XF.17a 107PHYS-107 (17) (Kaldon-37944) Name ____________S O L U T I O N ___________WMU - Fall 2004Final Exam - 200,000 points CHECK-OUT: T1 T2 _________Rev. 12/08/04 We.6State 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 EquationsShort Answers Should Be Short! – Feel Free to Ask Any QuestionsFINAL EXAM [FORM - A]PHYS-107 (KALDON-17)FALL 2004WMUThe Polar Express vs. The Space ExpressPhysics 107 / Final Exam [Form-A] Fall 2004 Page 2“Almost Over” (50,000 points) Multiple-Guess-Pick-The-Best-Answer-Fill-In-The-Bubbles1.) In parts (a)-(d), select which Newton’s Law or Conservation Law best applies.(a) Tom hits Harry and Harry hits Billy…A = Newton’s 1stB = Newton’s 2ndC = Newton’s 3rdD = Momentum E = Energy F = None of these(b) Tom sits on Billy, so Billy can’t move…A = Newton’s 1stB = Newton’s 2ndC = Newton’s 3rdD = Momentum E = Energy F = None of these(c) This is conserved in all collisions…A = Newton’s 1stB = Newton’s 2ndC = Newton’s 3rdD = Momentum E = Energy F = None of these(d) A perpetual motion machine, if it could exist, would violate…A = Newton’s 1stB = Newton’s 2ndC = Newton’s 3rdD = Momentum E = Energy F = None of theseIn parts (e)-(h), select which speed best represents the situation described.(e) “Let me take my pet cheetah out for a run.”A = 1.0 m/s B = 7.0 m/s C = 30. m/s D = 300 m/s E = 344 m/sF = 1,000 m/s(f) “I just had to run and catch the bus!”A = 1.0 m/s B = 7.0 m/s C = 30. m/s D = 300 m/s E = 344 m/sF = 1,000 m/s(g) “By this time tomorrow I’ll be flying on a Northwest flight to Europe.”A = 1.0 m/s B = 7.0 m/s C = 30. m/s D = 300 m/s E = 344 m/sF = 1,000 m/s(h) “Traveling faster than a speeding bullet – it’s SUPERMAN.”A = 1.0 m/s B = 7.0 m/s C = 30. m/s D = 300 m/s E = 344 m/sF = 1,000 m/sUse vector rA with an x-component Ax = 22.0 m and a y-component Ay = -24.0 m in the follow problems:(i) The magnitude of this velocity vector is A = __________ .A = 2.00 m B = 6.78 m C = 6.59 mD = 32.6 m E = 46.0 m F = 528 m(j) The Standard Angle for rA is ________.A = 42.6° B = 47.4° C = 132.6°D = 137.4° E = 312.6° F = 317.4° M (e=2.71828) r r y X mass < ==========OR“Merry Christmas!”! " # $ % id " # $ % &! " # g % &! " # $ h &! " f $ % &! e # $ % &! " # g % &! " # $ % i! " # g % &! " # $ h &Physics 107 / Final Exam [Form-A] Fall 2004 Page 3And the Winner of the $10,000,000 Ansari X-Prize is… (50,000 points)2.) On 4 October 2004, history was made when the privately funded, non-government Scaled Composites SpaceShipOne successfully flew into space twicein two weeks. Space is defined as an altitude of 100. km (100,000 m) above thesurface of the Earth. (a) Calculate the value of g, the acceleration due to gravity,100. km above the surface of the Earth. MEarth = 5.98 × 10 24 kg,REarth = 6.37 × 10 6 m, G = 6.67 ×10 -11 N·m2/kg2 .rR km m mmFGmMrmggGMrkgmmsEGENmkg=+ = +=====´´=-×100 6 370 000 100 0006 470 000667 10 598 10647000095282211 242222.,, ,,,..,,./( )dichbgstill close to 9.81m / s2(b) SpaceShipOne is carried aloft by another aircraft called the White Knight. Itwas released at h1 = 14,100 m (46,000 feet) above the Earth and climbed to analtitude of h2 = 115,200 m (377,591 feet or 71½ miles). Let’s treat this flight likea tossing a ball. What speed v1 would this ball need, if it traveled straight up fromh1 to h2 ?Going straight up, turning point at top, v = 0.Using Kinematics. . . Using Conservation of Energy. . .vv gyyvgyyvgyyvgyyms m mms msyyyyy202002002000222202222 9 81 115 200 14 1001 983 582 1408=- -=- -=-=-=-==()()()()(. / )( , , ),, / / mgh mv mgh mvgh v ghvghhvghhms m mms ms11212212221121221221121222222 9 81 115 200 14 1001 983 582 1408+=++==-=-=-==bgbg(. / )( , , ),, / /Physics 107 / Final Exam [Form-A] Fall 2004 Page 4(c) The rocket engine burned for 84.0 seconds. If SpaceShipOne started at a speed v0 = 200. m/s, find theconstant acceleration a needed to reach the speed v1. If you didn’t get an answer to (b), use v1 = 894 m/s(2000 mph, about Mach 3).avtvvtms msmsfinal initial==-=-=DD1408 20084 014 382/./.sec./Or about 1½ gee’s average acceleration.(Because mass isn’t constant in the real problem, theacceleration is not constant, and peaks at about 3 gee’s.) (d) How far did the rocket travel during these 84.0 seconds? It is possible to solve this without (c).Using answer to (c): Using average speed:avtvvtms msmsxx vt atms msmm mfinal initial==-=-==+ +=+ +=+=DD1408 20084 014 380200 0 1438 016 800 50 730 67 5302001221222/./.sec./( . / )(84. sec) . / (84. sec),, ,ch vvvms msmsdvt msmaveragefinal initial=+=+====21408 2002804 00067 540/././(804. / )(84. sec),(e) On the video of the flight, the announcer says that SpaceShipOne reached itsmaximum height when the rocket engine stopped. Is this TRUE or FALSE?What one short statement can you write to prove your case?FALSEBecause vy ¹ 0 when engine shuts down.Physics 107 / Final Exam [Form-A] Fall 2004 Page 5The Real Polar Express (50,000 points)3.) The Lima Locomotive Works Pere Marquette 2-8-4 Berkshire number 1225(really, no kidding!), considered one of the most attractive steam locomotives ofall time, was the model for the engine in the movie The Polar Express. Itweighs 443,000 lbs. (mass = 201,000 kg). (a) This engine is capable ofdelivering 4000. hp. Find the work Number 1225 can do in 10.0 seconds. 1h.p. = 746 W4000 746 2 984 0002 984 000 10 029 840 000./,,,, .sec,,hp W hp WPWtWPt WJ´=====bgbg(b) The 1225 has a maximum “drawbar tractive effort” or applied force of 69,400 lbs. (309,000 N). If thelocomotive only moves itself (no passenger cars in tow), how far can it move from rest in 10.0 seconds? Startwith Newton’s Second Law.FmaaFmNkgmsxx vt atmsm=== ==+ +==309 000201 00015371537 10076 852001221222,,././(.sec).ch(c) The main driving wheels of the 1225 are 69.0 inches in diameter (1.75 m). At 79.0 mph (35.3 m/s), find thecentripetal acceleration ac at the rim of the wheels.rDmmavrmsmmsc== === =217520875035 3087501424222..././bgWow! That’s equivalent to 145 gee’s!(No wonder we don’t strap ourselves to theoutside of locomotive wheels at 79.0 mph!)Physics 107 / Final Exam [Form-A] Fall 2004 Page 6(d) The boiler pressure is 245 psi (1,690,000 Pa) at a temperature of 400.°F = 204°C. A volume of steam, V1 =1.00 liters, inside the boiler at this pressure and temperature