Exam_2_solution_alex - ‘ l I 1 g I l f" v ‘ ‘ E Each?n>‘o\em is i 2 Dynamics'53 pts max EGN 3321 Section 4 C*‘ Pt be u us Exam 2

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Unformatted text preview: ‘ , l I 1 g I _ ( l ) f /,." v ‘ ‘ E \ Each ?n>‘o\em is _ i 2 Dynamics '53 pts max EGN 3321 Section 4 Feb 27, 2001 C *‘ Pt be u us) Exam 2 Name: 0 Social Security or Student Number: a nd 1. The spring has a stiffness of k = 100 N/ and an unstretched length of 1 meters. As shown, it is confined by the plate the wall using cables so that its length is at”? 0.5 meters: A 4 kg block is given an initial speed vA= 5 m/s when it is at A, and it 3,4 slides down the incline having a coefficient of kinetic friction 11150.2. If it strikes b, {\rv the plate and pushes it forward 0.25 meters before stopping, determine the $3) q rebound length, 5. Neglect the mass of the plate and spring. E“ ’\\ «M S \ xiv ’5 e given: 31-2: ' H‘VZUSY; ‘. €- Ut-n. '= 6+G+G> Nantes.- @ ii": *3 L; B (Wazfirmhfi = «my» h \n. 0. 25+ 6 h =0. \06? ‘TOJ-tZZé-VS sings); @ :- ’H‘qlm‘(O.1054+O-QQ26'6> : (“*- Héfle. 59,3 6» Z><Uxe2>swkm$1 a: k$T_Jik(SZZ §‘= [\‘Xu 1 =‘OH‘S'M @ :2 550003 (‘0}5)2_g: boo) (—0 g)? 32 = 12 'JL‘H . = 0.5 " : '5.$Z£_§' (N0‘\€‘- M $FT§fig : -—O,S SIDPS Cd: <35 8 5) (UszfiLfim-a — $.01 d: dis‘wma; ' ‘ = o.25+e 0'3 74 ‘ (Mn; N)‘ 0Q ‘ ~ 0 ) Ls) : =a0.2' ‘l fiflstbfl-Razs-rsl 3C :: -|.—:|-:F% "' 7.\\3'$ N K) ZUI—vzto hm Prev. ank «KEN-3‘0 O :. 0:6H"[email protected]?3' fi> + N : wtosufi) 8 + ‘5-625‘ +(—~\.Hg~1.n5.5) oza.}o\~13.co%s upkf’az o‘aogm! “Marxch sum=o, o 068;be 3 ' “ncdl Omsbfiv. ,7 3 flaive/‘k 9—7 9:1,"- : u?’ AL"? 9' “ OA/‘mdityon TfiV‘ ~ TS‘VV5'+ \0558$ v‘fivg’ o (a) 094W Q TIA '2. _ _\ Z__ / GD ‘2mv“2(\(53"5b 3;: h: @[email protected] = mahl: 0 26%) 6m(2‘5 '—‘ 3;: = Wits} W“ “F5 5“”(E) :: \.'Zé>%m / T :O ‘U = For “M093? i m f? @ 5 ( 5 0) 3s wkm (#625) ESSAJ’LtLogoz-K _ earn: 2.+L @@ V5 ‘ “‘35: ~ 49.8.!- [s-smpsfl Tu ‘r v.: ‘1 = \Co.‘$8'-t-$ Ts” ‘1 L. n C’s: 9: O osses— {asses duo. +0 fricfibfl 507:: \d 2‘ = JC'(dY$'t6mCL “traveled? gm H5) difit. “(TWzfl 2 5+ o.26‘+o.26‘+6 =35+s ‘:= 0.24 H '0le ~ Cosby) = 1mst GD L05$ES : 1HB~(35+6) = 2LL8C§+11H3~S ’PM ~ . 6: UW$ Mb 399 af bP 3 soWQ @ T\—VV\ :TS‘VVS‘V \OafiéS k5.) i g t; j 2. Two smooth disks A and B, having a mass of 1kg and 2kg respectively, collide ; - with the velocities shown in the figure. If the coefficient of restitution for he disks is e = 0.75, determine the x and y components of the final velocity of each , disk just afier collision. l Given: mp‘ : \wis 5) (Vex: \m/é m g ; 1 I. 4) = 3)in(30 A » A a 3) 4A Wmsmfeofiz . Em? A , 4k , 230° ‘ ' _\YI.2. o (1)?) = '1‘ [03398) n +%\n(qs)’c] q ‘qud: (,UA) 2 3 m B 8 = 0"}5‘ ' s . (Em. (U ’ l l / Flam. c ' RX wk '1) ‘ 'U Q Con 'Lact 7 3 \ 8x > 53 D :9va 1m fiaJL melt) cribfls 2* A _ _ )\ -—‘ {J UA’2.bC.)<5 v» -’r \,6l:, "DB: —0.?o=+?\1, -0313}? y M: :Deg" 0"7—07-m/SL C4D de‘mcfibn 0i i Uwa MOWfiMRK 4; "kaflfimdwafl‘ = MA 'UA/x 4"mgvéx (ZSWM 2-(-o.=tz>1)= \- bA’x +2418; [ma :vNQ +2qge>l< (ago I / / l / (uhx 'QJA/ ‘ 714x " Vex . I ( J ' 03$: WW “UM \ 25634-03013 I . (Dex fink): = ORS’ (2,598+030’fl ipax’~vA; = .232 84”) z_ I I < '2 @v "/5, 2W2énmdn659 30“: (gm 794x DVBX <:> '(UA; : /.Zénvg ‘?—- I {UBX = /.22 0/5 —---v anémer I . r 5 " gwen (0,43 : L’s M/3 f (“Wag M: 2+; hmchtZ W+H I - ’Ugg ~ 0.3}07‘l “WHAng 6.: amm: 2 3. In a game of billiards, ball A is given an initial velocity v0 of magnitude v0 = 3 ; m/s along line DA parallel to the axis of the table. It hits ball B and then ball "C, which are both at rest. Knowing that A and C hit the sides of the table squarely at points A’ and C’, respectively, that B hits the side obliquely at B’, and assuming g frictionless surfaces and perfectly elastic impacts, determine the velocities VA, VB, and vc with which the balls hit the sides of the table. Assume all balls have equal f mass and that the balls are moving freely on a horizontal plane). 0.5m 2,5,”,x 0.5m C f? A‘ ‘ T5 \/ Sweat) 1, l I '} 3 ._. c 02: \ 2‘ l ’00: —3*\ C. x \ ‘ ___‘ IL \m i \\A7:___3:‘.__"‘D @ 3A-(U5’6' I WV. ‘ A l ' v5/‘B Al ‘ 10.5w) “5“ 'UGX ' i ” BI A ‘~ ’UA i I/ O (—6 .4 J 2m e > ‘Dc / 1+3 sz 2-m ch no ng (tool-la crutckecQ *0 a 4:940. \e 3% : ‘DGX .M' ' M S‘W; Z 3 l3 -- W + vex 30 B ‘ii’éi‘fi‘iaili fiL’L‘b H.=H7 : ZA-(p V“) %E L ‘ Ti*‘i:;z¥“zs§ (. . \3 - Lz%_ ) Llj-O >Lz~gr~mV8 meA 1+?) D __ yr<v V47) 3 A 3 A3 “*3 “0% (23 2} Can sevkmkfifll 08 a 8 (sweat WM momentum/L bO'H‘Dm \Q coCAQJ» as 71er (,4 “A (Hf—"4») H ‘ )0 2 ( H230 U Uh‘momfis: IDA“ BUBX 37965 ) max :2) L‘ um Knon05 I ‘ Liv-05 H 8 eff. 3 Mflkwowos :fi ‘ mach/usan a, ’ M @ .az m 1’7 W’yi’i’fifigk vemtfl answer‘- ...
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This note was uploaded on 01/13/2011 for the course MMAE 0L02 taught by Professor Irinaionescu during the Fall '02 term at University of Central Florida.

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Exam_2_solution_alex - ‘ l I 1 g I l f" v ‘ ‘ E Each?n>‘o\em is i 2 Dynamics'53 pts max EGN 3321 Section 4 C*‘ Pt be u us Exam 2

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