Edexcel AS PhysicsAdvanced Institution of Physics Describing Motion Answers 1) 12.5 m s–1 2) a) 0.018 s b) 0.036 m a) Speed 3) Mamun sir 5 m s–1 0 Speed b) Time t 0 t = 5s Time t Distance and Displacement Answers 1) a) 380 km b) 34 km North 2) –2.5 m s–2 Mamun Sir -1- 1.1.1 Describing Motion there will be a collision. 1 m s–2.5 m s–2.f. –1 m s–2. 1 m s–2 Equations of Motion Answers 1) Unless the pedestrian gets out of the way. 2) a) 10 m s–2 b) i) 45 m ii) 3 s c) 15 s d) 24 m e) 120 m (to 2 s. –0.1 Describing Motion More Information from Graphs of Motion Answers 1) Speed /m s–1 a) 20 10 0 50 100 150 200 Time/s Mamun sir –10 –20 b) i. 1800 m forwards ii. 1400 m forwards iii.) Moving in More Than One Direction – Using Vectors Answers 1) The relative velocity against wind increases their wind speed for a comparatively Mamun Sir -2- .Edexcel AS Physics Advanced Institution of Physics 1.1. a puck on ice will eventually slow and stop.1. 3) 5. so a force is needed to overcome friction. 2) 58 cm (to 2 s.Edexcel AS Physics Advanced Institution of Physics 1. and a clock pendulum needs a weight or a spring to keep it ticking.) north of west.4 m s–1 with a bearing 30° east of north. Thus they don’t have to hit the ground so fast but still get enough lift from the wind passing over the wings.) at an angle of 37° (2 s. The scientific explanation is that a friction acts to oppose the motion.f. Mamun sir Newton’s First Law of Motion Answers 1) reaction force forward force from engine drag forces weight of racing car 2) reaction force centre of gravity of block weight of ruler Drag Forces Answers 1) weight Mamun Sir -3- . Causes of Motion Answers 1) Examples such as a ball that is kicked will stop rolling.f.1 Describing Motion low ground speed. 00 0.5 2.Edexcel AS Physics Advanced Institution of Physics 1.0 1.60 0.00 0.0 m s–2 N–1.15 a = 14/0.40 b) Acceleration a /m s–2 0.3 Force = 60 kg × 93.6 Force F/N 1. 1/mass d) Acceleration is proportional to the applied force for constant mass.5 1.15 seconds v = u + at (equation 1) 14 = 0 + a × 0.0 For part b the gradient is /kg 0. Newton’s Second Law of Motion Answers 1) Acceleration a /m s–2 Mamun sir a) 1.1 Describing Motion 2) At first the only vertical force acting on the skydiver is their weight. and also to the reciprocal of mass for a constant force (it is inversely proportional to the mass).50 c) For part a0 the gradient is 2.20 0 0 0. 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.20 1. 2) Mass of locomotive (m) = 70 tonnes = 70 000 kg Rate of acceleration of locomotive (a) = 1 m s–2.80 0. slower terminal velocity.5–1 m s–2 kg.3 0.15 = 93. 0 0.4 0.1 0.5 0. This slows the skydiver down until once again the air resistance balances the skydiver’s weight and the skydiver reaches a new. When the parachute is opened the air resistance increases hugely so there is a net force upwards.1.3 m s–1 = 5600 N Mamun Sir -4- .2 0. 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. as they experience the force as if their feet were being pulled out from under them. but for someone standing it must come from friction with the floor.1 Describing Motion Weight = 9.) b) 19. if the person is to accelerate with it a force must be applied. Inertia.5 s (2 s. For someone seated this comes from the reaction of the seat.1. while their body remains in its original position. 2) 19. This can result in the person being thrown forward or backward. Newton’s Third Law of Motion Answers 1) a) b) Statics Answers 1) 30 N Projectiles Answers 1) a) 1.6 m 3) 1. Sound would only take 0.9 s b) Yes.6 N.87 m Mamun Sir -5- .5 times as large as her weight. Mass and Weight Answers 1) When the bus accelerates.Edexcel AS Physics b) Advanced Institution of Physics 1.f. The force acting on the woman is approximately 9.9 m c) 2.6 N kg–1 Mamun sir 4) a) 6.69 s to reach the ground.81 × 60 kg = 588. 2) 96 J (2 s.4 m c) 15.2 kW 2) 2710 kW 3) 27.8 m s–1 2) Power Answers 1) 2.4 m s–1 (2 s. b) Putting a can of paint on a shelf. Energy Transformations Answers 1) a) Boiling water in a kettle.f.) Energy and Efficiency Answers 1) a) 4.) 3) 93.13 s Eureka Answers 1) 915 kg m–3 2) a) Mamun Sir 0. The energy is stored as gravitational potential energy of the water in the higher reservoir. stretching a spring.f.5 kW HSW The Mechanics of Hockey Answers 21 N 1.1 Describing Motion 4) a) 6.85 m a) 22.1.6 J b) 14.1 m s–1 b) 0. but simply allows energy generated while there is less demand to be used to provide a source of energy when demand is high.Edexcel AS Physics 2) 2.4 m s–1 Advanced Institution of Physics 1. but is transferred to other forms such as heat and sound.82 g cm−3 -6- . 2) This method Mamun does notsiractually ‘save’ energy.4 s b) 570 m c) 110 m s–1 The Concept of Energy Answers 1) Energy is not lost. but with the spring and the thaw of snow the creek is full.Edexcel AS Physics b) Advanced Institution of Physics 1. leaves floating on the water move following each other exactly.98 × 10–3 N Mamun Sir -7- . 4) 0. so a thinner coating could be applied. 2) For gases there is an increase in viscosity with rise in temperature. car body. a ship will sink its lowest in fresh water.18/8000 = 2. Terminal Velocity Answers 1) 1. For a room 3 m × 8 m × 8 m the mass of air would be 192 kg. 6) Volume = 0. For a certain load.7658 – 0.17658 = 1.17658 N Weight = 0.54 N 5) The line for fresh water is higher on the hull because fresh water is less dense than salt water.18 × 9.6 N Mamun sir Fluid Movement Answers 1) Hull of a racing yacht. Drag Act Answers 1) There is more resistance to movement in water than in air.7658 N Tension = Weight – upthrust = 1.25 × 10–5 m2 Upthrust = (800 × vol) × 9. allowing faster production.81 = 1. producing the eddies and currents described in the poem. 2) streamline flow turbulent flow 3) In summer the volume and rate of water flow is such that the creek flows smoothly with streamline flow.1.1 Describing Motion 820 kg m−3 3) Suitable estimates. The fast flowing water flows turbulently. so this line needs to show the lowest the ship can float safely. indicating streamline flow. racing bicycle. but for liquids viscosity decreases with rise in temperature.81 = 0. It would also be thinner. In winter there is no flow as the creek is frozen. In autumn. 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. Edexcel AS Physics Advanced Institution of Physics 1. c) 69. for example.1 Describing Motion 2) The cat does not have a fixed shape. 2) a) 9. 3) a) 3.7 × 1011 N m–2 a) A metre of the wire under test (1000 mm original length) should not stretch by more than 1 mm. The unit of stress is N m–2 (= Pascals) and strain is a ratio and has no unit. Stokes’ law does not apply for such a large object. Mamun sir The Physical Properties of Solids Answers 1) 80 N m–1 2) 3) a) F×x b) 330 N m–1 c) 34.1. b) plastic behaviour – the region of the graph where stress produces permanent deformation of the material. The limitation in the amount of strain is to ensure that the wire under test obeys Hooke’s law throughout the experiment. that the gravitational force is constant. that the meteorite has enough time to reach its terminal velocity. b) Refer to description on pages 66–67.5 mm d) 5.6 × 10–4 c) 1. All these assumptions are flawed. or at such high speeds. Stokes’ law only applies to small spheres moving at slow speeds.5 × 107 N m–2 b) 5. It may be moving and so the air resistance will be constantly changing. and that the temperature of all the objects involved is constant at 20°C.04 × 10–3 J 800 J Characteristics of Solids Answers 1) The Young modulus is stress/strain.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.8 × 109 m s–1 b) 6. Mamun Sir -8- .9 N 3) 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) You would want a material with a high breaking stress to protect against impact. 3) a) The maximum stress that can be applied to a material before it breaks. Stress the material shows plastic behaviour here elastic limit Mamun sir in this region the material obeys Hooke’s law 0 Strain 2) 40 cm 3) a) malleable – a material whose shape can be changed permanently and shows plastic deformation at low stress. Example: copper. used for bullet-proof vests. the lower the viscosity. d) ductile – a material that can be pulled into wires with small stress required. Example: Kevlar. Example: biscuits and crisps. e) brittle – a material that breaks without plastic deformation. e) compressive strain – the strain (deformation) when a material is squashed. and from this the terminal velocity (from distance ÷ time) and hence the viscosity can be calculated.1.1 Describing Motion c) Hooke’s law – the straight-line portion of the graph where stress is proportional to strain.. so the higher the terminal velocity. Extension is proportional to the applied force. c) hard – a material that cannot be scratched or dented easily. b) tough – a material that can withstand high impact forces and absorbs a lot of energy before breaking. Mamun Sir -9- . 2) See practical described on page 59 of the Students’ Book. which are designed to break with a snap! Materials in the Real World Answers 1) a) Terminal velocity is inversely proportional to the viscosity. to make jewellery.Edexcel AS Physics Advanced Institution of Physics 1. c) Manufacturers need to control flow rates of the liquid chocolate to achieve consistent products with as little waste and as possible. d) breaking stress – the stress at which the material breaks. used for electrical wiring. Example: gold. Hooke’s law is obeyed. Example: diamond. used for heavy duty cutting wheels. b) The time for a small ball of a known diameter to fall a given distance is measured. stiff – yes (keeps its shape and hard to bend). The Vital Statistics of a Wave Answers 1) 83 Hz 2) a) b) v= s t v = f v= 1. vibration of a stretched string Longitudinal waves: sound.45 × 1014 × 550 × 10−9 = 3 × 108 m s–1 3) a) 360° b) 180° c) (–)180° Mamun Sir . The movement of the b) A wave train has a definite beginning and an end.5 × 1011 = 3 × 108 m s–1 500 v = 5. hard – yes (durable and hard to dent). but a continuous wave goes on forever (it is infinite). strong – yes (hard to break).1 Describing Motion c) malleable – no. mechanical waves in a slinky pushed back and forth. 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. ripples in water. d) Types of Wave Answers 1) Transverse waves: light.1.10 - . brittle – no. seismic p waves. 2) a) Mamun sirenergy that causes a wave. tough – yes (will withstand impacts). ductile – no.Edexcel AS Physics Advanced Institution of Physics 1. The string is a physical object that cannot be in two places at once. On reflection at the end any vibration undergoes a 180° phase change. It physically pushes air molecules back and forth.1.9 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. third harmonic 123 Hz. Moreover. Mamun Sir . and so there must be zero displacement at the ends of the string.1 × 1010 Hz 3) The string is fixed at both ends. b) 86. the incoming and reflected waves will always be in antiphase and therefore completely cancel each other.1 Describing Motion Behaviour of Waves Answers 1) Mamun sir 2) a) Second harmonic 82 Hz. 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.8 cm b) 1. 2) a) 2.Edexcel AS Physics Advanced Institution of Physics 1. so a positive displacement would change to a negative displacement at the point of contact. The engineers would need to ensure that the design did not offer resonant frequencies that are likely to occur naturally.11 - . 12 - . Polarisation Answers 1) The signal is polarised. A graph of sin i against sin r should Mamun producesira straight line with the gradient equal to the refractive index.8° 2) 2. Mamun Sir . produce the strongest evidence for the veracity of scientific theories. the main peaks should appear at the same frequency.Edexcel AS Physics Advanced Institution of Physics 1.8°. so the aerial needs to be in the correct orientation to pick up the signal. oblivious of each other’s work. As per Student Practical 15 with the glass block resting underwater instead of in air. 2) Sound waves are longitudinal and so cannot be polarised. measure the angle of refraction within the glass. 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. Models of Waves and Their Properties Answers 1) Critical angles are: diamond 24. and hence indicate the same pitch. ice 49. 3) Scientists determining the same conclusions independently.4°. For each. benzene 41.1.0 × 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.1 Describing Motion 4) Capture the sound of both instruments playing the same pitch on an oscilloscope. at various angles of incidence. Although the shape of the waveform will be different. so could not evolve cells that respond to it. In addition. Difference: X-rays produced by decelerating electrons while gamma rays are produced as a result of energy change in the nucleus of an atom. the distance can be calculated. speed of movement of storm is away from the detector. 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. 3) Similarity: X-rays and gamma rays can have the same frequency. Knowing the speed of the pulse and the time taken. Mamun Sir . radar use Doppler shift in the reflected frequency to calculate the speed of the moving object.6 × 1020 Hz 2) He knew that light waves travelled through a vacuum.3 × 1014 Hz c) 4. 2) Both bat’s echolocation system and air traffic control radar use reflection from the object to locate it. b) The driver is not moving with respect to the sound. so there is no shift in frequency. 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. Hubble’s observations led to new idea of the origin and structure of the Universe. Applications of Electromagnetic Waves Answers 1) Mamun sir a) 100 m to 1 m b) To avoid reflection by the ionosphere and reach satellites which are outside the atmosphere. They are different in that the bat uses ultrasound and radar uses radio waves.1. Ultrasound Answers 1) Send a radio signal to the Moon and record the time taken for the reflected pulse to return.1 Describing Motion Light as a Wave Answers 1) a) 7. 2) The atmosphere absorbs this UV wavelength. and so humans have never been naturally exposed to it.5 × 1012 Hz b) 4. 3) Distance away is 51 km.Edexcel AS Physics Advanced Institution of Physics 1. and oscillating magnetic fields could move electric charges. and oscillating electric charges could create magnetic fields.13 - . 2) 120 Ω 3) An ohmic conductor obeys Ohm’s law. 1 V = 1 J C–1. the current through a wire is proportional to the potential difference across its ends.76 mm Wavelength method: = v/f = 1520/(3 × 10–6) = 0.5 V b) 8 V Resisting Current Flow Answers 1) Provided the temperature and other physical factors remain constant.76 mm Electric Current Answers Mamun sir 1) 240 C 2) 1. but a non-ohmic one does not – the current is not proportional to the potential difference across the conductor. 2) a) 1. b) The pd is the amount of energy supplied by each unit of charge – a measure of the work being done.1 Describing Motion 4) Pulse length method: l = v × t = 1520 × (1 × 10–6) = 0.14 - .1.Edexcel AS Physics Advanced Institution of Physics 1.6 × 10−19 C 3) conventional current electron movement A M Energy and Electricity Answers 1) a) The amount of energy supplied to each unit of charge in a circuit is the electromotive force.00152 Resolution = half pulse length = 0. 4) If the pd were plotted on the y-axis then the slope of the line would be equal to the resistance.51 mm So worst resolution = 0. Mamun Sir . There is no ‘leaking’ of energy from the circuit. The Transport Equation Answers 1) 0. so finding cross-sectional area is difficult. as the charge is conserved.065 W Circuits Containing Resistors Answers 1) The ammeters should show the same reading in all wires. so by substituting for V in the equation you can obtain P = I2R. the fact that a ring doesn’t usually have a rectangular cross-section. 3) Mamun Sir .1 Describing Motion 5) a) 4. P = VI 2) a) 0.15 - .065 W b) 2. 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.1.Edexcel AS Physics Advanced Institution of Physics 1.4 × 10−7 Ω m Answers include difficulty in positioning the contacts. and the physical size of the connecting crocodile clips.5 kW c) 0.92 × 1035 mMamun sir Power and Work in Electric Circuits Answers 1) You know that from Ohm’s law V = IR.94 A 2) −3 1. which would make determination of the area so inaccurate as to render the answer useless. 2 V b) 4 V across the 3000 Ω resistor. with the contact across 0.16 - .1. so the resistance increases. 1) a) Resistance Understanding Conduction Answers 0 Temperature/°C b) When the bulb filament becomes hot the lattice vibrates more and there are more collisions between the conduction electrons. 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. 2) Arrange this as a potential divider circuit. Sources of emf Answers 1) Because of the internal resistance of the power supply.Edexcel AS Physics Advanced Institution of Physics 1.62 of the length of the 80 Ω resistance wire. Mamun Sir .1 Describing Motion Mamun sir The Potential Divider Answers 1) a) 7. 3) In n-type semiconductors the doping element donates electrons to provide negative charge carriers.6 cm 4) The ultraviolet catastrophe was that at higher frequencies more and more energy would be radiated by a black body. This reduces the current as it is proportional to drift velocity.4 × 10−3 m2. in direct contradiction to the prediction of the particle theory. increasing current.37 × 10−15 J c) 3. Boyle. Newton Wave theory: da Vinci. Area needed = 4. Planck’s idea was that energy could only be absorbed or radiated in discrete quantities. whereas in a p-type semiconductor the doping element traps electrons and so introduces positive holes as charge carriers.6 × 10−19 J 2) 1. 2) Mamun Sir . 4) It provides more charge carriers over and above those present in the semiconductor lattice. 4) A theory of wave−particle duality. Wave or Particle? Answers 1) a) 2. the solar flux = 1000 W m−1.17 - . so the resistance effectively goes down. 3) Newton was a very influential scientist and his view was generally accepted. Side length = 6. Foucault 2) Foucault’s work showed that light must travel more slowly in water than air. Hooke. but the photons of ultraviolet light do. Mamun sir A Brief History of Light Answers 1) Particle theory: Democritus.65 × 10−18 J b) 2.1 Describing Motion c) The average velocity is reduced because of the increase in collisions. not in continuous amounts. 5) In some semiconductors a rise in temperature frees more charge carriers. Young. reaching infinity – clearly impossible. Grimaldi.1. 2) A positive charge that is the result of an electron leaving an atom. 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. Huygens.66 × 1020 3) From the worked example on this page. in which light behaves as a wave and a particle in different circumstances.Edexcel AS Physics Advanced Institution of Physics 1. d) By extrapolating backwards there could be a temperature at which there is no resistance. In chemistry the distinct colours given out by common elements when they are heated are used to identify them. Ionisation is when an electron absorbs enough energy to escape completely from the atom. 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. A 20 eV photon would ionise the hydrogen atom. is the work function of the metal surface.Edexcel AS Physics Advanced Institution of Physics 1.18 - . reliability. 10 eV does not correspond to any allowed transition. while potassium gives out lilac light. c) Excitation is when an electron in raised to a higher energy level around the nucleus of the atom.2 eV is needed to lift the electron into the next energy level. 2) A variety of factors.1 Describing Motion a) hf = + ½mv2 max Here h is Planck’s constant. Two 10 eV photons would have no effect. 2) 120 nm 3) Each element has a unique structure of energy levels. As transitions can only occur between these levels. 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. 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. Atomic Electron Energies Answers Mamun sir 1) a) Line spectra are made up of distinct lines of light with distinct frequencies. Mamun Sir . Solar Cells to Light the World? Answers 1) The energy from the Sun will not run out. legislation. b) 3. and ½mv2max is the maximum kinetic energy of the photoelectron. each transition gives rise to a unique frequency of light corresponding to the energy difference between two levels. 3) 96 W. for example cost. then nothing would happen.6 × 1014 Hz 3) As per the text and diagram on page 154. efficiency of cells. for example sodium gives out yellow light when it is heated. as the energy must be supplied by a single photon. 4) If a 10 eV photon was incident. availability.0 × 10−20 J c) 2. f is the frequency of the incident light.6 × 105 m s−1 d) 5.1. as exactly 10.