11 f5physics2-ans midyear tai

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Physics Name :««««««««««««««« Form 5 ( )Paper 2 May 2011

2 ½ Jam

SMJK CHAN WA, SEREMBAN,NEGERI SEMBILAN

2011 MID YEAR EXAMINATIONFORM 5

PHYSICSPAPER 2

Two hours and thirty minutes

(You are adviced to spend 90 minutes for Section A,

30 minutes for Section B and 30 minutes for Section C)

INFORMATION FOR CANDIDATES

For Examiner¶s Use

Section QuestionFull

MarksMarksGiven

A

1 4

2 8

3 8

4 7

5 8

6 9

7 7

8 9

B9 20

10 20

C11 20

12 20

Total

This question paper contains 13 printed pages

1. Write down your name and form in the space provided

2. T his question paper consists of three sections: Section A, Section B andSection C.

3. Answer all questions in Section A. Write your answers for Section A in

the spaces provided in the question paper.

4. Answer one question from Section B and one question from Section C.

Write your answers for Section B and Section C on your answer sheets.

5. S how your working, it may help you to get marks.

6. If you wish to change your answer, cross out the answer that you have

done. T hen write down the new answer.

7 . T he diagrams in the questions provided are not drawn to scale unless

stated.

8. A list of formulae is provided on page 2.

9. T he marks allocated for each question or part question are shown in

brackets.

10. You may use a non-programmable scientific calculator.



The following information may be useful. The symbols have their usual meaning.

1. a =t

uv

2. v2 = u

2 + 2as

3. s = ut +2

1at

2

4. Momentum = mv

5. F = ma

6. Kinetic energy =2

1mv

2

7. Gravitational potential energy = mgh

8. Elastic potential energy =2

1Fx

9. Power , P =ti e

Energy

10. =V

m

11. Pressure, p =h V g

12. Pressure, p =¡

¢

13. Heat, Q = mcU

14. Heat, Q = m

15. P 1V 1 = P 2V 2

16. 11

T

V

=22

T

V

16. 1

1

T

P =

2

2

T

P

17. T

PV = constant

18. n =

r

i

sin

sin

19. n =depthapparent

depthreal

20. vu f

111!

21. Linear magnification, m =u

v

22. P = f

1

23. v = f P

24. P = D

ax

25. Q = It

26. £

= V Q

27. V = IR

28. ¤

= V + Ir

29. Power , P = V I

30. g = 10 m s-2



Figure 1

Table 1

Figure 2

8 m

There is some string around the end of the rule and the rule has thickness

Scalar has magnitude only but vector has magnitude and direction

= 6 cos 300

= 5.2 N

= 6 sin 300

= 3.0 N

Section A [60 marks]

Answer all questions

1 A person winds some thread tightly 4 times round the length of a metre rule and cuts the ends off level with theleft-hand end of the rule, as shown in Figure 1.

(a) To the nearest meter , what is the length of the thread?

«««««««««««««««««««««««««««« ..««««««................... [2 marks]

(b) Is the actual length of thread slightly greater or slightly less than your answer to (a)?Tick one box and give your reason.

slightly greater slightly lessreason ....................................................................................................................... ...........................................

................................................... ............................. ............................................ ...................... .................[2 marks]

2 (a) Distinguish between scalar and vector quantities.

««««««««««««««««««««««««««««««««««««««««««

««««««««««««««««««««««««««««««««««««««««««««[1 mark ]

( b) Table 1 lists a number of physical quantities. Put a ( ¥ ) in the box next to the quantities that are vectors.

[1 mark ] (c) Figure 2 shows a force of 6.0 N acting at 300 to the horizontal.

Calculate the component of the force that acts

(i) horizontally, (ii) vertically

[2 marks]

Acceleration ¥ Momentum ¥

Density Power

energy velocity ¥



Figure 3

= 12 - 6 sin 300

= 9.0 N

T cos 300 = 9.0 NT = 10.4 N

2 (d) Figure 3 shows two strings supporting an object of weight 12 N. The tension in one of the string is 6.0 N.The tension in other string is T N.

(i) Calculate the magnitude of the vertical component of the tension T in order for the object to be inequilibrium.

T cos 30[2 marks]

(ii) Hence calculate the magnitude of T .

[2 marks]

3 Figure 4 shows a trolley of mass 0.80 kg, on a bench surface, connected to a mass M by a string. The mass M isreleased and the trolley moves along the surface. Figure 5 shows the variation of velocity v of the trolley withtime t for the motion from A to B.

Figure 4

Figure 5

v / m s -1

0

0.5

1.0

1.5

2.0

0 0.2 0.4 0.6 0.8 1.0t / sA

B



Figure 6

a =0.8

1.8= 2.25 m s-2

F = 0.8 x 2.25= 1.8 N

s = area under graph= 0.5 x 0.8 x 1.8

= 0.72 m

t =1.8

0.72-2= 0.71 s

F =3

360= 120 N

Vertically upwards

Use blocks of larger area

3 (a) (i) Calculate the acceleration of the trolley between A and B.

[2 marks] (ii) Calculate the resultant force acting on the trolley between A and B.

[2 marks]

(iii) Show that the distance from A to B is 0.72 m.

[2 marks]

( b) When the trolley reaches B the mass M has just reached the floor. Ignoring any resistive forces, calculate thetime it takes the trolley to travel from B to C.

[2 marks]

4 (a) A garden pot containing soil weighs a total of 360 N. The pot rests on three equally spaced blocks , so thatsurplus water can drain out of the holes in the base of the pot. The soil is uniformly distributed in the pot.The pot is shown in Figure 6.

(i) What is the force exerted by each block on the pot?

«««««««««««««««««««««««««««««««««« ............... [1 mark ]

(ii) State the direction of these forces.

..................................................... ....................................................................................................... [1 mark ]

(iii) The gardener finds that the blocks sink into the ground, but he must have the pot up on blocks to allow thedrainage. What can he do to reduce the sinking of the pot?

««««««««................................................................................................................ ......... ...............

««««««««............................................................. ........................................... ............................ .....[1 mark ]



Figure 7.2Figure 7.1

The base of dam in Figure 7.2 is broader

The pressure at point A is lower

Figure 7.2

The base of dam in Figure 7.2 can withstand stronger pressure

v =65

1800= 27.7 m s-1

PE = mgh = 900 x 550 = 495 kJ

W = Fs = 250 x 1800 = 450 kJ

K E = 495 ± 450 = 45 kJ

K E =2

1x m x v2 = 45 kJ

v = 31.62 m s-1

4 ( b) Figure 7.1 and Figure 7.2 show two dams with different shapes.

Based on Figure 7.1 and Figure 7.2(i) Compare the dams in Figure 7.1 and Figure 7.2.

«« ......................................................... ................................................ ................................................ ........

««. ....................................................................... ............................ ................................................. .............[1 mark ]

(ii) Compare the pressure at point A and point B in the lake.

««............................................. .......................... ........... ................................................................................[1 mark ]

(iii) Based on the answer in part (ii), which dam is stronger?

««...................................... ............................. ..............................................................................................[1 mark ] (iv) Explain the reasons for your answer part (iv).

««««««««««............................................................... ..............................................................

««««««««««.............................................................................................................................[1 mark ]

5 (a) In a downhill ski race the total distance between the start and the finish is 1800 m. The total vertical drop is550 m. The weight of a skier (including his equipments) is 900 N and the time of his descent is 65 s.

(i) Calculate the average speed of the skier for the race.

[1 mark ]

(ii) Calculate the loss of gravitational potential energy of the skier

[2 marks] ( b) The average resistive force acting against the skier is 250 N.

(i) Calculate the work done against the resistive force.

[2 marks] If the skier does no work , (ii) calculate his kinetic energy as he passes the finish,

[1 mark ] (iii) his speed as he passes the finish

[2 marks]



Figure 8

Total internal reflection

1. Light passes from denser medium to less dense medium //refractive index of core > r.i. of cladding

2. The incident angle is greater than critical angle

v =48.1

10x3

8

!n

c= 2.03 x 108 m s-1

t =810x03.2

900 = 4.43 x 10-6s

s = vt = 2.03 x 108 x 45 x 10-9

= 9.1 m

6 (a) Figure 8 shows a step index optic fiber with a ray of light striking the core/cladding boundary at angle of incidence U.

(i) State the name of the effect shown in Figure 8.

««.«««««««««««««««««««««««««««««««««««««««[1 mark ]

(ii) State the conditions necessary for the ray in Figure 8 to be completely reflected into the core.

.. «««««««««««««««««««««««««««««««««««««««««

««««««««««««««««««««««««««««««««««««««««««

.. «««««««««««««««««««««««««««««««««««««««««

[2 marks] ( b) The optic fiber is 900 m long and the core of the fiber has a refractive index of 1.48.

(i) Calculate the speed of light in the core.

[2 marks]

(ii) Calculate the minimum time taken for light to travel along the core of the fiber.

[2 marks] (iii) The ray following the path shown in Figure 8 takes approximately 45 ns longer than the minimum time to

travel along the fiber. Estimate, by calculation, the extra distance travelled by this way

[2 marks]



Figure 9

3.8 cm

- 3.4 cm

2.64 ± 2.68 ms

379 - 373 Hz

P =380

10x0.3 8

! f

v = 0.79 m

7 Figure 9 shows a displacement-time graph for a wave source.

(a) Use Figure 9 to determine for this wave source

(i) the amplitude

««««««««««««««««««««««««««««««««««««««««««[1 mark ]

(ii) the displacement when t = 1.80 ms

««««««««««««««««««««««««««««««««««««««««««[2 marks]

(iii) the period

««««««««««««««««««««««««««««««««««««««««««[1 mark ]

(iv) the frequency

««««««««««««««««««««««««««««««««««««««««««[1 mark ]

( b) The speed of the waves produced by this wave source is 3.0 x 102 m s-1. Calculate their wavelength.

[2 marks]



Figure 10

Interchange the connections to the ammeter

Current

Voltmeter

I =15

6 !

R

V = 0.4 A

0.4 A

8 Figure 10 shows an electric circuit.

(a) The lamp lights, but the ammeter needle moves the wrong way.What change should be made so that the ammeter works correctly?

««««««««««««««««««««««««««««««««««««««««««««

««««««««««««««««««««««««««««««««««««««««««««

[1 mark ] ( b) What does an ammeter measure?

««««««««««««««««««««««««««««««««««««««««««««[1 mark ]

(c) In the space below, draw a circuit diagram of the circuit in Figure 10, using correct circuit symbols.

[2 marks]

(d) (i) Name the instrument that would be needed to measure the potential difference ( p.d.) across the 15 resistor.

«««««««««««««««««««««««««««««««««««««««««««[1 mark ]

(ii) Using the correct symbol, add this instrument in your circuit diagram in (c), in a position to measure the p.d. across the 15 resistor.

[1 mark ] (e) The potential difference across the 15 resistor is 6 V.

Calculate the current in the resistor.

[2 marks] (f) Without any further calculation, state the value of the current in the lamp.

««««««««««««««««««««««««««««««««««««««««««««[1 mark ]

A V



Section B [ 20 marks ] Answer any one question.

9 Figure 11.1 shows an ice cube melting when heat is absorbed from the surrounding.Figure 11.2 shows a kettle of water boiling when heat is absorbed from the fire.

(a) What is meant by heat?[1 mark ]

( b) Using Figure 11.1 and Figure 11.2, compare the processes that take place in both situations and the typeof heat needed for both processes to occur.Relate the processes with the type of heat needed to deduce a relevant physics concept. Name the physics

concept involved. [5 marks]

(c) (i) Your body sweats when you are feeling hot. How does sweating helps to cool down your body?[ 2 marks ]

(ii) Explain why a scald from steam is more serious than the one from boiling water?[2 marks]

(d) Figure 11.3 shows a simple solar water±heating system. Energy from the Sun falls on the solar panel.Water is pumped around the system so that a store of hot water is made available in the tank.

Using suitable physics concepts, explain the required modification needed in designing an efficient solar water-heating system. The modification should include the following aspects:(i) pipes design(ii) material used(iii) heat absorption

[10 marks]

Figure 11.3

Figure 11.1

Ice cube

Figure 11.2

Kettle



Answer Mark 9 (a) Heat is the quantity of energy that is transformed from one hot object to a cold object. 1

( b) 1-Figure 11.1 shows a situation where a solid changes into liquid /Figure 11.2 shows a situation where a liquid change into gas

2-Both processes have a change in the state of matter.3-Both process require heat.4-The heat absorbed is not to raise the temperature, to overcome the force between the

molecule particles during the physical change.

5-Concept involved is Latent Heat

1

111

1(c)(i)

(ii)

1-Water evaporates from the skin when we sweat.2-for water to evaporate it use heat from the body

1-Steam condenses to form water so, 2-latent heat of vaporization of steam is let out more steam.

11

11

(d)Suggestion Explanation

Use insulator behind the absorber panel

To prevent the loss of heat energy

Use an absorber panel which is painted black.

A black surface is a good absorber of radiation so it will absorb heat faster

The pipe inside the plate must bemade of metal

Metal is a good heat conductor , so itwill transmit heat to water easily

Pipe embedded in plate must belong

Longer pipe will enlarge surface areawill absorbs heat faster

A storage tank must be place at ahigher level

To give higher pressure

Use glass cover on the top of the panel

To trap heat energy. (energy is radiatedin, but cannot radiate out again).

Any five suggestions and explanation [ 10 marks ]

2

2

2

2

2

Total 20



10 (a) Figure 12.1 shows water waves moving towards the shore.

(i) Name the wave phenomena in Figure 12.1.[1 mark ]

(ii) What happens to the waves as it approaches the headland? Give reasons.[3 marks]

( b)

Situation 1A student can hear the sound from the stopwatch at its loudest when cardboard tube B is at the position as shown inFigure 12.2

Situation 2

Figure 12.3 shows a man fishing. Eye A can see the man¶s image on the water¶s surface.

(i) Name the types of waves in Situation 1 and Situation 2.[2 marks]

(ii) Based on Figure 12.2 and 12.3, compare the directions of wave propagation[1 mark ]

(iii) Based on Figure 12.2 and 12.3, compare the angles of wave propagation and the normal.[1 mark ]

(iv) Based on Figure 12.2 and 12.3, relate the angles before and after the wave phenomena.[1 mark ]

(v) State the wave phenomena for Situation 1 and Situation 2.[1 mark ]

(c) To attract more tourist to the island in Figure 12.4, a contractor wants to build a beach resort. As a consultantyou are asked to give suggestions on the proposed project based on the following aspects:

(i) The location of the resort(ii) Features to reduce the erosion of the shore(iii) Features to enable children to enjoy swimming in calm water.

Figure 12.1

Cardboard tube A Cardboard tube B

Stopwatch Listener

Smooth wall

Soft wood

Figure 12.2

Eye A

Figure 12.3



10(a)

(i) Refraction of water waves 1

(ii) Waves bend 1

Ocean ± deep , shore ± shallow 1

Velocity of waves decreases 1

( b) (i) Sound and light waves 2

(ii) Changes at the boundary 1

(iii)

Angles and normal are on the same plane 1

(iv)

Angle of incidence = angle of reflection 1

(v) reflection 1

.(c) (i) Suggestion Reason

Build near bay Waves are calmer due to divergence of energyConvergence of waves at the capeThe bay is shallower .The speed of waves decreases.

The amplitude of waves at the bay is small.Build retainingwalls

Reduce direct impact of the waves on the shore. Toreflect the waves from the shore.Protect the area from large waves

Build concretestructures with agap in between atdesignated area for children

Waves passing through the gap will be diffracted in thechildren µs area .Smaller amplitude of the diffracted waves causes the seato be calmer there.Energy of waves decreases.

4

3

3

20

Section C [ 20 marks ] Answer any one question.

11

Cape Bay Chalet

Boats

etty

Figure 12.4

Figure 13



As a researcher in a boat manufacturing company, you are assigned to study metal characteristics used to makethe boat as in Figure 13.

(a) (i) State Archimedes P rinciple. [1 mark ]

( b) A wooden cube with sides of 30 cm floats in a tank of water with of its volume above the surface of the

water. [The density of water = 1000 kg m 3 .]

(i) Calculate the buoyant force acting on the block.

(ii) What is the density of the cube? [4 marks]

(c) You are given four choices of metals P, Q, R and S. The table below shows the characteristics for the four metals.

Metal Shape Density / kg m 3 Specific Heat Capacity Strength

P Streamlined 900 Low Low

Q Oval 452 High Low

R Circle 387 Low High

S Streamlined 500 High High

Based on Table 2, (i) Explain the suitable characteristics of the metal to be used as the material to make the boat.

[8 marks] (ii) Determine the most suitable metal to be used as the material to make the boat and give your reasons.

[2 marks]

11 (d) The forms of water transport are purposely designed so that the buoyant force produced is greater. Explain whythe captain of a ship should be careful not to overload the ship?

[2 marks] (e) Explain why a ship may sink when it sails from sea water to fresh water

[3 marks] 11 (a) (i) When an object floats in a liquid, the buoyant force equal to the weightof the fluid displaced

(ii) W = Vg = 1000 x 2/3 (0.3 x 0.3 x 0.3) x 10 = 180 N

m, cube= W/g = 180/10 = 18 kg

= m / V = 18 / (0.3 x 0.3 x 0.3) = 666.67 kg m-3

( b)

Characteristic Reason

1.Streamline 2. reduce the resistance of water 3. Density low 4. higher buoyant force5. Specific heat capacity high 6. absorbs heat slowly7. High strength 8. Difficult to damage

Table 2



Table 3

S, because its streamline, low density, high specific heat capacity and highstrength

(c) Buoyant force always greater than the load weight, so that the ship will notsink.

(d) 1. In the sea, buoyant force > weight of the ship2. When the ship in the river , density of water< density of sea, so

buoyant force decrease3. The weight of the the ship > the buoyant force, so the ship will sink

12 (a) What is meant by potential difference?[1 mark ]

( b) Explain why the bulb connected to two dry cells lights up brighter than one bulb connected to one drycell. Explain.

[4 marks] (c) Table 3 below shows the characteristics of four types of cables that have the same length. You are to choose

one of the cables to be used in the National Grid Network.

Cable Diameter/ cm Density/ kgm-3 Rate of expansion Melting point

W 2 4.50 x 103 Medium High

X 4 3.00 x 103 Low High

Y 3 5.45 x 103High Low

Z 1 2.50 x 103 Low Medium

Based on Table 3, explain the suitability of each of the characteristics of the cables. Choose the most suitablecable and justify your choice.

[10 marks]

(d) A power of 9.5 kW is transmitted from a small wind-powered generator to a village along 6000 m of cablesthat have a total resistance of 1 . The power is transmitted at the usual mains supply voltage of 240V.Calculate(i) the current in the cables

[1 mark ]

(ii) the power loss due to the heating of the cables[2 marks]

(iii) the percentage of loss of power [2 marks]

12. (a)

( b)

(c)

Potential difference is defined as work done in moving a unitelectric charge between two points.

The two dry cells are connected in parallelThe effective e.m.f. remains the sameThe effective internal resistance of the two cells is smaller A larger current will flow through the bulb to make it brighter

1. Diameter should be large2. to reduce the resistance of the cables3. Density should be low

1

1111



(d)(i)

(ii)

(iii)

4. the cables will be lighter and can be supported securely5. The rate of expansion should be low6. So there is less expansion and less sagging in the cables duringhot days.7 . The melting point should be high8. So the cables does not melting in high temperature.9. X is suitable

10. Diameter is large, density is low, the rate of expansion is lowand the melting point is high.

Current = P/V= 9500/240= 39.58A

Power loss , P = I2R

= 39.582 x1= 1566.6 W

Percentage of loss of power

= 1009500

6.1566 x

= 16.49 %

10

1

2

2

11 f5physics2-ans midyear tai

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