Solution Manual for Newtonian Tasks Inspired by Physics Education Research: nTIPERs, 1st Edition

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SOLUTIONSMANUALNEWTONIAN TASKS INSPIRED BYPHYSICS EDUCATION RESEARCHnTIPERsCurtis J. HieggelkeJoliet Junior College

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CONTENTSContents........................................................................................................................................................................1Answering Ranking Tasks ..........................................................................................................................................2nTex-RT1: Stacked Blocks—Mass of Stack ............................................................................................................2nTpractice-RT2: Stacked Blocks—Number of Blocks.............................................................................................3nTpractice-RT3: Stacked Blocks—Average Mass ...................................................................................................4nT1 Preliminaries ........................................................................................................................................................5nT1A-RT1: Cutting up a Block—Density ................................................................................................................5nT1A-WWT2: Cutting up a Block—Density ...........................................................................................................6nT1A-CCT3: Breaking up a Block—Density...........................................................................................................7nT1A-QRT4: Slicing up a Block—Mass & Density ................................................................................................8nT1A-QRT5: Cylindrical Rods with Same Mass—Volume, Area, and Density......................................................9nT1A-BCT6: Four Blocks—Mass and Density......................................................................................................10nT1B-CT7: Scale Model Planes—Surface Area and Weight .................................................................................11

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2ANSWERINGRANKINGTASKSNTEX-RT1:STACKEDBLOCKS—MASS OFSTACKShown below are stacks of various blocks. All masses are given in the diagram in terms ofM,the mass ofthe smallest block.ABM3M5MC3M3MD3MMEM3M5M3MMRank the total mass of each stack.Greatest1 _______ 2 _______ 3 _______ 4 _______5 _______ LeastOR, The magnitude of the total mass of each stack is the same but not zero.___OR, The magnitude of the total mass of each stack is zero.___OR, The ranking for the total mass of each stack cannot be determined.___Explain your reasoning.Example answer formatsStacks A and D have a total mass of 9M, C and E have a mass of 4M, and B has a mass of 6M or theranking is A = D > B> C = E. Thus the ranking task answer should be expressed either asGreatest1 ___AD__ 2 _______ 3 ___B___ 4 ___CE___5 _______ LeastorGreatest1 ___A___ 2 __D____ 3 ___B___ 4 ___C___5 ___E___ LeastNote the order of equals is not important but it is easier if people are encouraged to use alphabetical orderwhen possible.Other alternative formats are AD > B > CE or A = D > B > C = EAn alternative but not preferred format isGreatest1 ___AD__ 2 ___B___ 3 ___CE___4 ______ 5 ______ Least

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3NTPRACTICE-RT2:STACKEDBLOCKS—NUMBER OFBLOCKSShown below are stacks of various blocks. All masses are given in the diagram in terms ofM,the mass ofthe smallest block.ABM3M5MC3M3MD3MMEM3M5M3MMRank the total number of blocks in each stack.Greatest1 ___A___ 2 ___D___ 3 __B____ 4 ___C___5 ___E___ LeastOR, The total number of is the same blocks for each stack.___OR, The ranking for the total number of blocks in each stack cannot be determined.___Explain your reasoning.There are 3 blocks in cases A and D, and two in all other cases. So a correct ranking is A = D > B = C= E.

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4NTPRACTICE-RT3:STACKEDBLOCKS—AVERAGEMASSShown below are stacks of various blocks. All masses are given in the diagram in terms ofM,the mass ofthe smallest block.ABM3M5MC3M3MD3MMEM3M5M3MMRank the average mass in each stack.Greatest1 ___A___ 2 __B____ 3 __D____ 4 ___C___5 ___E___ LeastOR, The average mass is the same for each stack.___OR, The ranking for the average mass in these stacks cannot be determined.___Explain your reasoning.Adding the mass of the individual boxes in each stack we find 9M for cases A and D each with threeboxes, 6M for case B with two boxes, and 4M for cases C and E each with two boxes. The averagemass in each stack is the total mass of each stack divided by the number of boxes in each stack givingan average of 3M for stacks A, B, and D. The average mass is 2M for stacks C and D. Thus theranking is A = B = D > C = E.

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5NT1PRELIMINARIESNT1A-RT1: CUTTING UP ABLOCK—DENSITYA block of material (labeled A in the diagram) with a widthw,heighth,and thicknesst,has a mass ofModistributed uniformly throughout its volume. The block is then cut into three pieces, B, C, and D, asshown.h/32h/3w/32w/3ABCDRank the density of the original block A, piece B, piece C, and piece D.Greatest1 ______ 2 ______ 3 ______ 4 ______LeastOR, The density is the same for all these pieces.___OR, The ranking for the densities cannot be determined.___Please explain your reasoning.Answer: The density is the same for all four pieces.Since the mass is uniformly distributed, a piece with half the volume will also have half the mass butthe density, the ratio of mass to volume, remains the same.

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6NT1A-WWT2: CUTTING UP ABLOCK—DENSITYA block of material with a widthw,heighth,and thicknesst,has a mass ofModistributed uniformly throughout its volume. The block is then cut intotwo pieces, A and B, as shown. A student makes the following statement:“The density is calculated by dividing the total mass by the volume.Since the volume is in the denominator, a large volume will give asmall density. Therefore the block with the smallest volume, block B,will have the largest density.”What, if anything, is wrong with the above statement? If something iswrong, explain the error and how to correct it. If the statement iscorrect, explain why.Answer: The statement is incorrect.The smaller block has 1/3 the volume of the original block but also 1/3 the mass, so the ratio of mass tovolume is the same as the original block.2wht3w3

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7NT1A-CCT3:BREAKING UP ABLOCK—DENSITYA block of material with a widthw,heighth,and thicknesst,has a massofModistributed uniformly throughout its volume. The block is thenbroken into two pieces, A and B, as shown. Three students make thefollowing statements:Andy:“The density is the mass divided by the volume, and the volumeof B is smaller. Since the mass is uniform and the volume is inthe denominator, the density is larger for B.”Badu:“The density of piece A is larger than the density of piece Bsince A is larger, thus it has more mass.”Coen:“They both have the same density. It’s still the same material.”Which, if any, of these three students do you agree with?Andy_____ Badu _____ Coen _____ None of them______Please explain your reasoning.Answer: Coen is correct. Density is the ratio of mass to volume. The larger piece has twice the massbut also twice the volume of the smaller piece. Therefore the ratio of mass to volume is the same forboth and the same as the density of the unbroken block.2wht3w3

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8NT1A-QRT4:SLICING UP ABLOCK—MASS&DENSITYThe block of material shown below has a lengthLoand a volumeVo.An overall mass ofMois spreaduniformly throughout the volume of the block to give a densityρoand a linear density (in the direction ofthe measured lengthLo)ofλo.VoMoρoLoThree possible ways to slice the block into unequal pieces are shown below. In each case, the larger piecehas a volume 2Vo/3 and the smaller piece has a volumeVo/3.EFVo32Vo3CDVo32Vo3ABVo32Vo3Fill in the table below with the quantities indicated for the pieces of the block labeled A – F in termsof the variablesMo,λo, andρo.13λoMass per unit lengthMass per unit volumeλo23λo13λo23λoρoλoλoρoρoρoρoρoρo13MoMo23Mo13Mo23Mo23Mo13MoOriginal blockPiece APiece BPiece CPiece DPiece EPiece FMass

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9NT1A-QRT5:CYLINDRICALRODS WITHSAMEMASS—VOLUME,AREA,ANDDENSITYTwo cylindrical rods are shown. Rod A has a heightHand a radiusRandrod B has a height2Hand a radius2R.Both rods have the same totalmass. Rod A has a densityρAand volumeVA.(a) What is the volume of rod B in terms of the volume of rod A?(Your answer should look likeVB=n VA, wherenis some number.)Please explain.The volume of Rod B = 8 times the volume of Rod A since the volume isproportional to the product of the area (which is proportional to thesquare of the radius) and the height of the cylinder, and the radius and the height have doubled.(b) What is the surface area of rod B in terms of the surface area of rod A?(Your answershould look likeSAB=n SAA, wherenis some number.)Please explain.SAB=4 SAA,The surface of each rod is made up of the circles at the top and bottom and the ‘side wall.’ Thetop and bottom of the rod are circles, and the area of a circle is proportional to the square of theradius. Since the radius is twice as big for Rod B as for Rod A, the area of the top and bottom is4 times as big for Rod B. The ‘side wall’ of the cylinder has an area equal to the circumferencetimes the height, and the circumference of a circle is proportional to the radius. Since bothheight and radius have doubled, the area of the side wall has also quadrupled. The overallsurface area is 4 times as large for Rod B as for Rod A.(c) What is the density of Rod B in terms of the density of Rod A?(Your answer should looklikeρB=nρA, wherenis some number.)Please explain.ρA/8Since B has 8 times the volume of A with the same total mass, its density will be one-eighth thatof A.(d) What is the mass per unit length of Rod B to the mass per unit length of Rod A?(Youranswer should look likeλB=nλA, whereλis the mass per unit length andnis some number.)Please explain.Rod B has 8 times the mass of A and twice the length so the mass per unit length of Rod B is 4 timesthe density of A.ARadiusR,HeightHBRadius 2R,Height 2H

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10NT1A-BCT6:FOURBLOCKS—MASS ANDDENSITYThe block of material shown to the right has a volumeVo.An overall massMoisspread evenly throughout the volume of the block so that the block has auniform densityρo.For each block shown below, the volume is given as well aseitherthe mass orthe density of the block.2Vo2MoA2VoMoB2VoρoC2Vo2ρoDConstruct two bar charts for the mass and density for the four blocks labeled A – D and for thepieces of the blocks if they were cut in half labeled A/2 – D/2.The mass and density for the originalblock is shown to set the scale of the chart.(Block B)/2(Block C)/2(Block A)/2(Block D)/2MassDensityBlock ABlock BBlock CBlock DOriginal block(Block B)/2(Block C)/2(Block A)/2(Block D)/2Block ABlock BBlock CBlock DOriginal blockPlease explain.Answer The mass of blocks A and B are given. The mass of block C must be 2Mo, since it has thesame density as the original block and twice the volume. The mass of block D must be 4 times the massof the original block, since it has twice the volume and twice the density and mass is volume timesdensity. The mass of one-half of a block will be half the mass of the full block. The density of one-halfof a block will be the same as the density of a full block, since the volume is halved as well as the mass,and the density is the ratio of these quantities.VoMoρo

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11NT1B-CT7:SCALEMODELPLANES—SURFACEAREA ANDWEIGHTA woodworker has made four small airplanes and one large airplane. All airplanes are exactly the sameshape, and all are made from the same kind of wood. The larger plane is twice as large in every dimensionas one of the smaller planes. The planes are to be painted and then shipped as gifts.Case BCase Aa)The amount of paint required to paint the planes is directly proportional to the surface area.Will theamount of paint required for the single plane in Case A begreater than, less than,orequal tothetotal amount of paint required for all four planes in Case B?Please explain your reasoning.Answer: Equal to or the same.Since the larger plane is twice as big in every dimension, any feature of the surface of the larger planewill have double the width and double the height of the equivalent piece of surface of the smallerplane. So the area of any piece of surface of the larger plane is four times the area of the equivalentsurface of the smaller plane. Thus, the total surface area of four small planes is the same as the totalsurface area of one large plane.b)The shipping cost for the planes is proportional to the weight.Will the weight of the single plane inCase A begreater than, less than,orequal tothe total weight of all four planes in Case B?Please explain your reasoning.Answer: Greater than.Since the larger plane is twice as big in every dimension, any piece of the larger plane will have doublethe width, double the height, and double the length of the equivalent piece of the smaller plane. So thevolume of any piecelarger plane is eight times the volume of the equivalent piece of the smaller plane,and will weigh eight times as much. Thus, the weight of four small planes is only half the weight of onelarge plane.

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1CONTENTS(BOOK PAGES7-25)Contents ....................................................................................................................................................................... 1nT2 Vectors.................................................................................................................................................................. 2nT2A-QRT1: Vectors on a Grid I—Magnitudes...................................................................................................... 2nT2A-RT2: Vectors on a Grid I—Magnitudes ........................................................................................................ 3nT2A-QRT3: Vectors on a Grid II—Directions ...................................................................................................... 4nT2B-CCT4: Adding Two Vectors—Magnitude of the Resultant........................................................................... 5nT2B-QRT5: Vectors on a Grid III—Graphical Representation of Sum................................................................. 6nT2B-RT6: Vectors I—Resultant Magnitudes of Adding Two Vectors.................................................................. 7nT2B-CCT7: Combining Two Vectors—Resultant................................................................................................. 8nT2B-CT8: Combining Vectors—Magnitude of Resultant ..................................................................................... 9nT2B-QRT9: Vector Combination II—Direction of Resultant.............................................................................. 10nT2C-CCT10: Two Vectors—Vector Difference .................................................................................................. 11nT2C-CT11: Two Vectors—Vector Sum and Difference...................................................................................... 12nT2C-RT12: Addition and Subtraction of Three Vectors I—Magnitude............................................................... 13nT2C-RT13: Addition and Subtraction of Three Vectors II—Direction of Resultant ........................................... 14nT2C-RT14: Addition and Subtraction of Three Vectors II—Magnitude of Resultant ......................................... 15nT2D-QRT15: Vector Combinations III—Components of the Resultant Vector .................................................. 16nT2D-QRT16: Force Vectors—Properties of Components ................................................................................... 17nT2D-QRT17: Velocity Vectors—Properties of Components............................................................................... 18nT2D-CCT18: Vector—Resolution into Components ........................................................................................... 19nT2D-CT19: Vector on Rotated Axes—Components............................................................................................ 20nT2D-CCT20: Vector Components—Resultant Vectors ....................................................................................... 21nT2E-WBT21: Calculations with Four Vectors—Vector Operation ..................................................................... 22nT2E-WBT22: Vectors on a Grid—Scalar (Dot) Product Expression................................................................... 23nT2E-WBT23: Vectors on a Grid—Vector (Cross) Product ................................................................................. 24nT2E-QRT24: Vectors on a Grid—Product Expressions That Are Zero............................................................... 25

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nTIPERs2NT2VECTORSNT2A-QRT1:VECTORS ON AGRIDI—MAGNITUDESEight vectors are shown below superimposed on a grid.AEHBCFGDGGGGGGGG(a) List all of the vectors that have the same magnitude as vectorGA.Answer: D and G since their magnitudes are 3 units like A.(b) List all of the vectors that have the same magnitude as vectorGB.Answer: None of them, since none of the other vectors have length 5 units.(c) List all of the vectors that have the same magnitude as vectorGC.Answer: F since its magnitude is 4 units like C.(d) List all of the vectors that have the same magnitude as vectorGD.Answer: A and G since their magnitudes are 3 units like D.(e) List all of the vectors that have the same magnitude as vector−GA.Answer: D and G since their magnitudes are 3 units like A. Vector –A has the same length as vector A and so hasthe same magnitude.(f) List all of the vectors that have the same magnitude as vector−GB.Answer: None of them, since none of the other vectors have length 5 units.(g) List all of the vectors that have the same magnitude as vector−GC.Answer: F since its magnitude is 4 units like C and like –C.(h) List all of the vectors that have the same magnitude as vector−GD.Answer: A and G since their magnitudes are 3 units like D and like –D.

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3NT2A-RT2:VECTORS ON AGRIDI—MAGNITUDESEight vectors are shown below superimposed on a grid.AEHBCFGDGGGGGGGGRank the magnitudes of the vectors.Greatest 1_______ 2_______ 3_______ 4_______ 5_______ 6_______ 7_______ 8______ LeastOR, All of these vectors have the same magnitude.____OR, We cannot determine the ranking for the magnitudes of the vectors.____Please explain your reasoning.Answer: B > F = C > A = D = G > E = H based on the relative lengths of the vectors, which represent theirmagnitudes.

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