ℎ − 0 =1

2𝑚(𝑣2 − 𝑢2)

(9.81)(ℎ) = 1

2(112 − 22)

ℎ = 6.0𝑚

Q10. B Take downwards & right as +ve

1) Consider vertical motion, 𝑢𝑦 = 0

𝑠𝑦 = 𝑢𝑦𝑡 +1

2𝑎𝑦𝑡

2

0.8 =1

2(9.81)𝑡2 ⇒ 𝑡 = 0.4𝑠

2) Consider horizontal motion

𝑢𝑥 =𝑠𝑥

𝑡 ⇒ 𝑢𝑥 =

1

0.4

𝑢𝑥 = 2.5𝑚𝑠−1 ⇒ 𝑢 = 2.5𝑚𝑠−1 (∵ 𝑢𝑦 = 0)

Q11. B Option A: Weightless ⇒ Normal reaction = 0N , not related to G-force

Option B: In circular motion,

Centripetal force is provided by the 𝐹𝑛𝑒𝑡 towards center

𝐹𝑛𝑒𝑡 =𝑚𝑣2

𝑟 ⇒

𝐺𝑀𝑚

𝑟2=

𝑚𝑣2

𝑟

𝑎𝑐 =𝑣2

𝑟

Both astronaut and the spacecraft are moving with same 𝑎𝑐

Option C: Reaction force of the floor = normal reaction = 0N

Option D: 𝐹𝑐 = 𝐹𝑛𝑒𝑡 towards center

𝐹𝑐 is not an extra force!!!




7

Q12. B 1) G-field experienced on the surface of the Earth = g

𝑔 =GM

𝑅2 = acceleration due to gravity

2) Consider circular motion (with radius = 2R),

𝐹𝑛𝑒𝑡 =𝑚𝑣2

𝑟

𝐺𝑀𝑚

(2𝑅)2=

𝑚𝑣2

2𝑅 ⇒ 𝑣 = √

𝐺𝑀

2𝑅

3) Consider 𝑎𝑐

𝑎𝑐 =𝑣2

𝑟 ⇒ 𝑎𝑐 =

(√𝐺𝑀2𝑅 )2

2𝑅

𝑎𝑐 =𝐺𝑀

4𝑅2=

1

4𝑔

Q13. B

Note that the direction of wave is ⊥ to the wavefront

Q14. A

1) ✔

P is moving upwards

2) ✗

Q and S are moving in same direction

3) ✗

R is moving downwards

Only the particles at extreme points (max. / min.) will be

momentarily at rest




8

Q15. C By Snell's law & 𝑣 = 𝑓𝜆,

sin θ1

sin θ2=

𝑣1

𝑣2=

𝜆1

𝜆2=

𝑛2

𝑛1

θ ∝ 𝑣 ∝ 𝜆 ∝1

𝑛 ⇒ θ ∝ 𝑣 𝑖𝑛 𝑡ℎ𝑖𝑠 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛

∴ 𝑣𝐼𝐼𝐼 < 𝑣𝐼 < 𝑣𝐼𝐼

Q16. A Degree of diffraction ∝ 𝜆 ∝

1

𝑎

where 𝑎 is the gap width

Q17. B Let the distance travelled by car be x

1) Consider the speed of ultrasound,

By 𝑣 =𝑑

𝑡

340=64 + 64 + 2𝑥

(0.5 − 0.1) ⇒ 𝑥 = 4𝑚

2) Consider the speed of the car,

By 𝑣 =𝑑

𝑡

𝑣 =4

(0.5 − 0.1)/2 ⇒ 𝑣 = 20𝑚𝑠−1

Q18. C 1) For anti-phase,

Constructive interference occur if 𝑃𝐷 = 𝑛 +1

2𝜆

Destructive interference occur if 𝑃𝐷 = 𝑛𝜆

2) Path diff. at O = 0𝜆 ⇒ Destructive Interference

Path diff. at P = (3−2.8)

0.1= 1𝜆 ⇒ Destructive Interference




9

Q19. A 1) ✗

Sound wave are mechanical waves

2) ✔

Sound wave ⇒ mechanical wave ⇒ need vibration of paricle to

transmit ⇒ cannot transmit in vacuum

3) ✗

Sound can form stationary wave

Q20. C

1) Two sphere in contact ⇒ consider as same conductor

2) +ve charged rod near X ⇒ -ve charged induced in X near rod

⇒ +ve charged induced in Y

3) When X is earthed,

e- flow to Y (∵ -ve charged is attracted by rod) to neutralize the +ve

4) Y become neutral

X become –ve charged after separation & rod removal




10

Q21. D By Columb’s Law, 𝐹 =

𝑄𝐴𝑄𝐵

4𝜋𝜀𝑟2

Let distance between each particle be r

Option A ✗ 𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄1 ⇒ 𝑡𝑜 𝑙𝑒𝑓𝑡

𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄2 = 0𝑁

𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄3 ⇒ 𝑡𝑜 𝑟𝑖𝑔ℎ𝑡

Option B ✗ 𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄1 =(+2)(−1)

4𝜋𝜀𝑟2 +(+2)(+2)

4𝜋𝜀(2𝑟)2=

−1

4𝜋𝜀𝑟2 𝑁

𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄2 = 0N

𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄3 =(+2)(+2)

4𝜋𝜀(2𝑟)2 +

(+2)(−1)

4𝜋𝜀𝑟2 =−1

4𝜋𝜀𝑟2 𝑁

Option C ✗ 𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄1 ⇒ 𝑡𝑜 𝑟𝑖𝑔ℎ𝑡

𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄2 ⇒ 𝑡𝑜 𝑙𝑒𝑓𝑡

𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄3 =(+4)(−4)

4𝜋𝜀(2𝑟)2 +

(+4)(+1)

4𝜋𝜀𝑟2 = 0𝑁

Option D ✔ 𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄1 =(−4)(+1)

4𝜋𝜀𝑟2 +(−4)(−4)

4𝜋𝜀(2𝑟)2 = 0𝑁

𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄2 =(+1)(−4)

4𝜋𝜀𝑟2 +(+1)(−4)


𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑄3 =(−4)(−4)

4𝜋𝜀(2𝑟)2 +

(−4)(+1)


Q22. D 1) Electron is –ve charged ⇒ 𝐹𝐸 to left ⇒ Left plate is +ve

⇒ E-field point from P to Q

2) By 𝐸 =𝐹

𝑄

𝐸 =8 × 10−18

1.6 × 10−19 ⇒ 𝐸 = 50𝑁𝐶−1




11

Q23. B Let the distance between charge +Q & X be 3x

By 𝑉 =+𝑄

4𝜋𝜀𝑟

𝑉𝑋 =+𝑄

4𝜋𝜀(3𝑟)

𝑉𝑌 =+𝑄

4𝜋𝜀(2𝑟) ⇒ 𝑉𝑌 =

3

2(

+𝑄

4𝜋𝜀(3𝑟) )

∴ 𝑉𝑌 =3

2𝑉𝑋

Q24. C 1) When S is closed ⇒ 3Ω resistor is shorted

By 𝑉 = 𝐼R

𝑒.𝑚. 𝑓. = (3)(6 + 𝑟) … (1)

2) When S is opened,

𝑒.𝑚. 𝑓. = 𝐼(3 + 6 + 𝑟) … (2)

3) Combine (1) & (2)

(3)(6 + 𝑟) = 𝐼(3 + 6 + 𝑟)

18 + 3𝑟 = 9𝐼 + 𝐼𝑟 … (3)

By substitution I into (3), internal resistor (r) is calculated

Option A ✗ 𝑊ℎ𝑒𝑛 𝐼 = 1.6𝐴, 𝑟 = −2.57Ω (Not possible)

Option B ✗ 𝑊ℎ𝑒𝑛 𝐼 = 2.0𝐴, 𝑟 = 0Ω (Not possible)

Option C ✔ 𝑊ℎ𝑒𝑛 𝐼 = 2.4𝐴, 𝑟 = 6Ω (Possible)

Option D ✗ 𝑊ℎ𝑒𝑛 𝐼 = 3.2𝐴, 𝑟 = −54Ω (Possible)




12

Q25. D 1) For short-circuited,

𝑅 = 0Ω, 𝑉 = 0𝑉, 𝑖 = ∞

2) For open circuit,

𝑅 = ∞, 𝑉 = 𝑒.𝑚. 𝑓. 𝑖 = 0𝐴

3) The reading of voltmeter = 𝑉𝑃 (Potential difference across P)

Option A ✗

𝑊ℎ𝑒𝑛 𝑃 & 𝑄 𝑎𝑟𝑒 𝑠ℎ𝑜𝑟𝑡𝑒𝑑 ⇒ 𝑉𝑃 = 0V

Option B ✗

𝑊ℎ𝑒𝑛 𝑃 & 𝑄 𝑎𝑟𝑒 𝑖𝑛 𝑜𝑝𝑜𝑛 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 ⇒ 𝑉𝑃 + 𝑉𝑄 = 6𝑉

𝑉𝑄 𝑜𝑟 𝑉𝑃 ≠ 6𝑉

Option C ✗

𝑊ℎ𝑒𝑛 𝑃 𝑖𝑠 𝑠ℎ𝑜𝑟𝑡𝑟𝑑 ⇒ 𝑉𝑃 = 0V

𝑊ℎ𝑒𝑛 𝑄 𝑖𝑠 𝑖𝑛 𝑜𝑝𝑒𝑛 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 ⇒ 𝑉𝑄 = 𝑒.𝑚. 𝑓. = 6𝑉

⇒ 𝑉𝑃 = 0𝑉 (∵ 𝑉𝑄 + 𝑉𝑃 = 6𝑉)

Option D ✔

𝑊ℎ𝑒𝑛 𝑃 𝑖𝑠 𝑖𝑛 𝑜𝑝𝑒𝑛 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 ⇒ 𝑉𝑃 = 𝑒.𝑚. 𝑓. = 6𝑉

𝑊ℎ𝑒𝑛 𝑄 𝑖𝑠 𝑠ℎ𝑜𝑟𝑡𝑒𝑑 ⇒ 𝑉𝑄 = 0𝑉

𝑉𝑃 = 6𝑉 (∵ 𝑉𝑄 + 𝑉𝑃 = 6𝑉)




13

Q26. A 1) By right hand grip rule, consider force acting at point O

Force by charge Q ⇒ towards P

Force by charge P ⇒ towards S

Force by charge S ⇒ towards P

Force by charge R ⇒ towards Q

2) Force toward Q & S are balanced

∴ Force in OP direction

Q27. D 1) By right hand grip rule, 𝑖 → �⃑�

If 𝑖 = +ve (flow upwards), �⃑� into coil PQRS

If 𝑖 = –ve (flow downwards), �⃑� out of coil PQRS

2) Consider 1st phase, when 𝑖 = +ve & ↓ ⇒ �⃑� into coil ↓

By Lenz’s Law, induced 𝑖 flow in clockwise to generate �⃑� into

coil (to oppose the change ⇒ ↑ �⃑� into coil)

3) Consider 2nd

phase, when 𝑖 = -ve & ↑ ⇒ �⃑� out of coil ↑

By Lenz’s Law, induced 𝑖 flow in clockwise to generate �⃑� into

coil (to oppose the change ⇒ ↓ �⃑� out of coil)

Q28. D 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝐵 =

𝜇𝑜𝑁𝐼

𝐼

𝐵 ∝ 𝑁 ∝1

𝑙

∴ When 𝑙 ↓ & 𝑁 ↑ ⇒ 𝐵 ↑ (Independent of cross sectional area)




14

Q29. B 1) Metal block ⇒ 𝑒− is charge carrier

2) PQ is higher potential ⇒ +ve charged ⇒ 𝑒− is accumulated at SR

⇒ 𝐹𝐵 acting into block (towards SR)

3) By Fleming left hand rule,

𝑖 to the left, 𝐹𝐵 into paper ⇒ �⃑� upwards (Q to P)

Q30. B 1) In d.c. condition,

𝑃 =𝑉2

𝑅 ⇒ 𝑃 =

102

𝑅⇒ 𝑃 =

100

𝑅

2) In a.c. condition,

1

2𝑃 =

𝑉𝑟.𝑚.𝑠.2

𝑅 ⇒

1

2(100

𝑅) =

𝑉𝑟.𝑚.𝑠.2

𝑅

𝑉𝑟.𝑚.𝑠. = 5√2 V

Q31. D 1) M = Mass no. = no. of 𝑝+ + 𝑛

A = Atomic no. = no. of 𝑝+

2) Consider the decays equations,

𝑊𝐴𝑀 → 𝑋𝐴−2

𝑀−4 + 𝛼24

𝑋𝐴−2𝑀−4 → 𝑌𝐴−1

𝑀−4 + 𝛽−10

𝑌𝐴−1𝑀−4 → 𝑍𝐴

𝑀−4 + 𝛽−10

Statement (1) ✗ Y has 1 more proton than X

Statement (2) ✔ No. of 𝑛 = 𝑀 − 𝐴

Statement (3) ✔ Isotope = atoms of same element of same no. of 𝑝+,

different no. of 𝑛 (OR mass number)




15

Q32. A 1) α radiation blocked by paper

∴ x < 450

2) Aluminum is used to block 𝛽 radiation

No β radiation ⇒ no ↓ after passing through aluminum

⇒ no difference in x & y

∴ Only A & C are possible answers

3) γ willl be ↓ to 1

2 of initial intensity by 25mm lead

⇒ 2mm lead will block a small amount of γ

⇒ z < y

4) Blackgrond radiation (50) cannot be blocked by any material

⇒ y > 50

∴ Only A is possible answer

Q33. C 1) Consider the decay equation,

𝑅𝑎 → 𝑅𝑛 + 𝛼 + 4.9𝑀𝑒𝑉

2) Energy difference = 4.9𝑀𝑒𝑉

= 4.9 × 106 × 1.6 × 10−19 ⇒ 7.84 × 10−13

3) By ∆𝐸 = ∆𝑚𝑐2

7.84 × 10−13 = (∆𝑚)(3 × 𝑥108)2

∆𝑚 = 8.7 × 10−30𝑘𝑔

4) ∵ Energy is released in the product

∴ mass of product < mass of reactant

- The End -


1 2014 hkdse physics paper 1a suggested · pdf file2014 hkdse physics paper 1a suggested...

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