Exam1_03 - First Name: Last Name: Section: I 1 February 14,...

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Unformatted text preview: First Name: Last Name: Section: I 1 February 14, 2003 Physics 207 EXAM 1 Print your name and section clearly on all 5 pages. (If you do not know your section number, write your TA’S name.) Show all work in the space immediately below each problem. Your final answer must be placed in the box provided. Problems will be graded on reasoning and intermediate steps as well as on the final answer. Be sure to include units wherever necessary, and the direction of vectors. Each problem is worth 25 points. In doing the problems, try to be neat. Check your answers to see that they have the correct dimensions (units) and are the right order of magnitude. You are allowed only the equations and constants-in this exam booklet. The exam lasts exactly 50 minutes. (Do not write below) SCORE: Problem 1: Problem 2: Problem 3: Problem 4: TOTAL: Don't open the exam until you are instructed to start. Please relax! First Name: Last Name: Section: I 2 PROBLEM I In order to hit more home runs, Mark McGwire is allowed to bat from an elevated platform as shown. Because of the platform, the ball starts its trajectory 5.0 meters above the rest of the playing field. Suppose he hits a ball and gives it an initial speed V0. «.4359 —————«l l WldCowa S'OM t 5,0M m I 60 M 7 a.) The ball initially travels at an angle of 45° with respect to the horizontal direction. If the projectile is to land on a spot on the ground that is at a range of 160 m from the base of the platform, what value of V0 is required? (9 pts.) b.) How long is the .ball in the air before it hits the ground? (8 pts.) 0.) Now suppose McGwire hits the ball so that it travels initially at an angle of 0° with respect to the horizontal direction. What value of V0 is required so the ball just clears the top of a fence that is 160 meters from the base of the platform and is 3.0 m above the ground? (8 pts.) First Name: Last Name: Section: I 4 PROBLEM 3 A rocket is launched vertically from the earth (in the positive 2 direction). Shortly after launch there are three constant forces acting on the rocket: i) Gravity. The mass of the rocket is 120 kg. ii) The rocket engines, which are badly misaligned with the rocket axis and produce a force of 5300 N in the yz plane, 30° from z the axis. iii.) The wind, which exerts a force of 330 N in the x direction. a.)What are the three components of the resultant force vector F while the rocket is firing? ( 7 pts.) b.) What is the magnitude of F? (6 pts.) c.) The rocket engine fires for 7.8 seconds. Where is the rocket (x, y, z coordinates) when the engine shuts off? Assume the rocket starts at position (0,0,0). (6 pts.) d.) Consider this unrelated problem: An object moves along the track shown in the top-view diagram below. The track is curved ‘ near point B, straight near point C, and curved near point D. The object moves from point A to point E with constant speed. i. Draw arrows on the diagram to represent the direction of the acceleration of the object at points B, C, and D. If the acceleration is zero at a point, then state that A explicitly. ii. Is the magnitude of the acceleration at point B greater than, less than, or equal to the magnitude of the acceleration at point D? First Name: Last Name: Section: I 5 PROBLEM 4 The first two parts of this question are related to units and dimensional analysis, and require no knowledge of Newton’s Law of Gravitation (which we’ll cover soon). a.) Newton’s Law of Gravitation determines the force of gravity between two objects with masses m1 and m2. F is measured in Newtons and is expressed as _ Gmlm2 _ r2 where r is the distance between the objects. In terms of kilograms, meters and seconds, what are the units of G, the Universal Gravitational Constant? (6 pts.) b.) A neutron star has a radius of 15 km and a mass m = 1.4 x 1031 kg. What is the density of the star in metric tons per cubic centimeter? (6 pts.) The last two parts of this problem are related to circular motion. c.) Calculate the radial acceleration of a person standing on the Earth’s equator, due to the rotation of the Earth about its axis. Radius of the Earth = 6.4 x 106 m. (7 pts.) (1.) What tangential acceleration at the equator would be required to slow the Earth’s rotation to zero over a period of 1 day with constant deceleration? (6 pts.) PHYSICS 207 Exam 1 Summary of Equations 2/14/03 CHAPTER 2 — 1D Motion Displacement Ax = x2 — x1 Ax Avera e velocit U = — g y av . l' Instantaneous veloc1ty HQ) = $91 = £11 At —> 0 At dt Relative velocity UpB = UM + U AB Average Speed total distance = _A_s total t1me At AverageAcceleration adv = % 2 Instantaneous acceleration a = fl = d—f— dt dt Acceleration due to gravity g = 9.81m/sz= 32.2 ft/sz 2 Dis lacement and Velocit as inte als Ax = U.At. = dt p y gr At-—> 0 ‘ ' U 2 Au: Ara OZaiAti = [I adt Constant-Acceleration Equations Velocity U = 00 + at Displacement in terms of vav Ax = x — x0 = uavt = 5(00 + U)t . . 1 Displacement 1n terms of a Ax = x —- x0 = uot + E at2 Vin terms ofaandAx 02 = 1): +2an CHAPTER 3 — 2 and 3D Motion Components Ax = A cos 6 Ay= A sin :9 Magnitude A = ‘lA: + A; Unit vectors 171 = Ax i + Ay j+ AZ k Adding vectors using components? if C = A + 3 then CX = Ax + B)r and C y = Ay + By . a lim , A” d " Instantaneous - veloc1ty vector u = ——C = r At —> 0 At dt . - lim A " d ” Instantaneous — acceleratlon vector a = ——U— = _0 At —> 0 At dt Relative Velocity UPB = 13M + GAB Projectile Motion Equation 0 t = 00x X y(t)= yo +UOyt_%gt2 2 yo +(Uo Sin9)t—%gt2 Uy(t)= on —gt 2 ()0 $116.. gt x0 + ont = x0 + (00 COS Path = (tan 00) — x2 0 0 Range when initial and final elevations are equal R = — sin 260 CHAPTER 4 - Newton's Laws Newton's Laws :Second law if = F” or 2F = fine, = m5 m Mass E2- = 2L 1in] a2 Weight W = mg Hooke's law F = —kAx Kinetic friction fk = M}; Static friction i; S ,thE, ()2 2 Centripetal acceleration a = — a, = 0% a, = d V61 t Speed and Period 0 = Mathematics Quadratic equation: the solutions of the quadratic equation ax2+bx+c=0 are x=—2ii2iVbz—4ac a a Trigonometry sin(—6) = —sin 6 cos(—6) = cos 6 sin(7r/2 - 6) = cos6 c0s(7z/2 - 6) = sin6 sin(7r — 6) = sin 6 cos(7r - 6) = -cos6 Constants Acceleration due to gravity at the earth’s surface: g = 9.81 In/s2 Avogadro’s Number: NA = 6.02 x 1023 molecules/mole 1 metric ton = 1000 kg Radius ofthe Earth = 6.4 x 106 m ...
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This note was uploaded on 04/01/2008 for the course PHYS 207 taught by Professor Winnokur during the Spring '06 term at Wisconsin.

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Exam1_03 - First Name: Last Name: Section: I 1 February 14,...

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