EXERCISE PAST YEAR QUESTIONS (MODULE PAST YEAR PG 23-28)

2.10 WORK, EFFICIENCY

1(a) Change of energy. Elastic potential energy to kinetic energy

(b) If bigger mass, the distance of movement shorter// nearer

(c)i. Elastic potential Energy =?

E=1/2Fx

=1/2( 10 N )( 0.2m )

=1.0J// 1 Nm

ii. Velocity,v=?

E.P.E = K.E (conservation energy happen)

1 J = ½ mv2

v2= (1x2)/0.02

v= 10ms-1

2.(a) Work is done when a force causes an object to move in the direction of the force

(b)i. Work done by pulling the rope down=?

W= Fs

=220x0.5

=110 J

ii.Work done on the load to raise=?

W=P.E= mgh

=20x10x0.5

=100J

(c)i.Compare the work done between i and ii. Work done of b(i) greater than b(ii)

ii.Why the work done is different? Has frictional force

3.(a)Transformation of energy 2.10.3. Chemical energy to kinetic energy

Transformation of energy 2.10.4 Electrical energy to kinetic

energy

(b)Justify your answer. Energy can be changed from one form to another// conservation of energy( keabadian tenaga)

(c)Energy can transform because work is done

(d)Law involved?Conservation of energy

i.Calculate distance AB.

Since sin 30 =opposite /hypotenuse

Sin 30 = 5/AB

AB= 10m.

ii.Calculate Work done

W= Fs

=(800)(AB)

=800x10

=8000J

3.1 PRESSURE

1(a) Soft bed compressed because

pressure involved //force exerted on it

(b)Definition of pressure is Force acting per unit area

(c)i.Weight of block,W=?

W= mg density, p=m/V

= (pV)g

=(133 kgm-3 ) (2mx2mx3m)(10ms-2 )

= 15960 N

ii.Pressure acting on soft bed?

P= F/A

= pVg/A

=15960/(2x2) ………………………1mark

=3990 Pa//Nm-2 ………………….1mark

Pg26

2.(a)Its easier to press thumbtack because surface area of thumbtack is lesser.

(b)Conclusion is if the area is smaller, pressure is greater

(c)Pressure exerted by the coin?

P=F/A

=5/3.14x10-4

=1.59 x10-4 Pa//Nm-2

(d)Bulldozer has broad tyres because when the broad tyre is used, area is higher, then the pressure decrease.(1M) This is to avoid bulldozer sink into the soil. (1M)

3.2 PRESSURE IN LIQUIDS

1.(a) Pressure is Force acting perpendicularly to unit area of surface.

(b)Water pressure=? (Use density of water=1000kgm-3 and g=10ms-2 )

P= pgh

=(1000)(10)(3)

=30kPa//30 000Nm-2

(c)Explain why has different between pressure

Because height from water level P is lesser than

height from water level Q, (1m)

therefore pressure P is lesser than pressure Q(1m)

2.(a)i.Compare diagram 3.2.2 and 3.2.3.

Wall of dam at bottom 3.2.3 is bigger than 3.2.2// area 3.2.3 >3.2.2

ii.Compare pressure A and B.

Pressure A is lesser than B

(b)i.Which dam is stronger.

Dam 3.2.3 is stronger

ii.Why? When the depth increase, pressure increase.

Thicker at the base can withstand high pressure

(c)i.Apparatus is siphon system

ii.Water flow from beaker to cylinder because

water level in cylinder is lesser than beaker//

pressure in beaker is higher cylinder

iii. mark**************higher level of water in cylinder is higher than before

3.(a)Pressure is force exerted per unit area

(b)i.Compare depth of water 3.2.6 and 3.2.7. Depth of 3.2.6 is higher than 3.2.7

ii.Compare distance travelled by water to spurt out

Distance in 3.2.6 is further than 3.2.7

(c)Relationship between pressure and depth

The higher the depth, the higher the pressure

PAST YEAR QUESTIONS

3.4 PASCAL PRINCIPLE (PG 28)

1.a) i. Pressure is the force acting normally on a unit surface area

ii.2 similarities are

-Surface area of piston E and G are same

-Pressure of E =Pressure of G

b) Principle applied in jack system is Pascal’s Principle

c) Given FE = 2N, PE = PF = 40Nm-2

Therefore, FH = ?

FH = PH X AH

= 40X 0.8X10-4

=3.2 X 10-3 N

2.a)i. Liquid is used in the system because

Liquid cannot be compressed

ii.Given p=800kgm-3 , g=10ms-2

Find liquid pressure, Pliquid at y =?

P=pgh

=800x10x0.5

=4000Pa

iii. Magnitude of pressure acting on piston?

4000Pa

iv. No change in pressure// pressure is constant (1m)

because density will make it balance PA = PB //Satisfy Pascal’s

principle. (1m)

3.a)i. Calculate pressure on input piston 3.4.5.

Pinput =?

Pinput = F input/ Ainput

= 500 / (10-2 m2)

=50 000 Pa

ii. Calculate force output for both hydraulic lifts

Foutput A = ?

Foutput B = ?

Fouput A = Poutput x Aoutput

= 50 000x 2000x 10-4

=10 000N **LOOK AT VALVE A IN Diagram 4

Fouput B= Poutput x Aoutput

= 50 000x 4000x 10-4

=20 000N **LOOK AT VALVE B IN diagram 5

iii. Based on your answer, which is more suitable to raise van 1800kg. Piston in diagram 3.4.4 or 3.4.5

Diagram 3.4.5 (Force is higher)

b)Vehicle can be lowered down by

-releasing valve B (1m)

4.a) Physics principle involve is Pascal’s Principle

b)Chair can be lifted up by…

-Small piston is pressed down, the pressure is exerted

-Liquid is transmit uniformly to large piston, then force is produced

And pushed the chair up

c) If air bubble present, system is less effective. Its because…

same force is used to compress air

d)Given Abig = 100cm2 , Asmall = 20cm2 , mchair =20kg.

Fsmall =? If the mchild to be lifted up =30kg

Pbig = Psmall

Fbig = Fsmall

Abig Asmall

Fsmall = (20+30)(10) x 20 ***Fbig include with mchair and mchild

100 use W =mg

= 100 N

e)Suggest modification so that chair is safe for adult

Piston size: Area of big piston has to be increased

Reason: Force will increase

Seat size: Enlarge the size of seat

Reason:To accommodate the large of mass of adult without damage

Pg31

3.5 ARCHIMEDES’ PRINCIPLE

1a) Density is mass per unit volume

b)i. Comparing density

Density of A is lesser than B

ii. Comparing weight

Weight of A is lesser than B

iii. Comparing weight of water displaced

Weight of water displaced by A is lesser than B

iv. Relate the weight and weight of water displaced

The larger the weight of sphere,

the larger the weight of water displaced

v. Relate weight of water displaced and uptrust(keapungan)

Weight of water displaced = Upthrust//bouyancy foce

c)Phy.principle- Archimedes’ Principle

d)application in submarine

2.a) Density is mass/volume

b)i. Compare level of boat in sea and water

Level of boat in sea is higher than river

ii.Compare volume of water displaced of them

Water displaced in sea is lesser

iii. Compare their density

Density of sea water is higher (has salt in it,so its denser)

c)i. Relate volume of water displaced with density of water

The higher the density, the lower the water displaced

ii. Deduce relationship between weight of boat and weight of water displaced

Weight of boat = weight of water displaced

d) Name phy.principle- Archimedes’ Principle

e)How a submarine can submerge

-Ballast tank is filled by sea water(1m)

-Weight is more than upthrust, it will submerge(1m)

3.i. One function of plimsoll line is for safety purpose

ii. Bouyant force=?

F= mg

= 7500x10

=7.5 x104 N

iii.Mark the level of water when enter the river mouth (tanda lebih rendah daripada semasa kapal di laut)

iv. Why is that

Density of sea water is higher.

When density of liquid is decreased,

volume of water displaced increase.

b) Relationship between weight of balloon and upthrust

Weight of balloon = upthrust

i)When the load is dropped, air balloon will move upward

ii)Reason

-The weight of air balloon is decreased// upthrust higher than weight

-The balloon experiences unbalanced force

Pg 34

3.5 BERNOULLI’S PRINCIPLE

1.a)P: lifting force// upthrust

Q:weight

b)Equation is. P=Q

c)Force caused by Bernoulli’s Princ.? P

2a)(i) Princp.? Bernoulli’s princp.

(ii) region with low P?Y

b)Reason?The air moves with high speed

c)How it spray the liquid?

-The atmospheric pressure is low at Y but at X is high. (1M)

-It push the liquid up through the narrow tube (1M)

3. Mark all 2 forces

b)Princp.? Bernoulli’s principle

c)Use the rubber ball,what will happen

-Weight of rubber ball is higher than upthrust(1M)

-The rubber ball will drop (1M)

4.a)State 2 similar characteristics

-When there is movement of air, paper and tarpaulin will lift up(1M)

-When there is no air movement, the paper and the tarpaulin

will not lift up (1M)

b)(i)Compare air pressure inside and outside the tarpaulin.

The air pressure inside tarpaulin is higher than outside the tarpaulin

(ii)Relate speed and pressure

When the speed is low, pressure is high

c)Principle = Bernoulli’s principle

CHAPTER 4: HEAT(pg 36)

4.1 Thermal Equilibrium

a) Thermal equilibrium is no net flow of heat between 2 objects

and they have same temperature

b) 45 0 C

c) How P achieve thermal equilibrium

-Heat from block P transfer to the water

-until the rate of heat transfer between Pand water become equal

d) Amount of heat lost by P when achieved thermal equilibrium

Q = mc θ

= 0.3X 900 X (100-45)

= 14850 J

4.2 Specific heat capacity /haba pendam tentu(pg37)

1. a) Specific heat capacity,c is

(the amount of heat that must be supplied to increase)( the temperature

by 1 0 C )(for a mass of 1 kg) of the substance. Unit Jkg -1 K -1

b)i)Energy output of the electric kettle in 3.5min

Energyoutput = Power x time

Eo = Pt

=(3 x103W) x (3.5x60s)

=6.3x 105 J

b)ii) Energy required to raise temperature of 1.7kg of water from 20 0 C to 1000 C

Q = mc θ

=1.7 x 4.2x103 x (100-20)

=5.71 x105 J

iii) Energy required to boil 0.23kg of water at 100 0 C (kekal suhu; water->steam)

Q = ml

= 0.23 x 2.3 x 106

= 5.29x 105 J

2.a) Thermal equilibrium

b)Specific heat capacity of water is higher

c)Kinetic energy of particle in the spoon increase

d)i) Specific heat capacity of Q1 , cspoon= ?

mc θspoon = mcwater θwater

(0.1)cspoon (75-30) = (0.1)(4200)(80-75)

cspoon =(0.1)(4200)(80-75)

(0.1)( 75-30)

Cspoon = 466.7 Jkg-1 C-10

ii)assumption (andaian)-No loss of heat to the surrounding (1m)

PG 38

4.3 SPECIFIC LATENT HEAT( haba pendam pelakuran)

(a) 300s

(b) Pt = ml

l =100 (1050- 300)

0.5

l = 150 000 Jkg-1

(c) Heat supplied is used to break up bonds between molecules.

Heat is used to increase kinetic energy

2.(a) Evaporation

(b) Heat absorbed or released by the substance when its state changes, without any change in temperature

(c) Ether released heat from surrounding causing it to freeze

(d) Q = ml

= 0.15 (3.3x105)

=4.95 x 104 J

(e)Working of an refrigerator // air conditioner

3. (a) L?Quantity of heat required to change a substance from solid to liquid

without any change in temperature

(b) (i) B

(ii) The ice is exposed to the surrounding and absorbed heat

(c) Heat released, Q = mcθ

= 0.2 x 4500 x (75.5- 25)

= 45450 J

(d)(i) ml + mcθ = Heat released

m( 3.36x105 + (4200x25)) = 45450

m(441000) = 45450

m= 0.103 kg

(ii) No heat loss from coffee to surrounding

4. (a) Electrical energy heat energy

(b) Q = mcθ Energy, E= Heat, Q

= (0.5)(4200)(100-30)

=147 000J// 147kJ

4.4. Gas Law(pg 42)

1(a) Bourdon Gauge

(b) Pressure

(c)(i) Increase

(ii)Pressure law

2.(a) Form of energy

(b)(i) The air trapped in diagram 4.4.3 higher than 4.4.2

(ii) The temperature of air trapped in 4.4.3 is higher than 4.4.2

(c) Pt = ml

l= 500W x 60s

0.0125kg

=2.4 x 106 Jkg-1

(d)

(i) Pt = mcθ

t = (3)(3.9)(170)

600

=331.5 s

(ii) Pt = mcθ

t = (2.5)(900)(170)

1200

= 318.75s

(iii) Pt = mcθ

t = (4)(400)(170)

900

=302.22s

(e)(i)S

(ii)Heat up faster (time ,t <)

(iii)Volume

(iv)When volume is higher, temperature is higher

(f)Charles’ law

3.(a) (i)Increase

(ii)Pressure

(b)Pressure law

(c)P1 = P2

T1 T2

200kPa = 230kPa

300K T2

T2 = 230 x 300

200

= 345 K // 720 C

(d)Surface area = wider

Reason = The pressure exerted on road is reduced

Track of tyre= thick

Reason = To give better grip to the road

CHAPTER 5: LIGHT

1. (a) Reflection

(b) 8m

(c)

(d)

Object Image

Incident ray

Normal line

Student Plane mirror Reflection ray

2(a) Reflection of light

(b)virtual

(c)

C F Object Image

u<f

Concave mirror

3.(a) Convex mirror

(b) convex mirror has a wide view than plane mirror

(c)(i) dotdotkan hijau ehhh

Object C F F C

Image

Convex mirror

(ii)1. Virtual 3. Diminished (smaller)

2.Upright

4.(a) convex mirror

(b)Virtual (maya) // Upright//Diminished (smaller)

(c) object

image sentuh C

C F F C

u<f Convex mirror

(d) To increase field of vision

5.(a)(i)

P Q

(ii)Reflection

(iii) The driver in P cannot see the car Q

(b)(i) convex mirror

(ii)

P Q

(iii)Convex mirror increase the field of view

(c)(i) unchanged

(ii)The characteristics of image of a convex mirror not depends on the focal length

5.2 REFRACTION OF LIGHT

1.a) Refractive index is an indication of light bending ability of the medium

b)critical angle, c?

n= 1/ sin c

sin c=1/n

c= sin-1 (1/1.33)

=48.8o

c)(i) R 400 S

in water

(ii) 400 new path

d)

in water

U towards normal T

Outside aquarium

2.a)(i) Pencil in diagram 5.2.4 bending more

ii) Density of sea water is higher

iii)The higher the density, the greater the bending

b) The size of pencil in water is greater

c)Refraction

d)velocity eye

e)

image

object

5.3 Total internal reflection (pantulan dalam penuh)

1.a)Close to the sky// cool air

b)refraction

c)i)Total internal reflection

ii)The image of sky

d)-Light from sky to the earth (refracted)

-The light reach at point-X, (total internal reflection) occur

e)optical fibre

2.a)(i) Total internal reflection

(ii)The refraction index of Y is higher than X

b)Critical angle, c in core Y

sin c = 1/n

= 1/2.1

= 0.4762

c = sin-1 (0.4762)

= 28.440

c)Prism periscope// camera//endoscope

3.a) Prism periscope

b)(i)

box A

ii)-The prism of other optical instrument P is placed

in b(i) because (1M)

light ray allow from object to be reflected 900 onto lower prism (1m)

out to eyes of observer

c)(i) path? light ray

eye

(ii)Upright

d)critical angle

sin c = 1/n

c= sin-1 (1/n)

=sin-1 (1/1.52)

=41.140

4. a) The angle of incidence when the angle of refraction is 900

b) Angle of incidence at point A is 0, So angle of refraction is 0

c) (i) Prism P; Increase the angle of incidences

Prism Q; decrease the angle of incidences

ii) The density of air is less than glass

iii) 5.3.5; It travels along the glass-air-boundary

5.3.6; It travels along glass-air-boundary

d)Phenomenon is total internal reflection

5.a) Critical angle is the angle of incidence when the angle of refraction is 900

b)(i) Reflection of light from car headlight back to driver

giving indication of its position

ii)Path?

c)Angle of incidence 0 //

incidence ray is parallel to normal

d)Reflection

e) total internal reflection

f) n= 1/sin c

=1/sin 38

=1.62

5.4 Lenses ms 54

1. (a)

(b) real (nyata) // inverted (terbalik) // diminished (kecil)

(c)(i)the image distance, v

1u+ 1v=1f

130

+ 1v= 110

V = +15cm (real)

(ii)linear magnification

M= vu

= 1530

= 0.5 // 1/2

F F 2F

I

(d)(i)objective lens : X TELESCOPE

Eyepiece : Y

(ii) reason

-focal length of eyepiece smaller than objective lens

(e)

2.(a) refraction

(b)the size of the image will increase //larger

(c)(i)convex lens

(ii)calculate image distance

f = +8 cm (positive because convex lense)

u = +6cm

v = ?

1u

+ 1v=1f

1V

=18−16

= −124

v = -24CM (VITUAL IMAGE)

(iii)

F1 F2

2FF O

image

2FF O