Transcript
Page 1: Terangganu-Answer Physics P2-Trial SPM 2007

JABATAN PELAJARAN TERENGGANU

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA

2007

PHYSICS

MARKING SCHEMEPAPER 1, 2 and 3

Page 2: Terangganu-Answer Physics P2-Trial SPM 2007

PEPERIKSAAN PERCUBAAN SPM 2007SCHEME PAPER 1

QUESTION NO. ANSWER QUESTION NO. ANSWER1 A 26 B2 E 27 B3 B 28 A4 B 29 C5 B 30 C6 C 31 D7 C 32 D8 C 33 C9 B 34 C10 E 35 C11 C 36 A12 A 37 C13 B 38 C14 A 39 B15 D 40 B16 C 41 B17 B 42 A18 D 43 D19 A 44 C20 D 45 B21 D 46 C22 D 47 B23 D 48 B24 B 49 C25 C 50 D

Page 3: Terangganu-Answer Physics P2-Trial SPM 2007

SECTION A

Question 1Part Mark Answer Note(a)

(b)(i)

(b)(ii)

(c)

1

1

1

1

Name the container X correctlyLead container / concrete container

Underlining thee correct phrase in the box correctlya beta particle

Give the reason correctlyNegatively charge

Give the reason correctlyThe mass of ray P < ray Q // The speed f ray P > ray Q

Total 4

No 2 2 a)

1M Quantity matters in a material //An object at rest tends to remain at rest // an object in motion tends to stay in motion

2 b) 1M The car2 c) 1M Mass // Inertia is smaller2 d) 1M

1Mv2 = u2 + 2as02 = 152 + 2a(375)- 0.3 ms -2

Question 2Question 3

Section Marks Answer Catatan(a) 1 There is no net flow of heat between two objects and

they have same temperature

(b) 1 45 0C

(c) 11

The heat from block P transfer to the waterUntil the rate of heat transfer between P and water become equal

(d)11

E = mcӨ = 0.3 x 900 x 55 = 14 850 J

6 marks

Question 4No 44 a)

1M1M

v = 5 x 0.8 = 4 ms -1

4 b) 1M1M

The wave are circularWavelength remains unchanged

4 c) 1M Diffraction4 d) 1M Waves pass through an aperture or round a small obstacle

Waves spread out.

Page 4: Terangganu-Answer Physics P2-Trial SPM 2007

1M

Question 5

Q 5 / S 5(a)(i) 1 Density is the mass per volume (b)(i)

(ii)

(iii)

1

1

1

Level of the boat is higher in the sea than in the river.

Water displaced in the sea is less than in the river.

Density of sea water is higher than river water. (c)(i)

(ii)

1

1

The lower the density of water, the greater /higher the volume of water displaced.

Weight of the boat = Weight of the water displaced(d) 1 Archimedes’ principle // Bouyancy’s law(e) 1

2 Ballast tank filled by sea waterWeight of submarine > upthrust

Maximum 1 mark(Any one mark)

Tot/Jum : 8

QUESTION 6

Part Mark Answer Note(a)

(b)

(c)

(c)

(d)

1

1

11

1

11

1

Name the current I1 correctlyBase currentName the current I2 correctlyCollector currentState two observations correctlyI1 < I2 // I1 = 0 and I2 = 0The small changing of I1 give the large changing of I2

or I1 is directly proportional to I2

Give the reason correctlyThe mass of ray P < ray Q // The speed f ray P > ray Q

Complete the circuit correctlyThe two resistors in seriesOne of the resistor in base circuit

State the correct function of the two resistorsAs the voltage divider

Total 8

Page 5: Terangganu-Answer Physics P2-Trial SPM 2007

QUESTION 7

Part Mark Answer Note(a)(i)

(a)(ii)

(a)(iii)

(b)(i)

11

1

1

1

Complete the ray diagram in Diagram 7.2 correctlyTwo reflected rays are shown (diagram) Angle of incidence = Angle of reflection (diagram)

State the phenomenon involved correctlyReflection

State the problem that happens correctlyThe driver in car P cannot see car Q // field of view very small

Name the curve mirror correctlyConvex mirror

Page 6: Terangganu-Answer Physics P2-Trial SPM 2007

Part Mark Answer Note(b)(ii)

(b)(iii)

(a)(iii)

(b)(i)

(c)(i)

(c)(ii)

11

1

1

1

1

1

Complete the ray diagram in Diagram 7.3 correctlyTwo reflected rays are shown (diagram) Angle of incidence = Angle of reflection (diagram)

State the solution of the problem in (a) (iii) correctlyThe convex mirror increase the field of view

State the problem that happens correctlyThe driver in car P cannot see car Q // field of view very small

Name the curve mirror correctlyConvex mirror

Give the correct answerUnchanged

Give the reason correctlyThe characteristics of image of a convex mirror not depends on the focallength

Total 10

Page 7: Terangganu-Answer Physics P2-Trial SPM 2007

Question 8

Bhg Markah Jawapan Catatan(a) 1 Isotope that are not stable

Isotop yang nukleusnya tidak stabil

(b) 11

DThe reading of ratemeter is highestBacaan meter kadar paling tinggi

(c) (i) 11

Ra Rn + He Symbol alfa betul - 1Persamaan seimbang -1

(c) (ii) 1

111

100% → 50% → 25% → 12.5%Sodium-24 : 3T = 45

T = 15 hoursCobalt-60 : T = 15.9/3 = 5.3 yearsRadium-226 : T = 4860/3 = 1620 years

(d) 11

1

Sodium-24Half life is short @ activity decreases fasterSetengah hayat pendek @ keaktifan cepat berkurangEmmits beta rays @ can penetrate ground but can’t penetrate pipeMemancarkan sinar beta @ boleh menembusi tanah tetapi tidak boleh menembusi paip

12 markah

Page 8: Terangganu-Answer Physics P2-Trial SPM 2007

SECTION B

Question 9

Bhgn. Markah Jawapan Catatan

Soalan 9(a)(i)

(ii)

1

11

111

Area per unit square // P = F / A

200 / 0.2 = 1000 N m-2

800 / 0.8= 1000 N m-2

Pressure at P = Q =RForce is directly proportional to the surface area

Define symbols used

With unit

Accept without unit

(b) 1111

Pressure is applied to the toothpaste (tube)The toothpaste carry and apply the pressure of the equal magnitude to the whole tubePascal’s Principle

(c)

1,2

3,4

5,67,8

9,10

modification explanation

piston of bigger cross-sectional area

Can support greater force (weight)

Low density material Lightweight // easy to carry

Non-compressible liquid

Piston can be lifted up

Longer handle Less effort needed to press the small piston

Apply released valve between small and main reservoir

Liquid can flows into small reservoir

Accept any reasonable modification

Page 9: Terangganu-Answer Physics P2-Trial SPM 2007

Question 10

Q10/S10Sect. Mark(a)(i)

(b)

1

6

1 State the definition of current correctly The rate of charge flows.

1 State the correct comparison of the type of circuit connection Diagram 10.1 connected in series and Diagram 10.2 connected in parallel. 2 State the correct comparison of the ammeters reading The reading of ammeter in Diagram 10.2 is greater than in Diagram 10.1. 3 State the correct comparison of the voltmeters readings The reading of voltmeter in Diagram 10.1 > Diagram 10.2. 4 State the correct comparison of the effective resistance The effective resistance in Diagram 10.2 < Diagram 10.1.5 State the correct relation between the effective resistance and the current flows

Effective resistance increases, the current flows decreases.6 State the correct relation between type of circuit connection and effective

resistance. Circuit connected in parallel, the effective resistance decreases.

(c) 3 1 State the correct happens of the ammeter reading The ammeter reading increased2 State the correct happens of voltmeter reading The voltmeter reading decreased.3 State the correct reason Effective resistance in the circuit decreased.

(d) 10 State the correct extra device attached and correct reason 1 Attach one fuse to the live wire in the consumer unit/ fuse box. 2 To break/switch off the circuit when large current before the wire become

hotter and produce fire. State the correct choosing of wire/cable and correct reason3 Using the insulating wires // thicker wires 4 To prevent short circuit // To reduce resistance, improve efficiency. State the correct adjustment of each lamp and the correct reason 5 Attach switch for each lamp. 6 To allows each lamp to be switched on and off independently. State the correct modification of connection of metal fitting lamp and give the correct reason7 Connect the metal fitting lamp to the earth wire/cable. 8 To flows electron (extra) to earth to avoid lethal shock. State the correct using of bulbs and give the correct reason 9 Using only 240 V light bulb. 10 To ensure the bulbs light up with normal brightness.

Total : 20

Page 10: Terangganu-Answer Physics P2-Trial SPM 2007

SECTION CQuestion 11

Bhg MrkJawapan Catatan

Soalan 11(a) 1 Aerofoil / airfoil

(b)(i)

( ii )

1

2 3

1

The shape of cross section of the wing causes the speed of airflowAbove the wings to be higher than the speed of airflow belowWhen the speed of moving air is higher ,the pressure is lowerHence air pressure below the wings is higher compare to above the wingsBernoulli’s principle

( c) 1

2

34

56

789

10

A shape of cross section which is upper side is longer than the bottomTo produced the speed of airflow Above the wings to be higher than the speed of airflow below

The larger the area of the wingThe larger the lift force

The less density of the wing materialsLess weight // produce more upward resultant force

The higher the difference in speed of airThe higher the difference in pressureThe most suitable choice is P

Because it hasA shape of cross section which is upper side is longer than the bottomThe larger the area of the wingThe less density of the wing materialsThe higher the difference in speed of air

(d)(i)

(ii)

12

1

P= F/AF=PAF=500 x 40F = 20000NResultant force = 20000 – 900(9.8) = 11180NUpward

Page 11: Terangganu-Answer Physics P2-Trial SPM 2007

(iii) 2

1

F =maa=F/ma=12.42ms-2

Total 20

Question 12Section Mark Answer

12(a) 1 Electric to sound(b) 1

111

Current flows and soft iron core is magnetisedHammer hit the gongContacts openSoft iron core loses its magnetism

(c) 11

11

11

11

11

Size of the hammer is bigSurface area contact is big

Larger distance between the hammer and the gongThe force applied on the gong is greater

The number of turns of the coil is greaterThe strength of the magnetism is greater

High curvature of the gongThe area of air molecule vibrate is greater

QSize of the hammer is big, Larger distance between the hammer and the gong, The number of turns of the coil is greater, High curvature of the gong

(d)(i)1

1

I =

0.83 A (with unit)(ii)

1

1

1

PI =

= 12.5 W

I = / 0.052 A (accept e.c.f)

TOTAL 20 M

Page 12: Terangganu-Answer Physics P2-Trial SPM 2007

section A

1 (a) (i) Refractive index, type of transparent material / n 1

(ii) Refracted angle / r 1 (iii) Incident angle/ i 1 (b)

n r / 0 sin r

1.49 19.0 0.3256 3.071.92 16.5 0.2840 3.522.42 12.5 0.2164 4.622.91 10.0 0.1736 5.763.50 7.5 0.1305 7.66

1. columns n , r , sin r,

2. all the units of n , r , sin r, are correct

3. all the value of r +/ - 0.5 0

4. Value of r , constant at 1 decimal place5 Value of sin r , constant at 4 decimal places

6. Value of constant at 2 decimal places

7. All the value of r ,sin r, are correct

7

(c) score

1. n at y-axis and at x-axis

2. the units at both axis is correct 3. both axis has initial scale and not ganjil 4. 5 points ploted are correct 5. 3 points plotted are correct6. smooth line 7. the minimum size is 5 x 4 ( from origin to the last point)

score marks7 5

5-6 43-4 32 21 1

5(d) Directly proportional 1

Page 13: Terangganu-Answer Physics P2-Trial SPM 2007

Total 16

2 (a) Straight line from I= 0.40 A

Straight line to Y axis at V = 0.74

0.72 V (with unit)

11

1

(b) increase 1

(c) constant 1

(d) (i)

(ii)

Show a triangle (enough size)

Substitution

0.60

0.88

r= 0.68Ω

E = V + Ir

= .0.72 + 0.4 X 0.68 ( V = 0.72V and I =0.68 A)

= 0.9921 V ( 4 decimal point)

1st

2nd

3rd

1st &2nd

3rd

(e) eye position must be perpendicular to scale of ammeter/voltmeter./

The connection must tied

1

Total 12

Page 14: Terangganu-Answer Physics P2-Trial SPM 2007

Section BQuestionSoalan

AnswerJawapan

MarksMarkah

3. (a)

(b)

(c) (i)

(ii)

(iii)

(iv)

(v)

State a suitable inferenceAcceleration is influenced by the mass 1State a relevant hypothesisWhen the mass increased, the acceleration will be decreased. 1State the aim of experimentTo investigate the relationship between the acceleration and the mass. 1State the suitable manipulated variables and responding variable (Quantity that can be measured)Manipulated variable : massResponding variable : acceleration

1

State the constant variableForce applied 1State the complete list of apparatus and materials 5 Trolleys, ticker timer, ticker tape, a rubber band, a wooden runway, 12 V a.c power supply.

1

Draw the functional arrangement of the apparatus Trolley rubber band Ticker Timer Ticker tape Friction compensated runway Power supply 1State the method to control the manipulated variableThe apparatus is set up as shown in the diagram.The ticker-timer is switched on and a trolley is pulled using a rubber band. The extension of the rubber band is ensured to be of the same length 1State the method to measure the responding variableAcceleration of the trolley is calculated using the ticker-tape , a = ( v-u ) / t 1Repeat the experiment at least 4 times with the valuesProcedure 2 and Procedure 3 are repeated using 2, 3, 4 and 5 trolleys.(Note : Based on SPM standard , at least five manipulated values required.) 1State how the data tabulated with the title MV and RV

Mass / num. of trolley acceleration / cm s-2

12345

1

State how the data is analysed, plot a graph RV against MV acceleration / cm s-2

Mass / num. of trolley 1

Page 15: Terangganu-Answer Physics P2-Trial SPM 2007

Total 12

QuestionSoalan

AnswerJawapan

MarksMarkah

4. (a)

(b)

(c) (i)

(ii)

(iii)

InferenceThe rotation speed of the motor influeced on number of battery 1HypothesisIf the current increase , then the speed of rotation increase. 1AimTo investigate the relationship between the current and distance/speed 1

Variablesmanipulated :currentresponding : distance of the rodefixed :strength of magnet/ rod mass

1

1 List of apparatus and materials 2 bar magnet ,iron rod,batteries,ammeter, metere rule 1Arrangement of the apparatus

1

(iv)

(v)

Procedure of the experiment which include the method of controlling the manipulated variable and the method of measuring the responding variable. The current 0.5 A record from the ammeter.Measured the distance of rod movement .The experiment is repeated 4 times with the difference current, I = 1.0 A ,

1.2 A, 1.4A ,1.6A and 1.8A

1

11

(vi) Tabulate the dataCurrent,I/A 0.5 1.0 1.2 1.4 1.6 1.8Distance of the rod/cm

1

(vii) Analyse the data Distance/cm

Current,I/A

1

Total 12

Page 16: Terangganu-Answer Physics P2-Trial SPM 2007

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