ldse-phy--as 0 HKDSE PHYSICS Paper A Suggested Solutions Prepared by Andy Lai HKDSE Physics Teacher MC 係 分 ABC Grade 既 地 方, 越 出 越 煩, 越 出 越 深, 同 學 一 定 要 快 又 要 好 小 心! Enrollment Hotline: 677 300 Website: www.andylai.hk MSN: mrandylai@hotmail.com Address: Rm706, Prosper Commercial Building, 9 Yin Chong Street, Mong Kok, Kowloon, Hong Kong.
0 HKDSE Physics Paper IA Suggested Answers. C. A 3. A 4. D 5. A 6. D 7. A 8. C 9. B 0. C. A. C 3. D 4. D 5. B 6. D 7. B 8. D 9. A 0. A. A. D 3. C 4. B 5. B 6. B 7. B 8. B 9. D 30. C 3. B 3. A 33. C 34. C 35. C 36. D MC 係 分 ABC Grade 既 地 方, 越 出 越 煩, 越 出 越 難! 轉 數 快, 概 念 清! 缺 一 不 可! 同 學 一 定 要 快 又 要 好 小 心! Andy s predicted M.C. Grade boundaries: 5**: 35 / 36 5*: 3 / 36 5: 30 / 36 4: 6 / 36 3: 0 / 36 : 6 / 36
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS Section A. C. Let the mass of block X and block Y be m X and the specific capacity of block X c X and Y c respectively m Y respectively.. By Law of conversation of energy, Energy gained by block Y = Energy loss by block X m c ( 40 T ) = m c (40 T ) Y Y c ( T 30) = c (40 T ) Y X X X 3. Since c X > cy and so T 30 < 40 T gives T > 35 degree celcius.. A. The skin normally is hotter than the alcohol.. Therefore, Heat will transfer from skin to alcohol. 3. The alcohol gains enough K.E. to increase in temperature and gain enough P.E. to change from liquid to gas state, it will evaporate. 3. A. By pv=nrt gives p = (nr/v)t which is a straight line for p (Pa) and T (K).. However, T = θ + 73 3. Therefore, pv = nr( θ + 73) gives line with slope = nr 73 and y-intercept V VnR nr 73nR p = θ + which is a straight V V 4. Now, the number of gas molecules in the vessel is halved. Therefore, both the slope and y-intercept will decrease. 0 A Lai learning Center. ALL RIGHTS RESERVED. P.
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 4. D. Increase in temperature Increase in K. E. = 3RT N A K.E. since the molecular. Changing state Change in P.E. which is used to overcome the attractive force between molecules. 3. Therefore, when water boils at 00 C, there is no change in temperature, only changing from water to steam. 4. The potential energy of water molecules increases to overcome the attractive force between them and so they can increase the distance between water molecules. 5. A. When θ increases from θ 80 ο 0 ο cosθ + F ( F sinθ) F + Resultant Force = ( ) = F cos θ + F F cosθ + F + F sin θ = F F cos F + θ + F. Therefore, by substituting ο θ = 0, ο θ = 30, ο θ = 60, ο θ = 80, you will find the resultant force is decreasing. ο θ = 90, ο θ = 35 and P. 0 A Lai learning Center. ALL RIGHTS RESERVED.
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 6. D. By principle of moment, Sum of clockwise moment at any point = Sum of clockwise moment at that point.. T (.5 sin θ ) = Mg () 3. Therefore, the initial force required to raise the gang plank when it is horizontal is: T (.5 sin 45) = Mg () and gives T = 0. 94Mg 4. As the gangplank rise, θ increases and so tension will decreases and so smaller than 0.94Mg. 7. A. For free fall body, the acceleration due to gravity is always equal to g when take downward is negative, i.e. a = -g. For v-t graph, the slope = acceleration. 3. Therefore, () is the only answer for v-t graph. 4. For s-t graph, the slope = velocity. 5. Therefore, (3) is the best answer since the slope of s-t graph is negative and increasing in magnitude, which means the velocity is going downward and increasing. 6. However, do you when is the maximum point, the point of collision, the point of leaving the ground respectively representing on the v-t and s-t graphs respectively? 0 A Lai learning Center. ALL RIGHTS RESERVED. P. 3
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 8. C. Consider P and Q as a single system, By F net = ma, F F = 3ma (Do you know why not F F?) F F a = 3m. Consider either P or Q separately, For P: F T = ma or For Q: T F = ma ( F F F F T = F m ) or T = F + m( ) 3m 3m T = (F + F ) 3 9. B. By Power = Energy / time. P= mgh / t 3. P = (0.5)(9.8)(.5)/.5 = 8.75 W = 8. W (to sig. fig.) P. 4 0 A Lai learning Center. ALL RIGHTS RESERVED.
0 HKDSE Physics Paper IA Suggested Solutions 0. C DSE-PHY--AS. By Newton s st law, Net Force on the body = Zero Either at rest or uniform velocity. Therefore, the st statement is true.. The direction along the plane: mg sin θ = friction which gives friction = mg sin 30 ο = 0. 5mg Therefore, the nd statement is true. 3. The Net force is zero in all direction. Therefore, the acceleration is always zero. Therefore, the 3 rd statement is wrong.. A. By Newton s nd Law, F net = ma F net = ma F f = F f = a = ) ma ma ( F m f m. Therefore, slope = /m and y-intercept = -f/m. C. By considering vertical motion: u = 0, v =?, a = 9. 8, s = 000, t =?. By s ut + at 000 = ( 9.8) t gives t = 4.3s = gives 3. Remember, Vertical motion and Horizontal motion are independent. 0 A Lai learning Center. ALL RIGHTS RESERVED. P. 5
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 3. D. By Law of conversation of momentum, The total linear momentum of a system is conserved provided that there is no external force acting on the system. There is no external force in the system. Therefore, in all direction, the total linear momentum of a system is conserved. Statement is wrong.. For elastic collision, the total kinetic energy for the whole system is conserved. The total kinetic energy of the whole system before collision = mu Therefore, the total kinetic energy of the whole system after collision is also equal to mu Statement is true. 3. By Law of conservation of energy and elastic collsion: = mv P mv Q where v P and mu + collision. mu u = v = m( v P + v Q P + v Q ) v Q are the velocity of P and Q after Since v 0, therefore, v < u Q P Statement 3 is wrong. P. 6 0 A Lai learning Center. ALL RIGHTS RESERVED.
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 4. D. Consider a satellite moving around the Earth. As the satellite is in a circular orbit around the Earth, the gravitational force (weight) acting on the satellite provides the centripetal force needed for circular motion.. By Newton s Law of Gravitation and Newton s nd Law: F net = ma GM m r mv e = where s rs v a = r GM r e v = or s s e gοr GM e v = ( Q gο = = 9. 8ms ) r R 3. Therefore, the speed required for the satellite moving around the Earth is independent of the mass of the satellite, only depends on the radius of the orbit ( r s ) and the mass of the Earth (M) GMem mv = 4. Moreover, r r s s GM r s e = v GM e = v vt ( ) π π = v T GM e 3 Therefore, v 3 T = constant, which means same velocity implies same period. Therefore, Option D is the wrong option. 0 A Lai learning Center. ALL RIGHTS RESERVED. P. 7
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 5. B. In travelling waves, all particles will under vibration and will not stay at rest forever. They will only be at rest at amplitude.. In travelling waves, the particles with separation equal to one wavelength are in phase. Particles with separation equal to half wavelength are anti-phase. 3. The wavelength of the wave is equal to the distance between particle a and particle i, which is equal to 3 cm. 4. The frequency of the wave is unknown since the period of the wave cannot be determined. 5. Remarks: Please note that you cannot simply says the period is 0. x = 0. s since you do not know the vibration of the particles between t = 0 s to t = 0. s exactly. The particle may complete 0.5 cycle,.5 cycle,.5 cycles, etc. 6. D. When a pulse travel from less dense medium to denser medium, there will be a π phase change on the reflected pulse.. Therefore, the answer follows. P. 8 0 A Lai learning Center. ALL RIGHTS RESERVED.
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 7. B n. By ( n λ = ), the refractive index of medium X = (4/3) x =.33 λ. Please remember: n n = v v = λ = λ sinθ. sinθ 8. D. For stationary wave, Different particles have different amplitudes. In particular, amplitude is maximum at anti-nodes but minimum at nodes. All particles attain their amplitude at the same time.. Therefore, P attains its amplitude (crest) at t = 0. 3. The result follows. 9. A. In air, the wavelength of infra-red is longer than that of ultraviolet radiation.. Please remember the following Electromagnetic magnetic wave spectrum: f / λ : Radio waves Microwave Infrared Visible light UV light X-ray Gamma ray 0 A Lai learning Center. ALL RIGHTS RESERVED. P. 9
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 0. A. By d sin θ = nλ, For Red light: d sin θ = m(657) ---(), m is the order of diffracted red light. For Violet light: d sin θ = n(438) ---(), n is the order of diffracted violet light. () / () gives d sinθ m(657) = d sinθ n(438) m 438 = = n 657 3 m : n = : 3. A. The correct path for the light ray before and after passing through the concave lens L is shown as follows: P. 0 0 A Lai learning Center. ALL RIGHTS RESERVED.
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS. D. Pitch of the note Frequency of the note no. of waves per second. Loudness of the note Amplitude of the note 3. Note = Fundamental frequency + Overtones (Overtones = notes of frequency of multiples of fundamental frequency) Quality of the note depends numbers and amplitude of overtones accompanying the fundamental frequency. Because of quality, you can distinguish Andy Lau s voices and Andy Lai s voices singing the same song and pitch. Also, you can distinguish trumpet sound and guitar sound although they are performing the same kind of songs. 4. The result follows. 0 A Lai learning Center. ALL RIGHTS RESERVED. P.
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 3. C. Human beings Hearing of Short range of frequencies Audible Frequency Range! (0 Hz 0000 Hz) Common audible sound. Frequency of sound > 0000 Hz Ultrasonic waves 3. Applications of ultrasound: e.g. Sonar: v = fλ, Q v = constant, c degree of diffraction! 4. The results follows. P. 0 A Lai learning Center. ALL RIGHTS RESERVED.
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 4. B. Assume P and R is negatively charged Q and R repel instead of attract! P and Q R and S Q and R. Assume P and R is positively charged Q and R repel instead of attract! P and Q R and S Q and R 3. Assume P is positively charged and S is negatively charged Q and R attract! P and Q R and S Q and R 4. Remarks: Like charges repel, unlike charges attract. 0 A Lai learning Center. ALL RIGHTS RESERVED. P. 3
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 5. B Q. By E = 4 πε r and it is a vector quantity, the addition of E-field is calculated by vector-sum method and the direction is from positive charge to negative charge.. At either Point X or Y, Both Electric field due to +Q and Q is pointing to the right. Therefore, it is impossible for them to cancel each other. 3. At point Z, Electric field due to +Q is pointing to the right while the Electric field due to -Q is pointing to the left. The charge of +Q is larger than that of Q but the distance of Z from +Q is larger than that from Q, the magnitude of E-field may cancel each other. Q 4. By E = 4 πε r and it is a vector quantity, the addition of E-field is calculated by vector-sum method. 5. By V Q = 4 πε r and it is a scalar quantity, the addition of electric potential is calculated by scalar-sum method. 6. At Point X, the potential due to +Q is positive and that due to Q is negative. However, the distance between X and +Q is the larger than that between X and Q. Therefore, the magnitude of potential of +Q must be larger than that of Q. Therefore, the potential at point X cannot be cancelled. 7. At either point Y or Z, the potential due to +Q is positive and that due to Q is negative. However, the distance between them and +Q is longer than that between them and Q. Therefore, the magnitude of potential of +Q may be equal to that of Q. Therefore, the potential at point Y or Z may be cancelled. 8. By elimination and choosing the best answer, the result follows. P. 4 0 A Lai learning Center. ALL RIGHTS RESERVED.
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 6. B. The circuit diagram can be redrawn as follows:. Thinking Backward: Equivalent Resistance across terminal a and b = 6 ohms Equivalent resistance of the combination of all resistor R = ohms (Do you know why?) 3. Now, replace the -ohm resistor with 6-ohm resistor. The equivalent resistance across terminal a and b = [(6)()]/(6+) = 4 ohms 4. Remarks: For convenience: R R = if R = R = R3 =... = Rn n. 7. B. If the resistance of the rheostat becomes zero, the light bulb will become short-circuited.. Therefore, all the current will flow through the rheostat instead of flowing through the light bulb. 3. The light bulb will not light up. 0 A Lai learning Center. ALL RIGHTS RESERVED. P. 5
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 8. B. Equivalent resistance of 3-ohm resistor parallel to 6-ohm resistor = 3 x 6 / (3 + 6) = ohm. Therefore the potential difference across the internal resistor = x ( / 4) = 6 V 3. Therefore, the voltage across the 6-ohm resistor = 6 = 6 V 4. Therefore, the current across the 6-ohm resistor = 6 / 6 = A P. 6 0 A Lai learning Center. ALL RIGHTS RESERVED.
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 9. D. By Fleming s left hand rule (FBI), Thumb: Force First finger: B-field Middle finger: Current. The magnitude of the magnetic force per unit length experienced by wire X due to wire Z = F. 3. Force per unit length acting on Z by the B-field of X = F (To the left) (Do you know why?) 4. Force per unit length acting on Z by the B-field of Y = 3F (To the left) (Do you know why?) 5. The magnetic force per unit length on Z due to both X and Y = 4F = F (To the left) 30. C. By Fleming s left hand rule (FBI), Thumb: Force First finger: B-field Middle finger: Current. However, the direction of electron flow is opposite to the conventional current. 3. Therefore, the direction of force acting on the electron is downward. 4. If the electron keeps moving in the straight direction, there must be an electric force acting on it upwards. 5. Therefore, the electric field should be pointing from the top to the bottom. 0 A Lai learning Center. ALL RIGHTS RESERVED. P. 7
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 3. B. By Lenz Law: Induced current flows in a direction such that it opposes the change that causing it!. There is a change in magnetic flux linkage when the square metal frame with side L moving with uniform speed into and out of the magnetic field of side 5L, although the direction of current is different. 3. The total time period during which a current is induced in the frame = L/v. 3. A. According to Faraday s law of electromagnetic induction: The voltage induced in a conductor is directly proportional to the rate at which conductor cuts through the magnetic field lines.. Since the conductor rod PQ is cutting through the magnetic field of the Earth, there will be induced voltage across it. The st statement is true. 3. However, this is an open circuit. Therefore, there is no induced current. The nd statement is wrong. 4. There is a no current in the conductor rod PQ. Therefore, there is no magnetic force acting on conductor rod PQ. The 3 rd statement is wrong. P. 8 0 A Lai learning Center. ALL RIGHTS RESERVED.
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 33. C. The fuse blows if the current in live wire is much larger than normal operation.. If X and Y are connected, it is short-circuited and the resistance is nearly zero. Therefore, the current is too large and the fuse blows. 3. If Y and Z are connected, it is also short-circuited since the potential of the neutral and the earth wire is also 0 V, the resistance is nearly zero. Therefore, the current is too large and the fuse blows. 4. If X and Z are connected, it is no effect since the potential of the neutral and the earth wire is also 0V, there is no potential difference between them and so no current passing through them. 5. Remember: Potential of Live wire: +/- 0 V Neutral and Earth wire: 0 V 34. C. The filament F is heated by the negative terminal P, the electrons inside the filament can get enough energy to overcome the attractive forces between the free electrons and the positively charged nucleus. This emission of electrons is called Thermionic emission.. After the emission of electrons, they will be accelerated by the positive terminal Q. 3. When the electrons collide with the metal target T, the X-rays are emittied from T as a form of electromagnetic wave. 0 A Lai learning Center. ALL RIGHTS RESERVED. P. 9
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS 35. C. By ln t t N N e = ο, when t 0 and t = 0, = N = Nο e ln N = 0. 707N ο. Therefore, 3 > f >. 4 36. D. Let A be the mass number, N be the neutron number and P be the proton number.. By A = N + P, N = A P which a a straight line with slope = and y-intercept = -P. The end. P. 0 0 A Lai learning Center. ALL RIGHTS RESERVED.
DSE-PHY--AS 0 HKDSE Physics Paper IA Suggested Solutions 黎 sir 教 室 將 陸 續 推 出 下 列 新 高 中 香 港 中 學 文 憑 試 模 擬 試 題 系 列, 包 括 : 0 HKDSE Physics Mock Exam Set Paper (Heat and Gases) 0 HKDSE Physics Mock Exam Set Paper (Heat and Gases) 0 HKDSE Physics Mock Exam Set Paper (Force and Motion) 0 HKDSE Physics Mock Exam Set Paper (Force and Motion) 0 HKDSE Physics Mock Exam Set 3 Paper (Wave Motion) 0 HKDSE Physics Mock Exam Set 3 Paper (Wave Motion) 0 HKDSE Physics Mock Exam Set 4 Paper (Electricity and Magnetism) 0 HKDSE Physics Mock Exam Set 4 Paper (Electricity and Magnetism) 0 HKDSE Physics Mock Exam Set 5 Paper (Radioactivity and Nuclear Energy) 0 HKDSE Physics Mock Exam Set 5 Paper (Radioactivity and Nuclear Energy) 0 HKDSE Physics Mock Exam Set 6 Paper (Astronomy and Space Science) 0 HKDSE Physics Mock Exam Set 7 Paper (Atomic World) 0 HKDSE Physics Mock Exam Set 8 Paper (Energy and Use of Energy) 0 HKDSE Physics Mock Exam Set 9 Paper (Medical Physics) 特 別 為 應 考 香 港 新 高 中 中 學 文 憑 試 的 同 學 編 寫 的 模 擬 試 題 系 列. 深 淺 程 度 參 考 過 往 會 考 和 高 考 熱 門 試 題, 再 加 上 可 能 出 現 的 新 題 種. 每 條 試 題 結 尾 均 標 註 了 相 關 課 程 章 節, 提 醒 同 學 唔 會 讀 漏 書 ". 概 念 (Qualitative) 與 計 量 (Qualitative) 並 重, 讓 同 學 深 入 思 考, " 轉 多 幾 過 彎 ". 特 別 設 計 " 分 課 題 (By Topics) " 模 擬 試 題, 讓 同 學 可 以 按 課 題 熟 識 程 度 練 習. 同 學 完 成 這 一 系 列 模 擬 試 題 後, 希 望 能 幫 助 提 升 應 考 中 學 文 憑 試 的 信 心. Level 5**: 00% 0% (The Best of the Best!) Level 5*: 90% (Superb!) Level 5: 5 80% (Excellent!) Level 4: 4 70% (Quite good!) od!) Level 3: 60% (Not bad!) Level : 50% (Need improvement!) Level : 40% (Need revise again!) Unclassified: 40% (Zero Knowledge!) Remarks:. This is my opinion only, not official HKEAA standard. 0 A Lai learning Center. ALL RIGHTS RESERVED. P.
0 HKDSE Physics Paper IA Suggested Solutions DSE-PHY--AS P. 0 A Lai learning Center. ALL RIGHTS RESERVED.
黎 sir 簡 介 Andy Lai BEng CUHK, MIEEE 畢 業 於 香 港 中 文 大 學 電 子 工 程 學 系, 黎 sir 教 室 創 辦 人 之 一. 超 過 5 年 教 授 中 學 文 憑 / IB Diploma / GCE / HSC / SAT / AP / GCSE / IGCSE / IB MYP 課 程 經 驗. 為 了 與 學 生 一 起 面 對 新 制 度 的 中 學 文 憑 試, 黎 sir 親 身 上 陣, 以 實 力 於 物 理 科 奪 取 5**, 證 明 寶 刀 未 老. 熟 悉 出 題 趨 勢, 教 授 考 試 取 分 技 巧 ; 鼓 勵 同 學 獨 立 思 考, 增 強 同 學 理 解 能 力. 善 用 生 活 化 例 子 講 解, 教 法 生 動, 增 加 學 習 趣 味 ; 深 入 淺 出, 明 白 學 生 學 習 上 的 困 難 和 需 要. 精 心 編 制 筆 記, 適 合 中 文 和 英 文 中 學 學 生 就 讀 ; 精 心 編 制 練 習 和 試 題, 協 助 同 學 盡 快 掌 握 答 題 技 巧. 黎 sir 在 中 學 和 大 學 時 代 已 是 一 名 傑 出 學 生, 曾 獲 取 的 多 項 學 業 上 和 運 動 上 的 獎 學 金 及 獎 項. 曾 代 表 香 港 參 加 國 際 性 運 動 比 賽, 取 得 優 異 成 績, 又 讀 得 又 玩 得, 絕 不 是 死 讀 書 的 書 呆 子. 任 教 科 目 : 所 有 數 學 科, 物 理 科, 化 學 科, 生 物 科, 經 濟 科, 商 業 科. 黎 sir 教 室 學 生 佳 績 : 首 屆 香 港 中 學 文 憑 (HKDSE), 多 位 學 生 取 得 5/5*/5** 級 以 上 佳 績. 更 有 學 生 考 獲 5 科 5** 級 科 5* 級 科 5 級 優 異 成 績, 在 全 港 760 考 生 中, 排 名 8, 入 讀 港 大 醫 學 院. 英 國 高 考 (GCE AS/AL), 多 位 學 生 取 得 A*/A 最 高 級 別, 更 有 學 生 考 獲 5 科 A*. 國 際 文 憑 (IB Diploma), 多 位 學 生 取 得 6 / 7 級 別, 更 有 學 生 取 得 總 分 40 分 以 上. 英 國 會 考 (IGCSE / GCSE), 多 位 學 生 取 得 A / A* 成 績, 更 有 學 生 取 得 8 科 A* 加 拿 大 大 學 預 科 (CESI) 數 學 課 程 MCV4U, 取 得 98 / 00, 99 / 00 成 績 學 生 成 功 拔 尖 (EAS), 提 早 入 讀 港 大 理 學 院 和 中 大 法 律 學 院. 香 港 中 學 會 考 (HKCEE), 多 位 學 生 取 得 0 分 以 上 佳 績. 保 加 利 亞 國 際 數 學 競 賽 (BIMC 03) 隊 際 賽 金 牌. 奧 數 華 夏 杯 / 港 澳 杯 / 華 杯, 多 位 學 生 取 得 特 等 獎 / 金 獎 / 一 等 獎 / 全 港 第 二 名. 還 有 更 多, 怒 不 能 盡 錄, 詳 情 請 瀏 覽 以 下 網 址 : www.andylai.hk/result.htm 黎 sir 教 室 課 程 特 色 : 小 組 教 學 ( 6 人 ), 導 師 親 身 教 學 ; 照 顧 每 位 學 生 需 要, 事 半 功 倍. 精 心 編 制 筆 記, 練 習 以 近 30 年 本 地 和 外 國 公 開 試 題 為 藍 本. 概 念 理 解, 取 分 技 巧 並 重 ; 協 助 同 學 盡 快 掌 握 答 題 技 巧. 歡 迎 自 由 組 合 小 組 上 課, 時 間 及 課 程 內 容 編 排 更 有 彈 性. 詳 情 請 瀏 覽 以 下 網 址 : www.andylai.hk
黎 sir 簡 介 黎 sir 教 室 A Lai Learning Center HKDSE / IB Diploma / GCE AS AL / AP / SAT / HSC IGCSE / GCSE / IB MYP / KS3 / MO / F. F.6 / Y9 Y3 資 深 中 學 補 習 導 師 小 組 補 習 事 半 功 倍!!! 簡 介 Andy Lai BEng CUHK, MIEEE 畢 業 於 香 港 中 文 大 學, 黎 sir 教 室 創 辦 人 之 一. 超 過 5 年 教 授 中 學 文 憑 / IB Diploma / GCE / HSC / SAT / AP / GCSE / IGCSE / IB MYP 課 程 經 驗. 為 了 與 學 生 一 起 面 對 新 制 度 的 中 學 文 憑 試, 黎 sir 親 身 上 陣, 以 實 力 於 物 理 科 奪 取 5**, 證 明 寶 刀 未 老. 現 於 黎 sir 教 室 任 教 補 習 班, 學 生 就 讀 於 英 文 中 學, 中 文 中 學, 國 際 學 校 及 英 國 留 學 生. 熟 悉 近 年 出 題 趨 勢, 教 授 考 試 取 分 技 巧 ; 鼓 勵 同 學 獨 立 思 考, 增 強 同 學 理 解 能 力 善 用 生 活 化 例 子 講 解, 教 法 生 動, 增 加 學 習 趣 味 ; 深 入 淺 出, 明 白 學 生 學 習 上 的 困 難 和 需 要. 中 英 對 照 筆 記, 適 合 中 文 和 英 文 中 學 學 生 就 讀 ; 精 心 編 制 練 習 和 試 題, 協 助 同 學 盡 快 掌 握 答 題 技 巧. 黎 sir 在 中 學 和 大 學 時 代 已 是 一 名 傑 出 學 生, 曾 獲 取 多 項 學 業 上 和 運 動 上 的 獎 學 金 及 獎 項 ; 曾 代 表 香 港 參 加 國 際 性 運 動 比 賽, 取 得 優 異 成 績, 又 讀 得 又 玩 得, 絕 不 是 死 讀 書 的 書 呆 子. 黎 sir 在 就 讀 大 學 時 曾 於 全 球 最 大 美 資 電 腦 公 司 任 實 習 生 超 過 一 年, 大 學 畢 業 後 旋 即 於 全 港 大 型 英 資 電 腦 公 司, 負 責 主 理 該 公 司 所 代 理 的 全 球 大 型 美 資 電 腦 公 司 儲 存 系 統 銷 售 業 務. 於 短 短 半 年 內 將 該 產 品 線 銷 售 業 績 提 升 超 過 50%. 同 時 更 被 公 司 評 選 為 " 傑 出 表 現 員 工 Outstanding Performer", 成 功 將 書 本 上 的 知 識 靈 活 運 用 於 工 作 上. 黎 sir 為 了 教 學 理 想, 毅 然 辭 去 工 作, 全 身 投 入 教 學 事 業, 希 望 將 自 己 的 一 套 學 習 方 法 教 授 學 生 黎 sir 教 室 課 程 特 色 小 組 教 學 ( 6 人 ), 導 師 親 身 教 學 ; 照 顧 每 位 學 生 需 要, 事 半 功 倍. 精 心 編 制 筆 記, 練 習 以 近 30 年 本 地 和 外 國 公 開 試 題 為 藍 本. 概 念 理 解, 取 分 技 巧 並 重 ; 協 助 同 學 盡 快 掌 握 答 題 技 巧. 歡 迎 自 由 組 合 小 組 上 課, 時 間 及 課 程 內 容 編 排 更 有 彈 性. 時 間 及 課 程 請 瀏 覽 以 下 網 址 : www.andylai.hk 地 鐵 : 旺 角 E 出 口, 油 麻 地 A 出 口 小 巴 :, A,, 3C, 6, 6C, 6F, 9, 30X, 35A, 4A, 4A, 60X, 63X, 68X, 69X, 8S, 87D, 93K, 95, 04, 7, 03,, 30X, 34P, 34X, 38P, 38S, 59B, 70P, 8A 小 巴 : K, 74, 74S 上 課 地 址 : 查 詢 熱 線 : 電 郵 地 址 : 網 址 : 香 港 九 龍 旺 角 煙 廠 街 9 號 興 發 商 業 大 廈 706 室. 677 300 enquiry@andylai.hk www.andylai.hk