Answering Question Techniques Physics SPM (4531/2)Paper 2 (100 marks)

Time: 2 1/2 hours

Physics 4531/2 (Assessment Format)Subjective Questions (2 ½ Hours)3 Sections Section A: Structural QuestionsSection B: Restricted Respond & Extended Respond Questions Section C : Restricted Respond & Extended Respond Questions

Section A : 60% Answer All Questions (7-8 Questions)Section B : 20% (Two questions choose one) (Application & Problem Solving)Section C : 20% (Two questions choose One) (Conceptualisation & Making Decision)

Format of Question (Section A)Short structural questions: Give short definition, base on diagram state reading, give similarities,, state reason, state principle involve, calculation, label diagram (direction of forces), fill in the blanks (e.g half life____ days) underline words (step up, step down transformer) name an item, draw a picture (e.g ray diagram for magnifying glass), state function (e.g :oil in the hole of aluminium block in specific heat experiment )

Format of Question (Section B)i). Give a definition or explanation of terms or phrasee.g. : What is meant by semiconductor ? (Q9 P2 2007), meaning of elasticity (Q9 P2 2008) ii). Giving 2 situations, making comparison, state the relationship between 2 quantitiesiii). Explain working principle & suggest modification

Example:Q9- P2 SPM 2008

e.g. Q9 P2 SPM 2008a (i) What is the meaning of elasticity? [1 mark] (ii) Using diagram 9.1 & 9.2, compare the thickness of the spring wire & the maximum height reached by the balls. Relate the thickness of the spring wire with the max. height of the ball to make a

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deduction regarding the relationship between the thickness of the spring wire & the elastic potential energy of the spring [5 marks]

(b)(i) Compare F1 and F2 [2 marks](ii) Using Diagram 9.1 and Diagram 9.2, state the energy changes that take place from the moment the spring is compress until the ball reaches its maximum height [2 marks]c) Diagram 9.3 shows a pole vaulter performing a jump. Using appropriate concepts explain the use of suitable equipment and techniques to improve his performance. Your answer should include the following

Answers:(a) (i) Elasticity –the property of an object that enables it back to its original shape and size after the applied force is removed [1]ii) In Diagram 9.1, spring M is thinner (Smaller diameter of wire) compared to spring N which is thicker [2]In Diagram 9.2 Steel ball which was on spring M jumped to a lower height compared to the steel ball on spring N which bounced higher [2]

The maximum height of the ball increases as the thickness of the spring increases, it can conclude that the thicker the spring the bigger the elastic potential energy [1](b) (i) F1 is smaller than F2 .Spring 9.1(b) is thicker than spring 9.1 (a) [2]ii) Elastic potential energy to Kinetic energy to gravitational potential energy [2]

(c) Vaulter’s attire must be tight to reduce air frictionUse spiked shoes enable easy acceleration should sprint or move fast to acquire greater kinetic energy and momentumPole must be elastic and strong to increase the elastic potential energyThe mattress must be thick to lengthen the time of impact to reduce impulsive force and reduce injuries

Format of Question (Section C)a). Give a definition or name a certain concept or phenomenon (e.g.: TIR P2 Q11 2008) (1 mark)b) Explanation of the concept or phenomenon (4 marks)c) Explaining suitability and making a choice (e.g.: Choose the best Optical Fibre Q11 2008) (10 marks)d) Application or calculation (5 marks)(e.g.: c and draw light path Q11, 2008)e.g. Q11 P2 SPM 2008A light signal through an glass optical fibre

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i) Vaulter’s attireii) Vaulter’s movement

iii) Pole usediv) Safety [10 marks]

(a) Name the light phenomenon involved at Y [1 mark](b)(i) State 2 changes that happen to the light when it passes from air into optical fibre at X[2 marks](ii) Explain why the light ray follows the path shown in Diagram 11.1 when it hits the wall of the optical fibre at Y [2 marks](c) The optical fibre in Diagram 11.1 can be used in telecommunications and medicine. You are asked to investigate the characteristics of optical fibre for use in these fields as shown in Table 11

Optical Fibre Size Comparison between ni and no

Flexibility Purity of inner core

P Single fine optical fibre ni > no High Very High

Q Bundle of fine parallel optical fibres

no > ni Low Low

R Bundle of fine parallel optical fibres

ni > no High Very High

S Single fine optical fibre ni > no Low High

T Bundle of fine parallel optical fibres

no > ni High High

Explain the suitability of each feature of optical fibre in Table 11 for the use in telecommunications and medicine. Determine the most suitable optical fibre that is capable of carrying the largest number of signals simultaneously. Give reasons for your choice.

Answers:11 a) Phenomenon - Total internal reflectionb) (i) 2 Changes happen at X:- Speed of light is reduced, Wavelength is reducedii) Why follow path show at Y? Travel from medium of high n to low n (denser medium to less dense)

The incidence angle is greater than the critical angle of the medium)

Criteria Characteristic/Feature Explanation

Size Bundle of fine parallel optical fibre

Can carry large quantity of signals

Comparison between ni and

ni > no So that TIR can occur

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(d) Diagram 11.2 & 11.3 show a ray of light passing into glass & diamond respectively(i) Calculate the critical angle of diamond and glass [2 marks](ii) Copy the 2 diagrams and complete the path of the light ray in glass & diamond until it finally emerges from each object [3 marks]

no

Flexibility High flexibility Can be easily curved along path or patient’s body

Purity of inner core

High purity of inner core Signal can travel longer distance (Less attenuation)

The choice is R because it is a bundle of fine parallel optical fibre, ni > no high flexibility and high purity of inner core d(i) Calculate the critical angle of diamond and glass( Refractive index : Glass =1.50 Diamond = 2.42 )Answer: n = 1/ sin cFor glass sin c = 1/1.50 sin c = 0.67 c = 41.8c for glass is 42° and c for diamond is 24°

Tips for good achievement

1. Do not contradict answersGiving extra answers or points will not be penalised but answers must not contradict e.g.: State 3 characteristics of the image as seen by the observer

2. Understand the right concept and avoid misconceptione.g.: Mistaken negative velocity as slow down or undergone deceleration

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For diamond sin c = 1 / 2.42 sin c = 0.41

d (ii)

d (ii)d (ii)

Answer:Virtual Magnified Upright Inverted (WCR)

Answer:BC : Car velocity decreases & decelerates to rest (X)object travels with constant velocity in opposite direction (√ )

Correct concept:Negative velocity with negative acceleration is the acceleration in the opposite direction

3. Understand the command wordsState / write down - no writing or explanation neededSuggest, described- Detail explanation neededCalculate, find, show, solve- Some writing needed, include enough working to get answersSketch- show general shape of graphDraw-Plot accurately using graph paper

4. Pay attention to unit for physical quantityi). When necessary Unit should be Changed to SI

ii). Good understanding of unit conversionCommon wrong perception:1cm3 = 1 x10-2 m3

Should be 1cm3 = 1cm x 1cm x 1cm =(1x10-2 m)3 = 1x10-6 m3

e.g.: Convert 1g/cm3 to kg/m3

Answer: 1g/cm3 = 1g/1 cm3 = 1 x 10-3 kg / (1x10-2 m)3 = 1x 10-3 kg / 1 x 10-6 m3

= 1 x 103 kg / m3

iii). Gas Law –Conversion of °C to K (T= θ+273)e.g.: V gas in balloon = 0.04 m3. θ of the gas in balloon = 37° C. When θ was increased to 47° C, what will be the V of the balloon?

T1 = 37 + 273 = 310K V1 = 0.04 m3 ,T2 = 47 + 273 = 320K V1 / T1 = V2 /T2 Charles’ Law V2 = T2 /T1 x V1 = 320/310 x 0.04

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Lf = Pt/m ={50(3X60)}/(0.6)=1500 J kg-1A 600 g of solid substance is heated at a uniform rate by a 50 W heater. Calculate Lf of the substance

= 0.041 m3

5. Show steps and working (calculation)e.g.: An object of mass 2 kg accelerates from u=4m/s to v=6m/s, what is the work done for the car to accelerate ?Solution:-work done = Change in kinetic energy of the car = Final kinetic energy-Initial kinetic energy =1/2 mv2 - 1/2 mu2 = 1/2 x 2 x 36 - 1/2 x 2x 16 (√ ) 1 mark =20 J (√ ) 1 mark

6. Be careful of sign conventioni). Sign convention for lenses formulae.g. : An object is put at a distance of 5 cm from a concave lens which has a linear magnification of 0.4. Determine the focal length of the lens

Student’s answer

m=v/u, 0.4 = v/5image distance v =0.4 x 5 = 2 cm (x) u = 5 cm, v = 2 cm f = ?Using 1/u + 1/v = 1/f 1/5 +1/(2) = 1/f hence 1/f = 7/10 f = 10 / 7 cm Since concave lens, therefore f = -10/7 cm (X)

Correct answer:For concave lens m= -0.4m=v/u, -0.4 = v/5 hence, image distance v = -0.4 x 5 = -2 cm u = 5 cm, v = -2 cm f = ?Using 1/u + 1/v = 1/f 1/5 + -1/(2) = 1/f hence 1/f = -3/10 f = -10 / 3 cm (√ ) (Negative value indicates concave lens) { Alternative :- use 1/m = u/f –1 (m=negative) 1/(-0.4) =5/f-1 5/f = -1/0.4 +1 5/f = -1.5 f = 5/-1.5 = -50/15 = -10/3 cm }

ii) Sign convention in forces and motion

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7. Master general techniques

i. Spend the first few minutes looking through the whole exam. paperii). Plan the answers, jot down idea in points form before answeringiii). Use the mark allocation as guide to the number points needediv). Use simple and short points. Use a new line for each new pointv). Keep referring back to introduction for clues to the answers

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Prepared By: Mr. Chong Weng SungSMK St. Michael, Ipoh.(Guru Cemerlang FizikDaerah Kinta Utara)