perfect score sbp fizik spm 2011 answer

1

SECTION I

No UNDERSTANDING Total

mark

1

1. runner has Inertia

2. to continue moving forward

3. the legs stop, body continue moving forward

4. Unstable and fall

4

2

1. When the ball on one end is pulled up and let to fall, it strikes the second ball which is at

rest and comes to a dead stop.

2. The momentum of the ball becomes zero as its velocity is zero.

3. The Principle of Conservation of Momentum states that in a collision between two objects

the total momentum of the objects in the system remains unchanged.

4. The energy and momentum from the first ball is transferred to the second ball and then

transmitted through the balls at rest to the ball on the other end.

5. Because the momentum and energy is maintained in this system, the ball on the opposite

side will move at the same velocity as the ball that were in initial motion

[Any four]

4

3

1. The boat floats, so Weight of the boat = Weight f the water displaced = Buoyant force

2. As the weigh of the boat is the same so the weight of water displaced in the river and the

sea water also the same

3. Density of sea water is higher than river water

4. Volume of water displaced in the sea is less than in the river, Level of the boat is higher in

the sea than in the river

4

4

1. When the boiled water is poured onto the ping pong ball,

The temperature of the air/gas will increase/ the kinetic energy increase

2. The rate of collision between molecules and wall of the ball will increase so the pressure

will increase,

3. the ball will expand, so the volume will increase

4. when the volume increased, area of collision increased, so lastly the pressure will remain

the same

4

5

1. When water in tube pass through the engine it can absorb large amount of heat energy

2 Once water reach the radiator, the heat of the water absorbed by the fin blade of the

radiator

3 The same time the fan in the radiator push the heat out of the car.

4 Water has high specific heat capacity

4

6

1. For the fish, the light is refracted / change direction at B.

2. the light is refracted away from normal, towards the observer’s eyes

3. For the dragon-fly, the light is reflected by water surface at A.

4. Reflected angle = incidence angle, reflected towards the observer’s eyes

4

7

1. the air close to the surface is much colder than the air above it.

2. Light travels from denser to less dense medium

3. Light rays refracted away from normal line and bent downward toward the surface

4. thus tricking our eyes into thinking an object is located higher in appearance than it

actually is - the observer will see the image of the ship due to light travels in a straight line

4

8

From the ray diagram

[1.different medium (water/air), fish and

observer

2.Light refracted away from normal

3. Extrapolation to show position of the

observing image]

4. so, he should shoot the target at the lower position of the image

4

2


mark

9

1 - parallel ray between the condenser lens

2 - Two rays from condenser lens touching the end of the object

3 - two rays comes out from the convex lens reflected by the mirror

4 - image formed on the screen

5 - magnified, real, inverted

5

10

1. waves move from deeper to shallow area

2. the speed decreases

3. the wave refracted towards the normal line

4. the wave front which perpendicular to the direction of propagation of waves resulting the

wave front following the shape of the beach

4

11

1. waves move from deeper to shallow area

2 the speed decreases // the wave refracted towards normal line

3. the depth at cape decreases abruptly resulting waves focused at cape

4. the depth at bay decreases slowly resulting waves spread out

5. the energy of waves is smaller at bay resulting calmer region compared to at cape

4

12

1. Use ultrasound, ultrasound is transmitted to the sea bed

2. a receiver will then detect the reflected pulses

3. the time taken by the pulse to travel to the seabed and return to the receiver being

recorded, t

4. the depth of the sea can be calculated using the formula, d = vt/2

4

13

1. Wave length depends on depth of water

2. Increasing/decreasing of depth will cause the wave to refract

3. Refraction will increase/decrease the length of wavelength

4. The nodal/antinodal line will be affected

4

14

1. A parallel circuit can run several devices using the full voltage of the supply.

2. If one device fails, the others will continue running normally

3 A failure of one component does not lead to the failure of the other components.

4 More components may be added in parallel without the need for more voltage.

5. Each electrical appliance in the circuit has it own switch

6. Less effective resistance of the circuit

[any 4]

4

15

1. The lighted candle / the heat from the candle causes the air molecules to be ionized.

2. The positive charges would be attracted to the negative plate and/or the negative charges

would be attracted to the positive plate

3. The flame of the candle would be dispersed (flattened) into two parts // suitable diagram

4. Positive charges are heavier than negative charges

5. More of the flame is attracted to the negative plate //diagram

(any 4 correct)

4

3


mark

16

[ show structure diagram of transformer]

1. An alternating current flows through primary coil

2. The soft iron core is magnetised

3. The magnetic field produced varies in magnitude and direction

4. This causes a changing magnetic flux to pass through the secondary coil

5 An induced emf across the secondary coil is produced

4

17

1. When a high current flow, the magnetic field becomes stronger

2. Electromagnet attracts the soft iron armature.

3. spring P pulls the wire and break the contact, the current does not flow // the circuit is

disconnected.

4. When reset button is pressed, spring Q pulls the soft iron armature back to its original

Position

4

18

1. Magnetic field produced by the current in the coil

2. interact with the magnetic field of the permanent magnet

3. producing the catapult field

4. produces resultant force

4

19

1. intrinsic semiconductor such as silicon atom

2. doped with pentavalent atom such as phosphorous

3. each pentavalent atom contributes one free electron and there is an excess electron

4. the excess free electrons become negative charge carrier in n-type semiconductor

4

20

1. Anode of diode (p- type) connected to positive terminal of battery/ vice versa

2. Electron from n-type drift to p-n junction towards positive terminal

3. Holes from p-type drift across p-n junction towards negative terminal

4. Movement of electrons and holes produced current, thus the bulb lights up

4

21

1. Radioactive ray enter the tube through the mica window

2. ionizes argon gas under low pressure

3. The ions accelerate towards respective electrodes

4. Produce a current pulses

5. Pulses are recorded by rate meter

[Any 4]

4

4

SECTION II

Question 1 [Force and Pressure]

Characteristics Explanation

Streamlined shape To reduce water resistance

High strength of metal To withstand high water pressure

Wide base cross section area So that ship can float//prevent from overturn // ship more stable // ship not

sink deeper

High volume of air space in the ship Increase buoyant force

Structure U

Because it has streamlined shape, high strength of metal, wide base

cross section area, high volume of air space in the ship



Material made from glass Glass does not corrode with acid

Small diameter of stem To increase the sensitivity of the hydrometer

High density of shot Makes the hydrometer stays upright

Big diameter of bottom bulb To obtaine a bigger upthrust.

Choose N N is made from glass, has small diameter of capillary tube, high density of

shots and a big diameter of bottom bulb.


Specification Explanation

With ABS To reduce jerking when it is stopped immediately / can be controlled if

direction changes/ does not move side ways

Wide tyres Better support / more stable /safer when turn

Low mass Lighter, can move faster / low inertia.

Low seat height Lower centre of gravity/ more stable

C It has ABS, wide tyres, low mass, low seat height.


Specification Explanation

High specific heat capacity The rate of temperature increased caused by friction is low

High melting point Does not easily change in shape when the temperature is high

Difficult to compress Pressure will be transmitted uniformly in all directions

Use ceramic Can withstand high temperature // less dust produced

S Because it has high specific heat capacity, high melting point, difficult to be compressed and use ceramic.

5

Question 5 [Heat]


High boiling point It will not easily change into gas when absorb heat from the engine

Low viscosity It will not freeze during cold weather//can flow at low temperature

High specific heat capacity It can absorb a big quantity of heat with small rise in temperature

A low ability to react with metals The metal parts of the engine will not corrode easily

K is the most suitable liquid

Because of its low freezing point, high boiling point, high specific of heat capacity and low ability to react with metals

Question 6 [Light]

Specification Reason

Type of objective lens is convex lens To converge the light and produce real image

Focal length of the objective lens is

big To get higher magnification power

D < fo + fe To get virtual and magnified image

Diameter of the objective lens is

large More light can be captured , the image formed is brighter

Lens S Because lens used is convex lens, focal length of the objective lens is big,

D < fo + fe and diameter of the objective lens is large

Question 7 [Electromagnetism]

Characteristic Explanation

Low resistivity To reduce heat loss in the cables

Low density The cables will be lighter

Low rate of oxidation Not easily rust / corrode

Low rate of thermal expansion The cables will not expand under hot weather

Cable Q Low resistivity, low density, low rate of oxidation, low rate of thermal

Expansion

Question 8 [Electromagnetism/Electronic]

Characteristic Reason

Low density of the coil Lighter / less mass

High frequency of rotation The rate of change of magnetic field is higher, more current induced

Use 4 diodes Full wave rectification to get a d.c.

Has capasitor in the circuit To smooth the direct current obtained

Circuit I Because it has low density coil, high frequency of rotation, consist 4

diodes and has capasitor in the circuit.

6

Question 9 [Electronic]


The valency of the intrinsic

semiconductor is 4 When it is doped, the conductivity of the semiconductor increase

The valency of the doping substance

are 3 or 5

Valency of 3 is used to produce hole as majority charge carrier (P type)

and valency of 5 is used to produce electron as majority charge carrier (N

type)

Size of the atom of the doping

substance is almost same as the

size of the substance

Can maintain the crystalline structure of the substance/ Give good effect

in the doping process

Substance T

Because the valency of the intrinsic semiconductor is 4, The valency of

the doping substance are 3 or 5 and the size of the doping substance is

almost the same as the size of the substance

Type of pure semiconductor is Silicon

Because it has greater power handling (its not easy to get overheated)

Question 10 [Radioactivity]

Properties Reason

Type of radiation is gamma Has high penetrating power

Long half-life Long lasting

Solid Easy to handle

Low ionizing power Does not ionize healthy cells / does not cause cell mutation

K Because it radiates gamma ray, the half-life is long, the state of matter is

solid and it has low ionizing power

7

SECTION III - CONCEPTUALIZATION

Question Answer Mark

1

(a) 1.17 cm , 1.173 cm 2

(b) (i) The reading in Diagram 1.1 is less accurate than the reading in Diagram 1.2 1

(ii) The smallest scale in Diagram 1.1 is larger than the smallest scale in Diagram 1.2 1

(iii) The smaller the smallest scale of an instrument, the more accurate the measurement 1

(iv) The smaller the smallest scale of an instrument, the more sensitive the instrument 1

(v) The higher the sensitivity the instrument, the more accurate of the measurement 1

TOTAL 7


2

(a)(i) F1 = F2, F3 = F4 1

(ii) The direction of the two forces involved in both diagrams are opposite to each other 1

(iii) For any action, there is a reaction

which has the same magnitude but acts in the opposite direction.

1

1

(iv) Newton third Law of Motion 1

TOTAL 5


3

(a) (i) The total mass of the lorry and the load in Q is larger than in P 1

(ii) The difficulties in stopping the lorry Q is more than lorry P 1

(b)

When the mass of the lorry and the load is larger, the more difficult to stop it.

The difficulties in stopping the lorry is due to the inertia which tend to maintain the

state of motion.

When the mass of the object larger, the inertia is greater.

1

1

1

TOTAL 7


4

(a)(i) Density of salt solution is higher than density of water 1

(ii) The portion of the block immersed in salt solution is less than in water 1

(iii) Buoyant force acted in both liquid are same 1

(b)

Factors affect the buoyant force in diagram 4 are density and volume of liquid displaced

For the same buoyant force, when the density of the liquid is higher, the volume of liquid

displaced is less

1

1

(c) Archimedes’ Principle 1

TOTAL 6


5

(a) Temperature is the measurement of the average kinetic energy of the molecules in an

object / system

1

(b)(i)

(ii)

(iii)

Temperature of trap air in diagram 5.2 is higher than in diagram 5.1

Pressure of trap air in diagram 5.2 is higher than in diagram 5.1

Volume of trap air in diagram 5.2 and 5.1 are equal

1

1

1

(c) When the temperature of the air increase, the volume is also increase 1

(d) Pressure Law 1

TOTAL 6

8


6

(a)(i) Real image is the image that can form on the screen. 1

(ii) The object distance is longer than the focal length to form real image 1

(iii) Object distance in Diagram 6.1 is shorter than that in Diagram 6.2

Image distance in Diagram 6.1 is longer than that in Diagram 6.2

Size of image in Diagram in 6.1 is larger than that in Diagram 6.2

When the object distance is longer, the image distance is shorter

When the image distance is shorter, the size of image is smaller

1

1

1

1

1

TOTAL 7


7

(a)(i)

(ii)

(b)

In diagram 7.1, light propagates from low density medium to high density medium

In diagram 7.2, light propagates from high density medium to low density medium

In diagram 7.1 direction of light travel towards normal but in diagram 7.2 direction of

light is away from normal

When light travels from low density medium to high density medium, light bends towards

normal and when light travels from high density medium to low density medium, it will

bend away from normal

1

1

1

1

1

(b) Refraction of light 1

TOTAL 6


8

(a)(i)

In diagram 8.1, density of layer of air close to ground is lower than density of layer of air

at upper part. In diagram 8.2, density of layer of air close to ground is higher than layer

of air upper part.

1

(ii) Direction of propagation of sound waves in diagram 8.1 is away from the earth (upward)

but in diagram 8.2 it propagates toward the earth (downward) 1

(iii) The loudness of sound that can be heard in diagram 8.2 is greater than that in Diagram

8.1 1

(b)

If the density of air close to the ground is higher than the air at the upper part, the

direction of propagation of the sound is directed towards the ground,

therefore the sound can be heard louder / vice versa

1

TOTAL 4

9


10

(a) Negative 1

(b)(i)

(ii)

(iii)

In Diagram 10.1, there is no electric field while in diagram 10.2, the electric field exist

between the plates

The voltage of EHT between two plates in diagram 10.2 (a) is smaller than that in

Diagram 10.2 (b)

The angle of deflection of the cathode ray in diagram 10.2 (a) is smaller than that in

Diagram 10.2 (b)

1

1

1

(c) (i)

(ii)

When the voltage between EHT is higher, the strength of electric field is stronger

The stronger the electric field, the bigger the deflection of the cathode ray.

1

1

TOTAL 6

SECTION IV – PROBLEM SOLVING (QUALITATIVE)

QUESTION 1


The valency of the intrinsic

semiconductor is 4 When it is doped, the conductivity of the semiconductor increase

The valency of the doping substance

are 3 or 5

Valency of 3 is used to produce hole as majority charge carrier (P type)

and valency of 5 is used to produce electron as majority charge carrier (N

type)

Size of the atom of the doping

substance is almost same as the

size of the substance

Can maintain the crystalline structure of the substance/ Give good effect

in the doping process

Substance T

Because the valency of the intrinsic semiconductor is 4, The valency of

the doping substance are 3 or 5 and the size of the doping substance is

almost the same as the size of the substance

QUESTION 2

No Answer Mark

7(a)(i)

(ii)

(b)

c

(d)(i)

(ii)

Interference

Constructive interference

λ = 340

2000

= 0.17 m

Increase

f α 1/f and x α λ // x α 1/f

speaker

to convert and produce wave which has same frequency and amplitude

destructive interference

1

1

1

1

1

1

1

1

1

1 +

10

QUESTION 3

suggestion reason

Thin fuse has high resistance

low specificheat capacity It take shorter time to heat up/ to reach melting point /blow the fuse

Ceramic catridge Can withstand higher temperature

Fuse rating 13 A Normal current of device is 2400/240 = 10 A. Maximum rating must be

higher than normal current

Melting point must be low Easy to melt (blow faster)

QUESTION 4

suggestion reason

Attach one fuse to the live wire in the consumer unit/ fuse box.

To break /switch off the circuit when large current before the wire become hotter and produce fire.

Using the insulating wires // thicker wires

To prevent short circuit // To reduce resistance, improve efficiency

Attach switch for each lamp To allows each lamp to be switched on and off independently

Connect the metal fitting lamp to the earth wire/cable

To flows electron (extra) to earth to avoid lethal shock

Using only 240 V light bulb. To ensure the bulbs light up with normal brightness

QUESTION 5

suggestion reason

Handle made from insulator The rate of heat flow to the handle is less, does not easily heated

The body is made from low density

material Reduce mass, iighter

Coiled wire for the filament Can fit longer wire inside, more heat is released

Material for heating filament is

nichrome High melting point

Connection to earth // fuse Flow the excess /overload current to earth if there is a short circuit / blown

when it is excess of current flow

QUESTION 6

suggestion reason

Soft spring Give a greater sensitivity/ can detect small changes

Small density Small mass / ligh

Curve in shape of the magnet Radial magnetic field, create uniform strength of magnetic field around the

coil

Low resistance material of the coil To reduce energy loss

Place the seismometer in direct

contact with the earth to convert very small motions of the earth into electrical signals

11

QUESTION 7

suggestion reason

use thin diaphragm Easy to vibrate

Use strong material Not easy to break

More number of turns of coil Increase the rate of change of magnetic flux linkage // The magnitude of

the induced current or is also increased

Thicker diameter of wire of coil reduce the resistance of the coil

Using more powerful magnet to

increase the strength of the

magnetic field

Increase the rate of change of magnetic flux linkage //The magnitude of

the induced current or induced electromotive force is also increased

QUESTION 8

suggestion reason

OR gate The gate's output is ON if either one sensor is ON

One of the input X is connected to

+6 V

To supply an ON signal to gate X, so the current flow into the base of the

transistor

Place the resistor in base circuit To limit the current flow in the base circuit, The current amplification of the

transistor is higher

Use buzzer in the collector circuit It converts the electrical signal into sound energy

Relay switch To switch on the buzzer which is use a greater voltage

12

SECTION V – PROBLEM SOLVING (QUANTITATIVE)

QUESTION ANSWER MARK

1

(a)

a =

=

= -2 m s

-2

2

(b) s = ½ (0+4) x 2 = 4 m

2

2

(a)

m1u1+ m2u2= m1v1 + m2v2 (0.05)(0.8) + (0.03)(0) = 0 + (0.03) v2 v2 = 1.33 m s

-1

2

(b) m1v1 - m2u2 = 0 – (0.05)(0.8) = -0.04 kg m s

-1

2

(c) Impulsive force = -0.04 / 0.05 = 0.8 N

1

3

1 mark for correct parallelogram lines (measurement & angle). 1 mark for correct diagonal line. 1 mark for direction of the resultant force. 1 mark for magnitude with correct unit : 13.8cm X 2.5 = 34.5N

4

4

(a)

=

B = 375 N

2

(b) 2 x 21 = B x 15 B = 2.8 cm

2

5

(a)

P = F / A F = 400 x 50 F = 20000 N

2

(b)

Resultant Force = 20 000 – 900(10) = 11000 N Direction of force : upwards

3

6 (a)

Mass = density x volume Mass = 0.169 kg m

-3 x 1.2 m

3 = 0.20 kg

2

13


(b)

m = 1.3 kg m-3

x 1.2 m3 =1.56kg

Weight of displaced air = bouyant force = mg = 1.56 x 10 = 15.6 N

3

7

(a) V = 0.2 x 0.8 = 0.16 m

3

2

(b) B = Vρg = 0.16 x 1000 x 10 = 1600 N

2

(c) mass = 1600 ÷ 10 = 160 kg

1

8

=

x 100

= 35 oC = 308 K

4

9

(a)

Q = Pt = 48 x 5 x 60 // 14 400 J

Q = m c 14 400 = 500 x 10

-3 ( c ) ( 80 – 40 )

c = 720 J kg-1 o

C-1

(with unit )

3

(b)

Heat supplied by liquid = Heat received by water

( 500 x 10-3

)(720)(80 - ) = (1) ( 4200)( - 25 )

= 29.34 oC (with unit)

2

10

= 90o – 30

o

= 60o

1.33 = o40sin

sin

= 58.75o

4

11

(a)

1 = 1 + 1 f u v 1 = 1 – 1 v 5 400 v = 5.063 cm

2

(b)

h2 = v2 h1 v1 h2 = 5.063 100 400 h2 = 1.27 cm

2

12

I =

= 0.83 A R = V

2

P = 240

2 = 288 Ω

200

4

13

(a) 10A or 13 A

1

(b)

V = IR = 9 x 26.7 = 240.3 V P = IV = 9 x 240.3 = 2162.7 W

4

14


14

(a) 2 = I(1 +5) I = 0.33 A

2

(b)

R = ( 1/5 + 1/12)-1

= 3.53 Ω 2 = I( 1 + 3.53) I = 0.44 A

3

15

(a) 12 V

1

(b)

Np = Vp Ns Vs Np = 240 x 200 12 = 4000

2

(c)

Efficiency = Po x 100 Pi = 240 x 0.2 x 100 48 = 100 %

2

16

(a) I = 24/12 = 2A (with unit)

(b)

Efficiency = Output power x 100 % Input power Input power = 24 x 100 40 = 60W

17

v = 3.75 x 107

m s-1

3

18

(a)

The bulb not light up

3

(b)

R1 = 5500

2

19

(a)

+

1

(b) 3 alpha particles 2 beta particles

2

(c) t = 33.5 /6.7 = 5 T1/2 32 → 16 → 8 → 4 → 2 → 1 g

2

20 (a) M = (230.0331) - (226.0254 + 4.0026) = 0.0051 u = 0.0051 x l.66 x 10

-27 kg = 8.466 x 10

-30 kg

2

VQ = 500 = 0.48 V

12 12000 + 500

1 = 500

12 R1 + 500

15


(b)

E = mc 2

= 8.466 x 10-30

x (3 X 10 8

) 2

= 7.619 x 10-13

J

2

21

(a)

E = mc2

2.9 x 10 -11

= m x (3.0 x 108)2

m = 3.22 x 10-28

kg

2

(b)

Power obtained P = E/t = 2.9 x 10

-11 /1.5 x 10

-3

= 1.93 x 10-8

W

2

SECTION VI – PAPER 3

QUESTION 1

NO MARKING CRITERIA MARK REMARK

a

i Height, h 1

ii Acceleration, a 1

iii The frequency of the ticker timer, f 1

b

Calculate the acceleration, a

2

h = 20.0 cm a = 9.0/0.1 – 7.8/0.1 0.1

= 120 cms-2

h = 30.0 cm a = 9.8/0.1 – 7.4/0.1 0.1

= 240 cms-2

h = 40.0 cm a = 10.2/0.1 – 6.6/0.1 0.1

= 360 cms-2

h = 50.0 cm a = 11.0/0.1 – 6.2/0.1 0.1

= 480 cms-2

h = 60.0 cm a = 11.4/0.1 – 5.4/0.1 0.1

= 600 cms-2

Note: 1. All 5 values correct – 2 marks 2. 3 or 4 values correct – 1 mark

16


Height, h (cm)

Initial velocity, u (cm s

-1)

Final velocity, v (cm s

-1)

Acceleration, a (cm s

-2)

20.0 78.0 90.0 120.0

30.0 74.0 98.0 240.0

40.0 66.0 102.0 360.0

50.0 62.0 110.0 480.0

60.0 54.0 114.0 600.0

Give a tick () based on the following:

A a table which has h, u, v and a

B state name, symbol and correct unit for each column

C All values of h are correct

D All values of u are correct

E All values of v are correct

F All the values are consistent in 1 d.p or 2 d.p.

Marks awarded : Number of Marks

6 4

4 – 5 3

2 – 3 2

1 1

4

17


c

Give a tick () based on the following:

A Title of the graph

B a at the y-axis, h at the x-axis

C Name, symbol and unit for a

D Name, symbol and unit for h

E Uniform scale at both axis

F 5 points plotted correctly

[Note : 4 points plotted correctly : ]

G Straight line of best fit is drawn

H Size of the graph is ≥ ¾ size of the graph paper

Marks awarded : Number of Marks

9 6

8 5

6 - 7 4

4 - 5 3

2 - 3 2

1 1

6

d a increases linearly with h 1

e Avoid parallax error by making sure eyes are perpendicular to the scale of metre rule when measuring the height.

1

TOTAL 16

18

QUESTION 2


a T2 is directly proportional to m 1

b i

Draw line from m = 25 g to the given line and from the given line to the T2- axis

1

1

T2 = 0.30, T = 0.5477 s ( 2 – 4 d.p.)

ii Extrapolate the line to 0.75 or beyond 1

m = 62.0 g 1

c

i

Draw a sufficiently large triangle 1

Correct substitution (Follow candidate’s triangle) 0.60 / 50

1

State the correct value and unit 0.012 s2 g-1

1

ii

m / T2 = 1 / 0.012 1

k = 39.45 (1 / 0.012) 1

k = 3287.5 g s-2 1

d Avoid parallax error by making sure eyes are perpendicular to the scale of stop watch when measuring the height.

1

TOTAL 12

19

SECTION VII – PAPER 3

Question1 [Force and Motion]

No Answer mark

1 (a) State a suitable inference

The speed of the boy on reaching the ground depends on the height of the top of a slide

1

(b) States a relevants hypothesis

The higher the top of a slide the higher the speed of the boy on reaching the ground.

1

(c

)

State the aim of experiment

To study the relationship between the speed of an object on reaching the ground and the

height of the top of the slide.

State the manipulated variable and the responding variable

Manipulated variable : Height of slide ( the slooping runway)

Responding variable : Speed of the object

State ONE variable that kept constant

Fixed variable : Mass of the trolley / angle of inclination

of the runway board.

Complete list of apparatus and materials

A trolley, runway board, ticker- timer, ticker-tape, power supply,metre rule, cellophan tape

and wooden block

States the workable arrangement of the apparatus

a.c power supply

States the method of controlling the manipulated variable

The height of the trolley from the ground/ floor, h = 10 cm is measure using the metre rule.

States the method of measuring the responding variable

The power supply is switched on and the ticker-timer is started the trolley is released and

the final speed, v of trolley is determined by analysing the ticker-tape

Repeat the experiment at least 4 times

The experiment is repeated by using different values of h = 15 cm,

h = 20 cm , h = 25 cm and h = 30 cm. The final velocity of the trolley reaches the ground is

caculated from the dots made on ticker- tape.

Tabulating of data

Height h/cm Velocity of trolley , v /cms-1

10

15

20

1

1

1

1

1

1

1

1

20

Total

25

30

State how data will be analysed

Plot graph of velocity against height

v/cms-1

h/cm

1

1

12 marks


Question Answer Marks

2. (a)

(b)

(c) (i)

(ii)

(iii)

(iv)

State a suitable inference

The pressure exerted on the surface depends on the area of contact.

1

State a relevant hypothesis

When the area decrease, the pressure increases

1


To investigate the relationship between the area of contact and

pressure (depth of hole).

1

State the suitable manipulated variables and responding variable (Quantity

that can be measured)

Manipulated variable : Area of contact object.

Responding variable : pressure exerted (depth of hole)

1

State the constant variable

Force applied (using 1 kg load)

1

State the complete list of apparatus and materials

Vernier caliper, soft plasticine, ruler, Load 1 kg, the object with different surface

area but same weight.

1

Draw the functional arrangement of the apparatus

21

(v)

1

State the method to control the manipulated variable

Measure the bottom area of the wooden block with vernier caliper/ruler, A1 cm2.

Place the wooden block on the plasticine.

Place the 1 kg load on the top of the wooden block.

1

State the method to measure the responding variable

Remove the load and wooden block. Measure the depth of hole using vernier

caliper and record it.

1

Repeat the experiment at least 4 times with the values

Procedure 2 and Procedure 3 are repeated using A2,A3,A4,and A5

(Note : Based on SPM standard , at least five manipulated values required.)

1

State how the data tabulated with the title MV and RV

Area of contact/ cm2

Depth,x /mm

A1

A2

A3

A4

A5

1

State how the data is analysed, plot a graph RV against MV

Depth,x/mm

Area of contact,A/ cm2

1

TOTAL

12

22


3

(a)


The pressure of liquid is depends on its density.

1

(b) State a relevant hypothesis

The pressure of liquid increases as its density increases.

1

(c) State the aim of experiment

To investigate the relationship between the pressure and its density.

1


Manipulated : Density // ρ

Responding : The pressure of liquid

1


The depth of liquid/ /Volume of water

1


Beaker, thistle funnel, manometer, metre rule and salt

1

Arrangement of apparatus :

1

State the method of controlling the manipulated variable

1. Apparatus is set as shown in the above figure.

2. 20 g of salt is dissolved in 100 ml of water.

3. The thistle funnel is lowered into the water at constant h.

4.

State the method of measuring the responding variable

5. The value of y is measured by using metre rule.


6. The experiment is repeated using 30g, 40g, 50g and 60g

1

1

1

Tabulation of data:

Mass of salt/Density Pressure / y

1

23

Analyse the data .

Pressure/y

Mass of salt/Density

1

Total marks

12

Question 4 [Heat]

4

(a)


The rate of cooling of an object depends on its masses.

1

(b) State a relevant hypothesis

The rate of cooling of water increases as its mass decreases.

1

(c) State the aim of experiment

To investigate the relationship between the rat e of cooling of water and its mass.

1


Manipulated : mass of water // m

Responding : Rate of cooling

1


Initial temperature / Final temperature

1


Beaker 250 cm3, measuring cylinder, water, electric heater.stopwatch, thermometer.

1

Arrangement of apparatus :

1



8. Water is heated to 55oC.

9. 50 cm3 of water is placed in a 250 ml beaker with a thermometer immersed in the

water.


10. The stopwatch is started when the temperature of the water is at 50oC. The

stopwatch is stopped when the temperature reaches 35oC. The time, t is recorded.


5. The experiment is repeated using volumes of water 100 cm3, 150 cm

3, 200 cm

3and

250 cm3.

1

1

1

Thermometer

Beaker

water Stopwatch

24

Tabulation of data:

Mass, m (g) Time, t (s)

1

Analyse the data .

Time, t (s)

Mass, m (g)

1

Total marks 12

Question 5 [Heat]

5

(a)

1 State a suitable inference

The volume of gas depend on its temperature

(b) 1 State a relevant hypothesis

.The volume of gas increases as its temperature increases

(c) 1 State the aim of experiment

To investigate the relationship between the volume of gas and its temperature.

1 State the manipulated variable and the responding variable

Manipulated : Temperature

Responding : The volume of gas

1 State ONE variable that kept constant

Mass of gas

1 Complete list of apparatus and materials

Capillary tube, thermometer, water, metre rule and sulphuric acid

25

1

Arangement of apparatus :

1

1

1



12. Water is heated to 300C.


13. The vertical column of trapped air is measured by using metre rule.


4. The experiment is repeated using the temperature 400C, 50

0C, 60

0C and 70

0C.

1 Tabulation of data:

Temperature Volume of gas

1

Analyse the data .

Volume of gas

Temperature

26

Question 6 [Light]

Soalan mark Peraturan Pemarkahan

6 (a) 1 State a suitable inference

The size of the image depends on the object distance

(b) 1 States a relevants hypothesis

The longer the object distance, the smaller the image

(c )

1

1

1

1

1

1

1

1

1


To study the relationship between the object distance and the height of the image


Manipulated variable : Object distance

Responding variable : Height of the image


Fixed variable : Power of lens.


Convex lens, meter rule, screen, lens holder, object

States the workable arrangement of the apparatus

States the method of controlling the manipulated variable

The object distance is measured to be u = 20cm.

States the method of measuring the responding variable

The height of the image that formed on the screen is measured using the ruler.


The experiment is repeated by using different values of u = 25 cm, 30 cm, 35 cm, 40

cm dan 45 cm.

Tabulating of data

Object distance/cm Height of image/cm

20

25

30

35

40

27

Total

1

12

marks

State how data will be analysed

Plot graph ofobject distance against height of image

Height of image/cms-1

u/cm

7 (a) Inference : The Crane (b) had attracted more load of scrapped irons than the crane

because of its larger no. of coils/turns of the solenoid // the strength of the magnetic field

depends on the no. of turns/coils of the solenoid.

1

(b) Hypothesis : Bigger number of turns in the solenoid , the stronger will be its

electromagnetic field strength

1

(c) Aim : To investigate the relationship between the number of turns and the

electromagnetic field strength

1

Variables : Manipulated : number of turns in the solenoid

Responding :electromagnetic field strength / no. of pins attracted

1

Constant Variable : Current ; soft iron core . 1

List of apparatus : Solenoid, PVC tube / large iron nail , iron nails/ pins/paper clips,

ammeter , rheostat and power supply.

1

Arrangement of apparatus:

1

28

Question 8 [Electricity]

(a) The heating effect of a conductor is affected by magnitude of the current. 1

(b) The larger the current, the higher the temperature of the water which is being heated 1

(c) (i) To investigate the effect of current on heating 1

(ii) MV : current ,I

RV : temperature , 1

CV : volume of water 1

(iii) Beaker, ammeter, immersion heater, thermometer, connecting wire, rheostat and stop watch 1

(iv) Draws a labeled and functional diagram of the set up of the apparatus. 1

(v) Pour 200cm3 of water into the beaker and measure its temperature.

Switch on the circuit and adjust the rheostat until the reading of ammeter is 1.0 A. 1

The stop watch is started. The final temperature is recorded after 2 minutes. 1

Step repeated by adjusting the rheostat so that the ammeter readings are 2.0 A, 3.0 A, 4.0 A and

Control of Manipulated Variable : Wind the insulated wire around the large iron nail/PVC tube

, starting with 50 coils .

1

Measurement of RV : Lower the solenoild until it touches the iron nails

. Record the number of iron nails attracted by it.

1

Repeat the experiment & procedure 4 more times with no. of coils at 100 , 150 , 200 and 250. 1

Tabulation of data

No. of turns No of nails attracted

50

100

150

200

250

1

Analysis of data.

Sketch the graph of no. of coils against no. of nails attracted

1

TOTAL 12

29

5.0 A. 1

(vi) Tabulate the data 1

(vii) The graph of increased in temperature against current is drawn 1

Question 9 [Waves]

(a) The loudness of the sound depends on the distance (between the source and the observer) 1

(b) The smaller the distance (between the source and the observer), the louder the sound 1

(c) (i) To investigate the relationship between the loudness of a sound and the distance (between the

source and the observer) 1

(ii) Manipulated variable : distance, d

Responding variable : loudness of sound (amplitude, a) 1

Constant variable: wavelength or frequency

(iii) Audio signal generator, loudspeaker, cathode ray oscilloscope (CRO), microphone, meter rule /

measuring tape 1

(iv)

1v)

The microphone is placed at a distance, d = 20.0 cm from the loudspeaker

1

The amplitude, a, of the trace on the screen of the is measured. 1

The procedure was repeated for the values of distance, d = 30.0 cm, 40.0 cm, 50.0 cm and

60.0 cm 1

(vi)

1

30

(vii) A graph of a against d is drawn 1

Question 10 [Waves]

(a) The distance between 2 loud speaker affect the distance between 2 loud or soft sound 1

(b) When the distance between two coherent sources of sound is increase, the distance between two

consecutive constructive or destructive interference is decrease. 1

(c) (i) To investigate the relationship between two coherent sources and the distance between two

consecutive constructive and destructive interference.

(ii) Manipulated : Distance between two coherent sources, a

Responding : Distance between two consecutive constructive or destructive Interference 1

Constant : Distance between the source and the screen. 1

(iii) Loud speaker, audio signal/frequency generator, connection wire, power supply, measuring

tape. 1

(iv)

1

(v) By using a metre rule the distance between the listener from the loudspeaker is measured= D

The audio-frequency generator is switched on.

Use a distance between two loud speaker, a= 1.0m. 1

The listener is requested to walk in a straight path from left to right.

The distance between two successive loud regions is measured by a metre rule = x 1

The experiment is repeated using a distance between two loud speaker a=1.5m, 2.0m,

2.5m and 3.0m. 1

(vi) Tabulate the data 1

(vii) 1

perfect score sbp fizik spm 2011 answer

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