9106"xX` L4x1X` a"xGX` LXxU䅑X` "xhhhhhhhhhhhHHX` "hhhhhhh` "Ȅ؅ 񷊢ȑ` "ȱ` #UW0UU,15(UC,0:UN,0:UR$"":UI$"":UL0UL39$-UI$:UI$""45Y2UA(UI$):UR(UR$):UA13UR0UC,1:" ";:707UA20(UR0)45:UR$(UR$,UR1):" ";:45<URUL45A95:UR$UR$UI$:UI$;:45FUR$UX$" ";:145KUIUR11:(UR$,UI,1)" " PUR$(UR$,UI)::_(US$)0110JdUI1(US$):UI$(US$,UI,1):UI$""Pi]nUA46psUA32UR0xUA47UA58}UA64UA91UA193UA218UI$"":(UT)UU%UI$(UA128)170:251,0:252,US256:253,0:254,UF:UVUZ1:UV%(UT):300bUZ0:US$UJ$:UG$UK$:UT$UL$:UPUJ:ULUK:"";:UW0UY,6:UX,14UT,UV%:251,0:252,UF:253,0:254,US256:UV:UW0UU,15U1,U1%:U2,U2%:U3,U3%1:U4,U4%:35U1%(U1):U2%(U2):U3%(U3):U4%(U4)!UJ$US$:UK$UG$:UL$UT$:UJUP:UKUL:F,"":UT,UU%:UW0UY,12:UX,6f1" M E N U"6" --------------------------------";" 1. BEGIN MOMENTUM"@" 2. STUDENT INSTRUCTIONS" E" 3. TEACHER INSTRUCTIONS"$ J" 4. QUIT"P O" --------------------------------"{ T" PRESS A NUMBER KEY (1-4) "; YUL1:US$"1234" [35:US$"" ^(UR$)1000,400,500,390 c300 rUZ1UP%:155 w UE:"":UW0UY,6:UX,14:UU,14 :3 "":UT,UL%:UW0UY,11:UX,12i UT$" OMENTUM":UH1:UP0:UG$"TUDENT NST":850 " HIS PROGRAM WILL TEST YOU ON MOMENTUM" " QUESTIONS. OU WILL NEED A CALCULATOR " " AND A PENCIL. RITE DOWN YOUR ANSWER "' " WITH THE STEPS YOU USED TO REACH YOUR"U " ANSWER AND HAND IT IN TO THE TEACHER"q " WHEN YOU ARE DONE." " F YOU WANT TO RETURN TO THE MENU" " AFTER YOU HAVE RUN THE PROGRAM PRESS H" " FOR HELP." 835:850 " HEN ENTERING ANSWERS, YOU MAY"I " ROUND OFF YOUR ANSWERS TO ONE DECIMAL"Y " PLACE."l " XAMPLE:" " IVEN: =45 M=8" " HAT IS THE ACCELERATION OF THE OBJECT" " ALONG A HORIZONTAL, FRICTIONLESS PLANE?"; " NSWER = 5.6"+ " SING THE FORMULA =MA TO ARRIVE AT"S " A = /M ... A=45/8 = 5.6 M/S2"\ 835b  "":UT,UL%:UW0UY,2:UX,10 UT$" OMENTUM":UH1:UP0:UG$"EACHER NST":850 " O THE BEST OF OUR KNOWLEDGE,"" THIS PROGRAM IS IN THE PUBLIC"?" DOMAIN. F THIS IS NOT TRUE,"]" PLEASE CONTACT:"" ŠӠ"" 3370 HARMACY VE." GINCOURT, NTARIO"" ANADA, 1 24"" (416) 499-4292"%UH1:835:850J" ROGRAM AME: OMENTUMk" RITTEN BY : OSEN" EVISED BY : AURICE ELDHUIS"" HIS PROGRAM WILL RUN ON ANY" à2.0 OMPUTER" 4.0 ԠOMPUTER5" 8032 OMPUTERZ" OMMODORE 64 OMPUTERh 835:850" FTER YOU TEACH YOUR STUDENTS HOW TO"" SOLVE MOMENTUM PROBLEMS, THIS PROGRAM"" WILL TEST THEM WITH 11 PROBLEMS." " ACH STUDENT SHOULD HAVE A CALCULATOR"N" AND A PENCIL SO THAT THEY CAN WRITE "" THEIR SOLUTIONS TO THE PROBLEM.":835:850" HIS PROGRAM IS DESIGNED TO TEST" " THE STUDENT'S KNOWLEDGE OF THE" "" FORMULAS. HIS PROGRAM DOES NOT ALLOW"2$" ENTRY OF ANSWERS IN SCIENTIFIC"`&" NOTATION. NSWERS ARE TO BE ROUNDED"(" TO ONE DECIMAL PLACE."I835NUZ1UP%:155SUC$"BEFORE RUNNING THIS PROGRAM ON AN 8032" LOAD THE 'CBM 4032' PROGRAM":6 "";:UT,UU%:UW0UY,1:UX,14L!I138:" ";:d"I122:" ";:|#I137:" ";:$I122:" ";:%" &"  '"  O M M O D O R E(" 6)"EDUCATIONALV*"SOFTWAREt+"1983,(UD$,24);"PRESS SPACE BAR OR WAIT"UC$:UI0-UI$:UI$" "UIUI1:UI250813.300CUN,0:(UD$,25);"PRESS SPACE BAR TO CONTINUE";3HUI$:UI$" "8409MkRUPUP1:""(UT$UP$,21);(UP$UG$(UP),18)W" --------------------------------------"\UH0(UD$,25);(UP$,7);"HELP - PRESS H ";aUH0:"":k(UD$,24);" ONE MOMENT PLEASE... ":nUX$"H":US32768:U1196:U2197:U3198:U4216:UC167:UB1024:UD1084:UE1105UT59468:UU%12:UL%14:UP%1158:UN158:50003,0:UW(50003):UF(53)UW0US1024:U1209:U2210:U3211:U4214:UC204:UB2048:UX53280:UD2064OUW0UT53272:UU%21:UL%23:UP%2151:UN198:UY53281:UU2199:UE2087{UC$(155):UV1214:UW0UF196:UV2238UD$"":UP$" "UT$" ": PROGRAM TITLE900:"":"": SHIFTED SPACE/(US40)96700U((0)(((0))0)65536)500935UC$"THIS MACHINE DOES NOT HAVE ENOUGH""MEMORY TO RUN THIS PROGRAM !":UE:""UC$:UW0965UF60UF108UF12495052,0:53,UF4::9003UW1601103,88:1104,228:1107,85:1111,228:965pUW1"THIS PGM WILL NOT WORK ON THIS MACHINE!":UE:1103,49:1104,230:1107,46:1111,230:965UD800:300UZ1UP%:UP%:UZ0""UC$;::UW0UY,6:UX,14$V UT,UL%#`XX(((TI),2)):XX21120AjY(0):I1XX:Y(XX):IWL$,E$,P$,M$,F$,G("METERS","JOULES","KG M/S2","KG","NEWTONS",9.8-FF$"":GG$"":HH$""2M(10(TI)2):M21330<U(15(TI)2): U21340FA(10(TI)1.2):A1.51350PF(10MA)10/ZT(10(TI))2.5FdV(10(UAT))10`nS1((UV)10T2)10x" N OBJECT OF MASS M =";M;M$;" IS MOVING"" ALONG A HORIZONTAL FRICTIONLESS SURFACE"" WITH AN INITIAL SPEED VI =";U;"M/S."&" T IS ACTED UPON BY A HORIZONTAL FORCE"N" =";F;F$;" AND IS OBSERVED TO"|" TRAVEL FOR A TIME OF T =";T;"S.":2830 1470:1500"":860:" : M =";M;M$;" VI =";U;"M/S"" =";F;" T =";T;"S"" "FF$:GG$:HH$:" 1. HAT IS THE OBJECT'S";" ACCELERATION ? ";M3000:SA$UR$i(SA$)160"":1510CA:S(SA$):1470:2310:P01580P111560" O FIND A, MERELY USE: =MA ":P11:1500" INCE=MA, WE HAVE:""" A = /M = (";F;") / (";M;"KG)"3," A = ";A;" M/S2"O6P0P02P1P:P10:2830t@FF$"OUND: A ="(A)" M/S2"J 1470T" 2. HAT IS OBJECT'S FINAL SPEED? ";Y3000:SV$UR$^(SV$)160"":1620hCV:S(SV$):1470:2310:P01720rP111660,1680,1700K|" O FIND VF, MERELY USE DEFINITION"" OF ACCELERATION: A= (VF - VI)/T O":P11:1620" ROM A = (VF - VI)/T WE HAVE:"" VF= VI+ AT O:":P12:1620 " VF= VI + AT OR"- " VF =";U;" + (";A;" *";T;")"F " VF =";V;"M/S" 2830:P0P02P1P:P10:FF$FF$" VF ="(V)" M/S":1470 " 3. URING THE";T;"SECONDS, HOW FAR" " DOES THE OBJECT MOVE ? "; 3000:SD$UR$! (SD$)160"":1750.!CS1:S(SD$):1470:2310:P01840=!P111820\!" O FIND S, USE EITHER:"!" S= V * T WHERE V = AVERAGE SPEED OR:"!" S = VIT + 0.5AT2 O:":P11:1740!" ROM S = VT WITH V = (VI+ VF)/2")"&" WE HAVE: S = ( (VI+ VF)/2 ) * T)"X"0" S =";S1;L$;"":P0P02P1P:2830:P10":GG$"  S ="(S1)" M":1470"D" 4. HAT IS THE OBJECT'S INITIAL""N" INEAR MOMENTUM ? ";"S3000:SP$UR$"X(SP$)160"":18700#bLI(10MU)10:CLI:S(SP$):1470:2310:P01950?#lP111930]#v" ECALL: P = MV O"#" PI= MVI : O:":P11:1860#" PI = MVI OR:"#" PI =";M;" * ";U;""#" PI =";LI;P$:P0P02P1P:P10$GG$GG$" PI ="(LI)" KG M/S":835:1470@$" 5. HAT IS THE OBJECT'S"f$" FINAL LINEAR MOMENTUM ? ";x$3000:SF$UR$$(SF$)160"":1980$LF(10MV)10:CLF:S(SF$):1470:2310:P02050$P112030 %" P = MV O: PF = MVF O:":P11:19704%" INCE PF = MVF WE HAVE:"U%" PF =";M;" * ";V;" OR"%" PF =";LF;"KG M/S":P0P02P1:P10:2830% HH$"   PF ="(LF)" KG M/S":1470%" 6. HAT IS THE MAGNITUDE OF OBJECT'S"& " CHANGE IN LINEAR MOMENTUM ? ";$&%3000:SD$UR$@&*(SD$)160"":2080k&4CLFLI:S(SD$):1470:2310:P02150z&>P112130&H" HANGE IN MOMENTUM = P = PF- PI O:":P11:2070&R" P = PF - PI"&\" P =";LF;" - ";LI;""'f" P =";LFLI;"KG M/S"V'pP0P02P1P:2830:HH$" P ="(LFLI)" KG M/S"`'u1470i'zP10'" 7. HAT IS THE IMPULSE PRODUCED BY THE"'" FORCE ACTING ON THE OBJECT ? ";'3000:SI$UR$'(SI$)160"":2190 (IIFT:CII:S(SI$):1470:2310:P02260/(P112240^(" MPULSE = P = T O:":P11:2180(" INCE P = T WE HAVE:"(" P =";F;" * ";T;" OR:";(" P =";II;"S"(2830:P0P02P1P:P10:1470:2350 ) P0:P40.04:(SC)P4(C)2330 ) " ":23407) " NCORRECT":P1=)$ e). " 8. HAT IS THE OBJECT'S INITIAL")8 " KINETIC ENERGY ? ";)= 3000:SK$UR$)B (SK$)160"":2360)L KI(.5MU2):CKI:S(SK$):1470:2310:P02430)V P112410"*` " K = 0.5MV2 O:":P11:2350>*j " K = 0.5MVI2"g*t " K = 0.5 * ";M;"* (";U;") 2"*~ " K =";KI;E$:P0P02PP1:P10:2830:GG$"":HH$""* FF$": S ="(S1)" M VF ="(V)" M/S"+ GG$" K ="(KI)" J":1470.+ " 9. HAT IS THE OBJECT'S FINAL"M+ " KINETIC ENERGY ? ";_+ 3000:SL$UR${+ (SL$)160"":2470+ KF(.5MV2):CKF:S(SL$):1470:2310:P02540+ P112520+ " K =0.5MV2 O:":P11:2460 , " K' = 0.5MVF2 O:"9, " K' = 0.5 *";M;" * (";V;") 2"i, " K' =";KF;"JOULES":P0P02P1P:P10, GG$GG$" K' ="(KF)" J":2830:1470, " 10. HAT IS THE OBJECT'S CHANGE IN", " KINETIC ENERGY ? ";, 3000:SM$UR$- (SM$)160"":2570C- CKFKI:S(SM$):1470:2310:P02640R-( P112620-2 " HANGE IN K = K = K' - K O:":P11:2560-< " K = K' - K OR"-F " K =";KF;" - ";KI;"" .P " K =";KFKI;"JOULES":2830:P0P02P1P:P10<.Z GG$"   K ="(KFKI)" ":1470k.d " 11. OW MUCH WORK IS DONE BY THE FORCE".n " ON THE OBJECT ? ";.s 3000:SW$UR$.x (SW$)160"":2670. CKFKI:S(SW$):1470:2310:P02760. P112740/ " ORK = = *S*COS(,S) BUT THE"H/ " ANGLE BETWEEN & S = 0, SO WE HAVE:"y/ " נ= *S*COS(0) = *S O:":P11:2660/ " = *S "/ " =";F;" * ";S1;" OR"/ " =";KFKI;"JOULES":2830/ GG$GG$" ="(KFKI)" "0 P0P02PP1: 147050 " BSERVE: HE WORK DONE BY THE FORCE"d0 " IS EQUAL TO THE CHANGE IN THE OBJECT'S"0 " KINETIC ENERGY I.E. נ= K. HIS IS"0 " THE ORK-NERGY HEOREM.":2830:T022:28800 0 83506 "": 1@ " () OMENTUM & NERGY"1J Q(100P0T0)?1T " Ԡ =";Q;"% ";c1^ Q100" Ԡ":29701h Q90" ̠":29701r Q80" ":29701| Q70" נ":29701 Q60" -Π":T29702 " -"2 8352 P00:P10:@2 US$"0123456789.H":UL10:35F2 k2a" ROGRAM AME: OMENTUM2b" RITTEN Y : OSEN