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

2018 年香港中學文憑考試
HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION 2018

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

本評卷參考乃香港考試及評核局專為今年本科考試而編寫，供閱卷員參考之用。本評卷
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考試 成 績公 布 後， 將 本評 卷參 考 提供 任 教本 科 的教 師參 閱 。

- 在任何情況下，均不得容許本評卷參考之全部或其部份落入學生手中。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
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 2018


2018-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

2018-DSE-PHY 1B–2

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Solution Marks Remarks

1. (a) Amount of energy required E = mcT
= 6  4200  (50 – 15) 1M Correct sub.
= 882 000 J (or 882 kJ)
E 882000 1M Correct eqn.
Power P = 
t 60
= 14700 W (or 14.7 kW) 1A
3

(b) Let m kg per minute be the water flow rate m 1

mcT = Pt 1M OR 
t T
m(4200)(40 – 15) = 14700  60 m 50  15
m = 8.4 (kg min1 or kg) 1A  
6 40  15
2

2. (a) pV
n=  pV (for constant T)
RT 1M
nA ( p)(3V )

nB (2 p)( 2V )
nA = 0.75 × 0.80 mol
= 0.60 (mol) 1A
2

(b) (i) n = n A + nB 1M
p (3V + 2V) = p (3V) + (2p)(2V)
p = 1.4 p 1A
2

(ii) When the tap is open, no. of gas molecules in vessel A 1A

increases due to the (net) flow of molecules from B to A,
the frequency of collision of gas molecules with the 1A NOT accept:
vessel’s wall increases, pressure increases. collide more vigorously/violently
2

2018-DSE-PHY 1B–3

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Solution Marks Remarks

3. (a) If the maximum load is exceeded, the braking distance will 1A OR a larger friction is required in
increase if the friction provided remains the same. braking the vehicle within the same
Vehicles would not be able to stop quick enough in case of 1A distance, accident may occur if the
emergency. brakes cannot provide such frictional
forces.
2

(b) (i) If the brakes are applied continuously, thermal energy 1A

generated will heat up the brake pads / brakes to too high
a temperature that the brakes may fail.
1

(ii) Let D be the distance travelled along the ramp.

Kinetic energy of vehicle becomes its gravitational 1M
potential energy:
1 mass of vehicle = m
m(25) 2  m(9.81)( D sin 30 )
2
D = 63.710499 m 1A
 63.7 m (62.5 m for g = 10 m s2)
2

4. (a) (i)
K.E. + P.E. =
1
0.342  0.39.810.2 1M+1M
2
= 2.4 + 0.5586 = 2.9886 J = 2.4 + 0.6 = 3.0 J for g = 10 m s2
1A
 2.99 J (3.0 J for g = 10 m s2)
3

(ii) As the spring gun is fixed, there is external force acting 1M

on the system / the gun,
total momentum (of the spring gun and cannon ball) is 1A
not conserved.
2

(b) 1 2 2(usin ) 2(4sin50 )

Vertical : s = ut + at 1M OR t f  
2 g 9.81

0 = 4 sin50° tf –
1
9.81 tf2
2
tf = 0.624705 s (0.612836 s for g = 10 m s2)
 0.625 s (0.613 s for g = 10 m s2) 1A

Horizontal : R = 4 cos50°  tf = 4 cos50°  0.625 2(usin )

1M OR R  u cos  
= 1.606210 m g
 1.61 m (1.57 m for g = 10 m s2) 1A
4

(c) tf increases 1A
since the initial vertical velocity / component is greater. 1A
2

2018-DSE-PHY 1B–4

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Solution Marks Remarks
5. (a) (i) m (5.0 cm) = 50 g (10.0 cm) 1M
m = 100 g or 0.1 kg 1A
2

(ii) Counter-weight position : 10.0 cm  0.1 cm 0.1

OR 100  ( )  1g
0.1 10.0
Percentage error = 100%  ( ) 1% 1M
10.0
 m = 101 g to 99 g
i.e. maximum error =  1 g 1A
2

(b) Spring balance reading = mg = (0.1 kg) (9.81 N kg1)

= 0.981 N
 1.0 N 1A
1

(c) (i) counter-weight position spring balance reading

on beam balance

the same reading increases

1A+1A
2

(ii) The beam balance would fail to work / to measure the 1A

mass of the load,
as the apparent weight is zero, the counter-weight can 1A
take any position.
2

2018-DSE-PHY 1B–5

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Solution Marks Remarks

6. (a) Place the (cylindrical) lens on the piece of paper, trace the OR Using lens formula
outline of the lens and mark its principal axis on the paper. 1A

Direct a ray parallel to the principal axis and trace the path(s) of
the (emerging) ray(s) on the paper. 1A
Extend (the path of) the emerging ray backward and locate the
intersection point F (with the principal axis). 1A

Measure the distance from F to the centre of the lens which

gives the focal length of the lens. 1A

Source of error:
Scale uncertainty of the plastic ruler (read to nearest mm).
OR Unable to mark the path correctly because of the Any
thickness of the beam of light from the ray box ONE
OR The ray(s) is/are not parallel (to the principal axis)
OR Any reasonable answer 1A
5
L
(b)
O 1 cm representing 5 cm

A
principal axis
F

(i) L is diverging/concave. 1A
- only a diverging lens can produce a virtual image
between the object and the lens 1A

(ii) Focal length = 30 cm 1A

Correct ray 1A
2

(iii) Correct ray 1A

1

2018-DSE-PHY 1B–6

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Solution Marks Remarks

7. (a) (i) Increase the separation between the double slit and the 1A
screen, D.
1

(ii) The separation of the bright dots on the screen becomes 1A

larger, thus the percentage error in its measurement is
smaller.
1

(iii) The angular position of the 2nd order bright dot

 1.56 / 2 
 = tan 1   =29.124053 1M
 1.40 
10 3
Grating spacing d = = 2.5  106 m 1M
400

Applying d sin  = n,

2.5 10 6  sin 29.12
Wavelength  =
2 n=2
= 6.08378 107 m
 6.08107 m (= 608 nm) 1A
3

(b) (i) The equation can only be applied for Note : the order of magnitude of the
-   a (i.e. wavelength  separation of the two wavelength of sound is about 101 m
sources), OR  is comparable to a Any
- a << D (i.e. separation of the two sources << ONE
separation of the sources and detector, OR a is
much smaller than D 1A
1

(ii) For the 1st order maximum,

wavelength  = path difference PB  PA 1M
 (1  0.5) 2  2 2  (1  0.5) 2  2 2 1M
= 2.5 – 2.06155281 = 0.43844719 m  0.438 m

speed of sound:
v = f = 750  0.4384
= 328.835 m s1 1A
 329 m s1
3

2018-DSE-PHY 1B–7

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Solution Marks Remarks

8. (a) (i) To terminals X 1A
1

(ii) P=IV
800 W = I (220 V) 1M Correct sub.
I = 3.636364 A 1A
 3.64 A
2

(iii) 1
V 2 V 2 5V 2 1 1 
800    1M Heating : Req =   
R 4R 4R  R 4 R
2
V 1M Keeping warm : Req = 4R
Pkeepingwarm  1 
 1
   1  
4R
1 1A  R 4R  
 800( )  160 W
5  800 W      160 W
 4R 
 
 
3

(b) (i) Electrical energy (consumed) 1A

1

(ii) (1) only the fuse blows 1A

(2) only the RCCB cuts off the supply 1A
2

2018-DSE-PHY 1B–8

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Solution Marks Remarks

9. (a) (i) P
I
uniform
magnetic
field B
L v resistor

1A

Q 1

(ii) By Lenz’s law, a magnetic force FB acts on the rod that

opposes its motion, 1A
an external force F is needed to balance FB so as to
maintain uniform motion (or constant v) 1A

 F = FB = ILB (in magnitude) 1A

3

(iii) mechanical power input = Fv

= (ILB) v 1M

power input = power output (electrical)

ILB v = I  1M
 = BLv
2

(b) (i) Horizontal (component) is perpendicular to the mast. 1A

OR
Vertical (component) is parallel to the mast.
1

(ii)  = (B cos30°) L v 1M by (a)(iii) using the horizontal

= (50  106 cos30°) (20) (6) component of B
= 5.196152  103 V 1A
 5.20 mV

more electrons at end X 1A

3

(iii) No current. 1A
Both the cable and the mast cut the field lines in the same 1A
way, the e.m.f.’s produced are equal and thus oppose each
other.
2

2018-DSE-PHY 1B–9

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Solution Marks Remarks

10. (a) (i) 226  206 = 20 (multiple of 4 for ) 1M
206
82 Pb is the end product
1A
2

(ii) 50
 1  1600
 undecayed Ra-226 left =   1M
2
= 97.58721 % 1A
 97.6 %
2

(b) (i)  random process 1A

1

(ii) Some of the daughter products of Ra-226 are also

radioactive and may emit  particles. 1A
1

(iii) - raise the source to a distance greater than the

range of  (a few cm) will cease to have sparks. Any 1A
- insert a paper between the source and the gauze, ONE
sparks will cease to produce.

Reason: weaker ionizing power of  &  radiations 1A

2

2018-DSE-PHY 1B–10

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