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

2019 年香港中學文憑考試
HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION 2019

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 service agreement
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 2019


2019-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

2019-DSE-PHY 1B–2

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Solution Marks Remarks
1 (a) (i) (1.5)(4200)(60 – 10) = m(3.34  105) + m(4200)(10 – 0) 1M+1M (2 M)
m = 0.837766 kg  0.838 kg 1A

(ii) More ice is needed to cool the container as 1A

it will release heat/thermal energy as well. 1A
2

(b) (i) Conduction:

- Foam is a poor conductor (of heat) and it 1A
minimizes the transfer of heat/thermal energy
from the surroundings to the cool contents/ice
Any
cream (inside the bag). ONE
OR
Convection:
- The zipper prevents convection between the hot
air outside and the cool contents/ice cream
(inside the bag).
1

(ii) (Radiation) Make the outer surface (of the bag) shiny. 1A
1

2. (a) pV = nRT
(100 × 103)(0.52) = n (8.31)(273+15) 1M
n = 21.727504 (mol)  21.7 (mol) 1A
2

(b) (i) nRT

Since pV = nRT  V  / volume V of the balloon 1A
p
depends on both T and p,
Accept: volume V decreases as
the (fractional) decrease in pressure p (with height) is
1A temperature T decreases would be
greater/faster than the (fractional) decrease in
true only if pressure p is constant,
temperature T.
however, p decreases with height.
2

(ii) (1) pV
 constant
T
(100)( 0.52) p(8)
 1M
(273  15) 216
p = 4.875 kPa or 4875 Pa 1A
2

(2) p = p0ekx
4.875 = 100 e0.138 x 1M
x = 21.89166726 km  21.9 km 1A Accept 21.7 km to 22.0 km
2

2019-DSE-PHY 1B–3

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Solution Marks Remarks
3. (a) (i) (1) 1 2
mv  mgh 1M
2
v 2  2(9.81)(12)
v = 15.344054 m s1  15.3 m s1 1A
(v =15.491933 m s1  15.5 m s1 for g = 10 m s2)

(2) 1 2
s gt 1M
2
1
12  (9.81)t 2
2
t = 1.564124 s  1.56 s 1A
(t = 1.5491933 s  1.55 s for g = 10 m s2)
2

(ii)
F  mg  ma
1M+1M
70  (15.3 - 0)
F  70  9.81
0.3
= 4266.9793 N  4270 N 1A
(F = 4314.7845 N  4310 N for g = 10 m s2)
3

(iii) Elastic potential energy 1A

1

(b) (i) - Velocity is too high, hence the force for 1A

deceleration is too large. Any
- The life net can be torn. ONE
- The firemen are not able to hold the life net tight.
1

(ii) There exists a horizontal velocity when a person jumps and 1A

the horizontal displacement is very difficult to estimate as
it depends on the time of fall, which is usually long. 1A
2

2019-DSE-PHY 1B–4

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Solution Marks Remarks
4. (a) (i)  5.0
 1M/1A
2π 2π
= 0.795775 (rev s1)  0.80 (rev s1)
1

(ii)
O
23.1
1 m long inextensible string

FC
centripetal force pendulum bob
(tension component) of mass 30 g
FC correctly indicated. 1A

FC = mr2 Or Tcos = FC and Tsin = mg

= (0.03)(1 × cos 23.1)(5.0)2 1M mg
FC  cos  = 0.689866 N
= 0.689866 N  0.700 N 1A sin
(FC = 0.7033402 N  0.703 N for g = 10 m s2)
3

(iii) Horizontal component of tension provides the centripetal 1A

force, thus tension is larger than the centripetal force. 1A
OR Tcos = FC  T > FC
2

(b) (i) The gravitational force is perpendicular to the moon’s

1A
motion/displacement,
thus no work is done on the moon by this force (k.e./speed
1A
unchanged)
2

(ii) The claim is incorrect as, by Newton’s third law of motion, 1A

gravitational force of the same magnitude but in opposite
direction is acting on the Earth by the moon. 1A
2

2019-DSE-PHY 1B–5

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Solution Marks Remarks
5. (a) (i) 0.06
wavelength  =
7 1
= 0.01 m (= 1 cm) 1A
1

(ii) speed v = f = 10 × 0.01

= 0.1 m s1 (= 10 cm s1) 1M/1A
1

(b) (i) frequency = 10 Hz 1A

1

(ii)
shallow region P deep region Q
6 cm 1A Wavefronts bend away from the
normal of the shallow-deep
boundary.
1A Double (longer) wavelength
with wave fronts continuous at
boundary

(iii) Refraction. 1A
It is due to the change in wavelengths/wave speeds in 1A
different media/depth.
2

2019-DSE-PHY 1B–6

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Solution Marks Remarks
6. (a) L is diverging/concave. 1A
Only diverging/concave lens forms diminished, virtual 1A
image.
2

(b) L

(1) observer 1 cm representing 5 cm

(2)

F B F principal
axis

Correct position and height of object 2A

2

(c) Correct ray to locate F and focus F correctly marked. 2M Ray (1) or (2)
16.5 cm (Accept 15.5 cm – 17.5 cm) 1A
3

(d) Correct ray p from tip of object 2A Diverging and can reach the observer
2

2019-DSE-PHY 1B–7

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Solution Marks Remarks
7. (a)

1A Correct circuit
+  1A Correct polarity
V

Close the switch and record corresponding V and R readings 1A

Adjust the resistance R to lower/other value(s) and repeat the 1A
experiment

Precaution:
- First set the variable resistor to its maximum/a large value Any
1A
- Open the switch after each measurement ONE

(b) Terminal voltage V delivered increases with increasing (loading) 1A V

resistance R (or graphical representation) 
R 
V  V  - r
Rr OR Rr 1A
2 0 R

2019-DSE-PHY 1B–8

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Solution Marks Remarks
8. (a) Earth N L

X (Earth)

mains socket 1A
Z (L)

Y (N)
X Y Z
1

(b) (i) - If one of the lighting sets/circuits fails, the other 1A

(in parallel) can still operate, i.e. both work Any
independently. ONE
- Both can work at the rated power.
1

(ii)
P  IV
1M
(300  450)  I (220)
I = 3.409091 A  3.41 A 1A
Thus 5 A fuse should be used. 1M/1A Correct deduction
3

(c) Electrical energy used per day

= 0.500 kW × 8 h + 2 kW × 0.5 h + 3 kW × 2 h 1M
= 11 kW h
Cost = $0.9 / kW h × 11 kW h 1M
= $9.9 1A
3

2019-DSE-PHY 1B–9

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Solution Marks Remarks
9. (a) (i) By Lenz’s law, an e.m.f. would be induced such that it
opposes the change, i.e. decrease of magnetic flux (into the 1A
paper) by driving an induced current (clockwise) in the 1A
coil/circuit (complete).
2

(ii) NΦ  NBA

 (20)( 0.3)( 0.005)
 0.03 Wb 1M/1A
NΦ 0.03
  1M
t 0.5
 0.06 V (or 60 mV) 1A
3

(b) (i) (Magnetic) flux change is double that in (a)(ii), 1M/1A

i.e. 0.06 Wb.
1

(ii) Direction of current : PQRS 1A

1

(c) (i)

aluminium plate × × ×
1A Correct position (accept just within
× × ×
1A the magnetic field)
× × × Correct direction (clockwise)
× × ×
× × ×
2

(ii) Move/swing to the right initially/momentarily/briefly. 1A

1

2019-DSE-PHY 1B–10

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Solution Marks Remarks
10. (a) (i) x=3 1A
1

(ii) More neutrons are produced in each fission for triggering 1A

further fissions, i.e. x > 1.
1

(b) (i) m = m0e–kt

ln 2
k
t1
2
2109
 2 109  1 7.04108
ln 2    Or 0.06 = m0 ( )
8 1M 2
0.06 = m0 e  7.04 10 
m0 = 0.429882832 (kg)  0.430 (kg) 1A
2

(ii) 0.430
 0.03073691  3.1 % > 3%
13.556  0.430
Thus natural nuclear fission was possible. 1M/1A
1

(c) Underground water might run dry.

OR Energy released by fission drys up the underground water. 1A

Therefore, fission might stop without slow neutrons. 1A

2

2019-DSE-PHY 1B–11

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