SCHEME ANSWER-journey to success FINAL F4-SEM 2-2019 · Enjoy Learning Physics With Tcer Alina _ 3 Ø TRIAL TERENGGANU 2013 NO ANSWER MARK 3. (a) The ability to do work 1 (b) (i)

Dream big…aim high…never give up… physics is awesome!!

Answer Topic:

2.11 – 2.12 FORCES & PRESSURE

LIGHT WAVES

EnjoyLearningPhysicsWithTcerAlina_2

CHAPTER: FORCE & MOTION

Ø TRIAL PAHANG 2014

NO. ANSWER MARK 1. (a) The property (ability) of an object to return back its original shape when the

applied force is removed 1

(b) Shock absorber 1 (c) (i) Extension of the spring is directly proportional to the applied force when the

elastic limit is not exceeded 1

(ii)

2

TOTAL 5 MARKS

Ø TRIAL KEDAH 2013 NO ANSWER MARK

2. (a) Elastic Potential Energy 1 (b) Work done

1 1 (awu)

(c) Elastic Potential energy → kinetic energy 1 (d)

1

1 (awu) TOTAL 6 MARKS

1W = 2 elastic cord ( F x)2

1 = 2 ( x 6 x 0.2 )2

= 1.2 J

1 1or W = Fx = (12x0.2) =1.2 J2 2

2

2

-1

1 1 Fx = mv2 2

11.2 = (0.012) v2

v = 14.14 m s


Ø TRIAL TERENGGANU 2013 NO ANSWER MARK

3. (a) The ability to do work 1 (b) (i) Spring N is thicker than M 1 (ii) The distance bounced of the balls in Diagram 3.2 more than in

Diagram 3.1 1

(iii) The elastic potential energy stored by spring N is more than M. 1 (c) (i) Elastic potential energy increase, the distance bounced of the

balls increase 1

(ii) The thickness of material of a spring increase, the elastic potential energy increase

1

(d) The thickness of material of a spring increase, the elastic potential energy stored increase

1

(e) Elastic potential energy changes to kinetic energy 1 TOTAL 8 MARKS

Ø TRIAL KELANTAN 2014 NO ANSWER MARK

4. (a) A product of applied force and displacement in the direction of the applied force

1

(b) (i) Work = F x s = 100 x 12. = 1200 J

1 1 (awu)

(ii) Kinetic energy 1 (iii) Work done = Kinetic energy

1200 = ½ mv2= ½ x 3 x v2 v = 28.28 ms-1

1

1 (awu) (c) (i) Further // longer // more 1 (ii) Increase the force 1 (iii) small angle 1 (iv) Increase the forward velocity 1

TOTAL 10 MARKS

Ø TRIAL SBP 2013 NO. ANSWER MARK / NOTE

5. (a) The product of the applied force and the displacement in the direction of the applied force

1

(b) (i) W = Fs = 110 (0.5) = 55 J // Nm

1

1 (awu) (ii) F = ma

110 - 100 = 10a a = 1 ms-2

1



Ø TRIAL SBP 2015 NO. ANSWER MARK

6. (a) (i) Elasticity is the property of an object to return to its original length // shape after force exerted is removed

1

(ii) The spring is permanently deformed/damage // It has reached its elastic limit // Beyond the elastic limit, Hooke’s Law is no longer applied

1

(b) (i) Extension, x = 5 cm 1(awu) (ii) Upper spring, 100 g " x = 5 m

Two lower parallel springs, 100 g " x =2.5 m Total extension = 5 + 2.5 = 7.5 cm Total length, y = 10+10+5+2.5 = 27.5 cm

1 1 (awu)

TOTAL 5 MARKS

Ø TRIAL PERAK 2015

NO. ANSWER MARK 7. (a) (i) A product of force and displacement in the direction of the

applied force 1

(ii) E = m g h = 2 x 7 x 10 = 140 J

1

1(awu) (iii) Principle of Conservation of Energy

Gravitational Potential Energy " Kinetic Energy 1 1

(b) Ep = Ek 140 = ½ x 2 x v2 v = 11.83 m s-1

1

1(awu) TOTAL 7 MARKS


CHAPTER: FORCES & PRESSURE

Ø TRIAL KELANTAN 2014 NO ANSWER MARK

1. (a) Force per unit area // force acting perpendicular a unit area 1 (b) Density

depth // gravitational acceleration 1 1

(c) P = ρgh = 900 x 1.2 x 10 = 1.08 x 104 Pa

1 1 (awu)

(d) Increase / further 1 TOTAL 6 MARKS

Ø TRIAL SELANGOR 2010

NO ANSWER MARK 2. (a) To measure pressure 1 (b) (i) h1 more than h2 1 (ii) The difference in height of the water level in the manometer in

Diagram 6.1 more than in Diagram 6.2 1

(iii) liquid pressure 1 (iv) Depth increase, the difference in height of the water level in the

manometer increase 1

(v) Depth increase, liquid pressure increase 1 (c) • The difference in height of the water level in the manometer

increase • Density increase, pressure increase

1

1 TOTAL 8 MARKS

Ø TRIAL SBP 2014 NO ANSWER MARK

3. (a) Barometer mercury 1 (b) (i) 76 cm Hg 1 (ii) P = ρgh

= 1.36 X 104 X 0.76 X 10 = 1.034 X 105 Pa

1

1 (awu) (c) h decrease 1

TOTAL 5 MARKS


Ø TRIAL TERENGGANU 2012

NO ANSWER MARK 4. (a) Gravitational force acting on an object 1 (b) Weight = up thrust

Net force is zero 1 1

(c) Fb – W = ma 3200 – 2000 = 200 a 1200 = 200 a a = 6 ms-2

1


Ø Edited: 2016

NO ANSWER MARK 5. (a) Pascal’s principle 1 (b) • When the small piston is pressed down, the pressure is

exerted on the liquid and transmits uniformly to the large piston.

• The force is produced and pushes the chair up.

1

1 (c) Some of the force is used to compress air bubbles 1

TOTAL 4 MARKS

Ø TRIAL PAHANG 2016 NO ANSWER MARK

6. (a) Rate of change of distance 1 (b) (i) Diameter of horizontal tube at A > B 1 (ii) Speed of air at A < B 1 (iii) Air pressure at A > B 1 (c) (i) The higher the speed of air, the lower the pressure 1 (ii) The higher the speed of air, the higher the level of water in vertical

tube 1

(d) Bernoulli’s Principle 1 (e) Bunsen burner 1

TOTAL 8 MARKS


Ø TRIAL PAHANG SET B 2015 NO. ANSWER MARK

7. (a) (i) Force per unit area // force acting perpendicular a unit area 1 (ii) Pressure exerted in Diagram 4.1 > in Diagram 4.2 1 (iii) Cross sectional area increase, pressure decrease 1 (b) (i) 0.45 × 4 = 1.8 m2 1(awu) (ii)

1

1(awu) (c) Higher pressure exerted in a small cross sectional area

(car will sink) 1

TOTAL 7 MARKS

Ø EDITED 2017

NO. ANSWER MARK / NOTE 8. (a) (i) Pascal’s Principle 1 (ii) Transmission of pressure is slower // less effective // the force is

used to compress the bubble 1

(b) (i)

1

1 (awu)

(ii)

1 1 (awu)

TOTAL 6 MARKS

Ø TRIAL PERAK 2015 NO. ANSWER MARK

9. (a) (i) Q 1 (ii) high speed 1 (b) smaller nozzle

increase speed, reduce pressure 1 1

(c) Bernoulli’s principle 1 TOTAL 5 MARKS

F 76230P = =A 1.8

= 42350 Pa

-4

4

F 15P = = A 5.2 x 10

P = 2.88 x 10 Pa4 -4F =PA = 2.88 x 10 (6.2 x 10 )

F = 17.856 N


CHAPTER: LIGHT

NO. ANSWER MARK 1. (a) The bending of light when it enters medium where it’s speed is

different // The bending of light when it pass through the different medium or density

1

(b) (i) Angle of refraction in Diagram 1.1 is higher than angle of refraction in Diagram 1.2

1

(ii) Density in Diagram 1.2 is higher than density in Diagram 1.1 1 (ii) The higher the density, the lower the angle of refraction 1 (c)

1

1 (d) Decrease 1

TOTAL 6 MARKS

Ø TRIAL TERENGGANU 2013

NO. ANSWER MARK / NOTE 2. (a) Concave mirror 1 (b) Reflection 1 (c) Virtual, Upright, Magnified 1 (d) Same Size 1

TOTAL 4 MARKS

Ø AKADEMI JPNT 2014

NO. ANSWER MARK / NOTE 3. (a) Convex mirror 1 (b) Wider field of view // wider vision 1 (c)

2

TOTAL 4 MARKS

siniη =sinrsin30 =sin19

=1.54


Ø TRIAL SBP 2016

NO. ANSWER MARK / NOTE 4. (a) Reflection 1 (b) at P and is virtual 1 (c) Laterally inverted

Same size Upright

Max. 1 (d) 8.00 1

TOTAL 4 MARKS

Ø EDITED 2017

NO. ANSWER MARK 5. (a) Angle of incidence in denser medium when refracted angle is 90o 1 (b) (i)

2

(ii) Refraction 1 (c) (i)

1

1

(ii) Distance from the base of the beaker decrease Apparent depth increase

1 1

TOTAL 8 MARKS

H 20η = =h 12

η =1.67


Ø EDITED 2018

NO. ANSWER MARK 6. (a) (i)

(ii)

\

1

1

(b) (i) Reflection 1 (ii) All the light ray not totally reflected //

To prevent overlapping image 1

TOTAL 4 MARKS


CHAPTER: WAVES

NO. ANSWER MARK 1. (a) (i) Refraction 1 (ii) • Wavelength decrease

• Velocity decrease 1 1

(b)

2

TOTAL 5 MARKS

NO. ANSWER MARK 2. (a) Reflection 1 (b) High penetrating power // can be transmitted as narrow beam //

less diffract 1

(c) (i) t = 6 div x 2s/div

t = 12 s

1 (awu) (ii) d = vt/2

d = (1500) (12) / (2) d = 9000 m

1

1 (awu) (d) Ultrasonic imaging can be used to detect cracks and flaws inside

a block of metal using echo sounding method // Ultrasonic imaging is used to evaluate the structural aspects of the organs inside the body as well as the foetus of a pregnant mother // Ultrasonic ruler in ships uses ultrasonic echoes to measure distance. // Ultrasonic can be used to remove kidney stone // Ultrasonic can be used to clean jewellery

1 (max. 1)

TOTAL 6 MARKS


NO. ANSWER MARK 3. (a) (i) Distance between two successive crest or trough 1 (ii)

1

(b) (i) distance AB = 3/2 λ 3/2 λ = 21 m λ = 14 m

1

1 (awu) (ii) frequency, f =

= 2.5 Hz

1

1 (awu) (c) Vertical distance A to B = 2 amplitude = 1.0 m

amplitude = 0.5 m

1(awu) TOTAL 7 MARKS

NO. ANSWER MARK

4. (a) One color // One wavelength // lambda // frequency 1 (b) (i) a1 is shorter than a2 // a1 < a2 1 (ii) x1 is longer (than x2) // x1> x2 1 (iii) When a increase, x decrease 1 (iv) Wavelength λ // distance between double slit to screen, D 1 (c) Interference 1 (d) Constructive interference correspond to bright fringes

Destructive interference correspond to dark fringes 1 1

TOTAL 8 MARKS

2510


ESSAY SECTION B

NO. ANSWER MARK 1. (a) Total internal reflection 1 (b) (i) Density of Y is higher than X 1 (ii) Angle of ϴ and α are equal 1 (iii) • Light travels from denser glass into a less dense outer cladding

• The angle incidence is greater than critical angle • Angle of incidence equal to angle reflection

1 1

(max. 2) (c) Modification

Modification Explanation

Type of glass Fibre glass

Strong and not easily breake

Refractive index inner core more outer core

To produce total internal reflection when light travel inside the optical fibre

Density Low // small

Easy to handle // carry

Purity High

Signal can travel a long distance without losing information

Strength and Flexibility High

Not easily break and easy to bend at the curve part

10

(d) (i) 42o 1 (ii)

1

1 (iii)

1

(iv) Total internal reflection occur // The angle incidence is greater than critical angle

1

TOTAL 20 MARKS

o

1η =sinc

1 =sin42

= 1.49


NO. ANSWER MARK 2. (a) (i) To produce coherent sound 1 (ii) • The wavelength of the sound waves in Diagram 1.1 is greater than

that in Diagram 1.2. • The distance between two consecutive loud sounds in Diagram 1.1

is greater than that in Diagram 1.2 • The distance between two consecutive loud sound increases as the

wavelength of the waves increases • Interference of sound waves

1

1

1

1 (b) (i)

1

(ii) • During night time, the layer of air near to the ground is colder than the air layer above it. Cold air is denser than the warm air

• Hence, the sound waves propagate towards the normal as it travels downwards .Refraction occurs and the sound waves from the woman propagate towards her son

• During day time, the ground is heated faster. The layer of air near to the ground is warmer than the air layer above it.

• Hence, the sound waves propagate away from the normal as it travels downwards and further away from the son

1

1

1

1


(c) Modification

Modification

Explanation

Parabola

It acts like a concave reflector to reflect the signals

Focal length of the dish

So that all the signals converge at the focal point

Shiny surface So that all the signal are reflected effectively To face the direction of the wave signals from the station

So that the dish can receive the wave signals of the programs

Large diameter So that it is easy to receive the wave signals

10 TOTAL 20 MARKS


ESSAY SECTION C

NO ANSWER MARK 1. (a) When an object is partially or fully immersed in a fluid, the buoyant force

on the object is equal to the weight of fluid displaced by the object. 1

(b) • B is denser than A. • The weight of water displaced is the same of the weight of the rod • Weight of B is greater than weight of A • B will displace more volume of water

1 1

1 1

(c) Decision making

Characteristic Reason Material made from glass Glass does not corrode with acid Small diameter of capillary tube

To increase the sensitivity of the hydrometer

High density of shots Makes the hydrometer stays upright

Big diameter of bottom bulb To obtained a bigger up thrust. Choose N Material made from glass

Small diameter of capillary tube High density of shots Big diameter of bottom bulb

10 (d) (i) Weight of boat = weight of water displaced

W = ρVg 2500 N = 1000 x V x 10 V = 0.25 m3

1 1 (awu)

(ii) Maximum weight = maximum water displaced = 1000 x 4 x 10 = 40000 N maximum mass = 4000 kg Maximum load = 4000 – 250 = 3750 kg

1

1 1 (awu)

TOTAL 20 MARKS


NO. ANSWER MARK 2. (a) Gravitational force acting on an object 1 (b) (i) 2 N 1 (awu) (ii) Buoyant force // up thrust ( acts on the metal block ) 1 (iii) V × 1000 × 10 = 2

V = 0.0002 m3 1

1 (awu) (c) (i) Buoyant force

Weight // Gravitational force 1 1

(ii) • Valve release air from ballast tank. • Sea water flooded ballast tank. • The weight of water displaced is smaller. • Buoyant force < Weight of the submarine

1 1 1 1

Max. 3 (d) Decision making

Characteristic Explanation

Oil Incompressible // No air bubble

Small master piston Produce big output force Big salve piston Produce big force Steel Strong //

Does not brake easily // Non corrosive // Prevent leakage // Does not rust easily

P is the most suitable Oil Small master piston Big salve piston Steel

10

TOTAL 20 MARKS


NO. ANSWER MARK 3. (a) Distance between optical center and focal point. 1 (b) (i)

1

(ii)

1

1 (awu) (iii)

1

1 (awu) (c) • Parallel light rays from the hot sun at infinity pass through a convex

lens // labeled diagram • Light rays are focused after passing through the lens // labelled

diagram • Light rays are converged onto a very small area called the focal

point of the lens • At the point, the intensity of light is great (and the light energy)

causes an increase in temperature. When the spot on the paper become hot enough, the paper start to burn // light energy change to heat energy.

1 1 1 1

vM =

u

60 =

20 = 31 1 1= +f u v

1 1 = +20 60

f = 15 cm1P =f100 =15

P = 6.67 D


(d) Decision making

Characteristic Explanation Power of eyepiece: Low power

Focal length is longer // eyepiece must be longer focal length than objective lens.

Power of objective lens: High power

Focal length is shorter // objective lens must be more powerful lens than eyepiece // objective lens must shorter focal length than eyepiece

Distance between lenses: > fo + fe

To produce bigger image from the eyepiece // to increase the magnification

Position of the specimen: fo < u < 2fo

To produce real, inverted and magnified image

S is the most suitable

Focal length of eyepiece is longer than objective lens, distance between lenses is greater than (fo + fe) and the position of the specimen is between fo and 2fo.

10

TOTAL 20 MARKS


NO. ANSWER MARK 4. (a) Spreading of waves after passing through a gap or an obstacle 1 (b) (i)

• Diagram (a) shows the diffracted waves curving at the edge only. • Diagram (b) shows the diffracted waves becoming circular waves. • Wavelength of the diffracted waves remains unchanged.

3

(ii) • Diagram 4.2 will show a bigger effect of diffraction. • The narrower the gap, the more obvious is the spreading of

waves.

1 1

(c) (i)

1

1 (awu)

(ii) Frequency will remain the same Wavelength will decrease Speed will decrease

1 1 1

(d) Decision making

Characteristic Explanation Ultrasonic Suitable to detect soft organ Longer wavelength is used

Cut down diffraction and increase reflection

Higher amplitude More energy can be reflected Reflection Use different intensity of

reflection to map out the shape of the organ.

N is the most suitable Ultrasonic Longer wavelength is used Higher amplitude Reflection

10 TOTAL 20 MARKS

8

-2

10

v 3.0 × 10freqeuncy = =λ 3.0 × 10

f = 1.0 x 10 Hz

Documents

SCHEME ANSWER-journey to success FINAL F4-SEM 2-2019 · Enjoy Learning Physics With Tcer Alina _ 3 Ø TRIAL TERENGGANU 2013 NO ANSWER MARK 3. (a) The ability to do work 1 (b) (i)