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hw11 - PH 131 FALL 2005 HOMEWORK 11 Assigned Civfl...

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Unformatted text preview: PH 131 FALL 2005 HOMEWORK # 11 Assigned: 11/07/05 Civfl: 33IQQJHI131 CRIS! 711?)”C,?' 444:6.ckto. 23p. 3‘15 @ In the overhead view of Fig. 10-30. a 300 g ball with a speed : v of 6.0 mls strikes a wall at an angle 6 of 30" and then rebounds f With the same speed and angle. It is in contact with the wall for ' 10 ms. (at) What is the impulse on the ball from the wall? (b) What is the average force on the wall from the ball? V =' go art/)5 Fimol (a) 3-: Impfilse ME 2 Avg 19m. NOW V;K : Vcose Vi‘tRVéiMQ V41" : v C—DSfi AF .: o-l-Zrmusl‘ma : 2.1mué‘4u9 QCBwYIOU’D C9) (5tm50°) : |.¢g M5. :09 mm —— @E t) T; :Ar. = Lvs‘mfi At 4*: 2 (“‘3 Nb :- 104/» U fl L'J . - the first block, with mass 1.20 kg. and embeds itself in the second. In F1 . 10.325, a 3.50 g bullet IS fired honzomally a: two _ _ blocks t rgest on a frictionless tabletop. The bullet passes through wlth mass 1.8“) kg. Speeds 01:10.63? mfs and 1.40 mls. respectively. a q are thereby gwen to the blocks (Fig. 1032b}. Neglecting the mass Frictioniess _ removed from the firsl block by the bullet, find (a) the speed of {he huliet immediately after it emerges from the first block and (b) the hullcl's original speed. .. ' Gva'. b: bulie’r _ “7T 2 01152.55 a? bullé+zjvgofl met: {.20 kt}, m12l'go k? (b) —t> 0.630 m/s -—D 1.40 m/s YE“: ‘Im‘t-Haj- Vfil- A black 1 = 0 Fgmal; CD Valenti 4.x! V s : ~ u H II 2. SE 6 . 0% bul‘é‘l‘ 11-2 F: u u I '- 0' SIM/5 _ G—Her [5+ blockZVb§ v2; _~ .. .. u z : 1:40 Ina/5 9"“ 94- balm-l- beef-fife IS-Ih block=VLf I3 For band's Comm «J/ LloCk I'. ”"be == 01’1an + WM; U) Few bmflefl'fi. cpllifiaim w/ lochCKQi MW"? = (”VIN-Maw —-—@ From CD M (‘2); ”“5va :: (Mb+’m2.)v'z$ + WIUJF vb; = W -...L 92.? 079/5 ”“19 sm ‘2” M w m My Vine: MM ”’11. :3 WbeE 'H/MIUI 0% :: ¥me1/5 Luigi-37 S A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the on'ginal direction but with one-fourth of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body cctttcr of mass if the initial speed of the 2.0 kg body was 4.0 mfs‘? ssm mm (film? ”I? = Vii/L! v-Q DKQV ULL’ = film/b 007112. b) Vow: U‘ {—9. V1; :- 0 mu MW Be {we- AFJper Collision thu' ="MIVI£ +4111V'1fl ”“1 (WE‘VHL) = 0%ng #--—-—- (I) Emf‘b’fi Cmsewafian‘. «mg/1;"- : “fli-i WEE :27 «1 mm: Wu?)— ”Mi” M; (x) «1v; Ba ea. (23; em- %CVVMI+)CVH+VI() = Mimi: : Va“? /7 P223 Me ‘Is 8‘1 10-14: 6% +£x+ 1266+ (i125 and? defiveok Lu 1K0. Wk)‘ Fraana'): \JI-f~ (mw’mg 2 ”M‘vll‘ “WW“ m"; (we JFVIC) .2 /W\IVH: “-Vlf-(ml J'- mflL :2 UIIHVJ‘ (Ml _. VLiHQ/Ic/Zr)-AMI —. v. ,c +va ( vie/4y w 4Uli"vll: - a ma 1/13 + Am; 5% '5 ,4— D 3 (W'IVI. Vcw = ”MUN +Km1—Vé“ .k (M: L i '3; In a game of pool. the cue bail stn'kes another ball of the same mass and initially at rest. After the collision. the cue ball moves at 3.50 mfs along a line making an angle of 22. 0" with its original direction of motion, and the second ball has a speed of 2.00 mfs. Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball. (c) Is kinetic energy (of the centers of mass, don‘t consider the rotation} conserved? 55.11 I SO SHEETS 22 I42 100 SHEETS 22-144 200 SHEETS 22-14] “3‘06“?" A- meuéroré m a—b 60 we: 1.2552 7H5 r9 (.0 N52?» Knox; OF WW. x-‘D'T’n' i; I TO EH 4-}: l4 5 f9 COM‘EvE-z-VEJD E'MA1AJSCWSJ ANT CALL. 1"“ BEFDJLE w: AFTER Cow-15mg M COM?AE£'I Emmy Is. 35:37; magmas 90 77%; 3:5 m In £LA$TJ¢ COM-41.7! (HQ. CH. lS-‘i @ A body oscillates with simple harmonic motion according to the equation GuUEhH xm =(6-WJCOS an???” FIMD‘. (a) 23(5) wigs) a A00 4’ L6) {1 , (mi :13} T NOTE" awrrcH 7002. CALCULATOR To ITS ”PrfilbffiN MODE) For: "TH-Is PROBLEM. XC—F): Xm Co$(63-E+€P) NE 06E 'TH'E'QE to :: lv/T = 2V 3“ Dem m4w~ '5 - .r = (6.0 m) cos[(31r mdls): + «£3 rad]. At .f -—'- 2.0 5. what are (a) the diSplacemcm. (b) the velocity, (c) [he acceleration. and (d) the phase of the motion? Also. what are (e) the frequency and (0 the period of the motion? 55L 0) a: £15 - “(mi/6.0) coal—W (Mfg-1 0!) Pmlo MM IM 12—51 164% (rm lama. HIT JD ”(-7 6? In Fig. 16-3]. ablock weigh- ing 14.0 N. which slides without friction on a 40.0“ incline. is con- nected to the top of the incline by a massless spring of unstretched length 0.450 m and spring con- stant 120 NM. (3) How far from the top of the incline does the block stop? (b) If the block is 4 pulled slightly down the incline Fly. “-31 Problem 28. and released. what is the period of the resulting oscillations? Causing: THE FoU—ONWG THREE cllsas'. / - / 2/ \x 9 11:) Block legal 60W°+M° Wu) m Manda? SP1 imfi‘s free- 0419:0299. {-0 Arno-Hm . 6M '18 Q X0 M49. :10me M010 M X - ~L Fimoti 0») x 12,, Anmfim QT) - Molt: L): (pxo S‘FMQ «1 M h: KCX—Yo 9‘ “ CWQJ‘EM 3 8 T Chm J Cx_x;>1: -%-C“fi){" -_-. 2.0433040 53mg :. )(‘-Xv 9:}: ZMQS'WQ K w: J’ 01' hwy" Hui.) f] :rr (9" A: w A 40 kg block is suspended from a spring with a spring constant of 500 Nim. A 50 g bullet is fired into the block from a? q H directly below with a speed of 150 mis and becomes embedded COU5 lb E —.. in the block. (it) Find the amplitude of the resulting simple har- FD LLO U l U 61 TH EEC monic motion. (b) What fraction of the original kinetic energy of 5 I To A T f D N 5 . the bullet is transferred to mechanical energy of the harmonic oscil- lator? ssm m EHEM in, I) BUM- Inn Susi- ewlzts You Black (Assam. 1 “Matt I 1‘) 5491-6 13M d0£5¢}?f0;n:‘/‘2- 8”” 5‘0““ bod-ill m an: mew:— mbwlM-f in T2444?“ PBlaok @_ "[112 block) Boil? mam-€- lrum W. DUNK— veloct+fl,\f . r '- ' (tilled-lam ' 6}___v_____-_w AM.— LWfi W55, 005 la K2561) U/m up; [so 0% Fl‘md. - 6;.) \fm (TM. “maul-1M 111') b) WW: 0% BMW I]—-—>IL')L' a) Mame/«kw 15 can/sweat Pr : FIT: 0nU0+M (o) : (rm-HOV -‘~v: .__0“_"L._.:. Mos-Km“ M/b :- an+M (out) (’1 0) dug-H (Can’arnunol) W is Camera-wed 1M tin—45“) =K¢ :yn‘f' =CL5)V2‘ 7 : O 1 '9' K 250 m -B““5)"m"é' 60’ 1 a! .— Im’ ; (“0.09:3 - troy-nod e—anmla = M45 N M: CH‘O?)0'39)1”“"= m M '1. 2- : 351.13}: 32:91-4fw — CCme®F0°+9 (a a; = 252.9- =Ebnu£+ (wreak) 121: : Meal/\Wmi WM 01? 09(41de — Em; E —— E1“: :2. Mix/V?“ 1 35:9:- :: O 0! *3" ‘1 8| £31 £11 flmv} ”“on 50 SHEETS 22-142 [00 SHEETS 22-144 200 SHEETS 22-143 6 m 46 -\ pcrfmner sale—dpon a Impeze is swinging EEK anti 5 with a period of 8.85 s. lfshe stands up thus raising the / :_. g .85 5 mass of the trapeze + performer system by 35 0 cm, whfl the new period of the system? Tm: trapeze + peifiirmer simple pendulum. “)ng THE ngfioma‘c gwwbsx. 7-7,; New 4£m7h J’s: a? Wm" “Mm“ 23?. A 0.10 it; block oscillatcs back and forth along a straight line on a frictionless horizontal surface. Its displacement from the _ . origin is given by an,“ - x = (m cmlcoslllo mars): + «:2 mm. [— rad t «e c“ ' lo __.-v (a) What is the oscillation frequmcy? (b) What is the maximum x (5%.) '00 ) COS C F: ) speed acquired by theblock? At what value ofxdnes this occur? + 1'. «rod» .3 (c) What is the maximum acceleration of the block? At what .‘2. value of 1 does this occur? (d) What twee. applied to the block. results in the given oscillation? F: H0 91*.” la.) M..$W;Ufl I :9 - K a) WWW “MW 9-4“- X20 KG) 5%) F =1 '3 £143.: xc—e) e x... 695 (can 4) (E'q gig-IQ Co: 1.11/1- =2JIT§ \fm (Ea-5:151:10 (£53.;6-S16E) \ ”Ev—fl"? -- C '2 (3* , Um: wwxnsrmaH 4v) (Ea gig-5) (9‘? ’6‘“ ' W Ignace mere-examm 51m a) Wau- gfi‘vm g. [’3‘an 4241- leg-75'. to: lo “TM/5, x“: loam, 43:17/3. YAK/‘L' :4; uJ/zn z.- lo¥/1.n ,3 1 L) From 23 [pg v00 = — cone ssm 09H 4;) = ~14. 53m (mu-4a) van : we a Q» @5000”) =- - c.) From €91 16-63 6!. 2—0056. CofiCdJ++¢)-:“flw Cps(w++¢>) g a,“ 2 (01X... 2 Ogre-ova)“ Clo CM) 2 1m lOanvu 04:69,“, yam”. —--l‘m= Ann/u)“ =— -. ._ 'L 0‘) F 1". mat. _: m L—M'LEWE X33 mid X0“) 2 — (0-10 3:40 [loved/51:) (2.95 EClovfid/hy: + 313-de = “1000512001'2‘ + +lwa“ . ,5 2_ s93. HIdmplepmdulmwidnhgflllfiOmmakflflOmdl- lain-sin luamkflnmfimnifimfim? Gx'wm.‘ £21500” ...
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