The thickness of lens in diagram 2.1 is compare to lens diagram 2.2

The thickness of lens in diagram 2.1 is compare to lens diagram 2.2

Diagram 2.1 Diagram 2.2

The focal length in diagram 2.1 is compare to diagram 2.2

The size of image produced in diagram 2.1 is compare to diagram 2.2.

thicker

shorter

smaller

• Thickness of the spring wire

• Maximum height reached by the ball

• Relate :• Thickness and maximum height• Thickness and elastic potential energy

Compare depth of sinking :

Compare load :

Area on contact :

Relate depth of sinking and pressure

Relationship Pressure and area

Depth of sinking on diagram 3.1 is deeper than diagram 3.2

The load of diagram 3.1 is equal to diagram 3.2

The area of contact in diagram 3.1 is smaller than Diagram 3.2

The deeper the depth, the higher the pressure

When the area increaes,the pressure decreases

Explaination question

3 or 4 marks

Submarine on surface submerge ( 3 marks)

Ballast tank

Fill with water

Weight increases

Weight = buoyant force

Partly filled

Weight > buoyant force

Submarine submerge

Paper burns, convex lens, sun rayThe parallel rays of the sun will pass through the a convex lens

2. After entering the lens, the light rays is focused at the principal focus of the lens

3. At the principal focus, the light ray is focused on one small area

4 Heat increases, paper burn

Thermal equilibrium

• Heat flow• Hot to cold• Thermal equilibrium achieved.• Temperature of water = temp of thermometer• No more heat flow

High pitch , only C , low pitch until AHigh pitch high frekuensi

Low pitch low frekuensi long wave length

Short wave length will diffract until C only

Specific latent heat of vaporization of water is absorbed

Air movement velocity increases

Evaporation rate increases

Sweat cold rotating fan Sweat is being evaporated

Copper block volume bigBowl copper volume small

Small volume small Uptrust small

Block sink because weight > uptrust

Sheet float because weight = uptrust

• Diagram below shows a sailboat.

You are required to give some suggestions to design a sailboat which can travel faster. Using the knowledge on motion, forces and the properties of materials, explain the suggestions based on the following aspects:

• the surface of the board• the shape of the board• material used for the sail• the size of the sail

smooth surface / coat with wax

streamline shape

low density material

Wide sizewater-proof material

• Type of material used as the cap of the thermal flask• Type of inner wall and outer surface• The density of material used• Thermal strength of the flaskModification Reason

Plastic stopper Reduce heat loss through conduction

Vacum space Reduce heat transfer

•Comfort of the person who will use the lamp.

Reduce the heat from the desk lamp

Design of the lamp desk

Safety features of the lamp

Energy efficiency of the lamp

Use flourescent lamp

Bigger cover with white colour

Adjustable stand

Energy saver lamp

Connect earth wire

• Capacity• Safety features

• Power

• Stability and other relevant aspects•

Section B

variables

• Manipulated : depth of water ( real depth)• Responding : position of image (apparent

depth)

• Constant : density of water

• Inference :

Depth of water affect the position of image.

Hypothesis : when real depth increases, the apparent depth increases.

Aim :To investigate the relationship

between depth of water ( real depth ) and position of image ( apparent depth)

Apparatus and material :

tall beaker, meter rule, pins, cork, water, retort stand

procedure

a) Mention manipulated quantity :

Real Depth : start experiment with what depth?? Fill the beaker with water to a height of 20 cm

• b) Method of measuring responding variable : apparent depth

With meter rule, measure the apparent depth.

Repeat experiment : Repeat experiment with different depth such as : 30 cm, 40 cm, 50 cm and 60 cm.

Tabulate dataReal depth / D ( cm) Apparent depth / d (cm)

20

30405060

Analyse dataApparent depth / d ( cm )

Real depth / D (cm)

variable• Manipulated : mass

Responding :Period oscillationConstant : Number of oscillation

inference

• ____________________ affect

_______________________

massPeriod oscillation

hypothesis

When increases, increases

mass

period of oscillation

aimTo investigate the relationship between

and

massperiod of oscillation

Apparatus and arrangement

• Hacksaw blade, stop watch, plasticine and G clamp.

Method of controlling manipulated variable

• Manipulate variable : masscontrol : the first mass

Plasticine with a mass of 50 g is clamped.

Method of measuring responding variable

• Responding variable : period

• Measure pressure using : stopwatch

Using stop watch , measure the time taken for 20 complete oscillation, calculate the period and record the data

Analyse data

Mass / g Period / s203040

5060

Graph

Mass / g

Period / s

variable• Manipulated : force

Responding :Extension of thespring

Constant : Diameter of the spring

inference

• ____________________ affect

_______________________

forceextension of the spring

hypothesis

When increases, increases

force

extension of the spring

aimTo investigate the relationship between

and

forceextension of the spring

Apparatus and arrangement

• Spring, slotted weight, retort stand, meter rule

Method of controlling manipulated variable

• Manipulate variable : forcecontrol : the first mass of slotted weight

• Slotted weight of 50 g is attached to the spring.

Method of measuring responding variable

• Responding variable : extension of the spring

• Measure pressure using : meter rule

Using meter rule, measure the length of the spring .

Analyse data

203040

5060

Mass ( g) Force ( N) Length of the spring ( cm)

Extension of the spring ( cm)

50 0.5100 1.0150 1.5200 2.0250 2.5

Graph

Force / N

Extension of the spring / cm

Paper 3 Bahagian A

SECTION A : Question 1

1(a)(i) Manipulated variable : ____________________

1(a)(ii) Responding variable : ____________________

1(a)(iii) Constant variable : ____________________

EXAMPLE:

A student carries out an experiment to investigate the relationship between temperature, , and the volume, V, of trapped air. A beaker is filled with cold water until the air column in the capillary tube is totally immersed. A thermometer is put into the water to determine the temperature of the water. The arrangement of the apparatus for the experiment is shown in Figure 11.1.

Temperature,

Volume, V




1(a)(iii) Constant variable : ____________________

1

2

0

l1

Figure 11.2 : Temperature, = 30 C

length, l = ………… cm

m.vr.v

Temperature,

Length, l




1(a)(iii) Constant variable : ____________________ (depends on the experiment)

MISTAKE

S!!!

Which of the following is possibly a constant variable?

A) Mass B) Thermometer C) Air

A) Water B) Stopwatch C) Time

A) Length B) Air column C) Ruler


1(b) Based on Figures 11.2, 11.3, 11.4, 11.5, and 11.6, determine l when is equal to 30 C, 35 C, 40 C, 45 C and 50 C.

Tabulate your results for l and V for each value of in the space below.

/ C l / cm V / cm3

30

3540

45

50

SECTION A : Question 1 Spot the errors!

t1 t2 tmean T T2 l

17 17.6 17.3 0.87 0.76 20

22.4 22 22.2 1.11 1.23 30

25 24.6 24.8 1.24 1.54 40

28 28.8 28.4 1.42 2.02 50

30.8 31 30.9 1.55 2.4 60

No unit!

Not consistence!

SECTION A : Question 1 Spot the errors!Unit at the wrong place!

l t1 t2 tmean T T2

20.0 cm 17.0 s 17.6 s 17.3 s 0.87 s 0.76 s2

30.0 cm 22.4 s 22.0 s 22.2 s 1.11 s 1.23 s2

40.0 cm 25.0 s 24.6 s 24.8 s 1.24 s 1.54 s2

50.0 cm 28.0 s 28.8 s 28.4 s 1.42 s 2.02 s2

60.0 cm 30.8 s 31.0 s 30.9 s 1.55 s 2.40 s2

SECTION A : Question 1 Spot the errors!

Wrong unit!

Not consistence!

l/ cm t1 / s t2 / s tmean / s T / s T2 / s

20.0 17.0 17.6 17.3 0.865 0.748

30.0 22.4 22.0 22.2 1.11 1.232

40.0 25.0 24.6 24.8 1.24 1.537

50.0 28.0 28.8 28.4 1.42 2.016

60.0 30.8 31.0 30.9 1.545 2.387

SECTION A : Question 1 Spot the errors!No error!

l/ cm t1 / s t2 / s tmean / s T / s T2 / s2

20.0 17.0 17.6 17.3 0.87 0.76

30.0 22.4 22.0 22.2 1.11 1.23

40.0 25.0 24.6 24.8 1.24 1.54

50.0 28.0 28.8 28.4 1.42 2.02

60.0 30.8 31.0 30.9 1.55 2.40

GOOD ANSWER!


What is a GOOD graph?

1) Starting at origin (0,0)

2) Write quantities and units on x-axis and y-axis

3) Has uniform scale

4) All points are transferred correctly

5) Draw one straight line and intersect any of the axes

6) Distribute other points equally

• Plotting the graph: on the graph paper, plot graph T2 againts l Remember to write the quantity & units

T2 / s2

l / cm

Remember to choose a good scale

Carefully transfer all points

..

..

.Draw the best line


BAD GRAPH!


Accepted graph


Rejected graph

more than 1 cm


Rejected graph

more than 0.5 cm


Conclusion

a

b

A) a is inversely proportional to b

B) a is directly proportional to b

C) a is linearly increasing with b

a

b





Conclusion





B) a is directly proportional to 1/b

C) a is linearly increasing with 1/b

a

ba

1/b

SECTION A : Question 22(a)(i) Relationship




A) a is inversely proportional to b B) a is directly proportional to 1/b C) a is linearly increasing with 1/b

a

ba

1/b


2(a)(i) Relationship

a

b




a

b





2(a)(ii) Determine Corresponding Value : determine the value of P when h = 3.0 cm

h / cm

P / Nm-2

1.0 2.0 3.0

1.0

2.0

3.0

4.0 P = 4.0 Nm-2


2(a)(ii) Determine Corresponding Value : determine the value of m when is 4.0

1/m / g-1

/ C

0.1 0.2 0.3

1.0

2.0

3.0

4.0 1/m = 0.3 g-1

m = 3.3 g


2(b)(i) Determine Gradient

1/m / g-1

/ C

0.1 0.2 0.3

1.0

2.0

3.0

4.0

Gradient, l = 4.0 C 0.4 g-1

l = 10.0 C g

y

x

Triangle must at least 4 x 5 Checked substitution

Remember unit

Gradient must be in decimalnumber. No fraction.

0.4

Heat produce from an experiment can be determine by using the fomula

Q = ml,Where l is the gradient and m is the mass.Calculate the value of heat when mass is 2.0 kg.

2(b)(ii) Calculation problem (involving the gradient)

SECTION A : Question 22(c) Calculation problem

Reminder!1) Check for the conversion of

units2) Substitute figures at the correct places

3) Write the final answer in decimal number (Do not give the answer in fraction!!!)

4) Do not forget the UNIT

Q = ml

Gradient : l = 10.0 J g-1 Mass = 2 kg

Q = 2 x 10Q = 2 000 x 10

Q = 2 x 104 J = 20 kJ

2000 g

10 000 J kg-1

SECTION A : Question 22(d) Precaution

Reminder!1) It is NOT SAFETY PRECAUTION!!!

Take few readings and find the average

Position of eyes is perpendicular to the scale

method

reason

to reduce random error

to avoid parallax error

Chapter by chapter

2.1 Analysing LinearMotion : distance, displacement, speed, velocity, acceleration

2.2 Analysing MotionGraphs : interpret and analyse graph

2.3 Understanding Inertia : explain what inertia is.relate mass to inertia.give examples of situations involving inertia.suggest ways to reduce the negative effectsof inertia

• 2.4 Analysing Momentum : • define the momentum of an object.• define momentum, p as the product of mass,

m and velocity, v, i.e. p = mv.• state the principle of conservation of

momentum.• describe applications of conservation of

momentum.• solve problems involving momentum

Understanding the effects of a Force

• determine the relationship between force,mass and acceleration, i.e. F = ma.

• solve problems using F = ma.

• 2.4 Analysing Momentum• define momentum, p as the product of mass,

m and velocity, v, i.e. p = mv.• state the principle of conservation of

momentum.• describe applications of conservation of

momentum.

• describe the effects of balanced forces acting on an object.

• describe the effects of unbalanced forces acting on an object.

• determine the relationship between force, mass and acceleration, i.e. F = ma.

Understanding the effects of a Force

Analysing Impulse and Impulsive Force

• explain the effect of increasing or decreasing time of impact on the magnitude of the impulsive force.

• describe situation where an impulsive force needs to be reduced and suggest ways to reduce it.

• describe situations where an impulsive force is beneficial

Understanding Work, Energy, Power and Efficiency

• state that when work is done energy is transferred from one object to another.

• define kinetic energy • define gravitational potential energy • state the principle of conservation of energy.

Understanding Elasticity

Paper 3

Section B

Answer for question 1

• Manipulated variable : mass ( of the slotted weight)

• Responding variable : volume of the slotted weightConstant variable :density

• V0 = 27 cm3

• m = 50.0 kg, V1 = 33 cm3, V = 6 cm3

• m = 100.0 kg V1 = 39 cm3, V = 12 cm3

• m = 150.0 kg V1 = 45 cm3, V = 18 cm3

• m = 200.0 kg V1 = 51 cm3, V = 24 cm3

• m = 250.0 kg V1 = 57 cm3, V = 30 cm3

Mass / g Volume of water / cm3

Volume of slotted weight/ cm3

50.0 33.0 6.0

100.0 39.0 12.0

150.0 45.0 18.0

200.0 51.0 24.0

250.0 57.0 30.0

Answer for question 2 (a)

a)(i) a 1/x

(ii) a = 12, 1/x = 0.5. x = 2.0 cm

(iii) Gradient , m = 24 cm3

• 2(b) λ= m / l = 24 / 20.0 = 1.2 cm

2(c) v = λf = 1.2 x 12

= 14.4 cm s-1

2(d) The position of eye must perpendicular to the reading scale to avoid parallax error

Paper 3 Bahagian A

Graph

Force / N

Extension of the spring / cm

Documents

The thickness of lens in diagram 2.1 is compare to lens diagram 2.2