Page 1
902555
2SUPERVISOR’S USE ONLY
9 0 2 5 5 M
© Mana Tohu Mātauranga o Aotearoa 2011. Pūmau te mana.Kia kaua rawa he wāhi o tēnei tuhinga e tāruatia ki te kore te whakaaetanga a te Mana Tohu Mātauranga o Aotearoa.
Ahupūngao, Kaupae 2, 201190255 Te whakaatu māramatanga o te pūhanga manawa
2.00 i te ahiahi Rāapa 16 Whiringa-ā-rangi 2011 Whiwhinga: Ono
Tirohia mehemea e ōrite ana te Tau Ākonga ā-Motu kei tō pepa whakauru ki te tau kei runga ake nei.
Me whakautu e koe ngā pātai KATOA kei roto i te pukapuka nei.
Tirohia mēnā kei a koe te Puka Rauemi L2–PHYSMR.
Ki roto i ō whakautu, whakamahi māramahia ngā tātainga ā-tau, ngā kupu, ngā hoahoa hoki/ rānei e hiahiatia ana.
Me hoatu te wae tika o te Pūnaha Waeine o te Ao (SI) ki ngā whakautu ā-tau.
Ki te hiahia koe ki ētahi atu wāhi hei tuhituhi i tō whakautu, whakamahia ngā wāhanga wātea kei muri i te pukapuka nei.
Tirohia mehemea kei roto i tēnei pukapuka ngā whārangi 2–19 e raupapa tika ana, ā, kāore hoki he whārangi wātea.
HOATU TE PUKAPUKA NEI KI TE KAIWHAKAHAERE HEI TE MUTUNGA O TE WHAKAMĀTAUTAU.
MĀ TE KAIMĀKA ANAKE Paearu Paetae
Paetae Paetae Kaiaka Paetae KairangiTe tāutu, te whakaahua rānei i ētahi āhuatanga o ngā tītohunga ā-rongo, ngā ariā, ngā mātāpono rānei.
Te whakaahua, te whakamārama rānei e ai ki ngā tītohunga ā-rongo, ngā ariā, ngā mātāpono, ngā hononga hoki / rānei.
Te homai whakamārama poto, e whakaatu mai ana i te matatau ki ngā tītohunga ā-rongo, ngā ariā, ngā mātāpono, ngā hononga hoki / rānei.
Te whakaoti rapanga takitahi. Te whakaoti rapanga. Te whakaoti rapanga takimaha.
Whakakaotanga o te tairanga mahinga
See back cover for an English translation of this cover
Page 2
Kia 60 meneti hei whakautu i ngā pātai o tēnei pukapuka.
Kua hoatu ngā ture katoa i runga i te Puka Rauemi L2-PHYSMR motu kē.
PĀTAI TUATAHI: TE EKENGA PAHIKARA
(a) HekaiekepahikaraaJacquie.Itētahiatakatīmataiateekemaiitepūwāhiwhakatū,ā,kawhakatereite1.2ms–2mōte14hēkonatehaere.
Whakaatuhiakotōnaterewhakamutungaimuriite14hēkonakote16.8ms–1.
(b) Kaahuwhakaterāwhitiiaitētahitereaumouote16.8ms–1,ā,kahurikimauī(kaahukiteRaki),kāre e rerekē tana tere.
Tāiatētahihoahoa perekawhakamahikitetātaiitehuringa o tana tere(terahimeteahunga).
Raki
Tonga
Hauāuru Rāwhiti
2
Ahupūngao 90255, 2011
MĀ TE KAIMĀKA
ANAKE
Page 3
3
Physics 90255, 2011
ASSESSOR’S USE ONLY
You are advised to spend 60 minutes answering the questions in this booklet.
All formulae are provided on the separate Resource Sheet L2-PHYSR.
QUESTION ONE: THE BIKE RIDE
(a) Jacquieisabikerider.Onemorningshestartsridingfromrestand acceleratesat1.2ms–2for14seconds.
Showthatherfinalvelocityafter14secondsis16.8ms–1.
(b) SheheadsEastataconstantspeedof16.8ms–1,thenturnsleft(headsNorth),without changing speed.
Drawavector diagramanduseitcalculatethechange in her velocity (sizeanddirection).
N
S
W E
Page 4
(c) KaheriaJacquieitanapahikaramārungaitētahipotikawhakawhitiitētahiawae65mtewhānui.Kawhakawhititonuatutepoti(↑)iteawa,meteterengaote6.8ms–1irungaitewai.Nāteiaotewai(→)katautepoti15mkiraroatuiteawa.
Tātaihiatetereotepotiehāngaianakitetahataha(arā,tepapa).
Kataeatetīmataitōwhakautumātetāingāhoahoaperewhaitapangamōngātawhitimengāterenga.
(d) KātahikahaereaJacquiemātetahiaraporowhitahuapaemeteterengaaumou.
Whakaahuahiaheahaiatērāehoatuanaitetōpana e hiahiatia anakiahaereporowhitatonute pahikara.
Kōrerotiateahunga o tēnei tōpana.
4
Ahupūngao 90255, 2011
MĀ TE KAIMĀKA
ANAKE
Page 5
5
Physics 90255, 2011
ASSESSOR’S USE ONLY
(c) Jacquietakesherbikeonaboatacrossariver65mwide.Theboatheadsstraightacross(↑)theriver,withaspeedof6.8ms–1relativetothewater.Thecurrent(→)causestheboattoland15mdownstream.
Calculatethespeedoftheboatrelativetothebank(ground).
Youmaybeginyouranswerbydrawinglabelledvectordiagramsfordistancesandspeeds.
(d) Jacquiethenridesalongahorizontalcircularpathatconstantspeed.
Describewhatitisthatprovidestheforce neededtokeepthebikegoinginacircle.
State the direction of this force.
Page 6
PĀTAI TUARUA: TE ARAWHATA
EekepahikaraanaaJacquiemātētahiarawhataaumouetautokohiaanaingāpitoerua,pērāhokiitehoahoakeiraro.
5.0 m
25.0 m
A B
(a) E25.0mteroaotearawhata.KotepapatipuoJacquiemetanapahikarako72kg.Kotepapatipuotepiritiko760kg.
Tātaihiatetōpana tautoko (FA)etukunaanaetepito A, metetōpana tautoko (FB)etukunaanaetepito Botearawhatainae5.0maJacquiemaiitepitoA.
(b) Whakapuakinaōwhakautukitewāhanga(a)kitemahatikaongāmatitāpua.
Homaihepūtakemōtōkōwhiringaongāmatitāpuakirotoiōwhakautukitewāhanga(a).
(c) ItewāeekepahikaraanaaJacquieitetere16.8ms–1,katukiiakitētahipōropoiwhanaetakaporeanakiaiaitetereote8.0ms–1.Kapanawhakawahoatutepōropoiwhanakiteahungahāngaimeteterengaote5.0ms–1.
16.8 m s–1
8.0 m s–1 5.0 m s–1
i mua i te tutukinga i muri i te tutukinga
6
Ahupūngao 90255, 2011
MĀ TE KAIMĀKA
ANAKE
Page 7
7
Physics 90255, 2011
ASSESSOR’S USE ONLY
QUESTION TWO: THE BRIDGE
Jacquiecyclesalongauniformbridgethatissupportedatbothends,asshowninthediagram.
5.0 m
25.0 m
A B
(a) Thelengthofthebridgeis25.0m.ThemassofJacquieandherbikeis72kg.Themassofthebridgeis760kg.
Calculate the support force (FA)providedbyend A andthe support force (FB) providedby end BofthebridgewhenJacquieis5.0mfromendA.
(b) Expressyouranswerstopart(a)tothecorrectnumberofsignificantfigures.
Giveareasonforyourchoiceofsignificantfiguresinyouranswerstopart(a).
(c) WhileJacquieiscyclingataspeedof16.8ms–1,shecollideswithasoccerballthatisrollingtowardsherataspeedof8.0ms–1.Thesoccerballbouncesoffintheoppositedirectionwithaspeedof5.0ms–1.
16.8 m s–1
8.0 m s–1 5.0 m s–1
before collision after collision
Page 8
TātaihiatetereoJacquie(terahimeteahunga)whaimuriitetutukinga.
Mewaihoekoengāpāngaotewaku.
KotepapatipuoJacquiemetanapahikara=72.0kg.
Kotepapatipuotepōropoiwhana=0.430kg.
(d) Whakamāramahiaheahatetikangaotetutukinga tāwariwarimetetutukinga rorohakore.
Tukingatāwariwari:
Tukingarorohakore:
Whakaahuahiameriroiakoeteahakiawhakataumēnāhetutukingatāwariwari,rorohakorerāneitetutukingaiwaenganuiitepahikarametepōropoiwhana.
Kāorehetikangawhakaotitātainga.
8
Ahupūngao 90255, 2011
MĀ TE KAIMĀKA
ANAKE
Page 9
9
Physics 90255, 2011
ASSESSOR’S USE ONLY
CalculateJacquie’svelocity(sizeanddirection)afterthecollision.
Youmayignoreanyeffectsoffriction.
MassofJacquieandherbike=72.0kg
Massofsoccerball=0.430kg.
(d) Explainwhatismeantbyanelastic collision andan inelastic collision.
Elasticcollision:
Inelasticcollision:
Describewhatyouwouldneedtodoinordertodeterminewhetherthiscollisionbetweenthebikeandthesoccerballis elasticorinelastic.
Youarenotrequiredtocarryoutanycalculations.
Page 10
(e) WhakamāramatiateāhuaewhakamauruaitetōpanaotepōrokiaJacquiemetanapahikarakiteroaotewāitepānga,Ā,whakamāramatiatewhaipāngaotetōpanaotepōrokiaJacquiemetanapahikarakitetōpanaotepāngaoJacquiemetanapahikarakitepōro.
10
Ahupūngao 90255, 2011
MĀ TE KAIMĀKA
ANAKE
Page 11
(e) ExplainhowtheforceexertedbytheballonJacquieandherbikeisdependentonthedurationofthetimeonimpact,ANDexplainhowtheforceexertedbytheballonJacquieandherbikeisrelatedtotheforceexertedbyJacquieandherbikeontheball.
11
Physics 90255, 2011
ASSESSOR’S USE ONLY
Page 12
PĀTAI TUATORU: PŪNGAO ME TE NEKEHANGA TĪTERE
EpanatetungāneoJacquie,aErnie,itētahimīhinitapahiotaotametetōpanaote26Nmetekokiote34°kitepapa,pēneiitēneiiraronei.
www.treehugger.com/push-mower-jjh01.jpg
(a) Whakamāramakatoatiaheahaikoreaitekatoaotetōpana26NekarawhiuaanaeErnieewhakamahiakiapanaitemīhinitapahiotaotawhakatehuapaeitepapa.
(b) TātaihiatekahaehuaanaitāErniemahiinawhakatereiaitanamīhinitapahiotaotakite 4.0mirotoite3.0hēkona.
Homaingāwaeinetikamōtōwhakautu.
12
Ahupūngao 90255, 2011
MĀ TE KAIMĀKA
ANAKE
He tapu tēnei rauemi. E kore taea te tuku atu.
Aata tirohia ki ngā kupu kei raro iho i te pouaka nei.
Page 13
13
Physics 90255, 2011
ASSESSOR’S USE ONLY
QUESTION THREE: ENERGY AND PROJECTILE MOTION
Jacquie’sbrotherErnieispushingalawnmowerwithaforceof26Natanangleof34°totheground,asshownbelow.
www.treehugger.com/push-mower-jjh01.jpg
(a) Explainfullywhynotallofthe26NforceexertedbyErnieisusedtopushthelawnmowerhorizontallyalongtheground.
(b) CalculatethepowerproducedbyErniewhenheacceleratesthemowerthroughadistanceof4.0min3.0seconds.
Givethecorrectunitsforyouranswer.
For copyright reasons, this resource cannot be reproduced
here.
Page 14
(c) KawhanatetamaaErnie,aJacob,itētahipōrokiaErnieirotoitemāra.1.75mtetāroaroaoErnie.KawhanaaJacobitepōrometetere24ms–1itekoki36°kitepapa.E35mtanatawhitiatuotetūmaioErnie.
36°
35 m
1.75 m
Jacob
24 m s–1
Ernie
KĀORE i āwhatatia te
hoahoa
KapātepōrokiaErnie,kahiparāneiirungaakeitōnamāhunga?
Iōtātainga,tīmatamātewhakaatukotewāhangahuapaeoteteretīmataotepōroko 19.4ms–1.
14
Ahupūngao 90255, 2011
MĀ TE KAIMĀKA
ANAKE
Page 15
15
Physics 90255, 2011
ASSESSOR’S USE ONLY
(c) Ernie’ssonJacobkicksaballtowardsErnieinthegarden.Ernieis1.75mtall.Jacobkickstheballwithavelocityof24ms–1atanangleof36°totheground.Jacobisstanding35mawayfromErnie.
36°
35 m
1.75 m
Jacob
24 m s–1
Ernie
Diagram is NOT to scale
WilltheballhitErnieorgooverhishead?
Inyourcalculations,startbyshowingthatthehorizontalcomponentoftheinitialvelocityoftheballis19.4ms–1.
Page 16
Kahangawharetukutukutepūngāwerewereirotoitemāra,ā,kamautētahipepe.Katorowhakararotewharetukutukumāte0.065minamautepepee0.003kgtepapatipukiroto.
Keirarokotētahikauwhatamōtetōpanametetotorohokiotewharetukutuku.
tōpana (N)
totoro (m)
(d) WhakamāramahiaheahaikoreaietaeatewhakamahiitetureW=Fdheitātaiitepūngaomoekūtorotoroepuritiaanaitewharetukutukuinamautepepekiroto.
Mewhakaurukirotoitōwhakamāramatētahikōreromōngāmeamewhakamahiheitātaiitēneipūngao.
(e) Tātaihiatepūngaomoekūtorotoroepuritiaanaitewharetukutukuinamautepepeirotoitewharetukutuku.
16
Ahupūngao 90255, 2011
MĀ TE KAIMĀKA
ANAKE
Page 17
17
Physics 90255, 2011
ASSESSOR’S USE ONLY
Aspiderspinsawebinthegardenandamothgetscaughtintheweb.Thewebstretchesdownwardsby0.065mwhenthemothofmass0.003kgiscaughtinit.
Agraphforforceagainstextensionforthespider’swebisshownbelow.
force (N)
extension (m)
(d) ExplainwhytheformulaW=Fdcannotbeusedtocalculatetheelasticpotentialenergystoredinthewebwhenthemothgetscaughtinit.
Yourexplanationshouldincludeastatementofwhatshouldbeusedtocalculatethisenergy.
(e) Calculatetheelasticpotentialenergystoredinthewebwhenthemothiscaughtintheweb.
Page 18
18
Ahupūngao 90255, 2011
MĀ TE KAIMĀKA
ANAKETAU PĀTAI
He wāhi anō mēnā e hiahiatia ana.Tuhia te (ngā) tau pātai mēnā e hāngai ana.
Page 19
19
Physics 90255, 2011
ASSESSOR’S USE ONLY
QUESTION NUMBER
Extra space if required.Write the question number(s) if applicable.
Page 20
© New Zealand Qualifications Authority, 2011. All rights reserved.No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.
Level 2 Physics, 201190255 Demonstrate understanding of mechanics
2.00 pm Wednesday 16 November 2011 Credits: Six
Check that the National Student Number (NSN) on your admission slip is the same as the number at the top of this page.
You should attempt ALL the questions in this booklet.
Make sure that you have Resource Sheet L2–PHYSMR.
In your answers use clear numerical working, words and / or diagrams as required.
Numerical answers should be given with an appropriate SI unit.
If you need more room for any answer, use the extra space provided at the back of this booklet.
Check that this booklet has pages 2 – 19 in the correct order and that none of these pages is blank.
YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION.
ASSESSOR’S USE ONLY Achievement Criteria
Achievement Achievement with Merit Achievement with ExcellenceIdentify or describe aspects of phenomena, concepts or princi-ples.
Give descriptions or explanations in terms of phenomena, con-cepts, principles and / or relation-ships.
Give concise explanations that show clear understanding in terms of phenomena, concepts, principles and / or relationships.
Solve straightforward problems. Solve problems. Solve complex problems.
Overall level of performance
90
25
5M
English translation of the wording on the front cover