X2.6-8 107Physics 109 (Kaldon) Name ____________________________________________________WMU - Fall 1998 Book Title ________________________________________________Exam 2 - 100,000 points Section: 6a 6b 6c 6d 7a 7b 7c 7d 7e11:00Thu 12:00Tue 1:00Thu 4:00Wed 9:00Tue 10:00Wed 2:00Wed 3:00Tue 9:00ThuState 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 Questions“Me Tarzan – You Jane” (50,000 points)1.) In the movies, Jane is supposed to be a savvy city person, whileTarzan is Lord of the Jungle. One day Tarzan ( m = 125 kg ) sees alion ( m = 183 kg ) sneaking up on Jane ( m = 61.25 kg ). Tarzan’splan is to swing down on a vine and grab Jane to save her.(a) Tarzan starts from y = 9.50 m above Jane. What is his potentialenergy?(b) When he reaches Jane, y = 0, what is his speed? Use conservation of energy.(c) At the bottom of his swing, we can pretend that he is undergoing Uniform Circular Motion. Find the forcethat the vine exerts on Tarzan. HINT: Draw your Free Body Diagram first. Use r = 15.0 meters.If you don’t have an answer to (b), use v = 10.0 m/s.Physics 109 / Exam 2 Fall 1998 Page 2(d) Tarzan grabs onto Jane. Find the speed at which the two of thembegin to swing up on the vine. If you don’t have an answer to (b), use v= 10.0 m/s.(e) There is a tree branch at y = 3.50 m. If they can swing up at least that high, they are safe – if they can’t, thenthe lions eat tonight. Show whether the lions go hungry or not. If you don’t have an answer to (d), use v = 10.0 m/s.The Invincible Xena – Warrior Princess, Using the Forces of Good Against Evil (50,000 points)2.) Xena (m = 70.5 kg) is set upon by a bunch of cutthroat goons (m = 105 kg each),so in self-defense Xena beats them all up. She wants to drag them to town to seejustice served. (a) If Xena is walking at a constant 1.00 m/s while she is dragging thebody of the first goon, then the net force on the body of Goon #1 is zero. Right? Sotake the free body diagram below and label the four forces that are acting on Goon #1.(b) The coefficients of friction between the goon’s leather armor and the ground are 0.30 and 0.25 respectively.What is the value of the friction force between Goon #1 and the ground? If you do not have an answer for thispart, use Ff = 200. N as you need in later parts.Physics 109 / Exam 2 Fall 1998 Page 3(c) What is the value of the force that Xena pulls on the leg of Goon #1?The coefficients of friction between Xena’s boots and the ground are 0.80 and 0.75. (d) What is the value ofthe maximum friction force between Xena and the ground? If you don’t get an answer here, use Ff = 700. N asneeded in later parts.(e) What is the actual friction force between Xena and the ground, given that she is dragging Goon #1? CanXena drag more than one of the goons? If so, how many? Assume the goons are all identical.Physics 109 / Exam 2 Fall 1998 Page 4Studio 28: Twenty Movie Screens, Two Thousand Parking Spaces (50,000 points)3.) A few months ago, as we left Studio 28 in Grand Rapids, I heard a carracing along one of the side lanes. As a defensive driver, I slowed andwatched them shoot through the intersection without slowing/stopping.The kid’s eyes went wide as they saw us, because they were idiots andsomehow thought they could race through a parking lot without looking.But suppose I hadn’t been so wary… Let the two vehicles have thefollowing numbers: Car #1: 2125 kg, vx = 28.2 m/s (about 63 m.p.h.)SUV #2: 2700 kg, vy = 10.2 m/s (about 23 m.p.h.). (a) Find themagnitude (value) of the total momentum before the wreck. NOTE: Thisis a vector problem. If you cannot do this as a vector problem, you maycheck this box and do this problem as a head-on collision instead, with a 18,000 point penalty.(b) Find the magnitude (value) of the total momentum after the wreck. If you didn’t get an answer to (a), use px= 64,200 kg·m/s and py = 24,100 kg·m/s.(c) Find the final velocity vector, rv, of the wreck, after the crash. Give in Standard Form, of course. If youdidn’t get an answer to (a), use px = 64,200 kg·m/s and py = 24,100 kg·m/s.(d) The coefficients of static and kinetic friction between asphalt and rubber are 0.95 and 0.82 respectively.Find the friction force that will be causing the sliding wreck to come to a halt. Hint: the wreck is just the twovehicles combined.(e) How far does the wreck slide along the parking lot before it comes to rest? If you didn’t get an answer to(d), use Ff = 25,000 N. If you didn’t get an answer to (c), use vafter = 19.0 m/s. Assume that the wreck won’trun into any parked cars