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Ѐ0zP;}{3?;# 3""LHLꪪ~ @U7"GwvWbc]_/AIA`р"o{|~!3"7$/ 92-+>~|_@RL 䪨@}8o@db}ڊZ~ pGzcccժLDLD`    P2aL|LՅ>' ,,$$36..$$%%..">' ,,$$36..$$%%.." * MEDIA *JH * SERVICES *aR ****************9100:X244:9800:(#X,YX,Y6X3,Y6X4,Y5X4,Y4X3,Y3X,Y3:#X,Y6X4,Y6:X2,Y6X2,Y:A#X,Y:X,Y6:X4,Y6:X4,Y:X,Y1:X,Y5X,Y3X,Y3:A:100:8:" ************************":l" INELASTIC COLLISIONS - PART III:" ************************":255:A11000::: "";A$: ****************  * MODIFIED BY ** * TOM McCO,Y1:X1,YX3,Y:X1,Y3X5,Y3:JX,YX4,Y:X2,YX2,Y6X,Y4:X4,YX,Y:X,Y1:X1,Y2:X2,Y2:X3,Y3:X4,Y4:X4,Y5:X3,Y6X1,Y6:X,Y5:X2,YX,Y2X,Y4X2,Y6:X,YX2,Y2X2,Y4X,Y6: X2,Y3X2,Y3:X,Y6X4,Y6:X2,Y6X2,Y:X,Y:X,Y6:X4,Y6:X4,Y:X,Y1:X,Y5:X1,Y2:X1,Y4:X2,Y3:X3,Y2:X3,Y4:X4,Y1:X4,Y5:X,Y6X,Y4:X1,Y3X3,Y3:X4,Y4X4,Y6:X2,Y2X2,Y:$X,Y1X,Y5:X1,Y6X3,Y6:X4,Y5X4%b5:5,96276,96:133,22133,170:Yl3:X71:Y84:690:X81:720:X86:700:X94:730vX71:Y110:690:X81:720:X86:710:X94:730X221:Y105:690:X231:720:X236:660:X244:730:X,YX,Y6X3,Y6X4,Y5X4,Y4X3,Y3X,Y3:'S ALL FOR NOW. BYE!":20:780::20:"END":(4)"RUN MENU"DT5Q1:XT:Y.7X1:1:740:T3,Y:2:XT2:Y165.5X:740:T2,Y:0NXT:Y.7X1:740:XT1:Y165.5X:740::XT135Q2:XT1:Y.15X76:3:740:T3,Y:0:XT:Y.15X76:740:: WORK.":24:780a:7:"IF YOU'RE STILL CONFUSED, MAYBE"::"YOU SHOULD REVIEW THIS ENTIRE":&"PROGRAM. IF YOU WANT TO REVIEW"::"PART III, TYPE 1. TYPE 2 TO FINISH: ";A:A11000:8:"SEE YOUR INSTRUCTOR IF YOU NEED MORE":;:"HELP. THAT:" PT(Y) = 155 KG-M/S":f"6. PT = 1320 KG-M/S"::" THETA-T = 6.7 DEGREES NORTH OF EAST":"7. VT = 30.7 M/S AT 6.7 DEGREES N OF E"::"HOW DID YOU DO, "Z$"?"::"IF YOU GOT THE RIGHT ANSWERS, YOU'RE":"DONE. IF NOT, CHECK YOURTHE SOLUTION:":::"1. MT = 43 KG":["2. P1 = 330 KG-M/S"::" P2 = 1036 KG-M/S":"3. P1(X) = 310 KG-M/S"::" P1(Y) = -113 KG-M/S":"4. P2(X) = 1001 KG-M/S"::" P2(Y) = 268 KG-M/S"::24:780:3:"5. PT(X) = 1311 KG-M/S":LIDES":["INELASTICALLY WITH A 28 KG MASS"::"TRAVELLING 15 DEGREES NORTH OF EAST ":"AT 37 M/S. FIND THE MAGNITUDE AND"::"DIRECTION OF THE VELOCITY OF THE":"COMBINED MASSES.":::"SOLVE THIS PROBLEM AND ";:24:780%:3:"HERE'S E MORE"::"TIME BEFORE YOU DO ONE ON YOUR OWN.":24:780Y::580:600:610:24:780::3:"NOW HERE'S ONE FOR YOU. IT'S TURNED"::"AROUND--DON'T GET CONFUSED:":8"A 15 KG MASS IS TRAVELLING 20 DEGREES"::"SOUTH OF EAST AT 22 M/S. IT COL THAT THE ANGLE OF THE MOMENTUM"::"WILL BE THE ANGLE OF THE VELOCITY.":24:780h:5:"7. FINAL VELOCITY:"::" VT = PT/MT = 67.5/17.65":r" VT = 3.8 M/S AT 12.5 DEGREES S OF W":255:13:"THERE YOU HAVE IT!":9|"THAT'S NOT SO BAD. LOOK ONOTAL MOMENTUM:"::" PT = SQ RT "::" PT = SQ RT <65.9^2 + -14.6^2>":J" PT = 67.5 KG-M/S"::" THETA-T = ARCTANGENT ":T" THETA-T = ARCTAN <-14.6/65.9>"::" THETA-T = -12.5 DEGREES (S OF W)"::Q^"NOTE"::t""5. COMBINING X- & Y-COMPONENTS:"::" PT(X) = P1(X) + P2(X)"::" PT(X) = 36.5 + 29.4 = 65.9 KG-M/S":," PT(Y) = P1(Y) + P2(Y)"::" PT(Y) = -29.6 + 15.0 = -14.6 KG-M/S":6:"REMEMBER: P1(Y) IS NEGATIVE!";:31:780\@:3:"6. TP1 * SIN(THETA-1)"::" P1(Y) = 47 * SIN(39) = 29.6 KG-M/S":24:780:2:"4. COMPONENTS OF P2:"::" P2(X) = P2 * COS(THETA-2)"::" P2(X) = 33 * COS(27) = 29.4 KG-M/S":" P2(Y) = P2 * SIN(THETA-2)"::" P2(Y) = 33 * SIN (27) = 15.0 KG-M/SG":24:780S :3:"2. MOMENTUM:"::" P1 = M1 * V1 = 9.40 * 5.0 = 47 KG-M/S": " P2 = M2 * V2 = 8.25 * 4.0 = 33 KG-M/S":: "3. COMPONENTS OF P1:"::" P1(X) = P1 * COS(THETA-1)"::" P1(X) = 47 * COS(39) = 36.5 KG-M/S":E" P1(Y) = SHOULD HAVE:":::" M1 = 9.40 KG M2 = 8.25 KG"::" V1 = 5.O M/S V2 = 4.0 M/S"::" THETA-1 = 39 THETA-2 = 27":: "O.K. SO FAR, "Z$"?":::"NOW LET'S SOLVE IT:"::D13000::30 :"1. TOTAL MASS: MT = M1 + M2 = 17.65 KNELASTICALLY WITH AN 8.25 KG":v "MASS TRAVELLING 27 DEGREES NORTH OF"::"WEST WITH A SPEED OF 4 M/S. FIND THE": "MAGNITUDE & DIRECTION OF THE VELOCITY"::"OF THE COMBINED MASSES.":::"BEGIN BY FILLING IN YOUR CHART.":24:780 :3:"YOU :3:700:Y150:710: 580:X146:Y110:3:660:600:610p 620:22:" HERE'S THE SITUATION."::780 ::3:"HERE'S THE PROBLEM:":5:"A 9.40 KG MASS IS TRAVELLING 39 DEGREES": "SOUTH OF WEST WITH A SPEED OF 5 M/S. IT"::"COLLIDES I P OKLAHOMA CITY, OKLAHOMA ( Z750J d:6:"WELCOME BACK, "Z$"!": n9:"READY TO SOLVE A PROBLEM? SURE YOU ARE!"::"LET'S LOOK AT THE SITUATION BEFORE YOU": x"GET THE DATA. BE SURE TO MAKE A SKETCH!":15:780 Q1133:Q2275::X10:Y24b2 INELASTIC COLLISIONS PART III USING THE CONSERVATIVE LAW TO ANALYZE TWO-DIMENSIONAL INELASTIC COLLISIONS ( NSF LOCI PROJECT2 < SOUTH OKLAHOMA CITYF JR. COLLEGE             *******************":255:A11000:::>"";A$:U ****************l * MODIFIED BY * * TOM McCORD * * EDUCATIONAL *  * MEDIA * * SERVICES *  **************** ************************"Y4:X4,Y5:X3,Y6X1,Y6:X,Y5:GX2,YX,Y2X,Y4X2,Y6:hX,YX2,Y2X2,Y4X,Y6:X2,Y3X2,Y3:X,Y3X,Y3::100:8:" ************************":" INELASTIC COLLISIONS - PART II):" *****1:X4,Y5:PvX,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:X,YX4,Y:X2,YX2,Y6X,Y4:&X4,YX,Y:X,Y1:X1,Y2:X2,Y2:X3,Y3:X4,BNX173:Y124:600:X181:610:X189:670:X195:630:X203:680:yXX,YX,Y6X3,Y6X4,Y5X4,Y4X3,Y3X,Y3:bX,Y6X4,Y6:X2,Y6X2,Y: lX,Y:X,Y6:X4,Y6:X4,Y:X,Y1:X,Y5:X1,Y2:X1,Y4:X2,Y3:X3,Y2:X3,Y4:X4,Y0:670:X46:630:X54:680:M&3:191,105243,114241,110:243,114239,116o0X252:Y121:600:X261:610::5:189,104243,104:241,103:241,105:189,105189,114:187,112:191,112D3:X245:Y99:600:X253:610:X261:670:X267:620:X275:68092:670:X98:620:X106:680YX25:Y62:600:X33:650:X41:670:X47:630:X54:680:2:42,14266,142:64,141:64,143:42,14242,131:40,133:44,1333:X72:Y145:600:X80:660:X88:670:X94:620:X102:680X24:Y127:600:X32:660:X4236:610:X244:680:V3:41,3171,5266,52:71,5270,48:X78:Y62:600:X87:65042,14266,13160,131:65,13164,135:X76:Y129:600:X85:660:1:41,3171,3168,29:71,3168,33:41,3141,52:39,50:43,503:X76:Y34:600:X84:650:X:690::RT135Q2:XT1:Y.15X76:3:690:T3,Y:0:XT:Y.15X76:690::x5:5,96276,96:133,22133,170:3:X71:Y84:640:X81:670:X86:650:X94:680X71:Y110:640:X81:670:X86:660:X94:680X221:Y105:640:X231:670:XG|:8:"PLEASE WAIT WHILE PART III IS LOADED.":18:15)"-LOADING-D13500:::(4)"BLOAD CHAIN,A520":520"TWO-D COLLISIONS III":T5Q1:XT:Y.7X1:1:690:T3,Y:2:XT2:Y165.5X:690:T2,Y:0 XT:Y.7X1:690:XT1:Y165.5X3,91133,101:22:" FINAL VELOCITY = PT/MT"::730^::6:"THAT'S ALL FOR THE METHOD, "Z$".":::"WE'LL SOLVE A REAL PROBLEM IN PART III."::h"TO REVIEW THIS PART, TYPE 1."::"TYPE 2 TO CONTINUE: ";A:A140r:Q1133:Q2275:400:420:6:16304,0:550:5:189,105:22:" ADD PT(X) & PT(Y) FOR PT":24:730@::5:"7. FINALLY, DIVIDE PT BY THE TOTAL MASS"::" TO FIND THE FINAL VELOCITY.J13:"THAT'S ALL THERE IS TO IT!":18:7307T16304,0:420:5:125,96141,96:13 (PT(X)"::" SQUARED + PT(Y) SQUARED). AND ANGLE"::" THETA-T = ARCTANGENT OF PT(Y)/PT(X).":,:" PT = SQRT "::" THETA-T = ARCTAN .":::" OR, THE VECTOR SUM OF 2 RIGHT-ANGLE"::" VECTORS.":24:730ON YOUR CHART.":::730{:16304,0:570:21:" PT(X) = P1(X) + P2(X)":" PT(Y) = P1(Y) + P2(Y)"::730::3:"NOTICE THAT WE ADD X- & Y-COMPONENTS"::"SEPERATELY. NOW TO FIND THE TOTAL:"::c""6. TOTAL MOMENTUM = SQUARE ROOTAL MOMENTUM IN THE X-DIRECTION.":" DO THE SAME IN THE Y-DIRECTION.":::" PT(X) = P1(X) + P2(X)"::" PT(Y) = P1(Y) + P2(Y)"::"NOTICE THAT THE Y-COMPONENTS MUST BE"::"SUBTRACTED SINCE THEY ARE IN OPPOSITE"::"DIRECTIONS. FILL I":X"DIRECTION OF THE VECTORS! P1(Y) IS"::"NEGATIVE, P2(Y) IS POSITIVE.":20:730:16304,0:520:21:" P2(X) = P2 * COS(THETA-2)":" P2(Y) = P2 * SIN(THETA-2)"::730"::5:"5. COMBINE THE TWO X-COMPONENTS TO GET"::" TOT04,4:490:21:" P1(X) = P1 * COS(THETA-1)":" P1(Y) = P1 * SIN(THETA-1)"::730 ::5:"OBVIOUSLY, THE NEXT STEP IS TO FIND:":::"4. P2(X) = P2 * COS(THETA-2)":" P2(Y) = P2 * SIN(THETA-2)":::"BE SURE TO PAY ATTENTION TO THE 3. FIND THE X- & Y-COMPONENTS OF P1.":j " SINCE OUR ANGLES ARE MEASURED FROM"::" THE X-AXIS,": " P1(X) = P1 * COS(THETA-1)"::" P1(Y) = P1 * SIN(THETA-1)": :"FILL IN YOUR CHART AND THEN LOOK AGAIN.":20:730\ :163ATER"::"THAN P2, & THEY ARE NOT IN THE SAME":W x"DIRECTION. LET'S LOOK!":21:730 Q145:Q2191:::400:22:"HERE ARE THE PATHS.":24:730 :470:22:"HERE ARE THE MOMENTUM VECTORS.":24:730' ::3:"THAT'S EASY! NOW FOR STEP 3.":6:"FILL IN"::"YOUR CHART AND";:16:730 Z:5:"NOW LET'S LOOK AT WHAT YOU'VE DONE SO"::"FAR. WE'LL STOP THE MOVING BODIES &": d"SHOW THE MOMENTUM VECTORS. REMEMBER"::"THAT BOTH MAGNITUDE & DIRECTION ARE":0 n"IMPORTANT. NOTICE THAT P1 IS GRE 2"COLLIDING BODIES. FILL IN M1, M2, V1,"::"AND V2 ON YOUR CHART.": <"1. COMBINED MASS EQUALS M1 PLUS M2."::" MT = M1 + M2": F"2. MOMENTUM = MASS * VELOCITY"::" P1 = M1 * V1 & P2 = M2 * V2":& P:"SEE HOW EASY & LOGICAL IT IS? $ - INELASTIC COLLISIONS - IIQ700::6:"THANKS FOR WAITING, "Z$"!":::"READY TO LOOK AT THE PROBLEM SOLVING"::"METHOD? LET'S GO!":::730::(:3:"WE'LL ASSUME YOU'VE BEEN GIVEN THE"::"MASS, VELOCITY, & ANGLE OF THE TWO":H                * MODIFIED BY **> * TOM McCORD *AH * EDUCATIONAL *XR * MEDIA *o\ * SERVICES *f ****************.':" ************************"8'255:A11000:::(7):B'"";A$:Z"MOMENTUM & CONSERVATX2,Y4X,Y6:4X2,Y3X2,Y3:X,Y3X,Y3:h:100:8:" ************************": " INELASTIC COLLISIONS - PART I:" ************************":255:A11000::: "";A$:* ****************4Y1X,Y5:X1,Y6X3,Y6:X4,Y5X4,Y1:X1,YX3,Y:X1,Y3X5,Y3:qX,YX4,Y:X2,YX2,Y6X,Y4:X4,YX,Y:X,Y1:X1,Y2:X2,Y2:X3,Y3:X4,Y4:X4,Y5:X3,Y6X1,Y6:X,Y5:X2,YX,Y2X,Y4X2,Y6:X,YX2,Y26X4,Y5X4,Y4X3,Y3X,Y3:EX,Y6X4,Y6:X2,Y6X2,Y:X,Y:X,Y6:X4,Y6:X4,Y:X,Y1:X,Y5:X1,Y2:X1,Y4:X2,Y3:X3,Y2:X3,Y4:X4,Y1:X4,Y5:X,Y6X,Y4:X1,Y3X3,Y3:X4,Y4X4,Y6:X2,Y2X2,Y:KX,Y121:670:X261:680:d5:189,104243,104:241,103:241,105:189,105189,114:187,112:191,1123:X245:Y99:670:X253:680:X261:740:X267:690:X275:750X173:Y124:670:X181:680:X189:740:X195:700:X203:750:!X,YX,Y6X3,YɱFN2:42,14266,142:64,141:64,143:42,14242,131:40,133:44,133X3:X72:Y145:670:X80:730:X88:740:X94:690:X102:750bX24:Y127:670:X32:730:X40:740:X46:700:X54:750:l3:191,105243,114241,110:243,114239,116vX252:&42,14266,13160,131:65,13164,135:X76:Y129:670:X85:730:01:41,3171,3168,29:71,3168,33:41,3141,52:39,50:43,50:3:X76:Y34:670:X84:720:X92:740:X98:690:X106:750DX25:Y62:670:X33:720:X41:740:X47:700:X54:750:5:5,96276,96:133,22133,170:T3:X71:Y84:710:X81:740:X86:720:X94:750X71:Y110:710:X81:740:X86:730:X94:750X221:Y105:710:X231:740:X236:680:X244:750:3:41,3171,5266,52:71,5270,48:X78:Y62:670:X87:720C(4)"BLOAD CHAIN,A520":520"TWO-D COLLISIONS II":T5Q1:XT:Y.7X1:1:760:T3,Y:2:XT2:Y165.5X:760:T2,Y:0XT:Y.7X1:760:XT1:Y165.5X:760::T135Q2:XT1:Y.15X76:3:760:T3,Y:0:XT:Y.15X76:760:: "FOR SOLVING THE PROBLEM. IF YOU WANT":|"TO REVIEW THIS PROGRAM BEFORE YOU GO"::"ON, TYPE 1. TYPE 2 TO CONTINUE:";AA180::8:"PLEASE WAIT WHILE PART II IS LOADED."::5:"SEE YOU IN A FEW SECONDS!":20:14)"-LOADING-":D13800:6"2. DO NOT ADD THE VELOCITIES OF THE"::" BODIES: ADD THEIR MOMENTUMS!!":"3. MOMENTUM IS A VECTOR. ADD X- AND Y-"::" COMPONENTS SEPERATELY!!":::"LEARN THESE BASIC PRINCIPLES!!":24:800):6:"OK, "Z$", LET'S LOOK AT A METHOD":: PROBLEMS."::"YOU MUST KNOW AND REMEMBER THESE!!!!":21:800r:3:"1. IN INELASTIC COLLISIONS, MOMENTUM IS"::" CONSERVED: KINETIC ENERGY IS NOT.":|" THEREFORE, THE MOMENTUM BEFORE THE"::" COLLISION EQUALS THE MOMENTUM AFTER.":M TO SOLVE"::"WILL PROVIDE YOU WITH MUCH OF THE":T"INFORMATION--PROBABLY MASS, VELOCITY,"::"AND ANGLE FOR BODIES #1 & #2.":^"BUT BEFORE WE ACTUALLY BEGIN TO SOLVE"::"PROBLEMS, HERE ARE THE THREE BASIC":>h"RULES OF INELASTIC COLLISIONS *":"---------------------------------------},"X-MOM. * * *":"---------------------------------------6"Y-MOM. * * *":"---------------------------------------@24:8003J:4:"THE PROBLEM YOU ARE TRYING 2 T":"---------------------------------------"MASS * * *":"---------------------------------------"VELOCITY * * *":"----------------------------------------""MOMENTUM * * 1(Y) P2(Y) PT(Y)":" ANGLE THETA-1 THETA-2 THETA-T":24:800::"IT LOOKS LIKE A LOT! THAT'S ONE REASON"::"WHY STUDENTS SOMETIMES HAVE PROBLEMS.":"LET'S TAKE IT SYSTEMATICALLY--DRAW"::"THIS CHART:":=" 1 WE WILL NEED TO DEAL WITH X- AND"::"Y-COMPPONENTS OF MOMENTUM:": " MASS M1 M2 MT":" VELOCITY V1 V2 VT " MOMENTUM P1 P2 PT":" X-MOMENTUM P1(X) P2(X) PT(X)L" Y-MOMENTUM P :510:22:"AND HERE ARE THE IMPORTANT ANGLES.":24:800 ::4:"AS LONG AS WE'RE DISCUSSING NOTATION,"::"LET'S DO IT ALL. FOR EACH BODY, #1, #2,": "AND THE COMBINATION, THERE WILL BE A"::"MASS, A VELOCITY, AND A MOMENTUM.":D "ALSO,. ALSO, WE'LL MEASURE ANGLES FROM"::"THE X-AXIS, SIMPLY FOR CONVENIENCE.":Z 24:800 :11:"LET'S LOOK AT THE SKETCH AGAIN."::"MAKE A COPY TO REFER TO AS WE GO ALONG.":18:800 :16304,0:24:800::500:22:"HERE'S OUR AXES.":24:800< GOOD!"::= d"LET'S AGREE ON SOME SIMPLE NOTATIONS:":: n" 1 - THE FIRST BODY"::" 2 - THE SECOND BODY"::" T - THE COMBINED BODIES":: x"LET'S ALSO AGREE TO ADD A SET OF AXES"::"WITH THE ORIGIN AT THE POINT OF COL-":M "LISIONIC"::"COLLISIONS OCCUR WHEN TWO OBJECTS STICK":_ <"TOGETHER AFTER THEY MAKE CONTACT.":: F"LOOK AT ONE TO BE SURE YOU UNDERSTAND.":24:800 P:Q1133:Q2275::X10:Y24:3:720:Y150:730 Z470:X146:Y110:3:680:490:24:800::::"OK? - INELASTIC COLLISIONSP770::3:"HELLO. WHAT'S YOUR NAME? ";Z$::"WELL, "Z$", THIS PROGRAM SHOULD"::"GIVE YOU SOME HELP IN SOLVING PROBLEMS":("THAT DEAL WITH INELASTIC COLLISIONS."::2 2"YOU SHOULD ALREADY KNOW THAT INELAST         ****************URN>";A$:" THERMODYNAMIC PROCESSES"::" ***********************^255:A11000:::s"";A$: **************** * MODIFIED BY * * TOM McCORD * * EDUCATIONAL * * MEDIA * * SERVICES *X4,Y:X2,YX2,Y6X,Y4:zfX4,YX,Y:X,Y1:X1,Y2:X2,Y2:X3,Y3:X4,Y4:X4,Y5:X3,Y6X1,Y6:X,Y5:pX,Y6X3,Y6:X4,Y5:X3,Y4:X2,Y3:X3,Y2:X4,Y1:X,YX3,Y:z:100:8:" ***********************":E.01F:F1.545(M)195,F60(M)26:::X126:Y22:860:X148:870:~>X,YX,Y6X3,Y6X4,Y5X4,Y4X3,Y3X,Y3:HX,Y4X,Y6:X4,Y4X4,Y6:X1,Y2X1,Y3:X3,Y2X3,Y3:X2,YX2,Y1:RX,Y6X4,Y6:X2,Y6X2,Y:\X,Y8,90136,31:106,95170,40:112,102180,68:3:X132:Y27:k830:X140:860:X186:Y69:830:X194:880: 840:X140:860:X186:Y69:840:X194:880:*5:M4.9726.1.024:116(M)218,96(M)::3:X196:Y96:850:G43:F11.31.3:M4.75612:0850:X120:860:X172:850:X180:880:\840:X120:860:X172:840:X180:880:2:104,35182,35:104,62182,62:104,89182,89:3:X186:Y38:830:X194:880:Y92:860:X186:830:840:X194:880:Y92:860:X186:840:: 1:986:Y132:840:+830:X186:Y132:850:G840:X186:Y132:850:F13:M4.716.28(.24F):24F(M)184,21F(M)30::Y53:X188:850:X195:880:Y95:860:X188:850:1:119,30119,100:149,30149,100:179,30179,100:3:Y27:X1LL FOR THIS PROGRAM. BE SURE"::"TO WATCH:"::Wv8::"THERMODYNAMIC CYCLES":::"WE HOPE THIS PROGRAM HELPED YOU!"::" BYE, "Z$"!":20:920:20:"END":(4)"RUN MENU"::5:79,2079,120195,120:3:X68:Y30:830:X1DIABATIC ON P-V DIAGRAMjD" 2. ISOMETRIC ON P-V DIAGRAM":" 3. ISOTHERMAL ON P-V DIAGRAM?N" ANSWER 1,2, OR 3: ";T::T3610X21:"WRONG. TRY AGAIN.":24:920:570b21:"EXCELLENT, "Z$"!":24:9200l::5:"THAT'S AISOBARIC ON P-V DIAGRAM":" 3. ISOBARIC ON V-T DIAGRAM?o" ANSWER 1,2, OR 3: ";T::T1560&21:"THAT'S NOT RIGHT. TRY AGAIN.":24:920:520021:"YES. ISOMETRIC PROCESSES ON P-V.":24:920:660:670:700:21:"IS THIS 1. A" 3. ISOBARIC ON P-T DIAGRAM?V" ANSWER 1,2, OR 3: ";T::T251021:"SORRY, "Z$". TRY AGAIN.":24:920:470:21:"GREAT, "Z$"!":24:920660:670:720:740:21:"IS THIS: 1. ISOMETRIC ON P-V DIAGRAM?" 2. :"ADIABATIC PROCESSES ON P-V DIAGRAM.":24:920::3:"STUDY YOUR DRAWINGS FOR A MINUTE.":6:"WHEN YOU'RE READY, TEST YOURSELF.":9:920660:690:780:790:21:"IS THIS: 1. ISOTHERMAL ON V-T DIAGRAM&" 2. ISOBARIC ON V-T DIAGRAM": P = (UR/V)*T,"V1";A$:S ****************j  * MODIFIED BY * * TOM McCORD *  * EDUCATIONAL ** * MEDIA *4 * SERVICES *> ****************,Y6:X,Y5:ZX,Y6X3,Y6:X4,Y5:X3,Y4:X2,Y3:X3,Y2:X4,Y1:X,YX3,Y:X3,Y6X,Y3:X,Y2X4,Y2:X3,Y6X3,Y1::100:8:" ************************":" THERMODYNAMIC CYCLES"::" **********X,YX,Y6X3,Y6X4,Y5X4,Y4X3,Y3X,Y3:X,Y4X,Y6:X4,Y4X4,Y6:X1,Y2X1,Y3:X3,Y2X3,Y3:X2,YX2,Y1:X,YX4,Y:X2,YX2,Y6X,Y4: X4,YX,Y:X,Y1:X1,Y2:X2,Y2:X3,Y3:X4,Y4:X4,Y5:X3,Y6X1C CYCLES. THEN, FOR APPLICATIONS":fv"OF THE CARNOT CYCLE, WATCH:":::13::"HEAT ENGINES":::"BYE FOR NOW, "Z$".":23:760::17:"END"D11500:::(4)"RUN MENU"::5:79,2079,120195,120:3:X68:Y30:670:X186:Y132:680:5"::31:760:nX"IF YOU SAID THAT IT RESULTS IN THE"::"MAXIMUM POSSIBLE EFFICIENCY, CORRECT!"::31:760b:3:"THAT'S ALL FOR NOW. YOU SHOULD REFER"::"YOUR TEXTBOOK FOR PROBLEMS DEALING":#l"WITH FINDING THE NET WORK IN THERMO-"::"DYNAMI:760:s::3:"ONE SPECIAL CYCLE YOU SHOULD KNOW IS THE"::"CARNOT CYCLE";::". HOW IS IT MADE?"::31:760:D"IF YOU SAID BY COMBINING 2 ISOTHERMAL"::"PROCESSES AND 2 ADIABATIC PROCESSES,":N"YOU'RE RIGHT, "Z$"!":::"WHY IS IT SPECIAL?COND TERM IS ALREADY"::"NEGATIVE.)":::"USING THIS METHOD, WE CAN FIND THE WORK":"DONE IN MANY CYCLES.":18:"LET'S LOOK AGAIN.":::760::16304,0&1:T19:1174T,471174T,36(5.05T8.1)39:021:" NET USEFUL WORK":DONE IN COMPRESSION.":r"REMEMBER THAT THE INTEGRAL OF 1/V IS"::"NATURAL LOG OF V. SO, THE WORK DONE IS:":" W = URT * LN (V1-V3)"::760:3:"THE NET WORK IS THEN--"::" W = P1*(V2-V1) + URT*LN(V1-V3)":V"(NOTE THAT THE SE760r:3:"THE AREA UNDER 3-1 IS HARDER TO FIND:"::"REWRITE THE IDEAL GAS LAW--"::" P = 1/V * URT":"SINCE TEMPERATURE IS CONSTANT, WE CAN"::"INTEGRATE 1/V FROM V3 TO V1 AND":"MULTIPLY BY URT. THIS GIVES THE"::"NEGATIVE WORK ":O"2. FIND THE AREA UNDER 3-1 (COMPRESSION)":"3. SUBTRACT #2 FROM #1"::"ASSUME YOU ARE GIVEN THE PRESSURE,"::"VOLUME, & TEMPERATURE AT 1,2, & 3.":"NOTE THAT P1=P2, V2=V3, & T1=T3.":::"THE AREA UNDER 1-2 IS SIMPLY P1*(V2-V1)."::4.9(.04):46(T)164,36(T)40::118,47X3:X108:Y43:690:X163:700:Y83:71021:"THE NET WORK IS THE AREA INSIDE THE":"CYCLE. HOW WOULD YOU FIND IT?"::760::3:"TO FIND THE NET WORK:":::"1. FIND THE AREA UNDER 1-2 (EXPANSION)z#Xİ   Ǡ ծԠǮԠ  ϭĠӠɠ Π ű Ų îà Үʠ ˯ˠ ŭĠӠ-ͯŠŠɠ ͯŠŠɠ" àӠàӠ ϭĠӠɠ ϭĠӠɠ̠Ϡ ̠Ϯà!ŠŠŠՠ ٠4ԠӭӠ ԠӭĠ$@@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 P OKLAHOMA CITY, OKLAHOMA Z1790::4:"HELLO, PHYSICS STUDENT!"::"PLEASE GIVE ME YOUR NAME: ";Z$ d10:Z$", THIS PROGRAM WILL HELP YOU"::"LEARN TO SOLVE CALORIMETRY (OR MIXING)":; n"PROBLEMS. THE THEORY I2 CALORIMETRY ANALYZING CALORIMETRY (MIXING) PROBLEMS USING CONSERVATION OF ENERGY APPROACH. INCLUDES HEAT OF FUSION FOR WATER AND WATER EQUIVALENT OF CALORIMETER CUP.  ( NSF LOCI PROJECT2  < SOUTH OKLAHOMA                  ^D$"RUN HELLO" hD$"CATALOG( r ? | ****************V  * MODIFIED BY *m  * TOM McCORD *  * EDUCATIONAL *  * MEDIA *  * SERVICES *  ****************DLISIONS III"0 D$"RUN SHM/REFERENCE CIRCLE I"U "D$"RUN SHM/REFERENCE CIRCLE II"n ,D$"RUN CALORIMETRY" 6D$"RUN THERMODYNAMIC PROCESSES" @D$"RUN THERMODYNAMIC CYCLES" JD$"RUN HEAT ENGINES-METHOD" TD$"RUN HEAT ENGINES-APPLICATIONS" J 24:"TYPE IN THE NUMBER AND PRESS ";D:D180:D1380:34,0P D240,250,260,270,280,290,300,310,320,330,340,360,350 D$"RUN ONE-D COLLISIONS" D$"RUN TWO-D COLLISIONS I" D$"RUN TWO-D COLLISIONS II" D$"RUN TWO-D COL:2:"SOLUTION:"::" 1*.53*15 + .093*.56*TI =":b#" 1*.53*22.5 + .093*.56*22.5"::1820#::"THEN,"::" 7.95 + .0521*TI = 11.925 + 1.172":#"OR,"::" .0521*TI = 13.097 - 7.95"::1820: $:"THUS,"::" .0521*TI =HART."::1820@":1730:6:8:"1.0";:14:".5+.O3";:24:"15k"10:7:".093";:15:".56";:31:"22.5"18:"NOTE THAT WE ADD THE AMOUNT OF WATER &"::"THE WATER EQUIVALENT OF THE CUP."::""NOW, "Z$", YOU SOLVE THE PROBLEM!":30:18201#S 500 GM OF WATER AT 15 DEGREES C.":!"A 560 GM SAMPLE OF COPPER (C =.093) IS"::"ADDED. IF THE FINAL TEMPERATURE OF THE":!"WATER IS 22.5 DEGREES C, FIND THE"::"INITIAL TEMPERATURE OF THE COPPER."::""BEGIN BY PUTTING THE DATA IN YOUR CATER, ADD THE":k z"WATER EQUIVALENT TO THE AMOUNT OF WATER"::"IN THE CUP AND PROCEED WITH ONLY TWO": "OBJECTS. THAT MAKES IT SIMPLER!":::"HERE'S AN EXAMPLE:":23:1820%!:2:"PROBLEM:"::"A CALORIMETER (WATER EQUIVALENT =.03 KG)":"HOLDV-":U\"ALENT IS .01 KG WILL BEHAVE AS IF IT"::"WERE .01 KG OF WATER! "::1820f:5:"THEREFORE, THE MASS & SPECIFIC HEAT DO"::"NOT NEED TO BE GIVEN. SIMPLY TREAT THE": p"CUP AS IF IT WERE THAT MUCH WATER. IF"::"THE CUP IS FILLED WITH W:"SOME STUDENTS:"::O>"-THE WATER EQUIVALENT OF A CALORIMETER-":::1820H:5:"THE MAIN IDEA IS THAT A CALORIMETER"::"CAN BE DESCRIBED AS HAVING A THERMAL":R"EFFECT EQUIVALENT TO A GIVEN AMOUNT OF"::"WATER. THUS A CUP WHOSE WATER EQUI0C:2:"ANSWER:"::"IF YOU RECOGNIZED THAT THE EQUILIBRIUM": "TEMPERATURE WAS 21.1 DEGREES C, YOU"::"SHOULD HAVE FOUND THE SPECIFIC HEAT:":*6::"C = .033 KCAL/KG-DEGREE C"::::1820::4"NOW LET'S LOOK AT AN IDEA WHICH CONFUSES"ER AT 10"::"DEGREES C. AFTER MIXING, THE TEMPERATURE":"OF THE WATER HAS INCREASED 11.1 DEGREES.":"FIND THE SPECIFIC HEAT OF THE LEAD."::"(HINT: THE EQUILIBRIUM TEMPERATURE IS": "NOT 11.1 DEGREES C!)":::"WHEN YOU HAVE AN ANSWER, ";:182B"THAT'S EASY!":::"NOW HERE'S ONE FOR PRACTICE. ";:1820:2:"PROBLEM:"::"A 450 GM PIECE OF LEAD IS HEATED TO":"100 DEGREES C AND THEN DROPPED INTO A"::"50 GM COPPER (C =.093) CALORIMETER CUP":e"WHICH CONTAINS 100 GM OF WAT" 1.932 + 23 + 23*C3"::1820t:2:"COMBINING:"::" 16.26 + 37.5*C3 = 24.932 + 23*C3"::"OR,:" 14.5*C3 = 8.672":::"DO YOU AGREE, "Z$"?"::1820:"FINALLY,":::"C3 = 8.672/14.5 = .598 KCAL/KG-DEGREE C"::: WORKING BY"::"YOURSELF. THEN, ";:1820i:2:"SOLUTION:"::1780;:8:1820::"SUBSTITUTE DATA:":" .14*.30*30 + 1.0*.5*30 + C3*.5*75 ="::" .14*.30*46 + 1.0*.5*46 + C3*.5*46"::1820:"MULTIPLYING:"::" 1.26 + 15 + 37.5*C3 =":.OF ALCOHOL?":::"ENTER THE DATA IN YOUR CHART.Fl23:1820::1730v6:7:".14";:15:".30";:24:" 30":10:7:"1.0";:15:".50";:24:" 30";:32:"4614:15:".50";:24:" 75":18:"AGAIN, ONLY ONE UNKNOWN!"::)"SEE HOW WELL YOU CAN DOPROBLEM:"::"A 300 GM GLASS CONTAINER (C =.14) HOLDS":N"500 GM OF WATER AT 30 DEGREES C. WHEN"::"500 GM OF ALCOHOL AT 75 DEGREES ARE ":X"ADDED, THE MIXTURE REACHES A FINAL"::"TEMPERATURE OF 46 DEGREES C. WHAT IS":0b"THE SPECIFIC HEAT R 2 TO CONTINUE: ";A::A1460}&::"NOTICE THE SMALL CHANGE IN THE WATER"::"TEMPERATURE. THIS EFFECT IS DUE TO THE":0"LARGE AMOUNT OF WATER AND THE HIGH"::"SPECIFIC HEAT OF WATER.":::"LET'S DO ANOTHER PROBLEM. ";:18208D:2:" 30 + 1.32 =":C" .0558*TE + 1.5*TE + .011*TE"::1820:}"COMBINING:"::" 32.436 = 1.5668*TE"::1820:2:"FINALLY,"::3::"TE = 32.436/1.5668 = 20.7 DEGREES C":::"CONFUSED, "Z$"? IF SO, ENTER 1"::"TO REVIEW. ENTEVATION EQUATION:":O:1780:1820::2:"NOW SIMPLY SUBSTITUTE THE DATA:":" .093*.60*20 + 1.0*1.5*20 + .11*.10*120"::" =":" .093*.60*TE + 1.0*1.5*TE + .11*.10*TE"::1820:"THE REST IS ALGEBRA:"::" 1.116 +,205:2:X811:15,20X:l3,10:13,1914:14,1920:13,1919:21:"GIVEN A HOT SAMPLE, COOL LIQUID, & A36"CALORIMETER CUP."::18203@12:23,2416:23,2418:1:9,1029:9,1030:2:13,2027:X2832:15,20X::13,2032K4J:21:"THE LIQUID :1820X2:4:9::"ANSWER: .02 KG OF ICE"::9:"WELL, "Z$", THAT'S ALL FOR THIS":2"PROGRAM. SEE YOUR INSTRUCTOR IF YOU'RE"::"STILL HAVING TROUBLES."::::"BYE FOR NOW!":22:18202:20:"END":(4)"RUN MENU"3"::150:1:19,204:19F WATER AT 30 DEGREES C":E1"ARE COOLED TO 20 DEGREES C BY THE":1"ADDITION OF ICE. HOW MUCH ICE WAS"::"ADDED, ASSUMING NO HEAT EXCHANGE WITH":1"THE OUTSIDE ENVIRONMENT?":17:"WORK OUT THE PROBLEM AND THEN CHECK":2"YOUR ANSWER.":22"FINALLY,"::5::"M(ICE) = 1.5/80 = .01875 KG":::0"AGAIN, YOU CAN SOLVE MORE COMPLEX"::"PROBLEMS IF YOU REMEMBER WHICH POINT":0"OF REFERENCE YOU ARE USING."::::"NOW HERE'S ONE LAST PROBLEM:":24:18201:4:"PROBLEM:"::"200 GM OMASS)ICE =":@/" (C*M*TE)LEAD + (C*M*TE)WATER":::1820/::"BUT, THE EQUILIBRIUM TEMPERATURE IS"::"ZERO DEGREES C!"::/"THEREFORE, (C*M*T)LEAD = H.F.* M(ICE)":::1820/:" .03*.5*100 = 80*M(ICE)":24:182060:4:M OF LEAD (C =.03) PELLETS AT 100":z.n"DEGREES C ARE POURED IN A HOLE IN A"::"LARGE PIECE OF ICE. HOW MUCH ICE IS":.x"MELTED BEFORE THE LEAD COOLS TO ZERO?":::"TRY IT YOURSELF, "Z$".":23:1820 /:2:"SOLUTION:"::" (C*M*T)LEAD - (H.F.* 1820::?-F"FINALLY, "::8::"TE = 16.25 DEGREES C":::s-P:"JUST REMEMBER NEGATIVE ENERGIES!":24:1820-Z:2:"INPUT 1 IF YOU WANT TO SEE THE PROBLEM"::"AGAIN. INPUT 2 TO CONTINUE: ";A:::A11170$.d:"NOW HERE'S ONE MORE:"::"500 GS PROBLEM,"::" 1*.21*30 - 80*.03 = 1*.21*TE + 1*.03*TE"::,("NOTE THAT WE NOW TREAT THE MELTED ICE"::"LIKE ORDINARY WATER (WHICH IT IS!).":::1820,2:2:"THEN,"::" 6.3 - 2.4 = .21*TE + .03*TE":: -<"AND,"::" .24*TE = 3.9"::: = ENERGY (AFTER)"::"WHERE ENERGY = (C*M*T) EXECPT FOR THE":x+"ICE WHICH IS = -(HEAT OF FUSION * MASS)":24:1820+ :2:"SO WE GET:"::" (C*M*T)WATER - (H.F.* M)ICE =":+" (C*M*T)WATER + (C*M*T)MELTED WATER":::1820=,::"FOR THIEM:"::"A 30 GM PIECE OF ICE AT ZERO DEGREES C":*"IS PLACED IN 210 GM OF WATER AT 30"::"DEGREES C. NEGLECTING HEAT LOSSES TO":*"THE ENVIRONMENT, WHAT IS THE FINAL"::"TEMPERATURE? ";:1820:::?+"SOLUTION:"::" ENERGY (BEFORE)IN KCAL) FROM"::"THE INITIAL ENERGY IN THE SYSTEM.":g)"(THINK OF IT AS AN ENERGY DEBT.)"::1820)::"THEN, INCLUDE THE MASS OF THE ICE (NOW"::"MELTED) AS WATER IN THE FINAL CONDITION."::)"HERE'S AN EXAMPLE:":23:18202*:2:"PROBLRGY!":7("DON'T GET CONFUSED, "Z$"! ";:1820::("HERE'S HOW TO USE THE HEAT OF FUSION"::"IN A CALORIMETER PROBLEM:"::1820(:2:"IF THE ICE IS AT ZERO DEGREES C, WE"::"MULTIPLY BY THE HEAT OF FUSION AND":7)"SUBTRACT THE ANSWER (1820m't:2:"IT TAKES 80 KCAL TO CHANGE 1 KG OF ICE"::"AT ZERO DEGREES C TO WATER AT ZERO."::1820::'~"THEREFORE, THE ICE HAS LESS ENERGY THAN"::"THE WATER. USING OUR REFERENCE, THE ICE":("MUST BE THOUGHT OF AS HAVING NEGATIVE"::"ENEHING: ";:1820u&V::"** WE HAVE CHOSEN TO LET ZERO DEGREES C"::"MEAN ZERO ENERGY (E=C*M*T). OF COURSE":&`"THIS IS NOT TRUE, BUT A CONSEQUENCE OF"::"USING THE CELSIUS TEMPERATURE SCALE **":::1820'j::"YOU SHOULD KNOW.... ";:ING"::"ICE AND A LIQUID (NOT ALWAYS WATER)."::%8:"THE ICE MELTS & THE MIXTURE COMES TO"::"AN EQUILIBRIUM TEMPERATURE.":::1820%B:4:"THERE ARE SEVERAL WAYS TO SOLVE THESE"::"PROBLEMS. HERE'S AN EASY METHOD IF YOU":&L"REMEMBER ONE T 5.147":V$"AND,"::11::"TI = 98.8 DEGREES C":::"ISN'T THAT EASY?":::1820$:4:"NOW WE'RE READY FOR ONE LAST IDEA--"::"USING ENERGIES ASSOCIATED WITH PHASE":$$"CHANGES. ";:1820:2%.::"THESE PROBLEMS USUALLY INVOLVE MIX 1 KCAL/Q2 = 473/378":W@" 1 KCAL/Q2 = 1.25"::" Q2 = 1 KCAL/1.25":J" Q2 = 0.80 KCAL"::" Q2 = (0.80)(4186) = 3349 J"::T"SEE HOW EASY IT IS?":24:1230^:3:"LET'S DO ANOTHER:":::" AN IDEAL REFRB" EFF = W/Q1"::"WE DON'T USE THE CONCEPT OF COEFFICIENT":""OF PERFORMANCE SINCE THIS IS NOT A "::"HEAT PUMP.":24:1230,:3:"SOLUTION:"::"USE THE HEAT FLOW/TEMPERATURE RELATION":6"TO FIND Q2."::" Q1/Q2 = T1/T2"::" 200 DEGREES CELSIUS AND 105"::" DEGREES CELSIUS IF 1 KCAL OF ENERGY":i " IS SUPPLIED?":24:1230 :3:"HERE'S WHAT WE KNOW:"::" T1 = 200 + 273 = 473 K": " T2 = 105 + 273 = 378 K"::" Q1 = 1 KCAL": " Q2 = ?"::" W = ?":GO ON!":::"TYPE 1 TO REVIEW."::"TYPE 2 TO CONTINUE: ";A:A1130 :5:"NOW, HERE'S A SAMPLE PROBLEM. BE SURE "::"TO WRITE IT DOWN.":: " HOW MUCH USEFUL WORK MAY BE EXTRACTED"::" FROM AN IDEAL HEAT ENGINE OPERATING":I " BETWEEN W/Q1"::" K - HEAT PUMP PERFORMANCE = Q2/W":z " RELATIONSHIPS: W = Q1 - Q2"::" Q1/Q2 = T1/T2 24:1230 :7:Z$", YOU MUST KNOW THE VARIABLES"::"AND RELATIONSHIPS. IF YOU DON'T, LEARN":E "THEM BEFORE YOU TRY TO RST, LET'S REVIEW:"::" T1 - HIGH TEMP SOURCE (KELVIN)": " T2 - LOW TEMP SINK (KELVIN)"::" Q1 - HEAT FLOW IN/OUT OF HI-TEMP": " Q2 - HEAT FLOW IN/OUT OF LO-TEMP"::" W - WORK EXTRACTED OR USED":1 " EFF - HEAT ENGINE EFFICIENCY =ROGRAM IS DESIGNED"::"TO HELP YOU SOLVE PROBLEMS DEALING WITH": n"IDEAL HEAT ENGINES & HEAT PUMPS. IF YOU"::"HAVE YOUR CALCULATOR HANDY, TELL ME": x"YOUR NAME AND WE'LL GET STARTED."::"NAME? ";Z$:20:"THANKS, "Z$".":24:1230; :3:"FI HEAT ENGINES:APPLICATIONV ANALYZING IDEAL HEAT ENGINES AND HEAT PUMPS] x( NSF LOCI PROJECT2 < SOUTH OKLAHOMA CITYF JR. COLLEGEP OKLAHOMA CITY, OKLAHOMA Z1200B d:5:"HELLO AGAIN! THIS P          ORIMETRY"::" ***********************F:255:A11000:::[:"";A$:r:& ****************:0 * MODIFIED BY *:: * TOM McCORD *:D * EDUCATIONAL *:N * MEDIA *:X * SERVICES *:b ****************S:" * * * * *":"------------------------------------"::Z99" C1*M1*T1 + C2*M2*T2 + C3*M3*T3 = "::" C1*M1*TE + C2*M2*TE + C3*M3*TE":::9:100:8:" ***********************":-:" CAL-* *8" * * * * *":" 2 * * * * *":" * * * * *":"----------------------------* *T9" * * * * *":" 3 * * * * *"115707::f72:" C * M * T * TE "::"------------------------------------ 8" * * * * *":" 1 * * * * *":" * * * * *":"---------------------------e6:150:21:"FINALLY, EVERYTHING REACHES THE SAME":"EQUILIBRIUM TEMPERATURE (TE). ";:1820:06T1800::7:17,13:22,13:17,14:22,14:Y1520:17,22Y:T1150:T:Y7150:" INPUT 1 TO SEE AGAIN.":" INPUT 2 TO CONTINUE: ";A:255:AT1500:>5r21:"ENERGY FLOWS & TEMPERATURES CHANGE.":1820~5|0:T11000::11:19,2017:T160::19,2018:T1350:56:17,13:22,13:17,14:22,14:17,2215:17,2216617,1817:21,2217:17,1818:21,2218:17,2219:17,2220:T11200:& CUP HAVE THE SAME TEMP.":1820:"NOW THE SAMPLE IS PUT IN THE CUP.":18204T::1:9,1019:9,1020:2:13,2017:X1821:15,20X:4^13,2022:T11000::40:Y1116:1:19,20Y:0:19,20(Y2): 5hY1718:1:19,20Y:2:19,20(Y2)::150:"ENTER EITHER EFF OR K: ";E$:E$"EFF"950U::"INCORRECT, "Z$". THIS PROBLEM":z"DOESN'T CONCERN HEAT PUMPS.":::"USE EFFICIENCY FOR HEAT ENGINES."::24:1230 :3:"NOW FIND Q2 AND ENTER YOUR ANSWER."::"ANSWER? (USE NUMBERS ONLY)OU MISUNDERSTAND THE PROBLEM?":Cz:"Q1 = 500 KCAL"::24:1230:3:"IF YOU'VE COME THIS FAR, YOU KNOW THAT"::"Q2 IS NOT GIVEN."::"NOW TELL ME IF YOU WANT TO USE THE"::"CONCEPT OF EFFICIENCY OR COEFFICIENT":+"OF PERFORMANCE."::T2300850?>(7):"NO. HAVE YOU CONFUSED T1 & T2? OR, ":kH"DID YOU USE THE WRONG TEMP. SCALE?":R:"T2 = 300 K"::22:1230\:3:"WHAT IS Q1? (USE NUMBERS ONLY): ";Q1::Q1500890f:"WRONG. Q1 IS THE HEAT ABSORBED. DID": p"Y:"WHAT IS T1? (USE NUMBERS ONLY): ";T1::T1600810"SORRY. DID YOU MISS THE FACT THAT THE"::"TEMPERATURE WAS GIVEN IN DEGREES": "KELVIN OR DID YOU USE T2?":*:"T1 = 600 K"::22:1230 4:3:"WHAT IS T2? (USE NUMBERS ONLY): ";T2:: 600 DEGREES KELVIN AND EXHAUSTS":}" HEAT AT 300 DEGREES KELVIN. WHAT IS"::" THE EFFICIENCY AND HOW MUCH HEAT IS":" IS EXHAUSTED, ASSUMING AN IDEAL"::" ENGINE?"::"WHEN YOU'VE GOT THE PROBLEM COPIED,"::"PRESS ";:12305 :3223"::" EFF = 22.3 %"::xZ$", YOU SHOULD BE READY FOR THE "::"TEST. LET'S DO ONE MORE TO BE SURE.":24:1230:3:"COPY THIS ONE DOWN. THEN WE'LL CHECK"::"YOUR WORK, STEP BY STEP."::#" AN ENGINE ABSORBS 500 KCAL OF HEAT"::" ATR THAN COEFFICIENT OF PERFORMANCE?"::"NOW SOLVE THE PROBLEM AND CHECK YOUR":q"ANSWER BY PRESSING ";:1230:3:"SOLUTION:":::" EFF = W/Q1 = (Q1-Q2)/Q1":" EFF = (T1-T2)/T1"::" EFF = (448-348)/448":" EFF = 100/448 = 0."THE APPROPRIATE RELATIONSHIPS."::1230i:2:"YOU SHOULD HAVE:":::" T1 = 175 + 273 = 448 K":" T2 = 75 + 273 = 348 K"::" Q1 = ?"::" Q2 = ?"::" W = ?":" EFF = W/Q1":::"DID YOU REMEMBER TO USE EFFICIENCY":O"RATHESTEAM ENGINE IF THE TEMPERATURE"::" OF THE INPUT STEAM IS 175 DEGRES":b" CELSIUS AND THE TEMPERATURE OF THE"::" EXHAUST IS 75 DEGREES CELSIUS?":l:"SOLVE THIS ONE LIKE THE OTHERS. BEGIN"::"BY WRITING DOWN THE DATA IN A LIST WITH":*v"SO, EFF = (Q1-Q2)/Q1":::"THEN USE THE HEAT FLOW/TEMP RELATION:"::D" EFF = (T1-T2)/T1":::"LET'S DO ANOTHER PROBLEM. ";:1230N:3:"HERE'S ONE FOR YOU TO WORK OUT:":::" WHAT IS THE MAX. POSSIBLE EFFICIENCY":JX" OF A :"OF THE TEMPERATURES. THEN ";:27:1230:3:"HERE'S THE SUBSTITUTION FOR EFFICIENCY."::"AGAIN, DON'T MEMORIZE. YOU SHOULD BE":&"ABLE TO WORK THIS OUT WHEN YOU NEED IT."::0" EFF = W/Q1 AND"::" W = Q1 - Q2":K: Q2 = 0.112 KCAL OR 469 J"::"THE ONLY HARD PART IS THE SUBSTITUTION."::"DON'T BOTHER TO MEMORIZE, BUT DO BE":"ABLE TO FIND THAT:":::" K = Q2/W = Q2/(Q1-Q2) = T2/(T1-T2)":):"SEE IF YOU CAN FIND EFFICIENCY IN TERMS":Q1/Q2 = T1/T2 TO GET":A" K = T2/(T1-T2)"::"THEREFORE: K = 258/(313-258)"::" K = 4.69":24:1230:3:"FINALLY, SINCE K = Q2/W"::" Q2 = (K)(W)":)" Q2 = (4.69)(.0239)"::"N'T USE EFFICIENCY FOR HEAT PUMPS.":24:1230:3:"THE NEXT PART IS TRICKY. TO FIND Q2,"::"WE NEED TO FIND K. HERE'S HOW:"::" SINCE W = Q1 - Q2, AND"::" K = Q2/W ":" THEN K = Q2/(Q1-Q2)":::"NEXT USE 4:12309:3:"DATA:"::" T1 = 40 + 273 = 313 K":i" T2 = -15 + 273 = 258 K"::" Q1 = ?":" Q2 = ?"::" W = 100 JOULES"::" = 100/4186 = 0.0239 KCAL":" K = Q2/W":::"REMEMBER, T1 IS ALWAYS THE HIGH TEMP!":-"DOIGERATOR EXTRACTS HEAT":uh" FROM A STORAGE CHAMBER AT -15 DEGREES"::" CELSIUS & EXHAUSTS IT AT 40 DEGREES":r" CELSIUS. HOW MANY KCAL ARE EXTRACTED"::" IF 100 JOULES OF WORK ARE SUPPLIED?":|:"START BY WRITING DOWN WHAT WE KNOW.":2TEMP SOURCE TO THE":.,"LO-TEMP SINK!"::6"NOW, DON'T GET CONFUSED. WE WANT TO"::"LOOK AT A HEAT PUMP & IT'S MUCH THE":@"SAME. THE ONLY DIFFERENCE IS THE"::"DIRECTION OF THE HEAT FLOW!":22:1310J:3:"HERE'S THE BASIC IDEA OF A HETHIS IS A HEAT ENGINE. NOTE THE HI-TEMP":"SOURCE & THE LO-TEMP SINK. NOW LET'Sq"WATCH THE HEAT FLOW.":1310:22:" HEAT FLOW FOR HEAT ENGINES.":W15:1220::24:1310"::3:"BE SURE YOU SEE THAT THE HEAT IS"::"FLOWING FROM THE HI-LY FLOWS FROM A HIGH TEMP-"::"ERATURE SOURCE TO A LOW TEMPERATURE": "SINK. WHILE THIS HAPPENS, WE CAN"::"EXTRACT USEFUL WORK.":: "LET'S LOOK AT A SCHEMATIC DRAWING."::"COPY THE DRAWING FOR LATER REFERENCE.":24:1310O::1130:21:"OULD REMEMBER THAT THE CARNOT":w "CYCLE IS MADE FROM TWO ISOTHERMAL"::"PROCESSES & TWO ADIABATIC PROCESSES.":: "(BE SURE TO MAKE NOTES AS WE GO ALONG.)":22:1310 :3:"HERE'S THE BASIC IDEA OF A HEAT"::"ENGINE:"::F "HEAT NATURALVEN TEMPERATURE RANGE. HOWEVER, NO":~ "HEAT ENGINE/PUMP USES THE CARNOT CYCLE."::"THAT'S WHY WE SAY 'IDEAL' ENGINES.": 24:1310::3:"SO, REAL ENGINES/PUMPS ARE ACTUALLY"::"LESS EFFICIENT THAN THE ONES WE'LL": "CONSIDER.":::"YOU SH"STAND HOW TO SOLVE PROBLEMS DEALING":| "WITH 'IDEAL' HEAT ENGINES & HEAT PUMPS."::"FOR THESE PROBLEMS WE'LL USE A": "SPECIAL THERMODYNAMIC CYCLE: THE"::"CARNOT CYCLE. THIS CYCLE GIVES THE":% "MAXIMUM POSSIBLE EFFICIENCY FOR ANY"::"GIR d:5:"HELLO! WELCOME TO THE WORLD OF"::"THERMODYNAMICS, HEAT ENGINES AND": n"HEAT PUMPS.":12:"BEFORE WE BEGIN, TELL ME YOUR NAME:": xZ$:18:"EXCELLENT, "Z$". LET'S BEGIN.":24:1310' :3:"THIS PROGRAM SHOULD HELP YOU TO UNDER-"::" HEAT ENGINES: METHODu METHOD FOR ANALIZING IDEAL HEAT ENGINES AND HEAT PUMPS| ( NSF LOCI PROJECT2 < SOUTH OKLAHOMA CITYF JR. COLLEGEP OKLAHOMA CITY, OKLAHOMA Z1280          ***********":255:A11000:::6&"";A$:M& ****************d& * MODIFIED BY *{& * TOM McCORD *& * EDUCATIONAL *& * MEDIA *&  * SERVICES *& *****************&  * SERVICES * ' *******INSTRUCTOR IF YOU WANT MORE"::"HELP OR ADDITIONAL PROBLEMS TO WORK."::r%"BYE, BYE!":20:1230::20:"END"%(4)"RUN MENU"%:100:8:" *************************":%" HEAT ENGINES - APPLICATIONS!&:" **************10,000 JOULES OF WORK?":::"ENTER YOUR ANSWER: ";A[$t::A3.84A3.861150:"GREAT!";$~8:"THE ANSWER IS 3.85 KCAL.":::1230$:8:Z$", THAT'S IT FOR THIS PROGRAM."::"TYPE 1 TO REVIEW: TYPE 2 TO CONTINUE: ";A:A1130J%:8:"SEE YOUR VE FOR THE UNKNOWNS."::M#L"DO ONE MORE PROBLEM BEFORE YOU QUIT."::1230#V:3:"SOLVE THIS ONE BY YOURSELF:":::" A CARNOT ENGINE OPERATES BETWEEN 20":#`" AND 500 DEGREES CELSIUS. HOW MUCH"::" ENERGY MUST BE SUPPLIED TO PRODUCE":4$j" RE IS TO IT!":24:1230o".:5:Z$", YOU SHOULD UNDERSTAND THE"::"METHOD BY NOW. MAKE A LIST OF THE DATA":"8"(CHANGING THE UNITS OF ENERGY AND"::"TEMPERATURE AS REQUIRED). THEN WRITE":#B"DOWN THE APPROPRIATE RELATIONSHIPS."::"FINALLY, SOL"ANSWER IN DECIMAL FORM: ";E,!E.51050}!:(7):"OPPS! YOU'VE MADE AN ERROR."::"EFFICIENCY = W/Q1 = (Q1-Q2)/Q1":!:"TRY IT AGAIN. ENTER YOUR ANSWER IN"::"DECIMAL FORM. ";E:!::"EFFICIENCY = 0.50 = 50 %":::"$"THAT'S ALL THE: ";Q2 :Q22501000:i "INCORRECT. YOU SHOULD USE THE HEAT"::"FLOW/TEMP RELATION TO GET:":: " Q2 = (Q1*T2)/T1 "::" Q2 = (500*300)/600": :6::"Q2 = 250 KCAL"::24:1230!:3:"NOW FIND THE EFFICIENCY. ENTER YOUR"::::"* THE BASIC IDEA OF BOTH HEAT PUMPS &":n" ENGINES.":::"* THE MEANING & UNITS OF THE 5 FIXED":" VARIABLES (T1,T2,Q1,Q1,W).":24:1310:3:"* FOR HEAT ENGINES, EFF = W/Q1"::" HEAT PUMPS, K = Q2/W"::9 "* FOR BOTH:"PROCESSES:":\p" Q1 T1":" ---- = ----":" Q2 T2"::z"WRITE THESE DOWN!";:22:1310:3:"BOTH EQUATIONS ARE GOOD FOR BOTH HEAT"::"ENGINES AND PUMPS.":::."NOW LET'S REVIEW WHAT YOU NEED TO KNOW:":IONS.":`H"IT SHOULD BE OBVIOUS THAT THE WORK"::"IS EQUAL TO THE HEAT FLOW FROM THE HI-":R"TEMP SOURCE MINUS THE HEAT WHICH GOES"::"TO THE LO-TEMP SINK. THEREFORE:":\" W = Q1 - Q2"::f"IT'S NOT SO OBVIOUS, BUT FOR 'IDEAL'":20E*10:"THAT'S GREAT, "Z$"!"::"LET'S GO ON...";:31:1310:830410:"THAT'S NOT SO GOOD, "Z$"."::"PERHAPS YOU SHOULD WATCH THIS PROGRAM"::"AGAIN. TYPE 1 TO REVIEW, 2 TO GO ON: ";A:A1130>B$"Y"100::3:"LET'S GO ON TO THE MAIN EQUAT?";B$:B$"F"760FNN1::" NO. USE CALORIES OR JOULES."::770_:" VERY GOOD!":"3. K = W/Q2":" T OR F?";B$:B$"F"790 NN1::" WRONG. K = Q2/W"::800:" TERRIFIC!": 24:1310::4:"SCORE: "N" WRONG.":N08 (T) OR FALSE (F):":_N0:"1. USE EFFICIENCY FOR HEAT PUMPS.":" T OR F?";B$:B$"F"730NN1:(7):" SORRY. USE EFFICIENCY FOR HEAT":" ENGINES ONLY."::740:" EXCELLENT!":"2. Q2 IS MEASURED IN DEGREES KELVIN.":" T OR FR A ":6" GIVEN HEAT SUPPLY (Q1)"::"HEAT PUMP: K = AMOUNT OF HEAT"::" EXTRACTED (Q2) FOR A":" GIVEN AMOUNT OF WORK (W)":::24:1310:3:"LET'S SEE HOW YOU'RE DOING, "Z$"."::"ANSWRER TRUERECT":%v"ANSWERS!":24:20:1310:3:"USING THE WRONG EXPRESSION OR CONFUSING"::"THE TERMS IS THE BIGGEST SOURCE OF ":"ERROR. HERE IS A GOOD WAY TO REMEMBER:"::"HEAT ENGINE: EFF = AMOUNT OF WORK"::" PRODUCED (W) FO Q1":::CN"** FOR HEAT PUMPS ONLY:"::X" COEFFICIENT OF Q2":" PERFORMANCE = K = ----b" W"::l"IT'S VERY EASY TO GET THESE CONFUSED."::"DON'T DO IT, OR YOU'LL GET INCORUT OF"::"THE LO-TEMP SINK. IT'S EXACTLY THE SAME":\&"FOR HEAT ENGINES AND PUMPS."::0"NOW LET'S LOOK AT THE DIFFERENCE.":24:1310::3:"** FOR HEAT ENGINES ONLY:"::" W D" EFFICIENCY = EFF = ----":" SENT"::"ENERGY, THEY MAY BE EXPRESSED IN EITHER":z"CALORIES OR JOULES. JUST USE THE SAME"::"UNITS FOR EACH."::"CAUTION! THESE NOTATIONS DON'T CHANGE. "::"T1 IS ALWAYS THE HIGH TEMPERATURE, AND":5"Q2 IS ALWAYS THE HEAT FLOW IN OR OGH TEMP (KELVIN)":^" T2 - LOW TEMP (KELVIN)"::" Q1 - HEAT FLOW IN/OUT OF HI-TEMP":" Q2 - HEAT FLOW IN/OUT OF LO-TEMP"::" W - WORK EXTRACTED OR USED"::"COPY THESE NOTATIONS.":23:13104::"SINCE HEAT FLOWS AND WORK REPREP SOURCE (PUMP)."::dZ$", YOU SHOULD HAVE THE BASIC "::"IDEA BY NOW. LET'S GO ON.":24:1310:3:"IN ANY PROBLEM, YOU WILL HAVE TO WORK"::"WITH 6 VARIABLES. FIVE OF THESE ARE":"THE SAME FOR BOTH HEAT ENGINES & PUMPS:":::" T1 - HISED TO EXTRACT WORK"::" FROM HEAT FLOW."::"HEAT PUMPS - USED TO REMOVE HEAT"::" FROM A LO-TEMP SINK":" (REFRIGERATOR) OR TO"::" DELIVER HEAT TO A HI-":" TEM::"DIRECTION OF HEAT FLOW!":24:1310^|::1130:21:"THIS IS A HEAT PUMP. IT LOOKS JUST"LIKE THE HEAT ENGINE. BUT NOW WATCH":"THE HEAT FLOW.":1310:" HEAT FLOW FOR HEAT PUMPS.":W15:1250::24:1310<::3:"HEAT ENGINES - UAT"::"PUMP:"::kT"WE CAN USE WORK TO REVERSE THE HEAT"::"FLOW AND EXTRACT HEAT FROM THE LO-TEMP":^"SINK. THE EXTRACTED HEAT & WORK ARE"::"DELIVERED TOGETHER TO THE HI-TEMP ":h"SOURCE."::'r"LET'S LOOK AT IT. BE SURE TO NOTE THE"ERVED":::"** ELASTIC COLLISIONS --"::" MOMENTUM IS CONSERVED": " KINETIC ENERGY IS CONSERVED":::"LET'S LOOK AT AN EXAMPLE OF EACH.":24:1590 1320:1410 ::"QUESTION:"::"FOR WHICH TYPE OF COLLISION DO THE"::"BODIES STIC PAPER, PENCIL, AND"::"CALCULATOR READY, WE'LL BEGIN.":22:1590 :3:"TO BEGIN, YOU SHOULD KNOW THAT THERE"::"ARE 2 TYPES OF COLISIONS:":: "** INELASTIC COLLISIONS --"::" MOMENTUM IS CONSERVED":H " KINETIC ENERGY IS NOT CONSMA 6 Z1560::3:"HELLO! WHAT'S YOUR NAME? ";Z$:: d"WELL, "Z$", THIS PROGRAM WILL"::"HELP YOU SOLVE COLLISION PROBLEMS": n"IN ONE DIMENSION. YOU SHOULD ALSO WATCH:"::3::"2-DIMENSIONAL INELASTIC COLLISIONS":::B x"WHEN YOU HAVE YOUR*+  ONE-DIMENSIONAL COLLISION USING THE CONSERVATIVE LAWS TO ANALYZE ELASTIC AND INELASTIC COLLISIONS IN ONE-DIMENSION ( NSF LOCI PROJECT2 < SOUTH OKLAHOMA CITYF JUNIOR COLLEGE P OKLAHOMA CITY, OKLAHO               **************** SERVICES *9*d ****************26:27,2827:27,2829:29,28:28,30:29,31*x#27,3131:32,29:33,30:33,28:34,27:34,31:*#X1120:0:5,X:9,X:D1175::13:5,X:9,X: +#X2125:0:5,X:X12,X:D1175::13:5 HEAT ENGINES - METHOD"::" ***********************Y)255:A11000:::n)"";A$:)( ****************)2 * MODIFIED BY *)< * TOM McCORD *)F * EDUCATIONAL *)P * MEDIA *)Z * SERVICES **d *:5,X:D1175::13:5,X::K(X29261:0:5,X:D1175::13:5,X:(X25211:0:5,X:X12,X:D1175::13:5,X:X12,X:(X20111:0:5,X:9,X:D1175::13:5,X:9,X::(:100:8:" ***********************":@) " ,3124:22,2327:22,2331:27,3126:27,2827:27,2829:29,28:28,30:29,31'27,3131:32,29:33,30:33,28:34,27:34,31:'X1120:0:5,X:9,X:D1175::13:5,X:9,X:'X2125:0:5,X:X12,X:D1175::13:5,X:X12,X:(X2629:0617\&14,1728:14,1730:29,14:29,17:14,1632:16,1733:15,1634:16,1735:14,1636&13:X1120:4,10X::X2129:4,6X::7,1121:8,1222:9,1323:10,1324:11,1325&27,2915:29,3116:28,2917:29,3118:27,2919M'27,3121:27""%j1:X410:3,12X::5,917e%t2:X3036:3,12X::16,1813:18,14:17,15:16,16:16,1817%~0:4,65:6,95:5,99:4,631:32,355:8,1031:10,32:9,33:8,34:8,1035&12:20,217:20,2115:6,924:6,926:25,7:6,928:14,1724:25,2ENCY.";:20:1310i$B:7:"THAT'S ALL THERE IS TO THIS PROGRAM."::"WHEN YOU'RE READY FOR HELP WITH":$L"SOLVING PROBLEMS, LOOK AT:":::6::"HEAT ENGINES - APPLICATIONS.":::$V:"BYE FOR NOW, "Z$"!":20:1310::20:"END"%`(4)"RUN MENUW = 226.6 - 200"::" W = 26.6 CAL":y#$:"OF COURSE, WE COULD ALSO FIND THE"::"COEFFICIENT OF PERFORMANCE:":#." K = Q2/W"::" K = 200/26.6"::" K = 7.5 (NO UNITS)"::$8"SINCE THIS IS A HEAT PUMP, THERE IS "::"NO EFFICI = 25 + 273 = 298 K"::" T2 = -10 + 273 = 263 K":k"" Q1 = ?"::" Q2 = 200 CAL"::" W = ?"::""SINCE Q1/Q2 = T1/T2, FIND Q1."::" Q1/200 = 298/263"::" Q1 = 226.6 CAL":24:1310+#:3:"THEN, SINCE W = Q1 - Q2,"::" EES CELSIUS AND 25 DEGREES CELSIUS.":~!"HOW MUCH WORK MUST BE DONE TO EXTRACT"::"200 CALORIES OF THERMAL ENERGY FROM":!"THE LOW TEMPERATURE SINK?":20:"COPY DOWN THE PROBLEM.":24:1310!:3:"HERE'S HOW WE WOULD SOLVE IT:"::5"" T1, W = Q1 - Q2, AND"::" Q1/Q2 = T1/T2":: "THAT'S ALL THE THEORY YOU NEED TO"::"SOLVE PROBLEMS. JUST DON'T GET CONFUSED.": "NOW LET'S SOLVE A SIMPLE PROBLEM.":22:1310&!:3:"A HEAT PUMP IS TO OPERATE BETWEEN -10"::"DEGRTUM & CONSERVATION OF ENERGY."::"(MOST OF THESE PROBLEMS REQUIRE YOU TO":"SOLVE QUADRATIC EQUATIONS.)":::"HERE'S HOW TO FIND THE SOLUTION:":24:1590:2:"GIVEN:"::" M1 = 8 KG M2 = 2 KG":2" V1 = 2 M/S V2 = -2 M/S":2:"NOW WE'RE READY FOR ELASTIC COLLISIONS."::"THEY'RE MORE DIFFICULT SINCE THE BODIES":"HAVE DIFFERENT VELOCITIES AFTER THE"::"COLLISION."::"WITH 2 UNKNOWNS (V1' & V2'), WE NEED"::"TWO EQUATIONS. WE USE CONSERVATION OF":L"MOMEN"ANSWER: V(T) = +4 M/S"::::"IF YOU GOT THE RIGHT ANSWER, GO AHEAD.":"IF YOU DIDN'T GET IT RIGHT, CHECK YOUR"::"WORK. IF YOU WANT, REVIEW THE FIRST":"PART OF THIS PROGRAM.":::"ENTER 1 TO REVIEW, 2 TO CONTINUE: ";A:A1130[: M/S":::"FIND THE VELOCITY OF THE COMBINED ":l"BODIES. THEN, SEE IF YOU ARE RIGHT."::"ENTER YOUR ANSWER WITHOUT UNITS: ";A:::A4640v"SORRY, THAT'S NOT RIGHT. TRY AGAIN."::1590:650"GREAT WORK! LET'S GO ON..."::1590M:4:::::"BOTH MOVE TO THE LEFT BECAUSE THE":yD"MAGNITUDE OF M2'S MOMENTUM IS GREATER"::"THAN M1'S MOMENTUM. WATCH!!N24:1590:1360X::4:"HERE'S ONE FOR YOU TO WORK:"::" M1 = 8 KG V1 = 6 M/S":1b" M2 = 10 KG V2 = -45:"CONSERVATION OF MOMENTUM:"::1520:::1590&:3:"FOR INELASTIC COLLISIONS ONLY:"::" M1*V1 + M2*V2 = (M1+M2)*V(T)":0"THEN:"::" 10*2 + 5*(-6) = (10+5)*V(T)"::" 20 + (-30) = 15*V(T)":+:7::"V(T) = -10/15 = -.67 M/S": = 5 KG":`" V1 = 2 M/S V2 = -6 M/S":::"NOTE THAT M2 IS MOVING TO THE LEFT."::"AS ALWAYS, WRITE DOWN THE DATA. THEN"::"USE THE LAW OF CONSERVATION OF ":"MOMENTUM, AND TRY TO SOLVE THE PROBLEM"::"BEFORE YOU ";:15901:PUT IN THE DATA:"::" 10*3 + 5*0 = (10+5)*V(T)":" 30 + 0 = (15)*V(T)"::7::"V(T) = 30/15 = 2 M/S":::"THAT'S IT! NOTICE THE POSITIVE VELOCITY.":1590:1320 ::4:"HERE'S ANOTHER PROBLEM:"::" M1 = 10 KG M2:"SYSTEM BEFORE THE COLLISION IS EQUAL":t"TO THE MOMENTUM OF THE SYSTEM AFTER "::"THE COLLISION.":22:1590:2:"FOR INELASTIC COLLISIONS,"::" V1' = V2' = V(T)":"THUS:"::" M1*V1 + M2*V2 = (M1+M2)*V(T)"::1590:9"2= 0":::"NOTICE THAT BODY #2 IS STATIONARY.":Z:"WRITE DOWN THIS DATA, "Z$"."::"NOW, LET'S LOOK AT IT.":::1590:1320::4:"SOLUTION:"::"* LAW OF CONSERVATION OF MOMENTUM *"::1520*:"THIS MEANS THAT THE MOMENTUM OF THE":W WE'RE READY TO GO. WE'LL DO AN"::"INELASTIC PROBLEM FIRST (THEY'RE EASY).":r"ONLY MOMENTUM IS CONSERVED, AND THE"::"VELOCITIES AFTER COLLISION ARE EQUAL.":24:1590|:4:"GIVEN:"::" M1 = 10 KG M2 = 5 KG":1" V1 = 3 M/S VSION)":::"BE SURE YOU UNDERSTAND THE NOTATION!":24:1590T:4:"REMEMBER THAT VELOCITY & MOMENTUM ARE"::"VECTORS! TO AVOID ERRORS, LET'S AGREE":^"THAT MOTION TO THE RIGHT IS POSITIVE &"::"AND MOTION TO THE LEFT IS NEGATIVE."::Ph"NOCOMBINATION":a," P(T) = COMBINED MOMENTUM":::"AND, TO SHOW THE VELOCITIES BEFORE AND":6"AFTER THE COLLISION:":::" V1 = VELOCITY OF BODY #1@" (BEFORE COLLISION)"::" V1' = VELOCITY OF BODY #1<J" (AFTER COLLIVECTOR)"::" KE = KINETIC ENERGY"::"TO SHOW THE DIFFERENT BODIES, WE'LL"::"USE NUMBERS:":::" M1 = MASS OF BODY #1":" M2 = MASS OF BODY #2":23:1590":2:"FOR INELASTIC COLLISIONS, WE'LL USE:":::" V(T) = VELOCITY OF THE OST":N "WHEN THE BODIES STICK TOGETHER IN"::"INELASTIC COLLISIONS.":: "NOW LET'S SEE ABOUT SOLVING PROBLEMS.":23:1590 :2:"HERE'S THE NOTATION WE'LL USE:"::" " M = MASS"::" V = VELOCITY (VECTOR)":'" P = MOMENTUM (K TOGETHER?":\ " 1. INELASTIC 2. ELASTIC":::"ENTER 1 OR 2:";A::::A1210 (7):"SORRY, "Z$". YOR'RE NOT PAYING"::"CLOSE ATTENTION. LOOK AGAIN."::1590:170 "EXCELLENT, "Z$". BE SURE TO"::"REMEMBER THAT KINETIC ENERGY IS LES IT'S DIRECTION!"::n+ :"IF YOU NEED MORE HELP, SEE YOUR"::"INSTRUCTOR--THAT'S WHAT HE'S HERE FOR.":+"BYE FOR NOW, "Z$".":24:1590::22:"END"+(4)"RUN MENU"+(::21:" ONE-DIMENSIONAL INELASTIC COLLISION":2:10,1120B,2X01"GIVEN:"::" M1 = 6 KG M2 = 2 KG"::" V1 = 1 M/S V2 = -1 M/S":z*:"WHEN YOU HAVE AN ANSWER, ";:1590*:3::"ANSWERS:":::" V1' = 0"::" V2' = 2 M/S"::+"SO, M1 COMES TO REST IN THE COLLISION"::"AND M2 REVERSEW"::"THIS LAST SOLUTION."::T)"ENTER 1 TO REVIEW, 2 TO CONTINUE:";A:A1970):5:"THAT'S ABOUT ALL, "Z$".":::"HERE'S ONE MORE FOR YOU TO DO. WORK":)"THE SOLUTION BY YOURSELF. DON'T BE"::"ALARMED YOUR ANSWER IS SURPRISING!":P*:,"::" V2' = 1 - 2*V1'":\(" V2' = 1 - 2*(-1/3)"::7::"V2' = 1.67 M/S"::("SO, M2 MOVES OPPOSITE M1 AND AT A"::"FASTER RATE."::("LET'S SEE WHAT IT LOOKS LIKE.":22:1590:1410: ):5:"AGAIN, YOU HAVE THE CHANCE TO REVI't:4:"FACTORING:"::" (3*V1' + 1)(V1' - 1) = 0":'~"SO,"::" V1' = 1 (INITIAL VELOCITY)"::5::"V1' = -.33 M/S"::'"NOTICE THAT M1 MOVES TO THE LEFT.":::"NOW YOU SOLVE FOR V2'.":22:"ENTER YOUR ANSWER: ";A(:5:"FROM ABOVE24:1590Y&V:4:"IF YOU'VE DONE IT RIGHT:"::" 3 = 2*V1'^2 + 1 - 4*V1' + 4*V1'^2"::&`:"GROUP THE TERMS & DIVIDE BY 2:"::" 3*V1'^2 - 2*V1' - 1 = 0":::&j"FACTOR THIS EXPRESSION TO FIND THE"::"ANSWER. THEN CHECK YOUR WORK.":22:15909+ 1/2*1*(-1)^2 ="::" 1/2*2*V1'^2 + 1/2*1*V2'^2"::1590%8:"THEN,"::" 1 + 1/2 = V1'^2 + 1/2*V2'^2"::" 3/2 = V1'^2 + 1/2*V2'^2"::%B"MULTIPLY BY 2:"::" 3 = 2*V1'^2 + V2'^2":&L:"NOW ADD THE EXPRESSION FOR V2'^2.":1590::"ALSO, FIND V2'^2. WORK IT OUT, THEN":23:1590$:3:"YOU SHOULD HAVE:"::" V2'^2 = 1 - 4*V1' + 4*V1'^2"::1590$$::"NOW APPLY THE SECOND CONDITION:"::1530::"ADD THE DATA & CHECK.":23:1590@%.:2:"YOU GET:"::" 1/2*2*(1)^2 LAW. WRITE OUT THE"::"LAW, THEN CHECK YOURSELF.":24:1590z#:2:1520::"GET IT RIGHT?":::"NOW ADD THE DATA:":#" 2*1 + 1*(-1) = 2*V1' + 1*V2'"::" 2 - 1 = 2*V1' + 1*V2'":#" "(91)"V2' = 1 - 2*V1']"::7$NUE, ENTER 2: ";A:A1680W":4:"HOW ABOUT ANOTHER EXAMPLE, "Z$"?":::"GIVEN:":"" M1 = 2 KG M2 = 1 KG"::" V1 = 1 M/S V2 = -1 M/S":""COPY THE DATA.":::"AGAIN, WE'LL START WITH THE CONSER-":=#"VATION OF MOMENTUM"V2' = 4.4 M/S":::"SO M2 ALSO MOVES TO THE RIGHT."::i!"LET'S LOOK AT IT, "Z$"!":24:1590:1470!::6:"THIS PROBLEM IS FAIRLY COMPLEX. IF"::"YOU WOULD LIKE TO GO THROUGH THE":""SOLUTION AGAIN, ENTER 1. IF YOU ARE"::"READY TO CONTI*V1' - 2)(V1' - 2) = 0"::e "THUS, V1' = 2 (INITIAL VELOCITY)"::10::"V1' = 0.4 M/S"::: "WE SEE THAT V1' IS 0.4 M/S TO THE RIGHT.":24:1590::4:"TO FIND V2':": " V2' = 6 - 4*V1'"::" V2' = 6 - 4*(0.4)"::!9:: V2'^2:"::" 20 = 4*V1'^2 + 36 - 48*V1' + 16*V1'^2":f::"GROUP TERMS & DIVIDE BY 4:"::" 20*V1'^2 - 48*V1' + 16 = 0"::" 5*V1'^2 - 12*V1' + 4 = 0":p24:"SEE IF YOU CAN FACTOR THIS. ";:1590 z:6:"IF YOUR ALGEBRA IS OK,"::" (5NDITION:":::1530641590::"NOW ADD THE DATA:":>" 1/2*8*(2)^2 + 1/2*2*(-2)^2 ="::" 1/2*8*V1'^2 + 1/2*2*V2'^2":H23:1590R:2:" 16 + 4 = 4*V1'^2 + V2'^2"::"THEN:"::" 20 = 4*V1'^2 + V2'^2":6\::"NOW SUBSTITUTE = 8*V1' + 2*V2'"::"DIVIDE BY 2 AND REARRANGE:":" "(91)" V2' = 6 - 4*V1']":::"ALSO, LET'S FIND V2'^2:": " V2'^2 = 36 - 48*V1' + 16*V1'^2":::"WRITE THIS DOWN AS WE GO. ";:1590*:2:"NOW WE APPLY THE SECOND CO::"COPY THE DATA. THEN START AS USUAL."::1590b::"* CONSERVATION OF MOMENTUM *"::1520::"NOW PUT IN THE DATA AND ";:1590:2:1520::" 8*2 + 2*(-2) = 8*V1' + 2*V2'"::" 16 - 4 = 8*V1' + 2*V2'":7 " 12 CEMENT":5" - T PERIOD OF ROTATION":20:720bLL1::570:640:650:660:670:23:720:::"HERE ARE THE STEPS TO FOLLOW IN"::"SOLVING A SHM PROBLEM. COPY THEM:":79"1. IDENTIFY THE MASS OF THE SHM BODY,"::" THE AMPLITUDE (1/2"HERE'S THE NOTATION WE'LL USE:":5:" - V(RB) VELOCITY OF THE ROTATING BODY": " - A(RB) ACELERATION OF THE ROT. BODY"::" - V(O) VELOCITY OF THE SHM BODY": " - A(O) ACCELERATION OF THE SHM BODY"::" - R RADIUS"::" - X SHM DISPLATING BODY: V(RB) & A(RB)."::"** FIND THE VELOCITY & ACCELERATION": " OF THE SHM BODY BY TAKING THE"::" COMPONENTS OF V(RB) & A(RB) WHICH": " ARE PARALLEL TO THE PATH OF THE SHM.":::"WRITE DOWN THIS BASIC IDEA. THEN"::720P ::":N x"WITH THE ROTATING BODY (RB). THAT'S"::"THE TRICK TO THIS ANALYSIS! LL4:20:720:21:" REFERENCE CIRCLE SHM":570:23:720 ::4:" + THE MAIN IDEA +":::"** FIND THE VELOCITY & ACCELERATION":E " OF THE ROTA WE BEGIN THE ANALYSIS,":[ Z"LET'S LOOK AT AN EXAMPLE OF THE MOTION.":::::720::4 d"REMEMBER THAT WE ARE TRYING TO "::"ANALYZE THE MOTION OF THE BODY WHICH": n"IS OSCILLATING IN SIMPLE HARMONIC"::"MOTION. BUT NOTICE HOW IT KEEPS PACEP YOU SOLVE SIMPLE":g <"HARMONIC MOTION PROBLEMS. BY USING"::"A TECHNIQUE CALLED THE REFERENCE": F"CIRCLE, YOU CAN DETERMINE VELOCITY,"::"ACCELERATION, FORCE, ENERGY, ETC. FOR": P"A BODY TRAVELLING WITH SIMPLE HARMONIC"::"MOTION. BEFORE - SHM & THE REF. CIRCLEP(4)"BLOAD SHAPE1":232,0:233,64:690::6"** YOU WILL NEED PAPER, A PENCIL AND"::" YOUR CALCULATOR FOR THIS LESSON. **":13("HI! WHAT'S YOUR NAME? ";Z$:: 2Z$", THIS PROGRAM IS"::"DESIGNED TO HEL          * SERVICES *)3| ***************** *G3| **************** (7)3 X06.25:12:10,11154X:1:10,1115X:10,1114X:T135::0:10,11154X:10,1115X:10,1114X:4 23:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A18300:*********":42"" ONE-DIMENSIONAL COLLISIONSs2,:" ************************":255:A11000:::26"";A$:2@ ****************2J * MODIFIED BY *2T * TOM McCORD *2^ * EDUCATIONAL *2h * MEDIA *3rP2'"::" M1*V1 + M2*V2 = M1*V1' + M2*V2'"::1"* CONSERVATION OF KINETIC ENERGY *"::" KE1 + KE2 = KE1' + KE2'":1" 1/2*M1*V1^2 + 1/2*M2*V2^2 ="::" 1/2*M1*V1'^2 + 1/2*M2*V2'^2"::1 2:100:8:" ***************0:10,11X:10,11X1:10,1120X2:/0(7)0X06.25:12:10,11154X:1:10,1115X:10,1114X:T135::0:10,11154X:10,1115X:10,1114X:023:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A11470:11" P1 + P2 = P1' + X:T135::0:10,11154X:10,1116X:10,1115X:/15,1612:15,169:23:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A11410//::21:" ONE-DIMENSIONAL ELASTIC COLLISION%0X013:1:10,11X:10,11X1:12:10,1120X2:T140::SEE AGAIN, 2 TO CONTINUE?";A:A11360:Z.::21:" ONE-DIMENSIONAL ELASTIC COLLISION.X014:1:10,11X:10,11X1:12:10,1130X:T140::0:10,11X:10,11X1:10,1130X:.(7)7/X06.25:12:10,11154X:1:10,1116X:10,1115NAL INELASTIC COLLISION|-ZX19:2:10,11X:10,11X1:12:10,11383X:T140::0:10,11X:10,11X1:10,11383X:-d(7)-nX801:12:10,11X2:2:10,11X1:10,11X:T1150::0:10,11X2:10,11X1:10,1