2010 H2 Physics Ans - NYJC

2010 NYJC Prelim H2 Paper 1 Solutions

1 A 11 B 21 C 31 A

2 C 12 D 22 B 32 D

3 A 13 A 23 A 33 B

4 C 14 C 24 B 34 C

5 D 15 C 25 C 35 A

6 C 16 C 26 C 36 A

7 B 17 C 27 C 37 D

8 C 18 B 28 D 38 A

9 D 19 D 29 B 39 B

10 B 20 C 30 B 40 C

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1

Section A

Answer all questions. It is recommended that you spend about 1 hour 15 minutes on this section.

1 (a) State, in words, the 2 conditions that need to be satisfied in order to achieve static

equilibrium. Condition 1: [1] Condition 2: [1] (b) A uniform trapdoor of mass 12 kg and length 1.00 m is smoothly hinged to the wall as

shown in Fig. 1.1 (not drawn to scale). It is supported in equilibrium by a stay wire connecting the wall to a point on the trapdoor at a distance of 0.25 m from its free end. The stay wire makes an angle of 60° with the wall and the trapdoor makes an angle of 30° with the horizontal.

Show that the tension in the stay wire is 78 N.

[2]

0.25 m

60°

30°

Fig. 1.1

Taking moments about the hinge,

0 0

0

0

0

cos30 0.50 sin60 0.75

0.50 12 9.81cos30

0.75sin6078 N

mg T

T

ττττ ====

× = ×× = ×× = ×× = ×

× ×× ×× ×× ×====

====

∑∑∑∑

Resultant external force acting on the body is zero. Resultant torque about any point is zero.



(c) A 20.0 kg sphere of uniform density rests between two smooth planes as shown in Fig. 1.2.

Determine the magnitude of the force acting on the sphere exerted by each plane.

force due to plane A = N

force due to plane B = N [3]

2 The question is about reverse bungee jumping.

Fig. 2.1 shows the set up of a reverse bungee jumping. A capsule is connected by two identical elastic cords each attached to a tower 30.0 m tall. The mass of the capsule when fully loaded with three passengers has a total mass of about 300 kg. When released, the capsule will shoot up at high speed.

20 kg

70° 30°

Plane A

Plane B

Fig. 1.2

Let the force due to plane A be FA and the force due to plane B be FB. These 2 forces must be perpendicular to the sides of the respective planes. Resolving vertically,

FB cos 300 + FA cos 700 = mg

Resolving horizontally, FB sin 300 = FA sin 700

Solving simultaneously,

FA = 99.6 N FB = 187 N



(a) The original length of each of the elastic cords is 25.0 m with an elastic constant of

19 000 N m-1 and the capsule has an effective diameter of 2.0 m. Prove that the total elastic potential energy at the ground level = 510 kJ when the cord length is 30.2 m.

[1] (b) Fill in the blanks in the table below to determine the various amounts of energy when

the capsule starts from the ground level and shoots up to its highest point.

Total elastic potential energy /kJ

Gravitational potential energy of capsule /kJ

Kinetic energy of capsule /kJ

Ground level 510 0 0

30 m above the ground

0 88 422

Highest point 336 174 0

[2] (c) Use the value in (a) to determine the speed reached by the capsule when the cords

first become loose.

30.0 m

10.0 m

Elastic cords

capsule

Fig. 2.1

Ground level

Extension ≈≈≈≈ 30.2- 25 = 5.2 m Total EPE = 2 × ½ kx2 = (19000)(5.2)2 = 510 kJ (proven)

When the cord is unstretched,

Ht above the ground = 30 – √√√√(252 – 42) = 10.8 m above the ground. By conservation of energy, gain in KE + gain in GPE = loss in EPE ½(300)v2 + (300)(9.81)(10.8) = 510 000 v = 56.5 m s-1



speed = m s-1 [2]

(d) State and explain the position where the apparent weight of the passenger will be the

greatest. [2] 3 A flat horizontal plate is made to oscillate with simple harmonic motion in a vertical direction

as shown in Fig. 3.1. The plate starts its oscillation at the equilibrium position and moves downwards initially.

A graph of velocity against displacement for this oscillation is shown in Fig. 3.2. Point S marks the start of the oscillation.

plate

oscillator

Fig. 3.1

Fig. 3.2

0 0.01 0.02 0.03 0.04 - 0.04 - 0.03 - 0.02 - 0.01

0.4

0.2

0.6

0.8

- 0.6

- 0.8

- 0.4

- 0.2

S

velocity / m s-1

displacement / m

C

The apparent weight of the passenger will be the greatest at the lowest point immediately after the capsule is released. The EPE is the largest, this implies that extension is the largest which will results in the largest force.



(a) Deduce, from Fig. 3.2,

(i) the amplitude of the oscillation,

amplitude = 0.035 m [1]

(ii) the angular frequency ω of the oscillation.

ω = 19 rad s-1 [2] (b) A mass of 0.100 kg is placed on the plate before oscillation is started. (i) Determine the displacement of the plate when the mass just loses contact with

the plate.

displacement = m [3] (ii) Mark on Fig. 3.2 the point C when the mass just loses contact. [1] 4 (a) State two conditions that must be satisfied in order to obtain observable interference

patterns.

[2]

The sources must be coherent. The amplitude of the waves at the point of interference must be about the same. The distance between the two sources is much larger than the wavelength of the waves emitted.

0 0

1

0.66 (0.035)

19 rad s

v x

−−−−

= ω= ω= ω= ω

= ω= ω= ω= ω

ω =ω =ω =ω =

(Accept v0 to be 0.65 – 0.67)

( )

0 when a

F ma

mg N ma

N m g a

N g

Σ =Σ =Σ =Σ =

− =− =− =− =

= −= −= −= −

= == == == =

2

2

When ,

9.81 (18.86)

0.0276 m

a x

a g

x

x

= −ω= −ω= −ω= −ω

====

− = −− = −− = −− = −

====



The apparatus shown in Fig. 4.1 below (not to scale) is used to demonstrate two-source interference.

(b) The separation of the two slits in the double slit arrangement is a and the interference fringes are viewed on a screen at a distance D from the double slits. When light of

wavelength λ is incident on the double slit, the separation of the bright fringes on the screen is x.

(i) Write the equation that links the quantities described in the above paragraph, and state the assumption made in the use of that equation.

[2]

(ii) The slits are separated by a distance of a, with the screen at a distance of 1.00 m from the plane of the slits. The slits are illuminated by monochromatic light of wavelength 589.3 nm traveling perpendicular to the plane of the slits. It was observed that the distance between the two 4th order bright fringes are 20 mm. Calculate the separation of the slits, a.

a = m [2]

Double slit

Light,

wavelength λ

Screen

D

a

Fig. 4.1

Dx

a

λλλλ==== condition: D>> a

x = 20 / 8 = 2.5 mm

Dx

a

λλλλ====

(589.3 10 - 9)(1.00) 0.24 mm

2.5 10 - 3

Da

x

λλλλ ××××= = == = == = == = =

××××



(c) (i) Explain why the central fringe is always a bright one.

[1]

(ii) Explain why an experiment using two separate sources of light will not show interference.

[1]

5 (a) Define electric field strength. [1]

(b) Fig. 5.1 shows a stream of electrons entering a region between two parallel plates which have a potential difference.

(i) Draw on Fig 5.1, the electric field lines between the plates and the expected path

of the stream of electrons between and after the plates. [2] (ii) Calculate the distance between the plates given that the electric field strength

between the plates is 2.0 x 104 N C−1.

4

4000.020

2.0 10

V VE d m

d E= => = = == => = = == => = = == => = = =

××××

distance = m [1]

- 200 V

+200 V

Stream of electrons

Fig 5.1

The electric field strength at a point in an electric field is defined as the electrostatic force acting per unit positive charge on a test charge placed at that point.

Since the central fringe is equal distance from the double slits, path

difference = 0 λλλλ. Hence the waves from each slit must arrive in phase resulting in constructive interference.

Two separate sources of light will not be coherent and hence interference cannot take place.

Parabolic path between plates

Straight path after plates

Both blue and red paths are accepted



(iii) Calculate the acceleration on the electrons between the plates

acceleration = m s-2 [1]

(iv) Hence, given that the length of each plate is 0.040 m and initial horizontal speed of the electrons is 1.0 x 108 m s-1, calculate the vertical deflection of the electron at the end of the plates.

deflection = m [2]

6 A stationary Polonium-212 nuclide may undergo alpha decay spontaneously to produce the stable lead-208 daughter nuclide as shown in the equation below:

212 208 4

84 82 2Po Pb He→ +

The rest masses of these nuclei are

212

84Po : 211.9888 u; 208

82Pb : 207.9766 u and 4

2He : 4.0026 u.

(a) Calculate the total kinetic energy of the decay products.

total kinetic energy = J [2]

(b) Determine the ratio of kinetic energy of He-4

kinetic energy of Pb-208. Give your answer in 3 s.f.

Mass difference = 211.9888 u – (207.9766 u + 4.0026 u) = 0.0096 u Total kinetic energy = (0.0096 u) c2 = (0.0096)(1.66 × 10-27)(3.00 × 108)2 = 1.43 × 1012 J

19 4 15

1515 2

31

1.6 10 2.0 10 3.2 10

3.2 103.51 10 ( )

9.11 10e

F qE N

Fa m s downwards

m

− −− −− −− −

−−−−−−−−

−−−−

= = × × × = ×= = × × × = ×= = × × × = ×= = × × × = ×

××××= = = ×= = = ×= = = ×= = = ×

××××

10

8

2

15 10 2

0.0404.0 10

1.0 10

1

2

1(3.51 10 )(4.0 10 ) 0.00028 ( )

2

x

x

y y y

y

st s

v

s u t a t

s m downwards

−−−−

−−−−

= = = ×= = = ×= = = ×= = = ×××××

= += += += +

= × × == × × == × × == × × =



ratio = [2] (c) Hence, determine the kinetic energy of the alpha particle in MeV.

kinetic energy = MeV [2] (d) To escape from the nucleus, the alpha particle must overcome the Coulomb barrier of

26 MeV. Using your answer in (c), comment on how the alpha particle can penetrate the barrier.

[2]

The alpha particle’s 8.79 MeV of K.E. is unable to overcome the 26 MeV Coulomb barrier by classical physics. However, the alpha particle can behave as a wave and be associated with a wave function. The square of the amplitude of the wave function represents the probability of locating the particle at that point. Thus it is able to tunnel through the potential barrier and appear outside the barrier with non-zero amplitude.

2 2

0

1 1

2 2

. . 207.976652.0

. . 4.0026

Pb Pb He He

Pb Pb He He

Pb Pb Pb He He He

He Pb

Po He

M V m v

M V m v

M V M m v m

K E M

K E m

= −= −= −= −

====

====

= = == = == = == = =

12

12

. .51.9604

. .

. .51.9604

. . . .

. . 51.9604( . . . . )

52.9604 . . 51.9604(1.43424 10 )

. . 1.40716 10

8.79

He

Po

He

Total He

He Total He

He

He

K E

K E

K E

K E K E

K E K E K E

K E

K E J

MeV

−−−−

−−−−

====

====−−−−

= −= −= −= −

= ×= ×= ×= ×

= ×= ×= ×= ×

====



7 In the first half of the last century, numerous experiments were conducted to investigate the absorption and the scattering of X-ray by matter.

It was discovered that when a monochromatic beam of X-rays is incident on a light element

such as carbon, the scattered X-rays have wavelengths dependent on the angle of scattering.

Compton (1923) assumed that the scattering process could be treated as an elastic collision between an X-ray photon and a ‘free’ electron, and that energy and momentum would be conserved.

(a) Explain what is meant by a photon.

[1] (b) The elastic collision between a photon and a stationary electron may be represented as

in Fig 7.1.

The incident photon has momentum pi and energy Ei. The photon is scattered through

an angle θ and, after scattering, has momentum ps and energy Es. The electron of mass

m, which was originally stationary, moves off with speed v at an angle φ to the original direction of the incident photon.

(i) Write down equations, in terms of pi, ps, Ei, Es, m, v, θ and φ, that represent, for this interactions,

1. Conservation of energy, [1]

θ

φ

scattered photon momentum ps energy Es

electron mass m, speed v

incident photon momentum pi energy Ei

Fig. 7.1

A photon is a quantum of electromagnetic energy. The energy of a photon is proportional to the frequency of the electromagnetic waves.

Ei = Es + ½ m v2



2. Conservation of momentum along the direction of the incident photon, [1]

(ii) Suggest, with a reason, whether the scattered photon will have a wavelength that is greater or less than that of the incident photon.

[2]

(c) In an experiment to provide evidence to justify Compton’s theory, measurements were

made on the wavelength λi of the incident photon, the wavelength λs of the scattered

photon and the angle θ of scattering. Some data from this experiment are given in Fig. 7.2.

λλλλi / 10-12 m λλλλs / 10-12 m θθθθ / o ∆∆∆∆λλλλ=λλλλs - λλλλI /m

191.92 193.27 57 1.35 x 10-12

153.30 154.65 57 1.35 x 10-12

965.04 966.84 75 1.80 x 10-12

Use the data in Fig 7.2 to show that, when a photon is scattered, the change in wavelength produced is independent of the wavelength of the incident photon.

[2]

Fig. 7.2

Pi = Ps cos θθθθ + m v cos φφφφ

Since some energy of the photon is converted to the k.e. of electron,

Ei > Es, hfi > hfs, λλλλI < λλλλs , the scattered photon has longer wavelength.

When θθθθ is unchanged, ∆∆∆∆λλλλ is a constant. When θθθθ is changed, ∆∆∆∆λλλλ changes.

Hence ∆∆∆∆λλλλ is independent of λλλλI but dependent on θθθθ.



(d) In this experiment, the uncertainty in the measurement of θ is ±5o.

Determine the value of cos θ, with its uncertainty, for the angle θ = 75o ± 5o.

cos θ = 0.26 ± 0.09 [3]

(e) Compton’s theory suggests that the change in wavelength ∆λ is related to the angle θ of scattering by the expression

∆λ= k (1 – cos θ) where k is a constant.

Experimental data for the variation with cos θ of ∆λ are shown in Fig 7.3.

(i) On Fig. 7.3, draw the best-fit line for the points. [1]

(-0.50, 3.70)

(1.00, 0.12)

2.50

cos 80o = 0.1736 cos 75o = 0.2588 cos 70o = 0.3420

∆∆∆∆ cos θθθθ = (0.3420 - 0.1736)/2 = 0.084 = 0.08 or 0.09 (1 s.f.) cos 75o = 0.2588 = 0.26 (2 d.p.) or 0.2588 - 0.1736 = 0.0852 0.3420 - 0.2588 = 0.0832 Take max uncertainty,

∆∆∆∆ cos θθθθ = 0.085 = 0.09 (1 sig)



(ii) Use two different ways to determine the constant k from the graph of Fig. 7.3.

Find the average value of k. average value of k = 2.45 × 10-12 m [4]

(f) For a carbon atom, the binding energy of an electron is of the order of a few electronvolts. Compton's theory assumes that the electrons are not bound in the atoms but are free. Suggest whether, for 30 keV photons, this assumption is justified.

[1]

The binding energy of an electron of a carbon atom is very much smaller than the energy of the incident photons (30 keV), the binding energy of the electrons can be ignored. Thus it is justified to assume that the electrons are free.

∆∆∆∆λλλλ= k – k cos θθθθ (-k is the gradient and k is the intercept)

k is the y-axis intercept. k = 2.50 x 10-12

k = - gradient= 12

12(3.70 0.12) 102.39 10

( 0.50) 1.00

XX

−−−−−−−−−−−−

− =− =− =− =− −− −− −− −

<k> = 2.45 x 10-12 m



Section B

It is recommended that you spend about 30 minutes on this section. 8 Many musical instruments, such as organ pipes, flutes and clarinets, employ resonating air

columns to produce note of particular frequencies. The length of the resonating column may be changed to produce a note of a different frequency. It is suggested that different volumes of air in a container may resonate at frequencies which depend on the volume of the air.

Design an experiment to investigate how the resonant frequency of the fundamental mode of vibration of air in a container depends on the volume of the air. You may assume that the following apparatus is available with any other standard equipment which may be found in a school or college:

Flute

Measuring cylinder

Containers having different volumes

Signal generator

Microphone

Bunsen burner

Loudspeaker

Voltmeter

Ammeter

Oscilloscope

Bucket of water

Thermometer

Your answer should contain a diagram showing how the chosen equipment would be arranged, together with details of (a) the procedure to be followed, (b) the method by which the volume of the air and the resonant frequency may be

measured, (c) the control of variables, (d) any precautions you would take which may improve the accuracy of your experiment.

[12]



Diagram

1. Perform this experiment in a sound-proof room.

2. Fill up the resonance bottle completely with water. Pour the water from the resonance bottle into a measuring cylinder, pouring out in stages if necessary, until the total volume VT of water has been measured. Read all volumes at the lower meniscus.

3. Ensure room temperature is constant by monitoring using a mercury in glass thermometer.

4. Using a measuring cylinder, measure and pour a volume VW of 50 cm3 of water into the empty resonance bottle. The volume of the air in the resonance bottle is now V = VT – Vw

5. Set up the resonance bottle with the apparatus as shown in the diagram above. Throughout the experiment, clamp the resonance bottle and the measuring cylinder to avoid spillage or breakage.

6. Switch on the signal generator. Starting with the lowest possible value of frequency, gradually increase the frequency output of the signal generator. The loudness of the sound from the resonance bottle will increase gradually even though the amplitude of the signal is not adjusted. Stop increasing the frequency when the sound is perceived to be the loudest. Make small changes to this frequency to locate the frequency at which the loudest value is heard.

7. The amplitude of the sound wave to the speaker is kept constant.

8. The speaker is placed at the same position for every part of the experiment.

9. By multiplying the number of divisions on the x-scale occupied by a complete waveform in the oscilloscope screen with the time scale of the oscilloscope, the periodic time of the

sound wave can be found. Calculate the resonant frequency by f = 1

T.

10. Add another 20 cm3 of water to the resonance bottle. Repeat the experiment to find 8 sets of values of f vs V

11. Plot a graph of f vs V

Signal generator

Loudspeaker

Resonance bottle

Oscilloscope



OTHER ACCEPTABLE VARIATIONS

1. Use of constant frequency and gradually changing volume of water.

� Start with full container of water and gradually remove water, to ensure first resonance detected is fundamental mode. Use syringe or dropper for fine changes to the volume of the container

� Use of a container with some means of gradually draining water out.

2. Use of tuning forks instead of signal generator with loudspeaker

3. Use of ruler/vernier calipers to measure dimensions of resonance bottle and using calculations to find volume of air in container.

� V = π r2 d

4. Use of microphone and oscilloscope to detect loudest sound. Microphone must be place OUTSIDE the resonance bottle.





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2010 H2 Physics Ans - NYJC