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exam_2_soln_fall00 - bvwfmics 650504 2 Wt 02...

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Unformatted text preview: bvwfmics 650504 2 Wt. 02,. ,_ soLJflfioM -— Problem 1 A thin circular rod is supported in a verfical plane by a bracket at A. Attached to the bracket and loosely wound around the rod is a spring of constant k = 3 lb/fi and undeformed length equal to the arc of circle AB. An 8 oz. Collar C, not attached to the spring, can slide without friction along the rod. Knowing that the collar is released from rest when 6 = 30°, determine the maximum height above point B reached by the collar. In which position the collar has maximum velocity and what mechanical principle governs this fact? 4 . ' Hakka/arm new if 55 «1);; O (we) \._.——--—_. 0.090 0. an; mp: mrzo __., 7:70 vgemga - «Haywamea-Mfiade/Mam) aaa ,3; lap - 2mm .- 09-04%”; a [Z .7) [£2 0. 956% i m- ‘ ' ' a!” 7‘.- M’E #2 cog/aim 6J3 / __»{ce @1066M77442 W“? m if}: (4:) MWWL WOW W M \ Jam % 54 mgauui) afcp‘rfa amid W .65 WIIW. ) __ _ 7%“; a flea Macaw iroéocig’j A "ti—2mg. Ban ‘5‘ GfmGSS 31A- _—_~ 1 1:313 mfl‘flhg “Nth a ‘rielgcfl'v ‘r. E: 1 mfg having the dffPCflOfl indicated 1n the figure, when it is hit by ball 13 of mass mg = 1 kg Knowing that afier impact the velocity of ball A15 VA = 1/— m/s 1n the direction shown and that the coefiicient of restitution is e= 0.8, find the speed of ball B before end after impact (notice that the direction of the velocity of B before and after impact are given in the figure!). Indicate on the figure the line of unpact. oedema if irt PAOT W BEFORE MACT AFTER IMPACT ( U (wt 2 (£0 60" a: / (143,1 “Home" a“; AMA/1.1 (wt 2 161260. w --‘[;5:‘ flu/’3 (”A/4 2 3%: “/0 (”6% 1' - V6 (vat 1» e” (’12)é , 0 (5t .7 O éluaé'tv1 : of 9W1 Wigwam 14a 1/ 0&2CC52L fit eat—ti 5M 0751 0/);- 3‘ 0‘44 (’V ”)5 mm) fl/Ji .9 (1”,; Ive/o m'ti'CL__ W6€y6)¢ 2. M6 (4/5125 —-—7 64514, a: (1/5)}? WfIC-d ,_,_ , d5 04 .2) W’Vw a? 7009/ ()1’mntcotwci/f1L/OGJJ-Jtil WMn 1M M 1 W m)“ f 0015645)” 9 wk (ti/1+ Mimi’s/L ....~_‘.: ... _._.-.... “.17.“... ..7._ __ Name: 6043/,4 / SSN: 4/ —- —¥.s) ( m—b “fa/+1.5 7 0,471 0.7? ”/6 c0 5 ,7 (Ag) _”____....7-._} WIS/‘- 0.g Vg z'fwf ’* (.2) / V+’[email protected]?_ s — 'Vs’ + 0.ng 7.7.1.; / 4.30/62- 3,! mg, 9 gig“: 4.51222071/5 0/77]? 2.719212 0.247757 7.77/3 ...m- _ \ x7. .. 7.77. /. ...... L__,____.—.——-¥I/5ms. figs/70.23 077/3 —-—-)\ ._____.-..___~__ Problem 3 A 1 kg block B is moving with a velocity v0 of magnitude v0 = 2 m/s as it hits the 0.5 kg sphere A, which is at rest and hanging from a cord attached at 0. Knowing that 14k = 0.6 between the block and the horizontal surface and e = 0.8 between the block and the sphere, determine after impact (a) the maximum height h reached by the sphere, (b) the distance x traveied by the block. i dildo?” W {G/WCZKL £96 :k’: nae” 63.56” ‘HCflwW 01 % g’km_ 4WLQ—LLA‘4-u— J) 71—4-2 52144614.. ré—‘—- (m, p; + “(gift—J :a .31 «n 1-. 5 ._., I O +(’)(7‘) ., (0 6) (VA! +(’{>,V£/ —-—-u:> 27/3, 4.— 0.§.\/Al a L coca (aw af mmézfi ' ' "’— e Z \é/"l‘éi/ ’14 "1/6 'V’ i 4' l ‘1 ”(O/09%» e—VA #MMMJ, {1/25.— q/R ., _,é J. {AW/0931., awn/gym.“ 7 n p» In». n 01744 a“ Lads ,4 /K / »/~\ / L—vxj ...
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