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ACJC 2010 Prelims H2 P1 Ans H2P P2 Ans Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 A C B D D A D B C A B A C D D D D A B C Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40 A A B B D B C B A D B D C B D C D B C A 1(a)(i) 3 V d 6 π = × , where d = 22 cm (accept values from18 cm to 24 cm) = V 5.6 x 10 -3 m 3 (accept values from 3.0 to 7.5 x 10 -3 m 3 ); accept corresponding values if alternative units are used (e.g. mm 3 or m 3 ) 1(a)(ii) 2 2 1 = mv KE , where v = 10 m s -1 (accept values from 8 m s -1 to 12 m s -1 ) and mass = 60 kg (accept values from 40 kg to 100 kg) = KE 3000 J (accept values from 1900 J to 7200 J); accept corresponding values if alternative units are used (e.g. kJ) (iii) Density is between 800 to 1100 kg m -3 1(b)(i) The values are close to each other but very far from the expected value which should be less than density of water (1000 kg m 3 ). Therefore, set B is precise but inaccurate. 1(b)(ii) - mass of beaker is included in the measurement of mass - balance calibrated incorrectly - negligence in zero-ing the weighing balance (either one of the above) 2a(i) Graph is a straight line through the origin a is directly proportional x Graph is a straight line with negative gradient a is in the opposite direction of x (ii) ω 2 = 250 rad s -1 Τ = 2π/ω Τ = 0.40 s (iii) Cosine curve correct amplitude of 0.050 m and with 2 2 1 complete waves (b) ω = T π 2 = 2 2 π =π rad s -1 Apply x = x 0 sin ωt where x 0 = 50 mm and x = 25 mm Time interval when shuttle remains open , www.erwintuition.com www.erwintuition.com

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ACJC 2010 Prelims

H2 P1 Ans

H2P P2 Ans

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20

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

Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40

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

1(a)(i) 3V d6

π= × , where d = 22 cm (accept values from18 cm to 24 cm)

=∴V 5.6 x 10-3 m3 (accept values from 3.0 to 7.5 x 10-3 m3 ); accept corresponding values if alternative units are used (e.g. mm3 or m3)

1(a)(ii) 2

2

1= mvKE , where v = 10 m s-1 (accept values from 8 m s-1 to 12 m s-1) and

mass = 60 kg (accept values from 40 kg to 100 kg)

=KE∴ 3000 J (accept values from 1900 J to 7200 J); accept corresponding values if alternative units are used (e.g. kJ)

(iii) Density is between 800 to 1100 kg m-3

1(b)(i) The values are close to each other but very far from the expected value

which should be less than density of water (1000 kg m3).

Therefore, set B is precise but inaccurate. 1(b)(ii) - mass of beaker is included in the measurement of mass

- balance calibrated incorrectly - negligence in zero-ing the weighing balance (either one of the above)

2a(i) Graph is a straight line through the origin ⇒ a is directly proportional x

Graph is a straight line with negative gradient

⇒ a is in the opposite direction of x

(ii) ω2 = 250 rad s-1

Τ = 2π/ω

Τ = 0.40 s

(iii) Cosine curve

correct amplitude of 0.050 m and with 2

2

1complete waves

(b) ω =

T

π2=

2

Apply x = x0 sin ωt where x0 = 50 mm and x = 25 mm

Time interval when shuttle remains open ,

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ACJC 2010 Prelims

3 (a) force on magnet/ balance is downwards since reading increases (so by Newton’s third law)force on the wire is upwards (b) pole P is a north pole (using Fleming’s left hand rule) (allows ecf from (a)) (c) Egg shape field lines closer at the bottom and spread out at the top Direction is clockwise (d) F = B I L sin 90° and F = mg

2.3 x 10-3 x 9.8 = B x 2.6 x 4.4 x 10-2 ( g missing, then 0/2 in this mark)

∴ B = 0.20 T

⇒ 25 = 50 sin

2

2πt

t =

6

1s ⇒ t = 0.167s

4ai Draw a suitable circuit diagram for this investigation.

A

V

aii

I

V

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ACJC 2010 Prelims

5(a)(i) Vs is (stopping potential). Electrons with max KE cannot reach the anode/cannot be collected (as even they do not have enough energy to overcome the electrostatic repulsive force between cathode and anode.)

(ii) (Electrons are emitted with a range of KE), hence when anode is make less negative with respect to cathode, some electrons may be able to (overcome the electrostatic repulsive force and) reach the anode

(iii) Saturation not achieved immediately once V is +ve because the electrons are scattered randomly in different directions. (Hence with higher V, the path of more electrons may be altered so that it is able to reach the anode due to the increased in the magnitude of the electric force.)

(Saturation current is achieved at V1 when all emitted electrons are collected.)

5(b)(i) φ = 5.32 x 1.6 x 10-19

= 8.51 x 10-19

J

hc/λ = 8.51 x 10-19

J

λ= 2.34 x 10-7

m

(ii) p�

=h/λ

p�

=2.83 x 10-27

N s (allow ecf)

6(a)(i) For bullet : ∆x = 2.64 x 10-32

m (allow if given to 1 sf)

For electron ∆x = 1.95 x 10-3

m (allow if given to 1 sf)

Both correct

(ii) For bullet, HUP sets no practical limit to the locating of the position of the bullet accurately. About 10

-17 times diameter of nucleus

For electron, (the position of the electron is probabilistic.) Cannot determine its exact position at any instant of time. About 10

7 times diameter of an atom.

Explanation related to reasonable estimates on order of magnitude stated implicitly or explicitly.

(b)(i) Quantum tunneling which is the phenomenon where the probability of the alpha particle passing through the barrier is non-zero where classically it is unable to overcome the barrier as it has less energy than the barrier.

(ii) The long half-life implies low decay constant which is the probability of decay per unit time,

due to the error in the (huge) underestimate of the transmission probability of the alpha particle.

aiii As I and V increases, rate of atomic vibration increases

Number of free electrons remains the same, hence resistance increases

aiv

Know that R = l

A

ρ; or mention that R depends on ρ

Therefore ρ must have been unique as R is unique at each temperature.

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ACJC 2010 Prelims

(iii) breaking the barrier into segments and (multiplying the successive tunneling probabilities) or

Use T proportional to exp2

1

r

r∫ −2

2

2

8 ( ( ) )m U r E

h

π −dr

(Note r is d in the formula) to account for the change of U with r. (In this case r is from 9.01 to 26.90 fm

7(a)(i)

The body is a relatively good conductor. If the body is in contact with the ground, the body and the ground will tend to make one equipotential surface. So the body still have very nearly 0 potential difference between the head and the feet.

(ii) Current Density =

24

1800

=

( )231063804

1800

×××π (for correct substitutions)

= 3.5 x 10-12 A m-2

Hence current density near the surface of earth is in the order of about 10-12 A m-2

(iii) P = 4 x105 x 1800

= 720 MW (2 sf) (Accept 700 MW)

(b)(i) Volume of cloud =

4

2Hdπ

=

( ) ( )4

102102 323 ××π

= 6.283 x 109 m3

No. of droplets at lowest density = 50 x 106 m-3

Vol. of each droplet = ( )361010

3

4 −×π

= 4.189 x 10-15 m3

No. of droplets in the cloud = 50x106x6.283x109

= 3.14 x 1017

Total volume of water droplets = 3.14 x 1017x4.189x10-15

= 1.315 x 103 m3

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ACJC 2010 Prelims

Mass of cloud = 1.315x103 wρ

= 1.315x106 kg

(b) (ii) Rainfall =

Area

Volume

=

( )23

3

101

10315.1

×

×

π

= 0.42 mm

(c) (i)

VQW ×=

= 20 x 1 x 108

= 2 x 109 J

(ii) VIP = or

t

WP =

=108 x 20 000

9

3

2 10

10

x−

=

= 2 x 1012 W = 2 x 1012 W

Accept approx 109 / 10-3

(iii)

dr

VdE −=

07.0

1080 3×=E (Accept 60 x 103 / 0.07)

= 1.14 x 106 V m-1

Since the electric field at the top of the tower is greater than the order of 105 (a few hundred thousands V m-1), a return stroke is likely to occur.

(d) (i) The electrical discharge results in heating up of the atmosphere around the lightning channel.

The sudden rise in pressure causes the air around it to expand rapidly resulting in an explosion of the air.

(ii) Distance (m) = 330 (m s-1) x t (s) (Assuming time taken by light is

negligible)

= 330t m

= t

1000

330 km t =

8

10

(3 10 330)

9.9 10

x d

x

=

3

t km

(e)(i)

=

0

1log10120I

I

=

0

2log108.124I

I

=−

0

1

0

2 log10log101208.124I

I

I

I

I1 = 1.0 W m-2 and I2 = 100.48 W m-2 found separately

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ACJC 2010 Prelims

=

1

2log108.4I

I

=

1

2log48.0I

I

1

248.010I

I=

2

1

rIα

2

22

2

11 rIrI =

48.0

2

2

2

1

1

2 10==r

r

I

I

48.0

2

2

2

102

=r

15.12 =r km

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ACJC 2010 Prelims

.

Basic procedure

Mentioned that ionisation current and air pressure measured or State IV is P and DV is ionization current

B1

Explain how pressure is varied using vacuum pump B1

[2]

Diagram shows

Correct electrical circuit; power supply must be shown. B1

Alpha source between electrodes, Pressure gauge/Bourdon Gauge and Vacuum pump/ air pump shown connected correctly.

B1

[2]

Measurements

Measure ionisation current using milliammeter / microammeter / galvanometer B1

Measure air pressure using pressure sensor with datalogger or direct reading from pressure gauge / Bourdon gauge

B1

(accepted if shown on diagram) [2]

Control of variables

Measure that activity of source is ensured to be constant by using a source with long half-life / high activity / checked by using radiation detector.

B1

Separation of parallel metal plates is kept constant throughout the experiment B1

Position of source is kept constant wrt plates B1

[max 2]

Any further detail

Use Americium 241, 185 kBq as alpha particle emitter (a relatively pure alpha emitter) B1

Source of 185 kBq is also relatively safe to use in a school laboratory with standard basic precautions

B1

Use high voltage (50 V to 1000 V) power supply as current is very small B1

Place source close to plates as alpha particles have short range in air B1

Electrical insulation between electrode plates and walls of metal case B1

8 sets of data is collected (4 below and 4 above atmospheric pressure) B1

Count recorded over a fixed duration of 1 or 2 minutes if using detector to check for constant activity

B1

Source placed near to detector if using detector to check for constant activity. B1

Tap pressure gauge /Bourdon gauge when taking readings in case needle sticks B1

Plot suitable graph to verify proposed relationship (eg log I vs log P) with suggested conclusion given (Do not accept if mention plot I against P)

B1

[max 3]

Safety

Use of source handling tool B1

Source kept in lead container when not in use B1

Do not point source at anyone / do not look directly at source B1

Use safety goggles when working with low/high pressure (Do not award if no explanations given) B1

Use safety screen in case of explosions B1

[max 1]

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2010 H2 Prelims

H2 P3

1(a) Vertical component of tension = weight of stone Horizontal component of tension/centripetal force is perpendicular to the

velocity or direction of motion.

Horizontal component of tension/Centripetal force/constant net force points towards a fixed point.

Constant tension supplied to maintain constant centripetal force Horizontal component of tension/Centripetal force/constant net force just

sufficient to provide circular motion for a given v and r

Constant energy supplied to stone (Do not accept Apply const v)

(b) H = ½ g t2

T sin θ > mv2/r when v decreases, hence at this instant r decreases.

When r decreases, θ decreases. Hence from T cos θ > W, therefore T must decrease so that there will be

vertical equilibrium

(c) X = v t and H = ½ g t2 .

Use H = ½ g t2

H =

2

2

2v

Xg

t = 2.944 s

X = v

g

H2

= (22.8)

81.9

)5.42(2 (for correct

substitutions)

Use X = ut

= 67.1 m X = 67.1 m

θ

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2010 H2 Prelims

2a An elastic collision is a collision in which the (total) kinetic energy of system is conserved,

total linear momentum and total energy of the system are all conserved b(i)

For an elastic collision with a stationary piston, the speed of the gas remains the same when it bounce off in the opp direction

With a piston moving down, using the fact that relative speed of approach = rel speed of separation, its rebound speed will be greater than that earlier

Since for ideal gas, T is dependent on ave KE of molecule alone,

Higher ave speed implies higher ave KE, hence Temperature rises

(ii) Apply ∆U=Q+W where

∆U = Increase in internal energy, Q is heat supplied to system + W is WD on system

mentions that Q=0 and work done is done on gas(or W is +ve), hence higher internal energy,

For ideal gas, Internal energy depends on temp alone thus higher temperature

(iii)

m M

After collision v�

m M

Before Collision

u�

V�

v�

Process shown correctly

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2010 H2 Prelims

3(a)

There are standing waves produced in the microwave oven during the cooking process because incident wave from the left gets reflected by the reflective wall on the right and they superimpose /overlap /interfere

The conditions must be right such that the distance between the source and wall must be integral multiples of half the wavelength of the microwave.to form standing wave

as they have the same speed, frequency and almost the same amplitude

(b)(i) Intensity of the microwave is strongest (largest amplitudes) at the antinodes, hence the dry regions are the regions of antinodes

Wavelength = 3.0 x 108/2.45 x 109

= 0.1224 m = 12.2 cm

Hence distance apart is 6.1 cm

(ii)

D D D D D

Standing wave pattern drawn with ends as nodes (need not have 5 loops)

5 Ds shown

(c) So there are regions in the oven where the microwave has high amplitude (antinodes) and there are region where the microwave has no displacement (node). Thus ants can stay away from the regions of high amplitude which has high heat and thus stay alive.

Understand that the ants will stay away from the position of antinodes to stay alive.

(d) The turntable enables different parts of the food to move to the antinodes of the standing wave and get heated up. Thus it helps heat up the food more uniformly.

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2010 H2 Prelims

4a) F spring= kx

5 x 9.81 = k(0.05)

k=.981 N m-1

(No not penalize for inconsistent –ve sign that appears) – Pls annotate BOD

4b (i)

By conservation of energy, At a height 40cm, the contraction of spring is 10 cm.

Energy stored in spring = ½ (981)(0.1)2

= 4.905 J

Energy gain = Energy lost ½ mv2= mgh + 4.905

½ (5)v2= (5)(9.81)(0.1)+4.905 = 9.81 v= 1.98 m s-1

(ii) By conservation of energy, Let final extension be x (measured from natural length)

Energy at natural length= Energy at lowest point

9.81 = - (5)(9.81)x + ½ (981)x2

Initial energies = final energies

4.905 =-5 (9.81)(x+0.1) + ½ (981)x2

490.5 x2 – 49.05 x – 9.81 = 0

50 x2 – 5x – 1 = 0

Hence x = -0.10 (initial released point) or + 0.20 (lowest point)

Hence lowest point is (0.30 – 0.20) = 0.10 m above the ground.

4c In reality, it is larger. There will be some energy lost due presence of air resistance/friction in spring and thus less energy will be available for conversion to elastic potential energy.

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2010 H2 Prelims

5 (a)(i) | ε | = | - dΦ/dt |

= | - NA dB/dt |

= | - (80) * (π * 0.052) * (30-120)*10-3 / (0.16-0.12) |

= 1.41 V

(ignore –ve sign)

(a)(ii) When the magnet is oscillating, an emf will be induced in the coils and thus induced current will flow since circuit is closed resulting in

either electrical energy being dissipated or heating effect of coils

Hence total (mechanical) energy of the magnet-spring system decreases continuously as it is being converted into the electrical

energy in the coil

As amplitude of oscillations depends on amount of mechanical energy present,

Hence the amplitude decreases continuously

(b)(i) A lower amplitude indicates that more ME is converted into electrical energy over each cycle, (that can only be the result of a higher induced emf, hence induced current)

A lower amplitude indicates that there is greater damping /induced force,

(Since system has same frequency and started with same amplitude) and N and A are constant, higher induced emf must be due to higher rate of change of B.

(Since system has same frequency and started with same amplitude) This is due to greater induced emf/current which must be due to greater rate of change of B

Magnet field strength of Magnet B is stronger than that of Magnet A

(b)(ii) • Reduce the number of coil. Since | ε | = | - NA dB/dt |, a lower N will result in a lower induced emf, therefore higher amplitude for the oscillation

• Use a wire with higher resistance. Same emf will result in a lower induced current

• Use a resistor with higher resistance. Same emf will result in a lower induced current

• Use coil of smaller cross-sectional area

• Use spring of smaller k

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2010 H2 Prelims

c (i) As the current pass through the coil, the coil will generate a magnetic field and will behave like a magnet. As the current is alternating, the polarity of this coil will change as the direction of the current flow changes. The magnet will then be attracted and repelled by the coil as the current changes direction.

As the alternating current in sinusoidal, the force exerted on the magnet will also be sinusoidal and therefore, the magnet will be forced into a sinusoidal motion.

Idea that coil generates an alternating(sinusoidal) field

Idea that magnetic force created is hence also alternating

(and its magnitude is proportional to I at that instant.)

with the same frequency as the ac supply.

Hence magnet is undergoing forced oscillation.

(ii) • Amount of damping on the driven system

• The relative values of the natural frequency of the driven system and the value of the frequency of the external driving force

• Amplitude of the periodic driving force

• More turns used in the coil

(iii) Maximum energy transfer to driven system when driving frequency is equal to (approaches the) natural frequency of system.

(iv) Mention that method is to increase damping on system Any suitable method that increases damping, examples

• Attached an object of large surface area to the magnet

• submerged magnet in a viscous fluid

Alternative: Use AC of lower amplitude

This will result in lower amplitude for all frequencies, hence flatter peak.

(d)

-2

-1

0

1

2

0 0.5 1 1.5 2

Time / s

Displacement / cm

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2010 H2 Prelims

6 (a)(i) The distinct lines shows that only photons of certain frequency are emitted.

This took place for electron transition from higher energy levels to a lower energy

levels.

This electron transition results in the release of a photon from the atom of an

amount of energy (E = hν) equal to the difference in energy of the electronic energy

levels involved in the transition.

Hence showing the energy levels are quantised.

(ii) The high voltage applied broke the bonds of the hydrogen molecules into its

(isolated) atoms (and promote the electrons into higher energy levels.)

(iii) θ = 23.2o

λ = 656.6 nm

(b)(i) 1,2,3,4 L to R

(ii) 1 – 410.3 nm; 3 – 486.3 nm; 4 – 656.6 nm (no ecf)

(iii) All 3 points plotted correctly

Axes given to correct units

Wavelength read correctly as 435 nm (theoretical value 434 nm)

(accept 430 to 440 nm as read correctly from graph)

(c)(i) E = hc/λ

E = 6.63 x 10-34

x 3 x 108/λ

(correct sub for h and c)

E2 = 4.57 x 10-19

J(allow ecf)

E4 = 3.03 x 10-19

J

(credit given only if both calculations are correct)

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2010 H2 Prelims

(ii)

5 energy levels draw with decreasing spacing

Scale correct

Transitions all correct

Energy values all correct

(d)(i) For significant diffraction to occur so that higher wavelength photons could be

observed.

Accept (so that lines further away from straight through position are observable)

(ii) Brighter line spectrum / better contrast

2 (-5.44 x 10-19

J)

3 (-2.41 x 10-19

J)

4 (-1.35 x 10-19

J) 5 (-0.87 x 10

-19 J)

6 (-0.59 x 10-19

J)

1 2 3 4

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2010 H2 Prelims

7a Random Impossible to predict when and which individual nuclide will decay OR There is a constant probability or decay Or Unable to predict which atom and how many will decay at the next instant of time

Spontaneous: Decay is unaffected by environmental changes such as temp / pressure, external source of energy supplied etc.

7bi

( ) ( ) ( )

2

227 8

( )

226.0254 222.0176 4.0026 1.66 10 3 10

Ra RnQ M M M cα

= − −

= − − × × × ×

137.7688 10 J

4.86 MeV

−= ×

=

ii COM: MV + MαV = 0 --------- (1)

COE:

Q =

1

2MV 2 +

1

2MαVα

2 --------- (2)

From (1):

V = −Mα

M

Vα -------------- (3)

Subst (3) into (2)

Q =1

2M −

M

2

+1

2MαVα

2

=1

2MαVα

2 Mα

M+ 1

= Kα

M+ 1

iii 4.00264.86 1

222.0176Kα

= +

(correct substitution shown)

4.77 MeVKα =

The alpha particles carries away most of the energy – 98 %

7ci ln2- 2.5

105.20e 37 10A

×

≥ ×

11

0 5.16(3) 10 BqA ≥ ×

11

0 5.2 10 BqA = ×

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2010 H2 Prelims

ii λ = ln2 (5.2 365 24 60 60)× × × ×

= 4.2268 x 10-9

s-1

0 0A Nλ=

11

0

5.2 10 Bq

ln2 (5.2 365 24 60 60)N

×=

× × × ×

= 1.23 x 1020

20 231.23 10 6.02 10

60

m× = × ×

12.3 mgm =

iii Energy emitted per decay = 0.31+1.17+1.33=2.81 MeV Rate =( 2.81 x 1.6 x 10-13 J) x 37 x 1010 decay per second

= 166 mW

iv The β-decay energy is low compared to the γ-ray.

Hence the (2) strong γ-lines, could be used as a γ-ray source to check for uniformity of the thickness of metal sheets in the industries

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