JC2 Preliminary Examination 2015score-in-chemistry.weebly.com/.../jjc_2015_prelim.pdfJJC 2015 9646/JC2 Prelim Exam P1/2015 [Turn Over 2 Data speed of light in free space, c = 3.00

JJC 2015 9646/JC2 Prelim Exam P1/2015 [Turn Over

JURONG JUNIOR COLLEGE JC2 Preliminary Examination 2015

Name Class 15S

PHYSICS Higher 2

Multiple Choice

Additional Materials: Multiple Choice Answer Sheet Soft clean eraser Soft pencil (type B or HB is recommended)

9646/1

18 September 2015

1 hour 15 min

READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name and class in the spaces provided at the top of this page.

Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name, class and index number on the Answer Sheet in the spaces provided. There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate

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 booklet.

(This question paper consists of 23 printed pages)


2

Data

speed of light in free space, c = 3.00 108 m s1

permeability of free space, o = 4 107 H m1

permittivity of free space, o = 8.85 1012 F m1 = (1/(36)) 109 F m1

elementary charge, e = 1.60 1019 C

the Planck constant, h = 6.63 1034 J s

unified atomic mass constant, u = 1.66 1027 kg

rest mass of electron, me = 9.11 1031 kg

rest mass of proton, mp = 1.67 1027 kg

molar gas constant, R = 8.31 J K1 mol1

the Avogadro constant, NA = 6.02 1023 mol1

the Boltzmann constant, k = 1.38 1023 J K1

gravitational constant, G = 6.67 1011 N m2 kg2

acceleration of free fall, g = 9.81 m s2

Formulae

uniformly accelerated motion, s = ut + 12

at2

v2 = u2 + 2as

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

hydrostatic pressure, p = gh

gravitational potential, =

Gm

r

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

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

v = 2 2( )ox x

mean kinetic energy of a molecule of an ideal gas

E = 32

kT

resistors in series, R = R1 + R2 + . . .

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

electric potential, V = o

Q

ε r4

alternating current / voltage, x = xo sin t

transmission coefficient, T exp(2kd)

where k = 2

2

8 ( )m U E

h

radioactive decay x = xo exp(-λt)

decay constant λ =

1/2

0.693

t


3

1 Four students A to D were measuring the potential difference across a piezo-electric material. Each student obtained 5 sets of data for the potential difference.

If the actual potential difference is 1.65 102 V, which student’s measurement was accurate but not precise?

Potential difference / 102 V Mean / 10-2 V

A 1.61 1.65 1.65 1.58 1.61 1.62

B 1.69 1.70 1.69 1.70 1.69 1.69

C 1.66 1.68 1.63 1.64 1.65 1.65

D 1.65 1.64 1.65 1.65 1.66 1.65

2 A boat changes its velocity from 8 m s-1 due north to 6 m s-1 due east.

What is its change in velocity?

A 2 m s-1 at a direction of 37° east of north

B 2 m s-1 at a direction of 53° east of north

C 10 m s-1 at a direction of 37° east of south

D 10 m s-1 at a direction of 53° west of south

3 A marble rolls off the edge of a horizontal table with a speed of 2.0 m s -1. The edge of the table is 1.2 m above the floor.

What is the horizontal distance the marble is away from the edge of the table when it hits the floor?

A 0.49 m B 0.99 m C 1.9 m D 2.3 m


4

4 A wire W, initially of length L, is fixed at one end and pulled by a force F at the other end. The wire extends by x.

Three wires, all identical with wire W, are connected as shown to form a composite wire.

The composite wire is fixed at one end and pulled by force 2F at its other end.

What is the total extension of the composite wire?

A 3

2x

B 2x C 8

3x

D 3x

5 A small block of wood, of density 400 kg m−3, is held fully submerged in water. What is the initial acceleration of the block towards the surface when the block is released?

(Take the density of water to be 1000 kg m .)

A 1.5 m s-2 B 9.8 m s-2 C 15 m s-2 D 25 m s-2

6 When a 1000 kg car and a 10000 kg truck collide head-on, why is the passenger on the car more likely to experience a bigger force by the seat belt?

A The car experiences a bigger force than the truck.

B The passenger on the car experiences a larger impulse.

C The car undergoes a larger change of velocity than the truck.

D The force on the car is the same as the truck.


5

7 A student plotted a time-force graph as shown below.

Which of the following correctly show how change in velocity can be determined between time t = 0 s to t1?

A Gradient of the graph m

B

Area A

m

C AreaB

m

D AreaB m

8 A man intends to position a box by giving a push and letting go. He misjudges his push and the box goes only one-third of the way to the intended position.

If the initial speed of the box after the push is vo and the resistive force acting on the box

is constant, what should the initial speed of the box be for it to slide to the intended position?

A 2 ov

B 3 ov

C 6 ov

D 3 ov


6

9 A car of mass m is accelerated horizontally from rest to a speed v by a constant force F.

How much work is done on the car during this acceleration?

A 1

2Fv

B Fv

C mv2

D 21

2mv

10 The diagram below represents a cyclist making a left turn on a rough road surface at constant speed v, as viewed from behind. The total mass of the bicycle and rider is m

and their combined centre of gravity is at G.

If R is the resultant force of the normal reaction and the frictional force, f which vector

diagram below represents the directions of the forces acting on the bicycle and its rider?

A B

C D

G

R

mg

R

mg

2mv

r

2mv

r

R

mg

R

mg

f


7

11 A simple pendulum is released from rest at A. What is the tension in the string when the pendulum is at position B (where the string is vertical), given that the mass of the bob is m and the length of the pendulum is L?

A 0.87 mg

B 1.0 mg

C 1.3 mg

D 2.0 mg

12 The moon orbits the Earth once every 27.3 days, with a mean orbital radius of R. The

period of an Earth satellite with an orbital radius of 30

R is

A 0.17 hours

B 4.0 hours

C 68 hours

D 260 hours


8

13 P is a planet with centre O, as shown. X and M are two points of equal gravitational potential. Y and N are two other points of equal gravitational potential.

Which of the following statement(s) is/are incorrect?

I The work done by an external agent to move a mass from Y to X is negative.

II The work done by the gravitational field to move a mass from X to N is the same as that needed to move the same mass from M to N.

III The gravitational potential at X is smaller than the gravitational potential at Y.

IV

gravitational potential at X distance OX

gravitational potential at Y distance OY

A I, II

B II, III

C III, IV

D IV only

14 Which statement describes the electric potential difference between two points in a wire?

A The force required to move a unit positive charge between the two points.

B The ratio of the energy dissipated to the current flowing between the two points.

C The ratio of the power dissipated to the electric charge flow between the two points.

D The ratio of the power dissipated to the current flowing between the two points.


9

15 Two wires made of the same material and length are connected in parallel to the same voltage supply. Wire P has a diameter of 2 mm. Wire Q has a diameter of 1 mm.

What is the ratio current in P

current in Q?

A 14

B 12

C 2 D 4

16 In the circuit shown below, ammeter 1 and ammeter 2 have negligible resistance. The

source has internal resistance 0.5 and ammeter 1 reads 6 A.

Determine the reading of ammeter 2.

A 1.0 A B 2.7 A C 3.3 A D 5.0 A

17 A potentiometer consists of a 1.00 m long resistance wire XY in series with a battery of e.m.f. E1 9.00 V and internal resistance r 1.42 Ω. The resistance of wire XY is 8.30 Ω.

Determine the e.m.f. E2 if the balance length l is found to be 0.745 m.

A 0.636 V B 2.72 V C 5.73 V D 9.00 V


10

18 A fixed mass of ideal gas undergoes cyclical changes in pressure and volume as shown in the diagram below.

Which row of the table correctly describes the net work done on the gas and the heat supplied to the gas for one cycle?

Net work done on the gas Heat supplied to the gas

A Zero Negative

B Positive Negative

C Positive Positive

D Negative Positive

19 A fixed mass of an ideal gas absorbs 1000 J of heat and expands under a constant external pressure from an initial volume of 20 cm3 to a final volume of 50 cm3. The increase in the internal energy of the gas is 400 J.

Determine the constant external pressure.

A 13 MPa B 20 MPa C 33 MPa D 53 MPa


11

20 A mass hanging from a spring suspended from the ceiling is pulled down and released. The mass then oscillates vertically with simple harmonic motion of period T. The graph shows how its distance from the ceiling varies with time t.

What can be deduced from this graph?

A The amplitude of the oscillation is 70 cm.

B The kinetic energy is a maximum at t = T

2.

C The restoring force on the mass increases between t = 0 and t = T

4.

D The speed is a maximum at t = T

4.


12

21 The figure below shows two simple pendulums X and Y having the same lengths and same bob sizes and a simple pendulum Z whose length can be varied. (The density of polystyrene is less than the density of copper.)

The pendulum X and pendulum Y will oscillate after the pendulum Z is set into oscillations in a plane parallel to the wall.

Which of the graphs below shows the variation of the amplitude a for the pendulum X

and pendulum Y with frequency f of the pendulum Z when the length of pendulum Z is varied?

A

B

C

D

0 f

a

Y

X

0 f

a

X

Y

f

a

Y

X

Y X

a

f

Y


13

22 A sound wave of frequency 3600 Hz propagates from left to right through a gas. The diagram below shows the positions of some gas molecules at a particular instant of time.

The distance between particles P and Q is 0.500 m.

Which one of the following statements is true?

A The gas is not dry air at room temperature.

B The speed of the gas is 1800 m s-1.

C The pressure at P is always below atmospheric pressure.

D The sound wave is stationary.

23 An air column open at one end and closed at the other end is at resonance.

Which one of the following statements is always true?

A There is an odd number of nodes.

B There is an even number of nodes.

C The total number of nodes and antinodes is an odd number.

D The total number of nodes and antinodes is an even number.

24 In a Young’s double slit experiment, the intensity at the centre of the fringe pattern is I. If

one of the two identical slits is now closed, the intensity at the centre of the pattern is

A I

B

2

I

C

2

I

D

4

I


14

25 The arrows on the diagrams represent the movement of the air molecules in a pipe in which a stationary longitudinal wave has been set up. The length of each arrow represents the amplitude of the motion, and the arrow head shows the direction of motion at a particular instant.

Which of the following diagrams shows a possible stationary wave that could be set up in the pipe?

A II only

B III only

C II and IV

D I, II and IV

26 An electron travelling at constant speed enters a uniform electric field at right angles to the field. While the electron is in the field it accelerates in a direction which is in the

A same direction as the electric field.

B opposite direction to the electric field.

C same direction as the motion of the electron.

D opposite direction to the motion of the electron.


15

27 A +5.0 C point charge is initially stationary at a point P in a uniform electric field of strength 4.0 Vm−1. It is then moved 2.0 m parallel to the field and 3.5 m perpendicular to the field to a point Q as shown in the diagram below.

What is the work done by the electric field?

A 70 J

B 40 J

C 40 J

D 70 J

28 An electron is moving in air at right angles to a uniform magnetic field. The diagram below shows the path of the electron.

Which one of the following correctly gives the speed of the electron and the direction of the magnetic field?

Speed Direction of magnetic field

A decreasing into plane of paper

B increasing out of plane of paper

C increasing into plane of paper

D decreasing out of plane of paper


16

29 Three concentric coils X, Y and Z that are equally spaced apart, carry the same current in the directions as shown in the diagram.

Which of the following statement on coil X is correct?

A The directions of the forces acting on X are outwards, away from the centre of X.

B The directions of the forces acting on X are inwards, towards the centre of X.

C Coil X experiences forces that cause it to start rotating.

D Coil X experiences forces perpendicular to the plane of X, into the diagram.


17

30 A circuit containing a circular loop of wire connected to a low power light bulb is positioned around a solenoid connected to a sinusoidal AC source and a diode as shown in the diagram below.

Which of the following statements is true?

A The light bulb does not light up at all because the diode prevents current from flowing and thus producing any magnetic flux in the solenoid.

B The light bulb does not light up at all because the magnetic flux linkage through the loop does not reverse its direction.

C The light bulb lights up because the magnetic flux linkage through the loop reverses its direction every cycle.

D The light bulb lights up because the magnetic flux linkage through the loop varies with time.


18

31 The magnetic flux linkage through a coil varies with time as shown.

Which graph shows the variation with time of the e.m.f. generated by the coil?

A

B

C

D


19

32 The figure below shows the variation with time of a periodic current.

What is the root-mean-square value of the current?

A 3.35 A

B 3.75 A

C 3.97 A

D 4.50 A

33 A pure resistor has an alternating supply connected across it. The frequency of the supply is varied while the r.m.s. voltage is kept constant.

How is the mean rate of heat dissipated across the resistor related to the frequency?

A proportional to (frequency)2

B proportional to frequency

C proportional to (frequency)½

D Independent of frequency

34 The de Broglie wavelength of a particle that has kinetic energy K is λ. The wavelength λ is proportional to

A 1

2K

B 1K

C 1

2K

D 2K


20

35 Some energy levels for a hydrogen atom are shown in the diagram.

In the spectra below, the frequency scale is linear and increases to the right.

Which spectrum best corresponds to the possible transitions when the electrons of an excited hydrogen atom fall to the -3.41 eV level?

A

B

C

D


21

36 The diagram below shows the wave function ψ(x) of an electron.

Which statement about the electron is false??

A The probability of finding the electron is the lowest at O.

B The probability of finding the electron is zero at Q and R.

C The probability of finding the electron is higher at S than at P.

D The probability of finding the electron between P and S is less than one.

37 Which diagram correctly shows the three possible processes: excitation, stimulated emission and spontaneous emission between the energy levels in a 4-level laser system?

A

ground

metastablespontaneous

emission

spontaneous

emission

photonexcitationstimulated

emission

B

ground

metastable

spontaneous

emission

spontaneous

emission

excitation

photonstimulated

emission

C

ground

metastable

spontaneous

emission

spontaneous

emissionphoton

excitationstimulated

emission

D

ground

metastable

spontaneous

emission

spontaneous

emission

excitation photonstimulated

emission


22

38 Which of the following statements below on intrinsic semiconductors is true?

A The total current flow is the sum of both ‘hole’ and ‘electron’ currents.

B There are more electrons in the conduction band than there are holes in the valence band.

C The valence band is completely filled and the conduction band is partially filled.

D The valence band is completely filled and the conduction band is empty at room temperature.

39 Radiation from a radioactive source enters an evacuated region in which there is a uniform magnetic field directed into the plane of the diagram. This region is divided into two by a sheet of aluminium 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 Y

B alpha Z

C beta Y

D beta Z


23

40 To initiate nuclear fission on a nucleus, a particle is usually needed to knock onto the nucleus. The particle is suggested to be a

A neutron since the mass is comparable to that of the nucleons.

B neutron since it has no electrical interaction with the nucleons.

C proton since there will be strong nuclear forces with the nucleons.

D proton since the mass is comparable to that of the nucleons.

End of Paper

Page 1 of 4

JURONG JUNIOR COLLEGE PHYSICS DEPARMENT

JC2 Preliminary Examination 2015 H2 Physics Paper 1 solutions

Qn Ans Suggested solution (Font Arial, 11 pt)

1 C Student A’s results: mean value = 1.62 x 10-2 V, spread = 0.07 x 10-2 V Student B’s results: mean value = 1.69 x 10-2 V, spread = 0.01 x 10-2 V Student C’s results: mean value = 1.65 x 10-2 V, spread = 0.05 x 10-2 V Student D’s results: mean value = 1.65 x 10-2 V, spread = 0.02 x 10-2 V

The actual potential difference is 1.65 102 V. Hence, student C’s results are accurate (low systematic errors => mean value close to actual value) but not precise (high random errors => large spread).

2 C Change in velocity v = vf - vi = vf + (-vi)

3 B 2 21 12 2

1.2 9.81 0.495y y ys u t a t t t s

212

2.0 0.495 0.99x x x xs u t a t s m

4 D For spring in parallel, they share the load. Total extension is 2x+ x= 3x

5 C Resultant force = Upthrust – weight ρwVwa = ρfVwg - ρwVwg (400)a= 1000(9.81)- 400 (9.81) a = 14.7 m s-1

6 C Newton’s 3rd law will prove option A and B wrong. Option D is a correct statement but does not show that the passenger in the car will experience a bigger force. The car and thus the passenger will experience a larger change in velocity and the car has a much smaller mass.

7 B Area A will correctly depict the change in momentum under the F-t graph. Thus change in

velocity will be area A/ mass.

8 B

2

0

2

1

2 2 2

1 0 1 1 0

1Work done against friction =

2

1Required work done against friction = 3 3

2

1 1 13 3( ) 3

2 2 2

F D mv

F D FD mv

FD mv mv mv v v

9 D Work done on the car = Gain in KE = ½ mv2

10 C Centripetal force is the resultant force and should not be drawn in the free body diagram.

11 D

2 2

2 2

Loss in GPE = Gain in KE

1( - cos60 )

2

At point B, 2

omg L L mv mv mgL

mv mv mgLT mg T mg mg mg

R R L

6 m s-1 (vf)

8 m s-1 (-vi)

v

Hence, |v| = 2 26 8 = 10 m s-1

= tan-1 (8/6)

= 53

Angle = (90-53) east of south

Page 2 of 4




12 B

2 3

2 3

Using ,

27.3( ) ( ) 0.166 day 4.0 hour

30

T R

RR

T R

13 D Only statement (IV) is wrong.

gravitational potential at X distance OY

gravitational potential at Y distance OX

14 D By definition.

15 D Since the diameter of P is two times the diameter of Q, its cross-section area will be four times, its resistance will be one quarter and its current will be four times.

16 B Let the current through the 3 resistor be 1 part.

The current through the 1 resistor will be 3 parts.

The current through the 2 resistor will be 1.5 parts.

Ammeter 2 will read 2.5 parts, out of 5.5 parts, of the 6 A current 2.5

6 2.75.5

A

17 C Current in XY

9.000.926

1.42 8.30

A

p.d. across XY = (8.30)(0.926) = 7.69 V E2 = (0.745)(7.69) = 5.73 V

18 D For a full cycle, work done by gas (when expanding, XY and YZ) exceeds work done

on gas (when contracting, ZX) so net work done on gas W is negative.

For a full cycle, there is no change in internal energy so U is zero.

Using U = W + Q, the heat supplied Q is positive.

19 B 6 7400 1000 (30 10 ) 2.0 10 Pa = 20 MPaU Q W Q W p p

20 D At time t =0s, object is at the extreme end of its SHM. At time t = T/4, object is at equilibrium position. At time t = T/2, object is at the other extreme end. Since speed is maximum at equilibrium position, mass has maximum speed at time t = T/4.

21 A Pendulum Y (having less mass) will experience more damping than Pendulum X. (Air resistance has a greater effect on a smaller mass. Even though the same resistive force acts on both objects, the object with a larger mass will experience a smaller deceleration due to the resistive force.)

22 A B: The speed of the gas is (0.5/2)(3600) = 900 m s-1. C: The pressure at P is NOT always below atmospheric pressure. D: The sound wave is progressive.

23 D Factual Question

24 D When both slits are opened, 2 2( ) 4I k A A kA

When only one slit is opened,

2' / 4I kA I

Page 3 of 4




25 C Nodes are formed at closed end of a tube while anti-nodes are formed at the open end of a tube. At a node, the neighbouring air molecules are either all moving away from it or all moving towards it.

26 B The direction of electric force on the electron is opposite to the direction of the electric field. The electron accelerates in the same direction of electric force, which provides the resultant force (N2L). Hence, the electron accelerates in direction opposite direction to the electric field.

27 B Displacement of 2.0 m is in opposite direction to direction of electric force. Hence work done by electric field must be negative. Magnitude of work done = Electric force x displacement = (4.0)(5.0) x 2.0 = 40 J

28 D Using Fleming’s LHR, the magnetic field acting on the electron is out of the paper.

Using 2mv

Bqvr

The speed decreases as the radius of the path decreases. 29 A X will experience a repulsive force due to Y and an attractive force due to Z. As Y is closer,

the repulsive force is greater, thus X will experience a force away from its centre.

30 D When the diode is forward-biased, current flows through the solenoid. When the diode is reverse-biased, no current flows through the solenoid. Hence the current through the solenoid varies periodically leading to a magnetic flux linkage through the circular loop of wire that varies with time.

31 C By Faraday’s Law : E =

dt

dBNA

dt

d

E dt

dB (i.e. e.m.f. gradient of B-t graph)

32 C

r.m.s. current =

2 2(6 )(1) (3 )(3)

4

= 3.97 A

33 D Mean power <P> = Vr.m.s.2/R

Since r.m.s. voltage is kept constant, mean power does not change (i.e. independent of frequency).

34 A

2

22

Sub =2

2

h pp K

m

h

mK

into

Page 4 of 4




35 C

Photons emitted from transition 1 have a larger frequency than photons emitted from transition 4. The spacing between lines in the line spectrum is proportional to the difference in energy gaps. Hence the 2 lines representing transitions 2 and 3 must be closer to line for 1 than that for 4.

36 A The probability of finding the electron is the square of the amplitude of its wave function so the probability of finding the electron is the highest at O.

37 D The higher energy level for lasing transition (stimulated emission) must be in metastable state to give rise to net emission of photons. Hence only C is correct.

38 A B: A hole is formed in the VB for every electron thermally excited into the CB. Hence the number of electrons must be equal to the number of holes. C: The VB is also partially filled due to the presence of holes. D: Electron-hole pairs are formed even at room temperature.

39 C Alpha particles would have been stopped by the sheet of aluminium. Hence A and B are wrong. Since magnetic force provides the centripetal force,

2mvBqv

r

mvr

qB

The beta particles would lose energy after going through the aluminium sheet and their velocities would decrease. Smaller velocity results in a smaller radius of circular motion. The path closer to Z is of smaller radius and point of entry must be at Y.

40 B It is preferable to minimise any repulsive force between this incoming particle and the positive nucleus.

1 2 3 4

JJC 2015 9646/JC2 Preliminary Exam P2/2015 [Turn Over

JURONG JUNIOR COLLEGE JC2 Preliminary Examination 2015

Name Class 15S

PHYSICS Higher 2

Structured Questions

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

9646/2

2 Sep 2015

1 hour 45 mins

READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name and class in the spaces provided at the top of this

page. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Do not use paper clips, highlighters, glue or correction fluid. 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.

For Examiner’s Use

1

2

3

4

5

6

7

8

9

Total



2

Data














Formulae


at2

v2 = u2 + 2as




Gm

r



v = 2 2( )ox x


E = 32

kT




Q

ε r4



where k = 2

2

8 ( )m U E

h


decay constant λ =

1/2

0.693

t


3

1. A zoologist studying the wildlife in Africa decided to model the thigh bone of a male African Elephant as a physical pendulum pivoted at the hip joint of the elephant.

The period of the physical pendulum can be expressed as

IT=2π

mgh

where T is the period of oscillation,

I is the rotational inertia of the pendulum about the pivot which has a unit of kg m2,

m is the mass of the thigh bone,

g is the acceleration of free fall and

h is the distance of the centre of gravity of the bone from the pivot.

(a) Show that the equation is homogeneous in terms of base units. [1]


4

(b) If the value of I is (0.2085 0.0001) in kg m2, and m and h are measured to be

(4.1 0.5) kg and (47.0 0.1) cm respectively, calculate T and express it with its associated uncertainty.

T = ( ) s [4]

2 (a) Define electric potential at a point.

[1]


5

(b) Two charged spheres of radius 2.0 mm are held 4.00 cm apart (centre-to-centre). Fig. 2 shows the variation with displacement x, of the electric potential V, between the two charged spheres.

V / V

x / cm

Fig. 2

Position of charged

sphere


6

Using Fig. 2, determine

(i) the magnitude of the electric field strength at x = 1.00 cm and state its direction,

magnitude of the electric field strength = N C-1 [2]

direction of electric field strength is [1]

(ii) the charge on each sphere.

charge on each sphere = C [2]

3. Fig. 3 shows a region PQRS of a uniform magnetic field directed downwards into the plane of the paper.

Fig. 3


7

Electrons, all having the same speed, enter the region of the magnetic field.

(a) On Fig. 3, show the path of the electrons as they pass through the magnetic field, emerging from side QR. [2]

(b) A uniform electric field is also applied in the region PQRS so that the electrons now pass undeflected through this region. On Fig. 3, mark, with an arrow labelled E, the direction of the electric field. [1]

(c) The undeflected electrons in (b) each have charge - e, mass m and speed v. State

and explain the effect, if any, on particles entering the region PQRS of the same magnetic and electric fields as in (b) if the particles each have

(i) charge - e, mass m and speed 2v,

[2]

(ii) charge +e, mass m and speed v.

[2]


8

4 In a photoelectric effect experiment, the maximum kinetic energy Ek,max of the ejected

photoelectrons is measured for various wavelengths, of the incident light.

Fig. 4 shows the variation with wavelength , of the incident light, of Ek,max.

Fig. 4

(a) (i) From Fig. 4, state and explain one piece of evidence that supports the theory that light has a particulate nature.

[2]

Ek,max / eV

/ nm 0 50 100 150 200 250

5.0

10.0

15.0

20.0


9

(ii) Describe briefly how the maximum kinetic energy of the emitted electrons may be measured experimentally.

[3]

(b) Using a suitable point from Fig 4, determine the work function of the metal used.

work function = eV [3]

5 A junction is formed between slices of p-type and of n-type semiconductor material, as

shown in Fig. 5.

Fig. 5

(a) On Fig. 5, draw an arrow to show the direction of movement of holes as the two slices are brought into contact. [1]

p-type

material

n-type

material


10

(b) Describe the origin of the depletion region at the junction.

[4]

(c) On Fig. 5, draw the symbol for a battery connected so as to increase the width of the depletion region. [1]

6 (a) The tip of a scanning tunnelling microscope (STM) is brought close to but does not touch the surface to be scanned. A small current flows between the tip and the surface. Explain briefly why there is a current using the concept of quantum tunnelling.

[2]


11

(b) An electron is moving in the x-direction with a speed of 2.05 106 m s-1. Given that the precision in measuring the speed of the electron is 0.50 %, determine the minimum uncertainty with which the position of the electron along the x-direction can be simultaneously determined.

minimum uncertainty of electron’s position = m [2]


12

7 Many unstable parent nuclei undergo radioactive decay, a random and spontaneous

process, to give a daughter product which is itself radioactive. This may give rise to a

radioactive series where there may be ten or more different radioactive daughter

products.

The variation with time t of the number N of different nuclei in a radioactive sample is

illustrated in Fig 7. The parent nucleus P decays to a daughter nucleus D, which in turn,

produces a further daughter S which is stable as shown in the following process below.

P D + 0

-1e S + 4

2He +

Initially, there are 50 x 105 nuclei of P in the sample.

(a) (i) On Fig. 7, label with the letter S the line representing the variation with t of the number of nuclei N of the stable further daughter S. [1]

Fig. 7

(ii) Determine the half-life of product D.

half life = years [2]

N / 105

50

40

30

20

10

0

0 50 100 150 200 250 300 350 t / year


13

(iii) State the time at which the activity of the daughter product D is maximum.

time = years [1]

(iv) Calculate the maximum activity of the daughter product D.

maximum activity = yr-1 [2]

(v) Explain why the number of nuclei of the daughter product D increases initially.

[1]

(vi) The relative activities of the parent isotope and the daughter products may be used as a means of determining the age of the sample.

Suggest why this technique may provide reliable results for ages up to about 300 years but for an age of about 1000 years, the method would be far less reliable.

[1]

(b) Describe qualitatively the direct effects of the emitted alpha particles on cells.

[2]


14

8. The hull is the main body of a boat. The load waterline length LWL refers to the

horizontal length of a hull at the water’s surface when a boat is carrying a normal load. It is a significant factor in establishing how fast a boat can go.

As a boat first begins to move, rather small waves are generated at the bow – the front of the boat – with a pattern of these small waves moving back along the side of the boat, reaching the stern – the back of the boat. As the boat moves faster, the wavelength of the bow wave increases.

Hull speed vhull is the speed of the boat such that the wavelength of the bow wave of the boat is equal to its load waterline length LWL.

If the boat moves faster than the hull speed vhull, the wavelength of the bow wave

becomes even longer and the stern of the boat will begin to fall into the trough so that the boat will be angled upwards as shown in Fig 8.1.

Fig 8.1

(a) (i) On Fig. 8.1, label clearly the load waterline length LWL of the boat at hull speed vhull.

[1]

(ii) Give two reasons why it is inefficient for the boat to move above its hull speed vhull.

Reason 1:

Reason 2:

[2]


15

(b) The variation with LWL of vhull for a typical boat is shown in Fig. 8.2.

Fig 8.2

The relationship between vhull and LWL is thought to follow the expression

vhull =

1

nWLkL

where k is a constant and n is an integer.

6.0 4.0 8.0 10.0 12.0 14.0

LWL / m

vhull / m s1

16.0

2.5

3.0

3.5

4.0

4.5

5.0

2.0


16

Data from Fig. 8.2 are used to obtain values of lg vhull and lg LWL. These are plotted on the

graph of Fig. 8.3.

(i) Use Fig. 8.2 to determine lg vhull for LWL = 12.0 m.

lg vhull = [1]

(ii) On Fig. 8.3,

1. plot the point corresponding to LWL = 12.0 m,

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

lg (LWL / m)

lg (vhull / m s1)

Fig. 8.3

0.7 0.6 0.8 0.9 1.0 1.1 1.2

0.40

0.45

0.50

0.55

0.60

0.65

0.35

0.70


17

(iii) Use the gradient of the line drawn in (b)(ii)2 to determine n.

n = [2]

(iv) Using your answer in (b)(iii), determine the magnitude of k.

magnitude of k = [2]

(v) Determine the SI base units for k.

SI base units = [1]


18

(c) The Formidable-class frigates are the latest boats to enter into service with the Republic of Singapore Navy. They have a load waterline length of 0.11 km.

(i) Determine the hull speed of a Formidable-class frigate.

hull speed = m s-1 [1]

(ii) Suggest why the Formidable-class frigate is designed to achieve a higher maximum speed than its hull speed.

[1]

(d) Large ships such as cargo ships and oil tankers often have a protruding bulb at the

bow of the ship just below the waterline, known as a bulbous bow. A bulbous bow is shown in Fig. 8.4, and can generate waves similar to the bow of the ship.

Fig. 8.4 Use the principle of superposition, suggest how bulbous bows help ships to achieve

better fuel efficiency than similar vessels without them.

[1]

bow of ship

bulbous bow


19

9 As a bar magnet is dropped through a coil, an e.m.f. is induced in the coil. The maximum e.m.f. E is induced as the magnet leaves the coil with speed v. It is suggested that E is directly proportional to v.

Design a laboratory experiment to the relationship between E and v. You should draw a diagram showing the arrangement of your equipment. 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 relationship between E and V is determined from your readings

(e) any precautions that would be taken to improve the accuracy and safety of the

experiment.


20

Diagram


21


22

End of Paper

Page 1 of 10

JURONG JUNIOR COLLEGE

PHYSICS DEPARTMENT 2015 JC2 Preliminary Examination

9646 H2 Physics Paper 2 Suggested Solutions with Markers’ Comments

Qn Suggested solution (Font Arial, 11 pt) Remarks 1(a)

For an equation to be homogenous, all the terms in the equation must have the same base unit.

Units of T = s

Units of 2I

mgh =

2

2

2

kgm

kgms m

s s

Since units of T = Units of 2I

mgh , the equation is homogeneous.

[1] – correct

working for base units

of 2I

mgh

(b) 1 1 1

( ) ( ) ( )2 2 2

T I m h

T I m h

= ½ (0.0001/0.2085) + ½ (0.5/4.1) + ½ (0.1/47.0) = 0.0623

Mean value of period, T = 2

0.20852

(4.1)(9.81)(47.0 10 )x

= 0.660 s

Hence, absolute uncertainty in period, T = (0.0623)(0.660) = 0.04 s

Therefore, (0.66 0.04)T s

[1] correct

expression [1] ans for T [1] ans for

T (allow ecf) [1] final ans

(allow ecf)

2(a) Work done per unit positive charge in moving a small charge from infinity to that point.

[1]

Many students did not include “positive” charge.

Page 2 of 10

JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT

2015 JC2 Preliminary Examination 9646 H2 Physics Paper 2

Suggested Solutions with Markers’ Comments

Qn Suggested solution (Font Arial, 11 pt) Remarks (b)(i)

gradient of graph

dVE

dr

E

2

2.50 0.00=

0.00 2.50 10

-1100 N C

Acceptable range: 90 110E

positive x-direction / towards right / from left to right

[1] sub [1] answer [1] direction

For those who managed to give the correct working, many students did not convert the unit for the change in displacement.

(b)(ii) At x = 0.20 cm (sample point),

0 04 4 0.040

Q QV

x x

0

1 16.00

4 0.002 0.040 0.002

Q

121.27 10 CQ

[1] working [1] ans

Almost all students have no idea the effect on potential or electric field strength due to 2 charges and how to use the graph to solve this question.

3(a)

[1] curve down [1] straight path after exit

Curved DOWN

Straight line

(outside RQ)

Page 3 of 10




Qn Suggested solution (Font Arial, 11 pt) Remarks

Quite a number of students did not draw the path outside the field or the path is not straight. Many students also drew a path upwards.

(b) Towards the bottom of the paper

[1] e.c.f

(c)(i)

For undeflected electrons in the cross field, B

Ev

If the particles now have vnew = 2v, the magnetic force downwards (FB=Bqv) is doubled. Since the electric force (FE=qE) remains unchanged, there is a resultant force downwards. Hence the charge will be deflected downwards (towards SR).

[1] [1]

While many acknowledged an increase in FB, they did not remark on the FE being unchanged.

(c)(ii) The charge is now +e. The magnetic force is now upwards, while the electric force is downwards (direction of forces are opposite). However, since m, v and e remains the same, the 2 forces are equal in magnitude. Hence, the charge particle will pass through undeflected.

[1] [1]

Many did not mention about the magnitude of the two forces being equal and thus there is no resultant force.

4(a)(i) For every material of C irradiated, there is a maximum wavelength beyond which no electron would be emitted from the surface.

Since the energy of a photon is expressed as hf, hence the minimum photon energy required to liberate an electron from its surface must correspond to a maximum wavelength or minimum frequency.

[1] [1]

Poorly answered as many just described the relationship of KEmax and the wavelength using the graph.

(a)(ii) Apply negative potential at the collector and slowly increase potential difference until photo-current drops to zero.

The potential difference when the photo-current just reduces to zero is the stopping potential Vs.

and the maximum kinetic energy is determined by Ek, max = eVs.

[1] [1] [1]

Quite a number of students did not mention about applying negative potential to the collector with respect to the emitter.

Page 4 of 10




Qn Suggested solution (Font Arial, 11 pt) Remarks (b)

-34 8-19

-9

-19

(6.63 10 )(3.0 10 )= + (20.0 1.60 10 )

50 10

= 7.78 10 J = 4.86 eV

k,max

hc= + E

λ

[1] read off correctly from graph [1] sub [1] ans

A few students read the max. wavelength to be 250 nm.

5(a) and

(c)

(c) A significant number did not draw their answers on the figure but used the space below or did not draw a complete circuit with a proper symbol for a cell.

(c) [1] for correct connection of battery and complete circuit (a) [1] for arrow pointing from p- towards n-type

(b) Diffusion of any particles tends to be from a region of higher concentration to a region of lower concentration.

Holes (highly concentrated in the p-type region) would diffuse across the junction to the n-type region, and electrons (highly concentrated in the n-type region) tend to diffuse across the junction to the p-type region.

As the holes and electrons diffuse across the junction in opposite directions, most of them meet and combine and the region becomes virtually depleted of mobile charge carriers.

The junction electric field (due to the positive immobile ions in the n-type region and the negative immobile ions in the p-type region) also prevents further diffusion of the holes and electrons.

The majority focused on the migration of electrons only and did not mention the reason for the migration of electrons and holes or the occurrence of electron-hole combinations.

[1] [1] [1] [1]

A number of students explained how p-type and n-type semiconductors are formed. This is not necessary.

p-type

material

n-type

material

Page 5 of 10





6(a) In quantum mechanics, an electron has a wave function whose amplitude decreases exponentially through a potential barrier. Therefore there is a non-zero wave function on the other side of the potential barrier. Since the probability of finding an electron is the square of the amplitude of its wave function, there is a non-zero probability that the electron can be found on the other side of the potential barrier. (This current is the tunnelling current, due to the process of quantum tunnelling.)

[1] [1]

Many students did not explain that electron has wave nature and hence has a wave function. Many students just explained how quantum tunneling is applied to STM. They are not answering the question.

(b)

34

31 6

9 9

4

6.63 100.005 9.11 10 2.05 10

4

5.7 10 m (2 sig. fig.) or 5.65 10 m (3 sig. fig.)

x

hp x

x

x

[1] Sub [1] Ans

Students were very careless. E.g. they did not include the mass of electron; they treat 0.5% as 0.5.

7(a)(i)

[1] for label

(a)(ii) N falls from 20 x 105 at t = 125 yr to 10 x 105 at t = 200 yr Half-life = 200 – 125 = 75 yr

[1] working [1] ans

Many students calculated the half-life wrongly when the parent nuclides were still decaying. The correct period to calculate the period is after 120 years.

(a)(iii) 45 years [1]

(a)(iv)

4

1/2

5 -1

0.693=

0.693A = 34×10 = 3.14×10 yr

75

A N

A Nt

[1] sub [1] ans

Many students calculated the maximum gradient of the graph for the daughter nuclide. This is not accurate as this graph is the net value of the daughter nuclides (i.e. which is formed from P to D and decayed from D to S) The maximum gradient of the graph for the stable nuclide S is acceptable as the rate of formation of S is the rate of decay of D.

(a)(v) The number of product D increases when the rate of formation is higher than the rate of decay of D.

[1]

(a)(vi) After 1000 years, the activity will be too low to be detectable [1]

Many students just indicated the numbers of nuclides are very small, they should link to low activity of these nuclides which made it not reliable for analysis.

S

Page 6 of 10





(b) The alpha particles create ions which break the bonds within the DNA and cause mutations, which lead to

creation of tumour cells or cell death.

[1] [1]

8(a)(i)

[1] - LWL

represents the wavelength

(a)(ii) Reason 1:

Above hull speed, the bow will be tilted upwards and the driving force of the boat engine is no longer horizontal but directed at an angle above the horizontal.

Reason 2:

The drag force is directly proportional to the square of the speed and work done against drag force increases.

[1] – Ans

[1] - Ans

(b)(i) vhull = 4.35 m s1

lg vhull = 0.638

[1] - Ans

Most students were able to score this mark. A few students wrongly read off Fig. 8.3 instead of Fig. 8.2. for the value of lg(vhull).

LWL

Page 7 of 10

(b)(ii)1./

(b)(ii)2.

[1] – correct point plotted [1] – line of best fit ignoring anomalous data point

Many students are still unable to draw a good best-fit line. Also, many did not ignore the anomalous point when plotting the best-fit.

(b)(iii) Gradient = 1/n =

0.690 0.4200.500

1.180 0.640

n = 2

[1] – Relevant working [1] - Ans

Quite a few students do not know how to linearize the equation to obtain the graph in Fig. 8.3. Most students did not read the question carefully and hence did not leave the value of n as an integer.

(b)(iv) lg vhull = (1/n) lg LWL + lg k

Using (0.640, 0.420),

0.420 = 0.5 (0.640) + lg k

lg k = 0.100

k = 1.26

[1] – Relevant working [1] - Ans

lg (LWL / m)

lg (vhull / m s1)

Fig. ?.3

0.7 0.6 0.8 0.9 1.0 1.1 1.2

0.40

0.45

0.50

0.55

0.60

0.65

0.35

0.70

(1.180, 0.690)

(0.640, 0.420)

Page 8 of 10

Quite a significant number of students have problems manipulating logarithmic equations. Quite a number were able to write down the substitution of values correctly but still calculated the final answer for k wrongly. A few students went to log the values read off from the graph again (which were already log-values to begin with!). Quite a few students also went to read off the vertical-intercept value directly from the graph, even though the x-axis did not begin from zero.

(b)(v) Base units of k =

1

0.500

m s

m

= m0.500 s-1

[1] - Ans

Majority of students lost this mark. Many assumed that the unit of k is the same as the unit of the y-axis. Many attempted to but were still unable to solve this question by considering the base units on both sides of original equation.

(c)(i) vhull = k LWL1/n = 1.26 1100.5 = 13.2 m s1

[1] - Ans

(c)(ii) Warships are designed for combat operations. Having the highest possible maximum speed has priority over moving efficiently.

[1] – Relevant Ans

(d) The bulbous bow generates a wave that superposes destructively with the bow wave and reducing the amplitude of the bow wave. The drag due to the bow wave is reduced, increasing fuel efficiency.

[1] – Relevant Ans

Many students erroneously thought that the wave generated by the bulbous bow moves in the opposite direction from the bow wave, hence setting up a stationary wave.

9 Diagram: (A maximum of 1 mark: Labelled diagram showing magnet falling vertically through coil)

magnet

c.r.o

Height

Page 9 of 10

Objective

To investigate how the e.m.f, E generated by coil varies with speed, v

1. Independent variable: speed, v

2. Dependent variable: e.m.f, E

3. Controlled variables (A maximum of 1 marks) :

a. the initial speed of the magnet

b. the initial position of the magnet

c. the orientation of the falling magnet (i.e. falling vertically)

Instruments/ Methods/ Additional formulas (A maximum of 4 marks on the method/ instrument on collecting the independent and dependent variables)

1. The e.m.f of the coil can be measured using a c.r.o. connected to it or voltage sensor connected to datalogger.

Also accept:

using a voltmeter connected in parallel to coil

2. Calculate v using v = √(2gh)

Also accept:

v = 2h/t (for height method)

v = L/t for length of magnet or coil and by stopwatch, timer or lightgate(s) connected to datalogger.

v = gt for timing fall to bottom of coil

3. Vary speed of magnet by changing height.

4. Use metre rule to measure distance magnet falls to the bottom of the coil or metre rule to measure length of coil or metre rule to measure length of the magnet.

Also accept:

Allow timing instrument to measure the time of the fall from the start to the bottom of the coil

Procedures (A maximum of 2 marks on basic procedure of collecting repeated data for dependent and independent variables)

1. Set up the apparatus as shown above.

2. Measure, record the height and calculate the speed of the magnet.

3. Release the magnet.

4. Measure and record the e.m.f of coil.

5. Repeat steps 2 to 5 for different values of height and hence their corresponding speed and e.m.f for 6 sets of readings.

Analysis (1 mark)

Flat coil of wire

Page 10 of 10

1. Assume that the e.m.f. E and speed, v are related by the equation:

E = k vn ,

(where k and n are constants)

2. Taking lg on both sides, lg(E) = n lg(v) + lg (k).

3. Plot a graph of lg (E) against lg (v)

4. If the above relationship is true, a straight line graph will be obtained where the gradient is equal to n and the y-intercept is equal to lg(k).

Safety Precautions (A maximum of 1 mark)

1. Keep away from falling magnet/use sand tray/cushion to catch magnet

Other Additional details (A maximum of 2 marks)

1. Use coil with large number of turns/drop magnet from large heights/strong magnet

2. Detailed use of datalogger/storage oscilloscope to determine maximum E; allow video

camera including slow motion play back

3. Use of short magnet so that v is (nearly) constant

4. Use short/thin coil so that v is (nearly) constant

5. Use a non-metallic vertical guide/tube

6. Method to support vertical coil or guide/tube

7. Repeat experiment for each v and average

Do not allow vague computer methods.

Diagram: Equipment or set-up must be supported (and not seen to be hanging or suspended in the air).

More details must be provided if more sophisticated equipment such as data logger or high speed camera is used. E.g. the types of sensors or probes to be used with the data logger, how we can determine the velocity using info gleaned from images or playback from the high speed camera (which cannot indicate the velocity directly).

Adhere to the template/structure given by tutors to get most marks with the least words. E.g. no need to draw the table for data, explain instruments to be used for measurements separately from the procedures.

Procedures: focus on two main components. 1. List steps to measure and determine the 2 variables for 1 set of data. (no need to repeat instuments used if they have been listed separately) 2. “Repeat steps ? to ? by varying height and determine the corresponding induced emf and velocity for 6 sets of data.”

Need to read question carefully and be clear about the objective. In this case, to determine relationship between induced emf and velocity i.e. how to determine k and n from the graph. Many candidates took it for granted that induced emf is proportional to velocity.

Many of the safety precautions listed are trivial and overly generic to demonstrate evidence of thought.

JJC 2015 9646/JC2 Prelim Exam P3/2015

JURONG JUNIOR COLLEGE 2015 JC2 Preliminary Examination

Name Class 15S

PHYSICS Higher 2

Longer Structured Questions

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

9646/03

14 September 2015

2 hours

READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name and class in the spaces provided at the top of this

page. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Do not use highlighters, glue or correction fluid. Section A Answer all questions. Section B Answer any two 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.

For Examiner’s Use

1

2

3

4

5

6

7

8

Total



2

Data














Formulae


at2

v2 = u2 + 2as




Gm

r



v = 2 2( )ox x


E = 32

kT




Q

ε r4



where k = 2

2

8 ( )m U E

h


decay constant λ =

1/2

0.693

t


3

Section A

Answer all the questions in this Section.

1 (a) Define acceleration.

[1]

(b) A train driver Anthony spots another train, its tail 100 m ahead of him on the same track. Anthony’s train is moving at a speed of 72 km h-1 and the other train is moving at a constant speed of 54 km h-1 in the same direction.

(i) State the maximum speed of Anthony’s train, in m s-1, in order to avoid a collision at the time his train has closed the gap of 100 m between the two trains.

speed of Anthony’s train = m s-1 [2]

(ii) Sketch the distance-time graph of Anthony’s train. [1]


4

(ii) Determine the minimum deceleration Anthony’s train must have in order to just avoid a collision.

minimum deceleration = m s-2 [3]

2 (a) Define work done by a force.

[1]

(b)

Fig. 2.1 shows two blocks A and B connected by a light inextensible cord passing over a frictionless pulley. When both blocks are released, block A starts to move from rest along a rough plane which is inclined at 30o to the horizontal. The speed of block B just before hitting the ground is 4.0 m s-1

Calculate the average friction force acting on block A.

Fig. 2.1

2.0 kg

6.0 kg

5.0 m 30o

A B


5

frictional force = N [4]

3 (a) Fig. 3.1 shows a particle, of mass 0.30 kg, being suspended and rotated from a position A by an inextensible string of length of 2.0 m.

(i) In Fig. 3.1, draw the resultant force acting on the particle when it is at position C. [1]

(ii) Determine the minimum kinetic energy with which the particle should be projected from position B, in order for the string to be just taut at point D (i.e. highest position).

kinetic energy = J [3]

Fig. 3.1

D

A C

B

2.0 m


6

(b) A binary star system consists of two stars with masses M1 and M2 as shown in Fig. 3.2. The centres of the two stars are separated by a distance R = 1.2 x 1010 m.

The diagram is not drawn to scale.

Fig. 3.3 shows the variation of the gravitational potential ϕ with distance x from the centre of star M1.

A particle is launched with kinetic energy Ek from the surface of star with mass M2. The particle arrives at the surface of the star of mass M1.

- 0.5

- 0.7

- 0.9

- 1.1

- 1.3

ϕ / x 1012 J kg-1

x / x 109 m 4.0 2.0 6.0 8.0 10.0 0

Fig. 3.3

Surface of star with mass M1

Surface of star with mass M2

Fig. 3.2

R

x

M1 M2


7

(i) Explain whether the kinetic energy of the particle at the surface of M1 is less than, equal to, or larger than Ek,

[2]

(ii) 1. Determine the distance x at which the gravitational field strength due to the two stars is zero.

x = m [1]

2. Hence determine the ratio 1

2

M

M.

ratio = [3]

4 (a) Define specific heat capacity.

[1]


8

(b) A cake of mass 0.90 kg is baked in an oven at a temperature of 180°C. It is then taken out of the baking tin onto a rack to cool in a kitchen of 25°C. The oven of volume 0.10 m3 also cools down from 180°C to 25°C.

(i) Take the specific heat capacity of the cake to be 990 J kg-1 K-1.

Calculate the energy released from the cake in cooling.

energy released = J [2]

(ii) The pressure in the oven remains constant at the atmospheric pressure of 1.01 x 105 Pa. Assume the air behaves as an ideal gas. The molar mass of air is 0.030 kg mol-1.

Calculate the change in the mass of air in the oven between the two temperatures.

change in the mass of air = kg [3]


9

(iii) Calculate the root-mean-square speed of the air molecules at 25°C.

root-mean-square speed = m s-1 [2]

5 (a) (i) Define electromotive force of a source.

[1]

(ii) Distinguish between electromotive force and potential difference.

[2]

(b) Fig. 5.1 shows a circuit containing a thermistor and three fixed resistors of values

100 , 120 and 200 . The voltmeter connected between A and B has infinite resistance. The battery has e.m.f. 6.00 V and negligible internal resistance. The

voltmeter reads zero when the ambient temperature is 25 C.


10

Fig. 5.1

(i) State the resistance of the thermistor when the voltmeter reads zero.

resistance = [1]

(ii) As the temperature rises above 25 C, the voltmeter starts to register a reading. Deduce whether the potential at A is higher or lower than the potential at B.

[3]

(iii) Determine the two possible values for the resistance of the thermistor when the voltmeter registers a reading of 1.0 V.

resistance = or [3]


11

Section B

Answer two questions from this Section.

6 (a) State Newton's second and third laws of motion.

Newton’s second Law:

Newton’s third Law:

[4]

(b) Define force.

[1]


12

(c) Using sketches, with labelled arrows showing the directions of velocity v and acceleration a, describe situations in which an object

(i) has an acceleration at right angles to its velocity,

(ii) has an acceleration in the opposite direction to its velocity.

Add to your sketches a labelled arrow showing the direction of the resultant force acting on the object in each case. [4]

(d) (i) State the principle of conservation of momentum.

[1]

(ii) Does the principle apply in cases where two colliding objects lose kinetic energy as a result of sticking to one another at the point of collision?

Explain your answer with reference to Newton's third law.

[2]


13

(e) A particle of mass 1.20 kg collides head -on and elastically, with a ball of mass 0.60 kg moving with a speed 0.2 m s-1 in opposite direction. The ball moves off with a speed of 0.1 m s-1. Calculate for the particle

(i) its initial speed,

initial speed = m s-1 [4]

(ii) its final speed.

final speed = m s-1 [2]

(f) Suggest, with a reason, if the equations used in (e) would apply if the collision took place on a rough surface.

[2]


14

7 (a) Wave–particle duality is the concept that all matter and energy exhibit both wave-like and particle-like properties. An electron diffraction tube shown in Fig. 7.1 can be used to show the wave nature of particles. Electrons are accelerated from rest at the filament towards the target by a potential difference of 4.5 kV.

When the electrons pass through a graphite target, an interference pattern is observed on the screen as shown in Fig. 7.2. The first-order maximum of the

interference pattern occurs at an angle of 10 from the straight-through position.

(i) State the meaning of diffraction.

[1]

(ii) Show that the de Broglie wavelength of the electrons is approximately 1.8 x 10-11 m. [2]

(iii) Estimate the separation of the atoms in the graphite.

separation = m [2]

Fig. 7.1

4.5 kV

Fig. 7.2


15

(iv) Suggest why if the electrons are accelerated from rest at the filament towards the target by a potential difference much lower than 4.5 kV, no interference pattern is observed on the screen.

[2]

(b) The following experiment is set up to determine the frequency of a vibrating dipper. This vibrating dipper causes a water wave of small amplitude to travel in a tank of water d = 2.6 cm deep, as shown in Fig. 7.3.

The mean speed v of the travelling water wave in shallow water is dependent on

both d and the acceleration of free fall g and is given by v gd .

Fig. 7.3

(i) Calculate the frequency of the vibrating dipper, given that the distance

between 2 adjacent crests of the water wave is 0.025 m.

frequency = Hz [2]

(ii) Explain why the water wave needs to have a small amplitude.

[1]

vibrating dipper

d = 2.6 cm

water tank

C

direction of wave travel


16

(c) Two coherent sources A and B, which are in anti-phase with each other, emit wave

of wavelength 40.0 mm. (Not drawn to scale)

The amplitude of the wave from source B is twice that of source A.

A detector is placed at the point P where it is 1.00 m from A and 1.18 m from B as shown in Fig. 7.4.

Fig. 7.4

(i) State the type of electromagnetic radiation that is emitted by sources A and B.

radiation is [1]

(ii) Suggest a suitable distance between sources A and B.

distance = m [1]

(iii) State and explain whether constructive interference or destructive interference takes place at P.

[2]

(iv) Determine the ratio of the intensity at P to the intensity at O.

ratio = [2]


17

(ii) State how wind instruments A and B differ from wind instrument C, in terms of

the type of air column they possess.

[1]

(iii) Sketch the waveform of the 1200 Hz sound wave produced in the air column

of instrument B. Your answer should include a sketch of the air column of instrument B.

[1]

(iv) Sketch the waveform of the 1200 Hz sound wave produced in the air column of instrument C. Your answer should include a sketch of the air column of instrument C.

[1]

(d) Different musical instruments do not sound the same even if the same note is played. Fig. 7.5(a), (b) and (c) show the frequencies produced when the same note is played by wind instruments A, B and C, respectively.

f/ Hz

rela

tive

am

plit

ud

e

f/ Hz

rela

tive

am

plit

ud

e

500 1000 1500 2000 500 1000 1500 2000

Instrument A Instrument B

Fig. 7.1(a) Fig. 7.1(b)f/ Hz

rela

tive

am

plit

ud

e

500 1000 1500 2000

Instrument C

Fig. 7.1(c)

(i) State one similarity between the note produced by the three wind instruments.

[1]

Fig. 7.5(a) Fig. 7.5(b) Fig. 7.5(c)


18

8 (a) (i) State what is meant by simple harmonic motion.

[2]

(ii) A student ties a pendulum bob to a piece of long string and uses the pendulum to perform small oscillations. The position of the bob at 0.10 s intervals is shown in Fig. 8.1.

Fig. 8.1

Fig. 8.1 shows exactly one half-cycle of the bob’s motion.

If the motion is simple harmonic, determine the maximum speed vmax of the

bob.

vmax = m s-1 [3]


19

(b) A protective device in a mains circuit consists of a transformer with two primary coils A and B and a secondary coil, as shown in Fig. 8.2. The primary coils each

have the same number of turns and are wound in opposite directions on the core.

Fig. 8.2

The mains supply is connected in series with the two primary coils and the electrical appliance. The secondary coil is connected to an electromagnetic switch.

(i) State Faraday’s law of electromagnetic induction.

[2]

(ii) 1. At one particular moment, the live lead is positive with respect to the neutral lead.

On Fig. 8.2, mark arrows to indicate the direction of the magnetic field in the transformer core due to the primary coil A alone (label this

arrow A), and the direction of the magnetic field in the transformer core due to the primary coil B alone (label this arrow B). [1]

2. Hence state and explain if there is any e.m.f. induced in the secondary coil.


20

[3]

(c) Fig. 8.3 shows a region of magnetic field having a uniform flux density of 2.0 x 10-4 T and directed out the paper. A wire coil placed at P is in the plane of the

paper and away from the field. The coil has 200 turns, a total resistance of 2.0 and an area of 10 cm2.

The coil is moved from P to R in 0.20 s.

Fig. 8.3

(i) Calculate the amount of charge which flows in the coil.

charge = C [2]

(ii) State and explain if there would be any change in the amount of charge if the number of turns in the coil is increased.

[2]


21

(d) An ideal transformer has 5000 turns in its primary coil. It is used to convert a mains supply of 230 V to an alternating voltage having a peak value of 12.0 V.

(i) Calculate the number of turns in the secondary coil.

number of turns = [2]

(ii) The secondary coil is connected in series with a resistor R.

The variation with time t, in seconds, of the potential difference at the secondary coil is given by the expression

V = 12.0 sin(380t)

1. Determine the frequency of the supply.

frequency = Hz [1]

2. To prevent overheating, the mean power dissipated in R must not exceed 300W. Calculate the minimum resistance of R.

minimum resistance = [2]

End of Paper

Page 1 of 8

JURONG JUNIOR COLLEGE

PHYSICS DEPARTMENT 2015 JC2 Prelims 9646 H3 Physics


Qn Suggested solution Remarks

1(a) Acceleration is defined as the rate of change of velocity. [1]

MC

(b)(i) In order not to collide into the train in front at the time Anthony’s train close the 100 m gap, his train must be moving at less than or 54 km h-1.

(54)(1000)15

3600

m s-1

[1] deduce [1] convert

MC

(b)(ii)

[1] convex shape (finite, non-zero gradients at both ends)

MC

(b)(ii) Let the time taken be t.

In time t, distance moved by the other train = 15t

and the distance moved by Anthony’s train =

20 15

2t

In order to just avoid a collision,

20 1515 100

2t t

t = 40 s

minimum deceleration

50.125

40

m s-2

[2] Any correct method [1] Ans

MC

2 (a)

Work done by a force is the product of the force and the displacement in the direction of the force.

[1]

(b)

2

o 2

Lose in GPE = Gain in GPE Gain in KE Work done against friction

1m gh = m gh + (m m )v Work done against friction

2

1Work done against friction (6 9.81 5) (2 9.81 5sin30 ) (8 4 )

2

B A A B

B B A A A b

181.25

Average frictional force displacement = 181.25

181.25Average frictional force = 36.25 36.3 N

5

[1] for eqn

based on COE. [1] for

expression for work done against friction [1] sub [1] ans

Page 2 of 8


2015 JC2 Prelims 9646 H3 Physics



3 (a)(i)

[1]

(ii)

2

2

2

D

B D

mvAt highest point, T + mg

r

To be taut and at min speed,

mvT = 0 mg =

r

1 1 minimum KE mv = mg(2) = mg

2 2

KE = minimum KE Gain in GPE = mg + mg(4)

= 5 mg = 5(0.3)(9.81) = 14.7J

[1] for eqn for

centripetal force [1] for min. kinetic energy at point D or any correct working [1] ans

(b)(i) The kinetic energy will be greater than Ek.

The change in gravitational potential energy from the surface of M2 to M1 is negative and hence by conservation of energy, there is a gain in kinetic energy

[1] [1]

(ii) 4.8 x 109 m (allow 4.7 to 4.9 x 109 m) [1]

(iii)

9

1 2

1 2

2 2

2 9 2

1

2 10 9 2

2

At x = 4.8 10 ,

( )

(4.8 10 ) 0.444

( ) (1.2 10 4.8 10 )

g g

GM GM

x R x

M x

M R x

(Allow 0.415 to 0.476)

[1] for correct

equation [1] for correct

substitution [1] for correct answer

4(a) The specific heat capacity of a substance is defined as the thermal energy per unit mass of the substance required to cause a unit change in temperature.

[1]

MC

(b)(i) [Q = mc]

Energy released = (0.90)(990)(180 – 25) = 138105 = 138 kJ

[1] Sub [1] Ans

MC

(b)(ii) Let mi and mf be the initial and final masses of the air.

5

5

[ ]

(1.0 10 )(0.10) ( )(8.31)(180 273.15) 0.07967 kg0.030

(1.0 10 )(0.10) ( )(8.31)(25 273.15) 0.1211 kg0.030

ii

ff

pv nRT

mm

mm

Change in mass = 0.1211 – 0.07967 = 0.0414 kg

[1] initial mass [1] final mass [1] Ans

Page 3 of 8





MC

(b)(iii) 2 312 2

2 23 1 1312 223

[ ]

0.030( ) (1.38 10 )(27 3.15 25) 498 m s6.02 10

rms

rms rms

mc kT

c c

[1] Sub [1] Ans

MC

5(a)(i) The electromotive force of a source is defined as the non-electrical energy converted to electrical energy per unit charge delivered through the source.

[1]

MC

(a)(ii) Electromotive force is defined for a source whereas potential difference is defined across any two points in an electric circuit. Electromotive force defined using non-electrical energy converted to electrical energy whereas potential difference is defined using electrical energy converted to non-electrical energy.

[1] [1]

MC

(b)(i) 240 [1] Ans

MC

(b)(ii) As temperature rises, the resistance of the thermistor decreases. The potential difference across the thermistor decreases. Therefore the potential at A will be lower than the potential at B.

[1] [1] [1]

MC

(b)(iii) Assume the negative terminal of the battery is earthed (i.e. 0 V). Therefore the positive terminal of the battery will be at 6 V.

This means point A of the circuit will be at 100

6100 200

2 V.

Point B will therefore be 3 V or 1 V. The corresponding p.d. across the thermistor will therefore be 3 V or 5 V.

With point B at 3 V: 3 6 120120

TT

T

RR

R

With point B at 1 V: 5 6 600120

TT

T

RR

R

The corresponding resistance of the thermistor will be therefore be 120 or 600

.

[2] if 1 correct answer [3] if 2 correct answers

MC Accept any other correct methods. Also accept correct deductions without working.

6(a) The rate of change of momentum of an object is directly proportional to the resultant force acting on it. The change in momentum takes place in the direction of that force. When object A exerts a force on object B, then object B exerts a force of the same type that is equal in magnitude and opposite in direction on object A.

[2] [2]

(b) Force is proportional to the rate of change of momentum.

[1]

Page 4 of 8





Almost all students did not give force correctly as proportional to the rate of change of momentum, even though we still award the mark based on their notes.

(c)(i) In uniform circular motion, the acceleration of the object is always perpendicular to the velocity. This is known as the centripetal acceleration.

[1] for correct

situation [1] for correct

diagram

Many students drew the F diagonally to a and v. Many also did not describe the situation in words.

(c)(ii) An object thrown up will have a velocity UPWARDS, but the acceleration due to the gravitational pull is DOWNWARDS.

[1] for correct situation [1] for correct

diagram

Surprisingly, many students drew the resultant force and acceleration in opposite direction.

(d)(i) The Principle of conservation of momentum states that the total momentum of a system of interacting bodies is constant provided no external resultant force acts on it.

[1]

Many students did not mention “external”.

(d)(ii) Yes, the law will still apply. The force on the ball by the particle and the force on the particle by the ball are equal and opposite and thus there is no resultant force acting on the system. Despite the system losing kinetic energy, the total momentum of the system is still conserved.

[2]

Poorly attempted, and there is no link to zero resultant external force to the system.

(e)(i) Using C.O.M:

1.2 0.6( 0.2) 1.2 0.6(0.1)

1.2 1.2 0.18 (1)

u v

u v

Using relative speed of approach = relative speed of separation:

0.2 0.1 (2)u v

Solving (1) and (2) u = 0.025 m s-1

[1]

[1] [1] for sub [1] for ans

Many forgot about the negative sign for the direction to the left and many still use COE to craft out their 2nd equation and yet could not solve it.

(ii) Solving (1) and (2) v = -0.125 m s-1

[1] for sub [1] for ans

a

v

F

a v

F

Page 5 of 8





(f) The equations used in (e) will not apply as friction is an external force acting on the system and thus the total momentum before and after the interaction is not constant.

[1] [1]

Many mentioned since there is an external force, Principle of COM does not apply. Actually it still does.

7(a)(i) Diffraction is the spreading of the waves as they pass through an obstacle or aperture.

[1] - ans

(a)(ii) Using

2

2

peV

m and

hp

,

It can be shown that the de Broglie wavelength of the electrons is given by the expression

2

h

mVe

Thus, -34

-31 -19

6.63 x 10

2(9.11 x 10 )(4500)(1.60 x 10 )

-111.8 x 10 m

[1] – exp [1] – sub

(a)(iii) Using sin d n and the answer for (a)(ii),

It can be shown that for a 1st order maximum, -11sin10 1.8 x 10od

-101.1 x 10 md

[1] – sub [1] – ans

(a)(iv) Diffraction of electrons in the inner structure of graphite is pronounced only if the de Broglie wavelength of the electrons is comparable to the separation of the carbon atoms in graphite. If low energy electrons are used instead, the de Broglie wavelength of the electrons increases significantly and the amount of diffraction of the electron could be too low for observable interference to take place.

[1] – ans [1] – ans

(b)(i) frequency,

(9.81)(0.026)20.2 20

0.025

vf Hz

[1] – sub [1] - ans

(b)(ii) It is to ensure that the crest and the trough of the wave travel at the same speed so that the wavelength remains approximately constant during wave propogation.

[1] - ans

(c)(i) microwave [1] - ans

(c)(ii) 0.040 m – 1.0 m [1] - ans

Page 6 of 8





(c)(iii) Path difference: 0.18 m 4.5 x The waves arrived at P in phase and constructive interference takes place.

[1] – ans [1] - ans

(c)(iv) Destructive interference takes place at O.

2 2Intensity at P Amplitude at P 2

9.00Intensity at O Amplitude at O 2

A A

A A

[1] – sub [1] – ans

(d)(i) They have the same fundamental frequency. [1]

(d)(ii) Instruments A and B are each made of an air column which is open at both ends while instrument C is made of an air column which is open at one end and closed at the other.

[1]

(d)(iii)

[1]

(d)(iv)

[1]

8(a)(i) A periodic motion where the object’s acceleration is proportional and opposite to

its displacement from the equilibrium position.

[2] or 0

(ii) Since the graph shows only half a cycle, and there are 10 time intervals between the various positions of the bob each of 0.10 s: period of oscillation, T = (2)(10)(0.10) = 2.0 s From graph, amplitude x0 = 5.0 cm

Hence, vmax = x0 = (2π/T)(x0) = (2π/2.0)(5.0 x 10-2) = 0.157 ms-1

[1] correct T [1] subst

(practice e.c.f. for wrong T) [1] ans

(b)(i) Faraday's law states that the magnitude of the induced e.m.f. in a coil is proportional to the rate of change of magnetic flux linkage through that coil.

[2] or 0

Page 7 of 8




Qn Suggested solution Remarks (ii)1.

[1] for both arrows correctly drawn and labelled

2. At the instant the live lead is positive, using Right Hand Grip Rule, the magnetic

fields generated by primary coils A and B are in opposite directions to each other. Since both primary coils carry the same magnitude of current and have the same number of turns, the magnitude of the magnetic flux density produced by both coils are identical. This means that both magnetic fields cancel out and there is no resultant magnetic flux within the transformer core. Hence, there is no change of flux linkage in the secondary coil and therefore, by Faraday’s Law, there is no induced e.m.f. in the secondary coil.

[1] [1] [1]

(c)(i) Q = It = (e.m.f./R)(t) Hence, Q = (NBA – 0) /R = (200 x 2.0 x 10-4 x 10 x 10-4 ) / 2.0 = 2.0 x 10-5 C

[1] subs [1] ans

(ii) There would be no change since increasing the number of turns would increase the resistance of the coil by the same amount as the increase in the change in flux linkage.

[1] correct conclusion (no ecf) [1] correct reasoning

(d)(i)

Vs (rms) = 12.0/(2)

Vs/Vp = Ns/Np

12.0/[(2)(230)] = Ns/5000

Ns = 184

[1] for correct conversion of peak to rms voltage or vice versa [1] ans

A

B

Page 8 of 8




Qn Suggested solution Remarks (ii)1.

From the equation, V = 12.0 sin(380t)

= 380 = 2πf

Hence, f = 380/(2π) = 60.5 Hz

[1] ans

2. Mean power = ½ Peak power

= ½ [(12.0)2/R]

Since mean power must not exceed 300 W,

½ [(12.0)2/R] 300

R ½ [(12.0)2/300] = 0.240

[1] working [1] ans

Documents

JC2 Preliminary Examination 2015score-in-chemistry.weebly.com/.../jjc_2015_prelim.pdfJJC 2015 9646/JC2 Prelim Exam P1/2015 [Turn Over 2 Data speed of light in free space, c = 3.00