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9646/01/AJC2014/Prelim [Turn Over

2014 JC2 Preliminary Examination PHYSICS 9646/01 Higher 2 Paper 1 Multiple Choice Tuesday 2 September 2014

1 hour 15 minutes

Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name, PDG and NRIC/FIN and shade the 7 digits of your NRIC/FIN in soft pencil on the Multiple Choice Answer Sheet.

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ANDERSON JUNIOR COLLEGE

There are forty questions in this section. 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 separate Multiple Choice Answer Sheet. Read the instructions on the Answer Sheet very carefully. 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 question paper. The use of an approved scientific calculator is expected, where appropriate.

This document consists of 19 printed pages and 1 blank page

Candidate Name

PDG

( )

2

9646/01/AJC2014/Prelim

Data

speed of light in free space, c = 3.00 x 108 m s-1 permeability of free space, 0 = 4 x 10-7 H m-1 permittivity of free space, 0 = 8.85 x 10-12 F m-1 (1/(36)) x 10-9 F m-1 elementary charge, e = 1.60 x 10-19 C the Planck constant, h = 6.63 x 10-34Js unified atomic mass constant, u = 1.66 x 10-27 kg

rest mass of electron, me = 9.11 x 10-31 kg rest mass of proton, mp = 1.67 x 10-27 kg molar gas constant, R = 8.31 J K-1 mol-1 the Avogadro constant, NA = 6.02 x 1023 mol-1 the Boltzmann constant, k = 1.38 x 10-23 J K-1 gravitational constant, G = 6.67 x 10-11 N m2 kg-2

acceleration of free fall, g = 9.81 m s-2

3


Formulae

uniformly accelerated motion, s = ut + ½ at2

v2 = u2 + 2as work done on/by a gas, W = pV hydrostatic pressure, p = gh gravitational potential, -

Gm

r

displacement of particle in s.h.m., x = x0 sin t velocity of particle in s.h.m., v = v0 cos t 22

0 xx

mean kinetic energy of a molecule of an ideal gas

E = 2

3KT

resistors in series, R = R1 + R2 + … resistors in parallel, 1/R = 1/R1 + 1/R2 + … electric potential,

r4

Q = V

0

alternating current/voltage, x = x0 sin t

transmission coefficient, T exp(-2kd)

where k =

2

28

h

EUm

radioactive decay, x = x0exp(- t) decay constant,

12

0 .6 9 3=

t

4


1 Which of the following pairs of physical quantities are both vector quantities? A work done, electric current

B work done, electric field strength

C force, electric current

D force, electric field strength

2 The diameter D of a sphere is measured to be 5.0 cm with a fractional uncertainty of 0.02.

What is the absolute uncertainty and fractional uncertainty of the radius R of the sphere?

absolute

uncertainty of R fractional uncertainty

of R

A 0.05 cm 0.01

B 0.1 cm 0.01

C 0.05 cm 0.02

D 0.1 cm 0.02

3 The precision and accuracy of a measurement is affected by:

precision accuracy

A Systematic error Random error

B Random error Random error

C Systematic error Systematic error

D Random error Systematic error

4 A golf ball travels with an initial speed vo and it went only one-third of the way to the hole.

Assuming the force of resistance due to the grass remains the same, what new initial speed should the ball travel for the ball to go into the hole?

A vo B vo C vo D 3vo

5


5 The graph is a displacement-time (s-t) graph for a tennis ball during part of a game. Which part of the graph shows the highest speed?

6 The diagram shows a trolley moving on a frictionless horizontal table at a speed of 0.5 m s1. 500 g of sand is then released onto the trolley.

What is the change in the momentum of the trolley?

A zero B 0.15 N s C 0.20 N s D 1.80 N s

7

A pole of weight W is attached to a wall. It is held horizontal by a wire attached at point X of the pole where T is the force of the wire. The pole experiences a reaction force R from the wall. Which triangle of forces could correctly represent the three forces acting on the pole?

A

B C D

X

wire

wall pole

500 g of sand

0.5 m s1

A

B

C

D

t

s

T

W

R

T

W

R

T W

R

T

R

W

6


8 A ball bearing was released from rest in a viscous liquid. Which of the following graphs would represent the variation of the forces acting on the ball bearing with time?

A

B

C D

9 A man lifts a 10 kg sack of rice from a ground to above his head using his both hands. The

sack of rice does not experience a change in kinetic energy between the two positions. Which of the following statement is correct?

A The work done on rice sack by gravity is less than the work done on rice sack by man.

B The work done on rice sack by gravity is more than the work done on rice sack by man

C The work done on rice sack by gravity is equal to the work done on rice sack by man.

D There is no work done on the rice sack.

10 A motor driving a pump raises 0.10 m3 of water through a vertical height of 5.0 m in a time of

10 minutes. If the efficiency of the pump is 60%, what is the power generated by the motor? (Take the density of water to be 1000 kg m3)

A 4.9 W B 8.2 W C 14 W D 82 W

11 A body of mass m moves in a horizontal circle of radius r at constant speed v for one complete revolution. Which of the following statements is incorrect?

A The angular velocity of body is directed perpendicular to the plane of circular motion.

B The work done by the centripetal force is 2mv2.

C The change in linear momentum of the body is zero.

D The total energy of the body is constant.

force

time 0 0

weight

viscous force

upthrust

force

time 0 0

upthrust

weight

viscous force

force

time 0 0

weight

viscous force

upthrust

viscous force

upthrust

time 0 0

weight force

7


12 A straight length of tape unwinds from a roll rotating about a fixed axis with constant angular velocity, the radius of the roll decreasing at a steady rate. Which of the following graphs correctly shows how the variation of speed v at which the tape moves away from the roll with time t?

A B C D

13 The gravitational potential at point X due to the Earth is 7.2 kJ kg1. At point Y, the gravitational potential is 3.4 kJ kg1.

The change in gravitational potential energy of a 4.0 kg mass when it is moved from X to Y is

A 42.4 kJ B 10.6 kJ C +3.8 kJ D +15.2 kJ 14 Two satellites, A and B, orbiting around Earth have the same kinetic energy. Satellite A has

a larger mass than satellite B. Which of the following statements is correct? A Satellite A has the same total energy as satellite B. B Satellite A has a smaller orbital radius than satellite B. C Satellite A has a smaller period than satellite B. D Satellite A has a larger angular velocity than satellite B.

v radius

Earth

X

7.2 kJ kg1

Y

3.4 kJ kg1

v

t

v

t

v

t

v

t

8


15 A body moves with simple harmonic motion about a point P. The graph shows the variation with time of its displacement from P.

What is its angular frequency?

A 1.6 × 10-3rad s1 B 4.0 × 10-3rad s1 C 4.0 × 103rad s1 D 2.5 × 104rad s1

16 Two horizontal springs are attached to a trolley. The trolley is then displaced along the axis

of the springs and then released. The trolley undergoes simple harmonic motion. The variation of the kinetic energy EK of the trolley with horizontal displacement x from its equilibrium position is shown in the figure below.

The trolley loses energy so that its maximum kinetic energy is reduced by 50 mJ. What is the amplitude of the oscillations?

A 2.4 cm B 1.9 cm C 1.4 cm D 0.95 cm

displacement / m

time / ms

1.0

0.5

0.5

1.0

0 0.2 0.4 0.60 0.8

0 1 2 31 23 x/cm

10

20

30

40

50

60

70

80

9


17 Two bulbs X and Y containing air at different pressures are connected by a tube P which contains two mercury threads. The density of mercury is 13 600 kg m3. Which pair of values of h1 and h2 is possible?

h1 / cm h2 / cm

A 2.0 8.0

B 4.0 2.0

C 6.0 6.0

D 8.0 2.0

18 The intensity of a progressive wave, besides being dependent on the amplitude of the wave, is also proportional to the square of the frequency.

The diagram shows two waves X and Y.

The intensity of wave X is I0.

What is the intensity of wave Y?

A 0.028 I0 B 0.11 I0 C 0.44 I0 D 2.25 I0

h1 h2 h1 and h2 are not to scale

a 2a wave X

wave Y time

displacement

10


19 The diagram shows an ice cube floating in water. Both the ice cube and the water are at 0 °C. Which statement correctly compares the molecular properties of the ice and those of the water?

A The mean kinetic energies are the same for both the ice molecules and the water molecules. B The mean inter-molecular separations are the same for both the ice and the water. C The mean inter-molecular potential energies are the same for both the ice molecules and the water molecules. D The mean total energies are the same for both the ice molecules and the water molecules.

20 Polaroids P, Q and R are aligned such that the axes of polarisation of P and Q are aligned

to each other and the axis of polarisation R is perpendicular to that of P and Q as shown.

Polaroid Q is then rotated about its centre through an angle θ until its axis of polarisation is aligned with that of R.

Which of the following describes the change in light intensity seen by the observer?

A Light intensity remains zero all the time.

B Light intensity was maximum initially and decreases to zero.

C Light intensity decreases to zero and then increases back to maximum.

D Light intensity increases to maximum and then decreases back to zero.

P Q R

θ

light

Observer

11


21 A double-slit interference of green-coloured light was set up as shown and interference fringes are formed on the screen.

Which change would increase the distance between adjacent fringes? A Use orange-coloured light. B Reduce the width of each slit. C Use a double-slit where the slits are further apart. D Move the double-slit closer to the screen.

22 A speaker producing sound of frequency 2500 Hz is placed at the open end of a closed pipe containing a gas.

As the piston is moved along the pipe, a series of 6 loud sounds was heard. The first loud sound was observed when the piston was 4.0 cm away from the open end, and the sixth loud sound was observed when the piston was 37 cm away from the open end. What is the speed of sound in the gas?

A 270 m s1 B 330 m s1 C 370 m s1 D 400 m s1

speaker

piston

interference fringes

double-slit

screen

green-coloured light

12


23 Which one of the following statements about the electric potential at a point is correct? A The electric potential at a point due to a system of joint charges is given by the sum of

the potentials at that point due to the individual charges of the system

B The electric potential is given by the rate of change of electric field intensity with distance

C The unit of electric potential is either the joule or the volt

D Two points in an electric field are at the same potential only when a unit positive charge placed anywhere on the line joining them remains stationary.

24

Three identical electrical sources each with internal resistance r are used to operate a lamp of resistance R as shown in figure below.

What fraction of the total power is lost due to the internal resistance of the sources?

A

3R + r3R B

3R r3R C

r3R + r D

3R3R + r

25 Four resistors of equal values are connected as shown. How will the current through the resistors change when resistor W is removed? A The current through X will increase and the currents through Y and Z will decrease.

B The current through X will decrease and the currents through Y and Z will increase.

C The current through X will increase and the currents through Y and Z will remain unaltered.

D The currents through X, Y and Z will all decrease.

R

W

X Z

Y

13


26 The current I flowing through a component varies with the potential difference V across it as shown. Which graph best represents how the resistance R varies with V?

A

B

C

D

I

V

R

V

R

V

R

V

R

V

14


27 A cell of emf E1 = 6.0 V and negligible internal resistance is connected to a uniform resistance wire XY. Another cell of emf E2 = 3.0 V and negligible internal resistance is connected as shown. When the movable contact C is placed at B, there is no current in the branch CE. The contact C is then moved to point D.

What will be the direction of current in the branch CE and the change in current reading of the ammeter A?

direction of current in

branch CE change in current reading

in ammeter A

A C E decrease

B C E increase

C E C decrease

D E C increase

28 An -particle and a -particle both enter the same uniform magnetic field, which is

perpendicular to their direction of motion. If the -particle has a speed 15 times that of the -particle, what is the value of the ratio

?particle α on force of magnitude

particle β on force of magnitude

A 3.7 B 7.5 C 60 D 112.5

E

D X Y

C

E2 = 3.0 V

E1 = 6.0 V

A

B

G

15


29 A coil PQRS mounted on an axle, has its plane parallel to the flux lines of a uniform magnetic field B, as shown. When a current I is switched on, and before the coil is allowed to move, A there are no forces due to B on the sides SP and QR.

B there are no forces due to B on the sides PQ and RS.

C sides SP and QR tend to attract each other.

D sides PQ and RS tend to attract each other.

30 A conducting wire XY moves between the magnets as shown in the diagram.

Which of the following motions of XY will make the galvanometer deflect the most?

A It moves sideways along the magnetic field quickly.

B It moves sideways along the magnetic field slowly.

C It moves perpendicular to the magnetic field quickly.

D It moves perpendicular to the magnetic field slowly.

uniform magnetic field B

P Q

S R

I

I

axle

S

N

YX

0

16


31 A flowmeter is used in many industries to measure the volume of liquid passing through a pipe per unit time about the axis of the turbine. When the liquid flows through the turbine, the 8 magnets attached on it rotate.

Which of the following voltage-time graphs shows the signal detected by the coil?

A B

C

D

32 When a resistor is connected across an sinusoidal A.C. source of peak voltage 170 V, the

average power dissipated is 40 W. Two such identical resistors are now connected in series to the electrical mains of 220 V r.m.s.

What would be the total power dissipated?

A 33.5 W

B 67.0 W

C 80.0 W

D 134 W

Voltage / V

Time / s

Voltage / V

Time / s

Voltage / V

Time / s

Voltage / V

Time / s

17


33 In an experiment to investigate the photoelectric effect, monochromatic light is incident on a metal surface. The photoelectric current and the maximum kinetic energy of the photoelectrons are measured. Which one of the following correctly shows the change, if any, in the photoelectric current and in the maximum kinetic energy of the photoelectrons when light of the same intensity but higher frequency is incident on the same metal surface?

photoelectric current maximum kinetic energy

A decreases no change

B decreases increases

C no change decreases

D no change increases

34 The diagram below shows some possible electron transitions between three principal energy levels in the hydrogen atom. Which transition is associated with the absorption of a photon of the longest wavelength?

35 According to the de Broglie hypothesis, matter waves are associated with A

B

C

D

electrons only charged particles only neutral particles only all particles

0 eV

1.2 eV

13.6 eV

A

B

C

D

18


36 A semiconductor X is made by doping germanium crystal with arsenic (donor). Another semiconductor Y is made by doping germanium with indium (acceptor). The two are joined end to end and connected to a battery as shown.

Which of the following statements is correct? A X is p-type, Y is n-type and the junction is forward biased.

B X is n-type, Y is p-type and the junction is forward biased.

C X is p-type, Y is n-type and the junction is reverse biased

D X is n-type, Y is p-type and the junction is reverse biased.

37 Which one of the following statements best describes stimulated emission in a laser?

A Electrons collide with atoms in a metastable state and cause photons to be emitted.

B Atoms in a metastable state de-excite and cause electrons to be emitted.

C Photons interact with atoms in a metastable state and cause photons to be emitted.

D Photons interact with atoms in a metastable state and cause electrons to be emitted

38 Radiation from a radioactive source enters an evacuated region in which there is a uniform magnetic field perpendicular to the plane of the diagram. This region is divided into two by a sheet of aluminum about 1 mm thick. The curved, horizontal path followed by the radiation is shown in the diagram below. Which of the following correctly describes the type of radiation and its point of entry?

type of radiation point of entry

A alpha A

B alpha B

C beta A

D beta B

X Y

A B

19


39 The diagram shows a graph of the binding energy per nucleon for a number of naturally occurring nuclides plotted against their mass number. Which of the following statements is a correct deduction from the graph? A Energy will be released if a nucleus with a mass number less than about 80 undergoes fission as a result of particle bombardment. B Energy must be supplied for a nucleus with a mass number greater than about 80 to undergo fusion with any other nucleus.

C U23892 is the stable end-point of a number of radioactive series.

D Al2713 will spontaneously emit an alpha particle to become Na23

11 .

40 The probability of decay in one second of a radioactive nucleus is . During a particular one second interval, a nucleus does not decay. What is the probability of decay of this nucleus during the next one second interval?

A B C 2 D 2

1

mass number 240 22020018016014012010080 60 40 200

2

1

4

3

6

5

8

7

9

binding energy per nucleon / MeV

U23892

lA2713

Na2311

H21

1

Suggested Solution to 2014 H2 Physics Prelim Paper 1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

D C D B A C A A C C B A D A D C D D A D

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

A B A C A C A B B C D B B B D D C D B B

No Answer & Solution

1 Ans: D

Even though work done can be negative, it is a scalar since the negative does not represent its direction.

Electric current flows along the wire and therefore has no fixed direction in space.

2 Ans: C

The absolute uncertainty of the diameter is 0.02 x 5.0 = 0.1 cm.

The absolute uncertainty of the radius will be 0.1 cm / 2 = 0.05 cm

The fractional uncertainty of the radius will be 0.05 cm / 2.5 cm = 0.02.

3 Ans: D

Random error causes a scatter of points about an average which affects the precision.

Systematic error causes a shift in the value away from the true value which affects the accuracy.

4 Ans: B

0 = vo2 – 2a (

13s) => vo

2 = 2a(13s) ------------------- (1) OR using

12mv2 = Fx

0 = u2 – 2as => u2 = 2as --------------------(2)

(1)/(2) : vo

2

u2 = 13

u = 3vo

5 Ans: A

A has the steepest gradient.

2

6 Ans: C

Applying conservation of momentum to trolley.

mtrolleyvtrolley = (mtrolley+msand)vtrolley&sand

vtrolley&sand = (2.0)(0.5)/(2.5) = 0.4 m s1

ΔPtrolley = |(2.0)(0.4 0.5)| = 0.2 N s

7 Ans: A The three forces must through a common point. Hence direction of R must act along the dotted line from the wall.

8 Ans: A

Since object is released from rest, no viscous force would act on it at the start. Upthrust is dependent on the pressure difference acting on the object. As pressure difference remain the same as object falls, upthrust will remain unchanged.

9 Ans: C

Based on Energy Work Theorem and isolating rice sack as system,

Work done on rice sack by man + Work done on rice sack by gravity = Change in KE

Work done on rice sack by man = - Work done on rice sack by gravity

|Work done on rice sack by man| = |Work done on rice sack by gravity|

10 Ans: C

Pout = 0.6 x Pin

Vρght = 0.6 x Pin

Pin = 13.6 W = 14 W

X

wire

wall pole

3

11 Ans: B

A Correct. Using “right hand grip”, the direction of angular velocity is perpendicular to the plane of circular motion.

B Incorrect. The work done by the centripetal force is zero since centripetal force (ie resultant force) acting on the mass is always perpendicular to the displacement travelled by the mass.

C Correct. When the object is back to the same position in one complete revolution, the velocity is the same. Hence, change in linear momentum of the body is zero.

D Correct. The total energy (ie KE & GPE) of the body remains constant.

12 Ans: A

Using v = r, since angular velocity remains constant, hence speed v varies proportionally with r, which decreases at a constant rate.

13 Ans: D

There is work done on mass to move it from X to Y. Hence, a gain in its GPE.

Change in GPE = [3.4 (7.2)]x4 = +15.2 kJ

14 Ans: A

Since mAvA2/rA = GMmA/rA

2, mAvA2 = GMmA/rA

Similarly, mBvB2 = GMmB/rB

Since ½ mAvA2 = ½ mBvB

2, GMmA/rA = GMmB/rB, mA/rA = mB/rB,

Since mA > mB , rA > rB , and vA2 < vB

2, vA < vB,

Since v = rω, rA ωA < rB ωB , ωA < ωB , as T = 2π/ω, TA > TB.

Total energy E = -1/2[GMm/r],

Since GMmA/rA = GMmB/rB , EA = EB

15 Ans: D

angular frequency, = 2f =

31076.0

32 24800 rads-1

4

16 Ans: C

First method

Shift the graph down by 50 mJ.

Read off the x-intercept value.

x0 = 1.40 cm

17 Ans: D

For pressure in P, 16000 - ρgh1 = 8000 - ρgh2,

h1 – h2 = [16000 – 8000]/[13600 x 9.81] = 0.05996 m = 6.0 cm

since pressure in X > pressure in Y, h1 > h2 hence D is the answer.

18 Ans: D

Amplitude of X = 2 x Amplitude of Y

Period of X = 3 x Period of Y Frequency of X = 1/3 Frequency of Y

Hence, IX / IY = (2)2(1/3)2

IY = 9/4 IX = 2.25 I0

1.40

Second method

cm 39.1

4.2

5075

75

2

1

0

2

0

20max

20

2max

x

x

xKE

xmKE

5

19 Ans: A As the ice and the water are at the same temperature, they have the same mean kinetic energy. C is wrong as distance between molecules in ice is less than water. Hence inter-molecular potential energy of ice is less than water. B is wrong as explained in A. D is wrong as the ice and the water have different inter-molecular potential energy, they do not have the same total energy.

20 Ans: D

At the original position, the light from Q is polarized vertically. Hence no light will pass through R as the incoming light from Q is in the plane that is perpendicular to the axis of polarisation of R.

At the final position where the axes of polarisation of Q and R are aligned, the light that passes through P is polarized vertically. Hence no light will pass through Q as the incoming light from P is in the plane that is perpendicular to the axis of polarisation of Q. Therefore the intensity of light through R is zero.

When Q is rotated, the light intensity will increase until it reaches a maximum at an angle of 45°, and it reduces again as the angle is increased.

21 Ans: A

Fringe separation, x = λD/d, where λ is the wavelength of the light, D is the distance between slits and screen and d is the slit separation.

Options C and D reduces the fringe separation while option A increases fringe separation (as the wavelength of orange light is larger).

Option B has no effect on fringe separation.

22 Ans: B

Since 6 nodes form 5 segments,

Distance between adjacent nodes = (37 – 4.0) / 5 = 6.6 cm

Wavelength of sound = 2 x 6.6 = 13.2 cm

Speed of sound = 2500 x 0.132 = 330 m s1

Be careful: 4 cm λ4 because there is an end correction.

6

23 Ans: A

The electrical potential at a point due to a system of joints charges is given by the sum of the potentials at that point due to the individual charges of the system.

(A) Unit of potential cannot be Joule,

(B) V dEdr

(D) +ve charge may not remain stationary if there is presence of lower potential in the field.

24 Ans: C

Fraction of total power lost

= Power lost by internal resistance

Power lost by internal resistance+ Power of external resistor R

= i2r+ i2r + i2r

i2r+ i2r + i2r + (3i)2R = r

3R + r

25 Ans: A

Original distribution of resistance across the two loops in the circuit is R/2 to R/2 i.e. 1:1. The new distribution is R to R/2 i.e. 2:1.

Due to the new distribution of resistance, the voltage across X will increase (2/3 V) and the voltage across Y and Z will decrease (1/3 V). Since current is V / R, and R is

loud loud loud loud loud loud

(37 – 4.0 ) cm

End correction

(C)

7

constant, the increase in voltage across X will cause an increase in current through X while the decrease in voltage across Y and Z will cause a decrease in current through Y & Z.

26 Ans: C

Resistance is the inverse of I/V ratio.

27 Ans: A

When C is connected at D, the pd across XD will take on the pd of 3.0 V. The pd across DY will then also be 3.0 V as the pd across XY is 6.0 V. The current through the ammeter is VCY/RCY. When C is shifted from B to D, R of CY will increase, while the voltage across CY remain the same, the current through the ammeter will decrease.

VXD = VDY = 3 V. Since RXD < RDY => I3 > I1.

At node X, I1 and I2 combine to give a higher current I3.

Direction of current in branch CE is C E.

Extension question:

What if C is shifted to the right of B?

28 Ans: B F=Bqv

5715

2

15

2.

))((

)(

veB

veB

F

F

29 Ans: B

F = BIlsin

For sides PQ and RS, = 0o and 180o, hence F = 0

For sides SP and QR, = 90o, F = BIl.

Direction of the forces on SP and QR can be found by Fleming’s left-hand rule. The force on QR points out of the plane of the diagram and the force on SP points into the plane of the diagram.

A

G

D B I1 X

I1

I2

I3

I2

E2 = 3.0 V

E2 = 6.0 V

8

30 Ans: C

For the galvanometer to deflect the most, the rate of change of magnetic flux linkage in the wire is to be the largest. Hence, the wire moving perpendicular to the magnetic field quickly will give a larger induced emf than moving perpendicular slowly. For A & B, wire is moving sideways along magnetic field, this will not induce any emf.

31 Ans: D

When each magnet approaches the coil, there is an increase in the flux linkage in the coil. The maximum flux linkage in the coil occurs when the magnet is closest to the coil. The increase in flux linkage with respect to time results in an induced emf. When that magnet moves away from the coil, there is a decrease in the flux linkage in the coil. The minimum flux linkage in the coil occurs when the coil is at equidistant from the 2 magnets that are closest to the coil. Thus, the decrease in flux linkage with respect to time results in having an induced emf in the opposite direction.

Recall the Demo Lab EMI experiment “Activity B: EMI - Swinging bar magnet above solenoid”

32 Ans: B

Let peak voltage of 1 resistor be Vo.

<P> = 12

Vo2

R

Given <P> = 40 W,

40 = ½ (170)2/R

R = (170)2/80

Ptotal = 2202 / [2x(170)2/80] = 67 W

33 Ans: B

From photoelectric effect equation: hfelectrons of KEmax , using a light of higher

frequency, maximum energy of ejected electrons increases.

As intensity of light =Area

hfnp , with a higher frequency, the rate of number of

incoming photons decreases. Hence, the photoelectric current decreases. 34 Ans: B

For absorption of a photon to take place, the energy transition is from a lower energy level to a higher energy level. For absorption of a photon with highest wavelength, the

difference between the 2 energy levels is the smallest. i.e. hc

E

9

35 Ans: D

The de Broglie hypothesis is a fundamental principle exhibited by both radiation and matter. The de Broglie hypothesis can ‘predict’ wavelengths for all particles.

36 Ans: D

X being doped by Gp V arsenic, would have valence electrons as mobile charge carriers, hence n-typed. Y being doped by Gp III indium, would have holes as mobile charge carriers, hence p-typed. The connection of XY with battery increases the potential barrier, hence reverse biased.

37 Ans: C

A – Photons interact with electrons/atoms in metastable state, resulting in stimulated emission of photons. B- Electrons in a metastable state de-excited releasing photons. D- Photons and not electrons are emitted.

38 Ans: D It is beta as the particles are able to penetrate through the thin film of aluminum where alpha particles will be stopped by the aluminum film. Entry is at B because the beta particles lose energy after passing through the aluminum. Hence, they have a lower speed and hence, a smaller radius, greater

curvature of path under the influence of magnetic field. (Bqv =mv2

r => r = mv / Bq)

39 Ans: B

A Incorrect. When a nucleus with a mass number less than about 80 splits into smaller nuclei, there is a decrease in the binding energy per nucleon, hence, energy is required to trigger the fission process i.e. energy is absorbed.. B Correct. When a nucleus with a mass number greater than 80 fuses with another nucleus, there is a decrease in the binding energy per nucleon, hence, energy is required to trigger this fusion process.

C Incorrect. For U23892 to be the daughter nucleus of a radioactive decay, the parent

nucleus has to have a greater mass number than 238. However, from the graph,

U23892 is the nucleus with greatest mass number. Hence, it is not the stable end-point of

a number of radioactive series.

D Incorrect. Al2713 is more stable than Na23

11 . Hence, the former will not

spontaneously emit an alpha particle to become the latter. 40 Ans: B

Radioactive decay is a random process where the probability of decay per unit time is constant.

9646/02/AJC/2014Prelim [Turn Over

1 2014 JC2 Preliminary Examination PHYSICS 9646/02 Higher 2 Paper 2 Structured Questions Thursday 18 September 2014

1 hour 45 minutes Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, index number and PDG on the spaces provided at the top of this page. 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, paper clips, highlighters, glue or correction fluid. The use of an approved scientific calculator is expected where appropriate. Answer all questions. 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.


Candidate Name

PDG

( )

For Examiner’s Use

1

2

3

4

5

6

7

8

9

Deduction

Total

This document consists of 19 printed pages and 1 blank page

9646/02/AJC/2014Prelim

2

Data speed of light in free space, c 3.00 x 108 m s-1 permeability of free space 0 4 x 10-7 H m-1 permittivity of free space 0 8.85 x 10-12 F m-1 (1/(36)) x 10-9 F m-1 elementary charge, e 1.60 x 10-19 C the Planck constant h 6.63 x 10-34 J s unified atomic mass constant u 1.66 x 10-27 kg rest mass of electron, m

e 9.11 x 10-31 kg

rest mass of proton, m

p 1.67 x 10-27 kg

molar gas constant, R 8.31 J K-1 mol-1 the Avogadro constant, N

A 6.02 x 1023 mol-1

the Boltzmann constant, k 1.38 x 10-23 J K-1 gravitational constant, G 6.67 x 10-11 N m2 kg-2 acceleration of free fall, g 9.81 m s-2


3

Formulae

uniformly accelerated motion, s 221 atut

2v asu 22 work done on/by a gas, W Vp hydrostatic pressure, p gh

gravitational potential, r

Gm

displacement of particle in s.h.m., x = x0 sin t velocity of particle in s.h.m., v = v0 cos t

v 22 xxo

mean kinetic energy of a E kT2

3

molecule of an ideal gas, resistors in series, R R1 + R2 + … resistors in parallel, 1/R 1/R1 + 1/R2 + …

electric potential, V r04

Q

alternating current/voltage, x x0 sin t transmission coefficient, T exp(-2kd)

where k 2

28

h

EUm

radioactive decay, x x0 exp(- t)

decay constant, 21

0.693t


4

1 (a) A body has an initial velocity u and an acceleration a. After a time t, the body moved a distance s and has a final velocity v. One of its equation of motion is

s = ut + 12at2.

State the conditions that must be satisfied for the above equation to be valid.

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

(b) A hot air balloon was rising steadily at a speed of 10.0 m s-1 when weather conditions turned windy. A constant breeze of 3.0 m s-1 blew horizontally across the sky, which caused the hot air balloon to travel with a resultant velocity of vR at an angle to the horizontal, as shown in Fig. 1.1 below. Fig. 1.1

(i) Calculate the magnitude and direction of the resultant velocity vR.

= …………………….. o vR = …………………….. m s-1 [2]

(ii) A sandbag was dropped from the balloon. 1. In Fig. 1.2, sketch the path of trajectory of the sandbag as it drops from the

balloon. [1]

Fig. 1.2

sandbag

vR

vR


5

2. Determine how far below the balloon would the sandbag be after 4.0 s. You may assume that the sandbag had not landed on the ground, the dropping of sandbag did not affect the velocity of the hot air balloon and that air resistance on the sandbag is negligible.

distance = ……………………………. m [3] 2 Fig. 2.1 shows a signboard suspended by two elastic ropes of tension T1 and T2. The tension

in T1 is 300 N and the tension in T2 is 252 N.

Fig. 2.1

(a) (i) State the conditions for equilibrium.

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

(ii) On Fig. 2.1, mark the centre of gravity of the signboard with a dot and label the point as G. Show clearly your construction to determine the centre of gravity on Fig. 2.1. [1]

(iii) Determine the weight of the signboard.

weight = …………………………….N [1]

T1

T2 signboard

40 50


6

(b) The signboard is pulled vertically downwards with a force of 20 N so that the ropes are stretched. Determine the acceleration of the signboard immediately after it is released.

acceleration = …………………………….m s-2 [2] 3

The curve in Fig.3.1 shows the way in which the gravitational potential energy of a body of mass m in the field of the Earth depends on r, the distance from the centre of the Earth, for values of r greater than the Earth’s radius RE.

R

Energy A

B

r 0 RE

Fig.3.1


7

The body referred in Fig. 3.1 is a rock which is projected vertically upwards from the Earth. Assume air resistance can be neglected. At a certain distance R from the centre of the Earth, the total energy of the rock (i.e. its gravitational potential energy plus its kinetic energy) may be represented by a point on the line AB.

(a) For a rock that is momentarily at rest at the top of its trajectory, which is at the distance R from the centre of the Earth,

(i) mark its total energy on the line AB in Fig.3.1 with a cross and label it P. [1]

(ii) sketch the variation of its kinetic energy with distance from the centre of the Earth on Fig.3.1. [2]

(b) The escape speed of the rock is the minimum speed that the rock must possess when it is on the Earth’s surface so that it can escape to infinity.

(i) On Fig.3.1, mark on the line AB with a cross and label it Q to represent the total energy of the rock if it were projected from the Earth with the escape speed. [1]

(ii) Explain why the minimum speed for the rock to reach the moon from the surface of the Earth is less than the escape speed. ………………………………………………………………………………….................... ………………………………………………………………………………….................... …………………………………………………………………………………............... [1]

4 (a) Outline and explain experimental observation(s) which provided evidence for the wave nature of light. ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ……………………………………………………………………………………………………… [3]


8

(b) The photoelectric effect provides evidence for the particulate nature of electromagneticradiation. State three experimental observations that support this conclusion.

1.

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

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

2.

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

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

3.

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

………………………………………………………………………………………………………… ……………………………………………………………………………………………………… [3]

5 In Fig 5.1. below, an electrical device (load) is connected in series with a cell of e.m.f. 2.5 V

and internal resistance r. The current I in the circuit is 0.10 A. The power dissipated in the load is 0.23 W.

Fig 5.1

(a) S.I. unit for resistance is defined as the Ohm (Ω).

Define the Ohm. ……………………………………………………………..……………………………….…… ………………………………………………………………….……………………………..[1]

e.m.f. = 2.5 V

I = 0.10 A

load


9

(b) Show that the internal resistance r of the cell is 2.0 Ω. [1]

(c) A second identical cell is connected into the circuit in (a) as shown below in Fig.5.2.

Fig. 5.2 The current in this circuit is 0.15 A. Deduce that the load is a non-ohmic device. ……………………………………………………………………………………………….…… ………………………………………………………………………………………………... [3]

6 (a) Define the tesla.

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

I = 0.15 A

load


10

(b) A large horseshoe magnet produces a uniform magnetic field of flux density B between its poles. Outside the region of the poles, the flux density is zero. The magnet is placed on a top-pan balance and the wire XY is situated between its poles, as shown in Fig. 6.1. The wire XY is horizontal and normal to the magnetic field. The length of wire between the poles is 4.4 cm. A direct current of magnitude 2.6 A is passed through the wire in the direction from X to Y. The reading on the top-pan balance increases by 2.3 g.

(i) State and explain the polarity of the pole P of the magnet.

………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….…………........ ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [3]

(ii) Calculate the flux density between the poles.

flux density = ……………………………...T [2]

Fig. 6.1

pole P

magnet

top-pan balance

X

Y


11

(c) The direct current in (b) is now replaced by a very low frequency sinusoidal current of r.m.s. value 2.6 A. Calculate the variation in the reading of the top-pan balance.

variation in reading = …………………………….g [1]

7 (a) X-rays are produced when electrons are accelerated through a potential difference

towards a metal target such as tungsten. Fig. 7.1 shows a typical X-ray intensity spectrum that can be produced from an X-ray tube.

Fig. 7.1 Using conservation of energy, explain why there is a minimum wavelength for the emitted X-rays. ………………………………………………………………………………………………………. ………………………………………………………………………………………………………. …………………………………………………………………………………………………... [1]

Intensity, I

Wavelength,

K

K

L

L


12

(b) In a chest X-ray, a photographic film receives photons which have travelled through flesh and bone from a source.

(i) Estimate the area of a film which covers the chest of an adult. area = ……………………. m2 [1]

(ii) Assume that on average, 10 x-ray photons fall on each grain of the photographic film and the grains are about 1.0 µm apart. Use your estimate in (b)(i) to calculate the total x-ray energy falling on the film. Each x-ray photon has a quantum energy of 10-15 J.

total energy = …………………..…. J [3]

(c) When all possible photon paths are summed, the amplitude of the wave function for paths travelling through flesh is four times the size of the amplitude for paths travelling through bone. Calculate the ratio

bone through arrival photon ofy probabilit

flesh though arrival photon ofy probabilit.

ratio = …………………. [1]


13

8 A student is investigating the stopping distance for a motorcycle with high-performance brakes. A motorcyclist riding and stopping a motorcycle on a test track is recorded on film.

The stopping distance d is measured for different speeds v. The variation with speed v of the stopping distance d is shown in Fig. 8.1

It is suggested that v and d are related by the equation

d = v2

2A + Bv

where A and B are constants.

Fig. 8.1 v / m s-1

5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0

110.0

100.0

90.0

80.0

70.0

60.0

50.0

40.0

30.0

20.0

10.0

d / m


14

Data from Fig. 8.1 are used to obtain values for

dv and v. These are plotted on the graph of

Fig. 8.2.

(a) (i)

Determine the value of dv for v = 25.0 m s-1.

dv = …………………………….s [1]

(ii) On Fig. 8.2,

1. plot the point corresponding to v = 25.0 m s-1

2. draw the line of best fit for all the points. [2]

v / m s-1

5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0

Fig. 8.2

1.00

1.40

1.80

2.20

2.60

3.00

3.40

•

•

•

•

•

dv / s


15

(b) (i) Use Fig. 8.2 to determine the gradient of the line drawn in (a)(ii). gradient = …………………………… [2]

(ii) Hence find the value of A with appropriate units. A = …………………………….. [2]

(iii) Use Fig. 8.2 to determine the magnitude of B.

B =………………………s [1]

(iv) The stopping distance d consists of the thinking distance and the braking distance. Suggest which part of the equation given represents the thinking distance and which part represents the braking distance. Explain your answers clearly. ………………………………………………………………………………….................... ………………………………………………………………………………….................... ………………………………………………………………………………….................... ………………………………………………………………………………….................... ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [4]


16

(c) The motorcyclist riding at a speed of 50.0 m s-1 towards an obstacle which is 200 m away along the test track applies the brake. Determine the distance he will be from the obstacle when the motorcycle stops. distance =…………………….m [2]

(d) Explain why the stopping distance has to be increased when it is raining. …………………………………………………………………………………………….……….. …………………………………………………………………………………………….……….. ………………………………………………………………………………………….………….. …………………………………………………………………………………………….……….. …………………………………………………………………………………………….......... [2]


17

9 Two flat circular coils carrying current can be used as Helmholtz coils to produce uniform magnetic field between the coils. Fig. 9.1 illustrates the magnetic flux pattern due to the Helmholtz coils.

Fig. 9.1

The magnetic flux density B between the Helmholtz coils is thought to depend on the current I in the coils and may be written in the form, B = kIn where k and n are constants. You are provided with two flat circular coils. You may also use any of the other equipment usually found in a Physics laboratory. Design an experiment to determine the value of n. You should draw a labeled diagram to show the arrangement of your apparatus. In your account you should pay particular attention to (a) the identification and control of variables, (b) the equipment you would use, (c) the procedure to be followed, (d) how the magnetic flux density between the Helmholtz coils would be measured,

(e) any precautions that would be taken to improve the accuracy of the experiment. Diagram


18

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19

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………………………………………………………………………………………………………. [12]


20

BLANK PAGE

1

2014 Prelim H2 Physics Paper 2 Solutions

1a Conditions for equations to be valid:

Constant acceleration

Motion in a straight line

1b(i) 22

yxR vvv

= (10.02 + 3.02) = 10.44

10 m s-1

= tan-1 (103

)

= 73.3

73

1b(ii) 1. Initial velocity of sandbag is vR with angle to horizontal.

Path is symmetrical and parabolic

1bii 2. Taking downwards as positive,

Sbag = -10.0 (4.0) + 12(9.81)(4.0)2

= 38.48 m Sballoon = - 10.0 (4.0) = - 40.0m

Distance apart = 40.0 + 38.48 = 78.48 78 m

vR

vR

2

2ai There is zero net force acting on the signboard. There is zero net torque about any axis acting on the signboard.

ii G is at the intersection of lines of action of T1 and T2

iii Weight = T1 cos 40 + T2 cos 50

= 300 cos 40 + 252 cos 50 = 391.8 = 390 N

b Net force acting on the board immediately after released = 20 N Initial net acceleration = F/m

= 20/(391.8/9.81) = 0.501 0.50 m s-2 [allow ecf from (aiii)]

T1

T2 signboard

40

50

G

3

3ai aii 3bi

When the rock is momentarily at rest, KE is zero and total energy = GPE (ai) Hence, total energy of rock is the intersection of line AB and GPE graph (aii) At distance R, KE is zero. At distance RE, KE = total energy – GPE (bi) For a rock to escape to infinity with minimum speed, total energy at infinity is zero.

3bii Possible explanation includes

Moon would attract the rock

potential at Moon(’s surface) is o not zero o less than 0 o lower than at infinity

moon is nearer; less gain in GPE from Earth to moon Thus, escape speed would be lower.

Correct point P

Correct curve with:

KE = 0 at R and KE + PE = ET Roughly correct energy at RE

KE + PE = ET Roughly correct shape between RE and R Kinetic

energy

Correct point Q

X P

X Q R

total energy for

rock in (a)

Energy A

B

r 0 RE

4

4a Diffraction Diagram (if drawn) must show light source (e.g. light, laser), object causing diffraction (e.g. single slit) and means of observation (e.g. screen) For observable diffraction, the size of the slit should be of the same order as that of the wavelength. OR Description of setup must mention the above items. It was expected that the light on the screen would be the size of the slit but the width of the band of light seen on the screen is greater than the slit width. This demonstrates the spreading effect on a wave as it passes through an aperature. OR 2-Source Interference Diagram (if drawn) must show light source (e.g. light / laser), object causing interference (i.e. double slit) and means of observation (e.g. screen) Light source must be monochromatic and coherent OR Description of setup must mention the above items. Instead of two bright fringes of the screen, a fringe pattern containing light and dark fringes is observed. OR Difraction grating Diagram (if drawn) must show light source (e.g. light / laser), diffraction grating and means of observation (e.g. screen) Light source must be monochromatic and coherent OR Description of setup must mention the above items. A pattern of bright narrow lines/spots with dark regions between them are observed on the screen. The spacing between the lines/spots are not equal. Note: drawing of wavefronts / ripples is not accepted as there are multiple planes for light waves.

4b 1. For a given metal, there is a threshold frequency below which no emission of

photoelectrons occurs regardless of the intensity of radiation.

2. Emission of photoelectrons occurs with no observable time lag at frequencies of

greater than threshold frequency, even at very low intensity.

3. Increasing the intensity of the radiation has no effect on the maximum energy of

the electrons

4. The rate at which electrons are emitted increases proportionally with the intensity

of the radiation.

5. The maximum kinetic energy (not kinetic energy) of the emitted photoelectron is

dependent on / increases linearly (not proportional) with the frequency of the

incident radiation (above threshold frequency), even with low intensity.

Any 3 from the above.

5

5 (a) One ohm is the resistance of a conductor in which the current is 1 ampere when a

potential difference of 1 volt is applied across it.

OR

One ohm is the resistance of a conductor in which the ratio of potential difference

across it to the current flowing through it is 1 volt per ampere.

(b) Power dissipated in r = Total Power – Power dissipated in Load I2 r = ε I – PL

(0.10)2 r = (2.5)(0.10) – 0.23 r = 2.0 Ω

(c) Single cell

Resistance of Load, R1

Two cells 5.0 = 0.15R + 0.15 x 4.0 Resistance of Load, R2 = 29 Ω As an ohmic device has constant resistance, since the load’s resistance has changed/ increased, the load can be deduced as a non-ohmic device.

6

6a One tesla is the magnetic flux density which causes a force per unit length of one

newton per metre on a straight wire carrying a current of one ampere and is at

right angles to the direction of the magnetic field.

6bi As seen from the increased balance reading, there is a downward force on

magnet due to wire carrying current.

By Newton’s third law, there is an upward force on wire by magnet.

By Fleming’s left hand rule, pole P is a north pole

6bii By Newton’s 2nd law, W – BIL =0 W= BIL

2.3 × 10–3 × 9.81 = B × 2.6 × 4.4 × 10–2

B = 0.20 T (g = 10, loses this mark)

6c Reading for maximum current = 2.6 x 2 = 3.68 A

F = BIL =(0.20)(3.68)(4.4x10-2) =0.032 N

mg = 0.032 N, m = 0.032/9.81 = 3.3 g

total variation of mass = 2 x 3.3 = 6.6 g

OR

Ig

BLm , m I

dcdc I

I

m

m maxmax 62

262

32 .

.

.

max m

mmax = 2.3 2 = 3.3 g

total variation of mass = 3.3 x 2 = 6.6 g

7

7(a) The cut-off wavelength corresponds to the most energetic photon that can be produced. That happens when all the kinetic energy of an accelerated electron is lost in a single collision/interaction with the target atom in producing one photon.

(b) (i) area = 0.200 m x 0.300 m = 0.0600 m2 Accept a range of 0.0300 m2 ≤ area ≤ 0.120 m2

(ii) Area of a graph:

no of grains = 66 1010

0600.0

= 6.00×1010

no of photons = 6.00×1010 ×10 = 6.00×1011 energy = 6.00×1011 × 10-15 = 6.00 x 10-4 J

(c) implied probability of arrival proportional to square of amplitude ratio = 42 = 16

10-6

m

10-6

m

8

8ai From Fig. 8.1, d = 57.5 m, d/v = 57.5/25.0 = 2.30 s Accept 57.0 m < d < 58.0 m, 2.28 s < d/v < 2.32 s

aii Plot the point corresponding to v = 25.0 m s

-1

Draw the line of best fit for all the points (fair scatter of points about the best fit line)

bi Gradient = (3.00-1.40)/(36.0-11.5) = 0.065312 = 0.0653 Accept range : 0.062 < gradient < 0.069

bii

From d = v

2

2A + Bv,

dv =

v2A

+ B, gradient = 1/(2A)

A = 1/(2x0.0653) = 7.652 = 7.66 m s-2

range : 7.3 < A < 8.1

biii Sub. (11.5, 1.40) into the equation

dv =

v2A

+ B

1.40 = 0.0653(11.5) + B , B = 0.6489=0.649 s range : 0.61< B < 0.69

biv Since A has the unit of acceleration,

v2

2A is most likely representing the braking distance or

the stopping distance where A is the deceleration of the motion. The braking distance depends on the speed and the deceleration of the braking force applied. B has the unit of time. Bv will represent the thinking distance where B is the thinking time or reaction time of the rider. Before the braking force is applied, the thinking distance depends on the speed and the thinking time.

v / m s-1

5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 ˡ ˡ ˡ ˡ ˡ ˡ ˡ 1.00

1.40

1.80

2.20

2.60

3.00

3.40

•

•

•

•

•

d/v / s

(11.5,1.40)

(36.0,3.00)

•

9

c Using d =

v2

2A + Bv, d = 50.0

2/(2x7.656) + 0.6489x50 = 195.7= 196 m

Distance from obstacle = 200 – 195.7 = 4.28 m

d Friction between the tyres and track surface will be reduced, reducing the magnitude of the motorcycle’s deceleration. This increases the required braking distance. Visibility of motorist’s is reduced, increasing the reaction time and hence the thinking distance. Thus stopping distance has to be increased.

10

Suggested Solution

Aim To investigate how the magnetic flux density B between the 2 circular coils varies with the current I in the coils

Procedure

1) Set up the apparatus as shown above. 2) Switch on the power source to pass a current through both coils to produce a magnetic

field. 3) Record the value of I from the ammeter. 4) Place the hall probe between the coils. Record the value of the magnetic flux density B

from the datalogger. 5) Repeat steps 2 to 4 to obtain further values of B for different I by changing the resistance of

variable resistor.

6) From nkIB = kInB lg+lg=lg . Plot a graph of lg B against lg I. The value of n can be

obtained from the gradient of the graph.

Control of variables 1) The separation of the 2 coils should be kept constant by using the meter rule to check and

adjust the separation before every reading. 2) The position of the hall probe between the 2 coils should be kept constant by clamping the

hall probe to a fixed position between the coils.

Safety and Accuracy

1) The plane of the hall probe should be perpendicular to the magnetic field within the 2 coils. 2) The axes of the 2 coils must coincide to ensure that the magnetic field in the region

between the 2 coils is uniform. 3) The coils are to be close to one another to ensure that the magnetic field between the coils

is uniform. 4) The position of the hall probe is along the centre axis of the 2 coils and it is equidistant from

the coils. 5) The range of the current used should be kept small to reduce heating effect in coils by

using the variable resistor.

A

Power source

datalogger

Hall

probe

9646/03/AJC2014Prelim

Name: _________________________________ ( ) PDG: ______/13

2014 JC2 Prelim Examination PHYSICS Higher 2 9646/03 Paper 3 Longer Structured Questions Monday 15 September 2014

2 hours Candidates answer on the Question Paper. No Additional Materials are required.

READ THESE INSTRUCTIONS FIRST

Write your name and PDG in the spaces at the top of this page. 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, paper clips, highlighters, glue or correction fluid. Section A Answer all questions. Section B Answer any two questions. You are advised to spend about one hour on each 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.

For Examiner’s Use

Section A

1

2

3

4

5

6

Section B

7

8

9

Deduction

Total


This document consists of 22 printed pages and 0 blank page.

2 Data speed of light in free space, c = 3.00 x 108 m s−1 permeability of free space, 0 = 4 x 10−7 H m−1

permittivity of free space, 0 = 8.85 x 10−12 F m−1

(1/(36)) x 10−9 F m−1 elementary charge, e = 1.60 x 10−19 C the Planck constant, h = 6.63 x 10−34 J s unified atomic mass constant, u = 1.66 x 10−27 kg

rest mass of electron, me = 9.11 x 10−31 kg rest mass of proton, mp = 1.67 x 10−27 kg molar gas constant, R = 8.31 J K−1 mol−1 the Avogadro constant, NA = 6.02 x 1023 mol−1 the Boltzmann constant, k = 1.38 x 10−23 J K−1 gravitational constant, G = 6.67 x 10−11 N m2 kg−2

acceleration of free fall, g = 9.81 m s−2


3 Formulae

uniformly accelerated motion, s = ut + 12 at2

v2 = u2 + 2as work done on/by a gas, W = pV hydrostatic pressure, p = gh

gravitational potential, ϕ = −G mr

displacement of particle in s.h.m., x = x0 sin t velocity of particle in s.h.m., v = v0 cos t

v = ± ω 220 xx

mean kinetic energy of a E = 32 kT

molecule of an ideal gas, resistors in series, R = R1 + R2 + … resistors in parallel, 1/R = 1/R1 + 1/R2 + …

electric potential, V = Q

4 ε0 r

alternating current/voltage, x = x0 sin t

transmission coefficient, T exp(−2kd)

where k = 2

28

h

EUm

radioactive decay, x = x0exp(− t)

decay constant, = 0.693

21t


4


Section A Answer all the questions in this Section

1 Fig. 1.1 shows the cross-section of a symmetrical object that is partially submerged in water

and displaced from its equilibrium position. The thickness of the object is uniform throughout.

Fig. 1.1

The density of the object is uniform and is less than the density of water.

(a) On Fig. 1.1, draw arrows to show the forces acting on the object. [1]

(b) Describe the subsequent motion of the object. …...………………………………………………………………………….……………….......[1]

(c) On Fig. 1.2, draw the final equilibrium position of the object and indicate the forces acting on the object with arrows.

Fig. 1.2

[2]

(d) If the amount of water displaced by the object at equilibrium is the same as that shown in Fig. 1.1, determine the density of the object in terms of the density of water, w.

density of object = ……………………………. [2]

water

½h

h ½h

h

5

9646/03/AJC2014Prelim [Turn Over

2 (a) (i) State the first law of thermodynamics. ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [1]

(ii) Explain why, for an ideal gas, the internal energy is equal to the total kinetic energy of the molecules of the gas. ………………………………………………………………………………….................... ……………………………………………………………………………….…………....[1]

(b) A cylinder contains 1.0 mol of an ideal gas.

(i) The volume of the cylinder is constant. Calculate the thermal energy required to raise the temperature of the gas by 1.0 K. energy = …………………………….J [2]

(ii) The volume of the cylinder is now allowed to increase so that the gas remains at constant pressure when it is heated. Explain whether the thermal energy required to raise the temperature of the gas by 1.0 K is now different from your answer in (b)(i). ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….…………....[2]

(c) In order for an atom to escape completely from the Earth’s gravitational field, it must have a speed of approximately 1.1 x 104 m s-1 at the top of the Earth’s atmosphere.

(i) Estimate the temperature at the top of the atmosphere such that helium, assumed to be an ideal gas, could escape from the Earth. The mass of a helium atom is 6.6 x 10-27 kg. temperature = …………………………….K [2]

6


(ii) Suggest why some helium atoms will escape at temperatures below that calculated in (c)(i). ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….…………....[1]

3 (a) Explain what is meant by the principle of superposition.

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

(b) Light from a mercury vapour lamp is incident normally on a diffraction grating and the

diffraction pattern is observed on a screen placed 30 cm away from the diffraction grating. The positions of the first order orange spectrum of wavelength 580 nm are 20 cm apart, as shown in Fig. 3.1 below.

(i) Show that the diffraction grating has 545 number of lines per mm.

[2]

light from mercury

vapour lamp 20 cm

screen

30 cm

Fig. 3.1

first order orange

first order orange

diffraction grating

7


(ii) Fig. 3.2 shows some of the energy levels in a mercury atom. Another spectrum that was also emitted from the mercury lamp is shown by the transition A.

Calculate the wavelength of the spectrum associated with the transition A.

wavelength = ……………………………… m [2]

(iii) Hence, determine the highest order of diffracted beam that can be observed on the screen in Fig. 3.1 for the spectrum calculated in (b)(ii).

highest order = ……………………………… [2]

energy / eV

- 1.60

- 2.71

- 5.77

- 10.44

Fig. 3.2

A

8


4 A short magnet is released and falls straight through a long solenoid as shown in Fig. 4.1.

A voltage sensor and datalogger are used to measure the e.m.f induced in the solenoid.

Fig. 4.1

Fig. 4.2 shows how the induced e.m.f in the solenoid changes with time, t.

Fig. 4.2

With reference to Fig. 4.2, explain

a) why the second peak values of the induced e.m.f. is greater than the first peak value. ………………………………………………………………………………….……..................... ………………………………………………………………………………….……..................... ………………………………………………………………………………………...................... ……………………………………………………………………………………......................[2]

b) the difference in the signs of the induced e.m.f. …………………………………………………………………………………………….........…. ……………………………………………………………………………………….........………. ………………………………………………………………………………………….........……. ……………………………………………………………………………………….........…….[2]

short magnet

long solenoid

datalogger voltage sensor

t

9


5 Free neutrons (neutrons not contained within a nucleus) are unstable and decay by -emission with a half-life of 770 s.

(a) Explain what is meant by (i) the neutrons are unstable.

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

(ii) half-life. ………………………………………………………………………………….................... ……………………………………………………………………………….…………....[1]

(b) Write down a possible nuclear equation for the decay of a free neutron. ………………………………………………………………………..…………….…………....[1]

(c) Using your relationship in (b) and the following data, calculate the energy released in the decay of a free neutron.

rest mass of neutron = 1.008665 u rest mass of proton = 1.007276 u

rest mass of electron = 0.000549 u

[2]

(d) Sketch, on the axes of Fig. 5.1, the following graphs to show the variation with time of

(i) the number of undecayed neutrons and label as N, (ii) the number of -particles and label as B.

[2]

(e) Label on the axis of Fig. 5.1, the half-life of neutron. [1]

number of particles

time / s Fig. 5.1

10


6 The energy bands in an intrinsic semiconductor are illustrated in Fig. 6.1

Using band theory,

(a) describe how the electrical resistance of an intrinsic semiconductor material decreases with increase of temperature. ………………………………………………………………………………………….………….. …………………………………………………………………………………………….……….. ………………………………………………………………………………………….………….. …………………………………………………………………………………………….……….. ………………………………………………………………………………………….………….. …………………………………………………………………………………………….......... [3]

(b) suggest and explain a difference in the resistance of an intrinsic semiconductor compared to a p-type semiconductor. …………………………………………………………………………………………….……….. …………………………………………………………………………………………….……….. …………………………………………………………………………………………….……….. …………………………………………………………………………………………….……….. ………………………………………………………………………………………….………….. …………………………………………………………………………………………….......... [3]

valence band

conduction band

Fig. 6.1

11


Section B Answer two questions from this section

7 (a) (i) State Newton’s first law of motion.

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

(ii) State Newton’s second law of motion. ………………………………………………………………………………….................... ……………………………………………………………………………….…………....[1]

(iii) With a suitable definition of the unit of force, Newton’s second law can be written in the following relationship

force = mass x acceleration for a body of constant mass. Hence, together with Newton’s third law, derive the principle of conservation of momentum. ………………………………………………………………………………….................... ………………………………………………………………………………….................... ………………………………………………………………………………….................... ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….…………....[2]

(b) An alpha particle collides head-on with a stationary nitrogen-14 atom. The nitrogen atom moves off in the same direction as the approaching alpha particle with a speed of 0.005 c where c is the speed of light.

(i) Calculate the change in momentum of the alpha particle.

change in momentum = …………………………….N s [2]

12


(ii) Show quantitatively whether the interaction described in (b) is elastic in nature if the initial speed of the alpha particle is 0.02 c.

[3]

(c) Alpha particles of speed 4.41 x 106 m s-1 enter a region of uniform electric field at an angle 25 with the horizontal. The region of field is set up by two plates P and Q in a horizontal plane as shown in Fig. 7.1.

Fig. 7.1

(i) State and explain how the horizontal component of the velocity of the alpha particles would vary along the length of the plates. ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….…………....[2]

plate P

plate Q

region of uniform electric field

25 alpha particles

13


(ii) Fig. 7.2 shows how the vertical displacement of an alpha particle varies with time as it passed through the plates.

Fig. 7.2

1. With reference to the features of the graph in Fig. 7.2, state and explain the direction of the resultant force acting on the alpha particle. Ignoring the effect of gravity. .…………………………………………………………………………………............ …………………………………………………………………………………............. …………………………………………………………………………………............. .……………………………………………………………………………….……… [2]

2. Sketch on Fig. 7.3 how the vertical displacement of the alpha particles would vary with position along the length of plate Q as they travel between the plates. You may assume that the vertical displacement above plate Q to be positive.

Fig. 7.3 [2]

0 0

0.75

0.50

0.25

time

vertical displacement from plate Q / m

position


0

0.75

0.50

0.25

left edge of plate Q

right edge of plate Q

14


(iii) Using Fig. 7.2, determine the resultant force experienced by an alpha particle.

resultant force = …………………………….N [3]

(iv) The stream of alpha particles entering the region between PQ is replaced with a stream of protons with the same entry speed and angle of projection. On Fig. 7.2, sketch the variation of vertical displacement with time for a proton.

[2]

15


8 A toy car’s wheel is set up as a pendulum by hanging it vertically from a fixed support. The wheel oscillates about the fixed support as illustrated in Fig. 8.1.

The variation with displacement x of the acceleration a of the wheel is shown in Fig. 8.2.

Fig. 8.2

(a) (i) Use Fig. 8.2 to explain why the motion of the wheel is simple harmonic.

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

……………………………………………………………………………........................[3]

(ii) State a condition that must be satisfied for the oscillation of the wheel to be simple harmonic.

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

(b) The wheel in Fig. 8.1 is displaced and released at time t = 0. The oscillations of the wheel have amplitude 14.7 cm and angular frequency 2.40 rad s-1.

(i) Define angular frequency. .………………………………………………………………………………………………. ……………………………………………………………………………………………..[1]

Fig. 8.1

a

x 0

16


(ii) State an expression for the displacement x of the wheel in terms of time t. ……………………………………………………………………………………………..[1]

(iii) Use your expression in (b)(ii) to sketch the variation of the kinetic energy of the wheel EK with time t for one complete oscillation in Fig 8.3. [2]

Fig. 8.3

(c) The angular frequency value stated in (b) is calculated from the period of the simple harmonic motion. An accurate value for the period is found by timing a large number of oscillations.

(i) Explain why a large number of oscillations would help to achieve a more accurate value for the period. ……………………………………………………………………………………………….. ….………………………………………………………………………………………….[1]

(ii) Calculate the number of oscillations that are measured in a total time of 83.8 s.

number of oscillations = ………………………… [2]

(iii) A stopwatch with display provides timing to 1/100th of a second over a range of 9 hours, 59 minutes and 59.99 seconds was used to record the duration of the oscillations in (c)(ii). Explain why the duration of the oscillations in (c)(ii) is recorded to 0.1 seconds. ……………………………………………………………………………………………….. ……………………………………………………………………………………………..[1]

Ek/J

t/s 0

17


(d) In order to time the oscillations of the wheel in Fig. 8.1, a stopwatch is started when the centre of the wheel passes a marker. This marker is 2.7 cm from the equilibrium position as shown in Fig.8.4.

Fig. 8.4

(i) The wheel has a mass of 0.165 kg. Calculate the restoring force acting on the

wheel as it passes the marker.

restoring force = ………………………… N [2]

(ii) Calculate the speed of the wheel at the instant it passes the marker.

speed …………………………m s-1 [2]

equilibrium position

marker

14.7 cm

2.7 cm

18


(e) In an experiment to demonstrate resonance, the wheel in Fig. 8.1 is made to oscillate by an external periodic driving force of frequency f. Fig. 8.5 shows the variation with frequency f of the amplitude of the forced oscillations of the wheel.

Fig. 8.5

(i) Describe how the amplitude of oscillation depends on the forcing frequency.

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

(ii) Sketch in Fig. 8.5 the shape of the graph if the forced oscillations of the wheel were repeated in a vacuum. Label your sketch as A.

[2]

amplitude

f

19


9 (a) Two small charged metal spheres A and B are situated in a vacuum. The distance between the centres of the spheres is 12.0 cm, as shown in Fig. 9.1.

Fig 9.1 (not to scale)

The charge on each sphere may be assumed to be a point charge at the centre of the sphere.

Point P is a movable point that lies on the line joining the centres of the spheres and is distance x from the centre of sphere A. The variation with distance x of the electric field strength E at point P is shown in Fig. 9.2.

Fig 9.2

12.0 cm

P sphere A sphere B

20


(i) Define electric field strength. …………………………………………………………………………………..................... ……………………………………………………………………………….………….....[1]

(ii) State the evidence provided by Fig. 9.2 for the statements that 1. the spheres are conductors,

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

2. the charges on the spheres are both positive.

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

(iii) An -particle is placed at x = 4.0 cm on the line joining the centres of the spheres.

Describe what will happen to the -particle?

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

(b) Fig. 9.3 shows two charged parallel plates which are 12.0 cm apart. A flame probe

connected to an electroscope or electrostatic voltmeter can be used to measure the electric potential in the space between the plates. The probe is a tiny gas flame which makes the air near the probe conduct. By moving the probe about, between the plates, the equipotential surfaces can be mapped out. The dashed lines A, B, C, D and E represent some equipotential surfaces. (Not to scale)

Fig. 9.3

- 40 V

- 10 V

- 35 V

- 30 V

- 25 V

- 20 V

- 15 V

A

B

C

D

E

X

Y

21


(i) Define electric potential.

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

(ii) Suggest how the flame probe enables air between the plates to conduct electricity.

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

(iii) On Fig. 9.3, draw lines to show the electric field between the plates. [2]

(iv) On Fig. 9.4, sketch the variation of the electric potential, V, at different points along the line XY where X and Y are midpoints of the parallel plates. Label the positions of equipotential lines A, B, C, D and E clearly on your graph.

Fig. 9.4 [2]

V/ V

d / cm 0

22


(v) 1. A proton, which is initially at rest at P as shown in Fig 9.5, is moved within the electric field. Complete the table below by stating the work done in moving the proton along each of the following path.

Fig 9.5

[3]

Path taken Work done / eV

P → Q

Q → R

P → Q → R → P

2. The proton, back at position P, is released from rest. Describe the motion of the proton. ………………………………………………………………………………….............. ………………………………………………………………………………….............. .……………………………………………………………………………….…….…[2]

3. Give a quantitative explanation as to why gravitational force due to Earth is not considered when predicting the motion of the proton between the plates.

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

- 40 V

- 10 V

- 35 V

- 30 V

- 25 V

- 20 V

- 15 V

A

B

C

D

E

X

Y

P

R

Q

30o8.0 cm

14.0 cm

2014 AJC H2 Phy Prelim Solutions Paper 3 (80 marks)

1a

b Object rotates anticlockwise as weight of object and upthrust form a couple which causes an anticlockwise moment. OR Object will rock back and forth.

c - Weight and upthrust will lie along the same vertical axis along the geometric centre of the object when in equilibrium.

- Figure is drawn symmetrical as seen below and the 3 points A, B and C lie at the water surface.

d By principle of floatation and Archimedes’ principle, Upthrust = weight of object = weight of fluid displaced obj gVobj = w gVw where Vw is the volume of object submerged in water obj = w Vw / Vobj = w(0.5h2t)/ (0.75h2t) where t is the thickness of object = 0.67w

2ai The increase in internal energy of a system is equal to the sum of heat supplied to the

system and the work done on the system.

aii Internal energy is the sum of total kinetic energy and potential energy of the molecules of gas. For ideal gas, no intermolecular forces so no potential energy (only kinetic). Therefore, internal energy of ideal gas = total kinetic energy of molecules.

bi ∆U = Q + W, since W = 0, Q = ∆U = ∆Ek = (3/2)Nk ∆T ( or (3/2)nR∆T) Thermal energy required Q = (3/2) × 1.38 × 10–23 × 1.0 × 6.02 × 1023 = 12.5 J

bii Additional energy is needed to push back piston /do work against atmosphere besides the increase in internal energy of gas. So thermal energy required is greater.

weight

upthrust

A

B

C

weight

upthrust


ci Ek = (3/2) kT ½ x 6.6 x 10-27x(1.1x104)2 = (3/2)x1.38x10-23xT T = 1.93 x 104 K

cii At a given temperature, helium atoms have a range of kinetic energy around the mean value. Some helium atoms have kinetic energy higher than the mean kinetic energy of a temperature below that in (c)(i). These atoms have a velocity more than 1.1 x104 m s-1 and hence they have sufficient energy to escape.

3a The Principle of Superposition states that when two waves meet at a point, the

resultant displacement is equal to the vector sum of the individual displacements. Note:

- paraphrasing the question statement is not accepted (i.e. “when two waves superpose” is not acceptable)

- use of “amplitude” instead of “displacement” is not accepted.

3bi Angle of diffraction of the first order = tan-1 (10/30) = 18.43° Using d sinθ = nλ, we have d = (1)(580 x 10-9) / sin(18.43°) = 1.834 x 10-6 m Hence, number of lines per mm = 1 mm / 1.834 x 10-6 = 545 lines [shown]

3bii Using E = hf, we have [-2.71 – (-5.77)] x 1.6 x 10-19 = 6.63 x 10-34 x 3.0 x 108 / λ λ = 4.0625 x 10-7 = 4.06 x 10-7 m

3biii To determine the highest order, sinθ ≤ 1. Hence, using d sinθ = nλ, nλ / d ≤ 1 n ≤ 1.834 x 10-6 / 4.0625 x 10-7 = 4.51 Therefore, highest order = 4 e.c.f. for wrong values of λ in b(i) and b(ii) that are of the same order as light (i.e.10-7 m).

4a Due to the acceleration of free fall, the speed of magnet leaving the solenoid is higher

than when it is entering the solenoid. Hence, the rate of change of magnetic flux linkage with solenoid is greater when the magnet is leaving the coil, leading to a larger second peak value of the induced e.m.f. according to Faraday’s Law.

b When the magnet is entering the solenoid, the magnetic flux linkage with solenoid is increasing but as it is leaving the magnetic flux linkage is decreasing. By Lenz’s law, the induced e.m.f acts in a direction such that it tends to produce effects to oppose the change causing it. Hence the induced e.m.f have opposite signs when entering and leaving the solenoid.


5ai It means neutrons will spontaneously and randomly disintegrate/decay into smaller particles.

ii It is the average time taken for the activity of a particular radioactive nuclide to fall to half its initial value.

b Hen 11

01

10

c Mass defect = (mass of initial – mass of product) = (1.008665 u - 1.007276 u - 0.000549 u) = 8.4 x 10-4 u Energy released = (mass of initial – mass of product)c2 = (1.008665 u - 1.007276 u - 0.000549 u) c2 = 1.25 x 10-13 J

di, ii e

6a An intrinsic semiconductor has a small energy gap

(about 1eV) which decreases as temperature increases.

As temperature increases, more electrons from the valence band are excited to become free electrons in the conduction band while leaving more holes in the valence band.

Although there is also an increase lattice vibrations of the ions but this effect is outweighed by the large increase in the number of charge carriers (free electrons and holes).

Hence the electrical resistance of the intrinsic semiconductor will decrease.

b A p-type semiconductor has an acceptor energy level

very close to the valence band.

Hence it is easier for the electrons from the valence band to be thermally excited to the acceptor level, leaving behind more positive holes in the valence band.

Due to the increase in the positive charge carriers (holes) for the p-type semiconductor, p-type semiconductor has a lower resistance than intrinsic semiconductor.

770 (half-life)

number of particles

time /s

B

N

N0

½N0

Band theory

Band theory

(Describe how free electrons & holes are formed)

(Describe the band structure)

(Describe how more holes are formed)

(Describe the band structure)


7ai Newton’s first law states that a body stays at rest or continues to move at constant

velocity unless a resultant force acts on it. ii Newton’s second law states that the rate of change of momentum of a body is

proportional to the resultant force acting on the body and takes place in the direction of the resultant force.

iii From Newton’s third law, if body A exerts a force on body B, then body B will exert a force of the same type that is equal in magnitude and opposite in direction on body A. Hence, (ma)A = -(ma)B where m represents mass and a represents acceleration By Newton’s second law, mA(vA-uA)/t = - mB(vB-uB)/t For the same duration of interaction, the gain in momentum by body A is equal to the lost in momentum by body B. This shows principle of conservation of momentum.

bi By COM, lost in momentum of alpha = gain in momentum nitrogen = 14 u (0.005 c) = 3.486 x 10-20 = 3.49 x 10-20 N s

bii By COM, 4u(0.02c) + 0 = 4uva + 14u(0.005c) final speed of alpha, va = 0.0025 c (after collision, alpha continues in its original dirn) Relative speed of approach = ua – un = 0.02 c Relative speed of separation = vn – va = 0.005 c – (0.0025 c) = 0.0025 c Since relative speed of approach is not equal to relative speed of separation, the interaction is not elastic.

ci As there is no net horizontal force acting on the alpha particles, Hence, horizontal component of velocity will remain unchanged along the length of the plates.

cii1. From the gradient of the graph, velocity decreases till zero and then increases in the opposite direction. OR the vertical distance covered per second decreases and then increases. Hence, there is a resultant force acting vertically downwards towards plate Q.


cii2.

ciii yyy asuv 222

)70.0(a2)25sin10 x 4.41(0 26 a = 2.481 x 1012 m s-2 F = ma = 4(1.66 x 10-27) (2.481 x 1012) = 1.647 x 10-14 = 1.6 x 10-14 N

civ Since charge of proton is half that of alpha, net force on proton is half of alpha. Since mass of proton is a quarter of alpha, net acceleration of proton is 2 times that of alpha. Proton would have half the maximum vertical displacement of alpha. With the same initial speed and doubled acceleration, proton takes half the time of alpha, to reach maximum height. Hence range of proton is half of the range of alpha. Path of proton should be symmetrical as same net force acts on it throughout the region of field.

0 0

0.75

0.50

0.25

position


time

vertical displacement from Q/ m

0.75

0.50

0.25


8ai SHM is the motion of a body such that its acceleration is proportional to its displacement from a fixed point and is always directed towards that point. The oscillations are simple harmonic because: 1. The straight line passes through the origin indicating that the acceleration a of the

plate is directly proportional to the displacement x from its equilibrium position 2. The negative gradient of the graph indicates that its acceleration a is opposite in

direction to its displacement x, and is directed towards the equilibrium position. Note: acceleration and displacement not inversely proportional

ii The angle of oscillation must be small.

bi Angular frequency of a body undergoing simple harmonic motion is a constant of a given oscillator and is related to its natural frequency f by f 2

ii )40.2cos(147.0 tx OR )40.2cos(147.0 tx OR )40.2cos(7.14 tx OR )40.2cos(7.14 tx

iii

- Correct shape of cos2 function - 2 cycles within one period with Ek max at T/4, 3T/4 and Ek = 0 at 0, T/2, T.

ci To reduce fractional error of time measurement.

ii

s 6180.240.2

22

T

number of oscillations = 326180.2

8.83 (round to lower integer)

iii As the average human reaction time is between 0.2 s and 0.4 s, the timing for the

oscillations is recorded to the same precision as the human reaction time.

T


di

(3sf) N 0257.0

0270.040.2165.0 2

2

xmmaF

ii 1st method

22220 027001470402 ... xxv

= 0.347 m s-1 (3sf) 2nd method

1-20 ms 347.0)40.2sin()107.14(40.2sinv

s 57753.0

)40.2cos(7.147.2

cos

ttx

t

t

txx o

ei The amplitude of oscillation increases when f increases to natural frequency f0. When f is larger than f0, the amplitude decreases. Amplitude is maximum when the frequency is equalled to f0. (Learning point: f0 = 1/T or /2 = 2.40/2 = 0.382 Hz)

ii - greater values of amplitude for all values of f - the peak is shaper and at the right of f0

9ai Electric field strength at a point is defined as the force per unit positive charge acting

on a stationary charge placed at that point.

ii1. Electric field strength inside spheres is zero

2. There is a point between the spheres, where electric field strength is zero.

iii The particle will move towards sphere B and oscillate about x = 8.0 cm

bi The electric potential at a point is defined as the work done per unit positive charge in moving a point charge from infinity to the point.

amplitude

f f0 = 0.382 Hz

A


ii Heat from the flame creates ions in the air. These ions then act as charge carriers.

Iii -within plates vertical parallel lines with equal spacing pointing towards upper plate -near edge of plates, field perpendicular to potential lines.

iv

Shape of graph is correct Points A, B, C, D & E are labeled at the correct positions

v1.

Path taken Work done / eVP → Q -5 Q → R 15 P → Q → R → P 0

2. Proton moves upward in a vertical straight path towards -40 V plate with increasing speed at constant acceleration.

3. Correct values of FG = 1.6 x 10-26 N and FE = 4.0 x 10-17 N Since FE is about 109 times larger than FG, hence gravitation force is not taken into account.

anderson junior college -...

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