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HWA CHONG INSTITUTION

JC2 Preliminary Examinations

Higher 2

CANDIDATE NAME

CT GROUP 13S

CENTRE NUMBER

INDEX NUMBER

PHYSICS

Paper 1 Multiple Choice

Additional Materials: Optical Mark Sheet

9646/01

24 September 2014

1 hour 15 min

INSTRUCTIONS TO CANDIDATES

Write in soft pencil.

Write your full name, CT, NRIC or FIN number on the optical mark sheet (OMS). Shade your NRIC or FIN

in the spaces provided.

There are forty questions on this paper. Answer all questions. For each question, there are four possible

answers A, B, C and D.

Choose the one you consider correct and record your choice in soft pencil on the OMS.

Each correct answer will score one mark. A mark will not be deducted for a wrong answer.

Any rough working should be done in this booklet.

2

© Hwa Chong Institution 9646 / 01 / C2 Preliminary Examinations 2014

Data Formulae

speed of light in vacuum,

c = 3.00 108 m s

-1

permeability of free space,

o = 4 10-7 H m

-1

permittivity of free space,

o = 8.85 10-12

F m-1

= (1/(36)) 10-9 F m

-1

elementary charge,

e = 1.60 10-19

C

the Planck constant,

h = 6.63 10- 34

J s

unified atomic mass constant,

u = 1.66 10-27

kg

rest mass of electron,

me = 9.11 10-31

kg

rest mass of proton,

mp = 1.67 10-27

kg

molar gas constant,

R = 8.31 J K-1 mol

-1

the Avogrado constant,

NA = 6.02 1023

mol-1

the Boltzmann constant,

k = 1.38 10-23

J K-1

gravitational constant,

G = 6.67 10-11

N m2 kg

-2

acceleration of free fall,

g = 9.81 m s-2

uniformly accelerated motion, 2

2

1atuts

asuv 222

work done on/by a gas, VpW

hydrostatic pressure, ghp

gravitational potential, r

GM

displacement of particle in s.h.m., txx o sin

velocity of particle in s.h.m., tvv o cos

22xxv o

mean kinetic energy of kTE2

3

a molecule of an ideal gas

resistors in series, ...21 RRR

resistors in parallel, .../1/1/1 21 RRR

electric potential, r

QV

o4

alternating current / voltage, txx o sin

transmission coefficient, )2exp( kdT

where2

2 )(8

h

EUmk

radioactive decay, )exp( txx o

decay constant,

2

1

693.0

t

3


1 What is the best estimate for the value of the mass of a raindrop?

A 4 µg

B 4 mg

C 4 g

D 4 kg

2 A student measures two lengths as follows:

X = 15.0 0.2 cm

Y = 30.0 0.2 cm. The student calculates:

FX the fractional uncertainty in X

FY the fractional uncertainty in Y

FY-X the fractional uncertainty in (Y – X)

FX+Y the fractional uncertainty in (X +Y)

Which of these uncertainties has the largest magnitude?

A FX

B FY

C FY-X

D FX+Y

3 A raindrop, in still air, falls at a constant vertical terminal velocity of 1.8 m s-1. If the raindrop were to fall through air in which a horizontal wind of velocity 1.4 m s-1 is blowing, calculate the magnitude of the resultant velocity of the raindrop.

A 0.4 m s-1 B 1.8 m s-1 C 2.3 m s-1 D 3.2 m s-1

4 Two stones, X and Y, of different mass are dropped from the top of a cliff. Stone Y is dropped a short time after stone X. Air resistance is negligible. Whilst the stones are falling, the distance between them will

A Decrease if the mass of Y is more than the mass of X

B Increase if the mass of X is more than the mass of Y

C Decrease regardless if the mass of X more than or less than the mass of Y.

D Increase regardless if the mass of X is more than or less than the mass of Y

4


5 A bullet is aimed directly and horizontally at the centre of a target which is 100 m away. If the bullet is then fired at a speed of 200 m s-1, what is the distance the bullet will land from the centre of the target?

A 0 cm B 12 cm C 123 cm D 196 cm

6

A man of mass M is standing on a weighing scale in an elevator. The elevator undergoes several different motions as described below.

Elevator’s motion Weighing scale reading

Moving downwards and coming to rest with uniform deceleration N1

Moving downwards and speeding up with constant acceleration N2

Moving downwards with constant speed N3

Which of the following is correct?

A N1 = N2 = N3

B N1 = N2 < N3

C N1 < N3 < N2

D N2 < N3 < N1

7 A stationary radioactive nucleus decays into two nuclei, one of which is an alpha particle which has a mass of 4 units and the other, the daughter nucleus of mass 40 units. If the daughter nucleus has an initial kinetic energy of 200 eV, what is the kinetic energy of the alpha particle?

A 20 eV B 500 eV C 2000 eV D 20 000 eV

5


8

Two beakers of water are shown below. Beaker A is brim-full of cold water with a solid ice-block floating in it. Beaker B is also brim-full with an ice-block floating in it, but the ice-block has a trapped lead ball within it.

When both ice-blocks melt, what will happen?

A Both beakers will remain brim-full.

B Beaker A will remain brim-full whereas beaker B will spill over.

C Beaker A will spill over whereas the water-level in beaker B will get lower.

D Beaker A will remain brim-full whereas the water-level in beaker B will get lower.

9 A constant force of 100 N, parallel to a rough inclined plane, moves a body of mass 20.0 kg at constant speed of 5.0 m s-1 through a distance of 40 m along the plane. The body gains a height of 12 m. Determine the magnitude of work done by friction that is dissipated as heat.

A 1.6 kJ B 2.4 kJ C 3.7 kJ D 4.0 kJ

10 A car of mass m has an engine that exerts a force of F on it. In a time t, the car travels a distance s and its speed increases from u to v. What is the useful output power of the engine?

A

B C D

100 N

Beaker A Beaker B

Lead ball

Ice-blocks

6


11 A small sphere is travelling horizontally around the circumference of the bigger circular loop in the figure below with an angular velocity of 63.0 rad s-1.

The sphere then moves into the smaller loop and continues to move along its circumference. What will be the angular velocity of the sphere when it is moving in the smaller loop?

A 37.7 rad s-1 B 63.0 rad s-1 C 105 rad s-1 D 126 rad s-1

12 A pendulum bob of mass 2.50 kg is supported by a string so that the radius of its path is 0.800 m. It is moving with velocity 0.475 m s-1 horizontally at the centre of its motion when the string is vertical. What is the tension in the string at this instant?

A 0.705 N B 23.8 N C 25.2 N D 29.0 N

19.4 cm

32.4 cm

7


13 The figure below shows the variation of the gravitational potential between the surface of Earth and the surface of the Moon along the line joining their centres. The gravitational potential is maximum at point P.

What is the minimum speed for a projectile to be launched from the surface of the earth to reach the surface of the moon? (Ignore air resistance)

A 11.2 km s-1 B 11.0 km s-1 C 10.8 km s-1 D 2.8 km s-1

14 A satellite of mass m is in circular orbit with radius r around the Earth with a period of T. The satellite is moved to a new orbit with radius 4r around the Earth. What is the new period in terms of T?

A ¼ T B 2T C 4T D 8T

Potential / 106 J kg-1

- 62.3

- 3.9

- 1.3

P

Earth Moon

0 Distance / m

8


15 A rubber ducky of mass 300 g is floating in a fish tank. When it is pushed into the water without fully submerging it and released, it bobs up and down. The distance of the rubber ducky from the base of the tank is given by

0.25 0.03cos2.5d t

where d is given in metres and t is the time in seconds.

Which of the following graphs shows the variation with time of the rubber ducky’s kinetic energy?

A

B

C

D

Kinetic energy / mJ

t / s 0.4

8.3

0.0 0.8 1.2 1.6

Kinetic energy / mJ

t / s 0.8

8.3

0.0 1.6 2.4 3.2

Kinetic energy / mJ

t / s 0.8

33

0.0 1.6 2.4 3.2

Kinetic energy / mJ

t / s 0.4

33

0.0 0.8 1.2 1.6

9


16 A 0.50 kg mass is placed on a flat horizontal plate and the plate is made to oscillate in the vertical direction with simple harmonic motion as shown below.

The plate oscillates with a frequency of 13 Hz. Calculate the amplitude of oscillation of the plate so that the mass just loses contact with the plate.

A 0.74 mm B 1.5 mm C 6.1 mm D 12 mm

17 A loudspeaker at position S emits sound of a single frequency. The sound travels to Leo who is at position L, both through a straight path and after reflection from a wall as shown.

As Leo walks directly towards the wall, the sound alternates between loud and soft. Which of the following changes would result in an increase in the distance between loud and soft sounds?

A Increasing the frequency emitted by the loudspeaker.

B Moving the loudspeaker closer to the wall.

C Moving the loudspeaker towards L.

D Increasing the loudness of the sound emitted by the loudspeaker.

plate

oscillator

mass

10


18 A sodium lamp produces two distinct yellow lines in the visible part of the spectrum, at 589.0 nm and 589.6 nm. Two such lamps are separated by a screen. The light of both lamps pass through a slit each, as indicated in the figure. It turns out that no interference pattern can be obtained on the projection screen.

Which of the following explains the observation?

A The lamps are not point sources.

B The light from the lamps do not have exactly the same amplitude.

C The light from the lamps is not coherent.

D The light from the lamps is not monochromatic.

19 Two samples of guitar strings, one 60.0 cm long and the other 30.0 cm long are plucked in the presence of a tuning fork of frequency 256 Hz. The wave velocity v is 153.5 m s-1 in both strings.

Which of the following is true?

A Only the 30.0 cm string causes the tuning fork to vibrate.

B Only the 60.0 cm string causes the tuning fork to vibrate.

C Both strings cause the tuning fork to vibrate.

D Neither strings cause the tuning fork to vibrate.

11


20 The power of an electrical immersion heater is 22.0 kW. Water flows through the heater at a rate of 0.12 kg s–1 as shown below.

The temperature of the water flowing into the heater is 25.0 oC and the temperature of the water flowing out of the heater is 40.0 oC.

Given that the specific heat capacity of water = 4200 J kg–1K–1, calculate the rate of heat lost to the surrounding.

A 1.8 kW

B 5.6 kW

C 9.4 kW

D 14 kW

21 Two copper spheres of different radii are in thermal equilibrium as shown.

Which statement must be correct?

A Each sphere has the same internal energy.

B There is a no net transfer of thermal energy between the spheres.

C Both spheres have the same heat capacity.

D The larger sphere has a greater mean internal energy per atom than the smaller one.

Electrical immersion heater

Water in

Water out

12


22 A vapour lamp emits light when its temperature is 3000 K but not when its temperature is 300 K. Assuming the vapour to be a monoatomic ideal gas, which of the following statements is false?

A At 3000 K, the average kinetic energy per molecule is 10 times greater than at 300 K.

B At 300 K, molecules have insufficient kinetic energy to cause excitation by collision.

C The vapour molecules excited by collision release photons when they return to the ground state.

D The molecules in the heated vapour lamp have a range of speeds but a mean kinetic energy of 0.0388 eV.

23 The diagram below shows equipotential lines in the vicinity of two unequal charges.

Which of the arrows below best represents the direction of the force that would act on a small positive test charge placed at point P?

A

B

C

D

24 The graph below shows how the electric field strength varies with distance x.

Determine the work done to move a positive test charge q from x = 0 to x = R.

+ -

electric field strength

distance, x R 0

E

13


A qER B - qER C ½ qER D - ½ qER

25 Four different arrangements of identical resistors are connected to the same constant voltage power supply. An ammeter of negligible resistance is connected as shown in each arrangement. In which arrangement will the ammeter show the minimum reading?

A B

C D

14


26 For the bizarre circuitry shown below, what is the effective resistance between points X and Y? Assume all components are ideal.

A 0.750 Ω

B 0.625 Ω

C 0.600 Ω

D 0.500 Ω

27 A cell with substantial internal resistance is connected in series to a resistive load. Which of the following statements is true when the resistance of the load is halved?

A The current through the load must decrease.

B The potential difference across the load must increase.

C The total power delivered by the cell must decrease.

D The efficiency of energy transfer to the loads must decrease.

X

A

V

Y

1.0 Ω 1.0 Ω

1.0 Ω

1.0 Ω

1.0 Ω 1.0 Ω

1.0 Ω

15


28 The diagram shows a diode-resistor network. The network is connected to a 12.0 V source that changes its polarity at regular intervals. The direction of the current leaving the source is either left to right or right to left.

Which of the following correctly describes the current through the ammeter?

A Constant at 1.5 mA.

B Constant at 3.0 mA.

C Alternates between 3.0 mA and 1.0 mA at regular intervals.

D Alternates between 1.5 mA and 0.0 mA at regular intervals.

29 An ideal transformer is used to step down the a.c. voltage supply to a resistive load R. If the number of turns in the primary coil is doubled, what is the new current in the primary coil?

A Twice the original current in the primary coil

B Same as the original current in the primary coil

C Half the original current in the primary coil

D One quarter of the original current in the primary coil

R

Ip

6.0 kΩ

2.0 kΩ

2.0 kΩ

6.0 kΩ

16


30

A cosmic ray proton falls vertically downwards from sky at a speed of 4.0 x 105 m s-1 in a region where the magnetic field of the Earth has the magnitude 5.0 × 10-5 T and is directed towards the north and downward at 30° below horizontal. The magnitude and direction of the magnetic force on the proton is

Magnitude Direction

A 1.60 x 10-18 N Towards the East

B 1.60 x 10-18 N Towards the West

C 2.77 x 10-18 N Towards the East

D 2.77 x 10-18 N Towards the West

31

A current of 0.3 A flows in a conductor that lies on the plane of the paper as shown in the figure below.

The conductor is inside a region of space containing a uniform magnetic field of 1.5 T. AB and FE are parallel to the magnetic field. BC and ED are parallel to each other. Angle BCD is 60o. The lengths of segments BC, CD and DE are 1.2 m long each. Segment CD is perpendicular to the magnetic field. The magnitude of the resultant force on the conductor is

A 0.27 N

B 0.47 N

C 0.54 N

D 1.08 N

D

Magnetic field, 1.5 T

B

C

E

A

F

60o

1.2 m

1.2 m

1.2 m

0.3 A

17


32 A metal rod is moving at a constant speed perpendicular to a magnetic field as shown. A conduction electron in the rod is represented by the dot marked P.

After some time, the electron experiences (apart from gravity)

A both an electric and a magnetic force.

B neither an electric nor a magnetic force.

C a magnetic force only.

D an electric force only.

33 Two coils of different diameters are in the same uniform magnetic field B.

If the magnitude of the magnetic field is increasing with time,

A the induced emf is the same in each coil.

B the induced emf in coil 1 is greater.

C the induced emf in coil 2 is greater.

D the induced emfs in the two coils is in opposite directions.

X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X

P

18


34

In a photoelectric experiment, a 100 W point light source emits light of wavelength 600 nm to illuminate a small metal target of area 4.00 cm2. If the light source is placed 50.0 cm away from the metal target, how many photons are incident upon the metal target every second?

A 3.84 x 1016 B 1.51 x 1020 C 3.02 x 1020 D 3.20 x 1037

35

Buckyballs are spherical molecules consisting entirely of carbon (see diagram below). C60 and C70 buckyballs are composed of 60 and 70 carbon atoms respectively. They are one of the largest objects to have been shown to exhibit wave–particle duality.

In a diffraction experiment, C60 and C70 buckyballs are heated to the same temperature. A collimated beam of Buckyballs with the same kinetic energy is targeted at a slit to form an interference pattern. The C60 buckyballs are shown to have a deBroglie wavelength of 2.5 pm. What would be the deBroglie wavelength of the C70 buckyballs?

A 2.1 pm B 2.3 pm C 2.5 pm D 2.7 pm

36 An alpha particle is fired at a thin piece of gold foil and passes close to a gold nucleus. As it approaches the gold nucleus which one of the following is true about the electric potential energy and the magnitude of the momentum of the alpha particle?

potential energy magnitude of momentum

A decreases unchanged

B increases decreases

C decreases decreases

D increases unchanged

37 A sample of a radioactive nuclide X has the same initial activity as a sample of a radioactive nuclide Y. The sample of X contains twice the number of atoms as the sample of Y. If the half-life of X is T then the half-life of Y is

A 2T B 1.5 T C T D 0.5 T

19


38 A radioactive isotope can be used as a trace element in the fields of medicine, agriculture and industries. Radioactive isotope when used within a plant or animal allows an observer to follow the movement of certain chemicals. In the selection of such a radioactive isotope, which of the following is given the least consideration?

A Daughter nuclide of nucleus.

B Half-life of radioisotope.

C Intensity of radiation emitted.

D Mass of the radioactive isotope.

40 In a p-type semiconductor, the dopant

A donates an electron to the conduction band and the resultant vacancy acts as a mobile charge carrier.

B donates an electron to the conduction band to act as a mobile charge carrier.

C provides a vacancy in an energy level near to the top of the valence band to act as a mobile charge carrier.

D provides a vacancy in an energy level near to the top of the valence band to accept an electron from the valence band.

End of Paper

39

Which of the following diagrams best represents the highest two energy bands of a conductor at 0 K?

A

B

C

D

energy energy

energy energy

H2 Paper 1 Suggested Solution

QN Ans

1 B

2 C

3 C

4 D

5 C

6 D

7 C

8 D

9 A

10 D

11 C

12 C

13 B

14 D

15 A

16 B

17 B

18 C

19 C

20 D

QN Ans

21 B

22 D

23 B

24 D

25 B

26 B

27 D

28 C

29 D

30 C

31 C

32 A

33 B

34 A

35 B

36 B

37 D

38 D

39 B

40 D




Higher 2

CANDIDATE NAME

CT GROUP 13S

CENTRE NUMBER

INDEX NUMBER

PHYSICS

Paper 2 Structured Question Set A

Candidates answer on the Question Paper.

No Additional Materials are required.

9646/02

2 September 2014

1 hour 45 minutes


Write your Centre number, index number, full name and CT class clearly on all work you hand in.

Write in dark blue or black pen on both sides of the paper.

You may use a soft pencil for any diagrams, graphs or rough working.

Do not use staples, paperclips, highlighters, glue or correction fluid.

Answer all questions in the spaces provided in this Question Booklet.

INFORMATION FOR CANDIDATES

The number of marks is given in brackets [ ] at the end of each question or part question.

You are reminded of the need for good English and clear presentation in your answers.

For Examiner’s Use

1 / 9

2 / 10

3 / 11

4 / 9

5 / 7

6 / 14

Deductions

Total / 60

2


Data Formulae

speed of light in free space,

c = 3.00 108 m s

-1


o = 4 10-7 H m

-1


o = 8.85 10-12

F m-1

(1/(36)) 10-9 F m

-1

elementary charge,

e = 1.60 10-19

C


h = 6.63 10- 34

J s


u = 1.66 10-27

kg


me = 9.11 10-31

kg


mp = 1.67 10-27

kg

molar gas constant,

R = 8.31 J K-1 mol

-1

the Avogadro constant,

NA = 6.02 1023

mol-1


k = 1.38 10-23

J K-1


G = 6.67 10-11

N m2 kg

-2


g = 9.81 m s-2

uniformly accelerated motion, s = ut + 2

1at2

v2 = u

2 + 2as

work done on/ by a gas, W = p V

hydrostatic pressure p = gh


Gm

displacement of particle in s.h.m., x = xo sin t

velocity of particle in s.h.m., v = vo cos t

= )( 22xxo

mean kinetic energy of a molecule of an ideal gas

kTE2

3

resistors in series, R = R1 + R2 + . . . resistors in parallel, 1/R = 1/R1 + 1/R2 + . . .


QV

o4

alternating current / voltage, x = xo sin t

transmission coefficient, T exp(-2kd)

where 2

2 )(8

h

EUmk

radioactive decay, x = xo exp ( -t )

decay constant,

2

1

693.0

t

3


1 (a) State the principle of moments.

………………………………………………………………………………………………………. ………………………………………………………………………………………………………. ……………………………………………………………………………………………………….

[1]

(b) A uniform rigid rod of mass 30 kg is attached to a vertical wall by a hinge as shown in Fig. 1. The other end of the rod is held to the ceiling by a cable.

Draw the free body diagram of the forces acting on the rod in Fig. 1. Label and explain all the forces clearly.

[2]

(i) Show that the tension T in the cable is 127 N.

[2]

(ii) Determine the force acting on the rod by the hinge.

force = ………………………… N

direction = ……………………………………..…………….

[4]

60o

30o Fig. 1

4


2 Fig. 2.1 shows a device for measuring the frequency of vibrations of an engine. The rigid metal probe and case is held against the engine so that the engine vibrations induce forced vibrations of the strip in the direction shown.

Fig. 2.1

The length L of the strip which is free to vibrate can be varied until a maximum amplitude is observed. The frequency corresponding to L can then be read from the scale.

(a) (i) State what will be observed as the length L of the strip that is free to move is

gradually increased from a small value.

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

[1]

(ii) Explain why this happens.

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

[2]

(b) In one test, the end of the strip oscillated with simple harmonic motion of amplitude 5.0 mm at a frequency of 12 Hz.

(i) Define simple harmonic motion.

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

[2]

length of strip free to vibrate

L movable strip

record of frequency

probe

plate with markings to enable

amplitude measurements

direction of oscillation

engine

5


(ii) Determine for the end of the strip,

1. its maximum speed

maximum speed = ………………… m s-1

2. its maximum acceleration

maximum acceleration = ………………… m s-2

[3]

(c) The free end of the vibrating strip is an antinode and the end near the probe is a node. When the natural frequency of the strip is 12 Hz, the length L of the strip that is free to vibrate is 16 cm. This has been plotted for you in Fig. 2.2. Sketch a graph on Fig. 2.2, showing how the length L varies with natural frequency of vibration of the strip. [2]

Fig. 2.2

length free to

vibrate/ cm

frequency/ Hz

6


3 (a) While Alvin drives through a long underground tunnel in Singapore, it is obvious that nothing can be seen a few metres into the tunnel without additional sources of light. However, his radio continues playing songs from Gold 90.5FM, which is broadcast at a frequency of 90.5 MHz, and his Global Positioning System (GPS) continues to display his car position. Further into the tunnel, the GPS reports that it has lost its connection with the satellites but his radio continues to play.

(i) State the type of electromagnetic waves that Alvin’s GPS system uses to communicate with its satellites. Explain how you come to this conclusion.

…………………………………………………………………………………………............

…………………………………………………………………………………………............

…………………………………………………………………………………………............

………………………………………………………………………………………………….

[2]

One particular day, the traffic was particularly heavy in the tunnel. When driving very slowly, Alvin finds that there were many points in the tunnel where the radio connection appears to be lost. Alvin decided to count those points and found there were exactly 400 points.

(ii) Assuming that the tunnel is perfectly straight, calculate the distance between the first and the last point where the radio connection is lost.

distance = ………………… m

[2]

(b) (i) State two conditions that must be satisfied in order that two waves may interfere.

1. …………………………………………………………………………………………........

2. …………………………………………………………………………………………........

[2]

7


The apparatus illustrated in Fig. 3.1 is used to demonstrate two-source interference using light.

Fig. 3.1 (not to scale)

For a particular experiment, red light of λ = 690 nm is used, with a = 0.333 mm and D = 1.20 m.

(ii) Calculate the fringe separation.

fringe separation = ……………………. mm

[2]

The double slit in the experiment above is now replaced by a diffraction grating, which is labelled 300 lines mm-1.

(iii) Determine the highest order of light that is still visible on the screen, given that the screen is flat and 2.40 m wide.

highest visible order: ..…………….

[3]

8


4 The variation with potential difference V of current I for a light emitting diode (LED) is shown in

Fig. 4.1.

Fig 4.1

(a) (i) Use Fig. 4.1 to determine the resistance of the LED at 2.25 V.

resistance = …………….. Ω

[1]

(ii) Shade in Fig. 4.1 the area that represent the increase in power dissipation in the LED if the potential difference across the LED is increased from 1.50 V to 1.75 V. [1]

V/V

I/mA

9


Two of these LEDs are connected to a 3.0 V battery with negligible internal resistance and a 160 Ω resistor as shown in Fig. 4.2.

Fig 4.2

(b) (i) Suggest the purpose of the 160 Ω resistor.

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

[1]

(ii) Draw in Fig. 4.1 the current-voltage characteristic of the 160 Ω resistor. Label the line R.

[1]

(iii) Hence, or otherwise, estimate the current in LED1 when switch S1 is closed but switch S2 remains open.

current = …………….. mA

[1]

(iv) Explain why LED1 becomes less bright when S2 is closed.

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

[3]

(v) In the space below, draw a rearrangement of the circuit shown in Fig 4.2 with the

addition of another 160 resistor that allows LED1 to be the same brightness, regardless of whether LED2 is switched on or not.

[1]

3.0 V

160 Ω S1

S2

LED1 LED2

10


5 Fig. 5 below shows a current balance. The wire frame AXYC is balanced on 2 knife edges

through which a current from a battery can be passed through. The wire frame is horizontal

when there is no current. A magnetic force acts on the wire frame when a 3.0 A current flows

through.

Fig. 5

(a) Explain why the current in AC causes a turning effect.

………………………………………………………………………………………….....................

………………………………………………………………………………………….....................

………………………………………………………………………………………........................

………………………………………………………………………………………........................

………………………………………………………………………………………………………..

[3]

(b) Explain why the current in AX and CY do not contribute to the turning effect.

………………………………………………………………………………………….....................

………………………………………………………………………………………….....................

………………………………………………………………………………………….....................

[1]

M

11


(c) A 0.030 kg mass M, is needed on the right hand side to restore the frame to its horizontal

position again. The length of AX is 60.0 cm and AC is 20.0 cm. The horizontal distance

between load M and XY is 40.0 cm. Calculate the magnetic flux density in the solenoid.

magnetic flux density = ………………… T

[3]

12


6 This questions is about a cliff railway that is entirely powered by water. The rail line links a town at the top of a hill with another town at the bottom of the hill. The railway has two carriages running on parallel tracks. They are connected by a continuous cable running around the two pulley wheels mounted at the top and bottom of the track bed (Fig. 6). Brakes can be applied to the lower pulley wheel to control the speed of the carriages.

Fig 6

Each carriage has a tank beneath the passenger compartment which can hold 5.0 m3 of water. Before the start of each journey, both tanks are full of water. When the passengers are aboard, a fraction of the water is drained from the water tank of the lower carriage so that its weight is less than that of the upper carriage. The brakes on the pulley wheel are then released and the carriages start to accelerate towards the other station. When the speed of the carriages reaches 6.6 m s-1, the brakes are partially applied to maintain a constant speed.

When the carriages approach the stations, brakes are fully applied and the carriages slow down and stop. After the passengers alighted, the water tank beneath the carriage at the top station is refilled with water from a river. Passengers board both lower and upper carriages and the whole process is repeated.

Data:

mass of each carriage fully loaded including a full tank of water = 10 000 kg

volume of water tank = 5.0 m3

length of rails = 260 m

angle of incline = 35o

density of water = 1000 kg m-3

mass of each iron brake block = 25 kg

specific heat capacity of iron = 4.7 x 102 J kg-1 K-1

13


(a) Describe the energy changes that occur when the lower carriage is lifted to the upper station while the upper carriage moves to the lower station.

[3]

……………………………………………………………………………………………………….

……………………………………………………………………………………………………….

……………………………………………………………………………………………………….

……………………………………………………………………………………………………….

(b) At the start of a journey, just before the water tank in the lower carriage is drained, both carriages are carrying their full load and are stationary. The tension T2 in the lower cable is small enough to be ignored. Determine the tension T1 in the upper cable when both carriages are fully loaded.

T1 = ……..……………….. N

[2]

(c) When the brakes are released, the acceleration of both cars is 1.5 m s-2 and there is a resultant force of 8.7 x 103 N parallel to the track acting on the lower carriage.

Calculate the volume of water which has been released from the lower carriage.

volume = ………………………… m3

[4]

14


(d) At the start of one particular journey, both carriages are fully loaded. 3800 kg of water is drained from the lower carriage.

(i) Calculate the change in the gravitational potential energy of the system after both carriages complete their journey.

change in potential energy = ……………………….. J

[2]

Six iron brake blocks apply a force against the lower pulley wheel. This maintains a constant speed during the journey.

(ii) Calculate the rise in the temperature of the brake blocks at the end of this journey.

rise in temperature = ……………………… oC

[2]

(iii) In practice, the rise in temperature of the brake blocks is much less than the value calculated in (ii). Suggest a reason why this is so.

………………………………………………………………………………………………

……………………………………………………………………………………………….

……………………………………………………………………………………………….

[1]

[Continue with Question 7 in Set B]




Higher 2

CANDIDATE NAME

CT GROUP 13S

CENTRE NUMBER

INDEX NUMBER

PHYSICS

Paper 2 Structured Question Set B



9646/02

2 September 2014

1 hour 45 minutes


Write your name, CT class, centre number and index number clearly on all work you hand in.



Do not use staples, paperclips, highlighters, glue or correction fluid.

Answer all questions in the spaces provided in this Question Booklet.

INFORMATION FOR CANDIDATES

The number of marks is given in brackets [ ] at the end of each question or part question.

You are reminded of the need for good English and clear presentation in your answers.


Q7 / 12

2

© Hwa Chong Institution 9646 / 02 / C2 H2 Preliminary Examination 2014

7 With increasing levels of noise in the environment, it is recognized that measures need to be taken to reduce noise particularly in homes and working environment. Noise near busy roads or airports is of great concern, and the level of noise insulation provided by windows must be considered.

Double glazed windows consist of two glass sheets separated by a layer of air as shown in Fig. 7. They can be used as a sound as well as heat insulator.

Fig. 7

You are required to design an experiment to find the best air spacing to reduce sound transmission between the two panes of glass, for a specific frequency of sound within the audible range. In your account, you should give details of

(a) the source of the sound,

(b) how you would measure the frequency and amplitude of the transmitted sound,

(c) how you would set the spacing between the two sheets of glass,

(d) the procedure to be followed,

(e) the control of variables,

(f) any precautions that would be taken to improve the accuracy of the experiment

[12]

glass glass

air

3


………………………………………………………………………………………………………………………….

………………………………………………………………………………………………………………………….

………………………………………………………………………………………………………………………….

………………………………………………………………………………………………………………………….

…………………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………………

4


2014 C2 H2 Physics Prelim Suggested Solution

Paper 2

Question 1

1 (a) The principle of moments states that for a body to be in rotational equilibrium, sum of clockwise moments about any point equals sum of anticlockwise moments about that same point.

B1

(b)

W is the weight of the rod, T is the tension acting on the rod, R is the force acting on the rod by the hinge.

(i) Take moments about the hinge,

NT

LTLo

1274.127

)2

(30cos81.930

M1

C1

(ii) NTR oo

X 7.6330sin4.12730sin)(

NTR o

Y 18430cos81.930)(

NRRR YX 1957.1941847.63 2222

o9.707.63

184tan above the horizontal as shown in Fig. 1

M1

M1

A1

A1

- All 3 forces correctly labeled and

explained [B1]

- All 3 forces intersect to meet at a

common point. [B1]

- Appropriate magnitudes of forces.

[B1]

Any missing point deduct 1 mark.

60o

30o

R

W

T

Question 2

2 (a) (i) The amplitude increases to a maximum and then decreases. B1

(ii) The strip has a natural frequency which depends on its length.

Maximum amplitude when the frequency of engine vibration equals to the natural frequency of the length of strip free to vibrate.

B1

B1

(b) (i) The magnitude of the acceleration is proportional to the displacement from its equilibrium position and

is directed opposite to the displacement/ directed towards the equilibrium position

B1

B1

(ii) 1. maximum speed = 12 2 12 0.005 0.38 m so ox fx C1

A1

2. maximum acceleration =

22 2 22 4 12 0.005 28 m so ox fx A1

(c)

The length L is equivalent to ¼ wavelength constant = 16 124

vL Lf

f

Graph shows inverse relationship.

Curve must pass through at least 3 points with Lf = 16 x 12, such as (6, 32), (16, 12) and (24, 8).

M1

A1

Question 3

3 (a) (i) Microwaves.

The degree of diffraction of a wave depends on its wavelength relative to the opening/obstacle size.

B1

B1

(ii) λ = c / f = (3.00 x 108) / (90.5 x 10

6) = 3.315 m

400 nodes correspond to 399 λ/2 segments.

Hence, distance is 399 x 3.315/2 = 661 m (3 s.f.).

M1

A1

(b) (i) They must be waves of the same type.

They must overlap.

B2

(ii) Δy = λ D / a

Δy = (690 x 10-9

) x (1.20) / (0.333 x 10-3

) = 0.002486 m = 2.49 mm (2 or 3 s.f.)

M1

A1

(iii) Since the screen is 2.40 m wide and a distance 1.20 m away, the maximum diffraction angle is 45

o.

d sin θ = n λ

n = d sin θ / λ

n = (1 / 300,000) sin 45o / (690 x 10

-9) = 3.4

The highest visible order is 3

M1

C1

A1

Question 4

4 (a) (i) 2.25/0.0108 = 208 Ω C1

(b) (i) LEDs in forward-bias have very small resistances. The resistor limits the current through the LED so that the LEDs do not blow.

B1

(iii) 6.2 mA B1

(iv) With two LEDs, effective resistance of the circuit is lowered,

leading to higher current from battery.

With higher current, the potential difference across the 160Ω resistor is larger, leaving less potential difference across the LEDs.

OR

With two LEDs in parallel, effective resistance between the LEDs is lowered.

By Potential Divider Principle, the potential difference across each LED is also lowered.

From the graph, it can be seen the current decreases with voltage, so the power (P=VI) dissipated in each LED is lower, so brightness is dimmer.

B1

B1

B1

B1

B1

B1

(v)

B1

V/V

I/mA

(a)(ii)

R

3.0 V

160 Ω

S1

S2

LED1 LED2

160 Ω

Question 5

5 (a) A current through the solenoid produced a magnetic field inside the solenoid and parallel to the axis of the solenoid. A current carrying conductor perpendicular to this field will experience a force.

Based on Fleming left hand rule, the force is perpendicular to the directions of the magnetic fields and the current and points downwards.

A turning effect about the pivot along XY is produced since the line of action of the magnetic force is some distance away.

Marking points:

[1] Current in solenoid produced a magnetic (B) field along axis of solenoid

[1] Current along AC perpendicular to B-field produce a force pointing downwards

[1] Turning effect produced by the magnetic force because pivot is some distance away.

B1

B1

B1

(b) The directions of the currents in AX and CY are anti-parallel and parallel, respectively, as that of the magnetic flux density. Based on Fleming’s Left Hand Rule, the force is zero.

B1

(c) Let length of AX = u = 0.600 m

Length between M and XY = v = 0.400 m

Length of AC = L = 0.200 cm

Taking moments about XY:

Anti-clockwise moment = clockwise moment

B I L u m g v

Magnetic flux density m g v

I Lu

0.030 9.81 0.400

3.0 0.200 0.600

0.327 T

M1

C1

A1

Question 6

6 (a) The gravitational potential energy of the upper carriage is converted into the gravitational potential energy of the lower carriage, the kinetic energies of both carriages and heat/work done against friction.

B1

B1

B1

(b) Since the lower carriage is stationary, there is no net force acting on it.

Resolving forces parallel to the track:

4

1

0

10000 sin35 5.6 10 No

F ma

T g

M1

A1

(c) Applying Newton’s second law of motion:

3

3

8.7 105800 kg

1.5

10000 5800 4200 kg

42004.2 m

1000

F ma

m

m

mV

M1

C1

C1

A1

(d) (i) Change in GPE for lower carriage = + mL gh

Change in GPE for upper carriage = - mgh

Net change in GPE of system = -(m-mL)gh = -3800 x 9.81 x 150 = - 5.5 x 106 J

Note: answer must be negative

M1

A1

(ii) Applying conservation of energy:

Gain in thermal energy of blocks = Loss in GPE of system

6

6

5.5 10

5.5 1078 C

6 25 470

o

mc

M1

A1

(iii) When brakes are heated, it loses thermal energy to the surroundings by conduction, convection and radiation.

B1

Question 7

7 Diagram

Procedure

1. Cut two vertical slits about 1 cm apart on a large study cardboard box. Insert the 2 sheets of glass

into the slots. Measure the width of the air gap between the two glass sheets using the internal

calipers of a Vernier calipers. Record the width of air gap as d. Cut the two sides of the box parallel

to the glass sheets to prevent reflection of sound in the box during the experiment (Figure 2).

2. Set up the rest of the apparatus as shown in Figure 1. Both the loud speaker and microphone

should be facing perpendicularly into the glass sheets. They should also be as close to the glass

sheets as possible without touching them. This is to minimize errors caused by sound from the loud

speaker reaching the microphone via reflected paths (eg from room walls).

3. Adjust the frequency of the signal generator to produce a sound in the audible range (20 H to 20

kHz), eg 1 kHz.

4. Adjust the volume so that a clear signal can be picked up comfortably by the CRO.

5. Do not change the amplitude and the frequency of the sound throughout the whole experiment.

6. Adjust the Y-scale and Y-shift to maximize the waveform on the CRO display. Measure the

amplitude x (see Figure 3). Using the Y-scale, calculate the amplitude A in volts.

7. Adjust the time-base (X-scale) of the CRO to get a stable waveform of one or two cycles. Using the

time base, calculate the period, T (see Figure 3).

glass sheets

air gap of width, d

bench top

x

T

Figure 3 enlarged CRO screen

supporting cardboard

box with slits to hold the

glass sheets in place

sides are cut sides are cut

sides are cut sides are cut

vertical slits are cut into the box to act as slots

cardboard box

glass sheets

Figure 2 Top view of cardboard box

with glass sheets inserted

Figure 1

8. Calculate the frequency of the sound using fT

1

.

9. Make another two more vertical slits in the cardboard box and increase the width of the air gap d by

2 mm. Shift the loud speaker and microphone to maintain a constant distance with the sheets of

glass similar to step 2.

10. Repeat steps 6 to 9 until 10 sets of readings are collected.

11. Plot the graph of amplitude A against width of air gap, d. Draw a best fit curve through the plotted

data.

12. Read off the value of air gap d corresponding to the lowest value of amplitude A on the graph. This

width provides the best insulation for the chosen frequency.

Other reliability measures:

13. The experiment should be conducted in a quiet room or sound proof box to reduce background

noise literally.

14. Preliminary trial can be conducted to choose an appropriate volume of sound that will be

comfortably detected by the CRO for the range of air gaps used.

15. Appropriate choice of air gaps (5 mm to 5 cm)

16. Glass sheets used to be reasonably large (eg 1 m x 1 m) so that the direct detection of the sound

produced from the loudspeaker by the microphone can be minimized.

Safety precaution:

17. Wear thick gloves when handling the glass sheets to provide additional grip.




Higher 2

CANDIDATE

NAME

CT GROUP 13S

CENTRE

NUMBER

INDEX

NUMBER

PHYSICS

Paper 3 Longer Structured Questions Section A



9646/03

17 September 2014

2 hours


Write your Centre number, index number, name and CT class on all the work you hand in.



Do not use staples, paper clips, highlighters, glue or correction fluid.

Section A

Answer all questions.

You are advised to spend about one hour on this section.

At the end of the examination, fasten all your work securely together.

The number of marks is given in brackets [ ] at the end of each question or

part question.


Section A

1 / 11

2 / 10

3 / 8

4 / 11

Deductions

Total / 40

2

© Hwa Chong Institution 9646 / 03 / JC2 Preliminary Examinations 2014

Data Formulae

speed of light in vacuum,

c = 3.00 108 m s

-1


o = 4 10-7 H m

-1


o = 8.85 10-12

F m-1

(1/(36)) 10-9 F m

-1

elementary charge,

e = 1.60 10-19

C


h = 6.63 10- 34

J s


u = 1.66 10-27

kg


me = 9.11 10-31

kg


mp = 1.67 10-27

kg

molar gas constant,

R = 8.31 J K-1 mol

-1

the Avogrado constant,

NA = 6.02 1023

mol-1


k = 1.38 10-23

J K-1


G = 6.67 10-11

N m2 kg

-2


g = 9.81 m s-2

uniformly accelerated motion, 2

2

1atuts

asuv 222

work done on/by a gas, VpW

hydrostatic pressure, ghp


GM

displacement of particle in s.h.m., txx o sin

velocity of particle in s.h.m., tvv o cos

22xxv o

mean kinetic energy of kTE2

3

a molecule of an ideal gas

resistors in series, ...21 RRR

resistors in parallel, .../1/1/1 21 RRR


QV

o4

alternating current / voltage, txx o sin

transmission coefficient, )2exp( kdT

where2

2 )(8

h

EUmk

radioactive decay, )exp( txx o

decay constant,

2

1

693.0

t

3


Section A

Answer all the questions in this section.

1 Fig. 1.1 shows a toy gun with the spring, with spring constant 2.00 x 103 N m-1, in an unstretched state.

Fig. 1.1

As the trigger is pulled, the spring compresses by 2.00 cm.

(a) (i) Show that the elastic potential energy that is stored in the spring is equal to 0.400 J.

[1]

(ii) Instead of one spring, a gun is now designed where two springs of the same spring

constant are put in series. The new spring system is again compressed by 2.00 cm.

Explain whether the total elastic potential energy in the spring system is now more, less

or equal to 0.400 J.

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

…………………………………………………………………………………………..............

……………………………………………………………………………………………………

[2]

(b) The spring from (a) (i) is released and a 50.0 g pellet is shot away horizontally at a height of 1.50 m above the ground. The pellet slides through the barrel and experiences a friction of 0.100 N.

(i) Calculate the kinetic energy of the pellet when it leaves the barrel.

kinetic energy = …………………………………. J

[2]

4


(ii) Calculate the angle at which the pellet hits the ground.

direction = …………………………….………………….

[3]

(c) As the pellet hits the ground, it takes 0.10 seconds for it to come to a full stop.

Calculate the average force exerted by the floor onto the pellet.

force = …………………….. N

[3]

5


2 (a) Define electric field strength.

………………………………………………………………………………………………………

……………………………………………………………………………………………………… [2]

A small charged sphere is suspended from a fine spring (made of an insulating material) between

horizontal parallel metal plates as shown in Fig. 2.1.

Fig. 2.1 (not to scale)

Initially, the plates are uncharged. Switch S is then set to position X. This causes the sphere to

move vertically downwards so that eventually it comes to a rest 18 mm below than its original

position.

(b) (i) State the nature of the charge on the sphere.

………………………………………………………………………………………………….

[1]

(ii) Given that the magnitude of the charge on the sphere is 41 nC, determine the spring

constant of the spring.

5000 V

dc supply plates charged

sphere

fine spring

4.8 cm

10.5 cm

S X

Y +

-

small hole in

top plate

6


spring constant = …………………… N m-1 [2]

(c) Switch S is now moved to position Y.

(i) State and explain the effect of this on the electric field between the plates.

…………………………………………………………………………………………………...

…………………………………………………………………………………………………..

…………………………………………………………………………………………………..

…………………………………………………………………………………………………..

[2]

(ii) By considering the forces acting on the sphere, explain the subsequent motion of the

sphere.

…………………………………………………………………………………………………...

…………………………………………………………………………………………………..

…………………………………………………………………………………………………..

…………………………………………………………………………………………………..

…………………………………………………………………………………………………..

…………………………………………………………………………………………………..

[3]

7


3 Fig. 3.1 below show a simple electric generator that can convert mechanical energy into electrical

energy. A metal rod CD of mass M and is able to slide downwards while maintaining contact with two

long smooth vertical metal rails PQ and ST. The rods are connected at the bottom by a resistor R,

and the whole device is in a uniform magnetic field B perpendicular to the page.

When the rod is released from rest, it falls downwards and as a result, an electric current I flows

around the circuit CDTQ. The rod speeds up initially before reaching a constant downward speed.

Fig. 3.1

(a) Draw on the diagrams below clearly labelled arrows to represent the vertical forces acting

on the rod CD in the two cases.

(i) just as it was released

(ii) falling with constant speed

[2]

P S

Q T

8


(b) Fig. 3.2 below shows the rod descending a distance y in a time t at constant speed vT.

Fig. 3.2

(i) Write an expression for the change of magnetic flux as the rod CD falls through

distance y.

= ………………………

[1]

(ii) Hence show that the induced e.m.f. E is given by E = BLvT.

[1]

(iii) Using (a)(ii) and (b)(ii), show that the constant speed vT is given by

T

MgRv

B L

2 2

Explain your working clearly.

[3]

y

P

Q

S

T

L

9


(c) State a disadvantage of this type of generator compared to a conventional rotating

generator.

…………………………………………………………………………………………...................

…………………………………………………………………………………………...................

[1]

4 (a) Name any two quantities that are conserved in all nuclear reactions.

1. ……………………………………………………………………………………………….……

2. …………………………………………………………..………………………………………..

[2]

(b) The radioactive nuclide carbon-14 (C-14) is continually being produced in the upper

atmosphere by the bombardment of N-14 with high energy neutrons. Complete the

nuclear reaction equation below for the formation of C-14.

………..

[1]

(c) Some of the carbon atoms in a living tree consist of the radioactive nuclide C-14. Due to the

continuous exchange of carbon with the environment, the relative amount of C-14 in the living

tree remains constant throughout the life of the tree. When the tree dies, this exchange of

carbon with the environment ceases and the relative amount of C-14 in the tree decreases with

time.

(i) Define activity of a radioactive sample.

………………………………………………………………………………….................

………………………………………………………………………………….................

[1]

(ii) Explain why the amount of C-14 decreases with time.

………………………………………………………………………………….................

………………………………………………………………………………….................

[1]

(iii) Explain what is meant by the spontaneous nature of radioactive decay.

………………………………………………………………………………………………

………………………………………………………………………………………………

[1]

10


(iv) The activity of wood from a living tree is measured as 15.5 disintegrations per

minute per kilogram. A piece of burnt wood (charcoal) of mass 20.2 g found at an

ancient settlement has an activity of 0.267 disintegrations per minute. If the half-

life of C-14 is 5730 years, estimate the age of the settlement.

age = …………………….. years

[3]

(d) Carbon is also found in coal, which originates from wood. However, it is found that coal

shows practically no radioactivity. Suggest why this might be so.

……………………………………………………………………………………………..............

……………………………………………………………………………………………..............

……………………………………………………………………………………….....................

[2]

End of Section A




Higher 2

CANDIDATE

NAME

CT GROUP 13S

CENTRE

NUMBER

INDEX

NUMBER

PHYSICS

Paper 3 Longer Structured Questions Section B



9646/03

17 September 2014

2 hours


Write your Centre number, index number, name and CT class on all the work you hand in.



Do not use staples, paper clips, highlighters, glue or correction fluid.

Section B

Answer any two questions.

Circle the questions attempted on this cover page.

You are advised to spend about one hour on this section.

At the end of the examination, fasten all your work securely together.

The number of marks is given in brackets [ ] at the end of each question or

part question.


Section B

5 / 20

6 / 20

7 / 20

Deductions

Total / 40

2


Section B

Answer two questions from this section.

5 (a) (i) Define angular velocity for an object travelling in a circle.

……………………………………………………………………………………..…………

………………………………………………………………………………………..……… [1]

(ii) Calculate the angular velocity of the Earth in its orbit around the Sun. Assume that the orbit is circular.

Angular velocity = ……………….… rad s-1 [1]

(b) A satellite with a mass 6200 kg is at point X, a distance of 1.50 x 109 m from the centre of the Earth, as shown in Fig. 5.

X

1.496 × 1011

m 1.50 × 109

m

Earth Sun

Fig. 5

(not to scale)

Mass of the Sun = 1.99 × 1030

kg

Mass of the Earth = 5.98 × 1024

kg

Earth-Sun distance = 1.496 × 1011

m

3


(i) Show that the resultant force acting on the satellite due to the Sun and the Earth is 37.1 N. State the direction of this force.

Direction = ……………………………….…………….. [4]

(ii) The satellite is in a circular orbit around the Sun. Calculate the angular velocity of the satellite.

Angular velocity = ……………..…… rad s-1 [2]

(iii) Using your answer to (a) (ii) and (b) (ii), describe the motion of the satellite relative to the Earth.

…………………………………………………………………………….……..……………

………………………………………………………………………………..………………

………………………………………………………………………………………..………

………………………………………………………………………………………..………

[1]

4


(c) A satellite P of mass m orbits the Earth in circular path as shown in Fig. 5.2. The satellite P has a speed v and the radius of its orbit is r.

(i) The speed v is related to the radius r by the relation nv kr . Find the expression for k and n in terms of G, M and m, where M = the mass of the Earth and G is the gravitational constant.

k = …………………….. . [2]

n = ……………………….. [1]

m

r

Earth P

Fig. 5.2

v

5


(ii) Show that the total energy ET of the satellite P is given by the expression

2T

GMmE

r

[2]

(iii) The satellite P has a mass m = 2200 kg, and the orbital radius r = 6.70 x 106

m. The radius of the Earth is 6.40 x 106 m.

1. Find the minimum energy that must be supplied to satellite P to escape the Earth’s gravitational field.

Energy = ………………………….. J [2]

2. Find the minimum energy that must be supplied to satellite P to launch it from rest at the surface of the Earth into its circular orbit.

Energy = ………………………….. J [2]

6


(iv) If the satellite had a mass of 2.99 x 1024 kg, with the aid of a diagram, describe how the motion of the satellite differs from the one shown in Fig. 5.2.

…………………………………………………………………………………………….…

………………………………………………………………………………………….……

………………………………………………………………………………………………

…………………………………………………………………………………………….…

[2]

7


6 (a) (i) One of the assumptions of the kinetic theory of gases is that the motion of gas molecules is random. Therefore explain why the average momentum of the gas molecules in a container is zero.

[2]

(ii) According to the kinetic theory of gases, explain how a gas exerts a pressure on the walls of a container.

[3]

8


(b) A volume of ideal gas at 30.0 C is enclosed in a metal cylinder fitted with a 8.00 kg freely moving piston of cross sectional area 60.0 cm2 as shown in Fig. 6.1a. Atmospheric pressure is

100 kPa. When the gas is slowly heated from 30.0 C to 100.0 C, the piston rises 20.0 cm as shown in Fig. 6.1b. The piston is then fastened in place, and the gas is allowed to cool back to

30.0 C as shown in Fig. 6.1c.

(a) (b) (c)

Fig. 6.1

(i) Name the thermodynamic process that represent the state changes from

1. Fig 6.1a to Fig 6.1b

………………………………………………………………………………………

2. Fig 6.1b to Fig 6.1c

……………………………………………………………………………………..

[2]

(ii) Calculate the pressure p of the gas shown in Fig. 6.1b before the gas was cooled.

p = ………………………. Pa

[3]

(iii) Sketch the variation with volume of the pressure of the gas for the processes [2]

20.0 cm

30.0 C 100.0 C 30.0 C

9


described in (b)(i).

(iv) Using the first law of thermodynamics, determine the net heat gained by the gas for the processes shown in Fig 6.1a to Fig 6.1c.

net heat gained by the gas = ……………………. J

[3]

(c) Fig. 6.2 shows a model rocket for demonstrating the principle of rocket propulsion. Air is pumped into an upside-down plastic bottle that has been filled with water. When the pressure reaches 3.6 x 105 Pa, the air valve is forced out by the water pressure and the air in the bottle expands.

Fig. 6.2

p

V

10


(i) Explain how the plastic bottle is launched into the air.

[3]

Fig. 6.3 below shows the variation of pressure with volume for the air initially in the bottle as it expands from 3.6 x 105 Pa to atmospheric pressure.

Fig. 6.3

(ii) Using Fig. 6.3, estimate the work done by the expanding air.

work = …………………………………. J

[2]

11


7 (a) Fig. 7.1 below shows a basic setup to investigate the photoelectric property of iron.

The component in the XY portion of the circuit (dashed box) is a low resistance wire (bold line).

(i) Explain why a current flows in the circuit.

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

…………….……………………………………………………………………………………..

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

[3]

(ii) Using Fig. 7.2, draw a rearrangement of the circuit in the XY portion with the addition of a 5.0 V DC cell and a voltmeter to enable the setup to determine the work function of iron, which is believed to be slightly less than 5 eV.

[2]

iron

light

Fig. 7.1

X

collector

vacuum chamber

Y

nA

X Y

Fig. 7.2

12


(iii) With a correctly modified setup, the Vs – f characteristic of iron can be determined and plotted, as shown in Fig. 7.3. Vs is the stopping potential and f is the frequency of the

incident light. The x-intercept of the graph is 151009.1 Hz.

1. Describe how an experiment can be conducted to obtain Fig. 7.3. Assume that light sources of various frequencies are provided.

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

…………….…………………………………………………………………………………

………………………………………………………………………….……………………

[2]

2. Explain why no data values can be obtained for f below 151009.1 Hz.

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

[1]

3. Describe how Planck constant can be determined from the graph.

………………………………………………………………………………………………

……………………………………………………………………………………………...

[1]

4. Determine the work function of iron in eV.

work function = …………………………. eV

[2]

Fig. 7.3

Vs

f 0 Hz

13


(b) Silicon is a semiconductor with a band gap of 1.11 eV and can be doped to form p-type or n-type semiconductor by introducing a small amount of another element (dopant) to the silicon. A semiconductor diode is created when a p-type semiconductor interfaces with an n-type semiconductor, as shown in Fig. 7.4. The dopant concentration is lower in the n-type semiconductor. A semiconductor diode that is optimized to convert light into electrical energy is known as a photodiode.

(i) Indicate on Fig. 7.4 the effective charge of each region 1, 2, 3 and 4 of the photodiode

with the symbols +, - and 0 to represent positive, negative and neutral respectively.

[2]

(ii) Explain why the depletion region extends farther on the n side, as shown in Fig. 7.4.

…………………………………………………………………………………………….……

…………………………………………………………………………………………….……

…………………………………………………………………………………………….……

…………………………………………………………………………………………….……

……………………………………………………………………………………………….…

[2]

Fig. 7.4

depletion region

p-type n-type

1 2 3 4

14


(iii) When a photon of sufficient energy strikes the photodiode, it creates an electron-hole pair.

1. With the aid of a band diagram, explain how a photon of sufficient energy can create an electron-hole pair.

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

[2]

2. If the electron-hole pair is created in the depletion region, they immediately experience an electric force.

Draw on Fig. 7.4 an electron-hole pair labelled e and h respectively in the depletion region, and indicate with an arrow the electric force acting on each.

[2]

(iv) Suggest one practical use for the photodiode.

………………………………………………………………………………………………..…

…………………………………………………………………………………………………..

…………………………………………………………………………………………………..

[1]

End of Section B

H2 Physics Paper 3 Suggested Solutions

Question 1

1 (a) (i) EPE = ½ k x2 = 0.5 x 2000 x 0.0200

2 = 0.400 J

where EPE is the elastic potential energy, k is the spring constant and x is the extension (or, in this case, the compression) of the spring.

B1

(ii) Method 1:

For two equal springs in series, the new spring constant is half that of the individual springs.

Hence, less force is required to compress the spring and the potential energy in the spring system is now less than 0.400 J.

Method 2:

If the total compression is 2.0 cm, the compression per spring is 1.00 cm.

The total energy is then given as 2 (0.5 x 2000 x 0.01002) = 0.200 J, which is less than

0.400 J.

B1

A1

B1

A1

(b) (i) By the principle of conservation of energy,

Elastic potential energy is converted into kinetic energy and heat (by friction)

KE = EPE – WD = 0.400 – (0.100 x 0.170) = 0.400 – 0.017 = 0.383 J

where KE is the kinetic energy of the pellet as it leaves the gun, EPE is the elastic potential energy as the spring is fully compressed and WD is the work down against friction.

M1

A1

(ii) Initial KE = ½ m ux2 = 0.383 J.

Horizontal speed = (2 x 0.383 / 0.0500)1/2

= 3.914 m s-1

For the vertical motion, use e.g. vy2 = uy

2 + 2 a s, with uy = 0, a = -9.81 m s

-2 and s = -1.5 m

vy = (2 x 9.81 x 1.50)1/2

= 5.425 m s-1

Since tan θ = vy / vx,

θ = tan-1

(5.425 / 3.914) = 54.2o below the horizontal

M1

M1

M1

(c) Speed at the moment of impact is given by v2 = vx

2 + vy

2

v = (3.9142 + 5.425

2)1/2

= 6.6895 m s-1

The change in momentum is Δp = m Δv = 0.0500 x (6.6895 – 0) = 0.3345 kg m s-1

Since Favg = Δp / Δt,

Favg = 0.3345 / 0.10 = 3.345 N

This is the resultant force, due to the contact force and the weight of the pellet. The component of the weight in the direction of the motion is given by W// = 0.050 x 9.81 x sin 54.2

o = 0.397 N.

Hence, the force exerted by the ground on the pellet is 3.345 + 0.397 = 3.74 N.

M1

M1

A1

Question 2

2 (a) Electric field strength at a point is the electric force per unit charge acting on a small positive test charge placed at that point.

B2

(b) (i) positive B1

(ii)

9

1

3

0

41 10 50000.24 N m

18 10 0.048

F

kx qE

qE qVk

x xd

M1

A1

(c) (i) The electrons/charges flow from one plate to the other until there is no potential difference between the plates. (discharge)

Electric field is zero.

B1

B1

(ii) Only tension acts on the sphere.

The net force/tension is proportional to extension and directed opposite to displacement.

Sphere moves in simple harmonic motion.

B1

B1

B1

Question

3 (a) (i) B1

(a) (ii)

B1

(b) (i) =BLy B1

(ii)

B1

(iii) Induced current TBLvE

IR R

Net force = 0

B

T

T

F Mg

BIL Mg

BLvB L Mg

R

MgRv

B L

2 2

B1

B1

B1

(c) Non-continuous operation: rod would reach the bottom and then have to be returned to its original height to start again.

B1

weight

magnetic force

weight

Mark by concept and deduction for both (i)

and (ii)

Question 4

4 (a) In all nuclear reactions, the following quantities are conserved:

o nucleon number, A

o proton number, Z (charge) not number of protons

o total energy

o Total momentum

B1

B1

(b)

or p1

1 B1

(c) (i) Activity of a radioactive sample is the number of disintegrations per unit time. B1

(ii) Since C-14 is radioactive it will transmute/decay to another element. B1

(iii) independent of external stimulus B1

(iv) Activity of wood per min per unit mass =

0.26713.217

0.0202

13.217 ln2ln

15.5 5730

1317 =1320 years

t

oA A e

t

t

Students who used activity of wood as 13.22 will end up with 1315 years, rounding to the same final answer in 3 s.f.

Students who used 13.2 will however end up with 1327 years and end up with 1330 years. Deduct 1 mark for early rounding.

If attempt to answer by estimating the fraction of half-lives

years then

award a max of [2]: [1] for idea and [1] for correct arithmetic.

M1

C1

A1

(d) Coal is older than several half-lives;

so activity is too weak to detect.

B1

B1

Question 5

5 (a) (i) The rate of change of angular displacement of the object about the centre of the circle. A1

(ii) 2

365 24 60 60

d

dt

. rads 7 11 99 10 (3 s.f.) A1

(b) (i) 11 24

2 9

6.67 10 (5.98 10 )62001.10 N

1.5 10

Earth

GMmF

r

11 30

2 9 11

6.67 10 (1.99 10 )620036.0 N

(1.50 10 1.496 10 )

Sun

GMmF

r

Resultant force = gravitational force by earth + gravitational force by sun

= 1.10 + 36.0

= 37.1 N

Direction: towards the Sun or Earth

A1

A1

M1

A0

A1

(ii) 2

net F mr

11 237.1 6200(1.511 10 )

M1

7 -11.99 10 rad s A1

(iii) The satellite remains at a distance of 1.5 x 109 m from the earth, on the line joining the

centres of the Earth and Sun. (The satellite, Earth and the Sun are collinear). A1

(c) (i) 2

2

GMm vm

rr

1

2 GM

v GM rr

M1

k = GM A1

n = -0.5 A1

(ii) 2 11 1

Total KE 2 2

mv m GM r

Total KE + Total GPE

=

= -

TE

GMm GMm

r r

GMm

r

2

2

M1

B1

(iii) 1 Total energy must be zero at infinity. Hence the energy supplied is

GMm

r2

11 24

6

6.67 10 (5.98 10 )2200

2 2(6.70 10 )

GMm

r

106.55 10 J

M1

A1

2 EITHER:

Minimum energy supplied = increase in GPE + increase in KE

11 24

6 6

10

= GPE KE KE

= - 02

= -2

1 1 = (6.67 10 )(5.98 10 )2200

6.4 10 6.7 10

= 7.16 10 J

f i f i

E

E

GPE

GMm GMm GMm

r R r

GMm GMm

R r

OR:

Minimum energy supplied

= Final Total Energy (in orbit) – Initial Total Energy (at rest at surface)

= E

GMm GMm

r R

2

=

11 2410

6

6.67 10 (5.98 10 )22006.54 10

6.4 10

= 10 116.54 10 1.371 10

= 7.17 10 J 10

M1

A1

(iv) The satellite and the earth would now orbit with the same angular velocity about a point along the centre line joining the two, with the satellite and the earth opposite each other.

The radius of the satellite orbit is twice the radius of the earth’s orbit (or any indication that the centre of rotation is closer to the Earth and to the satellite).

Zero marks if unclear.

B1

B1

m r

Earth

½ r

P

v

Question 6

6 (a)(i) Random motion of a large number of molecules in random velocity means no preferred direction of movement or the molecules move in different/all directions with a root-mean-square speed.

Momentum is a vector given by the product of mass and velocity. Since the velocities cancel/ or idea of speed in opposite directions, the average velocity is zero and hence the average momentum of the gas molecules of same mass in a container is zero.

B1

B1

(a)(ii) When the gas molecules hit the wall of the container, they rebound off the wall with the same speed and experiences a rate of momentum change normal to the wall, implying there is force acting on the molecule by the wall. Other molecules also have a momentum change normal to the wall when they strike and rebound.

By Newton’s third law, the molecules exert an equal and opposite force on the walls.

Since there are a large number of molecules, the number of collisions at any instant in time is very large and practically constant, resulting in a constant gas pressure. The pressure on the walls of the container is the average force per unit area exerted by the molecules on the walls of the container..

B1

B1

B1

(b)(i) 1. Isobaric expansion, constant pressure

2. Isovolumetric cooling, constant volume

B1

B1

(b)(ii) F = mg = 8.00 (9.81) = 78.5 N

P = 78.5/(60.0 x 10-4)

+ 100000 Pa

= 1.13 x 105 Pa

M1

C1

A1

(b)(iii)

-1 mark for heating (horizontal line)

-1 mark for cooling (vertical line)

Not necessary to draw in the isotherm.

B1

B1

(b)(iv) For heating, U = Q1 + W = Q1 + pV

For cooling, W = 0 U = Q2

Since U is the same for both processes, Q1 + pV = Q2

Hence net heat gained by gas = Q1 - Q2 = pV = 113000 x 0.200 x 60.0 x10-4

= 136 J

OR

net net netU Q W

Since there is no change in the temperature between (a) and (c), Unet = 0

Qnet = -Wnet = 1.13 x 105 x 0. 200 x 60.0 x10

-4 = 136 J

B1

B1

A1

(c)(i) The expansion of the gas in the bottle pushes the water molecules downwards out of the bottle.

B1

By Newton’s third law, the water molecules exert an equal and opposite force (upwards) on the air in the bottle, providing the thrust for the bottle.

When the thrust exceeds the weight of the bottle, the bottle will be launched into the air.

B1 B1

(c)(ii) Work done = Area under the p-V graph

= [0.5(3.6 + 2.6)(0.3) + 0.5(2.6 + 1.7)(0.5) + 0.5(1.7 + 1.3)(0.5) + 0.5(1.3 + 1)(0.5)]x105 x 10

-3

= 333 J

Accepted range (316 to 350) J

M1

A1

Question 7 (Arthur)

7 (a) (i) An electron near the surface of the iron can absorb the energy of a photon.

When the photon energy is larger than the work function, electrons will be liberated with varying kinetic energy and some will reach the collector.

This set up an emf/potential difference/electric field which causes a current to flow in the closed circuit.

B1

B1

B1

(ii)

A1

A1

(iii) 1. For each light frequency f, vary the potential difference across the iron and the collector using the potentiometer.

Vs is the voltmeter reading when the nano-ammeter reads zero.

B1

B1

2. No electrons are emitted when f is below the threshold frequency because the photons do not possess enough energy each to liberate even the most loosely bound electrons from the iron.

B1

3. The gradient of the graph = h/e. Hence h = gradient x e. B1

4. minh . .

= 7.23 10 J

7.23 10 = . eV

.

f

34 15

19

19

19

6 63 10 1 09 10

4 521 6 10

M1

A1

(b)

[2]

[1]

[1]

(i) See diagram

(ii) The magnitude of the fixed charges in region 2 and 3 are the same.

Since the number of free charge carriers per unit volume is smaller in the n-type, a larger

B1

X Y

V

5 V

depletion region

p-type n-type

1 2 3 4

0 + 0 -

h e

volume of n-type is needed, depletion region extends farther on the n side. B1

(iii) 1.

B1

B1

2. See diagram.

The electron hole pair can be in either region 2 or 3.

(iv) Light intensity meter, photo detector, solar cell, Charged Coupled Device, light sensor. A1

conduction

band

valence

band

band gap

e

A photon with energy larger than the energy band gap can

excite an electron from the valence band to the conduction

band, leaving a hole in the valence band.

Note: not necessary to label 1.11 eV or mention in statement.

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