Trial Examination VCE Physics Unit 1. Written Examination. Suggested Solutions

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Trial Examination 2016 VCE Physics Unit 1 Written Examination Suggested Solutions Neap Trial Exams are licensed to be photocopied or placed on the school intranet and used only within the confines of the school purchasing them, for the purpose of examining that school s students only. They may not be otherwise reproduced or distributed. The copyright of Neap Trial Exams remains with Neap. No Neap Trial Exam or any part thereof is to be issued or passed on by any person to any party inclusive of other schools, non-practising teachers, coaching colleges, tutors, parents, students, publishing agencies or websites without the express written consent of Neap. Copyright 2016 Neap ABN 49 910 906 643 58 Pelham St Carlton VIC 3053 Tel: (03) 8341 8341 Fax: (03) 8610 0090 PHYU1_SS_2016.FM

Area of study How can thermal effects be explained? Question 1 (17 marks) a. 100 K is equivalent to 173 C so block A is much hotter. Atoms in block A have a higher average kinetic energy since it is directly proportional to temperature. b. The hotter block has particles with a greater average kinetic energy than the cooler block. Thus when they come into contact, the hotter block particles transfer kinetic energy to the cooler block particles, such that eventually the average kinetic energy of the particles in both blocks is the same. Therefore the blocks now have the same final temperature. c. The transfer of energy by conduction is proportional to the temperature difference and the surface area of the blocks. Initially both these quantities are equal for both blocks, so both will receive energy at the same rate. 2 marks d. The copper block (A) is hotter because copper has a lower specific heat capacity than steel. The same amount of energy will translate to a greater temperature increase in the copper. e. m c C c Δt c = m w c w Δt w m c 385 J K 1 ( 227 31) C = 1 kg 4180 J K 1 ( 31 26) C m c = 0.33 kg f. P 1 ----- = ----- 4 P 2 2400 W ------------------ 1200 W T 1 T 2 = 500 K -------------- 4 T 2 T 2 = 420 K (rounded from 421 K) g. Wien s law: 0.0029 λ peak = --------------- T 0.0029 = --------------- 500 K = 5800 nm h. The correct answer is D. 2 marks 10 6 m is in the infrared part of the spectrum. Question 2 (13 marks) a. Q = U + W Q is zero since no energy is transferred in or out of the balloon. W is positive since the gas does work by expanding into the balloon. U must therefore be negative; the gas cools. 2 PHYU1_SS_2016.FM Copyright 2016 Neap

b. Unlike CO 2, H 2 does not absorb infrared radiation, so H 2 will not re-emit radiation back towards the Earth s surface; it is not a greenhouse gas. c. Only half of the Earth s surface is exposed to sunlight at any one time. Only at the equator will the Earth receive the full intensity of the Sun. At higher latitudes, the intensity is less/is more spread out (or words to that effect). d. energy in = energy out = 343 W m 2 e. Energy used in our homes comes mostly from burning fossil fuels. This produces carbon dioxide that gets released into the atmosphere. Greater levels of carbon dioxide in the atmosphere increase the greenhouse effect and so the Earth must heat up to emit more energy. f. Thick curtains: The thicker the material, the less conduction there is from the hot windows to the cool room. Light-coloured curtains: The lighter colour will reflect more of the infrared radiation from the Sun. Pelmets: These disrupt the convection current that causes the hot air next to the window to rise up behind the curtains and enter the room.. Copyright 2016 Neap PHYU1_SS_2016.FM 3

Area of study How do electric circuits work? Question 3 (10 marks) a. 10 V + 5 10 10 2 marks for correct use of symbols. for correct set-up. b. R T = 5 + 10 + 10 = 25 Ω, or consequential from drawing. c. 5 Ω 10 Ω 10 Ω I (A) 1 0.5 0.5 V (voltage drop) (V) 5 5 5 Students may show that the parallel section has an effective resistance of 5 Ω, so: V 10 R T = 10 Ω and I = -- = ----- = 1.0 A. R 10 3 marks for each correctly filled column. Note: Consequential on answer to Question 1a. d. In series: P V 2 10 2 = ------ = ------- = 4 W. R 25 In parallel: The 2 10 Ω resistors have an effective resistance of 5 Ω. 1 1 R eff = ----- + ----- 1 R 1 R 2 1 1 = ----- + ----- 1 10 10 = 5 Ω When combined with the other 5 Ω resistor, the total effective resistance is 2.5 Ω. P V 2 = ----- R 10 2 = ------- 2.5 = 40 W A circuit set-up in parallel will produce maximum power. 4 PHYU1_SS_2016.FM Copyright 2016 Neap

Question 4 (9 marks) a. t = 5 mins = 300 s AndQ = E V -- Pt = ---- V 6 300 = ----------------- 12 = 150 C b. E= Pt = 6 300 = 18 000 J = 1.8 kj c. The energy changes occurring are electrical to thermal (or heat) energy. d. The ammeter provides little resistance and all the current goes through the ammeter. This bypasses the globe (that is, a short circuit) so it does not light up. e. The ammeter should be placed in series instead. This will allow the globe to light up. 12 V + A Copyright 2016 Neap PHYU1_SS_2016.FM 5

Question 5 (8 marks) a. The diagram must show the voltmeter placed correctly across either globe, as shown below. 80 W 80 W S 1 V 240 V S 2 b. If one globe goes out then so does the other, since they are in series. The solution is to wire the second globe in parallel. c. E= Pt = 80 ( W) 1.75 ( hr) 3600 14 (days) = 7.1442 MJ Since 1 kw h = 3.6 MJ, then E 7.1442 = --------------- = 1.98 kw h. 3.6 d. Emma is correct the earth is not needed to run devices. The earth wire connects the metal casing of the device to the ground. This safety feature prevents a faulty live device from potentially electrocuting someone. Question 6 (3 marks) For example: One ohmic device is the fixed-value resistor. One non-ohmic device is the LED. An ohmic device maintains a constant resistance (or the device does not maintain this constant ratio. V -- I ratio is constant), whereas a non-ohmic 2 marks for choosing correct ohmic and non-ohmic devices. for explanation. Note: There can be a number of different answers. 6 PHYU1_SS_2016.FM Copyright 2016 Neap

Area of study What is matter and how is it formed? Question 7 (2s) a. baryons: proton, neutron (lambda or sigma for example) 2 marks leptons: electron, neutrino (muon, tau) 2 marks b. Electrons can only be held by the nucleus below a certain temperature (about 3000 K). The universe first had to cool (and expand) below this temperature, which took many years. c. The universe became transparent ; photons were able to travel freely through space (or words to that effect). d. The corrrect answer is C. 2 marks e. The nucleus of Fe-56 has the highest binding energy per nucleon in the periodic table, and so is the most stable atom. This means you cannot add more nucleons and grow towards greater atomic mass without adding extra energy. f. An isotope is a species of nucleus that has the same atomic number as a given element, but a different neutron/mass number. g. 100 % of iron-53 remaining 50 53 0 0 10 20 30 40 50 60 70 80 time (min) 53 0 3 marks 2 marks for three or four correct points. for curve of best fit. Students are awarded for only including two correct points. h. 26Fe 25Mn + +1 β + 0 0ν 3 marks for atomic numbers for Fe and Mn. for positron (beta+). for neutrino. i. Mn-53 undergoes gamma decay. The nucleus gives out pure energy in the form of EM-radiation/gamma photon. Copyright 2016 Neap PHYU1_SS_2016.FM 7

Question 8 (9 marks) a. The strong nuclear force that holds nucleons together is much stronger than the electromagnetic force that holds electrons in atoms. This means much more work is required to undo those nuclear bonds. b. 235 U 141 0 n 92 Kr 3 92 + 56 + n 36 + 0 + energy 3 marks for incoming thermal neutron. for Kr-92. for the three neutrons produced. c. E = mc 2 3.2 10 11 J m = ------------------------------------ 9 10 16 m s 1 = 3.6 10 28 kg d. Electrons in such atoms are found in higher (excited) states (levels). When the atoms lose energy, the electrons return to lower states (levels), emitting photons of light in the process. 8 PHYU1_SS_2016.FM Copyright 2016 Neap