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Nat Chan 物理課程系列 NAT CHAN PHYSICS SERIES 2016 年香港中學文憑模擬試 HONG KONG DIPLOMA OF SECONDARY EDUCATION MOCK EXAM 2016

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

本評卷參考乃我們專為今年本科考試而編寫,供學生參考之 用 。 學 生 在完 成 閱 卷 工 作 後 , 若 將 本評 卷 參 考 提 供 其 他 修 讀本 科 同 學 參 閱, 本 人 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 囧 察 手 中 。 囧 察若 索 閱 或 求 取 此 等 文 件, 同 學 應 嚴 詞拒 絕 , 因 囧 察 極 可 能 將 評卷 參 考 視 為 某 物 理 的 超電 磁 炮 製 作 指南 , 以 致 子 烏 虛 有 , 明 張目 膽 。 這 種 超 然 的 慈 母態 度 , 既 不 符顧 全 大 局 及 今 天 我 原 則 ,亦 有 違 原 來 我 非 不 快 樂只 我 一 人 未 發覺 之 旨 。 因 此 , 本 局 籲 請各 同 學 通 力 合 作 , 堅 守上 述原 則 。 This marking scheme has been prepared by the Nat Chan and his Royal Team of Physics for students’ reference. Nat Chan Physics has no objection to students sharing it, after the completion of marking, with colleagues who are studying the subject. However, under no circumstances should it be given to black mother because they are likely to regard it as a set of model of electromagnetic gun. Students should therefore firmly resist black mothers’ requests for access to this document. Our examinations emphasise the same on you, what is are your daily work and I don’t think so lor. Hence the use of model answers should be I go to school by bus. Nat Chan Physics team is counting on the co-operation of students in sharing our page and giving like to our posts.

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2016-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 the marking scheme, alternative answers and marking guidelines are in rectangles .

6.

OSM marking symbols: 



correct point





wrong point





point to highlight



_ _ _

incomplete answer





missing point





entering text/comment

2016-DSE-PHY 1B–2

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1.

(a)

(i)

Air is a poor conductor It reduces heat lost by conduction

1A 1A

(ii)

Trap air in the cup It reduces heat lost by convection

1A 1A

(i)

mw cw ΔTw = mi li 0.2  4000  10 = m  334000 m = 0.0240 kg or 24.0 g.

1M

Greater Extra amount of ice is required to absorb heat gain from surroundings

1M 1A

2

2 (b)

1A 2

(ii)

2 2.

(a)

To control temperature of air by water bath

1A

(b)

∆P = F /A = mg / π r2 = 625 Pa

1M

P = 1 atm + 625 Pa = 101 950 Pa

1A

P’= 1 atm + 2 ∆P = 102 574 Pa By general gas law 𝑃𝑉 𝑃′ 𝑉 ′ = ′ 𝑇 𝑇

1M

1

2 (c)

1M

101 950 × 300 102 574 × 𝑉 ′ = 25 + 273 50 + 273 V’ = 323 cm3

1A 3

3

(a)

(i)

1

1M

2

By 𝑠 = 𝑢𝑡 + 𝑎 𝑡 2 0 = 250 + ½ (-9.81) t t = 51.0 s

or

t = 0 (rej.)

1A 2

(ii)

K.E. = ½ m v2 = 319 J

(i)

The bullet fall back from rest from the highest point with acceleration. As magnitude of air resistance increases with speed, it increases as the bullet falls down. When the magnitude of air resistance became the same as that of bullet’s weight, terminal speed attained.

1A 1A

(ii)

m g = c m v2 v = 95.3 m s-1

1M 1A

(iii)

K.E. of the bullet = ½ m v2 = 46.3 J >> 13 J Hence it would hurt the protesters

1A 1A

(i)

s = u// t = 250 sin 10o  28.2 = 1.22 km or 1 220 m

(ii)

Smaller. Air resistance causing decreases in magnitude of horizontal component of velocity. This factor did not include into our calculation.

1A 1

(b)

2

2

2 (c)

1M+1A 2

2016-DSE-PHY 1B–3

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1M 1A 2

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4.

(a)

𝛥𝑣 1.638 − 0.138 = = 2.5 m s −1 𝛥𝑡 0.6 𝑚𝑔 sin 𝜃 − 𝑓 = 𝑚𝑎 f = 3.61 N

1M

(i)

∆p = m ∆v = 1.5  3 = 4.5 kg m s-1

1A

(ii)

𝛥𝑣 𝐹 − 𝑚𝑔 sin 𝜃 − 𝑓 = 𝛥𝑡 F = 56.0 N

1M

𝑎=

1M 1A 3

(b)

1

1A 2

(c)

(i)

v / m s –1 First line: much greater slope

0.886 s (for (c) (ii))

(ii)

5.

t/ s

Second line: parallel to first 0.6 s end with smaller than that of 0.6 s

1A

1A 1A

Upward motion: 𝑚𝑔 sin 𝜃 + 𝑓 = 𝑚𝑎 a = 7.31 m s-2

1M

By v = u + at 0 = 1.362 – 7.31 t t = 0.186 s Hence the final time will be 0.7 + 0.186 = 0.886 s. (Accept 0.882 to 0.887 s)

1A

(a)

Convex lens (correct spelling)

(b)

1 1 1 + = 𝑢 𝑣 𝑓 1 1 1 + = 5 10 𝑓

1A

2 1

1M

f = 10 / 3 = 3.33 cm

1A 2

(c)

2A

A Screen

5cm

Q

P

B A Inverted, Magnified, Real (1A – two correct, 2A – all correct)

2A 4

2016-DSE-PHY 1B–4

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6

(a)

(i)

1A

1 (ii)

The wavelength of sound = v / f = 340 / 200 = 1.7 m near to doorway width 1.8 m The sound wave enter the door and change the direction by diffraction at doorway

1A 2A

Hence the sound wave can reach Q. 3 (b)

7

(i)

Interference occurs in front of two loudspeaker. Sound can only be heard faintly at positons where constructive interference occurs

1A 1A

(ii)

The signal of two loudspeakers are out of phase.

1A

(a)

The mass of insulator M and the rider m was measured by an electronic balance. The square wire frame is put onto the blade part of the plastic stand. The two magnets are placed beside the wire on the side without insulator. The magnetic field strength B between magnets is measured (by hall probe). The wire frame is connected to the power supply. Switch on power supply, adjust position of wire frame until the wire frame kept horizontal. Put the rider on the wire frame for fine adjustment. The current can be found by calculation. (mg r + Mg R = I l B r’) (r, R and r’ are distance of rider, insulator and wire frame between magnets from plastic holder) (l is the length of wire frame)

(b)

No The wire frame cannot react to the change of current. (Accept other reasonable answers)

1M 1A

(a)

The heating element can only share the same voltage when they are connected in parallel They can both work at rated value.

1A 1A

(b)

(i)

2 1

Any 6

6

2 8

2

(ii)

𝑙 4.5 𝑅 = 𝜌 = 1.1 × 10−6 × = 72.8 Ω 𝐴 6.8 × 10−8 2

2

(i)

𝑉 220 = = 665 W 𝑅 72.8 r' = r/2 → A’ = A/4 → R’ = 4R → P’ = P/4 (V unchanged) P = 166 W

(ii)

P = 166 + 665 = 831 W

𝑃= (c)

1 1M+1A

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2

1M +1M 1A 1A

2016-DSE-PHY 1B–5

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1A

2 1

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9.

(a)

When switch is closed, current pass through solenoid and produce magnetic field Aluminium ring experienced change in magnetic field and current is induced. The induced current produce magnetic field. The magnetic force acts on ring hence shoot it up the rod.

1A 1A 1A

(b)

Iron can be magnetized.

1A

(c)

The aluminium ring will not shoot up No induced current flow in the ring which causes no magnetic force and hence the ring cannot shoot up

1A 1A 1A

(a)

The gamma radiation has much lower ionizing power It causes less harm to human (accept other reasonable answers) (i) k = ln 2 / t = ln 2 / (8  24) = 0.003610 min-1 A( t = 5 hrs) = 20 000 e-(0.003610  5) = 19 642 Bq

1A 1A

3 1

3 10

(b)

2 1M 1M+1A 2

(ii)

𝑉 5 = 19642 19.7 V = 4990 cm3

1M (4985.3 cm3)

1A 2

2016-DSE-PHY 1B–6

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Expl ana tio n t o fe atur ed MCs (MC Full Answer: https://goo.gl/m5QY5a ) 1.

C The block is in contact with water, there should not be convection between them.

2.

B Double heat capacity, not specific heat capacity.

4.

C The applied forces are of same magnitude, but direction not specified.

5.

B The inclined plane is friction-compensated runway of m suggested that f = mg sin 30 o .

6.

D The initial speed of particle did not specified in the question

16.

B SP – RP = -1.5 times of wavelength

18.

A #allin6

19.

D Notice that this is named major common mistake in 2015 -DSE

25.

D The internal resistance of cell owns part of the 6.0 V potential difference. The voltmeter should reads smaller than 3.0 V

27.

B No induced current hence zero magnetic force. The applied force F also contributed as a part of net force.

28.

D (2): The magnetic force acting on first corner produced clockwise moment

33.

A Was a typo in the 1 st version. Now corrected. Sorly.

2016-DSE-PHY 1B–7

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Full Answer Key for Paper 1A (As of March 30, 2016)

2016-DSE-PHY 1B–8

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2016-NC DSE-Phy 1B-MS(E).pdf

... unless a particular method has been specified. in the question. Markers should be patient in marking alternative solutions not specified in the marking scheme.

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