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Explanations also address conceptual points like inelastic collisions, air resistance effects, torque conditions, and relativistic time dilation effects for muons, with numerical results shown for each 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methods are used to solve Section A physics questions?","Question",{"text":63,"@type":64},"Section A applies standard kinematics and energy/impulse ideas, reading quantities from graphs and using formula-based calculations to reach the listed answer choices.","Answer",{"name":66,"@type":61,"acceptedAnswer":67},"How are motion and forces handled in Section B answers?",{"text":68,"@type":64},"Section B separates vertical and horizontal components for projectile motion, resolves forces into components for circular motion, and uses centripetal-force and orbital-period relations to interpret results.",{"name":70,"@type":61,"acceptedAnswer":71},"What electromagnetic and relativistic concepts appear in the solutions?",{"text":72,"@type":64},"The solutions include magnetic force and torque on coils, slip-ring and oscilloscope waveform relationships, transformer/induction reasoning for induced current, and relativistic time dilation for muons in different frames of 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| 8 | 9 | 10 |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| C | B | A | C | A | A | D | C | B | C |\n\n\n| 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| B | C | A | D | B | B | B | D | D | B |\n\n1 1 2  \nQ4 Work = area under graph =  2 􀁵 40 􀁵 0.10 =  2 􀁵 0.05v Q9 2r~~n~~2ro~~ew~~ld~~ ~~ = FF~~old~~new~~ ~~ = 16 , rn2ew = ~~ ~~16 􀁵 d2 , rnew = ~~ ~~16~~ ~~ d 􀁼 0.41d  \nC  \nB  \nQ10 Same constant rate of change in magnetic flux for the first and third sections, greater constant rate of change for the middle  \nsection. C  \n1 1  \nQ15 􀁊 = ~~   ~~ .: v = c 1 − ~~ ~~ , when 􀁊 increases from 200  \n~~ ~~1 − ~~(v~~c~~ ~~ )2  􀁊 2  \nto 2000 , it has a very small effect on v . E = 􀁊m0c2 , when 􀁊 increases from 200 to 2000 , it has a very large effect on the electron’s total energy E  \nQ18 􀀧x 􀁶 Ld  \nSECTION B  \nQ1a a = F~~net~~m~~ ~~ = ~~ ~~(9.2(5−01+.26−01).4􀁵)1􀁵01054~~ ~~ = 6.0 􀁵10−2 ms-1 Q1b T −1.4 􀁵104 = 6.0 􀁵105 􀁵 6.0 􀁵10−2 , T = 5.0 􀁵104 N  \nB  \nD  \nmv 2 -1  \nQ2a Centripetal force = mg tan 33􀁱 = ~~ ~~ , v 􀁼 87 m s  \n1.2 􀁵103  \nQ2b The normal force is perpendicular to the track surface. It can be resolved into vertical and horizontal components. The horizontal component points towards the centre of the circular motion. It is the centripetal force on the car.  \nr 3 GM  \nQ2c ~~ ~~ = ~~ ~~ = constant. When r is smaller, T is shorter.  \nT 2 4􀁓2  \nQ3a Vertically: s= − h , a= − 9.8 , u=+ 48 sin 35􀁱 , t = 6.2  \n− h = 48 sin 35􀁱 􀁵 6.2 + ~~ ~~12􀁵 − 9.8􀁵 6.22 , h = 18 m Q3b Horizontally: s=+x , u=+ 48 cos35􀁱 , t = 6.2  \n+ x=+ 48 cos35􀁱 􀁵 6.2 , x 􀁼 2.4 􀁵102 m  \nQ3c Air resistance is always opposite to motion  . It opposes both vertical and horizontal components of  . Thus it will reduce the altitude and the range of the golf ball.  \nQ4a Impulse = area under the graph  \n= ~~ ~~12 􀁵 240 􀁵10−3 􀁵 24000 = 2880 Ns Q4b  \nQ5a 􀀧 = f − i = 0.63 􀁵 (+ 10− − 12 )􀁼 + 14 , upwards 14 kg ms-1  \nQ5b The collision is inelastic because the ball has a reduced speed immediately after collision. Hence its total energy at floor level just after collision is lower than that just before collision. The decrease in total energy changes to other forms of energy, e.g. heat and sound energy.  \nQ6a No, because force of gravity is far too weak.  \nFg = ~~ ~~6.67~~ ~~􀁵10−11~~ ~~􀁵~~ ~~4~~ ~~.50500􀁵2105~~ ~~􀁵1~~ ~~.50~~ ~~􀁵104~~ ~~ 􀁼 1. 8 􀁵10−6 N Centripetal force required = 1.50~~ ~~􀁵10540􀁵0~~ ~~0.4362~~ ~~ 􀁼 5.7 N Q7a mvr2~~ ~~ = qvB  \nqBr 1.6 􀁵10−19 􀁵 5 .00 􀁵10−3 􀁵1 .50 􀁵10−2 7 -1  \nv = ~~ ~~ = ~~ ~~ 􀁼 1.32 􀁵10 m sm 9. 1􀁵10−31  \nQ7b qV0 = ~~ ~~12 mv 2 , V0 = 9.1~~ ~~􀁵~~ ~~102311~~ ~~􀁵6􀁵~~(~~11.301􀁵9107~~ )~~2~~ ~~ 􀁼 4.95 􀁵102 V Q7c The two particles have equal charge in magnitude, but mproton = 1. 8 􀁵103 melectron .: a proton requires a much stronger magnetic field (1.8􀁵103 􀁵5.00 mT) to keep it in circular. A proton will travel straight ahead with little effect from the magnetic field.  \nQ8a  \nQ8b E = ~~ ~~Vd 􀁼 1.208􀁵50109~~ ~~ = 1.41􀁵106 V m-1 Q8c Energy transferred  \n= VIt = 1.20 􀁵109 􀁵 30 .0􀁵103 􀁵 60 .0 􀁵10−6 􀁼 2. 16 􀁵109 J  \nQ9a Side FG is parallel to the magnetic field, i.e. the single coil EFGH is horizontal as shown in the diagram provided.  \nQ9b Coil EFGH is vertical, i.e. side GH is directly above side EF. At this orientation the magnetic force on side GH is vertically down while the magnetic force on side EF is vertically up. The two forces are aligned resulting in zero torque of the coil.  \nQ9c Bring the two magnets closer together to provide a stronger magnetic field.  \nQ10a Slip rings provide contacts with the leads connected to the oscilloscope so that when the loop rotates the leads will not tangle around the rings.  \n 1   1   \nQ10b Vpeak− peak 􀁼 8.0 V, f = T 􀁼 0.2 = 5 Hz  \nQ10c By doubling the magnetic field using stronger magnets. By doubling the number of turns in the coil.  \n1  \n2024 V","cbCaioBawyHsDWzG","https://ap.wps.com/l/cbCaioBawyHsDWzG","pdf",157597,"English","# Section A\n## Questions 1-20\n# Section B\n## Questions 1a-15b","[{\"question\":\"What methods are used to solve Section A physics questions?\",\"answer\":\"Section A applies standard kinematics and energy/impulse ideas, reading quantities from graphs and using formula-based calculations to reach the listed answer choices.\"},{\"question\":\"How are motion and forces handled in Section B answers?\",\"answer\":\"Section B separates vertical and horizontal components for projectile motion, resolves forces into components for circular motion, and uses centripetal-force and orbital-period relations to interpret results.\"},{\"question\":\"What electromagnetic and relativistic concepts appear in the solutions?\",\"answer\":\"The solutions include magnetic force and torque on coils, slip-ring and oscilloscope waveform relationships, transformer/induction reasoning for induced current, and relativistic time dilation for muons in different frames of reference.\"}]","2024 VCAA Physics Examination Solutions - Physics exam answers | PDF",1789909628]