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© Mana Tohu Mātauranga o Aotearoa, 2012. 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 3, 201290521M Te whakaatu māramatanga ki ngā pūnaha

pūkahakaha

9.30 i te ata Rātū 27 Whiringa-ā-rangi 2012 Whiwhinga: Ono

Tirohia mehemea e ōrite ana te Tau Ākonga ā-Motu (NSN) kei tō pepa whakauru ki te tau kei runga ake nei.

Me whakautu e koe ngā pātai KATOA kei roto i te pukapuka nei.

Āta tirohia mēnā kei a koe te Pukaiti Rauemi L3–PHYSMR.

I ō whakautu, whakamahia ngā whiriwhiringa tohutau mārama, ngā kupu, hoahoa hoki / rānei ki hea hiahiatia ai.

Me hōmai te whakautu tohutau me tētahi waeine o te Pūnaha Waeine ā-Ao (SI) ki ngā tau tika o ngā tau tāpua.

Ki te hiahia koe ki ētahi atu wāhi hei tuhituhi i tō whakautu, whakamahia te wāhi wātea kei muri i te pukapuka nei.

Tirohia mehemea kei roto nei 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 tautohu, te whakaahua rānei i ngā āhuatanga o te tītohu ā-rongo, te ariā, te mātāpono rānei.

Te homai whakaahua, whakamārama rānei e ai ki te tītohu ā-rongo, te ariā, te mātāpono, te hononga hoki / rānei.

Te homai whakamārama hei whakaatu i te tino māramatanga ki te tītohu ā-rongo, te ariā, te mātāpono, te hononga hoki / rānei.

Te whakaoti rapanga māmā. Te whakaoti rapanga. Te whakaoti rapanga matatini.

Whakakaotanga o te tairanga mahinga

SUPERVISOR’S USE ONLY

See back cover for an English translation of this cover

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Kia 55 meneti hei whakautu i ngā pātai o tēnei pukapuka.

PĀTAI TUATAHI: NEKEHANGA HURIHURI

nōhanga pūnoa nōhanga toro waewae

Kei te noho Ira ki tētahi nōhanga pūnoa, i tētahi tūru hurihuri noa, ā, kāore i tau ōna waewae ki te papa. Ko te tūpuku hurihuri o Ira i tēnei nōhanga me tō te tūru he 5.45 kg m2.

Ka pana a Ira ki tētahi tēpu tūtata mā ōna ringa kia huri ai ia me tōna tūru i te tere koki o te 7.50 rad s-1.

(a) Tātaitia te maha o ngā hurihanga o te tūru i roto i te 5.00 hēkona (s), i te tere koki pūmau o te 7.50 rad s-1.

Te maha o ngā hurihanga =

(b) He 0.54 s te roa o te piki o tō Ira tere koki mai i te kore ki te 7.50 rad s-1.

Tātaitia te tōpana whakahuri toharite i puta i a Ira.

Tōpana whakahuri toharite =

You are advised to spend 55 minutes answering the questions in this booklet.

QUESTION ONE: ROTATIONAL MOTION

normal position legs-out position

Ira is sitting in a normal position, with her feet off the ground, on a freely rotating swivel chair. In this position the rotational inertia of Ira and the chair is 5.45 kg m2.

Ira pushes with her hands against a nearby table to make herself and the chair rotate at an angular speed of 7.50 rad s-1.

(a) Calculate the number of rotations the chair does in 5.00 seconds, at a steady angular speed of 7.50 rad s-1.

Number of rotations =

(b) It took Ira 0.54 s to change the angular speed from zero to 7.50 rad s-1.

Calculate the average torque that Ira applied.

Average torque =

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(c) Ka whakahurihuri a Ira i a ia anō me te toro i ōna waewae, ka whakatauritea tēnei ki tērā o ana hurihanga i te nōhanga pūnoa. Ka kite ia i te toronga o ōna waewae, me kaha ake tana pana ki te tēpu kia hurihuri ai te tūru ki te tere koki ōrite me te wā ōrite.

Matapakitia he aha i pēnei ai.

(d) E noho ana a Ira ki te nōhanga pūnoa anō me te hurihuri i te 7.50 rad s-1 i te āta takanga o tētahi pukapuka e tana hoa ki ōna pona. I taka poutū mai te pukapuka me te noho tonu ki ngā pona o Ira i a ia e hurihuri ana.

Whakamāramahia he pēhea te pānga o te taunga o te pukapuka ki te tere koki o Ira me te tūru.

(c) Ira makes herself rotate in the legs-out position, and compares this with making herself rotate in the normal position. She finds that when her legs are out, she has to push against the table with a greater force to get the chair spinning at the same angular speed in the same time.

Discuss why this is so.

(d) Ira is sitting in the normal position again and rotating at 7.50 rad s-1 when her friend drops a textbook gently onto her lap. The book falls vertically and then stays on Ira’s lap as she spins.

Explain how the arrival of the book will affect the angular speed of Ira and the chair.

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(e) Ka pupuri torotikatia e Ira te pukapuka i a ia e hurihuri ana ki te tere koki pūmau o te 5.00 rad s-1. Ka whakapiri wawe mai ia i te pukapuka ki tana tinana. Me whakaaro he papatipu tongi te pukapuka, ā, he 0.600 mita ki te 0.050 mita te nekenga mai i te pokapū o te hurihanga. Ka taea te tātai te tūpuku hurihuri o te pukapuka mā te whārite I = mr2.

(i) Whakamāramatia he aha i piki ai te tere koki o te pūnaha hurihuri.

(ii) Ko te papatipu o te pukapuka he 2.10 kg.

Whakatau tatatia te tere koki o te pūnaha i te whakapirihanga mai o te pukapuka ki te tinana o Ira, me te whakamārama anō, ina mahia, he aha te take i nui ake ai pea te tere koki i tēnei.

(e) Ira holds the book out at arm’s length while she is spinning at a steady angular speed of 5.00 rad s-1. She quickly moves the book close to her body. Assume that the book is effectively a point mass and that it is moved from 0.600 m to 0.050 m from the centre of rotation. The rotational inertia of the book can be calculated using the equation I = mr2.

(i) Explain why the angular velocity of the rotating system increases.

(ii) The book has a mass of 2.10 kg.

Estimate the angular speed of the system when the book is moved close to Ira’s body, and explain why, in practice, the angular speed could be larger than this.

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PĀTAI TUARUA: TE NEKENGA HAWARITE MĀMĀ

E pārekareka ana ki a Daniel te tārere i te taura e herea ana tētahi pito ki te peka rākau, me tētahi papa rākau e mau ana ki tētahi pito. Ka tāwēwē mai te taura i te taha parenga o te awa.

Ki te hīkina e Daniel ōna waewae me te kore pana, ka tārere atu, ki runga ake o te awa, i te nekenga hawarite māmā. Ko te auau koki o tōna nekenga he 1.45 rad s-1, ā, he 0.80 mita te tawhiti o te tārere i mua i te taenga atu ki te tūnga taurite.

(a) Tātaitia te whakaterenga o te nekenga hawarite māmā o Daniel i te wā tonu ka hīkina e ia ōna waewae.

Whakaterenga =

(b) Tātaitia te tawhiti o te tārere o Daniel i roto i te 1.8 hēkona i muri o tana hiki i ōna waewae.

Tawhiti =

(c) I tētahi atu wā, i pana kē mai a Daniel mai i te pūwāhi tīmatanga, kaua mā te hiki i ōna waewae.

Matapakitia he aha te pānga o tēnei ki te nekenga o tana tārere.

Me uru ki tō whakautu tētahi whakamāramatanga he aha i nui ake ai te hōkai o tana tārere.

te awa

QUESTION TWO: SIMPLE HARMONIC MOTION

Daniel is having fun on a swing that consists of a rope with one end tied to the branch of a tree, and a wooden bar attached to the other end. The rope hangs down over the edge of a bank of a river.

If Daniel lifts his feet without pushing, he swings out, over the river, in simple harmonic motion. The angular frequency of his motion is 1.45 rad s-1, and he travels a distance of 0.80 m before reaching the equilibrium position.

(a) Calculate the acceleration of Daniel’s simple harmonic motion at the instant he lifts his feet.

Acceleration =

(b) Calculate the distance Daniel travels in 1.8 s after he lifts his feet.

Distance =

(c) On another occasion, instead of lifting his feet, Daniel pushes off from the starting point.

Discuss what effect this could have on the motion of his swing.

Your answer should include an explanation for why the amplitude of his swing increases.

river

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(d) I ētahi wā ka noho a Daniel ki runga i te papa rākau, ā, i ētahi wā ka tū ia.

Matapakitia he pēhea te pānga ki te wā nekenga o te āhua e eke ai a Daniel i te tārere.

(d) Sometimes Daniel sits on the wooden bar and sometimes he stands.

Discuss how the period of the motion would be affected by the way Daniel rides on the swing.

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PĀTAI TUATORU: NEKENGA RĀRANGI

Ko tētahi mahi atua e hiahiatia ana mō tētahi whitiāhua ko te ‘whakapahū’ i tētahi pōro ina whiua ki te takiwā. Ka tutuki tēnei mā te whakamahi i tētahi pūnikoniko kaha hei pana i ngā wāhanga e rua o te pōro kia wehe.

A

I mua I muri

Te pōro me ngā wāhanga o te pōro, te tirohanga mai i runga.

B

θΑ

He 0.105 kg te papatipu o te pōro, ā, i te wā tonu o tana wehe, he 67.0 m s-1 te tere o tana haere whakatehuapae. I tētahi hautanga hēkona i muri o te wehenga, e whakaaturia ana kei te haere huapae tonu te aronga o A me B. He 0.0800 kg te papatipu o wāhanga A, ā, he 80.0 m s-1 te tere.

Ko te tō-ā-papa anake te tōpana ā-waho kei te pōro: kāore he tōpana huapae ā-waho atu anō.

(a) Matapakitia mēnā e noho pūmau te ānga, te pūngao neke hoki / rānei i te wā o te pahūtanga.

Ānga:

Pūngao neke:

(b) Tātatia te ānga huapae o te pūnaha o ngā wāhanga e rua whai muri tonu mai o te pahūtanga.

Ānga =

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(c) He 0.0800 kg te papatipu o te wāhanga A, ā, he 80.0 m s-1 te tere, i te koki o te θA, e ai ki te hoahoa.

Mā te tātuhi i tētahi hoahoa pere, whakaaturia mai ka pēhea tō tātai i te tere o te wāhanga B whai muri tonu mai i te pahūtanga.

(Kāore i te hiahiatia tētahi tātainga o te tere o B.)

QUESTION THREE: LINEAR MOTION

A special effect for a movie requires a ball to ‘explode’ in mid-air after it has been thrown. This is achieved by using a strong spring to push the two pieces of the ball apart.

A

Before After

The ball and ball pieces, viewed from above.

B

θΑ

The ball has a mass of 0.105 kg and, at the instant it splits, it is travelling horizontally at 67.0 m s-1. A fraction of a second after the split, A and B are effectively still travelling horizontally in the directions shown. Piece A has a mass of 0.0800 kg and is travelling at 80.0 m s-1.

The only external force on the ball is gravity: there are no external horizontal forces.

(a) Discuss whether momentum and / or kinetic energy is conserved during the explosion.

Momentum:

Kinetic energy:

(b) Calculate the horizontal momentum of the system of two pieces immediately after the explosion.

Momentum =

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(c) Piece A has a mass of 0.0800 kg and is travelling at 80.0 m s-1, at an angle of θA, as shown.

By drawing a vector diagram, show how you could calculate the velocity of piece B immediately after the explosion.

(A calculation of the velocity of B is not required.)

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Tēnā whakaaroarohia te pānga o te tō-ā-papa ki ngā wāhanga e rua i te wā e taka ana ki te papa.

E whakaaturia ana i te hoahoa i raro nei tētahi tirohanga ā-taha o te ara o te wāhanga A. Ina taka haere te wāhanga A mai i P ki Q, ka piki haere te wāhanga poutū o tana ānga mā te 3.1 kg m s-1.

te āhunga tīmatanga o te nekenga o te wāhanga A

te papa

P

Q

(d) Matapakitia te tōpana whakahāngai kei te wāhanga A i te takanga haere mai P ki Q, ka whakamārama i te pūtaketanga o te tōpana whakahāngai me te pānga o taua tōpana ki te ānga o A.

(e) Tātaitia te tere o A i te wā tonu ka tau ki te papa.

Rahinga tere =

Ahunga tere =

Now consider the effect of gravity as the two pieces fall to the ground.

A side view of the path of piece A is shown in the diagram below. As piece A falls from P to Q, the vertical component of its momentum increases by 3.1 kg m s-1.

initial direction of motion of piece A

ground level

P

Q

(d) Discuss the impulse on piece A as it falls from P to Q, explaining the cause of the impulse and the effect that the impulse has on the momentum of A.

(e) Calculate the velocity of A at the instant it hits the ground.

Velocity size =

Velocity direction =

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ANAKETAU PĀTAI

He puka anō mēnā ka hiahiatia.Tuhia te (ngā) tau pātai mēnā e hāngai ana.

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QUESTION NUMBER

Extra paper if required.Write the question number(s) if applicable.

© New Zealand Qualifications Authority, 2012. 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 3 Physics, 201290521 Demonstrate understanding of mechanical systems

9.30 am Tuesday 27 November 2012 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 Booklet L3–PHYSR.

In your answers use clear numerical working, words and / or diagrams as required.

Numerical answers should be given with an SI unit, to an appropriate number of significant figures.

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 principles.

Give descriptions or explanations in terms of phenomena, concepts, principles and / or relationships.

Give 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

English translation of the wording on the front cover

90

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