hw7 - PROBLEM 13.35_| r1 h'iTIth ﬂ epaeeeral'l is...

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Unformatted text preview: PROBLEM 13.35 .._| r1 h'iTIth ﬂ. epaeeeral'l is describing EL eireular erbjt at an altitude uf [5W km nbere the surf'ztee 01' the earth. As it passes through point A. 11:; speed is: redtteeti . ,5- h}-‘ 40 percent and it enters an elhpt'te erash trttjeeterj' with the rtpegee at x ‘ pn'trtt A. h'egleeLing tlir resistahee. determine the speed tnl‘ the spneeernft \J' when it reaches the earth‘e surihee ttt 11eint B. +.-|. SD LUTIDN ['.'ireLI1ut'ttr|:rit veleeity 3. (gt a ‘1 = J ff, LF;'|r.Ir=gJ" f f t 5..” £1.31 {9.etm.-'s:][e.3tedeem)" 'l f- .F' I'- {Mte 3-: 11)" m | Lite :4 1e" m} r5. = 5mm x m" “13:53 vi. = "H 12 ntt's Veleeily‘ redtleed te ﬂee/.5 ef 111- = 426? tit-"s {Tensewtttlnn ef energy: TI. + VI = + V54 gﬂgvi_enrm’_l /'- (3313' t” — :3 J'I'F'I'E. — ﬁr III - "'.'-' Harem A Hr]I — — = — — 4 {13m x. 10"} 2 [6.3m x ttf‘} 031(estttxttef w} 9.31[e.3re\tett II'I._4 —- {1.43 lint-’5 '1 ' PRDBLEM 13.124 .-"IL truck is Inn-fling can u Icwl mad at 11 5111:1541 01' Mil rni'h uhun iLS hi'ﬂkci are: apphcd to 51cm [I dmm In 30 mi h. Hm amulzakéd braking :syswm limits ‘— thu braking form to a value a: which the “heels 01' the truck an: just about In slink. Runwing 111:11 11'“: cuci‘ﬁuienl call" Sllllil: I‘riutiun hulwcun 1hr: mad and Lin: wheels is {1.65. dclcrminc the shm‘leat time needed for the truck to slam down. SOLUTION _____} H —+ 1' 't *" “1-1 ’1 ,r’ i J! \I .. hr .. Fr .- ;:__I,,-"r'I - #jngf | 1', = 60 mi-h = 83 ft}: 1': = 3“ mi 11 = 3931.? ﬂ 5 pﬂ', lumpfm = Mr; ,1, - 0.135 “"- 35; — Uﬁiiﬁllj: -— 29.333 —.:~ I — 2.803 it ' _ at a height of 1.5 m. The hall bounces at point A and rises to a maximum l' "' height of 1 m where the velocity it It] n'L-"s. Knowing that the tloretion of the impact is 0.004 5. determine the llltthlSl‘t-‘E: force exerted on the hall at PROBLEM 13.141 _ I A ploy-er hits a jig tennis. ball with El horizontal initial velocity of t ti 1T1.-"5 | pointxi. 1 " ] 1 1 Eon-f + mgi'il = Emhl' + 13?} I / --' ' - / - 1 what + manic—o —- 3mm; + on ' r“. : 5.425 til-"5 {Just before impact] Conservation ol" ene rgy {tiller impact} imp-"F .— = guilt-v3 + High: ﬂ;|:t-':IIL +[10}:] = gel/{iii}: + ﬁfteenth t 5 vi“. = 4.420 mfg {Just otter impact} H .l 1”. .-__ H’I _~-.I _ ___ .I Psi _- ll ._ . W1. ,r _ .‘.L-‘: - 3.2—...3: '- I I. _ . ._ - : u-II.I~LJI.J_-i—I.‘__ .._..._ . in .l. r ._ _.._p tl '3 - .IJ wax i: h}. . :'-' I. .r: 0.051118} — F”[0_0tH] = '[TL'EJ'STIIHlI]~ F” = I14 H l _1': —o.o51{5_425}+ F..{o.oo4} = common, t}. = [413.4 N F} = impulsive force F} = 130.9 N ‘_‘-. 30.0“ 4 PROBLEM 13.158 A EDD-g ball A is moving with a velocityr e4 of magnitude of = 8 nuts . when it is struck by a Llukg ball I? whieh 1135 a velocity vH. Knowing . that the velocity of ball A is zero after impact and that the coefﬁcient of -:.- _ 13 kg restitution is [1.2, determine the velocity of ball B {a} before impact, '- " {b} aﬁer impact. SDLU TIDN From conservation of momentum L m3“ + mgr}? : mg»; + my}; [0.5 kgHS me} - {1.2 kg}v,, = o + {1.2 kg] 9;? . . y“ ncsmmmn 0.2 = H , v3 = Le + ozv1r E + v” {a} Velocity of 3 before impact from Equations {l} and {2} 4.8 — 1.2vE =1.2{1.e + ozvﬁ} = 1.92 + 0.24%, 1.4413; = 2.33. v3 = 2 m-"sa— '1 {h} Velocity of B after impaet v'3=l.6+ﬂ.2[2}. v}; =2mrs—r4 PROBLEM 13.154 Ft. Nil—kg sphere A of radius 90 mm moving with a velocity v“ of magnitude v0 = 2 mfg strikes a. T201: sphere B of radius 40 mm 1which R was ill rest. Both spheres are hanging from identical light flexible cords. Knowing that the coefﬁcient of restitution is [1.3. determine the velocity of each sphere immediately alter impact. SOLUTION Angle of impulse force from geometry 110 'm m B B (Ln-TEFL 531ml _ was ‘33 HCENTEFL f? = cos 5% = 21.132" 13!} 'I'otal momentum conserved BullA: “L. W e «ﬁx “\ e r—sn~ j + {a 1'Kl-II-Iff = i Niki-Tn; u Hm! ﬂit x 1.: mite.r A lecosci} + [I = "if"; m BullB: : O E I Fh'f'rliﬂﬂ'lfg Fit: = triﬂe}J l2} Restitution {FA 2 en} PROBLEM 13.164 CONTINUED 3" .F'Ilrrpmarf: .w'c‘lmmm'rm r 1'I5 —1'_'II:L:\$H - (*[1',,c0\$ﬂ]: 1-1.: = [1='_| + («a-'Jiluuﬁt? Us,ng [Equations [ I} EmcHE‘J x "'-: m_l1~r = ri;l.¢1';5cusﬂ+ .I:.I_,-.-_'| m2 kguz=1m2 ham-l ' .‘.r _-' I. I- {1'1}: Sigh-'3 . . . . J _, _ 12"— F12" 912‘- 7”: huhxuluhng lur 1-”: 1.1.34 = [ME [1.6 F I — I--' F — -.L1', a .1 '.~ _"I .- I. _j . if, 1.7-‘41111-"5 + '1 = 3.03 [TL-"S x: 4| PROBLEM 111?? a_ Hlueks .at and 3 each have a mass of 0.4 kg and block I." has 11 mass of _ [.2 kg. The coefﬁcient nl‘ frietieurt between the hitieks and the plant: — ' ism. = {LEI}. Initially block A is meeting at a speed = 3 tit-"s and "-- '.a hlueks B and C are 11': rest (Fig. 1]. Aﬁer _-l strikes 8 and B strikes C'. 31]] three hlt‘tcks crime In [I stnp in the pusiliﬂns shuwn (Fig, 2}. [Itetermine tut the coefﬁcients efrestitutien between _-t and E and between B and t". (hi the displacement .t' enfhleek C. SOLUTION Eu} Wﬂrk and energy Velocity t2!er just before impact with B iris-31ml: (DE Va (9 ﬁg—P—JGm—e‘ -- I 3 .. 1 3- II. I] -‘- EHIJVU J3 — EH1_4(1’J )1 III .'I L-l : _—_ —#J'.t” lg“); TH] E... I", = JUAN -- mum's: TI + LII 2 T2: éw-‘t LEE-H3 W5}: '0-3W-4 kEHq‘ﬁl111-"5:_}{0-3In} = %{ﬂ-4 kEHJ'E'I 1t.- _ ' r = 13342 {Pr}; = 1689611155 _ k: | r Velﬂcity of .-I after impact with B [1'74], I 'i: “_tiu. ;. — tee-Pl liq“? J“: = J 4—. .—- r- I 1: 9.01:; It”! II". = -l-m {til-J T- : It} . 2 -| I E '- LII: _: = —,u;m_rg[ﬂ.ﬂ?5m] : PROBLEM 13.1?7 CONTINUED J a": + 4-3-.. = '0 50.4 kgﬂﬁﬁL — 0.3{04 kg}{0_01 m..-'.43J1{0.ms m} = 0 [15. = {3.0644 111.55 {fungcrvaiinn ofmnmcntum 0:; A hits I? I, . flw. I" Lilia-j; I ...._._..__*9. —45. [11,11 = 2.0000 m-"s {LII L = 0.6644 111-"5 I'H_I{|.‘J}1 + MEIER} = .u':__.[1':|:]_I — mlril'lrh} n:__. = Nil-5 2.0000 | 0 = 0.0044 + 0}. 4-“. J?! = 2.0253 m.-"s Relative velocities (A and B} J +- [0.]. — = v.1. — 04.}. [2.0000 00;... = 2.025: — 0.0044 .3... = 0.500 «I Wurk and energy Vulucily a!" B jug: b0 ﬂare impact with C £14 0 1 , z {14. ,.: T1 = Emesh'rrh = Tlll'lﬁJh. {2.0252}: = 0.0.203 2'1 II E :|_- ,——. f: L H HI“! — —II1A.FIPH£[1}.3[}} = —C|.353E6 I"; + U: 4 _ T4: 0.0203 — 0.35310 = 02:03.] .I. (10;. L = 1.5233 EFL-"S PROBLEM 13.1TT CONTINUED Conservation 0f mnmcntum as B hits C 9% I )5 1}}. = 0 HQ”)... 1;- ' E 2 E E] mr = [.2 kg. m5. = 0.4 kg 3"- mﬂfﬁi}; + "’0"? = maivgh " my“? 0.4{15233} + 0 = 0.00;)“: +1.20}. Velucity UfB after 3 hits C, (U; L = With {0;}: = 0-, 0.51152 :12»); :5 v} = 0.5094 mx's Relative velocities [B and C] [[vhh — vile“- : ft- — [1:34; {L5233 — New 2 {3.51394 - D Em: = ‘ {b} Work and energy — Block C T4 25120105004}2 = 0.15559 UH = —y,-.mgx = —0.5[1.2}{0.01}x = 4.55155 I; + UH = T5: 0.15569 — 5.5501x = 0 =5 1: = 0.044 m ...
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This note was uploaded on 01/09/2009 for the course ENGINEERIN 440:222 taught by Professor - during the Summer '06 term at Rutgers.

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hw7 - PROBLEM 13.35_| r1 h'iTIth ﬂ epaeeeral'l is...

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