05-06 mock 1 AL - PLK Vicwood KT Chong Sixth Form Collegeday.sfc-plk.edu.hk/Physics/exam paper/05-06/05-06 mock 1... · 2006-11-06 · 05-06/Mock Examination/AL Physics/Paper 1/P.1

05-06/Mock Examination/AL Physics/Paper 1/P.1

PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE Form Seven Mock Examination

2005 - 2006

Physics A-Level Paper 1 Question/Answer Book A

Time Allowed: 3 hours

This paper must be answered in English

INSTRUCTIONS 1. This paper consists of TWO sections,

A and B. Answer ALL questions in BOTH sections.

2. Questions for sections A and B are

printed in two separate Question-Answer Books A and B respectively.

3. Sections A and B each carries 60

marks. You should spend about 1 hour 30 minutes answering each section.

4. Write your Name, Class and Class

Number in the spaces provided on the covers of Question-Answer Books A and B.

5. Question-Answer Books A and B

must be handed in separately at the end of the examination.

6. Supplementary answer sheets will be

supplied on request. Write your Name and Class on each sheet and fasten them with string inside this book.

Name

Class

Class No.

Marker’s Use Only

1

2

3

4

5

Total

A

05-06/Mock Examination/AL Physics/Paper 1/P.2 SECTION A i. Answer ALL questions in this section. ii. Write your answers in the spaces provided in this question-answer book. In calculations you

should show all the main steps in your working. iii. Assume: velocity of light in air = 3 × 108 ms-1

acceleration due to gravity = 10 ms-2

Question No. 1 2 3 4 5 Marks 14 12 12 11 11

1. A project is undertaken by a student who wishes to design a catapult. The student begins by

measuring the displacement x of the catapult rubber band and the action of different forces F, as shown in Figure 1.1.

Figure 1.1

(a) The student obtains the following values of force F and displacement x during the extension

process:

Displacementx/mm

0 5 10 15 20 25 30 35 40

45 50 55 60

Force F/N 0 0.5 0.8 1.1 1.3 1.4 1.5 1.6 1.8 2.0 2.3 2.7 3.2 Plot a graph showing how force varies with displacement during the extension process. (3 marks)

12.0 cm rigid

supports

catapult rubber band

applied force F

xrule graduated

in mm


(b) The ‘stiffness’ of the catapult can be defined as the rate of change of force with displacement, i.e. stiffness = dF/dx. Use the graph to describe how the stiffness of the catapult varies with displacement during the extension process. Explain what feature of the graph enables you to make this description. (2 marks)

(c) Estimate the stiffness of the catapult when x = 20 mm. (2 marks)

(d) The original length of the rubber was 0.12 m and its original area of cross section was 10 mm2.

Suppose that the catapult is now stretched until the displacement is a maximum, x = 60 mm. (i) Calculate the tension in the catapult rubber band, showing your working clearly. (2 marks)

(ii) Assuming that the volume of the rubber band remains constant during the extension,

calculate the stress in the catapult rubber band. (3 marks)

(e) The student now progressively reduces the applied force F, and measures the corresponding

displacement, x. It is found that the relaxation curve did not coincide with the stretching curve. (i) Name this phenomenon. (1 mark)

(ii) On the graph you plotted in (a), sketch and label one of these curves that corresponds to the relaxation process. (1 mark)

05-06/Mock Examination/AL Physics/Paper 1/P.4 2. This question is about the sport of bungee jumping.

100 m

elasticrope

Figure 1.1

A girl of mass 50 kg jumps from a bridge 100 m above a river. Attached to her ankles is an elastic

rope of natural length 50 m. The rope extends as she falls and brings her to a momentary halt 10 m above the water surface. Assume the rope obeys Hooke’s Law.

(a) (i) Describe the energy transfers which occur from the time the girl jumps until she first

comes to rest. (2 marks)

(ii) In the space below, sketch a graph to show the variation of the acceleration of the girl with the position measured from the bridge. (Take the downward direction to be positive.)

(2 marks) (b) (i) At which point during the jump does the elastic rope exert the greatest force on the girl?

Explain your answer. (2 marks)

(ii) A relative of the girl, observing the jump, thought this activity was far too dangerous, since the decelerating force on the girl was so large.

(I) Estimate the energy stored in the rope when the girl first comes to rest 10 m above the water. (1 mark)

(II) Hence estimate the mean force exerted by the rope on the girl whilst bringing her to rest. (1 mark)

Figure 2.1

05-06/Mock Examination/AL Physics/Paper 1/P.5 (c) The organizers of the event have ropes with different values of force constant (k) where k is

defined by Hooke’s Law:

force = k × extension They match the rope to the weight of the person. (i) Find the force constant of the rope which is most suitable for the girl. (2 marks)

(ii) Discuss what would happen if the force constant of the rope were too large. (2 marks)

3.

motormotor

flywheelA

flywheelAflywheel

B

flywheelB

60rev/min

60rev/min

Figure 3.1 Figure 3.2 (a) Figure 3.1 shows two flywheels A and B. Flywheel A is kept rotating at 60 rev/min by

connecting to a motor. Flywheel B is initially stationary. When they come in contact, flywheel B slips initially but it undergoes uniform angular acceleration and eventually rotates at 60 rev/min after it has turned through an angle of 5 rad. (Figure 3.2) (Moment of inertia of flywheel B = 0.5 kg m2 ) (i) Find the angular acceleration of flywheel B. (2 marks)

(ii) What is the time required for B to acquire the same speed as A? (1 mark)

(iii)What is the value of the torque that acts on B during acceleration? (1 mark)


(iv) When flywheel B comes in contact with flywheel A, can we apply the principle of conservation of angular momentum to find the final angular speed of flywheel B? Explain briefly. (2 marks)

motor

flywheelA flywheel

B6 cm

Figure 3.3 (b) Flywheel B is then used to raise a mass as shown in Figure 3.3. (You may assume that the

friction acting on the interface between A and B can keep them rotating at the same speed.) The optimum power output from flywheel B is 8 W at 60 rev/min. (i) Calculate the torque that develops at flywheel B. (2 marks)

(ii) If the diameter of the shaft of flywheel B is 6 cm, what is the maximum load that could be

lifted, hanging from a cord wrapped round the shaft? (1 mark)

(iii)Suggest one arrangement for the system, still working under optimum conditions, to raise a load greater than the maximum load in (b)(ii)? (1 mark)

(c) When a mass of 2 kg is being lifted by the system at uniform velocity, the contact between flywheel A and B suddenly loses. Find how many turns can flywheel B be rotated before it comes to rest. (You may neglect the frictional torque that acts on the shaft of flywheel B.)

(2 marks)

05-06/Mock Examination/AL Physics/Paper 1/P.7 4. (a)

axisr

θ38o

optical fibre

Figure 4 A ray of light is incident centrally at the end face of a cylindrical-shaped optical light-guiding

fibre, at an angle of incidence of 38º with the central axis(see Figure 4.1). The refractive index n of this optical fibre varies with the radial distance r from the axis as

n = 1.45 - 0.09 r,

where r is expressed in mm.

(i) Calculate the angle of refraction θ of the incident light ray at the first end face.(2 marks)

(ii) At a certain point within the optical fibre, the light ray turns back towards the central axis.

Find the distance of this point from the axis. (3 marks)

(iii) Explain why it is more suitable to use an optical fibre to convey a ray of light than to use

a hollow tube of small cross-section with a reflective inner surface. (2 marks)

Figure 4.1

05-06/Mock Examination/AL Physics/Paper 1/P.8 (b) In a compact disc player, light of wavelength 635 nm is directed normally from air at the

underside of the compact disc and is reflected from the reflective coating, as shown in Figure 4.2. (refractive index of transparent plastic layer = 1.53).

Figure 4.2

(i) The upper surface is encoded with pits in sequence. As the disc is rotated, the reflected

light beam is encoded as the pits cross the beam. Calculate the wavelength of the light in the disc. (1 mark)

(ii) Light reflected from inside a pit at A interferes destructively with light reflected outside the pit at point B because it travels further. Calculate the minimum depth of the pit for this effect to occur. (3 marks)

5. (a) The graph (Figure 5.1) below shows how the gravitational potential difference between a

point on the earth’s surface and a distant point, distance x from the earth’s surface, changes near to the moon’s surface. The moon’s surface is 384000km from the earth’s surface.

Figure 5.1

05-06/Mock Examination/AL Physics/Paper 1/P.9 The gravitational potential difference first increases, then achieves a maximum value at point S

and finally decreases to a smaller value on the moon’s surface.

(i) Explain why there is a point of maximum on the graph. (2 marks)

A spacecraft has just finished its mission on the moon and is planning to return to the earth.

(ii) If the spacecraft just succeeds in reaching the earth, what is the minimum energy per unit

mass that has to be supplied? (1 mark)

(iii) What is the minimum speed of the spacecraft at take-off? (2 marks)

(b) Show that the radius r of a satellite’s circular orbit about a planet of mass M is related to its

period T as

TGM

r22

34=

π

where G is the universal gravitational constant. (2 marks)

05-06/Mock Examination/AL Physics/Paper 1/P.10 (c) Two graphs of T2 versus r3 are shown below (Figure 5.2). One of them is for the moons of Jupiter

and the other is for the moons of Saturn. r is the mean distance of a moon from a planet’s centre and T is its period. The orbits are assumed to be circular. The mass of Jupiter is 1.90x1027kg.

Figure 5.2 (i) Find a value for the universal gravitation constant G. (2 marks)

(ii) Find a value for the mass of Saturn. (1 mark)

(iii) The mass of the earth is 6.0x1024kg. What can you say about the slope of the T2 versus r3 graph for the moon of the earth when compared with those for Jupiter and Saturn?

(1 mark)

END OF SECTION A


PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE Form Seven Mock Examination

2005 - 2006

Physics A-Level Paper 1 Question/Answer Book B

Time Allowed: 3 hours

This paper must be answered in English

INSTRUCTIONS 1. Write your Name, Class and Class

Number in the spaces provided. 2. Question-Answer Books A and B

must be handed in separately at the end of the examination.

3. Supplementary answer sheets will be

supplied on request. Write your Name and Class on each sheet and fasten them with string inside this book.

Name

Class

Class No.

Marker’s Use Only

6

7

8

9

10

Total

B

05-06/Mock Examination/AL Physics/ Paper 1/P.12 SECTION B i. Answer ALL questions in this section. ii. Write your answers in the spaces provided in this question-answer book. In calculations you

should show all the main steps in your working. iii. Assume: velocity of light in air = 3 × 108 m/s acceleration due to gravity = 10 m/s2

Question No. 6 7 8 9 10 Marks 15 8 13 13 11

6. Figure 6.1 shows part of a linear accelerator, which accelerates ions along the axis of a line of hollow cylindrical electrodes (A-D). Alternate electrodes are connected together and an alternating voltage is applied to them such that the ions are accelerated by the electric field in between each adjacent pair of electrodes.

Figure 6.1

Positive ions are accelerated in the following sequence: Step 1 The positive ion accelerates across gap 1 in a very short time when the voltage is at a peak. Electrode B is negative with respect to electrode A. (Figure 6.2)

Figure 6.2 Step 2 The ion moves at constant speed in electrode B for nearly half a period of the alternating voltage. The polarity of the electrodes reverses whilst the ion is inside B.

Step 3 The positive ion emerges from B and accelerates across gap 2 because electrode C is now negative with respect to B. (Figure 6.3)

05-06/Mock Examination/AL Physics/ Paper 1/P.13

Figure 6.3

(a) Explain why the ions do not accelerate whilst they are inside the cylindrical electrodes. (1 mark)

(b) The following data relates to an experiment using mercury ions (Hg+): accelerating voltage between adjacent pairs of electrodes = 71 kV frequency of the alternating voltage = 4.0 MHz charge on a mercury ion = 1.6 × 10-19 C mass of mercury ion = 3.35 × 10-25 kg

(i) Show that each mercury ion gains kinetic energy of approximately 1.14 × 10-14 J as it accelerates between a pair of electrodes. (1 mark)

(ii) Ions are injected into electrode A with an initial velocity of 2.1 × 105 m/s. Find the velocity of a mercury ion as it enters electrode B. (3 marks)

(iii) At each gap ions are accelerated for a time equivalent to 5% of the alternating voltage period. Calculate the force acting on a mercury ion as it accelerates in gap 1. (3 marks)

05-06/Mock Examination/AL Physics/ Paper 1/P.14 (iv) Calculate the electric field strength across gap 1., assuming that the electric field is

uniform whilst the ion is accelerating. (2 marks)

(v) Calculate the length of gap 1. (2 marks)

(vi) Calculate the length of electrode B. (2 marks)

(vii) Explain why each electrode is longer than the one that precedes it. (1 mark)

7. A plane-polarised laser beam is incident on a sheet of polariod. The power of the transmitted laser

beam is controlled by rotating the sheet of polariod, as shown in Figure 7.1.

Figure 7.1

(a) Use the axes in Figure 7.2 to sketch a graph to show how the power of the transmitted beam

varies with the angle of rotation, θ, of the polariod from its position at maximum transmitted power for one complete rotation of the polariod sheet. (3 marks)

Figure 7.2

05-06/Mock Examination/AL Physics/ Paper 1/P.15 (b) The laser beam has a power of 0.50 mW and consists of monochromatic light of wavelength

650 nm. (Planck constant = 6.63 × 10-34 Js) (i) Calculate the energy of a photon emitted by the laser. (1 mark)

(ii) How many photons the laser emits each second? (1 mark)

(iii)The laser beam has an area of cross-section of 1.8 mm2. When it is directed normally at a polariod of thickness 1.0 mm, orientated for maximum absorption, almost all the incident light entering the polariod is absorbed. Estimate the average number of photons absorbed per second by an atom, of diameter 0.30 nm, of the polaroid material. (3 marks)

8. (a) A circuit for operating an electric heater in a small room is shown in Figure 8.1. A thermistor is placed in the room some distance away from the heater.

Figure 8.1 (i) The thermistor has a resistance of 680 Ω at 12°C and the resistance decreases to 240 Ω at

25°C. Calculate the voltages VA and VB at A and B respectively when the room temperature is 12°C and 25°C. Then complete the following table. (3 marks)

Room temperature / °C VA / V VB / V

12

25

coil

heater

relay

240 V

-9 V

+9 V

-9 V

+9 V

470 Ω 10 kΩ

10 kΩ

A

B+

thermistor


(ii) Describe what happens to the heater as the room temperature increases from 12°C to 25°C. (1 mark)

(iii) What will be the resistance of the thermistor for the heater to switch off . (1 mark)

(iv) Suggest one limitation of the circuit shown in Figure 8.1 (1 mark)

(b) The circuit of Figure 8.2 shows a microphone connected to an amplifier.

Figure 8.2

(i) The amplifier is described as a non-inverting type. What does this mean? (1 mark)

(ii) On Figure 8.2, indicate the inverting and the non-inverting inputs of the operational amplifier. (1 mark)

(iii) State why, if the output of the amplifier is not saturated, the potentials at point A and B

are almost the same. (1 mark)

Vout microphone

120 kΩ

1.8 kΩ

A

B 100 kΩ

05-06/Mock Examination/AL Physics/ Paper 1/P.17 (iv) The bandwidth of the amplifier is 18 kHz.

(I) Explain, with the aid of a labeled sketch graph, what is meant by bandwidth in this case. (3 marks)

(II) State one change that could be made to the circuit of Figure 8.2 to increase the bandwidth of the amplifier. (1 mark)

9.

S

LR

lamp Blamp A

Figure 4

L

Figure 5

50 Hza.c.

supply500 Ω

To Y1 trace

To Y2 trace

To oscilloscopeearth

(a) In the circuit in Figure 9.1, L is a very large inductance. With switch S closed, resistor R is

adjusted until the two lamps glow with equal brightness. Assume that this adjustment has been made and that switch S has been opened, and left open for some considerable time.

(i) If the switch is now closed, what will be observed? Explain briefly. (3 marks)

(ii) What is the purpose of leaving the switch S open for considerable time? (1 mark)

(iii) State the time required in terms of R & L. (1 mark)

Figure 9.1 Fig. 10.2Figure 9.2


(b) The large inductance is now connected in series with a 500 Ω resistor and a 50 Hz a.c. supply as shown in Figure 9.2. The Y1 and Y2 inputs of a double trace oscilloscope are connected as shown. (Note that the earth connection is common to both inputs.) With the time base set to 2 ms/cm the stationary pattern shown in Figure 9.3 is obtained. (Assume that the inductance has negligible resistance.)

Y1 traceY2 trace

1 cm

Figure 6 (i) Find the phase difference between Y1 and Y2. (2 marks)

(ii) Write down what Y1 and Y2 actually represent. (2 marks) Y1 represents Y2 represents

(iii) By drawing a relevant phasor diagram, calculate a value for L. (4 marks)

Figure 9.3

Remarks: Voltage gains of Y1 and Y2 are arbitrary.

05-06/Mock Examination/AL Physics/ Paper 1/P.19 10.(a) (i) State the function of the moderator in a thermal nuclear reactor. (1 mark)

(ii) By considering the free neutrons in a thermal nuclear reactor to behave like the atoms of an ideal gas, estimate the speed of free neutrons in the core of a thermal nuclear reactor when the core temperature is 700 K. (2 marks)

(Given: mass of neutron = 1.67 × 10-27 kg, Boltzmann constant k = 1.38 × 10-23 J/K)

(b) In the core of a nuclear reactor, a fission neutron moving at a speed of 3.90 × 106 m/s collides with a carbon 12 nucleus which is initially at rest. Immediately after the collision, the carbon nucleus has a velocity of 6.00 × 105 m/s in the same direction as the initial direction of the neutron.

(Given: Molar mass of carbon 12 = 12.0 g , Avagadro number = 6.02 × 1023 mol-1)

(i) Find the speed of the neutron after rebounds. (2 marks)

(ii) Calculate the percentage of the initial kinetic energy of the neutron that is transferred to the carbon nucleus. (2 marks)

(iii)Assume that for all such kind of collisions, the percentage of K.E. given to the carbon nucleus is the same. Estimate the number of collisions that occurs to the same neutron before it slows down to become a so-called thermal neutron(speed ~ 1 × 103 m/s).(4 marks)

END OF PAPER


Useful Formulae in Advanced Level Physics

A1. a = rr

v 22

ω= centripetal acceleration

A2. a = x2ω− simple harmonic motion A3. L = Iω angular momentum of a rigid body

A4. T = dtdL torque on a rotating body

A5. E = 221ωI energy stored in a rotating body

B1. v = mT velocity of transverse wave motion

in a stretched string

B2. v = ρE velocity of longitudinal wave

motion in a solid B3. n = tan θp refractive index and polarising

angle

B4. d = aDλ fringe width in double-slit

interference B5. d sinθ = nλ diffraction grating equation

B6. f ' = )(s

ouvuvf

−− Doppler frequency

B7. 10 102

1log ( )I

I definition of the decibel

C1. F = 221

rmGm Newton’s law of gravitation

C2. V = r

GM− gravitational potential

C3. r³/T² = constant Kepler’s third law

C4. E = 204 r

Qπε

electric field due to a point charge

C5. V = r

Q

04πε electric potential due to a point

charge

C6. E = dV electric field between parallel plates

(numerically)

C7. C = dA

VQ 0ε= capacitance of a parallel-plate

capacitor C8. Q = RCteQ /

0− decay of charge with time when a

capacitor discharges C9. Q = )1( /

0RCteQ −− rise of charge with time when

charging a capacitor

C10. E = 221 CV energy stored in a capacitor

C11. I = nAvQ general current flow equation

C12. R = Alρ resistance and resistivity

C13. F = BQv sin θ force on a moving charge in a magnetic field

C14. F = BIl sin θ force on a moving conductor in a magnetic field

C15. V = nQtBI Hall voltage

C16. B = rIπµ2

0 magnetic field due to a long straight

wire

C17. B = lNI0µ magnetic field inside long solenoid

C18. F = rII

πµ

2210 force per unit length between long

parallel straight current carrying conductors

C19. T = BANI sin φ torque on a rectangular current carrying coil in a uniform magnetic field

C20. E = BANω sin ωt simple generator e.m.f.

C21. p

s

p

sNN

VV

≈ ratio of secondary voltage to

primary voltage in a transformer C22. E = -LdI/dt e.m.f. induced in an inductor

C23. E = 221 LI energy stored in an inductor

C24. XL = ωL reactance of an inductor

C25. XC = Cω1 reactance of a capacitor

C26. P = IV cos θ power in an a.c. circuit

C27. B

L

in

outRR

VV

β−= voltage gain of transistor amplifier

in the common emitter configuration

C28. V0 = )(0 −+ −VVA output voltage of op amp (open-loop)

C29. A = i

fRR

− gain of inverting amplifier

C30. A = i

f1RR

+ gain of non-inverting amplifier

D1. pV = nRT = NkT equation of state for an ideal gas

D2. pV = 231 cNm kinetic theory equation

D3. Ek = kTNRT

23

23

A= molecular kinetic energy

D4. E = Lx

AF macroscopic definition of Young

modulus

D5. E = Fx21 energy stored in stretching

D6. F =drdU

− relationship between force and

potential energy D7. E = k/r microscopic interpretation of Young

modulus

D8. P + 221 vρ + ρgh Bernoulli’s equation

= constant D9. ∆U = Q + W first law of thermodynamics

D10. En = eV6.132n

− energy level equation for hydrogen

atom D11. N = kteN −

0 law of radioactivity decay

D12. t1/2 = k2ln half-life and decay constant

D13. Φ−υ= hmv2m2

1 Einstein’s photoelectric equation

D14. E = mc² mass-energy relationship


PLK Vicwood K.T. Chong Sixth Form College Form Seven Mock Exam Physics (AL) Paper 1 (05-06): Marking Scheme

1. (a) Labels and axes 1 Points correctly plotted 1 Shape of curve correct 1 (b) The slope of graph at a point represents the stiffness of the rubber at that point. 1

Stiffness decreases from 0 – 20 mm and becomes constant between 20 – 30 mm and then increases again when x > 30 mm. 1

(c) slope at x = 20 mm ≈ 30 N/m 1 (steps shown for calculation of slope from graph) 1 (d) (i) 2T cos(45°) = F 1 => 2T cos(45°) = 3.2 N => T = 2.26 N 1 (ii) length of rubber band = (12 ×√2) cm 1 volume is constant => 0.12 × 10 = (0.12 ×√2) × A => A = 7.07 mm2 1 stress, σ = T/A = 3.2 × 105 Pa 1 (e) (i) Elastic hysteresis or hysteresis 1 (ii) Any curves drawn that can show the hysteresis loop. 1 14 2. (a) (i) For the first 50 m below the bridge, P.E. lost by the girl changes to her K.E. 1 After 50 m, P.E. lost changes to K.E. of the girl and the elastic potential energy of the rope. 1 (ii) acceleration is constant and = g (cut x-axis at x = 50m) 1 acceleration is a straight line with negative slope afterwards 1

a

g

50 m Distance / m

0 10 20 30 40 50 60 x/mm

1

2

3

F / N

relaxation curve


(b) (i) At 10 m above the water surface 1 because the extension of the rope is the greatest. 1 (ii) (I) Energy = mgh = 50 x 10 x 90 = 45 kJ 1 (II) Average force = (45kJ)/40 = 1125 N 1

(c) (i) 2

21 kxmgh = 1

50 × 10 × 90 = 1/2× k ×(40)2 => k = 56.25 N/m 1 (ii) force exerted on the girl will be too large 1 and hurt the girl. 1 12 3. (a) (i) 1 rev/s = 2π rad/s 1 uniform angular acceleration ⇒ (2π)2 = 2αθ ⇒ α = 3.95 rad/s2. 1 (ii) time required = 2π/α = 1.59 s 1 (iii) τ = Iα = 0.5 × 3.95 = 1.98 Nm. 1 (iv) No 1 As flywheel A is connected to the motor => external torque exists! 1 (b) (i) Power = τω 1 ⇒ τ = 8/(2π) = 1.27 Nm. 1 (ii) τ = Fmax × (0.03) ⇒ Fmax = 42.3 N ⇒ max. load = 42.3 N 1 (iii) Reduce the radius of shaft for the same torque. 1 (c) We consider the total K.E. of the system 1

turns63.2

)103(2 turnsof no.

m 493.0

J 91.9

)22103)(2(21)2)(5.0(

21

21

21

2

22

22

=×

=⇒

=⇒==

×××+=

+

−

−

π

ππ

ω

hh

mgh

mvI

1 12 4. (a) (i) When the light ray incident at the end face, r = 0 1 => n = 1.45 = sin(38°)/sinθ => θ = 25.12° 1 (ii) Imagine that as the light ray propagate, it continuously refract through many layers (each of slightly different refractive index). Apply the law of refraction between every adjacent layers, we have, n1sinθ1 = n2 sinθ2 n2sinθ2 = n3 sinθ3 n3sinθ3 = n4 sinθ4 ncsinθc = n(c+1) sin(90°) 1 (the light will start to totally reflected back towards the axis of the optical fibre.) => n1sinθ1 = n(c+1) sin(90°) Hence, 1.45 × sin(90° – 25.12°) = nc sin(90°) 1 => nc = 1.31 => r = 1.52 mm. 1 (iii) To reduce energy loss due to reflection from reflective inner surface, 1 thus increase the intensity. 1


(b) (i) nm 41553.1

635==discλ 1

(ii) path difference = 2 × depth 1 path difference = 0.5 × λdisc for destructive interference 1 => depth = (415 nm) /4 = 105 nm 1 11 5. (a) (i) FG (due to earth) = FG (due to moon) => net FG = 0 1

⎟⎠⎞

⎜⎝⎛−=

dxdVmFG = 0 => dV/dx = 0 => V is at maximum 1

(ii) Energy per unit mass = (61.4 – 59) × 106 = 2.4 × 106 Jkg-1 1

(iii) 2

21 mvmV = => v = 2.19 km/s 1 + 1

(b) 32

22

24 rGM

Tr

mvr

MmG π==>= 1 + 1

(c) (i) Slope (Jupiter) => G = 6.93 × 10-11 Nm2kg-2 1 + 1

(ii) M = kgSaturnslope

2611

2

12

27

11

2

1043.51093.6

4101.2100.2

1093.64

)(1

×=×

×××

=×

× −−

ππ 1

(iii) Mass of earth << Mass of Jupiter slope for moon of the earth much greater 1 11 than that of the Jupiter or Saturn. 6. (a) No p.d. within electrode hollow cylinder 1 (b) (i) Energy gained = qV = 1.14 × 10-14 J 1 (ii) new k.e. = k.e. of injected ions + gain in k.e. 1 => new k.e. = 1.14 × 10-14 + ½(3.35 × 10-25)(2.1 × 105)2 = 1.88 ×10-14 J 1 => v = 3.35 ×105 m/s 1 (iii) ∆v = 1.25 ×105 m/s 1 T = 1/f = 2.5 × 10-7 s t = 0.05 × 2.5 × 10-7 s 1 F = 3.35 × 10-12 N 1 (iv) If electric field is uniform => E = F/Q 1 => E = 2.09 × 107 N/C 1

(v) xVE∆∆

= 1

=> x = 3.4 mm 1 (vi) Time in electrode B = 0.5 × (2.5 × 10-7) × 0.9 = 1.125 × 10-7 s 1 length = (3.35 ×105 ) × (1.125× 10-7) = 0.038 m. 1 (vii) As the time of travel through a electrode is the same but the speed of the ion is increasing, ∴ the length of the electrode will increase. 1 15


7. (a) max. power at 0°, 180° and 360° 1 zero power at 90° and 270° 1 correct shape and same maximum power at the maxima 1

(b) (i) J 1006.3 19−×==λhcE 1

(ii) number of photons/sec = 1519

3

1063.11006.31050.0

×=××

−

−

1

(iii) volume of atom = 2.7 × 10-29 m3 volume of Polaroid in path of beam = 1.8 × 10-9 m3 1 => number of Polaroid atoms in path of beam = (1.8 × 10-9 )/(2.7 × 10-29) = 7 × 1019 1 number of photons absorbed per sec per atom = (1.63 × 1015 )/( 7 × 1019) = 2 × 10-5 1 8 8. (a) (i)

Room temperature / °C VA / V VB / V

12 0 1.6

25 0 -2.9

VA = 0 (at T = 12°C AND 25°C) 1 VB = 1.6 V (at T = 12°C) 1 VB = -2.9 V (at T = 25°C) 1 (ii) The heater is operating when room temperature increases from 12°C, eventually, the heater switches off when the temperature keep increasing 1 (iii) Resistance of thermistor > 470Ω. 1 (iv) Rapid switch on/off of relay(heater) around temperature for which VB ≈ 0V 1

θ / degree 0 90 180 270 360

transmittedPower


(b) (i) A positive input voltage gives rise to a positive output voltage or there is no phase change of 180° 1 (ii) 1 (iii) The Op-Amp has an infinite/very large open loop gain. 1 (iv) (I) The bandwidth is the range of frequency over which the gain remains constant. 1 Correct gain-frequency sketch (shape) 1 Bandwidth of 18 kHz marked on sketch 1 (II) Any sensible change that reduces the voltage gain e.g. replace the 1.8 kΩ resistor by a larger value, etc. 1 13 9. (a) (i) To make sure that the current drops to zero in the L-R loop 1 (ii) ~ 5 × (time constant of the LR circuit) = 5L/R. 1 (iii) Lamp A comes on before Lamp B. 1 When a constant voltage V is applied across an inductance L, The rate of increase of current through the coil is V/L, i.e. The current will always take a finite time to reach its final value. 1 In the resistor the current change is virtually instantaneous. 1 (b) (i) From CRO, Y1 is faster than Y2 by 3 ms since period = 20 ms 1 => phase difference, φ = (3/20) × 2π = 0.94 rad. 1 (ii) Y1 represents applied voltage 1 Y2 represents voltage across resistor 1 (iii) 1 + 1

From the phasor diagram, RLωφ =tan 1

H18.2100

)94.0tan(5002tantan

=×

===⇒ππ

φω

φf

RRL 1 13

VL = IωL

VR = IR

VY1

VY2 φ

Vout microphone

120 kΩ

1.8 kΩ

A

B 100 kΩ

+

Frequency

Gain

18 kHz


10. (a) (i) to reduce the average speed (or kinetic energy) of the fission neutrons 1

(ii) kTmv23

21 2 = 1

=> m/s 1017.43 3×==mkTv 1

(b) (i) mass of carbon-12 nucleus = kg 100.21002.6

012.0 2623

−×=×

1

by conservation of linear momentum, we have, )100.6()100.2()1067.1()109.3()1067.1( 52621627 ×××+××=××× −−− v => v = 3.3 × 106 m/s 1 (ii) initial k.e. of neutron = 1.27 × 10-14 J final k.e. of carbon nucleus = 3.6 × 10-15 J 1

Percentage of k.e. transferred = %3.281027.1106.3

14

15

=××

−

−

1

(iii) Percentage of K.E. transferred to carbon nucleus is constant => percentage of K.E. lost during every collision is the same => KE(after collision) = KE(before collision) × (1 - 0.283) 1

KE of thermal neutron = J222327 1035.8)101)(1067.1(21 −− ×=×× 1

If the no. of collisions required = n Then, 8.35 × 10-22 = (1.27 × 10-14) × (1 - 0.283)n 1 Solving => n ≈ 50 1 11

End of Marking Scheme

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05-06 mock 1 AL - PLK Vicwood KT Chong Sixth Form Collegeday.sfc-plk.edu.hk/Physics/exam paper/05-06/05-06 mock 1... · 2006-11-06 · 05-06/Mock Examination/AL Physics/Paper 1/P.1