Fizik Skema Pahang 07

ANSWER FOR PAPER 1- PHYSICS TRIAL 2007 1C 2E 3C 4D 5B 6D 7D 8B 9C 10B

11A 12B 13B 14C 15D 16C 17B 18A 19C 20B

21C 22A 23C 24D 25D 26A 27A 28C 29C 30B

31D 32B 33B 34C 35C 36A 37B 38D 39D 40C

41C 42D 43A 44B 45B 46C 47C 48C 49B 50C

SKEMA SPM TRIAL 2007 1. a i) Vernier caliper / Angkup vernier ii) to avoid measured material damage / elak bahan yang diukur rosak b)i) N ii) can measure smaller values /ukur nilai yang lebih kecil 2 a) i) bend his knee / bengkokkan lututnya ii) 1. prolong time of impact / panjangkan masa perlanggaran , 2. to prevent his knee injured/ elakkan kakinya cedera b)

1. F = mv/t 2. 60 x 5/ 3, 300/3 3. 100 N

3. a) i) The adult ii) The adult iii) The mass of the adult is greater / jisim dewasa lebih besar b) i) The mass increases, the inertia increases / Jisim bertambah inertia bertambah ii) 1.Supertanker has greater mass, /jisim besar 2. inertia is greater / inertia besar iii) keluarkan sauce drp botol

4 a) i) Close to the sky / dekat dengan langit/ cool air b) Refraction / biasan c)i)

Total internal reflection

ii) The image of cloud / imej awan d)

1.Light from sky to the earth refracted / cahaya daripada langit ke bumi dibiaskan 2. The light reach at a point X, total internal reflection occurred / Tiba satu titik X pantulan dalam penuh berlaku

e) Optical fibre/ fibre optik

5. a) i) Tekanan daya yang bertindak seunit luas ii) 1. omboh E = ombohG // beban yang sama 2. Tekanan yang dihasilkan oleh omboh E = G b) i ) Pascal ii) 1. F = PA 2.40 x 0.8 3. 32 N 6 a) Thumb tack has smaller surface area / Luas permukaan paku tekan lebih kecil b) 1. If the surface area smaller/jika luas permukaan lebih kecil 2. the pressure is higher/ tekanan lebih tinggi c) 1. P = 5 N / 3.14 x 10 −4 m 2 . 2. 1.59 x 10 4 Nm −2 d) 1. Wider tyre, less pressure 2. To avoid the bulldozer sink into the soil / elak bulldozer tenggelam dalam tanah 7. a) i) B and D / B dan D ii) A and C / A dan C

b) same/ sama

c) i)

ii)

iii)

d) Convert AC to DC / tukar AC ke DC

8. a) i) Radioactive decay is the breaking up of unstable nuclei into more stable nuclei with the emission of alpha,beta and gamma particles. ii) Alpha particle/α b) i) 1. low ionizing power 2. small mass ii) detection of leakage in pipe c) i) 1. Nuclear fission is the splitting of nucleus with big mass into two or more smaller nuclides 2. with the release of energy ii) a) moderator/to slow down the fast neutron b) absorb neutron

d) 1. total mass of product = 236.85854 amu and total mass of reactant = 236.05787 amu 2. mass defect = 0.80007 amu 3. E= 0.80007 x 1.66 x 10 −27 kg x 9 x 10

16

/ 11.96 x 10

−11

J

SECTION B 9. a) The total amount of heat in a body b) 1. the temperature of oil droplet = the temperature of oil poured on the hand 2. the oil droplets have smaller heat capacity 3. the oil poured on the hand has greater heat capacity 4. the heat capacity of droplets transferred to the hand is smaller compared to that of the poured oil 5. The bigger the mass of an object, the bigger the heat capacity of the object c) 1. It doesn’t react to the inner part of the engine 2. high specific heat capacity 3. can absorb large amount of heat energy 4. cheap and easily available d) 1.& 2. : Q = 0.5 x 910 x 80 3 & 4 :. Q= 364 J e) Modification 1.low heat capacity 3. the lid of the pan designed to lower the air pressure inside the pan 5.the lid of the pan made of substance which has weak conductivity of heat

reasons 2. the pan can get hot quickly 4. the boiling point of water decreased 6. heat will not absorbed by the the lid,so heat will not lost to surrounding

10. a) i)The waves that have same frequency and sam phase / sama frekuensi dan sama fasa ii)1. λ diagram 10.1 bigger // λ diagram 10.1 smaller 2. x diagram 10.1 bigger // x diagram 10.2 smaller 3. λ increase , x is increase // λ decrease , x decrease 4. & 5 : x directly proportional to λ b) 1. Sound waves from two loudspeakers produced two coherent sources 2. Sounds wave interfere 3.Constructive interference produced loud sound 4. Destructive interference produced soft sound c)i)1. At the centre of the ocean the water waves travel at uniform speed as the depth of the sea is uniform 2. When the waves reache the coast ,the water is shallower,wave speed reduced,refraction occurs 3. Refraction causes the wavefront bend toward the normal 4. This results the wavefront following the shape of of coastline

ii) Ubahsuaian Sebab 1. dinding konkrit 2. lebih kuat 3. tembok sekatan bercelah 4. sebarkan tenaga gelombang 5. tambak laut supaya lebih cetek 6. halaju gelombang berkurang dan lain-lain yg sesuai.

11. a)i) Buoyant force equal to the weight of the fluid displaced ii) W = ρVg = 1000 x 2/3 (0.3 x 0.3 x 0.3) x 10 = 180 N m, cube= W/g = 180/10 = 18 kg ρ = m / V = 18 / (0.3 x 0.3 x 0.3 ) = 666.67 kg m −3 b) Characteristic/cirri

Reason/Sebab

1.Streamline / larus 3. Density /ketumpatan rendah 5. Specific heat capacity high / muatan haba tentu tinggi 7. High strength / Kekuatan tinggi

2. reduce the resistance of water/ kurangkan rintangan air 4. higher buoyant force/ daya apungan tinggi 6. absorbs heat slowly/serap haba dengan perlahan 8. Difficult to damage / sukar rosak

S, because its streamline,low density, high specific heat capacity and high strength S, sebab ia adalah larus,ketumpatan rendah,muatan haba tentu tinggi dan kuat. c) Buoyant force always greater than the load weight,so that the ship will not sink. Daya apungan sentiasa lebih besar daripada berat beban,supaya kapal tidak tenggelam d) 1. In the sea, buoyant force > weight of the ship Di laut, daya apungan > berat kapal ,kapal timbul 2. When the ship in the river,density of water< density of sea,so buoyant force decrease Bila kapal di sungai, ketumpatan sungai < ketumpatan laut,daya apungan menjadi kurang 3. The weight of the the ship > the buoyant force ,so the ship will sink Berat kapal > daya apungan ,kapal tenggelam 12 a) i) The total resistace in the circuit / jumlah rintangan dalam litar ii) 1. 1/3 + 1/ 2+4 2. total resistance = 2 ohm 3. I 1 = 2 A 4. I 2 = 1 A

b) i) Characteristic/ ciri-ciri 1. Smaller surface area 3.High melting point 5. Low specific heat capacity 7. Long (coiled) metal

Reasons/sebab 2. the resistance is higher 4. not easy to melt 6. the temperature arises shortly 8. to increase the resistance

ii) V, because it has,smaller surface area, high melting point, low specific heat capacity and long metal.

SPM TRIAL EXAM 2007 ANSWER SCHEME

3 (a)

The wall of the dam is thicker at the bottom of the dam.

1

(b)

Water pressure increases with depth

1

(c)

i) to investigate the relationship between the pressure in a liquid and the depth.

1

ii) Variables (a) Manipulated : Depth of liquid, y (b) Responding : Pressure in liquid, h (c) Fixed : Density, p iii) List of apparatus and materials. Measuring cylinder, thistle funnel, rubber tube, manometer, and retort stand.

1 1 1

iv) Arrangement of app. 1 1 1 1 1

v) i) The measuring cylinder is completely filled with water. ii) The thistle funnel is connected to the manometer with a rubber tube. iii) The thistle funnel is lowered with the water to a depth y = 10.0m The manometer reading, h is measured. iv) Step 3 is repeated with values of depth y = 20.0m, 30.0 cm, 40.0 cm and 50.0 cm.

1 1 1

1 Depth, y (cm) 10 20 30 40

h (cm) h min

1

SPM TRIAL EXAM 2007 ANSWER SCHEME Paper 3

NUM.

ANSWER

MARKS

1(a)

(b)

(c)

i)mass // m

1

ii)rise in temperature //Ө

1

iii)density//(rate of) energy supplied

1

V/cm3 50 60 70 80 90 √ M1 √ M2 √ M3 √ M4 √ M5 √ M6 √ M7

√ A √ B √ C √√D D √ D √ E √ F

m/g 1/m / g-1 T/ºC Ө/ ºC 50 0.020 80 55 60 0.017 75 50 70 0.014 66 41 80 0.013 60 35 90 0.011 51 26 column and label for v,m,1/m,T&Ө shown units correct all values for v correct all values for m correct all values for 1/m correct all values for 1/m 3 d.p. all values for Ө correct.

1 1 1 1 1 1 1

Num. of √ Marks 7 7 6 6 5 5 4 4 3 3 2 2 1 1 y axis labeled Ө and x axis labeled 1/m unit for y axis and x axis are correct even and constant scale for graph 5 points are plotted correctly 4 or 3 points are plotted correctly the best straight line is drawn size of graph is more than half of graph paper

1 1 1

Num. of √ 7 6 5 4 3 2 1

2(a)

(b)

Marks 5 4 4 3 3 2 1

1 1

- Ө is directly propotional to 1/m Or - Ө increases with 1/m (base on graph plotted by student)

1

i)

x is inversely proportional to a

1

ii)

If a = 2.0 m 1/a = ½ 0.5 m-1 From the graph x = 1.5 m

1 1 1

iii)

Triangle = 1.5/0.6 = 2.5

1 1 1

λ = ax /D = 2.5/D

m

Gradient = x/1/x = ax = 2.5/5 = 0.5 m This experiment is done in an open space to reduce the effect of reflection.

1 1 1 1 1