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香港考試及評核局
HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY

2017 年香港中學文憑考試
HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION 2017

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

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This marking scheme has been prepared by the Hong Kong Examinations and Assessment Authority for
the reference of markers. The use of this marking scheme is subject to the relevant appointment terms
and Instructions to Markers. In particular:

- The Authority retains all proprietary rights (including intellectual property rights) in this marking
scheme. This marking scheme, whether in whole or in part, must not be copied, published, disclosed,
made available, used or dealt in without the prior written approval of the Authority. Subject to
compliance with the foregoing, a limited permission is granted to markers to share this marking scheme,
after release of examination results of the current HKDSE examination, with teachers who are teaching
the same subject.

- Under no circumstances should students be given access to this marking scheme or any part of it.
The Authority is counting on the co-operation of markers/teachers in this regard.

 香港考 試 及評 核 局 保留 版 權
Hong Kong Examinations and Assessment Authority
All Rights Reserved 2017


2017-DSE-PHY 1B–1

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HKDSE Physics
General Marking Instruction

1. It is very important that all markers should adhere as closely as possible to the marking scheme. In many cases,
however, candidates may have obtained a correct answer by an alternative method not specified in the marking
scheme. In general, a correct answer merits the answer mark allocated to that part, unless a particular method has
been specified in the question. Markers should be patient in marking alternative solutions not specified in the
marking scheme.

2. In the marking scheme, answer marks or ‘A’ marks are awarded for a correct numerical answer with a unit. In case
the same unit involved is given incorrectly for more than once in the same question, the ‘A’ marks thereafter
can be awarded even for correct numerical answers without units. If the answer should be in km, then cm and m
are considered to be wrong units.

3. In a question consisting of several parts each depending on the previous parts, marks for correct method or
substitution are awarded to steps or methods correctly deduced from previous answers, even if these answers are
erroneous or for inserting values of appropriate physical quantities into an algebraic expression irrespective of their
order of magnitudes. However, ‘A’ marks for the corresponding answers should NOT be awarded (unless otherwise
specified).

4. For the convenience of markers, the marking scheme is written as detailed as possible. However, it is still likely that
candidates would not present their solution in the same explicit manner, e.g. some steps would either be omitted or
stated implicitly. In such cases, markers should exercise their discretion in marking candidates’ work. In general,
marks for a certain step should be awarded if candidates’ solution indicated that the relevant concept/technique had
been used.

5. In cases where a candidate answers more questions than required, the answers to all questions should be marked.
However, the excess answer(s) receiving the lowest score(s) will be disregarded in the calculation of the final mark.

6. OSM (On-screen marking) marking symbols:

 correct point
 wrong point
 point to highlight
_ _ _ incomplete answer
 missing point
文 entering text/comment

2017-DSE-PHY 1B–2

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Solution Marks Remarks

1. (a) A larger bulb improves the sensitivity of the thermometer.
OR
A larger bulb minimizes the effect on the temperature reading
due to the other parts of the thermometer stem that are exposed to
different temperatures. 1A
1

(b) (i) E = mcT

= 0.015 × (2.9 × 103) × (20 – 15) 1M
= 217.5 J 1A
2

(ii) Time taken to reach air temperature = 217.5 / 0.5 1M

= 435 s 1A
2

(iii) The thermometer would be in direct contact with the cooler

air and would cool down quickly. 1A
The temperature reading would be less than the actual soil 1A
temperature.
2

2. Measure the mass of a bullet m and the mass of the trolley with
plasticine M. 1A
Fire the bullet towards the plasticine. 1A
Read the speed of the trolley v immediately after the bullet hit the
plasticine. 1A
M m
The speed of the bullet u is given by u  v. 1A
m
Precaution:
The bullet should be fired close to the plasticine.
The bullet should be fired along the direction of travel of the trolley.
The track must be horizontal / friction compensated.
(either one, or other reasonable answers) 1A
5

3. (a) c rms  f Tf
 1M
c rms T
cr.m.s. at 350 K 350

cr.m.s. at 300 K 300
= 1.08 1A
2

(b) The speed of the gas molecule increases. They collide more
frequently and violently with the wall of the container. 1A+1A
Thus, the pressure increase.
2

2017-DSE-PHY 1B–3

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Solution Marks Remarks
4. (a) (i) 1 2
By s = ut + gt
2
1
0.11 = g(0.05 × 3)2 1M
2
1A Accept 9.6 – 9.8 m s2
g = 9.79 m s2
2

(ii) (1)
×
×

×
1A Correct horizontal positions
1A Correct vertical positions
2

(2) vx = 1 m s1
vy = uy + gt
= 0 + 9.79 × (0.05 × 3) 1M
= 1.47 m s1
v= v x2  v 2y
1M
= 12  1.47 2
= 1.78 m s1 1A Accept 1.56 – 1.78 m s1
3

(b) The air resistance acting on the ball increases as its speed
increases. 1A
The net force acting on the ball becomes zero when the air
resistance equals to the weight of the ball. 1A
By Newton’s first law, the ball travels with constant speed.
OR
By Newton’s second law, the ball will not accelerate further and
travels with constant speed. 1A
3

2017-DSE-PHY 1B–4

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Solution Marks Remarks

5. (a)
normal reaction

friction

weight 1A Weight and normal reaction

1A Friction
Deduct 1A for any wrong additional
force(s) drawn
2

(b)  =  s1 1A
F = mr2
= (1)(0.3)()2 1M
= 2.96 N (towards the centre of the turntable) 1A
3

(c) The initial linear speed of the teapot = r = 0.3 m s1 1M

f 10
Deceleration of the teapot a = = = 10 m s2
m 1
Distance travelled s is given by
v2  u2 = 2as
u 2 (0.3 ) 2 1M
s= =
2a 2(10)
= 0.044 m 1A

ALTERNATIVE
The initial linear speed of the teapot = r = 0.3 m s1 1M
K.E. of the teapot is dissipated in work done against friction
1 mu2 = fd
2

mu 2 (1)( 0.3 ) 2 1M
d= =
2f 2(10)
1A
= 0.044 m
3

2017-DSE-PHY 1B–5

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Solution Marks Remarks

6. (a) (i) v = f 
=5×4 1M
= 20 cm s1 1A
2

(ii) Y is moving upwards at t = 0. 1A

1

(iii) displacement

time / s

0.2 0.4
1A 2 periods
1A Correct phase

(b) (i) The water waves from A to B are in anti-phase at Q.

OR
The path difference at Q = (n + 1/2). 1M accept ½, 1½...
Destructive interfere occurs to form a minimum. 1A
2

(ii) Path difference at Q = 1.5 = 3 cm 1A for 1.5 = 3

 = 2 cm 1A
2

(iii)
amplitude

O distance
P
1A Decreasing amplitude

2017-DSE-PHY 1B–6

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Solution Marks Remarks
7. (a) (i) At the critical angle c,
sin 90o
n
sin c
1
 1.36 1M
sin c
c = 47.3° 1A
2

(ii) Angle of refraction at E = 90° – 47.33° = 42.67° 1A

By Snell’s law
sin
 1.36 1M
sin 42.67o
 = 67.2° 1A
3

(iii)

B C

E


A D 1A Larger i, larger r at E
1A Ray escapes the block on side BC
2

(b)

45°

(i) 1A Light ray from object escapes the

prism.
The angle of incidence of the light ray from the object is
(45°, which is) less than the critical angle of the plastic 1A
prism.
The light ray will be refracted instead of being reflected, 1A
and fails to travel through the periscope.
3

(ii) Glass prism (with critical angle less than 45°)

OR
Plane mirror (with coating on the front surface) 1A
1

2017-DSE-PHY 1B–7

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Solution Marks Remarks
8. (a)

+
A

1A Correct symbols of light bulb,

variable resistor and voltmeter
1A Correct positions
+ 1A Correct positive terminal connection
V for the voltmeter

(b) As the voltage across the light bulb increases, the temperature of 1A
the light increases,
thus the resistance of the light bulb increases. 1A
2

(c) R = V/I is the definition of resistance. 1A

It is applicable to all conductors.
1

(d) At V = 0.1 V
V 0.1
R  = 1.31  1A
I 76  10 3

At V = 2.5 V
V 2.5
R  1A for reading correct values from the
I 250  10 3 graph
= 10  1A
3

(e) l
R
A
RA
l




1.31  1.66  10 9  1M+1M 1M for suitable resistance from (d)
1M for correct substitutions
8
5.6  10 1A
= 0.039 m
3

2017-DSE-PHY 1B–8

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Solution Marks Remarks
9. (a) (i) The magnetic field at Q due to P points out of the paper. 1A
1

(ii)
wire P wire Q

1A force to the left

1

(iii) The magnetic field at Q due to P is

 I
BQ  o P 1M
2 r
For a certain length segment l of wire, the magnetic force is
F  BQ I Q l sin 
 I 1M
 o P I Ql
2 r
The magnetic force per unit length is
F o I P I Q
Fl  
l 2 r 1M
3

(iv) The two forces is an action and reaction pair, 1A “The force acting on P due to Q is
Thus the two forces are equal in magnitude. 1A o I P I Q
Fl  , thus the two forces
2 r
are equal in magnitude” (0A)
2

(b) (i) As current passes in the same direction between “As current passes in the opposite
neighboring wire segments, 1A direction between neighboring wire
the wire segments attract each other, segments, the wire segments repel
and the solenoid is compressed. 1A each other, and the solenoid is
stretched.” (1A)
2

(ii) Current is still flowing in the same direction between

neighboring wire segments at each instant, 1A
thus the solenoid will be compressed due to magnetic
force.
1

2017-DSE-PHY 1B–9

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Solution Marks Remarks

84 Po 82 Pb 2 He
10. (a) 210 206 4 2A
2

(b) The  particles ionize the air particles. 1A

The ions neutralize the charges on the dust/photo or film surface. 1A
2

(c)  particles has a range of only a few centimeters in air. 1A

1

(d) 365
 1  138
Activity after 1 year =  
2 1M
= 0.160 unit 1A
2

2017-DSE-PHY 1B–10

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