2007 C Mech Fr

2007 C Mech Fr - AP® Physics C: Mechanics 2007...

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Unformatted text preview: AP® Physics C: Mechanics 2007 Free-Response Questions The College Board: Connecting Students to College Success The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the association is composed of more than 5,000 schools, colleges, universities, and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in college admissions, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT®, and the Advanced Placement Program® (AP®). The College Board is committed to the principles of excellence and equity, and that commitment is embodied in all of its programs, services, activities, and concerns. © 2007 The College Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central, SAT, and the acorn logo are registered trademarks of the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. Permission to use copyrighted College Board materials may be requested online at: www.collegeboard.com/inquiry/cbpermit.html. Visit the College Board on the Web: www.collegeboard.com. AP Central is the official online home for the AP Program: apcentral.collegeboard.com. TABLE OF INFORMATION FOR 2006 and 2007 CONSTANTS AND CONVERSION FACTORS UNITS PREFIXES 1 u = 1.66 ¥ 10 -27 kg Name = 931 MeV c 2 1 unified atomic mass unit, meter m kilogram kg Proton mass, m p = 1.67 ¥ 10 -27 kg Neutron mass, mn = 1.67 ¥ 10 -27 kg Electron mass, me = 9.11 ¥ 10 -31 kg Boltzmann’s constant, second kelvin K 10 R = 8.31 J (mol iK) mole mol k B = 1.38 ¥ 10 -23 J /K hertz Hz h = 6.63 ¥ 10 -34 J is newton N pascal Pa joule J = 1.24 ¥ 103 eV i nm watt W 8.85 ¥ 10 -12 C2 N im 2 coulomb = 9.0 ¥ 109 N im 2 C2 hc = 1.99 ¥ 10 -25 J im Vacuum permeability, Magnetic constant, Universal gravitational constant, Acceleration due to gravity at Earth’s surface, 1 atmosphere pressure, mega 10 -12 m nano 10 -9 m micro -6 c milli -3 k centi -2 M kilo 3 n pico p VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES k = 1 4p 0 m0 = 4 p ¥ 10 -7 (T im) A k ¢ = m0 /4 p = 10 -7 (T ◊ m) A G = 6.67 ¥ 10 -11 m 3 kgis2 g = 9.8 m s 2 1 atm = 1.0 ¥ 105 N m 2 1 eV = 1.60 ¥ 10 -19 J q sin q C 0 0 1 0 volt V 30 1/2 3 /2 3 /3 ohm W henry H 37 3/5 4/5 3/4 farad F 45 2 /2 2 /2 1 tesla 0= = 1.0 ¥ 105 Pa 1 electron volt, 10 N 0 = 6.02 ¥ 1023 mol -1 = 4.14 ¥ 10 -15 eVis Coulomb’s law constant, 10 s 10 c = 3.00 ¥ 108 m s 9 106 A Planck’s constant, Vacuum permittivity, 10 ampere Speed of light, Prefix Symbol giga G Factor e = 1.60 ¥ 10 -19 C Electron charge magnitude, Avogadro’s number, Universal gas constant, Symbol cos q tan q T 53 4/5 3/5 4/3 degree Celsius ∞C 60 3 /2 1/2 3 electronvolt eV 90 1 0 • The following conventions are used in this examination. I. Unless otherwise stated, the frame of reference of any problem is assumed to be inertial. II. The direction of any electric current is the direction of flow of positive charge (conventional current). III. For any isolated electric charge, the electric potential is defined as zero at an infinite distance from the charge. -2- ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2006 and 2007 MECHANICS u = u0 + at x = x0 + u0 t + 12 at 2 u 2 = u02 + 2 a ( x - x0 )  F = Fnet = ma F= dp dt J = Ú F dt = Dp p = mv F fric £ m N W= ÚF K= 12 mu 2 P= dW dt ∑ dr P=Fv DUg = mgh ac = a F f h I J K k = = = = = = = = = L= m= N= P= p= r= r= T= t= U= u= W= x= m= q= t= w= a= ELECTRICITY AND MAGNETISM acceleration force frequency height rotational inertia impulse kinetic energy spring constant length angular momentum mass normal force power momentum radius or distance position vector period time potential energy velocity or speed work done on a system position coefficient of friction angle torque angular speed angular acceleration u = w2 r r Fs = - k x Us =  t = t net = I a I = Ú r dm =  mr 2 2 rcm =  mr  m E= F q ÚE T= 12 kx 2 L = r ¥ p = Iw 12 Iw 2 FG = - w = w0 + at UG = 12 at 2 r2 Gm1m2 r q n  rii 1 4p 0 V= i UE = qV = 1 4p 0 0A k C= d  Ci Cp = i 1 1 = Cs Ci i dQ dt r A ˆ r  Ri 1 = Rp 1 ÂR i P = IV FM = qv ¥ B -3- = = = = = = = V= u= r= fm = k= ÚB ∑ d = m0 I dB = m0 I d ¥ r 4p r 3 ÚI d ¥B Bs = m0 nI fm = Ú B ∑ d A d fm dt e i i area magnetic field capacitance distance electric field emf force current current density inductance length number of loops of wire per unit length number of charge carriers per unit volume power charge point charge resistance distance time potential or stored energy electric potential velocity or speed resistivity magnetic flux dielectric constant P Q q R r t U F= V = IR Rs = = = = = = = = = = = = = N= q1q2 r Q V C= I = Neud A g Gm1m2 dV dr E=- E = rJ m k Tp = 2 p F I J L 0 R= 2p 1 = f w Ts = 2 p e Q dA= ∑ A B C d E 1 1 Uc = QV = CV 2 2 2 u = rw q = q0 + w0t + 1 q1q2 4p 0 r 2 I= 2 t=r¥F K= F= =- e = -L UL = dI dt 12 LI 2 ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2006 and 2007 GEOMETRY AND TRIGONOMETRY Rectangle A = bh Triangle A= 1 bh 2 Circle A = pr2 C = 2p r Parallelepiped V = wh Cylinder A= C= V= S= b= h= = w= r= CALCULUS df d f du = dx du dx area circumference volume surface area base height length width radius dn ( x ) = nxn - 1 dx dx (e ) = e x dx d (1n x ) = 1 dx x d (sin x ) = cos x dx d (cos x ) = - sin x dx V = pr2 Sphere 43 pr 3 a c tan q = dx = e x dx = ln x x Ú sin x dx = - cos x a 2 + b2 = c 2 b c x Ú cos x dx = sin x Right Triangle cos q = dx = Ú S = 4p r 2 sin q = n Úe S = 2p r + 2p r 2 V= 1 x n + 1 , n π -1 n +1 Úx c a 90° q a b b -4- 2007 AP® PHYSICS C: MECHANICS FREE-RESPONSE QUESTIONS PHYSICS C: MECHANICS SECTION II Time—45 minutes 3 Questions Directions: Answer all three questions. The suggested time is about 15 minutes for answering each of the questions, which are worth 15 points each. The parts within a question may not have equal weight. Show all your work in the pink booklet in the spaces provided after each part, NOT in this green insert. Mech. 1. A block of mass m is pulled along a rough horizontal surface by a constant applied force of magnitude F1 that acts at an angle q to the horizontal, as indicated above. The acceleration of the block is a1 . Express all algebraic answers in terms of m, F1 , q , a1 , and fundamental constants. (a) On the figure below, draw and label a free-body diagram showing all the forces on the block. (b) Derive an expression for the normal force exerted by the surface on the block. (c) Derive an expression for the coefficient of kinetic friction m between the block and the surface. (d) On the axes below, sketch graphs of the speed u and displacement x of the block as functions of time t if the block started from rest at x = 0 and t = 0. (e) If the applied force is large enough, the block will lose contact with the surface. Derive an expression for the magnitude of the greatest acceleration amax that the block can have and still maintain contact with the ground. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). GO ON TO THE NEXT PAGE. -5- 2007 AP® PHYSICS C: MECHANICS FREE-RESPONSE QUESTIONS Mech. 2. In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data back to Earth. Assume a circular orbit with a period of 1.18 ¥ 102 minutes = 7.08 ¥ 103 s and orbital speed of 3.40 ¥ 103 m s . The mass of the GS is 930 kg and the radius of Mars is 3.43 ¥ 106 m . (a) Calculate the radius of the GS orbit. (b) Calculate the mass of Mars. (c) Calculate the total mechanical energy of the GS in this orbit. (d) If the GS was to be placed in a lower circular orbit (closer to the surface of Mars), would the new orbital period of the GS be greater than or less than the given period? _______Greater than _________Less than Justify your answer. (e) In fact, the orbit the GS entered was slightly elliptical with its closest approach to Mars at 3.71 ¥ 105 m above the surface and its furthest distance at 4.36 ¥ 105 m above the surface. If the speed of the GS at closest approach is 3.40 ¥ 103 m s , calculate the speed at the furthest point of the orbit. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). GO ON TO THE NEXT PAGE. -6- 2007 AP® PHYSICS C: MECHANICS FREE-RESPONSE QUESTIONS Mech. 3. The apparatus above is used to study conservation of mechanical energy. A spring of force constant 40 N/m is held horizontal over a horizontal air track, with one end attached to the air track. A light string is attached to the other end of the spring and connects it to a glider of mass m . The glider is pulled to stretch the spring an amount x from equilibrium and then released. Before reaching the photogate, the glider attains its maximum speed and the string becomes slack. The photogate measures the time t that it takes the small block on top of the glider to pass through. Information about the distance x and the speed u of the glider as it passes through the photogate are given below. Trial # Extension of the Spring x (m) Speed of Glider u (m/s) Extension Squared x2 m2 Speed Squared u 2 m 2 s2 1 0.30 ¥ 10 -1 0.47 0.09 ¥ 10 -2 0.22 2 0.60 ¥ 10 -1 0.87 0.36 ¥ 10 -2 0.76 3 0.90 ¥ 10 -1 1.3 0.81 ¥ 10 -2 1.7 4 1.2 ¥ 10 -1 1.6 1.4 ¥ 10 -2 2.6 5 1.5 ¥ 10 -1 2.2 2.3 ¥ 10 -2 4.8 () ( ) (a) Assuming no energy is lost, write the equation for conservation of mechanical energy that would apply to this situation. (b) On the grid below, plot u 2 versus x 2 . Label the axes, including units and scale. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). GO ON TO THE NEXT PAGE. -7- 2007 AP® PHYSICS C: MECHANICS FREE-RESPONSE QUESTIONS (c) i. Draw a best-fit straight line through the data. ii. Use the best-fit line to obtain the mass m of the glider. (d) The track is now tilted at an angle q as shown below. When the spring is unstretched, the center of the glider is a height h above the photogate. The experiment is repeated with a variety of values of x. i. Assuming no energy is lost, write the new equation for conservation of mechanical energy that would apply to this situation. ii. Will the graph of u 2 versus x 2 for this new experiment be a straight line? _____ Yes _____ No Justify your answer. END OF EXAM © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). -8- ...
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