1.1.1 describing motion describing motion answers 2) · edexcel as physics advanced institution of...

Edexcel AS Physics Advanced Institution of Physics 1.1.1 Describing Motion

Describing Motion Answers

1) 12.5 m s–1

2) a) 0.018 s

b) 0.036 m

3) a)

b)

Distance and Displacement Answers

1)

a) 380 km

b) 34 km North

2) –2.5 m s–2

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More Information from Graphs of Motion Answers

1) a)

b) i. 1800 m forwards

ii. 1400 m forwards

iii. 1 m s–2; –0.5 m s–2; –1 m s–2; 1 m s–2

Equations of Motion Answers

1) Unless the pedestrian gets out of the way, there will be a collision.

2) a) 10 m s–2

b) i) 45 m

ii) 3 s

c) 15 s

d) 24 m

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e) 120 m (to 2 s.f.)

Moving in More Than One Direction – Using Vectors Answers

1) The relative velocity against wind increases their wind speed for a comparatively low ground speed. Thus they don’t have to hit the ground so fast but still get enough lift from the wind passing over the wings.

2) 58 cm (to 2 s.f.) at an angle of 37° (2 s.f.) north of west. 3) 5.4 m s–1 with a bearing 30° east of north.

Causes of Motion Answers

1) Examples such as a ball that is kicked will stop rolling, a puck on ice will eventually

slow and stop, and a clock pendulum needs a weight or a spring to keep it ticking. The scientific explanation is that a friction acts to oppose the motion, so a force is needed to overcome friction.

Newton’s First Law of Motion Answers

1)

2)

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Drag Forces Answers

1)

2) At first the only vertical force acting on the skydiver is their weight. As the skydiver gains speed the air resistance increases until this drag force is equal to the weight and the skydiver reaches a constant terminal velocity. When the parachute is opened the air resistance increases hugely so there is a net force upwards. This slows the skydiver down until once again the air resistance balances the skydiver’s weight and the skydiver reaches a new, slower terminal velocity.

Newton’s Second Law of Motion Answers

1)

a)

b)

c) For part a the gradient is 2.0 m s–2 N–1.

For part b the gradient is 0.5 m s–2 kg.

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d) Acceleration is proportional to the applied force for constant mass, and also to the reciprocal of mass for a constant force (it is inversely proportional to the mass).

2) Mass of locomotive (m ) = 70 tonnes = 70 000 kg Rate of acceleration of locomotive (a ) = 1 m s–2. Force exerted by locomotive (F ) = 70 000 kg × 1 m s–2 = 70 000 N

3)

a) Mass of woman = 60 kg Acceleration = 14 m s–2 Time = 0.15 seconds

v = u + at (equation 1) 14 = 0 + a × 0.15 a = 14/0.15 = 93.3 Force = 60 kg × 93.3 m s–1 = 5600 N

b) Weight = 9.81 × 60 kg = 588.6 N. The force acting on the woman is approximately 9.5 times as large as her weight.

Inertia, Mass and Weight Answers

1) When the bus accelerates, if the person is to accelerate with it a force must be applied. For someone seated this comes from the reaction of the seat, but for someone standing it must come from friction with the floor. This can result in the person being thrown forward or backward, as they experience the force as if their feet were being pulled out from under them, while their body remains in its original position.

2) 19.6 m

3) 1.6 N kg–1

4) a) 6.9 s

b) Yes. Sound would only take 0.69 s to reach the ground.

Newton’s Third Law of Motion Answers

1)

a)

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b)

Statics Answers

1) 30 N

Projectiles Answers

1) a) 1.5 s (2 s.f.)

b) 19.9 m

c) 2.87 m

2) 2.4 m s–1 (2 s.f.)

3) 93.4 m s–1

4)

a) 6.4 s

b) 570 m

c) 110 m s–1

The Concept of Energy Answers

1) Energy is not lost, but is transferred to other forms such as heat and sound.

2) This method does not actually ‘save’ energy, but simply allows energy generated while there is less demand to be used to provide a source of energy when demand is high. The energy is stored as gravitational potential energy of the water in the higher reservoir.

Energy Transformations Answers

1) a) Boiling water in a kettle.

b) Putting a can of paint on a shelf; stretching a spring.

2) 96 J (2 s.f.)

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Energy and Efficiency Answers

1) a) 4.1 m s–1

b) 0.85 m

2) a) 22.6 J

b) 14.4 m

c) 15.8 m s–1

Power Answers

1) 2.2 kW

2) 2710 kW

3) 27.5 kW

HSW The Mechanics of Hockey Answers

21 N

1.13 s

Eureka Answers

1) 915 kg m–3

2) a) 0.82 g cm−3

b) 820 kg m−3

3) Suitable estimates. For a room 3 m × 8 m × 8 m the mass of air would be 192 kg.

4) 0.54 N

5) The line for fresh water is higher on the hull because fresh water is less dense than salt water. For a certain load, a ship will sink its lowest in fresh water, so this line needs to show the lowest the ship can float safely.

6) Volume = 0.18/8000 = 2.25 × 10–5 m2

Upthrust = (800 × vol) × 9.81 = 0.17658 N

Weight = 0.18 × 9.81 = 1.7658 N

Tension = Weight – upthrust = 1.7658 – 0.17658 = 1.6 N

Fluid Movement Answers

1) Hull of a racing yacht; racing bicycle; car body.

2)

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3) In summer the volume and rate of water flow is such that the creek flows smoothly with streamline flow. In autumn, leaves floating on the water move following each other exactly, indicating streamline flow. In winter there is no flow as the creek is frozen, but with the spring and the thaw of snow the creek is full. The fast flowing water flows turbulently, producing the eddies and currents described in the poem.

Drag Act Answers

1) There is more resistance to movement in water than in air.

2) For gases there is an increase in viscosity with rise in temperature, but for liquids viscosity decreases with rise in temperature.

3) Warmer water would be less viscous so swimmers could travel faster through the water.

4) The chocolate would flow more quickly at a higher temperature, allowing faster production. It would also be thinner, so a thinner coating could be applied.

Terminal Velocity Answers

1) 1.98 × 10–3 N

2) The cat does not have a fixed shape. It may be moving and so the air resistance will be constantly changing. Stokes’ law only applies to small spheres moving at slow speeds.

3) a) 3.8 × 109 m s–1

b) 6.0 × 108 m s–1

4) The answers are clearly wildly wrong – the meteorite is travelling faster than light! We have assumed that the weight stays constant, that the gravitational force is constant, that the meteorite has enough time to reach its terminal velocity, and that the temperature of all the objects involved is constant at 20°C. All these assumptions are flawed. Stokes’ law does not apply for such a large object, or at such high speeds.

The Physical Properties of Solids Answers

1) 80 N m–1

2) a) F × x

b) 330 N m–1

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c) 34.5 mm

d) 5.04 × 10–3 J

3) 800 J

Characteristics of Solids Answers

1) The Young modulus is stress/strain. The unit of stress is N m–2 (= Pascals) and strain is a ratio and has no unit.

2)

a) 9.5 × 107 N m–2

b) 5.6 × 10–4

c) 1.7 × 1011 N m–2

3)

a) A metre of the wire under test (1000 mm original length) should not stretch by more than 1 mm, for example.

b) Refer to description on pages 66–67. The limitation in the amount of strain is to ensure that the wire under test obeys Hooke’s law throughout the experiment.

c) 69.9 N Characteristics of Solids II Answers

1)

a) elastic limit – the point on a stress–strain (or force–extension) graph beyond which the material will not return to its original size when the stress is removed.

b) plastic behaviour – the region of the graph where stress produces permanent deformation of the material.

c) Hooke’s law – the straight-line portion of the graph where stress is proportional to strain. Extension is proportional to the applied force. Hooke’s law is obeyed.

d) breaking stress – the stress at which the material breaks.

e) compressive strain – the strain (deformation) when a material is squashed.

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2) 40 cm

3)

a) malleable – a material whose shape can be changed permanently and shows plastic deformation at low stress.. Example: gold, to make jewellery.

b) tough – a material that can withstand high impact forces and absorbs a lot of energy before breaking. Example: Kevlar, used for bullet-proof vests.

c) hard – a material that cannot be scratched or dented easily. Example: diamond, used for heavy duty cutting wheels.

d) ductile – a material that can be pulled into wires with small stress required. Example: copper, used for electrical wiring.

e) brittle – a material that breaks without plastic deformation. Example: biscuits and crisps, which are designed to break with a snap!

Materials in the Real World Answers

1)

a) Terminal velocity is inversely proportional to the viscosity, so the higher the terminal velocity, the lower the viscosity.

b) The time for a small ball of a known diameter to fall a given distance is measured, and from this the terminal velocity (from distance ÷ time) and hence the viscosity can be calculated.

c) Manufacturers need to control flow rates of the liquid chocolate to achieve consistent products with as little waste and as possible.

2) See practical described on page 59 of the Students’ Book.

3)

a) The maximum stress that can be applied to a material before it breaks.

b) You would want a material with a high breaking stress to protect against impact.

c) malleable – no; ductile – no; tough – yes (will withstand impacts); brittle – no; strong – yes (hard to break); hard – yes (durable and hard to dent); stiff – yes (keeps its shape and hard to bend).

d)

Types of Wave Answers

1) Transverse waves: light, ripples in water, vibration of a stretched string

Longitudinal waves: sound, mechanical waves in a slinky pushed back and forth, seismic p waves.

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2) a) The movement of the energy that causes a wave.

b) A wave train has a definite beginning and an end, but a continuous wave goes on forever (it is infinite).

3) For a large enough circle the curve approximates to a straight line.

4)

5) In reality a wave cannot be infinite – it must have a beginning and an end.

The Vital Statistics of a Wave Answers

1) 83 Hz

2)

a)

b)

3)

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a) 360°

b) 180°

c) (–)180°

Behaviour of Waves Answers

1)

2)

a) Second harmonic 82 Hz; third harmonic 123 Hz.

b) 86.9 m s–1

3) If a bridge starts to vibrate at its resonant frequency the vibrations could become very large and tear the structure apart. The engineers would need to ensure that the design did not offer resonant frequencies that are likely to occur naturally.

Reflection at the End of a String Answers

1) The loudspeaker vibrates back and forth in the same direction as the propagation of the sound. It physically pushes air molecules back and forth.

2) a) 2.8 cm

b) 1.1 × 1010 Hz

3) The string is fixed at both ends. On reflection at the end any vibration undergoes a 180° phase change, so a positive displacement would change to a negative displacement at the point of contact. The string is a physical object that cannot be in two places at once, and

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so there must be zero displacement at the ends of the string. Moreover, the incoming and reflected waves will always be in antiphase and therefore completely cancel each other.

4) Capture the sound of both instruments playing the same pitch on an oscilloscope. Although the shape of the waveform will be different, the main peaks should appear at the same frequency, and hence indicate the same pitch.

Models of Waves and Their Properties Answers

1) Critical angles are: diamond 24.4°, ice 49.8°, benzene 41.8°

2) 2.0 × 108 m s–1

3) Place a semicircular glass block in a tank of water and shine a single ray of light through the water into the glass block, at various angles of incidence. For each, measure the angle of refraction within the glass. A graph of sin i against sin r should produce a straight line with the gradient equal to the refractive index. As per Student Practical 15 with the glass block resting underwater instead of in air.

Diffraction and Interference Answers

1)

2) There is interference between the signals from the two transmitters and she is sometimes in places where the signals cancel out and in other places where there is reinforcement.

3) Scientists determining the same conclusions independently, oblivious of each other’s work, produce the strongest evidence for the veracity of scientific theories.

Polarisation Answers

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1) The signal is polarised, so the aerial needs to be in the correct orientation to pick up

the signal.

2) Sound waves are longitudinal and so cannot be polarised.

Light as a Wave Answers

1)

a) 7.5 × 1012 Hz

b) 4.3 × 1014 Hz

c) 4.6 × 1020 Hz

2) He knew that light waves travelled through a vacuum, and oscillating electric charges could create magnetic fields, and oscillating magnetic fields could move electric charges.

Applications of Electromagnetic Waves Answers

1)

a) 100 m to 1 m

b) To avoid reflection by the ionosphere and reach satellites which are outside the atmosphere.

2) The atmosphere absorbs this UV wavelength, and so humans have never been naturally exposed to it, so could not evolve cells that respond to it.

3) Similarity: X-rays and gamma rays can have the same frequency.

Difference: X-rays produced by decelerating electrons while gamma rays are produced as a result of energy change in the nucleus of an atom.

4) (Students' own answers)

Pulse-echo Detection Answers

1) The fact that light from other galaxies is Doppler shifted towards the red end of the

spectrum shows they are moving away from us. That virtually all galaxies show this red shift indicates that the whole Universe is expanding.

2) a) The sound is Doppler shifted. As the car approaches the frequency is raised and as it

moves away the frequency drops.

b) The driver is not moving with respect to the sound, so there is no shift in frequency.

Hubble’s observations led to new idea of the origin and structure of the Universe.

Ultrasound Answers

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1) Send a radio signal to the Moon and record the time taken for the reflected pulse to return. Knowing the speed of the pulse and the time taken, the distance can be calculated.

2) Both bat’s echolocation system and air traffic control radar use reflection from the object to locate it. They are different in that the bat uses ultrasound and radar uses radio waves. In addition, radar use Doppler shift in the reflected frequency to calculate the speed of the moving object.

3) Distance away is 51 km; speed of movement of storm is away from the detector.

4) Pulse length method: l = v × t = 1520 × (1 × 10–6) = 0.00152

Resolution = half pulse length = 0.76 mm

Wavelength method: λ = v /f = 1520/(3 × 10–6) = 0.51 mm

So worst resolution = 0.76 mm

Electric Current Answers

1) 240 C

2) 1.6 × 10−19 C

3)

Energy and Electricity Answers

1) a) The amount of energy supplied to each unit of charge in a circuit is the electromotive

force. 1 V = 1 J C–1.

b) The pd is the amount of energy supplied by each unit of charge – a measure of the work being done.

2) a) 1.5 V

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b) 8 V

Resisting Current Flow Answers

1) Provided the temperature and other physical factors remain constant, the current through a

wire is proportional to the potential difference across its ends.

2) 120 Ω

3) An ohmic conductor obeys Ohm’s law, but a non-ohmic one does not – the current is not proportional to the potential difference across the conductor.

4) If the pd were plotted on the y-axis then the slope of the line would be equal to the resistance.

5) a) 4.4 × 10−7 Ω m

Answers include difficulty in positioning the contacts, the fact that a ring doesn’t usually have a rectangular cross-section, so finding cross-sectional area is difficult, and the physical size of the connecting crocodile clips, which would make determination of the area so inaccurate as to render the answer useless.

The Transport Equation Answers

1) 0.94 A

2) 1.92 × 1035 m−3

Power and Work in Electric Circuits Answers

1) You know that from Ohm’s law V = IR , so by substituting for V in the equation P = VI

you can obtain P = I 2R .

2) a) 0.065 W

b) 2.5 kW

c) 0.065 W

Circuits Containing Resistors Answers

1)

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The ammeters should show the same reading in all wires, as the charge is conserved.

2) Around a circuit the drop in potential energy where energy is supplied from the flowing charges is matched by the rise in potential energy where energy is supplied to the charges. There is no ‘leaking’ of energy from the circuit.

3)

The Potential Divider Answers

1)

a) 7.2 V

b) 4 V across the 3000 Ω resistor.

2) Arrange this as a potential divider circuit, with the contact across 0.62 of the length of the 80 Ω resistance wire.

Sources of emf Answers

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1) Because of the internal resistance of the power supply.

2) Plot voltage across the power supply against the current flowing will give a graph with a gradient of −r (where r is the internal resistance of the supply) and an intercept on the voltage axis of the emf of the power supply.

Understanding Conduction Answers

1)

a)

b) When the bulb filament becomes hot the lattice vibrates more and there are more collisions between the conduction electrons, so the resistance increases.

c) The average velocity is reduced because of the increase in collisions. This reduces the current as it is proportional to drift velocity.

d) By extrapolating backwards there could be a temperature at which there is no resistance.

2) A positive charge that is the result of an electron leaving an atom.

3) In n-type semiconductors the doping element donates electrons to provide negative charge carriers, whereas in a p-type semiconductor the doping element traps electrons and so introduces positive holes as charge carriers.

4) It provides more charge carriers over and above those present in the semiconductor lattice.

5) In some semiconductors a rise in temperature frees more charge carriers, increasing current, so the resistance effectively goes down.

A Brief History of Light Answers

1) Particle theory: Democritus, Newton

Wave theory: da Vinci, Grimaldi, Huygens, Hooke, Boyle, Young, Foucault

2) Foucault’s work showed that light must travel more slowly in water than air, in direct contradiction to the prediction of the particle theory.

3) Newton was a very influential scientist and his view was generally accepted.

4) A theory of wave−particle duality, in which light behaves as a wave and a particle in different circumstances.

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Wave or Particle? Answers

1)

a) 2.65 × 10−18 J

b) 2.37 × 10−15 J

c) 3.6 × 10−19 J

2) 1.66 × 1020

3) From the worked example on this page, the solar flux = 1000 W m−1. Area needed = 4.4 × 10−3 m2. Side length = 6.6 cm

4) The ultraviolet catastrophe was that at higher frequencies more and more energy would be radiated by a black body, reaching infinity – clearly impossible. Planck’s idea was that energy could only be absorbed or radiated in discrete quantities, not in continuous amounts.

The Photoelectric Effect Answers

1) The photons of red light do not have enough energy to release an electron from the

surface of the zinc, but the photons of ultraviolet light do.

2) a) hf = φ + ½mv 2 max Here h is Planck’s constant, f is the frequency of the incident

light, φ is the work function of the metal surface, and ½mv 2max is the maximum

kinetic energy of the photoelectron.

b) 3.0 × 10−20 J

c) 2.6 × 105 m s−1

d) 5.6 × 1014 Hz

3) As per the text and diagram on page 154.

4) Such a camera could be used to monitor dark areas (at night for example) and the results used to provide evidence of need for policing.

Atomic Electron Energies Answers

1)

a) Line spectra are made up of distinct lines of light with distinct frequencies. A continuous spectrum is made of light of all frequencies.

b) Emission spectra show the radiation given out by electrons of an element as they move from an excited state to one of lower energy. Absorption spectra are produced when electrons absorb energy from light incident on the atom to move from a lower energy level to a higher level.

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c) Excitation is when an electron in raised to a higher energy level around the nucleus of the atom. Ionisation is when an electron absorbs enough energy to escape completely from the atom.

2) 120 nm

3) Each element has a unique structure of energy levels. As transitions can only occur between these levels, each transition gives rise to a unique frequency of light corresponding to the energy difference between two levels. In chemistry the distinct colours given out by common elements when they are heated are used to identify them, for example sodium gives out yellow light when it is heated, while potassium gives out lilac light.

4) If a 10 eV photon was incident, then nothing would happen, as exactly 10.2 eV is needed to lift the electron into the next energy level. 10 eV does not correspond to any allowed transition. A 20 eV photon would ionise the hydrogen atom. Two 10 eV photons would have no effect, as the energy must be supplied by a single photon.

Solar Cells to Light the World? Answers

1) The energy from the Sun will not run out.

2) A variety of factors, for example cost, efficiency of cells, availability, reliability, legislation.

3) 96 W.

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1.1.1 describing motion describing motion answers 2) · edexcel as physics advanced institution of...

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