98826 - Problem 1(25 Points(a lOpts A bar of aluminum has a...

Unformatted text preview: Problem 1 (25 Points) (a lOpts) A bar of aluminum has a length of 2 m and a cross-sectional area of 1 x 10‘4 m2. If the macroscopic spring stiffness of the bar is equal to 3.1 x 106 N/m, what is the Young’s modulus of aluminum. [AT/1’ A (leq M2) n... y, F/A : kL , (slump/Mm”) i=éZGa (b lOpts) The density of aluminum is 2.7 g/cm3 and the atomic mass is 27 g/mol. What is the interatomic spring stiffness for aluminum? (c 5pts) If you hit one end of the aluminum bar with a hammer, how long does it take before the other end of the bar experiences this perturbation? If you were unable to complete part (b), express your answer in terms of ksfl. (ZSSXIo'mm) Problem 2 (25 Points) A block of mass m hangs from two identical springs each with spring stiffness ks, as show in the accompa- nying figure. How much does each spring stretch when the block is in static equilibrium? Problem 3 (25 Points) A Ferris wheel is a vertical, circular amuse- ment ride with radius 5 m. The Ferris wheel rotates clockwise at a constant rate, going around once in 10 s. Consider a rider whose mass is 75 kg and sitting in the fourth seat. Note that the figure is not to scale and the an- gle between seat four and seat five is actually 6 = 40°. (a 4pts) In the space provided draw a diagram indicating the forces acting on the rider in the fourth seat. Note that the Ferris wheel rotates at a constant rate. \P/ F\ K Refit,“ :0 g) E,” has “i0 be 013 iohj 0L3 Lima)“ Rem to owl MW cod” 50 F5]; is iOMJU ‘H‘Mn (mg); 5i} , (b 4pts) What is the parallel component of the rate of change of the riders momentum in the fourth seat? Explain how you know this. <03; :0 L/C \/ '«ComS‘i‘ml/xi' ll (c 4pts) What is the perpendicular component of the rate of change of the riders momentum in the fourth seat? Explain how you know this. oli3> Mvz : mom)? : mg (2:: HmeR 0'7 E— R T1 H (d 8pts) What is the parallel and perpendicular component of the contact force exerted by the fourth seat on the rider? 1/ lefl _"'E ~> at = meoséuma /[email protected] T7. =.(-75L5)(‘) gus‘l) MW + wgrd @ Em = WIND (e 5pts) Now consider a rider who is momentarily at the top of the Ferris wheel (i. e. riding in seat one). With what angular speed must the Ferris wheel be turning so that this rider feels weightless? Rwy Leah WWW» NM 5:0 A+ m +0? ' 'E +th : ”ARV2 ’ V W :J;ems~1)<gm‘): 1mg Ur): V —‘ 75w: "' rm: I q d a Sm {‘0‘ S For F5 Problem 4 (25 Points) (a 6pts) An electron of mass m is traveling in a straight line in the positive x—direction at a speed vi = 0.990, where c is the speed of light. A constant force brings this electron to a complete stop. What is the change in kinetic energy of the electron? " 2 ,—~ - _ t; KL+E ‘d/Mcl t4: Eo'r’lc)’ s AK: AE =—(2I-Ilmc‘ :<\I—‘——1’ ‘(l—lljm.l9(/0;3IL5> (BX/DQM§~' 2 1_ .‘M -. :lf‘ifi‘l XII)"5 j l (b 6pts) Determine the x—component of the constant force responsible for bringing the electron to a stop if the deceleration took place over a distance of 3219 m. 43 W= AK: Eat a F = _V_d_ : ”WW" J” teeming) M 62M». = LSSXIO‘” {Hf —I SSon"’°I\le .... -1...“ ... (c 6pts) After the electron comes to a stop it is then accelerated up to a speed of, at which point the speed remains constant. Once the electron is moving at this final speed, it enters a region containing a strong magnetic field. The result of the interaction between the electron and the magnetic field is that the path of the electron begins to curve. Since the speed remains unaltered, which components of the rate of change of momentum change and why? ' e -‘ A _-“' " v\ A cl cl lrl lFlI h alt: BIT? R 1 (2:39: CGMSIlM‘l": O fish/c “ t/z (ousted ‘ 1: 1" § _ Chfihjei 9M2. h (“Vol.pa'tem’hhny'lz 12 "C B W MIL Co‘mS‘lfih‘P'l Changes.” L (d 7pts) At some point along this curved trajectory, the osculating (or kissing) circle has a radius R— — 0.5 m. If the magnitude of the perpendicular component of the force acting on the electron at this instant is F = 1.62 x 10‘10 N how fast is the electron traveling? ORA ’Tlm’: o‘PPmecL 75 inQoI‘I‘fiClL. i ”F“ : INN/2 lo/C [A WWS i.,;‘/._:>C‘. “I. ’E I! a v: Fa mush) AMOS“) 7"; Ol~leO'nlq9 ‘l-‘l xIOqu” \ [l \L‘ 522 ”ex-l» Pmyfi Qt correfl‘ OIMSW Comet" ginHOM ‘3“ F=1EMIC KW‘VZ: ‘ < 5m (mom '14 cu [u I/ Y P J 6x CAICU (A'Hr 4'th him you‘ H J05" ...
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