marking scheme perfect score physics module 2008

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SEKOLAH

SEKTOR SEKOLAH BERASRAMA PENUH BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN KLUSTER KEMENTERIAN PELAJARAN MALAYSIA

MARKING SCHEMEPERFECT SCORE MODULE

PHYSICS

2008

SBP 2008

1

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MODEL ANSWER PHYSICS PERFECT SCORE 2008SECTION 1 - CONCEPTUAL QUESTION (a) (b) (c) (d) (a) (b) i ii iii (c) i ii (d) (e) (a) i ii (b) i ii iii (a) (b) 4 (c) (d) (e) (a) (b) i ii iii Iv (c) i ii (a) (b) (c) (d) (e) (f) ANSWER Weight/ air resistance Bigger Different: Velocity (terminal) water droplet > velocity (terminal) feather Velocity (terminal) increase when the decrease of surface area / Velocity (terminal) inversely proportional with the surface area. Density is the mass per volume 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. The lower the density of water, the greater /higher the volume of water displaced. Weight of the boat = Weight of the water displaced Archimedes principle // Bouyancys law Ballast tank filled by sea water Weight of submarine > upthrust The temperature of air increase The volume of air increase The mass of air constant When the temperature increase the volume of air increase Charles Law Temperature increase, the kinetic energy increase The rate of collision increase/ pressure increase and expand the ballon/ball Length between the two crest / trough / consicutive wave front The water wavelength remains constant after went through the gap. The wavelength in 4.2 is greater than 4.1 Curvature of the wave pattern in fig. 4.2 is more greater / Diffraction of waves is more in 4.2 The greater the wavelength, the wave pattern more spread or curve after went through gap. Diffraction same M1 brighter than M1 V 1 is bigger than V2 V 3 = V 1 +V 2 A1 = A2 + A3 Series circuit Not suitable If one bulb blown up ( does not light up) all the bulb will not funtion Diagram 6.1 has less number of turns compare to Diagram 6.2 Diagram 6.1 has less deflection of pointer compare to Diagram 6.2 P = North Q = South The larger the number of turns, the higher the induced current produced. Faradays Law Increase the speed of the relative motion between the coil and the magnet // Use a stronger magnet Direct Current Generator (D.C. Dynamo) // Alternating Current Generator (A.C. Dynamo) // Moving coil microphone MARK 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1

1

2

3

5

6

SBP 2008

2

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QUESTION i

7

ii

ANSWER Buoyant force is an upward force acting on an object immersed in a fluid The mass of the catch does not change through out the whole act of lifting the net The volume of the catch still in the water is getting smaller/is the least in Figure 9.3 and the largest in Figure 9.1 The force needed to lift the catch is getting bigger. / in Figure 9.3 is the greatest The weight of water displaced is getting smaller / the greatest in Figure 9.1 and the least in 9.3 Buoyant force is equal to weight of water displaced The greater the weight of water displaced the greater the buoyant force

MARK 1 1 1 1 1 1 1 (Any 5) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

i 8

Ii

(a) 9 (b) (c) (a)

10

(b)

(a)

11

(b)

12

(a)

13 (b)

Distance from the optical centre to a focal point. Lens K is thicker than lens J Light ray is refracted more in lens K than lens J. Focal length of lens K is shorter than lens J. Therefore the thicker the lens, the greater the refraction of light and with that the shorter the focal length of a lens will be magnetic field region The number of turns of the coil in diagram 3.2 is more // vice versa The arrangement pattern of the iron filings in diagram 3.1 is further apart. The angle of deflection of diagram 3.2 is more. The closer the pattern of iron filing the greater the strength of magnetic field. The strength of magnetic field increases as the number of turns increases// vice versa. Electric current is the rate of flow of charge. The voltmeter in Figure 10.1(a) is parallel to the resistor while in Figure 10.2(a) the voltmeter is parallel to the battery. The graph in 10.1(b) shows that potential difference is directly proportional to current Obeys Ohms law Graph in Figure 10.2(b) shows that the potential difference across the cell decreases when the current flowing through it increases.// V decreases linearly with I Voltage drop is due to the internal resistance of the battery The rate of charge flows Diagram 11.1 connected in series and Diagram 11.2 connected in parallel The reading of ammeter in Diagram 11.2 is greater than in Diagram 11.1. The reading of voltmeter in Diagram 11.1 > Diagram 11.2. The effective resistance in Diagram 11.2 < Diagram 11.1. Effective resistance increases, the current flows decreases. Circuit connected in parallel, the effective resistance decreases The light bulb in diagram 12.2 does not light up while the light bulb in diagram 12.1 light up In diagram 12.2 the negative terminal cell is connected to the anode. This increase the junction resistance in diagram 12.2 The current cannot flow through the junction. The diode in diagram 12.2 is reversed bias. Concept Diode allows current to flow in one direction. Nuclear fission is the splitting of a heavy nucleus into two lighter nuclei, which subsequently emit either two or three neutrons and release of large amounts of energy. // the high number mass/ high nucleon number to small nucleus Difference: State that nuclear fission involves the fission of heavy nucleus. State that nuclear fusion involves the combination of lighter nuclei Similarities: State that decrease/loss of mass after the process occurs. State that the neutrons are produced. State that nucleon number/ mass number/ atomic number before and after are the same 2 E = mc // loss of mass/ mass defect directly proportional to the energy released .

1

1 1 1

1

SBP 2008

3

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SECTION 2 PROBLEM SOLVING (QUALITATIVE) QUESTION 1 Elastic potential energy (i) [1 mark] (iii)

1

(ii) a (iii)

k=

F , x

k=

3.6 4

1-1

k = 0.9 N cm E = Fx

2 = x 3.6 x 4/100 = 0.072 J

(iv) (i) b (ii) (iii)

x=

5 0.9

2

l = 12 5.56 = 6.44 cm The compression of the spring is directly proportional to the load Exceed the elastic limit of the spring/ Spring does not return to its original condition after the applied force has been removed. Connect the spring in parallel Use spring with larger spring constant. Use spring with smaller diameter. TOTAL

1 1 2

10

QUESTION 2 (i) Pascal Principle a (ii) Oil cannot be compressed (i) 5 000 Nm F = PA b (ii) = 10 000N m = 1 000 kg (iii) move downward Increase diameter of the piston c Use thicker wall cylinder d Hydraulic brake TOTAL

1 1 1 1 1 1 1 1 1 1 10

QUESTION 3 a n-p-n transistor (i) 12 V (ii) 27 + 3 = 30 k b (iii) V YZ =

3 x 3 + 27

1 1 1 1 12 1 1 1 1 1 1 10

c d e

= 1.2 V 1- Lamp L will not glow 2- because the voltage across YZ (i.e 1.2 V) is less than the base-emitter potential difference of 2V Resistor R1 and R2 are swapped places 1-Resistor R1 is replaced by a thermistor. 2- Lamp L is replaced by an electric bell. TOTAL

QUESTION 4 (i) 0111 (ii) 1000 a (iii) 0001 (iv) 1110 b (i) 0001

1 1 1 1 1

SBP 2008

4

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(ii) c

AND

1

Q

Q

1

Qd

P RTOTAL

3

10

QUESTION 5 Suggestion 1 2 3 4 5 6 7 8 9 Balloon should be large size Balloon material is made of light weight material like nylon Balloon material should also have a high melting point. The part of the balloon (the skirt) near the burner must be fire resistant /coated with fire resistant material The burner burns (liquefied) propane/gas A large fan is needed initially The basket must be made off light and flexible/safe material (e.g. rattan or cane woven) There must be a line or rope from the balloon to the ground Best times to launch the balloon are early morning and late afternoon when the air is cooler Rationale To create sufficient buoyant force due to greater weight of surrounding air displaced. The total weight of the balloon is less than the buoyant force//reduce weight It will not disintegrate when exposed to hot air So that it doesnt catch fire easily Warms up the air in the balloon To blow enough air into the balloon Prolong the collision time between basket and ground// reduce impulsive force when basket hits the ground Prevents the balloon from being swept away from the carnival site. Cool air is denser, providing more buoyant force TOTAL QUESTION 6 Suggestion Rationale Choose lens S as objective lens Choose lens Q as eyepiece Lens S is placed in front of lens Q The two lenses are adjusted so that they are in normal adjustment where distance between the two lenses is equal to (fo + fe) Modification Explanation Use low power convex lens as the objective lens. (ii) Use high power convex lens as the eye lens Use bigger diameter of objective lens Magnification of telescope = 1 1 1 1 10 MARK

Any 5 correct

(i)

fo fe

,

1 1 1 1 1 1 10

Low power lens has a longer focal length, fo , magnification High power lens has a shorter focal length, fe , magnification More light permitted to enter the telescope and a clearer image is seen TOTAL

SBP 2008

5

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QUESTION 7 Suggestion Rationale A periscope/box is constructed by using cardboard / p.v.c. / etc Light from the object will pass through an opening at the top of the box and reflected by plane mirror A to plane mirror B / figure Both plane mirrors are placed towards the hole / figure The plane mirror is placed at 45C at the corners of the cardboard / figure 1 1 1 1

(i)

(ii)

because of refraction That causes images to overlap Modification Using 45C prism lens to replace the plane mirror Using a convex lens Explanation To prevent overlapping images To magnify the image that is produced TOTAL

1 1 1 1 1 1 10

(iii)

Question 8

(ii)

2

SBP 2008

6

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Use two separated convex lenses of short focal length, which are mounted at either end of a tube.

To form a magnify image

The first lens(objective lens) has a very short focal length, and the object is placed in front of this lens just beyond the focal distance.

To form a real, inverted and magnified image at a distance L behind the lens.

The second lens (eyepiece lens) has a longer focal length Use light source e.g a mirror, which would reflect light from outside to the object The distance between final image and the eyepiece lens, D 25 cm Eye piece lens is often equipped with an adjustment that can move it in and out along the length of the axis of the instrument.

Further enlarge the image. Helps to an extent to produce an enlarged image of that image that is already magnified by the objective lens The mirror can reflect light rays through the object. The lens that is near to the object (objective lens) will help to illuminate the object that has to be seen. The image formed at near point, the microscope is in normal adjustment This adjustment compensates for the different strengths of the eye lenses of different users. TOTAL

8 (any 4)

10

QUESTION 9 Suggestion using a hairspring of lower stiffness// increasing the strength of the permanent magnet // increasing the number of turns of the coil // smaller mass of the coil and indicator Using a pair of concave permanent magnet // Diagram The core used is cylinder in shape. // Diagram Using the soft iron core.// Diagram Using the linear scale.// Diagram. Rationale To produce a larger of rotation of the coil. // increase the rate of rotation. The sensitivity of the ammeter can be increased To produced a radial magnetic field.// Diagram. To supply the total magnetic flux uniformly to the coil. To concentrate the magnetic flux.// Diagram Because the angle of the rotation of the indicator is linearly increase with the current flow TOTAL Any 5 correct

1 2 3 4 5

10

QUESTION 10 Suggestion The main body of the kettle is made up of high specific heat capacity materials. The body of the kettle must be made from insulator like materials The material should also have a low density The handle must be made of insulators Rationale to make sure that the outer body does not heat up too fast. that reduces heat loss to the surroundings so that it is lighter to carry So that it does not get heated. 1 1 1 1 1 1 1 1 1 // 3.33 A 1 TOTAL 10

(i)

(ii)

I =

P V

=

800 240

Most suitable: 5 A fuse

SBP 2008

7

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QUESTION 11 Suggestion The pipe inside the plate must be made of metal Rationale Metal is a good heat conductor, so it will transmit heat to water easily Longer pipe - enlarge surface area will absorbs heat faster To prevent the loss of heat energy A black surface is a good absorber of radiation so it will absorb heat faster To trap heat energy. (energy is radiated in, but cannot radiate out again). TOTAL QUESTION 12 Suggestion The electric appliances are connected in parallel//diagram Attach switch to each electric appliance Rationale Allow each electric appliances to be switched on and off independently To allows each electric appliance to be switched on and off independently. To stop the flow of current by melting when a high voltage of electric current flows through the circuit // switches itself off very quickly if the current exceeds e.g. 15A (for 15A mcb). It protects electric appliances from damage Earth wire connected to earth, so that when a fault occurs and a current flows through the live wire and the earth wire, the fuse in the live wire will blow and cut off the supply. It will protect a person who may touch a faulty or live appliance It prevents electric shock 1 1 1 1 1 1 1 1 1 1 1 1

Coil pipe embedded in plate Use insulator behind the absorber panel

Use an absorber panel which is painted black.

1 1 1 1 10

Use glass cover on the top of the panel

Fit fuse at the live wire in the fuse box //diagram// Use miniature circuit breakers (mcbs)

1 1

Earth connection to the metal case of electrical appliances

Use low power lamps / install fluorescent lamp Regularly cleaning and removing dust from the air filters of air conditioners Do not put more water in a kettle than needed for hot drinks

To save the energy. Increase energy efficiency

1 1 10

TOTAL QUESTION 13 During hot weather, resistance at thermistor decreases, potential difference across it decreases Potential difference across R will increase This will produce bigger base current , and will increase the collector current Finally, electric relay will switch on the air conditioner During cold weather, resistance at thermistor increase. Potential difference across thermistor will increase. Potential difference across R will decrease. This will produce smaller base-current and no current flow in collector circuit. Finally, electric relay will switch off the air conditioner.

1 1 1 1 1 1 1 1

(ii)

SBP 2008

8

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(ii)

2

TOTAL

10

SECTION 3 DECISION MAKINGQUESTION NO. (a) (b) ANSWER Resultant force / net force is zero W = mg = 60 x 10 = 600 N MARK 1

2

3 (c)

1.

(d) (e)

(f)

2T2 = W 2T2 = 600 T2 = 300 N Method in Diagram 8.2 , tension is lower 2 mv = mgh 2 V = 2 gh = 2 x 10 x 2 = 40 -1 V = 6.32 ms Total

2 1 1 1 1 12

SBP 2008

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QUESTION NO. (a)

ANSWER Atmospheric pressure is the weight of air acts on a surface. //It caused by air molecule collision on the surface.. Air pressure in the sticker decrease. Have the different between pressure in the pump and the air pressure surrounding. Pressure outside higher than inside. The force is produce to pull the mirror Pulling Force > mirror weight ( Use F = PA ) 4 Force is produced by pump S = 4.0 x 10 x 0.0025 = 100 N 4 Force is produced by pump M = 4.5 x 10 x 0.0040 = 180 N Force is produced by pump T = 6.0 x 104 x 0.0035 = 210 N mirror weight = Vg 3 = 2.5 x 10 x 1.5 x 0.5 x 0.01 x10 = 187.5 N Pump T is choosed because force is produced bigger than weight of the mirror. Total ANSWER Electrical energy to heat energy E = pt = 220 x (5x 60) // 220 x 5 4 = 6.5 x 10 J Beaker P, Heat = mc = 0.8 x 4200 (38 30) = 26880 J Beaker Q, Heat = 0.8 x 4200 x (46 30) = 53760 J Beaker R , Heat = 0.8 x 4200 x (40 30) = 33600 J Beaker Q The increase of water temperature is higher with the same energy supplied Immerse the immersion heater fully in the water // wrap the beaker// // cover th beaker // use beaker of higher specific heat capacity 0.8x 4200 x (46-T) = 0.2 x 4200 x (T-28) T = 42.4 C Total

MARK 1 1 1 1 1 1 1 1 1 1 1 1 12 MARK 1 1 1

(b)

2.

( c) (i)

(ii) (iii)

QUESTION NO. (a) (b)

1 1 1 1 1 2 1 1 12 MARK 1 1 1

(c ) 3. (d) (i) (ii) (e) (f)

QUESTION NO. (a) (i) (ii)

ANSWER Electromagnetic induction At (a) flux changes in a while (when switch is on) // the coil cuts the magnetic field in a while // The coil experience flux increases. At (b) flux changes continously / always happened // solenoid always cuts the magnetic feild VpIp = 36 Ip =

(iii) 4. (b) (i)

36 240

1 1 1 1 1 1 1

(ii)

= 0.15 A D U shape can produce surface / lower part that consist of two polars: North and South. Electromagnetic strength is higher. Iron core is easy to magnetised and demagnetised. U shape core is wound with the number of turns of wire and the direction of current flow to produce both poles North and South at the end. North and South pole are labelled correctly follow the turns and direction of the current flow without consider the shape.

SBP 2008

10

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(iii)

Increasing the number of turns. Increase the magnitude of current flow inside the solenoid. Total

1 1 12

QUESTION NO. (a) (b)(i) (ii) 5. (iii) (iv) V =

ANSWER

MARK

240

1200 800

= 160V P = 0.25 x 240 = 60 W X Transformer = 36 W Y Transformer = 48 W Z Transformer = 54 W Z transformer Highest in Output power / Lowest in power lost

1 1 1 1 1 1 1 1

Kuasa output Kuasa input

100% Highest

1 1 1 1 12 MARK 1 1 1 1

(c)(i) (ii) (iii)

60 W Resistance of coil wire / Heat produced / Flux leaking / Eddy current produced / Lost of energy in term of magnetised and demagnetised Use the low resistance wire / design the efficient core shape / Use laminated iron core/ Use soft iron core Total ANSWER To limit the current through the transistor As the brightness increases, resistance of X decreases The greater the resistance, the higher the potential difference across BC// potential difference across BC increases with RX 12 V 1. V BC =

QUESTION NO. (a)(i) (ii) (iii) (b)(i)

(ii) 6. (iii) Yes 1. 2. 3. 1. 2. 3.

11 11 + 11

x 12

2. 6 V (answer and units)

1 1

1 V BC is greater than 4 V Base current flows Transistor is activated //Collector current flows// Relay will switched on lights (Any two correct) Resistor 11 k is replaced by termistor X replaced resistor bulbs replaced by alarm Total

(iv)

2

(c)

3

12

SBP 2008

11

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QUESTION NO. (a) (i) number of proton = 32 17 = 15

ANSWER

MAR K 1

1 (ii)

7. 1 (iii) Phosphorus -32 situated outside the graph that shown the stabile atomic nucleus. Phosphorus nuclei are not stable The atomic nomber element P incease 1. (b) (i) Nucleon no. Doest change Element P activity change from origin activity. The activity of Element Q activity doest change The mass of element P decrease to half from original mass. 1 1 1 1 1 (iii)

(ii)

(c)(i)

Penetration power Penetration power Q element Longer half life

ray is weak and cant penetrate aluminium plate. ray is too high and effect our health.

1 1

(ii)

1 1 Total 12

QUESTION NO. (a) (b)

ANSWER Isotope that are not stable/unstable nuclei/unstable isotope D The reading of rate meter is highest226 88

MAR K 1 1 1

(c) (i) 8.

Ra

222 86

Rn +

4 2

He

Correct symbol for alfa - 1 Equation balance -1 2 1 1 1

(ii)

100% 50% 25% 12.5% Sodium-24 : 3T = 45 T = 15 hours Cobalt-60 : T = 15.9/3 = 5.3 years

SBP 2008

12

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Radium-226 : (d)

T =

4860/3 = 1620 years

1 1 1 1 Total 12 MAR K

Sodium-24 Half life is short @ activity decreases faster Emmits beta rays @ can penetrate ground but cant penetrate pipe

QUESTION NO. Characteristics High spring constant

ANSWER Explanation Stiffer, more potential energy stored and converted to kinetic energy.The kinetic energy provided pushes the competitors higher up in the air. Stronger, can support the competitor Allows for higher bounces and more complicated stunts.

(i) 1

Steel frame More coiled springs

Q is chosen because it has the highest spring constant, it is made of steel and has the

2 2 2 2

most number of coiled spring.

(ii)

1. 2.

Jump harder/ push harder To increase reactive force/ to store more energy Total

1 1 10

QUESTION NO. Characteristics Type of liquid:- oil Thickness of brake line: -high

ANSWER Chose n brake system : S The ratio of cross sectional area Reaso for wheel piston and master n: it piston: -high uses The chosen of the type of brake Disc brake is more efficient than drum oil as for front and rear wheel.: brake the Front wheel : disc brake Rear wheel : drum liquid, has high thickness of the brake line, the ratio of cross sectional area for wheel and master piston is high, uses disc brake for front wheel and drum brake. Explanation uncompressible To counter the high pressure from the liquid To produce larger force on wheel piston Total

MARK

2 2 2

2

2 1 1 10 MARK

QUESTION NO. Characteristics Need safety valve

ANSWER Explanation To release extra steam so that the pressure in the cooker does not reach a dangerous stage To withstand high pressure Heats up quickly and food will be cooked faster Heats up slowly and can be held with bare hands S is chosen Reasons: It has safety valve, high thickness , low Total

2 2 2 2 1 1 10

3

High thickness of the pot Low specific heat capacity of the pot High specific heat capacity of the handle

specific heat capacity of the pot, high specific heat capacity of the handle.

SBP 2008

13

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QUESTION NO. Characteristics Shape should be convex Good weather resistance 4 High impact resistance High reflectivity

ANSWER Explanation Gives wider field of view and upright image. So that mirror will not get blur under heat, light and rain. prevent damage due to strong force cause by accident or vandalism. Produce brighter image under dim light.

MARK 2 2 2 2 1 1 Total 10 MARK

Mirror R is the most appropriate, because it is convex. It is made from excellent weather resistant material and has good impact resistance and high reflectivity.

QUESTION NO. Characteristic Low density High tension Smaller length of the string 5 tring material : Stainless Steel

ANSWER Reason Lighter/less massive string, wave travel faster and frequency higher High frequency hence high pitch Produce higher frequency - smaller length of string has low wave length Produce bright sound/ high corrosion resistance/Lasting and does not break easily/Prolong and retain their tone longer

2 2 2 2 1 1

Q is chosen Because it has low density, high tension, smaller length of the string and its made of steel. Total QUESTION NO. ANSWER

10 MARK

Use of instrument / component Voltmeter to measure the potential difference Ammeter to measure current 6

Way connected in thr circuit Connected in parellel to the constantan wire Connected in series to the circuit

3 3 2 1 1 Total 10 MARK 2 2 2 2 1 1 10

Rheostat to adjust/control current in the circuit The most suitable circuit is R Because the voltmeter is connected in parallel to the constantan wire, the ammeter is connected to the circuit and has a rheostat to adjust the current in the circuit.

QUESTION NO. Characteristics Low density High boiling point High resitivity Low rate of corrosion Reason

ANSWER

7

So that the loop will be light So that it cant easily freeze So that the current high// high heat energy produced Cant easily rust

Substance T is chosen Reason: Low density, high boiling point average, high resistivity and rate of corrosion low Total

SBP 2008

14

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SECTION 4 (UNDERSTANDING) No Answers When the boy jumps onto the river bank, his momentum is forward. Using the Principle of conservation of momentum, 1 the total momentum before and after jumping is equal. The boat moves backward to balance the forward momentum. When the girl jumps on the trampoline bed, the force of the jump stretches the springs. As the springs are stretched, they store elastic potential energy. 2 At that point, the potential energy in the springs is converted to kinetic energy, and the springs begin to restore themselves to their initial position. The kinetic energy provided by the springs pushes the girl up into the air and change to potential energy. The area in contact with the ground for the shoes in figure 3.2 is larger than the the shoes in figure 3.1. The pressure exerted by the shoes in figure 3.2 is lower than the shoes in figure 3.1 The shoes in figure 3.2 is difficult to sink into the ground compared to the shoes in figure 3.1. So, it is easier to run using the shoes in figure 3.2. A cargo ship displaces a larger volume of water. 4 The buoyant force acted on the ship is higher compare to iron nail B is denser than A.. The weight of water displaced is the same of the weight of the rod. 5 Weight of B is greater than weight of A B will displace more volume of water when the small piston, X is pressed down, the pressure is exerted under the piston ,X. the liquid transmit pressure to all directions and to large piston, Y. 6 when the pressure acted on larger piston, Y it will produced a large force. the large force will push the load up The shape of cross section of the wing causes the speed of airflow Above the wings to be higher than the speed of airflow below 7 When the speed of moving air is higher ,the pressure is lower Hence air pressure below the wings is higher compare to above the wings States that the related principle is the Archimedes Principle. 8 Draw a diagram of the hydrometer correctly (small and long stem, with big bulb with lead or steel balls) When a hydrometer floats in a liquid, the buoyant force is equal to the weight of the hydrometer Buoyant force = V g / depends on density of liquid. The submarine has a ballast tank in front and at the end of submarine. 9 A water is filled in the ballast tank to increase the density of the submarine // the weight of submarine The submarine will sank in the sea when it weight is bigger than buoyant force. Mark 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

3

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No

Answers In the sea, buoyant force > weight of the ship

Mark 1 1 1 1 1 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1

10

When the ship in the river,density of water< density of sea,so buoyant force decrease The weight of the the ship > the buoyant force ,so the ship will sink The specific heat capacity land less than sea water. During the day time, the land is warmer than the sea.

11

Air above the land is hot and less dense, so it will move up. The cooler air from sea more dense move to land. Water evaporates from the skin when we sweat. (i) for water to evaporate it use heat from the body Steam condenses to form water (ii) latent heat of vaporization of steam is let out more steam. Concave surface gives inverted image Sketch a ray diagram to show a diminished inverted real image

12

13

Convex surface give upright image Sketch a ray diagram to shown diminished upright virtual image. the layer of air near the ground are hotter and less dense medium // the layer of air higher up are cooler and denser medium light from the sky in refracted towards normal after passing through less dense medium from denser medium. Near the ground, the angle of incidence is greater than the critical angle the total internal reflection occur and the light is reflected to the eyes observer. the layer of air near the ground are hotter and less dense medium // the layer of air higher up are cooler and denser medium Sound waves from two loudspeakers produced two coherent sources Sounds wave interfere

14

1 1 1 1 1 1 1 1 1 1 1 1 1

15

Constructive interference produced loud sound Destructive interference produced soft sound At the centre of the ocean the water waves travel at uniform speed as the depth of the sea is uniform reduced, refraction occurs When the waves reach the coast ,the water is shallower, wave speed Refraction causes the wave front bend toward the normal This results the wave front following the shape of of coastline State the situation when touching the live wire correctly Our bodies are at earth potential (0V) If we touch the live wire, there will be a large potential difference (p.d) the live wire and our body. A large current flow through it, probably fatal. State the situation when touching the neutral wire correctly The neutral wire stays at earth potential (0V), roughly at the same potential as our bodies. If we touch the neutral wire, there is no p.d. across us and so no current flows. The lighted candle / the heat from the candle causes the air molecules (surrounding it) to be ionized.

16

17

18

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No

Answers The positive charges would be attracted to the negative plate and/or the negative charges would be attracted to the positive plate The flame of the candle would be dispersed (flattened) into top parts // suitable diagram More the flame is attracted to the negative plate //diagram Positive charges are heavier than negative charges. ( any 4 correct) Electric current flow trough copper strips from ZY to WX. Magnetic field produced in the copper strips when the current flows.// Diagram.

Mark 1 1 1 1

1 1 1 1 1

19

The direction of the magnetic fields is the same. // Diagram The copper rods repel to each other// the copper rods bent// diagram. 1. @

20

2. when current flow, (capasitor) charged up 3. when no current flow, (capacitor) is discharged 4. capacitor connected parallel // diagram Connect the dry cell terminal to the Y-input of CRO.

1 1 1 1 1 1 1 1 1

21

The Y-gain is set to a value so that the direct current wave form displayed on the screen CRO Determine the distance / part of y-axis. Potential different = ( Y-gain scale) x (Vertical distance of direct current wave)

22

Penetration power Penetration power

ray is weak and cant penetrate aluminium plate. ray is too high and effect our health.

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SECTION 5 : PROBLEM SOLVING ( QUANTITATIVE) QUESTION (i) 1 (ii) (iii) Distance = 20 x 5 = 100 m 500 N a = F / m = -500 800 -2 = -25 ms ANSWER MARK 2 2 2

(a) 2 (b)

The spring constant = gradient of the graph -1 For P , spring constant = 8 / 0.5 = 16 N cm -1 For Q , spring constant = 3 / 0.5 = 6 Ncm Elastic potential energy = x 8 x 0.5 x 10 =0.02 J Useful energy output =0 .8 x 10 x1.5 =12 Joule Energy input = Pt = 1.2 x 5 x 4 = 24 J Efficiency = 12/24 x 100 % = 50 %-2

3

2

(i) 3

2

(ii)

2

(iii)

2

v 0 u (i) 4

2

= u + 2as 2 = u + 2(-10)(5) = 10 m s-1 mv = mgh 2 v = 2 gh -1 v = 10 m s2 2

2

2

or

(ii) (iii)

E = mv 2 = ( 100/1000) (10) = 5J 2 kx = 5 J -1 k = 1000 N m P= F/A F=PA F=500 x 40 F = 20000N Resultant force = 20000 900(9.8) = 11180N Upward F =ma a=F/m -2 a=12.42ms Px = Fx / Ax = 15 N / 0.02 m2 -2 = 750 Nm F = PA = (750)(0.28) = 210 N Volume of liquid transferred is unchanged A2 x2 = A1 x1 x2 = ( A1 / A2)(x1) = (0.02 / 0.28) (21) = 1.5 cm

2

1

2

(i) 5

(ii)

2

(iii)

1

(i) (ii) 6 (iii)

2 2 2

SBP 2008

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QUESTION

ANSWER Volume = Mass Density = 30.2 / 10 460 -3 3 = 6.56 x 10 m Buoyant force = weight of water displaced = Vg -3 = 1000 x 6.56 x 10 x 9.8 = 64.3 N Buoyant force = weight of the board + tension in the spring Therefore, Tension in the spring = 64.3 - 30.2 = 34.1 N 100C m=V = (1) (100) = 100g .2 x 379 ( 100-T) = 0.1 x 4200 x (T-28) T = 39 C Total amount of heat falling in 8 hours 7 = 5x200J x 8 x 60 x 60 = 2.88 x 10 J Energy absorbed by the water 7 = 0.3 x 2.88 x 10 J = 8640000 J// 8640 kJ. u = 10 cm v/u = 2 , v = 2u v = 2x10 = 20 cm v = -20 cm 1 f = 1+1 v u

MARK

(i)

2

7 (ii)

2

(iii)

2

(i) 8 (ii) (iii)

1 2

2

(i) 9 (ii)

2 2

(i)

2

10 (ii)

1 = -1 + 1 f 20 10 f = 20 cm Convex Image formed on the screen (real) u + v = 60 cm v = 2u 1 = 1+1 f u v (ii) 1 = 1 + 1 f 20 40 f = 13 cm 2

(i)

1 1 1

11

1 1

(i) 12 (ii) PI =

I=

10 12

2

= 0.83 A (with unit)

10 x100 80

3

SBP 2008

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QUESTION = 12.5 W I=

ANSWER

MARK

12.5 / 0.052 A (accept e.c.f) 240

(i) 13

(ii)

(iii)

I = P = 20 kW V 4 kV = 5A 2 P= IR = 5x5x16 = 400 W 2 V = PR = 400x16 V = 80 V

2

2

2

(i) 14 (ii) (iii)

x 6 = 2.2 3500 V (3500 + 6000) x 6 = 3.8 V Potential difference across Q, 6000 (3500 + 6000) Ring because the voltage across Q exceed / greater than 3.2 V. Potential difference across P =

2 2 2

SBP 2008

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SECTION 6 DESIGN EXPERIMENT Q 1 Experiment Mass and acceleration Manipulated Variable Mass/ no of trolley Mass of water Responding Variable acceleration Rise in temperature/ Final temperature Resistance force Fixed Variable Topic Force and motion

2

Mass of water and rise in temperature Resistance and thickness of water Current and time taken for the rod to travel Depth of water and pressure

Time to heat up the water Length of wire

Heat

3

Thickness of wire Current/ no of battery Depth of thistle funnel Height of magnet Temperature

Electricity Catapult field Interaction between two magnetic field Pressure in water {manometer)

4

Time taken

Distance

5

Different depth of water level

Density of water

6

Height of magnet and induced current Temperature and volume of gas

Current

Number of coil

Induced Current

7

Volume

Pressure

Charles Law

8

Mass and period of oscillation

mass

Period of oscillation Depth of straw sink Number of pin attract Depth of straw sink Different depth of water level

Length of jigsaw blade Density of water

inertia

9

Mass and volume of water displaced Current and electromagnetic strength Density of water and volume of water displaced Density of water and pressure

Mass of sand

Hydrometer

10

Current Mass of salt dissolved in water Mass of salt dissolved in water Depth of rod sink in water

Number of turn of solenoid Mass of straw

Electromagnetic effect Buoyant Force

11

12

Depth of thistle funnel

Pressure in liquid

13

Volume of water displaced and buoyant force

Apparent weight

Density of water

Buoyant Force

14

Length of wire and resistance

Length of wire

Resistance

Thickness of wire

Electricity

15

Resistance and temperature

Temperature

Resistance

Length of wire / thickness of wire Spring constant

Electricity

16

Mass and extension of spring

mass

Extension of spring Pressure

Hookes Law

17

Volume and pressure

Volume

Temperature

Boyles Law

SBP 2008

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