Physics 2012 DSE

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May 4, 2012

香港考試及評核局
HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY
2012 年香港中學文憑考試
HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION 2012

物理 香港中學文憑考試 試卷一乙
PHYSICS HKDSE PAPER 1B

本評卷參考乃考評局專為今年本科考試而編寫,供閱卷員參考
之用。閱卷員在完成閱卷工作後,若將本評卷參考提供其任教
會考班的本科同事參閱,本局不表反對,但須切記,在任何情
況下均不得容許本評卷參考落入學生手中。學生若索閱或求取
此等文件,閱卷員/教師應嚴詞拒絕,因學生極可能將評卷參
考視為標準答案,以致但知硬背死記,活剝生吞。這種落伍的
學習態度,既不符現代教育原則,亦有違考試著重理解能力與
運用技巧之旨。因此,本局籲請各閱卷員/教師通力合作,堅
守上述原則。
This marking scheme has been prepared by the Hong Kong Examinations and Assessment Authority for markers’ reference. The Authority has no objection to markers sharing it, after the completion of marking, with colleagues who are teaching the subject.
However, under no circumstances should it be given
…show more content…
Because energy loss by the steam is also gained by the surroundings including the air/the jug (usually metallic) that holds milk.

1
)
V
4
4 p1 ( π × (0.8) 3 ) = (1.01 × 10 5 Pa )( π × (1.0) 3 )
3
3 p1V1 = p2V2 (Or p ∝

5

p1 = 1.97 × 10 Pa

1M correct eqn. or with sub.

1A

3

= 45600 J 1 A
1M
1A

2

1A
1A
2

1M

1A correct answer, accept 2 sig. fig.
[Accept without

4 π] 3

1A

2

(Accept 1.97 × 105 Pa to 2.00 × 105 Pa)
(b)

3.

(a)

Volume increases as bubble rises but the speed / k.e.
1A speed / k.e. of molecules of gas molecules remains unchanged unchanged therefore frequency of collision of molecules on
1A frequency of collision on inner bubble’s inner surface decreases, gas pressure decreases. surface/wall decreases AND pressure decreases

1A

(i)

1A

Friction f between the tyres and the road. mv 2 f = r 8000 N =

(1200 kg)v 2
45 m

1A Friction
1M correct eqn. or with sub.

2
1A

1M

1A correct answer

v = 17.3 m s −1
(Accept 17.0 to 17.4 m s–1)
(ii)

(b)

1A

3

Smaller
For the same f, v2 ∝ r, r decreases, v decreases.

1A
1A

2

(Max) friction / coefficient of friction reduced,
Not enough to provide the centripetal force / acceleration required for circular motion.
Or Tracking or allowed speed lowered.

2012-DSE-PHY 1B–3

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FOR MARKERS’ USE ONLY

1A
1A

2

只限閱卷員參閱

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Marks

4.

(a)

(i)

v = u + at

1M correct sub. into correct eqn.

= 60 + (– 4)5

1A correct answer

= 40 m s –1
(ii)

1M
1A

2

v / m s –1

60
40

A

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