NCEA Level 3 Physics (91524) 2014 page 1 of 9 Assessment Schedule 2014 Physics: Demonstate undestanding of mechanical systems (91524) Assessment Citeia Achievement Achievement with Meit Achievement with Excellence Demonstate undestanding equies witing statements that typically show an awaeness of how simple facets of phenomena, concepts o pinciples elate to a descibed situation. Fo mathematical solutions, elevant concepts will be tanspaent, methods will be staightfowad. Demonstate in-depth undestanding equies witing statements that typically give easons why phenomena, concepts o pinciples elate to given situations. Fo mathematical solutions, the infomation may not be diectly useable o immediately obvious. Demonstate compehensive undestanding equies witing statements that typically give easons why phenomena, concepts o pinciples elate to given situations. Statements will demonstate undestanding of connections between concepts.
NCEA Level 3 Physics (91524) 2014 page 2 of 9 Evidence Statement Q Evidence Achievement Meit Excellence ONE (a)(i) ω = Δθ = 360 Δt ω = 360 = 24.49 days 14.7 = 24.49 24 60 60 = 2.11 10 6 s 360 = 24.5 days 14.7 Coect ω: = 1.7014 10 4 s 1 ω = 2.96 10 6 ad s 1 (witten with ω o with units Coect woking showing convesion of days to seconds. Uses ω = Δθ Δt = 1 f + popotional cicle 360 86 400 (shows subs but not 14.7 equation) Coect woking showing convesion of days to seconds. = 360 / 2! ω Has descibed meaning of intemediate step, eg: 24.48 days fo full otation O 1.7 10 4 s 1 O 2.96 10 6 ad s 1 O show atios, eg14.7 day 360 24.4 O f = 4.73 10 8 Hz) Can go backwads, but has to be clea. (a)(ii) v = 2! = 2! 6.96 108 2.12 10 6 = 2.066.7 = 2070 m s 1 (Answe: 2060 2070 dependent on ounding.) Uses v = ω with incoect ω o and coect. Uses v = 2! coect. Coect answe. with incoect and Coect subs with wong answe. Coect speed 2060 2070 m s 1 with some woking (subs o eqn).
NCEA Level 3 Physics (91524) 2014 page 3 of 9 (b) When the coe of the Sun collapses, this will cause the adius of the paticles in the coe to otate with a smalle adius. Angula momentum will be conseved, so if the otational inetia deceases, the coe will have to otate at a highe angula velocity. Angula momentum is conseved. Rotational inetia of coe will get smalle. Angula velocity inceases because mass is close to the cente / axis. Accept inetia in place of otational inetia. Angula momentum is conseved, theefoe if otational inetia deceases, angula velocity inceases L = Iω theefoe, if I deceases, ω inceases. ω inceases because I deceases due to mass close to the cente / axis. Accept inetia in place of otational inetia. Radius smalle is acceptable fo mass close to the cente / axis. Angula momentum is conseved L = Iω I deceases due to mass close to the cente / axis HEREFE ω inceases. Accept inetia in place of otational inetia. adius smalle is acceptable fo mass close to the cente / axis. (c) Fo geostationay motion, the peiod of the satelite has to be equal to the peiod of Mecuy And F c = GMm 2 = mv2 3 = GM 2 4! 2 v = 2! = 6.67 10 11 3.30 10 23 (5.067 10 6 ) 2 3 4! 2 = 2.43 10 8 m F c = mv 2 v = = GMm 2 GM v = 2! Meges (F c = mv2 And v = 2!, o v = ω,ω = 2!f with eaanging incoect = GMm 2 ) Using F c = to deive = GM v 2 Full answe going backwads. Coect eaangement and substitution fo 3 o cube oot of (F c = ) v = 2! MAXIMUM 2As
NCEA Level 3 Physics (91524) 2014 page 4 of 9 (d) Intepeted as if otal pobe: he mass that is lost will have angula momentum, so the angula momentum of the space pobe will decease. he otational inetia of the space pobe will also decease. he effect of these two changes will be that they cancel and the angula velocity will emain constant. Intepeted as if Patial pobe: Angula momentum is conseved because of no extenal toque, theefoe angula speed stays the same. / Instument does not apply toque to the est of the pobe,so angula speed does not change. OAL pobe: Loss of instument means decease in otational inetia o angula momentum of pobe PARIAL pobe: No toques theefoe angula speed stays the same. MISINERPREAION: Idea that obital speed is independent of mass of satellite. Has to show idea of mass cancelling fom F c o o not in OAL pobe: instument takes L away and I away, theefoe angula speed stays the same. L = Iω PARIAL pobe: instument does not apply toque to the est of the pobe so the est of the pobe does not change angula speed. v = GM Q1 Not Achieved Achievement Achievement with Meit Achievement with Excellence N0 N1 N2 A3 A4 M5 M6 E7 E8 No esponse, no elevant evidence 1A 2A 3A 4A 2M + 1A 3M O 1E + 1M 1E + 2M 2E + 1M
NCEA Level 3 Physics (91524) 2014 page 5 of 9 WO (a) = 2! l 1.55 = 2! g 9.81 = 2.50 s Coect peiod. t = 1 2 = 1.25 s (b) ension and gavitational foce add to make the estoing foce towads the equilibium. he bob is stationay at point of elease. his estoing foce makes the bob speed up fom as it goes towads the middle. he estoing foce deceases and goes to zeo as the bob goes to the middle, so the acceleation deceases to zeo and the bob has constant speed at the equilibium position. Accept F F Res hoizontal Gavitational and tension foces identified. Foce is towads the equilibium position. Foce popotional to displacement At equilibium no foce at end points maximum foce. At elease point v = 0, and at equilibium v is maximum. Speeds up / acceleates as it goes towads the cente. MAXIMUM 2As Accept net /total / estoing foce in place of foce, but not gavitational foce o tension foce. F F Res Idea of diection of tension and gavitational foce (eithe stating the diections o saying they don t cancel, o coect ideas of components) (F es towads equilibium SO that bob speeds up / acceleates. Restoing foce deceases so acceleation deceases At equilibium no F es so constant speed / no acceleation.) Accept net foce /total foce in place of estoing foce. ( F F Res Idea of diection of tension and gavitational foce (eithe stating the diections o saying they don t cancel, o coect ideas of components).) F es towads equilibium SO that bob speeds up / acceleates. Restoing foce deceases so acceleation deceases At equilibium no F es, so constant speed/no acceleation. Accept net foce /total foce in place of estoing foce.
NCEA Level 3 Physics (91524) 2014 page 6 of 9 (c)(i) = 0.290 m L = 1.55 m sinθ = L = 0.290 1.55 θ = 10.78 o 0.188 ad = mg Coect angle Find = 17.658 and use INCREC angle to find F Coect tension foce showing woking fo coect angle. evidence of coect tig used. F = cosθ = 17.658 o evidence of mg used. = 1.8 9.81 = 17.658 F t = cosθ 1.8 9.81 cos10.78 = 17.975 N ALER: is wong answe (ii) cosθ = F C = 0.290 F 1.55 F C = F C = mv2 0.290 17.975 = 3.3631 1.55 v = v = 0.74 m s 1 3.3631 0.290 1.8 = 0.7361 Obtain F c as numbe o equation Eg F c = tanθ o F c = F sinθ o F c = F 2 2 Uses F c = mv2 Follow on eo accepted. to find v with wong F. (Evidence of use of tig o pythagous in attempt to get F c uses F c = mv2 t o find v (Eg mistakes doesn t use sq oot, uses wong tig o sin in pythagous, o e-aange incoectly)) Coect answe with insufficient woking Some woking shown ( F c = mv2 plus coect tig o pyth) and consistent answe and unit. Follow on eo accepted. Follow on eo accepted.
NCEA Level 3 Physics (91524) 2014 page 7 of 9 Q2 Not Achieved Achievement Achievement with Meit Achievement with Excellence N0 N1 N2 A3 A4 M5 M6 E7 E8 No esponse, no elevant evidence 1A 2A 3A 4A 2M 1E+2A 3M 1E + 2M 2E +1A
NCEA Level 3 Physics (91524) 2014 page 8 of 9 HREE (a) p system = (m A + m B ) v com = (0.517 + 0.684) v com p system = p A + p B = 0.517 1.21+ 0 = 0.62557 1.21 0.517 v com = = 0.52087 = 0.521 m s 1 0.517 + 0.6841 = 0.521m s 1 P = mv = 0.62557 evidence of coect numbe calculated eg 5.17 1.21 0.517 + 0.684 = 0.52087 p = mv m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v COM p COM o OAL = p A + p B Substitution into equation to find coect answe. (b)(i) Momentum is conseved, so the size of the change of momentum on disc A is equal to the size of the change in momentum of disc B. As B had no momentum to stat with, its final momentum must be equal to the change in momentum. Δp = mv v = 0.250 = 0.365 m s 1 0.684 Momentum is conseved. Δp B = Δp A (In wods, o equations, o by stating Δp B = 0.250 ) Momentum is conseved Δp B = Δp A (In wods, o equations, o by stating Δp B = 0.250 ) Statement of Newton 3d law, Δp = FΔt Statement of Newton 3d law, Δp = FΔt p = mv and 0.250 0.684 = 0.365497 p = mv and 0.250 0.684 = 0.365497 (b)(ii) p system = 0.62557 kg m s 1 p system = p A + p B p A = 0.62557 2 0.25 2 = 0.57344 p = mv equation used coectly fo some momentum. ONLY if no cedit given fo (a) o (b)(i). Shows Pythagous with momentum, wong answe. Coect vecto diagam with labels o numbes - diagam eithe Δp o total p. (Aows not needed) wong answe. Shows evidence of Pythagous o vecto diagam. Coect answe. Cay on eo fom wong p in 3a o 3b(i) OK. = 0.573 kg m s 1
NCEA Level 3 Physics (91524) 2014 page 9 of 9 (c) he only foces that will be acting on the discs will be tension foce in the cod. As this is an intenal foce acting within the system, neithe the momentum of the system no the velocity of the cente of mass of the system will change. No fiction theefoe no extenal foce acting on the system thus momentum is conseved. No fiction / fictionless. No extenal foces. ension / foce on sting is an intenal foce. Foces due to cod oppose / cancel. No fiction, no extenal foces. No extenal foces because foce due to sting is intenal. No extenal foces because sting foces oppose / cancel each othe. No fiction theefoe no extenal foces theefoe p is conseved Foce of the sting is intenal. sting foces oppose / cancel each othe. (Accept closed system in place of no extenal foces.) (Accept closed system in place of no extenal foces.) (Accept closed system in place of no extenal foces ) Q3 Not Achieved Achievement Achievement with Meit Achievement with Excellence N0 N1 N2 A3 A4 M5 M6 E7 E8 No esponse, no elevant evidence 1A 2A 3A 4A 2M +1A 1E+2A 3M 1E + 1M 1E+1M +1A 2E + 1A Cut Scoes Not Achieved Achievement Achievement with Meit Achievement with Excellence Scoe ange 0 6 7 13 14 18 19 24