Answers to Revision Questions

Answers to Revision Questions 1 Force and Motion 1 a) i) The path is a semi-circle of length πr = 3.14 100 = 314 m ii) Speed = distance time = 314 20 = 15.7 m/s iii) When the cyclist reaches the point B her displacement is 200 m to the west of A. Did you notice the north, south, east and west arrows on the diagram? They were supplied to help you give the direction in a) iii). Parts a) i) and a) iii) are examining the difference between distance and displacement. The most common error involved finding the distance travelled many candidates found the circumference of the circle! b) i) Distance travelled = area under graph = ½ base height = ½ 8 12 = 48 m ii) Acceleration = gradient of graph = v t = (12 0) 8 = 1.5 m/s 2 2 a) 0.4 m/s 2 b) 7.2 m The essential idea in a) is that you quote and apply the correct equation for acceleration. You must realise that the initial velocity is 2.4 m/s and the final velocity is zero. In b) you need to apply the idea that the area under the graph (the area of the triangle) represents the distance travelled. Remember also to use the numbers on the axes (not your ruler!) to find the height and width of the triangle. 3 2 m/s B corresponds to time t = 50 s and distance = 80 m. C corresponds to time t = 80 s and distance = 140 m. Subtract to find the distance BC and the time taken to go from B to C. Then use the equation for speed. 4 a) 0.2 s b) 6.4 s For a) Use the figures for a car at 20 m/s together with the equation: thinking time = thinking distance speed. b) Acceleration is uniform, so the velocity time graph is a straight line. Since starting speed is 20 m/s and final speed is 0 m/s, the average speed is (20 + 0) 2 = 10 m/s. Use this idea with the braking distance for a car at 20 m/s, together with the equation: braking time = braking distance average speed, to calculate the braking time. 5 a) Velocity is always positive b) 0.2 m/s 2 c) 95 s d) AB: 250 m; BC: 450 m; CD: 650 m e) 6 m/s Part b) requires you to use the definition for acceleration. The Chief Examiner commented on the number of occasions that candidates misread the graph here. Note that each tiny square corresponds to 5 s 1 Book Interior Layout.indb 1

and that point C, where the speed first begins to fall, is 5 s before 100 s. d) Here you need to find three areas. e) Use the formula for average speed and your answer from d). 2 Motion under Gravity 1 a) Gravitational force between Earth and Sun b) Towards the Sun c) Arrow at the Earth, vertically DOWN at tangent to circle 2 a) 3.5 kg b) Tension (or centripetal force) c) D d) A e) The direction of the motion is constantly changing. 3 Solids, liquids and gases are made up of molecules. In solids the molecules are vibrating. Heat gives them more kinetic energy. This causes more vigorous vibrating. When some of the molecules overcome the strong bonds between them so that they are no longer held in fixed positions, the solid turns to liquid. As more energy is given to the liquid, the remaining molecules overcome the bonds between them completely and a gas is formed. 4 a) The density of an object is its mass divided by its volume. b) Use the water displacement method as described on page 13. Subtract the volume of the water only from the volume when the bracelet is in the water to get the volume of the bracelet. c) Density = mass volume = 46 = 19.2 g/cm3 2.4 d) Gold 5 a) Arrow, labelled F, from object towards (but not beyond) the centre of the circle. b) Arrow, labelled v, from object, tangential to circle and vertically upwards. c) The force increases. 3 Momentum 1 a) Friction: acts parallel to and up the slope. Weight: acts vertically downwards from centre of car. (v u) (0 27) b) a = t, so 6 = t, giving t = 4.5 s c) Force = ma = 1000 6 = 6000 N, so additional force = 6000 5000 = 1000 N 2 0A: Constant acceleration because of a constant resultant force equal to the weight minus air resistance. BC: Acceleration is zero because the weight and the air resistance cancel each other out, so the resultant force is zero. DE: Constant deceleration because the upward resultant force due to the air resistance provided by the parachute is greater than the weight. (Note use of specialist terms underlined.) 2 Answers to Revision Questions Book Interior Layout.indb 2

3 a) Force = momentum change = (50 20) = 10 000 N t 0.1 b) Seatbelts increase the time taken to come to a stop, reducing the force. They could also prevent the passenger from hitting the dashboard. 4 Energy 1 a) i) Work = force distance = 400 N 3 m = 1200 J work done ii) Power = time taken 200 = 1200 1200, so time = time 200 = 6 s b) i) KE = ½mv 2 1 = ½ 0.005 v 2 = 0.0025 v 2 So v 2 = 1 0.0025 = 400 v = 400 = 20 m/s ii) Loss in KE = work done against friction Loss in KE = friction force distance 1 = friction 0.005 1 Friction = 0.005 = 200 N 2 a) Position of parcel Potential energy in J Kinetic energy in J Total energy in J b) 18 m highest point, A 3600 0 3600 at B 2400 1200 3600 just before impact, C 3 a) 500 J b) 800 W 0 3600 3600 4 b) W = Fd c) 300 000 J d) 0.8 e) P = W/t f) 20 000 W 5 Moments 1 a) P marked in the lower right corner of beam. b) Arrow vertically downwards from the centre of the beam. c) By Principle of Moments, ACWM = CWM so F 1 d 1 = F 2 d 2 20000 2.5 = F 4 F = 50 000 = 12 500 N 4 2 a) 600 N m anticlockwise b) 600 N c) 900 N upwards Note: The distance of the effort from the pivot is (0.6 + 0.4) m it is vital that you use the correct distances in part b). And remember, you always take moments about the pivot. 3 Book Interior Layout.indb 3

3 5400 N Note: The pivot is at the centre of the large wheel. Did you get the clockwise moment correct, as 600 N 4.5 m = 2700 N m? 4 a) i) effort pivot load force ii) The effort is much further from the pivot than the load, so can produce a large moment for a small force. b) i) It will topple, because the weight acts outside the short base. ii) 2.25 N Hint: the centre of mass of the ruler is at the 50 cm mark and the pivot is at the 30 cm mark. Take moments about the pivot. 5 a) By Principle of Moments, ACWM = CWM so F 1 d 1 = F 2 d 2 5000 5 = F 3 25 000 So F = = 8333 N 3 b) Weight = 30 000 weight of load and counterweight = 30 000 (5000 + 8333) = 16 667 N 6 Waves 1 a) Arrow in box points horizontally to the right b) X vibrates back and forth parallel to the axis of the slinky c) In a transverse wave the particles vibrate at 90 to the direction in which the wave is travelling. d) Water wave or any electromagnetic wave 2 a) 15 cm b) 2.5 c) 2.5 Hz d) v = f λ = 2.5 15 = 37.5 cm/s 3 a) amplitude decreases b) pitch increases c) tonal quality increases The common mistake is to write that the wavelength (or pitch or quality) changes. To be sure of gaining the mark you should say how it changes. 4 Answers to Revision Questions Book Interior Layout.indb 4

7 Light 1 a) The distance from the optical centre of the lens to the principal focus is 4 cm. b) In this question you were expected to draw a vertical line to represent the lens and a line perpendicular to it representing the PA. Then: Draw a vertical line 2 cm tall, with one end touching the PA, at a point 2 cm from the lens. This represents the object. Label principal focus (F) on the right-hand side of the lens at a point 4 cm from the lens on the PA. Draw ray from top of object to lens parallel to PA and refract it through F. Draw ray from top of object through optical centre and extend it without refraction. Extend diverging rays (on right-hand side of lens) backwards to converge at point (I) on left-hand side. Draw vertical line from point of convergence to lens and label it image. c) 4 cm (±0.2 cm) d) 4 cm (±0.2 cm) you would be expected to get the position and height of the image to within a single square. e) Move object away from lens f) Virtual, erect, magnified 2 a) No refraction at the vertical face (zero angle of incidence). The un-refracted ray must be continued straight to the sloping surface. Here you should draw a normal to the glass. Refraction at sloped surface away from the normal. b) Speed decreases, wavelength decreases 3 a) 60 The common incorrect answer is 30. However, the angle of incidence is the angle between the normal and the ray so in this case the angle of incidence is 90 30 = 60. b) 60 This question shows again why you should always give an answer to a numerical question. The angle of reflection is equal to the angle of incidence, so any answer equal to that given in part a) would gain credit. c) Examiners want to see a ray reflected from mirror A with angle of reflection 60. At mirror B the angle of incidence is 30. Examiners want to see a ray at mirror B with an angle of reflection of 30. You do not need to use a protractor to draw the rays. Examiners will judge your ray diagram by eye. The ray incident at A and the ray reflected at B are therefore parallel. normal 5 Book Interior Layout.indb 5

4 a) Ray striking surface at 90 is not refracted. Ray striking surface obliquely refracts away from normal. b) The speed increases. 5 a) X-rays, ultraviolet, visible light, microwaves b) 3 m radio, 1 10 12 m gamma, 1 10 4 m infrared Notice that it was not necessary to know the wavelengths of the waves just that radio waves have the longest wavelength and gamma rays the shortest. Infrared rays have a wavelength between these extremes. 6 air water normal A normal B C air/water boundary water drop Ray B is striking the surface normally (at 90 ), so it continues without bending. This ray was awarded 1 mark. Ray A was much more difficult and required candidates to show deeper understanding. The normal to the surface at the point where ray A enters the drop is the radius of the circle. Light travels slower in water than in air, so the light is refracted towards the normal. 1 mark was awarded for getting the ray on the correct side of the normal; the other was for the bending towards the normal. There is no need to show dispersion. Interestingly, this is how rainbows form! 7 a) Ray from top of O parallel to principal axis refracted through F Ray from top of O through centre of lens undeviated Rays intersect to the right of the lens Image marked at the intersection of rays O F L F 1 mark 1 mark 1 mark b) 3 cm c) Real, inverted, diminished d) See the method on page 59 using the metre ruler, white screen and convex lens in holder. 6 Answers to Revision Questions Book Interior Layout.indb 6

8 a) i) Gamma, ultraviolet, visible, infrared, radio ii) λ = v f = 3 10 8 2.45 10 = 0.12 m 9 b) With P waves the particles vibrate parallel to the direction of propagation. With S waves the particles vibrate at right-angles to the direction of propagation. 9 a) When the angle of incidence in the glass is 41, the angle of refraction in air is 90. b) Ray bent towards the normal as it enters the glass. Ray strikes the side BC and is reflected. Ray travels to side CD, bending away from normal as it enters the air. c) The angle of incidence is greater than the critical angle, so total internal reflection occurs. 8 Radioactivity 1 a) Alpha He nucleus/two protons combined with two neutrons; beta electrons; gamma electromagnetic radiation b) i) 238 92 U 238Np + 93 1 0 β ii) 92 days c) 13 cm d) Gamma it is the only radiation that can penetrate the metal of the pipeline. e) Repulsion between nucleus and the alpha particles so the nucleus is positively charged. Most of the alpha particles passed through without deflection so the nucleus is small. A few alpha particles were scattered back so the nucleus is more massive (than the alpha particle). 2 Remove all sources from the laboratory, set the counting device to zero and switch it to count. Determine the count after 30 minutes and divide the count by 30 to determine the background count rate in counts per minute. Place the beta source very close to the detector. Place successively sheets of aluminium tight between the source and detector and measure the count rate using the counter. Continue until the count rate is equal to the measured background count rate. Using a ruler, measure the thickness of aluminium absorber required to reduce the count rate to the background count rate this thickness is the approximate range of beta particles in aluminium. 7 Book Interior Layout.indb 7

3 a) i) Small particle (left) is an electron and larger particle in centre is a proton. ii) 2 3 He b) i) Nucleus ii) Alpha a particle consisting of two protons and two neutrons; beta a fast electron; gamma a high-energy electromagnetic wave. 4 a) i) Nucleus ii) Protons and neutrons iii) A positively charged helium nucleus consisting of two protons and two neutrons iv) A fast electron v) Alpha radiation b) Industry controlling the thickness of steel/aluminium/paper in a rolling mill; OR detecting leaks in underground pipes Medicine killing/detecting cancer cells; OR sterilising equipment Agriculture studying the uptake of nutrients into plants; OR killing bacteria on foods c) 13 X and 12 X are isotopes (of hydrogen). Each has one proton in the nucleus, but 13 X has two neutrons in its nucleus while 12 X has one neutron. 5 a) The half-life of a radioactive substance is the time taken for its activity to fall to half of its original value. b) 1000 c) 500 9 Fission and Fusion 1 The uranium nucleus absorbs a neutron and becomes very unstable. The uranium nucleus then splits (fissions), releasing a vast quantity of energy. In the process two or three neutrons are released as new particles. These neutrons strike other uranium nuclei and cause more fission, thus producing a chain reaction. 2 The process is nuclear fusion, in which light nuclei, such as hydrogen, fuse (combine) to form a heavy nucleus, such as helium. In the process vast quantities of energy are released. The nuclei that combine are positively charged and normally repel each other. To bring about fusion, the hydrogen nuclei must be moving at enormous speeds to overcome this repulsive force. This means that fusion can only occur at very high temperatures. 3 a) i) Nuclear fission ii) Nucleus iii) It splits into two or more fragments iv) 90 90 0 Sr 38 Y β 39 1 b) i) Nuclear fusion ii) Hydrogen nuclei (protons) combine to form a helium nucleus 8 Answers to Revision Questions Book Interior Layout.indb 8

4 a) i) Any ONE from: radiation from the Sun; leaks from nuclear power stations and nuclear weapons; radiation from hospital X-ray and nuclear medicine departments; radiation from granite rocks. ii) The true count rate is the measured count rate minus the background count rate. iii) Thin paper has almost no effect on the radiation count rate, so α-particles are absent. Thin aluminium reduces the count rate by about 50%, so β-radiation is present. Even 5 mm lead is not enough to stop all the radiation, so γ-radiation is present. b) i) 195 ii) 195 79 = 116 neutrons iii) Activity in cps 512 256 128 64 32 Time in half-lives 0 1 2 3 4 Time in minutes 0 0.5 1.0 1.5 2.0 iv) The radioactive isotope would remain active for too long, causing a radiation poisoning hazard for the patient and possibly destroying healthy cells in the patient s body. c) i) 226 222 Ra Rn + 88 86 The mass number 226 has decreased by 4 to 222. Only α-particles have a mass number of 4, so they must be the radiation emitted. α-particles also have an atomic number (relative charge) of 2. The atomic number of radon (86) is obtained by balancing the equation (88 = 86 + 2). ii) Gas can be breathed into the lungs. Radon is an alpha-emitter so it has very high ionising power and can cause serious damage to living cells (e.g. causing cancer). d) i) Nuclear fusion involves the combination of light nuclei (not atoms) to form a heavier nucleus. ii) Nuclear fission is used in the generation of electricity. Nuclear fusion is the source of energy in the stars. 10 Electricity 1 500 mc = 500 1000 C = 0.5 C I = Q t = 0.5 25 = 0.02 A Make sure you understand submultiples of the coulomb. 2 Here the examiner is testing that you know the symbol μ means micro or 10 6 and the symbol m means milli or 10 3. He or she is also testing your use of the calculator. That is why it is so important to get practice using your calculator in preparation for the examination. Q = It = (20 10 6 ) (10 10 3 ) = 20 0 10 9 = 2 10 7 C 4 2 α 9 Book Interior Layout.indb 9

3 a) 0.032 coulombs. Notice that you are expected simply to write down your answer. You have to remember that the current in the circuit is the amount of charge flowing every second. Now 32 ma = 32 mc per second, so the charge involved is 32 mc and the answer required is 0.032 coulombs. b) We have to find out how many electrons, each with a charge of 1.6 10 19 C, are needed to make up a total charge of 0.032 C. the total charge The number of electrons = charge on 1 electron = 0.032 1.6 10 19 = 2 10 17 electrons Note that this requires you to know about standard (index) form. A list of the full mathematical requirements is given near the end of the CCEA Specification for your subject. 11 Circuit Diagrams and Symbols 1 a) i) Electrons move from balloon to hair leaving balloon with a shortage of electrons (positive charge). b) i) Negative charges on the paint droplets repel each other. ii) Positive iii) A positive charge on the car will attract the negatively charged paint drops to give a thin, even covering of paint. c) 2 Look carefully at the text in the question. What stands out is the word identical. On both occasions it is in bold type. It is as if the examiner is shouting it! And it is the key to the solution of the problem. The currents leaving and returning to the power supply are always the same. The current splits equally at the parallel junction since the resistors are identical. So the solution is: The voltage across each individual resistor is 24 = 6 V, since the resistors 4 are identical. The middle combination has two resistors, so the PD there is 6 + 6 = 12 V. Current in A 1 Current in A 2 Current in A 3 60 ma 30 ma 60 ma Voltmeter V 1 Voltmeter V 2 Voltmeter V 3 Voltmeter V 4 24 V 6 V 12 V 6 V 3 a) Series combination: R total = R 1 + R 2 + R 3 = 6 + 6 + 6 = 18 Ω b) Parallel combination: R total = R 2 = 6 2 = 3 Ω 10 Answers to Revision Questions Book Interior Layout.indb 10

4 a) Number of resistors, N 2 3 4 5 Total resistance, R, in Ω 6 4 3 2.4 b) R N = 12 c) R = 12 N = 12 24 = 0.5 Ω 5 a) 0 A (no current because circuit is incomplete) b) 0 V c) R total = R 1 + R 2 = 6 + 3 = 9 Ω d) I = V R = 18 9 = 2 A e) V = IR = 2 3 = 6 V To do part d) requires knowledge of the circuit resistance found in part c). The number found in c) is therefore carried forward to d). Any error in c) is carried forward to d) without penalty. Similarly, an error in d) is carried forward without penalty to e). 6 a) Note the position of the switch! It is not in the parallel loop. b) i) The strip is made of an insulator rather than a conductor. The base is made of a conductor rather than an insulator. ii) Closing the plastic strip does not complete a circuit, because plastic is an insulator. The strip should be made of a conductor. OR The base is copper, a conductor, so the switch is permanently on. The base should be made from an insulating material. c) An electrical discharge between the cloud and the ground results in a lightning flash. This lightning is a very large electrical current. The danger is that the lightning might strike the tall buildings and damage them. d) i) Both balloons are charged with charge of the same sign. ii) The balloons are repelling each other. Charges with the same sign repel; opposite charges attract. 12 Wiring a Plug 1 a) i) In normal operation the TV must be connected to a 240 volt power supply. When this happens the TV converts 80 joules of electrical energy into other energy forms every second. ii) I = P V = 80 = 0.33 A. 240 iii) 1 A this fuse has the lowest rating above the maximum safe current. iv) R = V = 240 = 720 Ω. I 0.33 11 Book Interior Layout.indb 11

2 v) See the diagram of the fused plug on page 97 vi) Live (brown) vii) When the TV is switched off, no part of it will be connected to a high voltage, so the TV will be safe. viii) The TV is double insulated. The casing is made of an insulator. All metal parts are in a separate plastic box, so that the user can never touch a conductor at a high voltage. b) Number of kwh = no. of kw no. of hours = 8 2 = 16 kwh Cost of 16 kwh @ 11p each = 16 11 = 176 pence = 1.76 mains ~ switch switch In this question candidates lost marks by finishing their wiring at the switch boxes. The examiners insisted that the wires terminated at the circular terminals inside the switch boxes. Note that the wire immediately to the left of the lamp is the neutral wire. The switches are always on the live side. 3 a) The examiner is looking for six important points and expects you to use the appropriate specialist language. The fact that the motor has a metal case is the clue that the examiner wants you to talk about earthing and fuses. The material below would be expected. You might write: The earth wire is connected to the metal case. If, by accident, the live wire comes into contact with the metal case the earth wire offers a low resistance path (to earth). So, by Ohm s law, the current becomes very large thus blowing the fuse and disconnecting the appliance from the mains supply. b) You need to remember that the earth wire is connected to the metal frame. So to obtain full marks, you could write: The bulb should light because there is a conducting path from the earth pin to the metal kettle. 13 Magnetism 1 a) Candidates should note carefully that the answer must be in ma. So, I = P V = 0.120 W = 0.030 A = 30 ma 4 V b) N s N p = V s V p So, N s 18 000 = 4 240 N s = 4 18 000 240 = 300 turns 12 Answers to Revision Questions Book Interior Layout.indb 12

2 a) A is a step-up transformer; B is a step-down transformer b) A increases the voltage so that, for a given amount of power, the current is reduced. Reducing the current means less energy is lost as heat in the transmission lines, saving the electricity company money. B decreases the voltage to a much safer level, suitable for household appliances. c) i) An alternating electric current is induced in the coil. ii) Electromagnetic induction iii) 0 ac iv) The output from a battery is direct current. Direct current always flows in the same direction, but the alternating current from the generator reverses its direction regularly. 3 Part a) is a straightforward application of the transformer turns-ratio equation. Note that candidates are expected to recognise the symbol for a transformer. a) N s N p = V s V p So, N s 6000 = 20 240 6000 20 N s = = 500 turns 240 b) Candidates are unlikely to have seen the diagram given and undoubtedly this would put some off. But look closely. We are told that between A and K there is a PD of 20 volts. We are also told that between A and B there are 50 turns, between B and C there are 50 turns, and so on. This tells us that the voltage increases in equal steps. Since there are 10 steps (AK, BK, CK up to JK), the voltage increases by 20 = 2 V in each step. 10 14 The Solar System 1 asteroid, planet, star, galaxy, Universe 2 a) The light from distant galaxies has a longer wavelength than expected. b) These galaxies are moving away from us. 3 Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune 4 The Universe began with an enormous explosion (the Big Bang) from a tiny point that physicists call a singularity. Only then did matter, energy and time come into existence. 5 Nuclear fusion is the combining of light nuclei (such as hydrogen) to give a heavier nucleus (such as helium) with the generation of huge amounts of energy. It happens in stars. 6 Over a very long period of time, clouds of gas and dust (a planetary nebula) clumped together due to gravity in a process called accretion. 13 Book Interior Layout.indb 13

7 a) i) and ii) planet (i) Sun Moon (ii) b) Planets orbit the Sun in the same direction OR Planets orbit the Sun in the same plane. c) Light from the galaxy is red-shifted. d) i) x-axis: Distance; y-axis: Speed ii) Examiner would expect to see: both axes labelled all points, including (0,0), plotted correctly straight line of best fit drawn Speed in km/s 16000 14000 12000 10000 8000 6000 4000 2000 0 0 200 400 600 800 1000 Distance in millions of light years iii) Yes, Hubble s law is valid because the graph is a straight line passing through the origin. 8 a) i) Nebula ii) Hydrogen and dust iii) Particles come together due to gravity, the nebula becomes a spinning ball of very dense gas, and the temperature rises. iv) A star b) i) Nuclear fusion ii) Visible light (or any other member of the electromagnetic spectrum) 9 a) Universe b) i) Neptune ii) C iii) D and E iv) E is Jupiter, F is Saturn, G is Uranus and H is Neptune c) Nuclear fusion d) The distance is so great that the time required to get there would be very long. The problems of carrying enough fuel, food and water to last for the journey have so far been insurmountable. 14 Answers to Revision Questions Book Interior Layout.indb 14

10 a) The light we observe from distant galaxies has a longer wavelength than expected. The phenomenon is called red-shift. b) These galaxies are moving away from us. c) No matter where we point radio-telescopes in the sky, we observe microwaves that correspond to the radiation emitted by a black body at a temperature of about 270 C or 3 Kelvin. This cosmic background microwave radiation is the remnant or echo of the Big Bang. Its existence is evidence for the Big Bang, as its presence can only be explained by the Big Bang Theory. 15 The Structure of the Earth 1 At plate boundaries, plates can move parallel to each other. The plates often stick and pressure builds up. The sudden release of the plates causes a jerking movement and an earthquake occurs. 2 When an oceanic plate collides with a continental plate, the oceanic plate is always forced underneath. As the oceanic crust is pushed down, friction causes the rocks to melt and pressure builds up. The molten rock is forced to the surface and a volcano can form. 3 a) Mantle; Outer core liquid, iron and nickel; Crust solid b) Northen Ireland is far from tectonic plate boundaries. 15 Book Interior Layout.indb 15