XF.14a 205HPHYS-205(H14) (Kaldon-15032) Name ________S O L U T I O N_______WMU-Spring 2004 Final Exam - 200,000 points + 40,000 ¶ points Check-Out: Q X T HNRS-290___04/18/2004•Rev.8aState Any Assumptions You Need To Make – Show All Work – Circle Any Final AnswersUse Your Time Wisely – Work on What You Can – Be Sure to Write Down EquationsFeel Free to Ask Any Questions¶2a ¶2b ¶2c ¶2eFINAL EXAM [FORM - A]PHYS-205H (KALDON-14)SPRING 2004WMUCSI : Miami in Helsinki, Finland (!)Physics 205H/ Final Exam [Form-A] Spring 2004 Page 2As The World Turns (50,000 points)1.) (a) The Earth has a mass of 5.98 × 1024 kg and an average radius of 6,370,000 meters. Find theaverage mass-to-volume ratio of the Earth.Vr m mmVkgmkg m== =´==´´=43343321 32421 336 370 000 1083 10598 101083 105522ppr,, .../bg(b) Find the moment of inertia, I, of the Earth. Assume it is a uniform solid.IMR kgmkg msolid sphere==´=´ ×2522524 237 298 10 6 370 0009706 10(5. )( , , ).(c) Find the period (T), frequency (f) and the angular velocity (w) in SI units of the Earth as it spins on itsaxis.T hours hourfTHzfHzrad==== === =24 0 3600 86 4001186 4000 000011572 2 0 00001157 0 00007270. ( sec/ ) , sec,sec.../secwp pbg(d) Find the angular momentum (L) and kinetic energy (Krot) of the Earth as it spins on its axis.LI kgm radkg mKI kgm radJrot== ´ ×=´ ×== ´ ×=´ww9 706 10 0 000072707 056 109 706 10 0 000072702 565 1037 237 21221237 2229../sec./sec../sec.chbgchbgPhysics 205H/ Final Exam [Form-A] Spring 2004 Page 3(e) The Earth is located an average distance of 1.496 × 1011 meters from the Sun, which has a mass of1.991 × 1030 kg. Starting from Newton’s Law of Universal Gravity, find the tangential speed, v, of theEarth as it goes in Uniform Circular Motion in orbit around the Sun. G = 6.67×10 -11 N·m²/kg²FGmMrmamvrGMrvrvGMrrvGMrNm kg kgmmsGc=======´× ´´=-22222211 2 2 3011667 10 1991 101496 1029 790;./..,/chchchPhysics 205H/ Final Exam [Form-A] Spring 2004 Page 4Survivor: All-Star Points (50,000 points)2.) ¶(a) For the following x(t) find the acceleration as a function of time. All other constants ofintegration are zero.xt m m st m s t() . ( . / ) . /=+ +100 200 4001222chxt m m st m s tvdxdtddtmmst mstms ms t ms tadxdtdvdtddtms t ms() . ( . / ) . /.(./) ./(./)()./ ././ ./=+ +== + +=+ + ==== =100 200 400100 200 4000 2 00 2 4 00 4 00400 4001222122212222222chchejchchchej¶(b) For a () . /t rad s= 4002, find the angle as a function of time. All other constants of integration arezero.waqw== = +=== = +=zzzzd t rad s d t rad s t Crad s tdt rad s tdt rad s t Crad s t400 400400400 4002002222122222./ ./././ ././chchchchchch¶(c) A plate of mass m = 5.00 kg has dimensions a = 1.00 m and b = 0.480 m.Find the center of mass coordinates xcm by integrating xMxdmcm=z1. Theorigin is located at the lower left corner.lllll== = === ==FHGIKJ=-FHGIKJ== =zz zMLkgmmadm dxxMxdmMxdxMxdxMxmamaammcmaaa5001001122021002050000202..;..Physics 205H/ Final Exam [Form-A] Spring 2004 Page 5(d) A 3.00 meter long teeter-totter has a 25.0 kg kid on the left end and a 40.0 kg kid on the right end. Thepivot is located a distance d from the left end. Find d. Ignore the weight of the teeter-totter itself.tå=- -==-=-+=+==+=+=mgd mg L dmgd mg L dmd m L mdmd md m LmmdmLdmLmmkg mkg kgm121212212 212 2212040 0 30025 0 40 01846()()()().....bgbgbg¶(e) A mass on a spring has its position as a function of time given as xt A t B t() cos( ) sin( )=+ww.Find the speed v and acceleration a at time t = 0.xt A t B tvdxdtddtAtBt AtBtvA B BBadvdtddtAtB t A t BtaA B AA() cos( ) sin( )cos( ) sin( ) sin( ) cos( )( ) sin( ) cos( )sin( ) cos( ) cos( ) sin( )() cos() sin()=+== + =- +=- + = + === - + =- -=- - =- - =-wwwwwwwwww wwwww w w ww www wwbgbg000000002222 22Physics 205H/ Final Exam [Form-A] Spring 2004 Page 6CSI: Crime Scene Investigators (50,000 points)3.) On April 8th, 2004, CBS had a rerun of a CSI episode. The “B-story” was about an idiot target shooting a 9mmautomatic pistol in his backyard. He was “distracted” when the neighbor next door yelled at him to stop – and oneshot went high. A woman several blocks away was killed when the wayward bullet penetrated 4". Forensicanalysis leads them to determine that a 9mm bullet penetrating 4" is traveling at only 550 ft/sec (168 m/s), while afull-load shot at 1100 ft/sec (336 m/s) will penetrate 12". “How far does a 9mm bullet have to travel to slowdown from 1100 ft/sec to 550 ft/sec?” “1800 feet.” (a) Find the constant acceleration a that will slow abullet from 336 m/s to 168 m/s in 1800 feet (549 meters).†vv axx xvv axax v vavvaxms msmms2020020220220222220222168 336254977 11=+ - ==+=-=-=-=-();//./bgbgbg “Can a 9mm bullet even travel 1800 feet? That’s three football fields.” “600 feet to drop to ground iffired horizontally. But it could have been fired at an angle.” (b) Find the time t it takes for a bullettraveling at an average speed of 252 m/s to fall from a horizontal gun held 1.50 meters above the ground.yy vt gt y vygtygttygmmsyy=+ - = ==-=== =00122001220122020002215098105530;,(. )./.sec(c) Find the horizontal distance d traveled for a bullet traveling at an average horizontal speed of 252 m/sfor the time t from part (b). If you didn’t get a time in (b), use t = 1.50 seconds.dvt ms m== =252 0 5530 139 4/. sec .bgb g(d) Modern electronics allow one to measure the muzzle speed of a bullet as it leaves the barrel of a gun.In the “old days”, however, one used a ballistic pendulum – basically a block of wood hanging by a cable. † Actually, the bullet would be slowed by the high-speed drag force, F = -cv², but you do not want to work this problem with avariable, velocity dependent force. (grin)Physics 205H/ Final Exam [Form-A] Spring 2004 Page 7Suppose a 100. kg cube-shaped block of wood (r = 650. kg/m³) is suspended from the ceiling by a cable1.75 meters long. Find the period T of this simple pendulum if it is displaced slightly from rest (nofirearms involved).But wait – the length of the cable is to the center-of-mass of the bob…so a better answer is:rrwwpppp=== ==== ==+ = + ====== = =mVVmkgkg mmVd d V m mLL d m m mgLfgLTfLgmms;../.;.... .;../.sec1006500153801538 0 5358175 0 5358 2 01821212220189812850333333012122bg(e) The recoil of the pistol when it is fired can be described by two different, but important Physicsconcepts. 5000 points for the Law and 5000 points for the Conservation