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BYE!":D12000:CE":" DIVIDED BY ELECTRIC CHARGE)":640i:"9. ENERGY":620:"10.ANGULAR VELOCITY":640:D1300::D7:Z$", YOU HAD "M" WRONG ANSWERS."::M1610N"TERRIFIC. DO ANOTHER.":X"DO YOU WANT TO CONTINUE (Y/N)?";B$:(B$,1)"Y"230:660<bITIES":640#"3. VELOCITY":640{"4. ANGULAR MOMENTUM (MOMENT OF INERTIA":" TIMES ANGULAR VELOCITY VECTOR)":640"5. DISPLACEMENT":640:"6. MASS":620&"7. WORK (DOT PROD.-FORCE & DISAPLM'T)":620*0"8. ELECTRIC FIELD INTENSITY (FOR:"YOU CAN DO BETTER. READ YOUR TEXT!":D12000::660g:2:" VECTORS AND SCALARS, QUIZ 3":"****************************************"::M0::"1. DISTANCE":620 "2. INTENSITY OF E/M WAVES (CROSS PROD.-":" ELECT & MAG FIELD INTENS640:"9. TIME":620b"10.ANGULAR MOMENTUM (CROSS PROD.-RADIUS":" AND LINEAR MOMENTUM)":640D1300:::7:Z$", YOU GOOFED ON "M"."::M1480"EXCELLENT. DO ANOTHER.":"DO YOU WANT TO CONTINUE (Y/N)?";B$:(B$,1)"Y"230:6607***"::M0N:"1. ACCELEREATION":640:"2. DENSITY":620:"3. VOLUME":620"4. KINETIC ENERGY":620:"5. ELECTRIC CHARGE":620"6. ANGULAR VELOCITY (CROSS PROD.-RADIUS":" AND LINEAR VELOCITY)":640"7. SPEED":620:"8. DISPLACEMENT":M1370:"VERY GOOD! DO ANOTHER.":bh"DO YOU WANT TO CONTINUE (Y/N)?";B$:(B$,1)"Y"230:660r:"NOT SO GOOD. I THINK YOU NEED TO STUDY!":D12000::660|:2:" VECTORS AND SCALARS, QUIZ 2": "*************************************. TORQUE (CROSS PROD.-RADIUS & FORCE)":640:"7. MASS":620l@"8. IMPULSE (PRODUCT OF FORCE & TIME)":640J"9. MOMENTUM (PROD. OF MASS & VELOCITY)":640T"10.POWER (DOT PROD.-FORCE & VELOCITY)":620:D1300::#^7:Z$", YOU MISSED "M"."::RROR!"::240= :2:" VECTORS AND SCALARS, QUIZ 1": "****************************************"::M0::"1. AREA":620 ""2. VELOCITY":640:"3. TIME":620 ,"4. WORK (DOT PROD.-FORCE & DISPLACEM'T)":620:"5. TEMPERATURE":620;6"6ECTOR OR SCALAR BY CHOOSING ":h "EITHER 'V' OR 'S'.":::"WHEN YOU ARE READY, PRESS : ";A$ ::::"THERE ARE 3 QUIZES FOR YOU TO USE.": :"CHOOSE QUIZ 1, 2, OR 3."::"WHICH DO YOU WANT?";X:260 X270,380,490 :"INPUT E$", YOU NEED TO REVIEW.":J "STUDY YOUR TEXT SOME MORE!":::710:660s "VERY GOOD, "Z$"!":D11500:::5 "HERE'S A CHANCE TO SEE IF YOU CAN"::"IDENTIFY VECTORS AND SCALARS. 9:"FOR EACH QUANTITY LISTED, INDICATE IF": "IT IS A V"GREAT! NOW HERE'S ANOTHER."::D11000:x :::"TRUE OR FALSE? A DOT PRODUCT OF TWO"::"VECTORS IS A SCALAR.": :"ANSWER T OR F";::B$:::(B$,1)"T"180:(B$,1)"F"160 "YOU HIT THE WRONG KEY.":D11500::130 :9:"SORRY, "Z *":E F:"ENTER T OR F";::B$::(B$,1)"F"120:(B$,1)"T"90m P"YOU HIT THE WRONG KEY."::::50 Z:8::"YOU'RE NOT READY FOR THIS QUIZ!": d"STUDY THE DEFINITIONS IN YOUR TEXT."::"COME BACK WHEN YOU'RE READY. n:710:660+ x:s - VECTOR/SCALAR QUIZI680::5:"HI! WHAT'S YOUR NAME? ";Z$:8"WELL, "Z$", I'M HERE TO HELP YOU."::"SO, LET'S GET STARTED!"::(D12000::132"IS THIS STATEMENT TRUE OR FALSE?": <::"* A SCALAR HAS MAGNITUDE & DIRECTION.      MODIFIED BY *%". * TOM McCORD *<"8 * EDUCATIONAL *S"B * MEDIA *j"L * SERVICES *"V ****************,34:,!10,315:6,11:7,12:4,11:3,12:2!f!:100:8:" ************************":!" VECTOR RESOLUTION"::" ************************":!255:D11500:::!"";A$:! ****************"$ * V11,V2:V11,V24:R T021:T6,31T::24,2710:10,1427:V132:V25:900: 27,17:27,21:27,25:27,29:8,10:11,10:14,10:17,10:20,10:23,10: 1,2932:V132:V230:910: 8,395:V13:V21:920:!6,2732:26,31:25,30:26,33:253:V13,V22:V14,V21:V14,V2:V1,V2:V11,V21:V12,V22:V13,V23:V14,V24:V1,V24:V11,V23:V13,V21:V14,V2:V1,V2:V11,V21:V12,V22:V12,V23:V12,V24:V13,V21:V14,V2: V2,V24V1:V2,V24V12:V11,V13V22:RE STILL HAVING ":7f"TROUBLE, SEE YOUR INSTRUCTOR.p20:"TO REVIEW THIS PROGRAM, TYPE 1."::"TYPE 2 TO CONTINUE: ";A:A170z:10:"BYE FOR NOW, "Z$"!":20:"END":(4)"RUN MENU""V1,V2:V1,V21:V11,V22:V11,V23:V12,V24:V13,V2= V * COS(THETA)":" V(N) = 400 * COS(40)"::" V(N) = 306 POUNDS"::\H23:1040R:6:"HOW DID YOU DO? IF YOU GOT THE RIGHT"::"ANSWERS, YOU'RE DONE WITH THIS LESSON.":\"IF NOT, YOU SHOULD RUN THE PROGRAM ONE"::"MORE TIME. IF YOU' 400 POUNDS AND"::"THE ANGLE THETA IS 40 DEGREES."::j " FIND V(P) AND V(N). THEN,":16:4:1040*:6:"HERE ARE THE ANSWERS:"::4" V(P) = V * SIN(THETA)":" V(P) = 400 * SIN(40)"::" V(P) = 257 POUNDS":::N>" V(N) WER P OR N:";::B$#B$"N"780~:(7):21:"SORRY. REMEMBER THAT THE COMPONENT":"NEXT TO THE ANGLE USES THE COSINE.24:1040::740 ::5:"GREAT, "Z$". YOUR'RE GETTING"::"THE IDEA! NOW SOLVE THE PROBLEM":8"IF THE VECTOR ISNES":"PERPENDICULAR TO THE AXES.j24:1040::20,2:20,3:T013:.65T21,T3:.65T10,16T::18,213,1629:27,2916:13,169:9,1116:21:"FINALLY, THE COMPONENTS ARE FOUND.":24:1040:21:"WHICH ONE USES THE COSINE?":::"ANS:21:"HERE ARE THE AXES WE'LL USE--THE P & N":"AXES WHICH ARE AT RIGHT-ANGLES.24:1040:15:V122:V214:930::21:"AND, HERE'S OUR ANGLE THETA.24:1040:1:15,25:13,22:11,19:23,25:25,22:27,19!:21:"NATURALLY, WE NEED TO SKETCH LI29:17,28:V118:V233:900e:21:"HERE'S THE VECTOR WE'LL CONSIDER.":24:1040:12:18,0:T124.65T19,T::33,3425:22,2535:27,3235:27,3238:36,28:37,29:3T025:.65T4,25T::4,525:22,253:27,321:27,293:2,27:2,29Q[X"A VECTOR WHEN THE AXES ARE NOT THE X-"::"AND Y-AXES. IT REALLY DOESN'T MAKE ANY":b"DIFFERENCE--THE STEPS ARE THE SAME."::l"*** AS YOU DO THIS EXAMPLE, TRY TO"::" PREDICT EACH STEP!!":24:1040v:2:2,3019:20,29:21,28:18," V(X) = 8.19 METERS":::" V(Y) = V * SIN(THETA)":" V(Y) = 10 * SIN(35)":" V(Y) = 10 * .574"::" V(Y) = 5.74 METERS"::D"IT'S EASY!":24:1040N:3:"LET'S LOOK AT ANOTHER COMMON SITUATION."::"WE'LL CONSIDER THE COMPONENTS OF":G, FIND":T"THE X- & Y-COMPONENTS IF V = 10 METERS"::"AND THETA = 35 DEGREES.22:"WHEN YOU THINK YOU HAVE THE ANSWER,"::1040&:3:"HERE'S THE ANSWER:"::0" V(X) = V * COS(THETA)":" V(X) = 10 * COS(35)":" V(X) = 10 * .819"::"::1040b::3:"NOTE THAT THE X-COMPONENT WILL NOT"::"ALWAYS USE THE COSINE TERM! THE":"COMPONENT NEXT TO THE MARKED ANGLE "::"WILL ALWAYS USE THE COSINE.":12 "NOW LET'S CONSIDER A REAL PROBLEM."::"FOR THE DRAWING WE'VE BEEN USINEN THE VECTOR AND THE X-AXIS."::"NATURLLY, THEN:"::b" V(Y) = V * SIN(THETA)":24:1040:2:940:12:960:3:970:0:5,3215:V117:V225:930:1:950:980:990 :21:" V(X) = V * COS(THETA)":" V(Y) = V * SIN(THETA)(1) = V * COS(THETA), AND":G" V(2) = V * SIN(THETA)":24:1040:3:"FOR EXAMPLE, IN THE DRAWING WE'VE BEEN"::"LOOKING AT, THE X-COMPONENT WOULD BE:"::" V(X) = V * COS(THETA)":::"BECAUSE THE ANGLE THETA IS SHOWN ":8"BETWEY"::" (MEASURE THEM), OR USE TRIGONOMETRY."::"** HERE'S AN EASY TRICK: IF THE ANGLE"::"(THETA) IS BETWEEN THE VECTOR AND":"AXIS #1, THEN THE COMPONENT IN THE"::"DIRECTION OF #1 WILL USE THE COSINE":"OF THE ANGLE.**":::" V VECTOR.":_T"2. MEASURE THE ANGLE BETWEEN THE VECTOR"::" AND ONE AXIS (EITHER ONE).":^"3. SKETCH LINES FROM THE 'HEAD' OF THE"::" VECTOR PERPENDICULAR TO THE AXES.h24:1040:Q2:120r::31|"4. FIND THE COMPONENTS GRAPHICALL;,X17:980:990:1:940:980:990:2:940::24:1040j6:::"ISN'T THAT NEAT? OF COURSE IT IS!@5:"NOW HERE ARE THE SPECIFIC INSTRUCTIONS:":::"1. SKETCH THE RIGHT-ANGLE AXES WITH THE": J" ORIGIN (0,0) AT THE 'TAIL' OF THE"::" S, ONCE":"YOU HAVE THE COMPONENTS.P1:5,2732:10,315:24:1040::2:94021:"THAT'S BECAUSE THE VECTOR AND ITS":"COMPONENTS REPRESENT THE SAME THING.24:1040:"21:"THEY CAN BE EXCHANGED FOR EACH OTHER":"WHENEVER IT'S CONVENIENT!THODS OR TRIG-":"ONOMETRY TO FIND THE TWO COMPONENTS.d 24:1040::0:940:950:V117:V225:930 21:"ONCE YOU HAVE THE COMPONENTS, YOU DON'T":"NEED THE ORIGINAL VECTOR ANY MORE! 24:1040::960:970#21:"IN FACT, YOU DON'T NEED THE AXESTEP 2 - MEASURE THE ANGLE BETWEEN THE":"VECTOR AND ONE AXIS (THETA).^ 24:1040::1:950 21:"STEP 3 - DRAW PERPENDICULAR LINES FROM":"THE HEAD OF THE VECTOR TO THE AXES. 24:1040::Q2370:1:980:9906 21:"STEP 4 - USE GRAPHICAL MEER ANY VECTOR , POINTING":"IN ANY DIRECTION.Z 24:1040::12:960:3:970:0:5,32 21:"STEP 1 IS TO ADD A SET OF AXES. YOU DO":"NOT HAVE TO USE X & Y, BUT THE AXES "SHOULD BE AT RIGHT-ANGLES.":24:1040::V117:V225:15:930F 21:"LSO BE POSSIBLE TO REPLACE ONE VECTOR": Z"BY TWO EQUIVALENT VECTORS. THIS TECH-"::"NIQUE, CALLED RESOLVING A VECTOR INTO": d"COMPONENTS, IS VERY USEFUL IN PHYSICS."::"LET'S LOOK AT THE PROCESS: n24:1040 x:2::9400 :21:"CONSID"TRIG BEFORE YOU START THIS LESSON. BE"::"SURE YOU HAVE YOUR CALCULATOR HANDY!": <:"WHEN YOU'RE READY TO BEGIN,"::1040 F:4:Z$", THE BASIC IDEA IS"::"VERY SIMPLE. SINCE TWO VECTORS CAN BE":( P"ADDED TOGETHER TO MAKE ONE, IT MUST"::"A - VECTOR RESOLUTIONI1010::3:"HELLO! WHAT'S YOUR NAME?";Z$6:"THIS PROGRAM IS FOR YOU, "Z$"."::"IT SHOULD HELP YOU UNDERSTAND HOW TO":("RESOLVE A VECTOR INTO TWO COMPONENTS."::"YOU SHOULD BE ABLE TO DO RIGHT-TRIANGLE":S 2          CES *4 ****************O" VECTOR ADDITION - PART II"::" ************************h255:D11000:::}"";A$: **************** * MODIFIED BY *  * TOM McCORD * * EDUCATIONAL *  * MEDIA ** * SERVI3:X2,Y1X2,Y5:r221,126215,126219,121221,126:X226:Y130:620:X238:690:X242:630:X172:Y92:660:0:134,82134,126:5:133,82133,126:131,123135,123:135,81220,81:217,79217,83::100:8:" ************************":4:X4,Y1:X4,Y5:YX,Y6X,Y4:X1,Y3X3,Y3:X4,Y4X4,Y6:X2,Y2X2,Y:X,Y1X,Y5:X1,Y6X3,Y6:X4,Y5X4,Y1:X1,YX3,Y:X1,Y3X5,Y3:X2,YX,Y2X,Y4X2,Y6:X,YX2,Y2X2,Y4X,Y6:X,Y3X4,Y7:Y158:3:630:QlX,YX,Y5:X1,Y6X3,Y6:X4,Y5X4,Y:X,Y3X4,Y3:vX,YX,Y6X3,Y6:X4,Y5X4,Y4:X3,Y3X,Y3:X4,Y2X4,Y1:X3,YX,Y:X,Y:X,Y6:X4,Y6:X4,Y:X,Y1:X,Y5:X1,Y2:X1,Y4:X2,Y3:X3,Y2:X3,Y:20:"END":(4)"RUN MENU"R:3:15,81266,81:133,12133,155:134,12134,155wDX270:Y84:640:X132:Y9:650:N1:135,80250,60:249,61243,64:249,61243,59:243,61XX254:Y64:3:620:b2:132,82102,146102,141:102,146106,142:X9M, REVIEW THE METHOD (RERUN THIS":X"PROGRAM). THEN, SEE YOUR INSTRUCTOR.":24:750:8:"IF YOU WANT TO REVIEW THIS PROGRAM,"::"TYPE 1. TYPE 2 TO CONTINUE: ";A:A150&:9:"THAT'S THE END FOR NOW, "Z$".":12:20:"GOOD BYE!":20:75003:"5. R+S(X) = -33.4":" R+S(Y) = -5.9"::p"6. R+S = 33.9 NT":::"7. THETA = 7.9 DEGREES S OF W"::"YOU SHOULD HAVE THE CORRECT ANSWERS."::"IF YOU DON'T, CHECK EACH STEP FOR MATH":#"ERRORS. IF YOU STILL CAN'T FIND THE"::"PROBLES N OF W"::"3. R(X) = 34 * COS(70) = 11.6r" R(Y) = 34 * SIN(70) = 31.9"::" S(X) = 52 * COS(30) = 45.0" S(Y) = 52 * SIN(30) = 26.0":"4. R(X) = 11.6":" R(Y) = -31.9":" S(X) = -54.0":" S(Y) = 26.0":24:750/:> = 52 NT AT 30 DEGREES N OF W"::z"WORK OUT THE ANSWER COMPLETELY. THEN"::"CHECK YOUR WORK. THE ANSWERS ARE ON":"THE NEXT PAGE.":20:750:3:"1. MAKE A CHART."::"2. = 34 NT AT 70 DEGREES S OF E+" = 52 NT AT 30 DEGREEr"IF YOU USE ARCTAN < A+B(Y)/A+B(X) >,"::"THE ANGLE THETA WILL BE MEASURED FROM":s|"THE X-AXIS.":24:750:3:"THAT'S ALL THERE IS TO IT!"::"NOW HERE'S ONE TO DO ON YOUR OWN:"::#"ADD-- = 34 NT AT 20 DEGREES E OF S"::" ":}^" = SQRT <1436.4 + 78.1>"::" = 38.9 CM"::h"7. THETA(A+B) = ARCTAN < -8.84/-37.90 >"::" = ARCTAN < .233 >"::" = 13.1 DEGREES S OF W"::W" A(X) = -5.13"::" A(Y) = 14.10"::j6" B(X) = -32.77"::" B(Y) = -22.94"::@"5. ADD X- & Y-COMPONENTS:"::" A+B(X) = -37.90"::" A+B(Y) = -8.84":J"ARE YOU DOING O.K., "Z$"?":24:750>T::"6. ADD AXIS."::W"3. A(X) = 15 * COS(70) = 5.13"::" A(Y) = 15 * SIN(70) = 14.10"::" B(X) = 40 * COS(35) = 32.77"::" B(Y) = 40 * SIN(35) = 22.94":24:750"::"4. ADD MINUS SIGNS. IF YOU'VE MADE A"::" SKETCH, IT WILL HELP.":2,EM. THEN WE'LL"::"SOLVE IT TOGETHER.":24:750d::"1. WE'LL ASSUME YOU'VE MADE A CHART."::"2. DATA: = 15 CM AT 70 DEGREES"::" = 40 CM AT 35 DEGREES": " CHANGE THE ANGLE OF SO IT IS"::" MEASURED FROM THE X-AE"::" PYTHAGOREAN THEOREM."::Z "7. FIND THE ANGLE USING TRIG (ARCTAN).":24:750 :6:"NOW WE'RE READY. HERE'S A PROBLEM:":::"ADD-- = 15 CM AT 20 DEG. W OF N": " = 40 CM AT 35 DEG. S OF W"::0"WRITE DOWN THE PROBLS ":- " ARE FROM THE X-AXIS!":24:750 :3:"4. ADD MINUS SIGNS TO COMPONENTS IN"::" THE NEGATIVE DIRECTION.": :"5. ADD THE X-COMPONENTS TOGETHER. DO"::" THE SAME FOR THE Y-COMPONENTS."::# "6. FIND THE TOTAL VECTOR USING TH::" CHANGE ALL ANGLES SO THEY ARE":{ " MEASURED FROM THE X-AXIS.":::"3. FIND THE X- & Y-COMPONENTS OF EACH": " VECTOR:"::" A(X) = A * COS(THETA-A)": " A(Y) = A * SIN(THETA-A)"::" THESE FORMULAS WORK ONLY IF ANGLE-----------------------------m Z"X-COMPONENT * * *":"--------------------------------------- d"Y-COMPONENT * * *":"---------------------------------------": n24:750( x:3:"2. FILL IN THE GIVEN DATA. REMEMBER TO" KEEP YOU ON"::" THE RIGHT TRACK:":} <" A B A+B":"--------------------------------------- F"MAGNITUDE * * *":"--------------------------------------- P"DIRECTION * * *":"---------- - VECTOR ADDITION - II^720::5:"HELLO AGAIN!"::"REMIND ME OF YOUR NAME? ";Z$:"GOOD. NOW, "Z$", WE'LL TRY TO SOLVE"::"A FEW PROBLEMS. FIRST, A QUICK REVIEW":("OF THE BASIC METHOD.":20:750( 2:3:"1. START WITH A CHART TO     OM McCORD *#H * EDUCATIONAL *:R * MEDIA *Q\ * SERVICES *hf ****************1220,81:217,79217,83:M:100:8:" ************************": " VECTOR ADDITION - PART I"::" ************************255:D1900::: "";A$:* ****************4 * MODIFIED BY * > * T2,YX,Y2X,Y4X2,Y6::X,YX2,Y2X2,Y4X,Y6:`X,Y3X4,Y3:X2,Y1X2,Y5:221,126215,126219,121221,126:X226:Y130:670:X238:740:X242:680:X172:Y92:710:0:134,82134,126:5:133,82133,126:131,123135,123:135,86:X4,Y6:X4,Y:X,Y1:X,Y5:X1,Y2:X1,Y4:X2,Y3:X3,Y2:X3,Y4:X4,Y1:X4,Y5:X,Y6X,Y4:X1,Y3X3,Y3:X4,Y4X4,Y6:X2,Y2X2,Y:X,Y1X,Y5:X1,Y6X3,Y6:X4,Y5X4,Y1:X1,YX3,Y:X1,Y3X5,Y3:XX254:Y64:3:670:]2:132,82102,146102,141:102,146106,142:X97:Y158:3:680:X,YX,Y5:X1,Y6X3,Y6:X4,Y5X4,Y:X,Y3X4,Y3:X,YX,Y6X3,Y6:X4,Y5X4,Y4:X3,Y3X,Y3:X4,Y2X4,Y1:X3,YX,Y:aX,Y:X,YIS LOADED.":20:15:"-LOADING-":D11500:ib(4)"BLOAD CHAIN,A520":520"VECTOR ADDITION - PART 2":l3:15,81266,81:133,12133,155:134,12134,155vX270:Y84:690:X132:Y9:700:1:135,80250,60:249,61243,64:249,61243,59:243,6150:760:24:8007:::620:135,82220,126:750:760gD22:" HERE IT IS, SIMPLIFIED."::800N::6:Z$", YOU'VE FINISHED PART I.":::"TYPE 1 TO REVIEW THIS PROGRAM."::"TYPE 2 TO CONTINUE: ";A:A150,X::"PLEASE WAIT WHILE PART II " = 11.7 NT":::" THETA = ARCTAN < A+B(Y) / A+B(X) >" = ARCTAN < -5.9 / 10.1 >":" = ARCTAN < -0.584 >"::" = -30.3 DEGREES (S OF E)":&"THAT'S THE ANSWER. WE'RE DONE!":24:8000:16304,0:3:135,82220,126:7-ANGLE VECTORS"::"THAT MUST BE COMBINED USING THE ":s"PYTHAGOREAN RULE.":24:800::16304,0:760:24:800::3:"HERE IT IS:"::" A+B = SQRT < A+B(X)^2 + A+B(Y)^2 >" = SQRT < 10.1^2 + -5.9^2 >":" = SQRT < 136.82 ><:00:X160:730!24:800:::3b"THE NEXT STEP IS TO COMBINE THE X- &"::"Y-COMPONENTS:":" A+B(X) = A(X) + B(X)":" = 13.7 - 3.6 = 10.1":" A+B(Y) = A(Y) + B(Y)":" = 2.9 - 8.8 = -5.9":7:"NOW WE HAVE TWO RIGHT133,81102,81:104,79104,83:133,82133,146:131,143137,1433:X138:Y63:670:X146:720:X152:700:X160:730:X242:Y93:670:X250:720:X256:690:X264:730 X96:Y76:680:X104:720:X110:690:X118:730:X138:Y151:680:X146:720:X152:7:"NEGATIVE SIGNS WHEN THE COMPONENTS ARE":|"IN A NEGATIVE DIRECTION!":::"ARE YOU FILLING IN YOUR CHART? GOOD!"::"NOW LET'S LOOK AT IT.":24:800:::620:6401:135,81251,81:247,79247,83:134,80134,60:130,63136,63>660:2:PONENTS ARE IN A NEGATIVE":5^"DIRECTION!"::800h:3:" A(X) = 14 * COS(12) = 13.7"::" A(Y) = 14 * SIN(12) = 2.9":r:" B(X) = 9.5 * COS(68) = -3.6"::" B(Y) = 9.5 * SIN(68) = -8.8":,|"AGAIN, USING THIS METHOD, YOU MUST ADD": THE ANGLE WITH THE X-AXIS, "::"THIS PROCESS IS EASY:":@" A(X) = A * COS(THETA-A)"::" A(Y) = A * SIN(THETA-A), AND":J" B(X) = B * COS(THETA-B)"::" B(Y) = B * SIN(THETA-B)":T"JUST BE SURE TO USE NEGATIVE SIGNS"::"WHEN THE COM WE'LL FILL IN THE DATA.*200:24:800]12:16:"14. * 9.5":14:16:"12 * 68"21:"ANGLES ARE MEASURED FROM THE X-AXIS!":24:800,::"SO FAR, SO GOOD! NOW WE NEED TO FIND"::"THE X- & Y-COMPONENTS OF & ."::6"BY USING *":"---------------------------------------"X-COMPONENT * * *":"---------------------------------------"Y-COMPONENT * * *":"---------------------------------------":X3280:X4300:24:800::5:"NOW "KEEP THINGS STRAIGHT. COPY THIS ONE:"::8 XX1 10:" A B A + B":"--------------------------------------- "MAGNITUDE * * *":"---------------------------------------3"DIRECTION * *ON'T GO WRONG LATER! NO MATTER"::"HOW THE PROBLEM IS STATED, CHANGE ALL": "ANGLES SO YOU MEASURE FROM THE X-AXIS.":19:800 :16304,0:24:800:X0 ::3:"NOW TO SOLVE THE PROBLEM. AN EASY"::"METHOD IS TO USE A CHART. IT WILL HELP":.7308 xX205:Y77:710:X213:720:X219:670:X227:730 :21:" HERE ARE THE ANGLES. YOU SHOULD":" MAKE A SKETCH FOR YOUR NOTES."::800 ::7:"NOTE THAT WE HAVE MEASURED THE ANGLES"::"FROM THE X-AXIS. ADOPT THIS METHOD &":L "YOU WM:":::"ADD: = 14 NT AT 12 DEG. N OF E": Z" = 9.5 NT AT 22 DEG. W OF S":24:"WRITE DOWN THE PROBLEM & ";A$ d::620:640:660:22:" HERE ARE THE VECTORS.":24:800 nX97:Y96:710:X105:720:X111:680:X119:ELP YOU"::"LEARN HOW TO ADD VECTORS GIVEN IN": <"THE MAGNITUDE/DIRECTION FORM, I.E."::5::" = 23 METERS AT 30 N OF E":: F"YOU SHOULD KNOW HOW TO DO RIGHT-ANGLE"::"TRIGONOMETRY BEFORE YOU BEGIN."::1 P"LET'S LOOK AT A TYPICAL PROBLEi - VECTOR ADDITION - Ib770::5:"HELLO, PHYSICS STUDENT!"::"WHAT'S YOUR NAME? ";Z$:10:Z$", YOU WILL NEED PAPER, PENCIL,"::"AND YOUR CALCULATOR."::("IF YOU'RE READY, LET'S GO!":20:8001 2:3:"THIS PROGRAM IS DESIGNED TO H               ************************816255:A11000:::M1@"";A$:d1J ****************{1T * MODIFIED BY *1^ * TOM McCORD *1h * EDUCATIONAL *1r * MEDIA *1| * SERVICES *1 ****************BSTITUTING--": FOR PRACTICE. LOOK"::"IN YOUR TEXT OR SEE YOUR INSTRUCTOR.":0"GOOD LUCK. BYE FOR NOW!":24:1600::20:"END":(4)"RUN MENU"012:2,3737:2,3738:0":100:8:" ************************":1," PROJECTILE MOTION"::" 2*9.8*(1.3)^2"::C/"SOLVING--"::7::"Y = 3.67 METERS":::/"THAT ALL THERE IS TO IT! TYPE 1 TO"::"REVIEW, 2 TO CONTINUE: ";A:A11270/:6:"THAT'S THE END OF THIS PROGRAM."::Z$", YOU SHOULD PROBABLY WORK":@0"SOME EXTRA PROBLEMS;A::^."REMEMBER THAT X & Y MOTIONS DO NOT"::"AFFECT EACH OTHER, BUT THE FLIGHT TIME":."IS THE SAME. SO, T= 1.3 SECONDS."::1600.:3:"THEREFORE--"::" Y = V(0)Y*T - 1/2*G*T^2":/:"SUBSTITUTING--"::" Y = 9.192*1.3 - 1//7.714":/-5::"T = 1.3 SECONDS":::1600-:4:"HOW DO WE FIND THE ELEVATION WHERE THE"::"ARROW HITS THE CLIFF? ";:1600::-"HINT: HOW MUCH TIME DOES THE ARROW HAVE"::"TO GO UP AND DOWN?":."INPUT THE ARROW'S FLIGHT TIME: ")Y--":_,x" V(0)X = 12*COS(50) = 7.714 M/S"::" V(0)Y = 12*SIN(50) = 9.192 M/S"::1600,::"SINCE WE KNOW THE RANGE (10 METERS) AND"::"AND THE HORIZONTAL VELOCITY (7.714 M/S), -"WE CAN FIND THE FLIGHT TIME--"::" T = RANGE/V(X) = 107:29,20>+PT11400::22:" TIME MARKERS SHOW THE MOTION.{+ZY501:7:203Y,11Y2::Y03:7:203Y,11Y2:+dT11000::24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A11320,n::2:"HERE'S THE SOLUTION:":::" FIND V(0)X & V(0N WE'LL WATCH.":24:1600G*(::1560:1:X3035:14,36X::7:5,36*222:" PRESS TO SHOOT THE ARROW.";A$:T11200:::(7)*<Y50.5:7:203Y,11Y2:0:203Y,11Y2::YO3.5+F7:203Y,11Y2:0:203Y,11Y2::(7):W IS AIMED 50 DEGREES ABOVE THE"::"HORIZONTAL AND FIRED WITH AN INITIAL":) "VELOCITY OF 12 M/S TOWARD A VERTICAL"::"CLIFF 10 METERS AWAY. HOW HIGH ABOVE":)"THE GROUND DOES THE ARROW STRIKE THE"::"CLIFF?"::*"COPY THE PROBLEM AND THE4 SECONDS, WE FIND THE RANGE--"::" RANGE = V(X)*T":(" RANGE = 12 * 4"::8::"RANGE = 36 METERS"::::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A1990(:3:"LET'S DO ONE MORE PROBLEM, THIS TIME"::"IN REVERSE:"::L)"AN ARRO= V(0)Y*T - 1/2*G*T^2":Q'"OR,"::" -78.4 = 0*T - 1/2*9.8*T^2"::1600:x'"THEN,"::" T^2 = 78.4/4.9":'"AND,"::5::"T = 4 SECONDS":::1600:1250':10:"GREAT WORK, "Z$"!"::"KEEP GOING...":22:1600<(:5:"SINCE T = TIME OF FLIGHT--":::"USE: Y = V(0)Y*T - 1/2*G*T^2":::1600&::"REMEMBER THAT Y = -78.4 METERS, SINCE"::"THE FINAL POSITION IS BELOW THE START."::&" INPUT THE FLIGHT TIME: ";A::A41240':3:(7):"SORRY, "Z$".":::" Y M%t25::"V(0)X = 12 M/S":::"BECAUSE THE BALL IS KICKED HORIZONTALLY.":}%~" ENTER V(0)Y: ";A::A01160:"SURE!";%25::"V(0)Y = 0 M/S":::"SINCE THE BALL HAS NO INITIAL VERTICAL"::"VELOCITY. ";:1600C&:4:"NOW FIND THE, 2 TO CONTINUE: ";A:A11030T$L::2:"NOW FOR THE SOLUTION. YOU SHOULD BE ABLE$V"TO GET THIS ONE ON YOUR OWN.":::"START BY FINDING THE X & Y COMPONENTS":$`"OF THE INITIAL VELOCITY."::" ENTER V(0)X: ";A::A121140$j"OF COURSE!";7)<#X02:9:13X,11:T150:T:0:13X,11::Y05.5#$9:162Y,11Y2:0:162Y,11Y2::(7):9:26,36:T11200:#.22:" TIME MARKERS SHOW THE MOTION.#89:14,11:Y05:162Y,11Y2::T11000:$B24:" TYPE 1 TO SEE AGAIN THE BUILDING"::"DOES IT LAND IF THE INITIAL VELOCITY IS":""12 METERS/SECOND?":::"COPY THE PROBLEM. THEN WE'LL LOOK AT IT."::1600"::1560:3:X515:12,36X::9:13,11#22:" PRESS TO KICK THE BALL.";A$::T1800::(0:"EXCELLENT WORK, "Z$"!"::"IT'S A PRETTY BIG CANNON TO SHOOT THAT"::"HIGH.":22:1600!:3:"HOW ABOUT ANOTHER PROBLEM?"::!"A MAN KICKS A BALL HORIZONTALLY OFF THE"::"ROOF OF A BUILDING 78.4 METERS TALL.":=""HOW FAR FROM THE BASE OF:" V(Y)^2 = V(0)Y^2 - 2*G*Y":r " 0^2 = (173.2)^2 - 2*9.8*Y"::" Y = (173.2)^2/(2)(9.8)": "AND,":9::"Y = 1530.5 METERS"::: "IT'S A BIG CANNON!":::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A1680 990[!:1NT: THE FACT THAT THE SHELL MOVES IN"::"THE X DIRECTION HAS NO EFFECT ON HOW":"HIGH IT GOES.":::"INPUT YOUR MAX. HEIGHT: ";A::A1530A1531980:4:(7):"SORRY. HERE'S THE CORRECT SOLUTION:":# "AT MAXIMUM HEIGHT, V(Y) = 0. SO,":HAT'S NOT SO BAD, IS IT?"::::"IF YOU WOULD LIKE TO REVIEW THIS":p"PROBLEM SOLUTION AGAIN, TYPE 1."::"TYPE 2 TO CONTINUE: ";A:A1680z::"AS A CONTINUATION OF THIS PROBLEM, FIND"::"THE MAXIMUM HEIGHT OF THE SHELL."::1600Q:3:"HI173.2*T = -1/2*9.8*T^2"::1600[>:"FINALLY,"::6::"T = 173.2/4.9 = 35.35 SEC":::H"* NOW YOU CALCULATE THE RANGE."::" INPUT YOUR ANSWER (NO UNITS):";AR:5:A3535860:"GOOD!";\10::"RANGE = V(X)*T = 3535 METERS":::Ef"TTIME OF FLIGHT.":n:"REMEMBER THAT THE SHELL HITS THE GROUND,":"SO: Y = Y(0) = 0 METERS ";:1600 :3:"USE THIS EQUATION:"::" Y = V(0)Y*T - 1/2*G*T^2"::1600*:"THEN,":" 0 = 173.2*T - 1/2*9.8*T^2":4"AND":" -RIZONTAL VELOCITY."::V"INPUT YOUR VALUE FOR V(0)Y: ";A::A173.2760:"YES!";9::"V(0)Y = V(0)*SIN(60) = 173.2 M/S":::"NOW USE THE VALUE OF V(0)Y AND THE"::"EQUATIONS OF MOTION FOR GRAVITATIONAL": "ACCELERATION TO FIND THE PY THE PROBLEM. THEN TRY TO FIND THE"::"X & Y COMPONENTS OF V(0)."::"INPUT YOUR ANSWER FOR V(0)X: ";A::A100730:"GOOD!";10::"V(0)X = V(0)*COS(60) = 100 M/S":::1600:3:"REMEMBER THAT THIS IS THE VALUE OF THE"::"CONSTANT HOYOU NEED TO."::"THEN, ";:1600q:3:"NOW HERE'S A PROBLEM:":::"A CANNON SHELL IS FIRED WITH A MUZZLE":"VELOCITY OF 200 M/S AT AN ANGLE OF"::"60 DEGREES ABOVE HORIZONTAL. FIND THE":"RANGE OF THE SHELL. ";:1600::F"COND LEFT ARE NEGATIVE.":s"> THE STARTING POSITION (EVEN IF ABOVE"::" THE GROUND IS X(0) = 0, Y(0) = 0.":"> THE STARTING TIME IS T(0) = 0."::"> THE ANGLE (THETA) IS MEASURED FROM ": " THE X-AXIS.":::"COPY THESE CONVENTIONS IF TIME OF":" FLIGHT (T)":kl" 5. FIND RANGE = V(X) * T"::"COPY THE METHOD FOR REFERENCE: ";:1600v:2:"THE CONVENTIONS WE USE ARE THE SAME AS"::"THOSE ESTABLISHED IN PREVIOUS PROGRAMS:"::"> UP AND RIGHT ARE POSITIVE."::"> DOWN A" INITIAL VELOCITY--"::" V(0)X = V(0)*COS(THETA)N" NOTE: V(X) = CONSTANT V(0)X!"::" V(0)Y = V(0)*SIN(THETA)X" NOTE: V(Y) WILL CHANGE!"::" 4. APPLY EQUATIONS OF MOTION IN THEb" Y DIRECTION TO FIND THE AND ANGLE, FIND THE":o&"INITIAL VELOCITY). BUT, LET'S START WITH":"THE EASY ONE! ";:16000:2:"METHOD:"::" 1. ESTABLISH CONVENTIONS"::" 2. DETERMINE KNOWN & UNKNOWN VALUES"::" 3. FIND THE X & Y COMPONENTS OF THE>D0::a"IN THE BASIC PROBLEM, YOU ARE GIVEN THE"::"INITIAL VELOCITY AND ANGLE (THETA) OF":"THE PROJECTILE. YOU ARE TO SOLVE FOR THE":"RANGE (MAX. HORIZONTAL DISTANCE)."::"OF COURSE, THE PROBLEM CAN BE CHANGED"::"AROUND (GIVEN RANGE "THE AMOUNT OF TIME THE PROJECTILE HAS"::"TO MOVE SIDEWAYS."::"DO YOU WANT TO WATCH AGAIN, "Z$"?"::"TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1250:3:"NOW LET'S LOOK AT A METHOD FOR SOLVING"::"PROJECTILE PROBLEMS. ";:160SEE THAT THE PROJECTILE"::"POSITION IS DETERMINED BY THE CONSTANT":"VELOCITY IN THE X DIRECTION AND BY THE"::"ACCELERATED MOTION IN THE Y DIRECTION."::"THEREFORE, THE TIME IT TAKES THE PRO-"::"JECTILE TO GO UP AND DOWN IS EXACTLY":E1:T11000:TW11:10,36203Y:7:4,2011Y2:1:203Y,11Y2::7:21,3611oY15:T11000:T11:10,36203Y:7:21,3611Y2:1:203Y,11Y2:24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1400F::3:"YOU SHOULD 1000::22:" TIME MARKERS SHOW THE MOTION.l|Y501:13:203Y,11Y2::Y05:13:203Y,11Y2:24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1250::1560:1:5,36:22:" HERE IT IS IN SLOW MOTION.":T11500:Y50RS SHOW THE ACCELERATION.=J7:Y05:4,11Y2::24:1600T:22:" NOW COMBINE THE MOTIONS.":13:5,36:T11000:^Y50.5:13:203Y,11Y2:0:203Y,11Y2::YO5.5h13:203Y,11Y2:0:203Y,11Y2::13:35,36-rT1STANT VELOCITY. 911:Y010:53Y,10::24:1600":22:" HERE'S A VERTICAL ACCELERATION.":7:4,36:T11000:,Y50.5:7:4,11Y2:0:4,11Y2::Y05.567:4,11Y2:0:4,11Y2::7:4,36@T11000::22:" TIME MARKELET'S SEE HOW THE MOTIONS COMBINE."::"WATCH CAREFULLY, "Z$"!":::1600 ::1560:22:" HERE'S A CONSTANT HORIZONTAL VELOCITY. 11:5,10:T11000::Y010.1:11:53Y,10:0:53Y,10::11:35,10T11000::22:" TIME MARKERS SHOW CONME"::"TIME, AND THE RESULT IS PARABOLIC":X "PROJECTILE MOTION. ";:1600:: "THE TRICK IN SOLVING PROBLEMS IS TO"::"CONSIDER THE PROJECTILE MOTION AS TWO": "SEPERATE MOTIONS WHICH LAST FOR THE"::"SAME AMOUNT OF TIME.":I :" :"INDEPENDENT":Z 12:13:"EXECPT FOR THE TIME FACTOR.":::"THE MOTIONS ARE:": " 1. HORIZONTAL - CONSTANT VELOCITY"::" 2. VERTICAL - ACCELERATION DUE TO " GRAVITY":::1600, :3:"THESE TWO MOTIONS OCCUR AT THE SATICS & GRAVITATIONAL ACCELERATION":::"WHEN YOU'RE READY, ";:1600 :3:"THE BASIC IDEA IS SIMPLE, BUT YOU MUST"::"UNDERSTAND IT WELL TO SOLVE PROBLEMS.":: "PROJECTILE MOTION CAN BE CONSIDERED AS"::"TWO SIMULTANEOUS MOTIONS WHICH ARE":YOU, "Z$". THIS PROGRAM"::"IS DEVOTED TO THE SOLUTION OF PROJECTILE n"PROBLEMS. ALTHOUGH THEY'RE REALLY NOT "::"TOO HARD, STUDENTS HAVE DIFFICULTY WITH": x"THEM--SO PAY CLOSE ATTENTION.":::"YOU SHOULD HAVE ALREADY WATCHED:":D :"KINEMA) PROJECTILE MOTION( 7/23/83@SOLVING PROJECTILEU(MOTION PROBLEMSk2NSF LOCI PROJECT<SOUTH OKLAHOMA CITYFJR. COLLEGEPOKLAHOMA CITY, OKLAHOMA Z1570::3:"HELLO!"::"PLEASE TELL ME YOUR NAME: ";Z$E d8:"THANK                              * TOM McCORD *(>4 * EDUCATIONAL *?>> * MEDIA *V>H * SERVICES *m>R ****************12:X04>#7:10,12122XX2:T160:T:0:10,11122XX2:>#22:10060:A19100::?#::9:10,120:21:"A NEGATIVE ACCELERAT ************************":m=" ONE DIMENSIONAL KINEMATICS"::" ************************=255:D11000:::="";A$:= " TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:= ****************=  * MODIFIED BY *>*PRESS .";A$><12:14,180:T1800:::(7):X01<9:10,129X:T1100:T:0:10,119X::12:18,1419:X301<9:10,1235(2XX2):T1100:T:0:10,1135(2XX2)::9:10,1237<22:2060:A11970:=:100:8:" 0:::(7):X05U;7:10,122X:T1100:T:0:10,112X::13:18,1412:X04;7:10,12122XX2:T160:T:0:10,11122XX2:;22:2060:A11920:<::9:10,120:21:"A NEGATIVE ACCELERATION ADDED TO A"::"CONSTANT VELOCITY. TURN> TO START.";A$=:l7:14,180:T1800:::(7):X05~:v12:10,12X22X:0:10,11X22X::22:2060:A11890::::7:10,120:21:"A POSITIVE ACCELERATION ADDED TO A"::"CONSTANT VELOCITY. PRESS .";A$;13:14,180:T180I9N" T - FINAL":::" VELOCITY V(0) - INITIAL":9X" V - FINAL":::" ACCELERATION A - ACCELERATION":::2050:::b::12:10,120:21:"AN OBJECT AT REST IS GIVEN A POSITIVE"::"ACCELERATION. PRESS TO CHECK YOUR WORK.";A$v):2:1750:5:26:"40":9:19:"6":11:22:"+2)15:"O.K? NOW LOOK AT THE EQUATIONS. WHICH ":"ONE IS BEST FOR FINDING THE TIME?";A:::A11280)"SURE!";*11::"#1 CAE 2 TO CONTINUE: ";A:A1900j(:3:"HERE'S ANOTHER PROBLEM. IT'S A LITTLE"::"MORE COMPLICATED:"::("AN OBJECT TRAVELLING AT 6 M/S IS GIVEN ":"AN ACCELERATION OF +2 M/S^2. HOW LONG":("DOES IT TAKE TO TRAVEL 40 METERS?":::A)"TRY T HOW TO FIND REVOLUTIONS FROM"::"RADIANS?";:31:2050::w'" REVOLUTIONS = RADIANS/2*PI":::"THEREFORE--":'" REVOLUTIONS = 10/2*3.14"::6::"REVOLUTIONS = 1.59":::31:2050(:10:"TO REVIEW THIS LAST PROBLEM, TYPE 1."::"TYPHETA = 4/0.4 = 10 RADIANS"::2050:1150i&t10:"VERY GOOD!";:12:"THETA = 10 RADIANS":::2050&~:3:"DID YOU NOTICE THAT THE PROBLEM ASKED"::"FOR THE NUMBER OF REVOLUTIONS AND NOT":&"NUMBER OF RADIANS?";:31:2050::;'"REMEMBERB10::"#1 OR #3 ARE GOOD.":::2050::a%L"YOU USE #3 AND SOLVE FOR THE DISTANCE."::2050%V:3:"INPUT YOUR ANSWER: ";A::A101140%`::"NOPE!"::" OMEGA^2 = OMEGA(0)^2 + 2*A*THETA":.&j" (2)^2 = (0)^2 + 2*0.2*THETA"::" T'S THE SOLUTION--":l$" 2 = 0 + ALPHA*10"::6::"ALPHA = 2/10 = 0.2 M/S^2":::2050:1070$$6:"GREAT! ALPHA = 0.2 M/S^2"::2050$.13:"NOW FOR THE DISTANCE. WHAT EQUATION DO ":"YOU WANT TO USE? ";A::A21090$8"O.K!";+%ITION. REMEMBER THAT WE'RE WORKING INy#"THE ROTATIONAL SYSTEM, SO--"::" OMEGA = OMEGA(0) + ALPHA*T"::2050::#"YOU SUBSTITUTE THE DATA AND SOLVE FOR"::"ALPHA. INPUT YOUR ANSWER (NO UNITS): ";A::A.21060$3:"WRONG, "Z$"."::"HERE YOUR ERRORS IS A GOOD WAY TO ":"LEARN. O.K?"::31:2050":3:"NOW SELECT THE BEST EQUATION TO USE FOR ":"FINDING THE ACCELERATION: ";A::A21000""YES.";"11::"#2 IS THE BEST"::%#"WE CHOOSE #2, SINCE #1 AND #3 NEED THE"::"POSROBLEM AND TRY TO FILL IN THE ":"CHART YOURSELF. WHEN YOU'RE READY TO":t!"CHECK YOUR WORK, ";:2050::2:1750!7:26:"10":9:19:"0";:27:"2!15:"DID YOU GET IT RIGHT, "Z$"?"::"IF NOT, THINK ABOUT WHAT YOU DID WRONG.":;""ANALYZINGBLEM:":::"A WHEEL, INITIALLY AT REST, IS GIVEN AN "ANGULAR ACCELERATION. AFTER 10 SECONDS ":"IT IS ROTATING AT 2 RAD/SEC. WHAT WAS "THE ACCELERATION? HOW MANY REVOLUTIONS ":"DID IT MAKE IN THE FIRST 10 SECONDS?"::I!:"COPY THE P + A*T"::"SUBSTITUTING THE DATA--":Pf" V = 0 + 3*5"::"THEN--":p5::"V = 15 M/S (MOVING TO THE RIGHT)"::::"EASY! IF YOU WANT TO REVIEW THIS PROBLEMz"TYPE 1. TYPE 2 TO CONTINUE: ";A:A16606 :3:"HERE'S A ROTATIONAL PRO::2050:830=4"GOOD!":12::"Y = 37.5 METERS":::2050>:3:"NOW FOR THE SECOND PART--TO FIND THE ":"VELOCITY. WHICH EQUATION DO YOU WANT?";A::A1850H"O.K!";R10::"#3 IS FINE, BUT #2 IS EASIER.":::2050&\:" V = V(0)"USE EQUATION #1:";:31:2050:f" Y = V(0)*T + 1/2*A*T^2"::"NOW WE SUBSTITUTE THE DATA--": " Y = 0 * 5 + 1/2*3*T^2"::"WHAT'S THE ANSWER (WITHOUT UNITS)?";A::A37.5820 *"WRONG, "Z$"."::" Y = 0 + 1/2*3*25 = 37.5 METERS"A1750"GOOD!";?11::"#1 IS THE BEST";::31:2050:2:"SINCE WE'RE LOOKING FOR DISTANCE (X), WE":"WOULD LOOK AT EQUATIONS #1 & #3 (SINCE "#2 DOES NOT INCLUDE X). BUT EQUATION #3 ":"NEEDS THE FINAL VELOCITY 'V', SO WE WILL .";:31:2050@::1750:7:27:"5":9:19:"0":11:22:"+314:"NOTICE THAT WE HAVE ONLY TWO UNKNOWNS! "::"NOW LOOK AT THE EQUATIONS. INPUT THE":"NUMBER OF THE EQUATION YOU THINK WOULD ":"BE MOST HELPFUL TO FIND THE DISTANCE: ";A::INITIALLY AT REST, IS GIVEN ":"A POSITIVE ACCELERATION OF +3 M/S^2. HOW"FAR HAS IT GONE AFTER 5 SECONDS? HOW ":"FAST IS IT GOING AT THAT TIME?"::"COPY THE PROBLEM.";:31:2050:: "NOW LET'S FILL IN THE CHART WITH WHAT WE":"KNOW.."BY NUMBER.";:31:2050w:5:"THAT'S THE DISCUSSION ON METHOD. IF YOU ":"WOULD LIKE TO REVIEW IT BEFORE WE BEGIN "TO SOLVE PROBLEMS, TYPE 1. TYPE 2 TO ":"CONTINUE: ";A:A1190:3:"NOW HERE'S OUR FIRST PROBLEM:"::I"AN OBJECT, VE\X"OBTAINED ALL THE INFORMATIION IN THE"::"PROBLEM STATEMENT. FOR EXAMPLE, A BODY":b"'INITIALLY AT REST' MEANS V(0) = 0."::2050l:3:"NOW WE'RE READY FOR THE EQUATIONS."::2050::1830v"COPY THE EQUATIONS. WE'LL REFER TO THEM ":E THAT THERE ARE FIVE"::"EMPTY PLACES IN THE CHART? GOOD!"::2050::D"IN ANY PROBLEM, YOU WILL BE GIVEN THREE"::"OF THE VALUES. THEN YOU MUST SOLVE FOR":N"THE OTHER TWO. SO, BEFORE YOU BEGIN ANY"::"CALCULATIONS, CHECK TO SEE THAT YOU HART IS A HANDY WAY TO KEEP TRACK"::"OF THE INFORMATION.":::s"HERE'S ONE YOU CAN USE..."::2050::2:1750&17:"NOTICE THAT TWO VALUES ARE FILLED IN."::"THIS WILL ALWAYS BE TRUE. COPY THE":0"CHART AND ";:2050I::3:"DID YOU NOTICMBER THAT IF THE FINAL":u"POSITION WORKS OUT TO HAVE A NEGATIVE"::"VALUE, IT IS LOCATED TO THE LEFT OF THE":"INITIAL POSITION.":::2050:6:"IN ORDER TO ORGANIZE YOUR THINKING,"::"IT IS HELPFUL TO ORGANIZE YOUR DATA.":?"A CHAWHILE ACCELERATION TO THE LEFT IS":?"NEGATIVE.":31:2050:3:"** INITIAL CONDITIONS:":::" THE INITIAL TIME, T(0), AND THE":" INITIAL POSITION, X(0), ARE ZERO."::"THIS CONVENTION SIMPLIFIES THE EQUATIONS":"WE USE. JUST REMETHE FIRST STEP IN OUR"::"METHOD IS TO ESTABLISH CONVENTIONS.":"HERE ARE TWO USEFUL ONES:":::"** DIRECTIONS:"::" AND ARE POSITIVE. ":" AND ARE NEGATIVE."::$"THUS, VELOCITY TO THE RIGHT IS POSITIVE,":"METHOD WE USE WORKS FOR BOTH"::"SYSTEMS, WE'LL EXPRESS MOTION IN TERMS":"OF 'X', 'V', AND 'A'. JUST REMEMBER THAT":"YOU CAN SUBSTITUTE 'THETA', 'OMEGA',":"AND 'ALPHA' FOR ROTATIONAL PROBLEMS.":22:2050::2:1850A:3:"REMEMBER THAT :2050::dh"IN THE ROTATIONAL SYSTEM, WE MUST USE:"::" OMEGA - ANGULAR VELOCITY (RAD/SEC)":r" ALPHA - ANGULAR ACCELERATION (RAD/S/S)"::2050|::"BE SURE TO WRITE DOWN ANYTHING YOU DON'T":"KNOW, "Z$"!";:31:2050K:7:"SINCE THE Z-AXIS":I@:"** ROTATION"::" THETA - FOR MOTION AROUND AN":xJ" AXLE (IN RADIANS)"::2050T:3:"IN THE TRANSLATIONAL SYSTEM, WE USE THE"::"FAMILIAR 'V' FOR VELOCITY AND 'A' FOR": ^"ACCELERATION. ";SLOW DOWN, AND REVERSE."::2050v":4:"SO, IF WE WANT TO REPRESENT THE POSITION":"OF AN OBJECT, WE CAN USE:"::,"** TRANSLATION"::" X - FOR MOTION ON THE X-AXIS": 6" Y - FOR MOTION ON THE Y-AXIS"::" Z - FOR MOTION ON THEL MOTION,"::"THE OBJECT ROTATES AROUND A FIXED AXLE":"WITH A CONSTANT RADIUS--FOR EXAMPLE,"::"A WHEEL. IT CAN SPEED UP, SLOW DOWN, ":"AND REVERSE ROTATION DIRECTION JUST"::"LIKE AN OBJECT ON A STRAIGHT LINE PATH": "CAN SPEED UP, VE IN ONE DIMENSION ONLY.":v "FOR TRANSLATIONAL MOTION, WE SAY THAT"::"THE OBJECT MAY MOVE BACK AND FORTH ALONG "A STRAIGHT LINE--USUALLY THE X-AXIS."::"(WE COULD ALSO CHOOSE THE Y-AXIS, ETC.)"::20508:6:"FOR ROTATIONAL ONE DIMENSIONAe :"WRITE DOWN THESE STEPS. BE SURE TO NOTE"::"THAT YOU MUST LEARN THE EQUATIONS!"::2050:: :5:"BEFORE WE LOOK AT THE EQUATIONS, LET'S"::"DISCUSS THE VARIABLES. ";:2050:: "ONE DIMENSIONAL MOTION MEANS AN OBJECT"::"CAN MOOBLEMS.b "BE SURE TO COPY DOWN THIS METHOD AS WE"::"GO ALONG. YOU'LL WANT TO REFER TO THE": "METHOD WHEN WE SOLVE PROBLEMS."::2050: :"NOW LET'S LOOK AT A FEW EXAMPLES."::2050:1890:1920:1970 ::3:"HERE'S THE METHOD:"::1800ASE":^ "WHERE ACCELERATION IS CONSTANT. YOU"::"WILL NEED TO KNOW A FEW EQUATIONS AND": "SOME BASIC ALGEBRA.":::"GET YOUR CALCULATOR READY AND ";:2050 :4:"THE MOST IMPORTANT PART OF THIS PROGRAM"::"DESCRIBES A METHOD FOR SOLVING PRLAHOMA CITY, OKLAHOMA T d2020::3:"WELCOME!"::"PLEASE GIVE ME YOUR NAME: ";Z$ n8:"THANK YOU, "Z$". IN THIS PROGRAM"::"WE'LL SPEND SOME TIME LEARNING A METHOD": x"OF SOLVING ONE-DIMENSIONAL MOTION "::"PROBLEMS. WE'LL ONLY LOOK AT THE Cn6 ONE-DIMENSIONAL /KINEMATICS 7/27/83 SOLVING ONE-DIMENSIONAL, CONSTANT ACCELERATION KINEMATICS PROBLEMS, TRANSLATIONAL AND ROTATIONAL( 2 NSF LOCI PROJECT< F SOUTH OKLAHOMA CITYP JR. COLLEGE Z OK                       TOM McCORD *%7 * EDUCATIONAL *<7 * MEDIA *S7 * SERVICES *j7 ****************YE!":20:10050::D13:(7)::20:"END7D11500:::(4)"RUN MENU":8@:" *INITIAL* FINAL *":" ------*-------*-X515:12,36X::G6:100:8:" ************************":6"KINEMATICS & GRAVITATIONAL ACCELERATION"::" ************************6255:D11000:::6"";A$:6 ****************6 * MODIFIED BY *7 * " T * 0 * *":" ------*-------*-------*5" V * * *":" ------*-------*-------*5" A * -9.8 M/S^2 *":" ------*---------------*"::512:5,3437:5,3438:612:STRUCTOR IF YOU'RE STILL":f4h"HAVING PROBLEMS. BYE,BYE!":20:1750::20:"END":(4)"RUN MENU"4r:" *INITIAL* FINAL *":" ------*-------*-------*4|" Y * 0 * *":" ------*-------*-------*C5122.5 METERS"::20:"TYPE 1 IF YOU WANT TO REVIEW THIS"::"PROBLEM. TYPE 2 TO CONTINUE: ";A:A11210q3J16303T:10:"OUTSTANDING WORK, "Z$"!"::"YOU GET HIGH MARKS FOR THAT.":22:17504^:8:"THAT'S ABOUT ALL FOR NOW, "Z$".":::"SEE YOUR INBSTITUTE--"::" 0 = 39.2^2 - 2*9.8*Y":k2"" THEN--"::" Y = 39.2^2/19.6":2," AND, "::15::"Y = 78.4 METERS":::175026:8:"DON'T FORGET TO ADD THE HEIGHT OF"::"THE BUILDING.":g3@6::"Y = 78.4 + 44.1 = T YOU CAN'T DIVIDE THE TIME"::"BY 2 FOR THIS ONE. TRY EQUATION #3.1::"NOW SOLVE AND ENTER YOUR ANSWER: ";A:A122.51620:A78.415901:3:"SORRY, "Z$". HERE IT IS--":61" EQUATION #3--"::" V^2 = V(0)^2 - 2*G*Y":62" SU"GROUND REACHED BY THE ARROW IN THE"::"LAST PROBLEM?"::0"TYPE 1 IF YOU NEED TO SEE THE PROBLEM"::"AGAIN, 2 TO CONTINUE: ";A::A112100:"WHAT EQUATION WILL YOU USE? ";A::A315200"YES. THAT'S THE BEST ONE.":1530D1"NOTICE THA"! YOU'VE GOT"::"THE IDEA.w/16:"DON'T EXPECT ALL PROBLEMS TO FACTOR"::"NEATLY. YOU MIGHT HAVE TO EMPLOY THE":/"QUADRATIC FORMULA. BE SURE YOU KNOW IT!":::1750/:3:"ONE FINAL PROBLEM:"::"WHAT IS THE MAXIMUM HEIGHT ABOVE THE":<0B.x" REARRANGE--"::" 4.9*T^2 - 39.2*T - 44.1 = 0"::1750.:3:" DIVIDE BY 4.9--"::" T^2 - 8*T - 9 = 0":." FACTOR--"::" (T - 9)(T + 1) = 0":." FINALLY--"::9::"T = 9 OR -1 SECONDS"::1450/9:"TERRIFIC,"Z$ GO AHEAD AND FIND THE TIME."::"ENTER YOUR ANSWER WITHOUT UNITS:";A::A91440-Z:"SORRY! HERE'S THE CORRECT SOLUTION:"::-d" EQUATION #1--"::" Y = V(0)*T - 1/2*G*T^2":.n" SUBSTITUTE VALUES--"::" -44.1 = 39.2*T - .5*9.8*T^2":NG IN YOUR CHART:"::1750D,(1650:8:25:"-44.1":12:17:"+39.2,217:"WHICH EQUATION SHOULD YOU USE TO FIND"::"THE TIME OF FLIGHT? (ENTER THE NUMBER):";A::A11350,<"VERY GOOD, "Z$"!";,F25::"USE EQUATION #1":::1750Q-P:3:"OK,700::Y40.5:9:15,0Y2:0:15,0Y2::Y06.5+ 9:16,Y2:0:16,Y2::(7):9:16,36:T11200::Y06:13:31,Y2:+22:"TIME MARKERS SHOW ACCELERATION."::"TYPE 1 TO SEE AGAIN, 2 TO CONTINUE:";A:A11270,::2:"BEGIN BY FILLI44.1 METERS TALL. IF THE ARROW"::"MISSES THE BUILDING AND HITS THE GROUND,*"HOW LONG IS IT IN THE AIR?"::"LET'S WATCH. ";:1750*::1700:X515:12:17,36X::9:15,16:22:" PRESS TO SHOOT THE ARROW.";A$7+:T1FUL"::"WITH SIGN CONVENTIONS, YOU GET THE RIGHT_)"ANSWERS? IT'S REALLY NEAT! ";:1750)::"IF YOU CAN STAND IT, WE'LL DO ONE MORE."::)"A MAN SHOOTS AN ARROW UPWARDS WITH A"::"VELOCITY OF 39.2 M/S FROM THE TOP OF A":L*"BUILDING z#?Xİ    ҠΠԠҠΠүҠڠΠŠǠ ծԠ  ӠӠͯĠ͠ԠӠͯĠ͠ԯӠӠ Ӡ̮֮1Š̠Ӡ8 ŠΠ+ ҠΠԠ̠Ϡ ̠Ϯà!ՠҠέǠҠέԠέҠͮ έҠͮ-@@YY3fP*U*U*U*U*U*U*U*U*U*U*U`<P*U*U*U*UP*U*U*U*U`P*U*U*U*U*U@*U*U*U*U*U`F `P*ysdy?N3N3~9~O9fys($'<.'*** WRITTEN BYT8'*** LYNELL JACKSONnB'*** JANUARY 14, 1987tL'V'*** LAST MODIFIED BY`'*** LYNELL JACKSONj'*** JANUARY 27, 1987t'~'<*** "ISL HELLO" ***>DMONDAY-SATURDAY." #""F#"TECHNICAL/SOFTWARE SUPPORT HOURS ARE"w#"8:30 AM --> 4:30 PM PST MONDAY - FRIDAY." #""#"CALL US FOR A FREE CATALOG OF"#"COMPUTER SUPPLIES AND THE LATEST AND"#"MOST UP TO DATE PUBLIC DOMAIN SOFTWARE." Z#"A DIVISION OF U.S.COMPUTER SUPPLY INC."O d#"511-104 ENCINITAS BOULEVARD"j n#"ENCINITAS, CA 92024" x#"(800) 992-1992 (USA) #"(800) 992-1993 (FOR CA) #"(619) 942-9998 (FOREIGN) #"" #"ORDER HOURS ARE"#"6:00 AM --> 4:30 PM PST l24:7110 vGO$: # : TB%(21(PR$)2)I TB%:PR$;O U (#r +#<*** DATA LOCATION ***>x /# 2#"FOR OTHER PUBLIC DOMAIN AND" <#"USER-SUPPORTED PROGRAMS PLEASE CONTACT:" F#"" P#"THE INTERNATIONAL SOFTWARE LIBRARY",16297,1:16300,1:16301,1:16304,12 70108 @ :M I1ND%_ PR$:I:7110f Ip 7010v  RN$""ė:D$;"CATALOG": (:D$;"RUN ";RN$ 2 X [<*** SUBROUTINES ***> _ bPR$"< PRESS ANY KEY TO CONTINUE >" GF%11110 :E PR$" U.S. COMPUTER SUPPLY INC. PRESENTS:"S 10:7110 PR$"THE INTERNATIONAL SOFTWARE LIBRARY" $13:7110 .7010:1210 L V: `PR$"* LOADING PROGRAM *" j11:7110: ~D$;"BLOAD ISL HELLO.GRAPHIC"( CING FOR PRINTING DATA LINE6 * I = LOOP VARIABLE S * GO$ = CONTINUE STRING GF%1:* DISPLAY GRAPHIC FLAG, 1=YES, 0=NO ND%20:* NUMBER OF DATA LINES RN$"":* NAME OF THE PROGRAM TO RUN NEXT  <*** MAIN PROGRAM ***>  <*** "ISL HELLO" ***>">*** PROGRAM WRITTEN BYV(*** LYNELL JACKSONp2*** JANUARY 14, 1987vdg<*** VARIABLES ***>knD$(4):* CONTROL-DxH$(8):* CONTROL-H* PR$ = DATA LINE FOR PRINTING * TB% = TAB SPA     P*Os?~~?O`P*U*U*U*U*U@*U*U*U*U*U``P*?U`<P*U*U*U*U T*U*U*U*U`P*U*U*U*U*U*U*U*U*U*U*U`L}|L1`><P*ysd?N3N?~9rO9fIs`P*U*U*U*U*U@*U*U*U*U*U``s3O{qc<P*?U`<P*U*U*U*UP*U*U*U*U`P*U*U*U*U*UP*U*U*U*U*U` `P*ysx?N30~9O9fp ";DQ$:D(DQ$):D1DN50:DN160:DN70::D$"RUN "PR$(D) <D$"CATALOGp F:"RUN HELLO": VECTOR RESOLUTION,VECTOR RESOLUTION, VECTOR ADDITION - PART 1,VECTOR ADDITN F$(16),PR$(16)::D$(13)(4):16368,0:2)"SOUTH OKLAHOMA CITY JUNIOR COLLEGE":8)"PHYSICS TUTORIAL PROJECT":"BY STEVEN D. KAMM W/SUPPORT OF THE NSF"::"WHICH PROGRAM WOULD YOU LIKE TO VIEW ?"::N16:I1N:F$(I) PR$(I):I;:I10ĺ". "P*U*U*U*U*U*U*U*U*U*U*U`xP*U*U*U*U**U*U*U*U*U``@P*U*U*U*U*U*U*U*U*U*U`xP*U*U*U*U**U*U*U*U*U`@1@P*L3O?NsOsLgyg9U`~P*U*U*U*U*U*U*U*U*U*U*U``P*?`<P*U*U*U*U**U*U*U*U*U``P*y|?&bD"";A$:H ****************R * MODIFIED BY *\ * TOM McCORD *f * EDUCATIONAL *p * MEDIA * z * S,13:<:"ANSWER TRUE OR FALSE:";B$:::(B$,1)"F"760X"ERROR!":(7):GG1:j"GREAT!":::"ANSWER TRUE OR FALSE:";B$:::(B$,1)"T"790 "ERROR!":(7):GG1:"GREAT!":: :100:8:" ************************1N):NN.5!T14.85670:65017186,41:C1:0:T3.14712875.5(T),8365(T):712875.5(T),8365(T)TT.2:T14.85710:6907186,41:1175,21:5185,21:3191,21:6198,21:182,40151,13151,18:151,13157 NOW, "Z$".":::" BYE!":20:830Mb:20:"END":(4)"RUN MENU"tl:3:128,80132,80:130,78130,82vT3.143.14.1:13069.6(T),8060(T)::1:1:T3.14:N0 712875.6(T),8365(T):712875.6(T),8365(T):TT(."EXCELLENT, "Z$"! A PERFECT SCORE."::"IT LOOK LIKE YOU REALLY UNDERSTAND":b:"THE MATERIAL.oD24:830N:10:"IF YOU WANT TO REVIEW THIS PROGRAM,"::"PLEASE TYPE 1. IF YOU WANT TO CONTINUE,"::"TYPE 2: ";A:A160,X:10:"THAT'S ALL FORN CIRCULAR MOTION. THEN"::"SEE YOUR INSTRUCTOR IF YOU'RE STILL":_"HAVING TROUBLE.":580:"WELL, "Z$", YOU ONLY MISSED ONE."::"THAT'S PRETTY GOOD. BUT, YOU MIGHT":&"WANT TO WATCH BOTH CIRCULAR MOTION"::"PROGRAMS AGAIN.":580N0::ENTIAL ACCELERATION?"::740::830::6:G0560>G1540"WELL, "Z$", YOU MISSED "G"."::"THAT'S NOT SO GOOD. DID YOU UNDERSTAND":"THAT IN THE FIRST FEW SECONDS THE FAN"::"WAS SPEEDING UP? YOU HAD BETTER REVIEW":C"BOTH PROGRAMS O"::"ANGULAR MOMENTUM. WE'LL BE LOOKING AT":y <"SIMPLE 1-DIMENSIONAL PROBLEMS INVOLVING"::"ONE OR TWO BODIES.":: F"LET'S LOOK AT THE VARIABLES INVOLVED:"::780 P:4:" OMEGA = ANGULAR VELOCITY (RAD/SEC)"::" V = TANGENTIIAL VELOCITY"::+) - CONSERVATION OF ANGULAR MOMENTUMW750:A(V)140P1(V):B(V)80P2(V):4:"HELLO! WHAT'S YOUR NAME? ";Z$:7("WELL, "Z$", THIS PROGRAM WILL"::"HELP YOU LEARN TO SOLVE BASIC PROBLEMS":. 2"WHICH INVOLVE THE CONSERVATION OF         :" ************************":255:D1900::::O"";A$:f ****************} * MODIFIED BY * * TOM McCORD *  * EDUCATIONAL * * MEDIA *  * SERVICES ** ****************7186,41:;1175,21:5185,21:3191,21:6198,21:f182,40151,13151,18:151,13157,13:780,75:888,75:996,75:10104,75:2112,75::100:8:" ************************":" CIRCULAR MOTION - CONSTANT SPEED:EE YOU LATER, "Z$"!":24:740:20:"END":(4)"RUN MENU"_l:3:128,80132,80:130,78130,82vT3.143.14.1:13069.6(T),8060(T)::1:0:T3.14712875.5(T),8365(T):712875.5(T),8365(T)TT.2:T14.85670:650 "IF YOU WOULD LIKE TO REVIEW, TYPE 1."::"TO CONTINUE, TYPE 2: ";A:A170TD:7N"THIS CONCLUDES OUR DISCUSSION OF"::"CIRCULAR MOTION WITH CONSTANT SPEED.":X"YOU SHOULD NOW WATCH:":::3::"CIRCULAR MOTION - CHANGING SPEED":::8b"SRO"::" - RADIAL ACCELERATION = V^2/R":::740x620:640:680:700:2160,71:5169,71:4176,71:6184,71:5:6902:182,41156,60156,55:156,60162,60&21:"COPY THIS DIAGRAM FOR ROTATIONS":"WITH CONSTANT SPEED.":0740:::10K: OF THE CURVE. **"::740c:3:"TO SUMMARIZE;"::"WHEN THE ROTATION SPEED IS CONSTANT:"::0" - ANGULAR VELOCITY IS CONSTANT"::" - TANGENTIAL VELOCITY IS NOT CONSTANT"::" - ANGULAR ACCELERATION IS ZERO"::" - TANGENTIAL ACCELERATION IS ZETION."::" ** THIS ACCELERATION, WHICH IS":" ASSOCIATED WITH A CHANGE IN"::" DIRECTION BUT NOT MAGNITUDE, IS":" CALLED A RADIAL OR CENTRIPETAL"::" ACCELERATION. A(R) = V^2/R ."::" IT ALWAYS POINTS TOWARDS THE CENTER"::"TANGENTIAL VELOCITY IS CONSTANT,"::" THERE CAN BE NO ACCELERATION IN"::" THE DIRECTION OF THE TANGENT: A(T)=0"::740::30"3. SINCE THE DIRECTION OF THE TAN-"::" GENTIAL VELOCITY, V(T), IS CHANGING,"::" THERE MUST BE AN ACCELERARATIONS WHEN"::"THE SPEED OF ROTATION IS CONSTANT."::"1. SINCE THE ANGULAR VELOCITY (OMEGA)"::" IS CONSTANT, IT MUST BE TRUE THAT"::" THE ANGULAR ACCELERATION (ALPHA) IS"::" ZERO."::740:"2. SINCE THE MAGNITUDE OF THE"::" ."::::7406620:640:680:700:5:690:22:740::7:"TO SUMMARIZE;"::"WHEN THE ROTATION SPEED IS CONSTANT:"::" - ANGULAR VELOCITY IS CONSTANT"::" - TANGENTIAL VELOCITY IS NOT CONSTANT"::::7409:"NOW LET'S CONSIDER ACCELEr"VELOCITY (SPEED) IS CONSTANT, THE"::"DIRECTION (TANGENT TO THE CIRCLE) IS":~|"CONSTANTLY CHANGING."::::740:340"VERY GOOD! SINCE THE DIRECTION OF THE"::"TANGENT IS CHANGING, THE VELOCITY IS": "NOT CONSTANT. LET'S WATCH AGAIN0:290@J"GOOD. THE ANGULAR VELOCITY IS CONSTANT."::::740sT:5:"IS THE TANGENTIAL VELOCITY CONSTANT?":^:"ANSWER YES OR NO.";B$:::(B$,1)"N"390h"SORRY. REMEMBER THAT VELOCITY IS A"::"VECTOR. ALTHOUGH THE MAGNITUDE OF THE":RSTANT."::::740D":7::"IS THE ANGULAR VELOCITY CONSTANT?":w,:"ANSWER YES OR NO.";B$:::(B$,1)"Y"3306"WRONG. SINCE THE OBJECT SWEEPS OUT"::"EQUAL ANGLES IN EQUAL TIMES, THE":@"ANGULAR VELOCITY (OMEGA) IS CONSTANT."::::74 WHEELS."::780=b::"ONE MORE PROBLEM: ";:780l:5:"A 50 KG MASS IS ATTACHED BY A 20 METER"::"CORD TO A CENTRAL HUB (I = 4000 KG-M^2).":v"BOTH ROTATE AT 5 RAD/SEC. WHAT WILL BE"::"THE NEW VELOCITY IF THE CORD IS REDUCED":DLAR VELOCITY"::"IS BETWEEN THE ORIGINAL VALUES."::D"WHAT WOULD YOU HAVE DONE IF THE TWO"::"WHEELS HAD BEEN ROTATING IN OPPOSITE":N"DIRECTIONS? ";:780:X:"YOU'RE RIGHT! SIMPLY ASSIGN A NEGATIVE"::"VELOCITY TO ONE OF THEMEGA2' (SINCE THE TWO":@"WHEELS ARE JOINED). ";:780&::"THEN SUBSTITUTE--"::".05*20 + .15*2 = .05*OMEGA' + .15*OMEGA'":" 1 + .3 = .20*OMEGA'":0:"AND, OMEGA' = 6.5 RAD/SEC"::7806::3:"NOTICE THAT THE FINAL ANGU:780:c:"THIS PROBLEM IS LIKE THE 1-DIMENSIONAL"::"INELASTIC COLLISION PROBLEMS YOU'VE":"DONE BEFORE. ";:780:2:"SOLUTION:"::" I1*OMEGA1 + I2*OMEGA2 =":" I1'*OMEGA1' + I2'*OMEGA2'":::"WHERE OMEGA1'=OI = .05 KG-M^2) ROTATES AT 20":y"RAD/SEC. IT IS SUDDENLY JOINED TO A"::"SECOND WHEEL (I = .15 KG-M^2) WHICH IS":"ROTATING AT 2 RAD/SEC. WHAT IS THE FINAL":"SPEED OF THE TWO WHEELS IF THEY WERE":"BOTH ROTATING IN THE SAME DIRECTION?":OW HERE'S SOMETHING YOU MIGHT NOT KNOW.":"THE MOMENT OF INERTIA OF TWO BODIES":"(WITH THE SAME CENTER) MAY BE ADDED.":::"THIS ALLOWS US TO WORK 2-BODY PROBLEMS":"IF WE KEEP THEM ON A SINGLE AXIS."::780 :2:"PROBLEM:"::"A WHEEL (ORTANT! THE ANGULAR"::"VELOCITY DECREASES TO 1/4, WHILE THE":"TANGENTIAL VELOCITY ONLY DECREASES TO"::"1/2 (WHEN THE RADIUS IS HALVED)."::780:3:"PLEASE, "Z$", DON'T GET TANGENTIAL"::"& ANGULAR VELOCITIES CONFUSED!"::780::Q"N.0 TO .75 RAD/SEC. ";:780jr:3:"WHAT HAPPENS TO THE TANGENTIAL"::"VELOCITY? ";:780:|"THE ORIGINAL VELOCITY IS 6 M/S. THE"::"FINAL VELOCITY = R * OMEGA'":" = 4 * .75 = 3 M/S"::780@:"THIS IS VERY IMP OMEGA' = ?"::780zT::"APPLY THE CONSERVATION LAW--"::" L1 = L1'"::" I1 * OMEGA1 = I1' * OMEGA1'":^" 20 * 3 = 80 * OMEGA'"::" OMEGA' = .75 RAD/SEC"::780 h:"SO THE ANGULAR VELOCITY DECREASES FROM"::"35 * 2^2 = 20 KG-M^2":A," I' = 5 * 4^2 = 80 KG-M^2"::780:6"THE MOMENT OF INERTIA INCREASES WHEN THE":"RADIUS INCREASES."::@"NOW FIND THE ANGULAR VELOCITY--"::" OMEGA = V/R"::780J:3:" OMEGA = 6 M/S / 2 M = 3 RAD/SEC"::"ENED"::"TO 4 METERS?":::"COPY THE PROBLEM AND ";:780:::"NOTICE THAT WE HAVE ONLY ONE BODY, SO:"::" L1 = L1'":::780:3:"SOLUTION:"::"FIND THE MOMENT OF INERTIA--":"" I = M*R^2 (FOR A POINT MASS)"::" I = " TYPE 1 TO SEE AGAIN, 2 TO CONTINUE? ";A:A1200::2:"HERE'S THE PROBLEM:"::"A 5 KG MASS TRAVELS AT 6 M/S IN A":"CIRCULAR PATH OF RADIUS 2 METERS. "::"WHAT ARE THE INITIAL AND FINAL ANGULAR"::"VELOCITIES IF THE RADIUS IS LENGTHT AN EXAMPLE:":::780a :P150:P210:P3.5:P44:P50:P$" ROTATING POINT MASS":710 P1100:P220:P3.1:P42:P5.3:P$" THE RADIUS IS LENGTHENED":710 P150:P210:P3.6:P46:P$" RETURN TO THE ORIGINAL RADIUS":710424: ANGULAR":3 "MOMENTUM = (I * OMEGA).":::780 :5:"NOTICE THAT CHANGING EITHER THE MASS OR"::"THE RADIUS OF A ROTATING BODY WILL ": "CHANGE THE MOMENT OF INERTIA, AND WE"::"EXPECT THE ANGULAR VELOCITY TO CHANGE. 16:"LET'S LOOK A::"MOMENTUM:": " L1 + L2 = L1' + L2'"::" I1*OMEGA1 + I2*OMEGA2 ="::" I1'*OMEGA1' + I2'*OMEGA2'":: "WITH NO EXTERNAL TORQUES, ANGULAR "::"MOMENTUM BEFORE AN EVENT = ANGULAR": "MOMENTUM AFTER THE EVENT WHERE" V = (R * OMEGA)":w Z" I = MOMENT OF INERTIA"::" 1/2*M*R^2 - DISK"::" M*R^2 - POINT MASS": d" M = MASS"::" R = RADIUS":: n"WRITE THESE DOWN FOR REFERENCE.":::780 x:3:"NOW THE LAW OF CONSERVATION OF ANGULAR"" P(X)=M*V(X) P(Y)=M*V(Y)":24:1630||:3:" C. APPLY THE CONSERVATION LAW"::" P1(X)+P2(X)=P1(X)'+P2(X)'":" P1(Y)+P2(Y)=P1(Y)'+P2(Y)'":::" D. SOLVE FOR UNKNOWNS (USUALLY A":" VELOCITY).":::"BE SURE TO WRITE DOWNTION"::" IS NEEDED (UP & RIGHT ARE POSITIVE,":fT11:" WHILE DOWN & LEFT ARE NEGATIVE.)"::^"3. STEPS TO SOLUTION:"::" A. RESOLVE VELOCITIES INTO X & Y":h" COMPONENTS - V(X) & V(Y)."::" B. FIND MOMENTUM COMPONENTS -":)r::" THE CHANGE.":L,1590::" WHERE (') MEANS AFTER THE CHANGE.":z623:"WRITE DOWN THESE IDEAS AND ";:1630@:3:"2. MOMENTUM IS A VECTOR. THEREFORE, WE"::" MUST WORK WITH X & Y COMPONENTS":3J" SEPERATELY. ALSO, A SIGN CONVENT DOWN TO BUSINESS!":23:1630w:2:"HERE'S WHAT YOU NEED TO KNOW:":::"1. IN THE ABSENCE OF EXTERNAL FORCES,":" MOMENTUM IS CONSERVED. THEREFORE, THE":" TOTAL MOMENTUM BEFORE SOME CHANGE IS":"" EQUAL TO THE TOTAL MOMENTUM AFTER" SITUATIONS, MOMENTUM WILL BE"::"CONSERVED--THAT'S THE BASIS FOR SOLVING":f "PROBLEMS."::1630 12:"IN THIS PROGRAM, WE'LL ONLY LOOK AT "::"RECOIL TYPE PROBLEMS. YOU SHOULD ALSO":"WATCH THE PROGRAMS ON COLLISIONS.":::"NOW, LET'S GELEMS INVOLVE 2 OR "::"EVEN 3 DIMENSIONS. ";:1630 :6:"DON'T GET NERVOUS, "Z$". WE"::"WILL WORK ONLY IN 1 & 2 DIMENSIONS! 13:"NOW LET'S LOOK AT AN EXAMPLE OF EACH"::"TYPE OF SITUATION.":20:1630 1430:1490M ::3:"IN BOTH TWO OR MORE BODIES"::" ARE BLOWN APART.": " 2. COLLISIONS, WHERE TWO OR MORE "::" BODIES RUN INTO EACH OTHER."::1630 ::"SIMPLER PROBLEMS INVOLVE ONLY ONE"::"DIMENSION--MOTION ALONG THE X-AXIS.":9 "MORE COMPLEX PROBOMENTUM LAW'"::e "IF YOU KNOW WHAT THE LAW SAYS, AND IF"::"YOU HAVE YOUR PAPER, PENCIL AND": "CALCULATOR HANDY, LET'S GO!":24:1630 :3:"MOST MOMENTUM PROBLEMS IN BASIC PHYSICS"::"FALL INTO TWO CATEGORIES:":4 " 1. RECOIL, WHEREOKLAHOMA 0 Z- CONSERVATION OF LINEAR MOMENTUM` d1600::3:"HI! WHAT'S YOUR NAME? ";Z$:: n"WELCOME, "Z$". IN THE NEXT FEW"::"PROGRAMS WE'LL BE LOOKING AT PROBLEMS": x"WHICH CAN BE SOLVED MOST EASILY USING:":::" 'THE CONSERVATION OF M\+5 CONSERVATION OF LINEAR MOMENTUM 7/27/83 USING THE CONSERVATION LAW TO ANALYZE ONE AND TWO DIMENSIONAL RECOIL PROBLEM ( NSF LOCI PROJECT2 < SOUTH OKLAHOMA CITYF JR. COLLEGE P OKLAHOMA CITY,             * SERVICES **R ***********************":5" CONSERVATION OF ANGULAR MOMENTUMt:" ************************":255:A11000::: "";A$: ****************  * MODIFIED BY ** * TOM McCORD *4 * EDUCATIONAL *> * MEDIA *HA(V):YB(V)V3:740:1:140,80X,Y:0:740:140,80X,Y:2:140,50140,110:::" 1250*OMEGA' + 4000*OMEGA'":::" 0MEGA' = 60000/5250 = X2,Y2X2,Y2X2,Y2X2,Y2X2,Y2: :100:8:" *****************:" 100000 + 20000 = 5250*OMEGA'"::b:"FINALLY, OMEGA' = 22.86 RAD/SEC":::780:12:"WELL, "Z$", THAT'S IT FOR NOW."::"BYE, BYE!":20:780:20:"END":(4)"RUN MENU":21:P$:2:140,40140,120:L1P4:VP56.28P3:X"TO 5 METERS?":::"WORK IT OUT, THEN CHECK YOUR ANSWER."::780z:2:"SOLUTION:"::" I1*OMEGA1 + I2*OMEGA2 =":" I1'*OMEGA1' + I2'*OMEGA2'"::::" (50*20^2)*5 + 4000*5 =":." (50*5^2)*OMEGA' + 4000*OMEGA'":::" 600*V2(X)' = -1040":H!"AND,":8::"V2(X)' = -1.73 M/S":::!"SO THE GUN RECOILS BACKWARDS WITH A"::"VELOCITY OF -1.73 M/S. WHAT ABOUT THE":!"Y DIRECTION? ";:1630&":3:"SINCE THE SHELL HAS AN UPWARD MOMENTUM,"::"TH"NOW LET'S CONSIDER ONLY THE X-AXIS:"::" M1*V1(X) + M2*V2(X) =":} " M1*V1(X)' + M2*V2(X)'"::1630 :3:"SUBSTITUTING:"::" 4*0 + 600*0 = 4*260 + 600*V2(X)'"::1630:!" 0 + 0 = 1040 + 600*V2(X)'"::"OR,"::";:1630dp:3:"SINCE THE SHELL'S VELOCITY IS AT AN"::"ANGLE, THE FIRST STEP IS TO FIND THE":z"X & Y COMPONENTS:"::" V1(X) = V1*COS(30) = 260 M/S":" V1(Y) = V1*SIN(30) = 150 M/S"::"YOU SHOULD VERIFY THESE VALUES."::1630::I NSIONAL PROBLEM:"::"A 600 KG GUN IS MOUNTED ON WHEELS. IT":R"FIRES A 4 KG SHELL WITH A MUZZLE"::"VELOCITY OF 300 M/S AT AN ANGLE OF 30":\"DEGREES ABOVE HORIZONTAL. WHAT IS THE"::"RECOIL VELOCITY OF THE GUN?f24:"COPY THE PROBLEM AND ER WITHOUT UNITS: ";A:A1.5830/*::(7)4"SORRY!"::" .015*0 + 6*0 = .015*600 + 6*V2'"::" 0 + 0 = 9 + 6*V2'"::1630>:3:"THE CORRECT ANSWER IS:"::10::"V2' = -1.5 M/S":::1630::>H"NOW LET'S DO A 2-DIME00*V2'"::" 400*V2' = -100"::13::"V2' = -.25 M/S"::::1630:4:"LET'S CONTINUE:"::"A 6 KG RIFLE FIRES A 15 GM BULLET AT A":! "MUZZLE VELOCITY OF 600 M/S. WHAT IS "::"THE RECOIL VELOCITY OF THE GUN?"::"ENTER YOUR ANSW:1630:770f"EXCELLENT, "Z$"! THE RECOIL VELOCITY"::"IS -.25 M/S, ASSUMING YOU SAID THE MAN":"HAD A POSITIVE VELOCITY.":::1630:790:4:"SOLUTION:"::1590::" 50*0 + 400*0 = 50*2 + 400*V2'":Q " 0 + 0 = 100 + 4:"CORRECT.":720^"YOU SEEM TO UNDERSTAND THAT THE"::"VELOCITY OF THE BOAT SHOULD BE":"NEGATIVE, BUT YOU HAVE THE WRONG VALUE.::"TRY AGAIN.":::1630:620"YOU GOT THE RIGHT VALUE, BUT SHOULDN'T"::"THE SIGN BE NEGATIVE? :NN1:(7)"SORRY, "Z$", YOUR ANSWER IS WRONG."::N3770RA.25730:A0700"SINCE THE INITIAL SYSTEM MOMENTUM IS"::"ZERO, AND THE MAN HAS A (+) VELOCITY,"::"THE BOAT SHOULD HAVE A (-) VELOCITY.":"ALSO, YOUR NUMERICAL VALUE IS NOT":AT WITH A VELOCITY OF 2 M/S."::"WHAT IS THE RECOIL VELOCITY OF THE BOAT?":"RECOGNIZE THAT THE TOTAL MOMENTUM IS"::"INITIALLY ZERO, AND THAT YOU MAY WORK":"IN ONE DIMENSION.":20:"ENTER YOUR ANSWER WITHOUT UNITS: ";A::4:A.25750:UM OF THE SYSTEM WAS ZERO, THE"::"'POSITIVE' MOMENTUM OF THE 15 KG MASS":b"MUST BE BALANCED BY THE 'NEGATIVE'"::"MOMENTUM OF THE 5 KG MASS.) ";:1630:N0l:4:"NOW YOU TRY ONE:"::"A 50 KG MAN DIVES HORIZONTALLY FROM A":Mv"400 KG BO"OR,":" 0 + 0 = 45 + 5*V2'":21:1630g::3:"FINALLY,"::" 5*V2' = -45":D"AND,":15::"V2' = -9 M/S":::1630::N"YOU SHOULD RECOGNIZE THAT THE 5 KG MASS"::"MOVES TO THE LEFT. (SINCE THE INITIAL":LX"MOMENTVE ASSUMED THE INITIAL"::"VELOCITY IS ZERO (SINCE IT ISN'T GIVEN),|"AND THE 15 KG MASS HAS A (+) VELOCITY.":23:1630:3:"NOW THE CONSERVATION LAW:"::1590::1630::&"SUBSTITUTE THE DATA:"::" 15*0 + 5*0 = 15*3 + 5*V2'":30 PROBLEM YOU HAVE":D"SEEN?--OF COURSE YOU DO!":22:1630:1430u::5:"HERE'S HOW TO SOLVE THE PROBLEM:":"DATA:"::" M1 = 15 KG V1 = 0 M/S V1' = +3 M/S":" M2 = 5 KG V2 = 0 M/S V2' = ?"::1630::D"NOTICE THAT WE HARE PLACED":h"ON A FRICTIONLESS SURFACE. AN EXPLOSION"::"DRIVES THEM APART, AND THE 15 KG MASS":"MOVES TO THE RIGHT AT 3 M/S. WHAT IS"::"THE VELOCITY OF THE 5 KG MASS?"::"COPY THE PROBLEM. DO YOU REALIZE THAT"::"THIS IS THE RECOIL THE STEPS!":G"ENTER 1 TO REVIEW, 2 TO CONTINUE? ";A:::A1270"NOW WE'RE READY TO SOLVE A SIMPLE"::"PROBLEM. WE'LL BEGIN WITH 1 DIMENSIONAL":"MOTION (NO Y COMPONENT). ";:1630 :3:"PROBLEM:"::"A 5 KG MASS & A 15 KG MASS A:"2-DIMENSIONAL RECOIL":15:19,13:T1800::(7)(7)y1"X115.5:1:19X,13:2:19,131.60X:12:19X,13.75X:1,23:"TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A11560:26" P1 + P2 = P1' + P2'"::" M1*V1 + M2*V2 = M1*V1' + M2*0,1130X:)0(7):21:15:"COLLISION0X06.25:12:10,11154X:1:10,1116X:10,1115X:T135::0:10,11154X:10,1116X:10,1115X:015,1612:15,169:23:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A11490081::21:102.7X):XH/23:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A11430:N//::21:15:"COLLISION":1:10,110:10,111:12:10,1130/T1800::(7):(7) 0X014:1:10,11X:10,11X1:12:10,1130X:T140::0:10,11X:10,11X1:11630&.:20:"END":(4)"RUN MENU"e.::21:17:"RECOIL":1:10,1110:10,1111:13:10,1112.T1800::(7):(7).X010.25:1:10,11(10.9X):10,11(11.9X):13:10,11(122.7X) /0:10,11(10.9X):10,11(11.9X):10,11(12Ⱥ:"(SO THE RECOIL MASS IS 100 KG).":::1630|-n:4:"ANSWERS:"::" 1. V2' = -.90 M/S"::" 2. V2' = -.045 M/S"::-x"THE DIFFERENCE EXPLAINS WHY A SHOOTER"::"SHOULD HOLD THE RIFLE FIRMLY!":::."THAT'S IT FOR NOW. SEE YOU LATER!":23:ST PROBLEM FOR PRACTICE:"::y,P"A 5 KG RIFLE FIRES A 15 GM BULLET WITH"::"A MUZZLE VELOCITY OF 300 M/S. FIND THE":,Z"RECOIL VELOCITY OF THE RIFLE. ALSO,"::"FIND THE RECOIL WHEN A 95 KG MAN HOLDS":.-d"THE RIFLE FIRMLY AGAINST HIS SHOULDER": THETA IS THE ANGLE ABOVE THE"::"NEGATIVE X-AXIS. ";:1630N+(1560:+2:6:"IF YOU WOULD LIKE TO SEE THE LAST "::"PROBLEM AGAIN, TYPE 1. TYPE 2 TO":+<"CONTINUE: ";A:A11030,F:4:"WELL, "Z$", THAT'S ABOUT IT.":::"HERE'S ONE LAE PYTHAGOREAN THEOREM:":}*1630::" V3 = SQRT"(91)" V3(X)^2 + V3(Y)^2 ]"::" V3 = SQRT"(91)" -6.8^2 + 4.267^2 ]":* 3::"V3 = 8.03 M/S":::"AND,":*" THETA = ARCTAN (4.267/-6.8)"::8::"THETA = 32.1 DEGREES"::B+"WHERE*V3(Y)'"::1630P):"FINALLY,"::" 0 + 0 = -64 + 15*V3(Y)'":)"AND,"::" V3(Y)' = 64/15 = 4.267 M/S":::1630):"SO M3'S Y COMPONENT IS POSITIVE.":24:1630*:3:"NOW WE MUST FIND THE TOTAL V3. WE USE"::"TRIG & TH/S"::1630::a("SO THE X COMPONENT OF M3'S VELOCITY IS"::"IN THE NEGATIVE DIRECTION."::("NOW FOR THE Y DIRECTION:"::1630(:" M2*V2 + M3*V3(Y) = M2*V2' + M3*V3(Y)'"::1630):4:"SUBSTITUTING:"::" 8*0 + 15*0 = 8*(-8) + 15(X) = M1*V1' + M3*V3(X)'"::1630:f'"SUBSTITUTING:"::" 17*0 + 15*0 = 17*6 + 15*V3(X)'":'"OR,":" 0 + 0 = 102 + 15*V3(X)'"::1630':4:"THEN,":" 15*V3(X)' = -102":("AND,":" V3(X)' = -6.8 MTIONS & WILL FLY":/&V"OFF AT AN ANGLE."::y&`"FIND THE MASS OF PART #3."::"ENTER YOUR ANSWER WITHOUT UNITS: ";A&j:3:"MASS OF #3:"::" M3 = 40 - 17 - 8 = 15 KG"::&t"NOW LET'S LOOK AT THE X DIRECTION:"::1630#'~:" M1*V1 + M3*V3:3:"YOU SHOULD NOTICE THAT PART #1 HAS ONLY"::"AN X COMPONENT WHILE PART #2 HAS ONLY A":%B"Y COMPONENT. SINCE THE INITIAL TOTAL"::"MOMENTUM IS ZERO, PART #3 MUST BALANCE":&L"BOTH PARTS--IT WILL HAVE MOMENTUM COM-"::"PONENTS IN BOTH DIRECX$"(M = 8 KG) MOVES IN THE NEGATIVE Y"::"DIRECTION AT 8 M/S. WHAT IS THE VELOCITY$$"(MAGNITUDE & DIRECTION) OF THE THIRD"::"PART IF THE SHELL HAD A TOTAL MASS OF":$."40 KG?":22:"COPY THE PROBLEM, MAKE A SKETCH, AND"::1630:1560:^%8#20:"GETTING TIRED, "Z$"? WELL, WE'RE"::"ALMOST DONE!":24:1630#:3:"HERE'S A GOOD 2-DIMENSIONAL PROBLEM:"::"A SHELL EXPLODES INTO 3 PARTS. ONE PART":$"(M = 17 KG) FLIES OFF IN THE POSITIVE X"::"DIRECTION AT 6 M/S. THE SECOND PART":ERE MUST BE A BALANCING MOMENTUM DOWN.""BUT IT IS THE GUN AND THE EARTH WHICH"::"RECOIL IN THE Y DIRECTION. AND BECAUSE":""THE EARTH IS SO MASSIVE, IT'S RECOIL"::"VELOCITY IS VERY, VERY SMALL. WE USUALLY""NEGLECT THIS EFFECT."::1630GIRECTION OF X?"::"IF YOU SAID DOWN, YOU'RE RIGHT!"::" X = - X":::"LET'S LOOK AT IT.":24:800::660:680:690:21:"HERE'S X.":24:800X120:3:74,87173X,54X:5:61,90X61,903X:X4Ē1:21,11X,75X::5:59,3063,30:57,3265,32P3:X70:Y31:700:X78:760:X86:710:21:"TYPE 1 TO WATCH AGAIN.":"TYPE 2 TO CONTINUE. ";T:T2430:660:680:690:370::3:"THUS, THE DIRECTION OF X IS UP."::>"WHAT IS THE DE CROSS"::"PRODUCT ANSWER MUST BE PERPENDICULAR TO":fh"THE PLANE OF & . WATCH!":24:800r:16304,0:21:"ROTATE TO WITH YOUR RIGHT HAND.":24:800|X120:3:74,87194X,75X:5:61,8861,883X:D150:D&0:74,87194 108.1 FT-LBS":::"THAT'S NOT SO BAD, IS IT? ";:800J:3:"NOW LET'S SEE HOW TO FIND THE DIRECTION"::"OF X."::T"SINCE & LIE IN THE X-Y PLANE,"::"THE ANSWER MUST BE A VECTOR IN THE +Z":7^"OR -Z DIRECTION. REMEMBER, THGAIN.":24:800:210i"::"GOOD! NOW CALCULATE X."::"ENTER YOUR ANSWER WITHOUT UNITS: ";T::,T107T110320::"ERROR! HOW DID YOU MESS UP?":6" X = 12 * 11 * SIN(55)"::" X = 12 * 11 * 0.819"::9@" X ="HERE'S . BOTH VECTORS ARE IN THE":"X-Y PLANE."::800:X160:Y71:750:21:"HERE'S THE ANGLE BETWEEN & .":24:800::3:"WHAT IS THE ANGLE THETA?"::"ENTER THE NUMBER OF DEGREES? ";TT55290::"OOPS! THAT'S INCORRECT. TRY A 20 DEGREES N OF E":" = 11 POUNDS, 15 DEGREES E OF N":::"WRITE DOWN THE PROBLEM. THEN WE'LL"::"LOOK AT IT.":24:800::660:21:"HERE'S OUR COORDINATE SYSTEM.":24:800:680:21:"HERE'S VECTOR .":24:800::690:21: TO":W " THROUGH THE SMALLEST ANGLE. YOUR"::" THUMB SHOWS THE RESULT.":: "THIS RULE IS FOR X. TWIST TO"::" FOR X. MAKE A NOTE OF THIS!":24:800:3:"HERE'S A PROBLEM. FIND X IF:":::" = 12 FEET,:"WHERE THETA IS THE SMALLEST ANGLE":q "BETWEEN & .":::"THE HARD PART IS FINDING THE DIRECTION": "OF THE RESULTING VECTOR. WE USE THE"::"RIGHT HAND RULE.":24:800 :3:"RIGHT HAND RULE:"::" USE YOUR RIGHT HAND AND TWIST :3:"NO? THEN YOU HAD BETTER WATCH THE"::"PROGRAM: 'DOT PRODUCTS - M/D FORM'."::"LOAD THAT PROGRAM BEFORE YOU GO ON.": :800:660 :3:"GREAT! THE MATHEMATICS OF CROSS "::"PRODUCTS ARE EASY:"::' " X = A * B * SIN(THETA)"::CTS, YOU KNOW":l Z"THE NOTATION WE USE & YOU KNOW THAT THE"::"CROSS PRODUCT OF TWO VECTORS PRODUCES": d"A THIRD VECTOR AT RIGHT ANGLES TO THE"::"PLANE FORMED BY THE FIRST TWO.": n"RIGHT, "Z$;:"? (TYPE YES OR NO): ";T$:T$"NO"140| x" 'DOT PRODUCTS - M/D FORM'":z <:"DON'T START THIS PROGRAM WITHOUT SEEING"::"THE OTHER ONE FIRST!":24:800 F:3:"WHAT'S YOUR NAME? ";Z$:::"THANK YOU, "Z$". P::"LET'S GET STARTED. SINCE YOU'VE DONE"::"THE PROGRAM ON DOT PRODU! -CROSS PRODUCTS - M/D FORMe770::4:"HELLO!"::"THIS PROGRAM IS DESIGNED TO HELP YOU":"FIND THE CROSS PRODUCT OF TWO VECTORS"::"EXPRESSED IN MAGNITUDE/DIRECTION FORM.":("YOU SHOULD HAVE ALREADY WATCHED THE"::"PROGRAM:":& 2         3 * EDUCATIONAL *-3 * MEDIA *D3 * SERVICES *[3 ****************10,1115X:T135::0:10,11154X:10,1116X:10,1115X:3D 15,1612:15,169:23:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A182003N <4 ::21V2'"::<2@:100:8:" ************************":f2J" CONSERVATION OF LINEAR MOMENTUM2T:" ************************":255:A11000:::2^"";A$:2h ****************2r * MODIFIED BY *2| * TOM McCORD * "Z$"!"::180d :" * OF COURSE! THE VECTOR PRODUCT IS THE":" SAME AS THE CROSS PRODUCT.":: "3. THE VECTOR X LIES IN THE SAME":" PLANE AS & ."::" ANSWER T OR F?";T$ T$"F"2407(7):" * NO. X IS ALWAYS PERPTHE DOT PRODUCT IS ANOTHER"::" NAME FOR THE SCALAR PRODUCT."::140{ :" * GOOD! A DOT PRODUCT GIVES A SCALAR.":: "2. A CROSS PRODUCT IS THE SAME AS A":" VECTOR PRODUCT."::" ANSWER T OR F? ";T$ T$"T"200::" * YOU ARE SLEEPING,TO ALWAYS USE THE SMALLEST"::"ANGLE BETWEEN THE VECTORS!":24:840 :3:"JUST TO BE SURE YOU'RE AWAKE, ANSWER"::"THESE QUESTIONS:":: "1. A DOT PRODUCT GIVES A SCALAR ANSWER."::" ANSWER T OR F? ";T$ T$"T"170G (7):" * SORRY. NOTATION WE'LL USE:":::" = VECTOR A":x Z" = VECTOR B"::" . = DOT PRODUCT OF & ": d" X = CROSS PRODUCT OF & "::" THETA = SMALLEST ANGLE BETWEEN THE": n" VECTORS"::D x"REMEMBER CH:":::" 'CROSS PRODUCTS - M/D FORM'":::"LET'S WORK ONLY IN THE MAGNITUDE/": <"DIRECTION (M/D) FORM. IN OTHER WORDS,"::"WE'LL LOOK AT VECTORS IN THE FORM:":: F(2)::" = 15 KM AT 30 DEGREES N OF E"::24:8402 P:3:"HERE'S THE}  - DOT PRODUCTS - M/D FORM)810:9:"HELLO, PHYSICS STUDENT!"::"WHAT'S YOUR NAME? ";Z$::::"THANK YOU!":24:840(:3:Z$", THIS PROGRAM IS DESIGNED TO"::"HELP YOU FIND THE DOT PRODUCT OF TWO":W 2"VECTORS. YOU SHOULD ALSO WAT     255:D11000:::* "";A$:A* ****************X4 * MODIFIED BY *o> * TOM McCORD *H * EDUCATIONAL *R * MEDIA *\ * SERVICES *f ****************X4,Y1:X1,YX3,Y:X1,Y3X5,Y3:{X,Y5:X1,Y4:X2,Y3:X3,Y2:X4,Y1:X4,Y5:X3,Y4:X1,Y2:X,Y1::100:8:" *************************": " CROSS PRODUCTS - M/D FORM"::" *************************4,Y:X,Y1:X,Y5:X1,Y2:X1,Y4:X2,Y3:X3,Y2:X3,Y4:X4,Y1:X4,Y5:X,Y6X,Y4:X1,Y3X3,Y3:X4,Y4X4,Y6:X2,Y2X2,Y:X,Y6X4,Y6:X4,Y5:X3,Y5X,Y1:X,Y1:X,YX4,Y:(X,Y1X,Y5:X1,Y6X3,Y6:X4,Y5206:Y81:700:N2:65,89176,51169,51171,55176,51:3:X181:Y50:710:X,YX,Y5:X1,Y6X3,Y6:X4,Y5X4,Y:X,Y3X4,Y3:X,YX,Y6X3,Y6:X4,Y5X4,Y4:X3,Y3X,Y3:X4,Y2X4,Y1:X3,YX,Y:RX,Y:X,Y6:X4,Y6:X"TYPE 1 TO REIVEW THIS PROGRAM."::"TYPE 2 TO CONTINUE: ";A:A1140f:20:"END":(4)"RUN MENU"1:61,361,156:10,90229,90:21,110181,303:X224:Y102:720:X49:Y10:740:X170:Y27:730:2:65,89200,76196,74196,79200,76:3:XдT$"N"620lb(7):"INCORRECT. SINCE & ARE IN THE"::"X-Z PLANE, X IS EITHER +Y OR -Y.":l:" ** THE DIRECTION IS NORTH! **":24:800v:6:"THAT'S ALL FOR NOW. SEE YOUR INSTRUCTOR"::"IF YOU'RE STILL HAVING TROUBLE."::E"::540&0:"GOOD, "Z$".":24:800r::3:"NOW SOLVE THE PROBLEM. ENTER YOUR"::"ANSWER WITHOUT UNITS. ";TD::"X = 62 * 15 * SIN(40) = 393 NT-M"::N"WHAT IS THE DIRECTION OF X?"::"ENTER N, S, E, W, UP, OR DOWN: ";T$: XN.":::"HERE'S ONE FINAL PROBLEM. FIND X:":" = 62 NT, 25 DEGREES ABOVE EAST"::" = 15 M, DUE EAST"::"WHAT IS THE ANGLE BETWEEN & ?"::"ENTER THE ANGLE IN DEGREES: ";T:&T25560:(7):"SLEEPING? TRY AGAIN.0181,30OD150:D:0:74,87173X,54X::5:59,14863,148:57,14665,146z3:X70:Y150:710:X78:760:X86:700:21:"TYPE 1 TO WATCH AGAIN.":"TYPE 2 TO CONTINUE. ";T:T25204604::3:"THUS, THE DIRECTION OF X IS DOWX4,Y5X4,Y1:X1,YX3,Y:X1,Y3X5,Y3:b*:100:8:" ***********************":4" DOT PRODUCTS - M/D FORM"::" ***********************>255:D11000:::H"";A$:\ **************** f * MODIY6:X4,Y:X,Y1:X,Y5:X1,Y2:X1,Y4:X2,Y3:X3,Y2:X3,Y4:X4,Y1:X4,Y5: X,Y6X,Y4:X1,Y3X3,Y3:X4,Y4X4,Y6:X2,Y2X2,Y:X,Y6X4,Y6:X4,Y5:X3,Y5X,Y1:X,Y1:X,YX4,Y:/ X,Y1X,Y5:X1,Y6X3,Y6:6:3:X206:Y81:750:U2:65,89176,51169,51171,55176,51:3:X181:Y50:760:X,YX,Y5:X1,Y6X3,Y6:X4,Y5X4,Y:X,Y3X4,Y3:X,YX,Y6X3,Y6:X4,Y5X4,Y4:X3,Y3X,Y3:X4,Y2X4,Y1:X3,YX,Y:YX,Y:X,Y6:X4,:L"TYPE 1 TO REVIEW THIS PROGRAM."::"TYPE 2 TO CONTINUE: ";A:A180m:20:"END":(4)"RUN MENU"1:61,361,156:10,90229,90:21,110181,303:X224:Y102:770:X49:Y10:790:X170:Y27:780:2:65,89200,76196,74196,79200,7E CORRECT ANSWER IS 429 FT-LBS **"::S" DID YOU LEARN ANYTHING?":20:840:5:"THAT'S THE END OF THIS PROGRAM."::Z$", YOU REALLY SHOULD WATCH"::"THE PROGRAM:"::" 'CROSS PRODUCTS - M/D FORM'":::"YOU'LL FIND IT HELPFUL."::0FX::"YOU'RE SLEEPING, "Z$"!"::"THE ANGLE IS 40 DEGREES.":620]b::"GREAT, "Z$"!l::"NOW SOLVE AND ENTER YOUR ANSWER: ";L::3:L427L430650v:"NOPE. WHAT IS 40 * 14 * COS(40)?":"ENTER A CORRECT ANSWER! ";L::&"** THHER. THIS TIME"::"WE'LL USE A VERTICAL PLANE."::o0"FIND . IF:"::" = 40 POUNDS DUE WEST"::" = 14 FT, 40 DEGREES UP FROM WEST"::D"WHAT IS THE ANGLE BETWEEN & ?"::"ENTER THE NUMBER OF DEGREES: ";LNL4061921L925530Y::"INCORRECT. REMEMBER THAT:"::" . = A * B * COS(THETA)":"TRY AGAIN. ENTER YOUR NEW ANSWER: ";L::" . = 923.5 NT-M"::"SO MUCH FOR THAT ONE. LET'S GO ON.":24:8404&::"WE'LL DO ONE MORE TOGETH"::"OF THE X-Y PLANE AND TRY AGAIN."::}"WHAT IS THE ANGLE BETWEEN & ?"::"ENTER THE NUMBER OF DEGREES: ";L:3:"** THE ANGLE IS 55 DEGREES. **":::"NOW FINISH THE SOLUTION AND ENTER YOUR":"ANSWER: (DON'T TYPE UNITS) ";L:LLF:":::" FIND . IF":w" = 70 METERS, 10 DEGREES E OF N"::" = 23 NT, 45 DEGREES N OF W"::"WHAT IS THE ANGLE BETWEEN & ?"::"ENTER THE NUMBER OF DEGREES: ";LL55490,:"** THAT'S NOT RIGHT. MAKE A SKETCLUTION:":::" . = A * B * COS(THETA)":e" = 12 * 11 * COS(65)":" = 12 * 11 * 0.423"::" = 55.8 NT-M":::"EASY, ISN'T IT, "Z$"?":24:840:3:"NOW SOLVE THIS ONE YOURSE:9:"DRAW A SKETCH. THEN FIND THETA. YOU"::"SHOULD SEE THAT THE ANGLE BETWEEN THE":r"VECTORS IS 65 DEGREES. RIGHT?":20:840|16304,0:3:X160:Y71:800:21:"HERE'S THE ANGLE BETWEEN & .":24:8408::5:"NOW PROCEED WITH THE SOTWEEN THEM. FOR EXAMPLE:":::"FIND . IF:":T" = 12 METERS, 10 DEGREES N 0F E"::" = 11 NT, 15 DEGREES E OF N":24:840^:16304,0:730:740:21:"HERE'S WHAT IT LOOKS LIKE. BOTH":"VECTORS ARE IN THE X-Y PLANE."::840Xh:840O,::710:21:"REMEMBER:":" X = EAST Y = NORTH Z = UP":24:8406::3:"FOR DOT PRODUCTS,":::" . = A * B * COS(THETA)"::@"SO, SIMPLY MULTIPLY THE MAGNITUDES OF"::"THE VECTORS AND THE COSINE OF THE ANGLE":3J"BEDY TO CONTINUE. BE":k"SURE YOU UNDERSTAND THE MEANING OF"::"VECTOR PRODUCTS BEFORE YOU GO ON."::15:"IF YOU'RE READY, LET'S DO DOT PRODUCTS."::"THEY'RE VERY EASY SINCE THE ANSWER IS":""A SCALAR. HERE'S THE COORDINATE SYSTEM.":24:ENDICULAR":" TO THE PLANE MADE BY & ."::210:" * EXCELLENT! THE VECTOR X IS":" PERPENDICULAR TO THE PLANE FORMED" BY VECTORS & .":::840:3:"IF YOU HAD TROUBLE WITH THE QUESTIONS,"::"YOU MIGHT NOT BE REA- USE RIGHT-HAND RULE:"::" = 0 = <-K> = "::" = = 0 = <-I>"::" = <-J> = = 0"::"YOU MUST BE ABLE TO DETERMINE THE DOT"::"AND CROSS PRODUCT OF UNIT VECTORS TO"::"SOLVE> + FB + FC":i"DO NOT REVERSE ANY I,J,K TERMS! ":"IS NOT NECESSARILY EQUAL TO .":r870:"STEP 3 - EVALUATE TERMS:"::"*DOT PRODUCTS - ALL TERMS EQUAL ZERO"::"EXCEPT: = = = 1"::"*CROSS PRODUCTS 1, TOP 2 BY BOTTOM 2, ETC.":"THE SEQUENCE IS: S1R1, S1R2, S1R3":" S2R1, S2R2, S2R3":" S3R1, S3R2, S3R3":"FOR THIS PROBLEM:"::" DA + DB + DC"::" EA + EB + EC"::" FA + E + F":" TIMES = A + B + C":{|"BE CAREFUL! REVERSING GIVES ERRORS.":::::870:"STEP 2 - CROSS MULTIPLYING"::"MULTIPLY TOP 1 BY BOTTOM 1, TOP 1 BY"::"BOTTOM 2, TOP 1 BY BOTTOM 3, TOP 2 BY"::"BOTTOM ECTORS OF THE FORM:"::" = A + B + C"::" = D + E + F":i^::870h:"STEP 1 - SET UP VECTORS:"::"TO FIND TIMES , WRITE R ABOVE S."::"FOR TIMES , WRITE S ABOVE R."::Cr"EXAMPLE: OUR TEXT BEFORE CONTINUING.&6:870@:"NOW WE'RE READY TO BEGIN MULTIPLYING."::"THERE ARE FOUR STEPS TO THE PROCESS:":J" 1. SET UP VECTORS"::" 2. CROSS MULTIPLY"::" 3. EVALUATE TERMS"::" 4. COMBINE LIKE TERMS":\T:"ASSUME 2 V,300^:"NO. SINCE YOU HAVE & , THE"::"ANSWER MUST BE EITHER OR <-J>."::250":"WRONG. DID YOU USE THE RIGHT HAND RULE?"::250,:"YES! IF YOU GOT ALL THE ABOVE RIGHT,"::"YOU'RE READY TO GO ON. IF YOU MISSED"::"SOME, STUDY YCANNOT GIVE A VECTOR ANSWER. TRY AGAIN."::190~:"YES! A DOT PRODUCT OF A UNIT VECTOR"::"WITH ITSELF IS ALWAYS ONE."::"4. WHAT IS THE RESULT OF X?":" (1) <-I> (2) <-J> (3) ":::"ANSWER 1,2OR 3";::A:A280,290 (3) ":::"ANSWER 1,2 OR 3";::A= A220,240,230 :"INCORRECT. THE DOT PRODUCT USES THE"::"COSINE TERM. SINCE THE ANGLE BETWEEN A"::"VETOR AND ITSELF IS 0, THE COSINE IS 1."::190/:"SORRY. YOU MUST KNOW THAT A DOT PRODUCT"::"R THE CROSS PRODUCT"::"IS THE VECTOR PRODUCT.":c :"TRUE OR FALSE";::B$::(B$,1)"T"180 :"INCORRECT. TRY IT AGAIN."::150 :"VERY GOOD, "Z$". LET'S CONTINUE.": :"3. WHAT IS THE RESULT OF .?":* " (1) 0 (2) 1 HE DOT PRODUCT . IS A SCALAR.":::"TRUE OR FALSE";::B$W x:(B$,1)"T"140 :"WRONG. YOU SHOULD KNOW A DOT PRODUCT"::"ALWAYS GIVES A SCALAR. TRY AGAIN."::110 :"EXCELLENT! DOT PRODUCTS YIELD SCALARS.":2 :"2. ANOTHER NAME FO29:28,6:27,6:26,5:6,28:7,28e P13:3,3519:3,712:15,3:14,4:13,5:14,6:15,7:16,34:17,34 Z"THIS IS THE AXIS SYSTEM WE'LL USE.":"COPY IT DOWN. NOTE <-J> IS TOWARDS YOU.": d870:::"NOW WE NEED TO CHECK YOUR KNOWLEDGE.":@ n:"1. T" MEANS THE VECTOR A"::" MEANS THE UNIT VECTOR I"::" . MEANS THE DOT PRODUCT OF A&B"::" X MEANS THE CROSS PRODUCT OF A&B": 2870 <:1:7,3019:33,3516:16,2034:33,3520:4,17:5,17 F12:X928:X,38X::2,5f- 840:"HELLO! WHAT'S YOUR NAME? ";Z$:"WELL, "Z$", THIS PROGRAM IS"::"DESIGNED TO HELP YOU FIND THE"::"DOT & CROSS PRODUCT OF TWO VECTORS.":"LET'S WORK WITH VECTORS IN UNIT VECTOR"::"FORM ONLY. WE'LL USE THESE NOTATIONS:":: (