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# hw9 - PRDBLEM 14.38 3 In a game of pool ball A is moving...

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Unformatted text preview: PRDBLEM 14.38 3 In a game of pool, ball A is moving with a velocity v“. = vﬂi when it _ strikes balls 3 and C, which are at rest side by side. Assuming frictionless surfaces and perfectly eiastic unpaet {that is. conservation of energy]. determine the ﬁnal velocity of each ball, assuming that the path of A is . {a} perfectly centered and that A strikes B and C simultaneously, {b} not perfectly centered and thaw! strikes 3 slightly before it strikes C. SOLUTION {a} A strikes 3 and Csimattcmeomfy. During the impact1 the contact impulses make 3W angles with the velocity v.) , @‘w '36" I we: Thus, it a = v5 {cos 3ﬂ°l + sin 3W1] '4": = vctlcosE-ﬂ‘ﬁ — sin 30°j} By symmetry. VA. = vdi Conservation ol'memcntum: men = va + mm + myt- y component: i) = {I + mvﬂ sin 311° — mvr sin 311° uc = v3 1' component: my, = mud + ”W” cos3ﬂ“ + m1? cos 3|)“ I’D—11A 2 1’3 + l-‘r : c0530 = Ell?“ — FA}, . 1 ~ 1 1 l 1 Conservation of energy: Emmy; = 3-va + Emvﬁ + Erma;- PROBLEM 14.33 CONTINUED VA = 112901!” -— '1 v3 = \$693sz 41 3t)“ '1 vc- = sis-govﬂ “I: see 1 {b} A strikes 3 before if straits C. First impact; A strikes B. During the impact. the contact impulse makes a 3i)“ angle 1with the 1.relocitgir v“. mﬂ' ”Hi 'Is : FAN ‘h '5._ i' ' l \h‘; \__ J! Thus. '93 = v5{co53ﬂ°i + sin 30°j} Conservation ofmomenturn. men = me}, + mes y comPonenl: I] = mlvﬂl + mug sin 30" (it; it = -vysii'13[l° 1' component: vn = mlmﬁ it + mug c0530“ (In; L = '11 — v5 cosﬂﬂ” Conservation of energy: 1 ‘r l 2 l I l g Emit; = Emliﬂh + ‘mllirih. + EMVB - lmiv —- v 00530“): + liv shim“): + iv: 2 ll If! 2 H 2 H = é—mhﬁ r lvnva + vices2 3D“ +1’;5il‘l2 30° + vi] V}: = v.) cos3ﬂ° = git“, {whit = vﬂ sinIJD" 2&1)”, 5 {iii}? ‘-' ﬁt.) \$05313” sin3ﬂ° = -7“) Second impact: A strikes C. During the impact, the contact impulse makes a 30" angle with the velocity v“. I‘ my“ K..-.. (”E KI: '54.“. #1! GE]? ms; ... mg, m— = 12C [ccsﬁﬂ‘ti — sin30°j} PROBLEM 14.38 CONTINUED Conservation nfmomentmn: mv'A = va + mvc x component: m [vﬂx = JI1I9~[I.:*A}I + mvc ms 3L1“, {VAL = {“1111 _ ”I: 90330“ = in; — v: 00531)“ 3; component: m [1.93. L = m {VA}; — misc sin 30" (”up + ”(*Sinmc’ = —%vn + unsin3ﬂl° Canservatinn «of energy: I 1 am“; 1|: + Em£ 1 2 1 l 1 3 1 1 1 ' 1E . 1 —m —u“ —r' =—m —v —v 005312)“ + ——v +v sm3ﬂ° + - 2 [16” 16”] 2 [[4” C )[4‘3 C j v“ 1 1 1 1 1 1 —m —v ——vv c033ﬂ°+v' :03'3'0“ 2 [“5 n 2 n r: r 3215 + Eva — 1—1901? sin 30“ + vf1sin2 31')” + v31] 1 3 . u = —vnv£1[§c0330° +§sm3ﬂn] + 21:51 15 Ji w. = vﬂ[iL‘H53ﬂ°+ TsimSﬂa] = T1"; I 1E l [VAL = Ev” - Tvﬂcnﬂﬂ‘“ = —Evﬂ «E 15 J? v =——v +—v sin3ﬂ“ =-—-v (.111 4 1 4 1 3 1 1;. = 0.250111 7 4500 «1 v3 = 0.356112” ‘5 3D“ ‘1 1,1. = 10.433111, 11: an“ «I PROBLEM ‘MAZ In a game of pool. ball A is moving with a velocity,r v.3 of magnitude v“, = 5 mi's when it strikes balls 3 and C. which are at rest and aligned as shown. Knowing that alter the collision the three bells move in the directions indicated and assuming frictionless surfaces and perfectly elastic impact {that is, conservation of'enetgy}, determine the magnitudes ofthe velocities v1. v3, and Vi"- SOLUTION Velocity vectors: vﬂ = vu[cos45°l + sin 45°j} em = 5 ma’s VA = ‘51] v3 = vﬂilsinoﬂﬂ - eosﬁﬂﬁ] Vs: = vt-[coslSDDi + sinﬁrﬂnj] Conservation of momentum: mvu =mvl+mv3+mvf Divide by m and resolve into components. 1: v9 cos45° = v3 sin ﬁll“ + vc. cosoﬂ" j: vusin45" = \$54 — vﬂcosﬁﬂ“ + vt-sinﬁﬂ“ solving for v3 and vf, v3 = 0.2538215: — 0.5m Vt" = 0.96593“; - ﬂﬁﬁéﬂfivd . l 1 Conservation ofenergv: Emvﬁ = inn-f, + émtﬁ + Emvﬁ Divide by m and substitute for v3 and we. vﬁ = vi. + {oessszvn + o.5v,,}2 + {osssssvﬂ - osssosvdf = is + Loom.f + 21:3. pa, = orilm lvu = 3.54 ms v}, = 3.54 ms 4 v5 = [iﬁlZﬁ'i't-D = 3.06 Inn's v3 = 3.156 mfs *1 of = ill..'.’r5351.'?rvﬂl = 1.168 rnis vi. = LTﬁB this '1 PROBLEM 14.43 horizontal surface. Spheres A and B are attached to an inextensihle, inelastic cord of length i and are at rest in the position shown when sphere B is struck squarolj,r by sphere C which is moving with a velocity v”. Knowing that the cord is taut when sphere B is struck by sphere C.‘ and assuming perfectly.r elastic impact between B and C', and thus conservation of energy for the entire system, determine the velocity of each sphere immediately after impact. \$0 LUTIDN JE cost‘il = ?. e =19.4Tl° Velocity vectors v.“ = —vﬂj ".4 = v_4{eoso"| — sinﬂj} vb“ = nB{—sint?i — eosti‘j} 1"'s = v, + Vii-vi W: E ”(ii Conservation of momentum: mvu = ”We + mvﬁ. + mvc = 2mm + mv3._, + new Divide by m and resolve into components. i: t] = EvAcosﬂ — nasint? —j: v” = +2vA sint‘ir + a,_,.costiI — v.5. - ' Three spheres, each of mass m, can slide freelyr on a frictionless, I J Solving for ”.4 and 1:3, v, = item + Va} Hy = ﬂ.94281{vﬁ + v5] . 1 1 1 '." I. 'I I. Conservation of energy: Emvﬁ = Emu; + 3“? + Emir} 1 2 1 s 2 i 2 = Ens-VA + —m[".1 +1£3)+ Emitt- Divide by in: and substitute for v, and n3. v5 = Eta—{Va + 1t’c i1 +iﬂ-9423'i2ivc + VG]: + ”3' o I. v,-, — v5. = 0.94445“, + v._.] v... = casein, v,. = iiﬂEEﬁt-‘u 11 v, = c.1i14sv, v, =[o.1r14sv,‘“:1a4ne], v, = 0.1?14vn “\$19.5“! a, = casein, v ,,._,, [c.sesisv, f 19.4710] vﬂ = v,, + vﬁm vﬁ = [L935 q: 80.!” 4 PROBLEM 14.49 Two small spheresA and B, with masses of 2.5 kg and 1 kg, respectively, are connected by a rigid rod of negligible mass. The two spheres are resting on a horizontal, frictionless surface when A is suddenly given the velocity vﬂ = [3.5 Inl's}i. Determine {o} the linear momentum of the system and its angular momentum about its mass center G, on the 21 H mm velocities of A and B after the rod AB has rotated through 180°. SOLUTION Locate the mass center. Let I he the distance hetweon A and 3. my?! = {md + ”EVA ifim—s md+mﬂ ".I' 5 I J's-JFE I (a) Linear momentum. | L = m_,,v,, = {2.s}{s.s} = 8.15 L = ens kgalm's - 1 it“ | angular momentum shout G: -I,";.e,, = em, = Elmo, = E[o.s1o}[2.s}[3.s} = 0.525 I- 7 ? HG = 0.525 kg-mio :34 [b] There are no resultant external forces acting on the system; therefore, L and HG are conserved. i. L: myr1 +mﬁv3 = L 2.5V2f +1.13%. =3.T5 {1} . \ ix Ho: llamal’s ‘ {imam = Ht: %{U.21U}[1.ﬂ}vﬁ —%[o.2ie}{as}v, = 0.525 ﬂlﬁvﬁ. — [HSvA = 1152.5 [2} Solving [l] and [2} simultaneously, v, = 1.5 1nfs, v3 = 5 mfs v_,d = 1.5m} mils —- ‘1 v3 = Sﬂﬂmls _. *1 ...
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hw9 - PRDBLEM 14.38 3 In a game of pool ball A is moving...

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