14_15_h2_j2_ct2_p3

JJC 2015 9646/JC2 Common Test 2 P3/2015 [Turn Over

JURONG JUNIOR COLLEGE JC2 Common Test 2 2015

Name Class 15S

PHYSICS Higher 2

Longer Structured Questions

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

9646/3

3 July 2015

1.5 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 on both sides of the paper You may use a soft pencil for any diagrams, graphs or rough working. Do not use 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 section B. 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

Total

(This question paper consists of 19 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

Section A Answer all questions in this section.

1 (a) In an experiment to observe atomic spectra of gases, some mercury vapour is placed in a discharged tube as shown in Fig. 1.1.

Fig. 1.1

When the tube is connected to a high voltage supply, electrons are generated by the electrodes through thermionic emission. The mercury atoms become excited and emit electromagnetic radiation. Explain

(i) how the mercury atoms are excited in the discharge tube,

[2]

(ii) why the excited mercury atoms emit electromagnetic radiation of discrete frequencies.

[3]


4

(b) Two spectral lines of wavelengths 436 nm and 546 nm are observed in the atomic spectrum of mercury. These two lines are the result of electronic transitions from energy level A to two lower energy levels B and C as shown in Fig. 1.2.

(i) State the region of the electromagnetic spectrum in which the two spectral lines lies.

region is the range [1]

(ii) Draw an arrow on Fig 1.2. to represent the transition which gives rise to the spectral line of wavelength 436 nm. [1]

(iii) Calculate the value of the energy level A.

value = J [2]

Fig. 1.2


5

(c) Fig. 1.3 shows a beam of electrons, moving with speed 6.0 x 105 m s–1 in the x-direction, passing through a single slit of width 1.0 nm.

x

y

px

Fig. 5.1

(i) Calculate the momentum px of one of these electrons in the x-direction.

momentum px = N s [1]

(ii) Hence, show that the de Broglie wavelength of the electron is 1.2 x 10-9 m. [1]

(iii) With reference to wave theory, state why the electron diffraction would be prominent in this situation.

[1]

Fig. 1.3


6

2 (a) In a fission process, a neutron collides with a uranium-235 nucleus and causes a nuclear reaction summarised by the following equation.

1 235 P 90 1

0 92 Q 38 0n + U Xe+ Sr+3 n +energy

(i) Determine the numerical values of P and Q.

P = [1]

Q = [1]

(ii) State the feature of this equation that indicates that a chain reaction may be possible.

[1]

(iii) By reference to binding energy per nucleon, explain why energy is released in this fission reaction.

[2]

(b) A sample contains N nuclei of strontium-90 at time t. At time (t + ∆t), the sample contains (N - ∆N) nuclei of strontium-90.

Give expressions, in terms of N, ∆N, t and ∆t,

(i) the activity of the sample,

[1]


7

(ii) the probability of decay of a strontium nucleus in time ∆t,

[1]

(iii) The decay constant λ, for strontium-90.

[1]

(c) Strontium-90 undergoes radioactive decay by emission of a β particle and has a half-life of 28 years.

(i) Explain what is meant by radioactive decay.

[1]

(ii) Show that the decay constant of strontium-90 is 7.85 x 10-10 s-1. [1]

(iii) In a laboratory source of strontium-90, the number of radioactive atoms present in the year 2015 is 2.36 × 1013.

Calculate the number of radioactive atoms left in the source in the year 2115.

number of radioactive atoms left = [2]


8

Section B

Answer two questions in this section.

3 (a) (i) Define one coulomb and one volt.

One coulomb:

[1]

One volt:

[2]

(ii) Using the definition of volt or otherwise, show that one volt is also equivalent to one watt per ampere.

[2]

(b) Referring to Fig. 3.1, the 15 V source is of negligible internal resistance.

Fig. 3.1

9.0 Ω 6.0 Ω

6.0 Ω

6.0 Ω

6.0 Ω

2.4 Ω

15.0 V

A C

D

I

B

Switch K


9

(i) With switch K open, show that the effective resistance between point C and point D is 3.0 Ω.

resistance= Ω [2]

(ii) With switch K closed and given that the potential at point A is –3 V,

1. determine the potential at point C, and

potential = V [2]

2. calculate the current I.

I = A [3]


10

(c) In the circuit shown in Fig. 3.2, cell A has an e.m.f. of 2.0 V and negligible internal resistance. Wire XY is 1.0 m long with a resistance of 5.0 Ω.

Fig. 3.2

(i) Calculate the current flowing from X to Y when the galvanometer registers null deflection.

current = A [1]

(ii) Cell B has an e.m.f. of 1.5 V and internal resistance r.

1. Calculate the length XP required to produce zero current in the galvanometer.

length = m [2]

2. A resistor R of 10 Ω is now placed parallel to cell B. The length of wire XP required to produce zero current in the galvanometer is 0.75 m.

Calculate the value of r.

r = Ω [3]


11

3. State and explain one advantage of using Fig. 3.2 to measure the e.m.f. of B compared to using a laboratory voltmeter.

[2]


12

4

(a) A tuning fork is shown in Fig. 4.1 and 4.2 below. It is made of a handle and two tines. It can be made to vibrate by knocking one of the tines sideways against a hard object.

At time t = 0, the tuning fork is knocked against a hard object such that it vibrates about its equilibrium position with a frequency of 128 Hz. Fig. 4.1 Fig. 4.2

Tine A’s vibrations subsequently causes the neighboring air molecules to vibrate such that a longitudinal wave of the same frequency is formed. Fig. 4.3 below shows the positions of the air molecules around the tuning fork at a particular instant.

(i) By considering the movement of air molecules, state and explain the pressure experienced by the air molecule at Y.

[2]

X Y Z

Fig. 4.3

handle

tines

frequency = 128 Hz

A A


13

(ii) Given that the distance between X and Z is 5.2 m, calculate the speed of the longitudinal wave between X and Z.

speed of wave = m s-1 [2]

(iii) Determine the phase difference between X and Y.

phase difference = rad [1]

(b) The longitudinal waves created by tine A above can be assumed to have a power of 0.72 W and is equally generated in all directions. A microphone whose circular cross-section has radius 4.0 cm is placed 5.0 m away from the tine as shown in Fig. 4.4 (not to scale).

4.0 cm

5.0 m

Fig. 4.4


14

(i) Determine the power received by the microphone.

power received = W [3]

(ii) The microphone is replaced by a bigger one whose radius is twice of that of the previous one. At what distance away from the tine must the bigger microphone be placed so that it still picks up the same power?

distance from the tine = m [3]

(iii) Which of the following is a measure of the loudness of the sound? Circle it.

Power / Intensity / Amplitude [1]

(c) Explain how interference contributes to the alternating bright and dark fringe pattern observed on the screen in Young’s double slit experiment.

[2]


15

(d) A double slit with slit separation d = 0.800 mm is situated a distance 3.2 m from the screen as shown in Fig. 4.5. The double slit is illuminated with coherent light of

wavelength .

Fig. 4.5

The light intensity pattern along XY is shown in Fig. 4.6. The relative positions of the peaks are as indicated with respect to O, the central maximum.

Fig. 4.6

(i) From Fig. 4.6, show that the wavelength of the light is7

7.50 10 m

. [2]

O

P

X

Y D = 3.2 m

S1

S2

light d = 0.800 mm mm

0.0 3.0 -3.0 6.0 -6.0

intensity

distance / mm


16

(ii) With reference to Fig. 4.6, state the phase difference between the two waves at detector P, when P is at a distance of

1. 6.0 mm from O, and


2. 1.5 mm from O, and


(iii) Using data from (d)(ii)1 or 2, verify that the distance between detector P and O, Δy has the relationship of

Δy = (path difference) x D

d [2]


17

5. (a) A washing machine has a maximum spin speed of 600 revolutions per minute. Express this angular velocity in radians per second.

angular velocity = rad s-1 [2]

(b) A mass of 0.500 kg is attached to a string of length 0.400 m which will break if the tension in it exceeds 15.0 N. The mass is whirled in a horizontal circle with the

string inclined at 60 to the vertical as shown in Fig. 5.1.

Fig. 5.1

(i) Determine the radius of the horizontal circle.

radius = m [1]

(ii) Calculate the speed of the mass in the horizontal circle.

speed = m s-1 [3]


18

(iii) The speed of the mass is slowly increased. At the same time, the angle of

inclination of the string from the vertical increases to 70.9. Calculate the greatest number of revolutions that the mass makes per unit time just before the string breaks.

number of revolutions per unit time = Hz [3]

(c) Explain what is meant by a gravitational field.

[2]

(d) Satellites are artificial bodies placed in orbit around the earth in order to collect information or for communication purposes. It is generally possible to have a satellite orbit the earth from east to west (retrograde orbit) as well as west to east (prograde orbit).

(i) A satellite is launched in the west to east (prograde orbit) direction from a space station on the equator to the geostationary orbit. Explain why it is normally preferred to launch satellites in this direction.

[2]


19

(ii) Taking the Earth to be a uniform sphere with a radius R of 6400 km and mass M of 6.0 x 1024 kg, determine r, the radius of orbit of the geostationary satellite.

r = M [3]

(iii) The mass of one such geostationary satellite is m. Derive an expression in terms of G, M, m and r for the kinetic energy Ek of this satellite when it is in geostationary orbit. [2]

(iv) Hence, or otherwise, show that the total energy of the satellite in (d)(iii) is

given by Et = 2

GMm

r

. [1]

(v) Determine the total energy of the satellite in (d)(iii), given that the mass m of the satellite is 1000 kg.

total energy = J [1]

End of Paper


20

JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT

2015 JC2 CT2 Paper 3 9646 H2 Physics

Suggested Solutions with Markers’ Comments

Qn Suggested solution Remarks

1(a)(i) Fast-moving electrons in the discharge tube collide with the vapour atoms and transfer their (kinetic energy) to the electrons in atom.

[1] [1]

(a)(ii) Upon de-excitation, the electrons in the atoms move from higher to lower energy levels by emitting photons (EM radiation). Since the energy levels are of discrete values, the photon energies (which is equal to the difference in energies of any 2 energy levels) must also be discrete. Since photon energy is discrete and given by hf, the frequencies of the EM radiation would be discrete.

[1] [1] [1]

(b)(i) Visible (light) [1]

(b)(ii)

[1] arrow in correct direction

(b)(iii) Using transition for 546 nm,

energy of 546 nm photon = hc

λ=

-34 8

-9

(6.63 x 10 )(3.00 x 10 )

546 x 10 = 3.64 x 10-19 J

Value of energy level of A = (3.64 x 10-19) + (- 7.94 x 10-19) = - 4.30 x 10-19 J

[1] for correct photon energy [1] ans with negative sign

(c)(i) px = mv = (9.11 × 10-31)(6.0 × 105) = 5.5 × 10-25 kg m s-1

[1] ans

(c)(ii)

[1] working

(c)(iii) The slit width and wavelength of the electron have the same order of 10-9 m.

[1]

34

25

6.63 10

5.5 10

h

p


21





2(a)(i) Equating nucleon number, 1 + 235 = P + 90 + 3(1) P = 143 Equating atomic number, 0 + 92 = Q + 38 + 3(0) Q = 54

[1] ans [1] ans

(a)(ii) More neutrons are produced than the number required to cause the reaction. [1]

(a)(iii) The binding energy per nucleon of the products is greater than the binding energy per nucleon of the reactants. Since the number of nucleons of the reactants and products are approximately equal, the total binding energy of the products is greater than that of the reactants. The difference between these total binding energies is released.

[1] [1]

(b)(i)

N

t

[1]

(b)(ii) N

N

[1]

(b)(iii)

N

N( t)

[1]

(c)(i) Process by which a nucleus of an unstable atom loses energy by emitting an alpha or beta particle, and usually accompanied by the emission of a gamma ray photon.

[1]

(c)(ii) 0.693

(28 × 365 ×24×60×60)

[1]

(c)(iii) N = N0 e-λt

= (2.36 × 1013) e∧(- 0.693/28)(100) = 1.99 × 1012

[1] sub [1] ans


3 (a)(i) One coulomb is the quantity of electric charge that passes through a cross section of a circuit when a steady current of one ampere flows for one second. One volt is the potential difference between two points in a circuit in which energy converted from electrical to non- electrical form per unit charge passing from one point to the other is one joule per coulomb.

[1] [2]

(ii) 1 joule 1 joule/1 s 1 watt1 volt= = =

1 coulomb 1 coulomb/1 s 1 ampere

PUsing P =IV, V =

I

1 watt1 volt =

1 ampere

or

[1] for showing 1J/1s to 1 W [1] for showing 1C/1s to 1 A


22





(b) (i) Resistance across 6.0 & 9.0 resistors: 6.0 // 9.0 => R = 3.6

6.0 // (3.6+2.4)

Hence, Reff =

13.0

1 1

6 6

[1] for correct method [1] for correct answer

(ii)1. 151

15

VI

R

Vac = (1)(6) = Va Vc

Hence, Vc = 9 V

[1] for correct method [1] for correct answer

2. Vd = 9 – (1)(3) = 12 V By principle of potential divider,

Ved =

3.6 3.6( )( ) ( )(3.0) 1.83.6 2.4 6.0

cdV

1.80.30

6.0

VI

R

A

[2] for any correct method [1] for correct answer

(c) (i)

[1]

(ii)1.

[1] for any correct method [1] for correct answer

2.

[1] [1] [1] for correct answer


23





3. Advantage: The measurement of e.m.f is more accurate/ reduces systematic error Reason: There is a potential drop across the internal resistance of the battery when the voltmeter is used.

[1] [1]

Qn Suggested solution Remarks 4(a)i) Particle Y experiences lowest/minimum pressure

Since its neighboring particles move away from it simultaneously.

[1] [1]

(a)ii) 2 full cycles are found between X and Z This means that the wavelength λ = 5.2 / 2 = 2.6 m Since v = f λ = 128 x 2.6 = 333 m s-1

[1]- Wavelength [1]-speed

(a)(iii) Between a rarefaction and compression, half a wave is found. Hence, phase difference = π rad

[1] correct

(b)(i) Intensity of wave at position of microphone = P / 4πx2 = 0.72/[4π(5.0)2] = 2.29 x 10-3 W m-2 Power picked up by microphone = I x A = (2.29 x 10-3) x [π(0.04)2] = 1.15 x10-5 W.

[1]- correct working [1] – intensity [1] – power

(b)(ii) Surface area of bigger microphone = 22 = 4 times of original microphone. In order to pick up the same power, microphone must be shifted to a new location where the intensity of the wave is 1/4 times of the original value. Hence, I/4 = P/(4πd2) where d is distance to new position of bigger microphone Thus, d = √4 x = √4 x 5.0 = 10.0 m

[1]- intensity [1]- working [1]- distance

(b)(iii) Intensity [1] (c) When the waves arrive in phase, there is constructive interference of the light

from both slits, a bright fringe is observed on the screen. When the waves arrive in anti-phase, there is destructive interference of the light from both slits, a dark fringe is observed on the screen.

[1]- phase [1]- CI and DI

(d)(i)

3

3

7

3.23.0 10

0.800 10

7.50 10 m

Dy

d

[1] for getting data from graph of fringe separation [1] for application of formula

(d)(ii)1. By observation, it is 4 radians as it is the second order maximum. For

constructive interference, phase difference is of the order 2n radians. (Accept

zero)

[1]


24

Qn Suggested solution Remarks 2. By observation, it is radians as it is the first order minimum. For destructive

interference, phase difference is of the order (2n+1) radians.

[1]

(d)(iii) Since phase difference at P when OP = 6 mm is 4 radians

path diff

path diff

( )2 4

2

Corresponds to 6 mm which is (2 x d

D)

OR For L = 1.5 mm, phase difference at P when OP = 1.5 mm is radians

path diff

path diff

( )2

1

2

Corresponds to 1.5 mm which is (1/2 x d

D)

Therefore (path diff) x d

D= L (shown).

[1] for using phase difference to show path difference [1] for final evidence

5(a) Angular velocity = (600 x 2π)/(60)

= 62.8 rad s-1 [1] - subst [1] - Ans

(b)(i) Radius = 0.400sin60 = 0.346 m [1] - Subst

[1] - Ans

(ii) Tcos60 = mg

=> Tcos60 = (0.500)(9.81) = 4.905 --------------------------------(1) Horizontal component of tension provides the centripetal force.

Tsin60 = mv2/r

= (0.500)v2/(0.400sin60) = 1.4434 v2 ----------------------------------------------------(2) Taking (2)/(1):

tan60 = 1.4434v2/4.905 v = 2.43 m s-1

[1] – expression [1] – expression [1] – ans


25

Qn Suggested solution Remarks (iii) Horizontal component of tension provides the centripetal force.

Tsin70.9 = mr2 When tension, T = 15.0 N,

(15.0)sin70.9 = (0.500)(0.400sin70.9)2

= 8.66 rad s-1

Recall that = 2πf Hence, greatest number of revolutions per unit time, f = 8.66/2π = 1.38 Hz

[1] – Subt [1] – ans for ω [1] – final ans

(c) Gravitational field is a region of space where an object of mass m will experience

a (gravitational) force when placed in that region. [2] or 0

(d)(i) Direction is in the same direction as the Earth’s rotation.

Hence, the satellite would gain some rotational velocity/ kinetic energy so that less fuel would be required.

[1] [1]

(ii) Since gravitational force provides the centripetal force,

2

2

GMmmr

r

(centre of satellite’s orbit is the centre of Earth)

2

3

GM

r

2 11 24

3

2 2

(24 60 60) (6.67 10 )(6.0 10 )

(2 )

GM x x x xr

r = 4.23 x 107 m

[1] - expression [1] – subst [1] - ans

(iii) Since gravitational force provides the centripetal force,

2

2

GMm mv

r r

=> Ek =

21

2 2

GMmmv

r

[1] – statement [1] - Ans

(iv)

2 2t k p

GMm GMm GMmE E E

r r r

[1] – working

(v) 11 24

7

9

6.67 10 6.0 10 1000

2 2 4.2 10

4.764 10

t

t

GMmE

r

E J

[1] - ans

14_15_h2_j2_ct2_p3

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