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PHYSICS

Unit 2 – Written examination 2

2011 Trial Examination

SOLUTIONS SECTION A – Core Area of Study 1 – Motion Question 1 Answer: 7 m s-2 south Explanation: Acceleration is equal to the gradient of the velocity-time graph, which is steepest in the section from 5 – 6 seconds

a = ∆vt

a = 71

a = 7ms−2

Question 2 Answer: 44.5 m Explanation: Displacement is equal to the area under the graph, using a combination of triangles and rectangles to make calculations easier.

d = Area = 12×10×3+ 2×10+

10+3( )2

×1+ 12× 2×3

d = 44.5m

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Question 3 Answer: 5.56 ms-1 Explanation: Average speed is equal to total distance over time taken. The total distance is equal to the

vav =dt

vav =44.5

8vav = 5.56ms−1

Question 4 Answer: 46.9 J Explanation: Kinetic energy =

KE = 1

2mv2

KE = 12×0.15×252

KE = 46.875J

Question 5 Answer: 312.5 N Explanation: Work is equal to change in kinetic energy and also the area under the force-distance graph. W = ∆KE = 46.875J12× 0.3×Fb = 46.875

Fb = 312.5N

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Question 6 Answer:

Fair on ball

B Fearth on ball

Fearth on ball

Fair on ball

Explanation: Note that the air resistance acts in the opposite direction to motion. Question 7 Answer: 39.2 m Explanation: Use SUVAT equations, with u = 28 m s-1, a = - 10 m s-2, v = 0 v2 = u2 + 2as

s = −282

2×−10s = 39.2 m

Question 8 Answer: 58.8 J Explanation: GPE = mghGPE = 0.15×10×39.2GPE = 58.8J

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Question 9 Answer: 4.2 kg m s-1 Explanation: Use change in momentum: ∆p = m∆v∆p = 0.15× 28∆p = 4.2kgms−1

Question 10 Answer: 52.5 N Explanation: Use impulse equals change in momentum F∆t = m∆v

F = m∆v∆t

F = 4.20.08

F = 52.5N

Question 11 Answer:

Fearth on ball

Fground on ball

Explanation: Fground on ball is the reaction force which acts on the ball. F earth on ball is the weight force. Fground on ball is significantly larger as it must act to change the direction of the ball on impact.

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Question 12 Answer: 6.7 x 104 N m-1 Explanation: The gravitational potential energy at 5 m ( mgh = 0.15 x 10 x 5 = 7.5 J) is all converted to kinetic energy, then to elastic potential energy as the ball compresses. Then:

E = 12

kx2

k = 2×Ex2 = 2× 7.5

0.0152 = 6.7×104 N m−1

Question 13 Answer: 0.6 J as heat, sound. Explanation: Any losses are explained by the difference in gravitational potential energy. The energy is converted to heat and sound in the course of the collision. ∆E = mg∆h∆E = 0.15×10×0.4∆E = 0.6 J

Question 14 Answer: 400 N Explanation: Due to Newton’s Third Law – action reaction pair. Question 15 Answer: Newton’s Third Law Explanation: As above. Question 16 Answer: 0.54 m s-2 Explanation: Use F = ma, Newton’s Second Law, for the forces acting on the car.

F∑ =ma1200−150− 400 =1200×a

a = 0.54m s −2

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Question 17 Answer: 184 N Explanation: Use F = ma, Newton’s Second Law, for the forces acting on the trailer, knowing the acceleration from Question 16.

400 0.54 400184

X

X

F maF

F N

=∑× = −=

Question 18 Modern bicycle helmets are designed to crush and thus increase the time taken for the head to come to rest in an accident. This means that, although the impulse and change in momentum remain constant, the net force acting on the head is less. Another way of explaining the situation is to use Newton’s Second Law. An increased stopping time (and distance), decreases the average acceleration experienced by the head and thus the net force is reduced. Area of Study 2 – Wavelike Properties of Light Question 1 Answer: 1 x 10-14 s Explanation: One cycle of the wave covers 10 x 10-15 s = 1 x 10-14 s. Measured between two equivalent points (e.g. crest to crest) Question 2 Answer: 1 x 1014 Hz Explanation:

f = 1T

= 11×10−14

=1×1014 Hz

Question 3 Answer: 3000 nm Explanation:

86

143 10 3 10 30001 10

v f

v nmf

λ

λ −

=

×= = = × =

×

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Question 4 Answer: 4 ms Explanation:

v = dt

t = dv

t = 1200×103

3×108

t = 4×10−3 s

Question 5

Behaviour Wave Model Particle Model

Rectilinear Propagation X X

Interference X

Polarisation X

Refraction X

Reflection X X

Diffraction X

Question 6 Answer: i = 65o, r = 65o Explanation: Angle of incidence and reflection are measured relative to the normal and should be equal. Question 7 Answer: 1.47 Explanation:

1 2sin sinsin82 1.46sin88

1.473A

A

n i n rnn

==

=

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Question 8 Answer: 82.4o Explanation:

1 2

1 2

1

1

sin sin90

sin

1.46sin1.473

82.4o

n i n

nin

i

i

−

−

=

=

=

=

Question 9 Total internal reflection requires the light to be bending away from the normal to such an extent that it reflects back into the original medium. If light were to travel from B to A, it would bend towards the normal and thus never head towards internal reflection. Question 10 Answer: A A is most likely to be red because it is refracted least of all the colours in the visible spectrum. The splitting of white light into its spectral components is called dispersion. Question 11 Answer: Cyan. Explanation: By standard colour addition. Question 12 Answer: Black. Explanation: Although an orange filter may also allow some yellow and red to pass, the blue pants are reflecting only blue (along with small amounts of green and violet). Thus the filter blocks all of the light from the blue pants and they appear dark (black). Question 13 Answer: None – black Explanation: Although the orange filter may allow some red light to pass, the green filter certainly blocks all of it, so no light would pass through and emerge on the other side.

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Question 14 The observation of polarising behaviour for light requires a transverse wave model for light. Polarisation is the process whereby particular planes of a wave are filtered. Longitudinal waves and particles would pass through unaffected, so these models are insufficient. Question 15 Answer: As per diagram.

Red Green

A

Explanation: Green light has a higher refractive index than red, so will refract further when it meets the lens. Note that there should be separation both in the lens and further beyond it. Question 16 Answer: 900nm – infrared. Explanation: v = f λ

λ = vf= 3×108

3.3×1014= 9×10−7 = 900nm

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SECTION B – Detailed Studies Detailed Study 1 – Astronomy Question 1 Answer: B Explanation: Azimuth is measured clockwise from North, which is taken as 0o Question 2 Answer: A Explanation: Altitude is measured from the horizon, once azimuth is established. 0o = Horizon, 90o = Zenith. Question 3 Answer: B Explanation: By definition. Question 4 Answer: D Explanation: Stars will rotate about the zenith, so all will remain at a constant altitude. The same hemisphere of stars will remain visible at all times. Question 5 Answer: C Explanation: Alpha, beta, gamma, delta, epsilon in decreasing order of brightness as per the Bayer method of naming constellations. Question 6 Answer: D Explanation: Heliocentric model implies that the Sun is at the centre of the solar system, orbited by all planets.

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Question 7 Answer: B Explanation: By definition. Question 8 Answer: D Explanation: Many objects do emit radiation outside the visible spectrum, making radio telescopes ideal for observing otherwise unseen bodies. Images cannot be “seen” and resolution is poorer, but radio telescopes can be constructed in arrays to gather sufficient data. Radio telescopes can also double as radar. Question 9 Answer: A Explanation: Apparent magnitude refers to the brightness of a star when viewed from earth (negative means brighter) Question 10 Answer: B Explanation: Absolute magnitude refers to the brightness/luminosity of a star when viewed from 33 light years (i.e. distance is removed as a variable).

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Detailed Study 2 – Astrophysics Question 1 Answer: A Explanation: Option A refers to the steady state model. Other answers match The Big Bang Theory, which implies that all matter began from a single point, so galaxies were indeed closer earlier in time. Quasars were more common, background radiation would have a longer wavelength. Question 2 Answer: B Explanation: The wavelength of the background radiation is sufficiently lengthened due to the expansion of the universe to put it into the microwave region of EMR. Question 3 Answer: A Explanation: All galaxies are receding from us, with more distant ones moving fastest and thus receding at a greater rate. Red shifting means that the frequency of light from the galaxies is increased (as per Doppler Effect) and this would be more prominent for faster, more distant galaxies. Question 4 Answer: B Explanation: The horizontal axis is hottest on the left (Class O). V stars (Red dwarfs) are furthest right, so coolest. Question 5 Answer: D Explanation: W – White Dwarf stars Question 6 Answer: C Explanation: The parallax method is useful for measuring stars within about 150 light years (approx. 45 parsecs). It requires a reference point which remains comparatively stationary and must thus be significantly further away.

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Question 7 Answer: C Explanation: A white dwarf star is the remnant of a main sequence star of relatively low mass (less than 4 solar masses). Once the white dwarf cools, it becomes a red dwarf. Question 8 Answer: A Explanation: A large supernovae applies to heavier stars, resulting in a black hole or neutron star in a massive explosion. Smaller stars (such as our sun) become red giants then white dwarfs. Question 9 Answer: A Explanation: Standard shape - elliptical Question 10 Answer: A Explanation: Yellow star on the main sequence.

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Detailed Study 3 – Energy from the Nucleus Question 1 Answer: B Explanation: Matter must be “conserved” – at least by a nucleon count. Question 2 Answer: A Explanation: U-235 fission involves splitting of the atom into a variety of fragment pairs. Question 3 Answer: B Explanation: E = mc2

E = 3.7×10−28 × 3×108( )2

E = 3.33×10−11 JE = 208MeV

Question 4 Answer: D Explanation: Fission of U-235 requires a slow moving neutron to be absorbed. Question 5 Answer: A Explanation: Fusion involves the combining of lighter nuclei to form a heavier one. Question 6 Answer: D Explanation: E =18.3MeVE =18.3×106 ×1.6×10−19 JE = 2.93×10−12 J

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Question 7 Answer: D Explanation: The strong nuclear force is required to overcome the electrostatic repulsion between protons. Question 8 Answer: B Explanation: Y points to the fuel rods, where enriched fissile material (e.g. U-235) undergoes fission, producing large amounts of heat energy. Question 9 Answer: A Explanation: X points to the cadmium control rods, which absorb neutrons to control the rate of fission in the reactor core Question 10 Answer: D Explanation: Uranium enrichment is required because U-238 is not fissile, but U-235 certainly is.

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Detailed Study 4 – Flight Question 1 Answer: C Explanation: Vertical forces must be balanced: Lift =Weight +FT

FW =1.3×104 ×10+ 6×104

FW =1.9×105 N

Question 2 Answer: D Explanation: Torque must be balanced about the Centre of Gravity: FT × 23- 6.9( ) =1.9×105 × X −6.9( )X =12.0m

Question 3 Answer: B Explanation: Thrust must equal drag at constant speed. 2 x 85 = 170 kN Question 4 Answer: B Explanation: Increasing the angle of attack will increase the lift, but only until laminar flow is disrupted and the wing stalls. Question 5 Answer: A Explanation: Kinetic energy will increase slightly as the air moves faster. Static pressure must reduce to compensate and satisfy the overall constant.

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Question 6 Answer: C Explanation: Lift is generated perpendicular to the top wing surface, which is not at 90o to the incoming airflow. In fact, it is directed slightly backwards, hence the “induced” drag.

LiftVertical component

Horizontal component = Induced drag

Question 7 Answer: A Explanation: Skin friction is reduced by a laminar flow. Question 8 Answer: B Explanation: Streamlining of a fuselage is designed to reduce pressure (form) drag. Question 9 Answer: A Explanation: The rudder modifies airflow about the tail of the plane and causes left/right directional changes (yaw) Question 10 Answer: B Explanation: Newton’s Third law concerns action-reaction pairs, not acceleration (C) or intertia (D).

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Detailed Study 5 – Sustainable Energy Sources Question 1 Answer: C Explanation: Renewable resources must be replaceable or regrown to ensure sustainability. Question 2 Answer: C Explanation: C is clearly problematic and not a long term solution. Question 3 Answer: C Explanation: E = mgh1.2×109 = 2×106 ×10×hh = 60m

Question 4 Answer: B Explanation:

Pin =1.2×109

30= 40 MW

Pout = 33 MW

%Efficiency = 3340

= 83%

Question 5 Answer: B Explanation: E =14%×15×60×60×3E = 22680 JE ~ 23kJ

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Question 6 Answer: D Explanation: Incandescent globes produce excessive heat, which is lost to the surrounding air, reducing efficiency. Question 7 Answer: A Explanation: In fact, the majority of the electrical energy just heats the surroundings, rather than lighting it. Question 8 Answer: C Explanation: E = I × A× t9×106 = 500×5× tt = 3600 st =1hr

Question 9 Answer: B Explanation: E = 70%×9×106

E = 6.3×106 JE = m× c×∆t6.3×106 = m× 4200×1m = 2140kgm = 2140 L

Question 10 Answer: D Explanation: E = m× c×∆t6.3×106 ×2 = m× 4200× 22m =195kgm =195L

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Detailed Study 6 – Medical Physics Question 1 Answer: D Explanation: Gamma emitters are preferred due to their high penetration (can be detected) but low ionizing power (less damage). Alpha and beta do ionize, but have low penetration. Gamma radiation has no mass. Question 2 Answer: B Explanation: Coherent bundles are only necessary for image formation. Light guides tend to be incoherent. Question 3 Answer: B Explanation: X-rays produce remove electrons – this is ionisation. Question 4 Answer: A Explanation: Higher frequencies penetrate more effectively than lower frequencies and are thus far more useful for internal organs such as the liver. Question 5 Answer: D Explanation: Piezoelectric crystals are used and the transducer is designed to both send and receive ultrasound signals to form and image. Question 6 Answer: B Explanation: Ultrasound tends to reflect at significant bone/tissue boundaries, so image formation of the brain is difficult to achieve.

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Question 7 Answer: D Explanation: Rotation of the X-rays is required to form a cross-sectional image of the patient. Question 8 Answer: B Explanation: Lasers have very low divergence, so are in fact useful for treating isolated regions rather than large areas. Question 9 Answer: B Explanation: The protons resonate and when released, emit a short burst of energy which is analysed. Question 10 Answer: C Explanation: β+ represents a positron, which is emitted during the decay of isotopes such as Carbon-11 (to Boron-11). A proton in the carbon nucleus changes to a neutron and a positron (which is emitted).