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Ch15_Sec4_Ex-59-69-77-87-89

Ch15_Sec4_Ex-59-69-77-87-89 - 5 9 If two disksA and u nave...

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Unformatted text preview: 5- 9. If two disksA and u nave me same ulna-a a..- ..... ntral impact such that the collision ' ub acted to direct ce '5 rfectly elastic (e = 1), prove that the kinetic energy Is the kinetic energy after collision. efore collision equa 6 surface upon which they slide is smooth. (J19)3 va‘ =2mv; ”Mm )1 + "In (Va )l = "H (‘M )2 + my (Va )2 ‘ ’"A[(VA)I"(VA)2]=ml[(VB)1’(VB)l] (I) ~ i (Vim-(VA): ‘ l =—-——---=1 (a): ‘ (mi—(mi (Va )2 - (VA )2 = (VA )l - (Vn)i (2) Coinlbintng Eqs. (1) and (2) : MA EM )‘l - (VA )2][(VA)I + (VA )2] = my [(VB )1 ' (VI )i][(vl )1 + (V: )1] * l l i 1 d~ultilb~z ExpWanm pyy2 l i l l l -Mli (VA )i +51". (V- )i = EMA (‘M )i + Ema (V: )i Q-E-D- 15-69. ‘e slider block B“ _ IS confined to ' ' 2:00:11: t. It IS connected to two spring:I 2:312:33: l ess ofk=30 N/m The " . y are 71:22:; .5; when 3 = O as shown. Detergiillgienatiiz block A hl's hnice, 3'3“? block B moves after it is hit b Take e = 04m lS ongtnally traveling at (22,01 = 8 m/y and the mass of each block to be 1.5 kg s. (L’) 2’"?! “3M2 (letsxm = (1.5m ), 41.5“.” (s) .- l. — ( A)l -(v,), Ont = £F£8:(()VA )2 l -(l.5 5.601 ‘ —l ' 2 )( ifiz[2(3oxo.5)’]=o+2[§(30)(,/:,2...+2145)? ’mu = 1.53111 1 Ans 15-77. \ e cue ball A is given an initial velocity (“)1 = 5 m/s. If i makes a direct collision with ball 8 (e = 0.8), determine the velocity of B and the angle 0 just after it rebounds from the cushion at C(e' = 0.6). Each ball has a mass of 0.4 kg. Neglect he size of each ball. ‘9 Conserleiori of Momentum : When ballA strikes ball B. we have lmAthh-i-matva), =mA(uA):+mB(L),,)I (+ ) 1 0.4(5)+O=O.4(v‘)2+0.4(vfl)2 Co efficient 0y Restitution .- e= (”5): “(1,4): (1,4)] "(Dah + Jinn-(vi): ('7) 0'8‘ 5—0 Solving Eqs.[l] and [2] yields (19‘)2 =0.500 mls (u,,)2 =4.50 mls Conservation of 'y ” Momentum : When ball B strikes the cushion at C. we have “19(09): =m, (99,), (+ i) 0.4(4.50sin 30°) = 04(1),), sin e (1)5)3 sin 0 = 2.25 Coefficient of Restitution (x) : e_(vc)z '(”s,)3 (05,)2'(vc)| (*1) o‘6=0—[-(v,)3cosBl 4.50am 30° - 0 1 Solving Figs. [13) and [2] yields (v5)3 = 3.24 m/s 9 = 43.9° 296 15-87. Two {iisks A and B each have a weight of 2 lb and the initia velocities shown just before they collide. If the coefficient of restitution is e = 0.5, determine their speeds just after impact. \ i i 2 ‘ 3 2 4 2 2 353‘3l(3)‘5r2.—2“’(§)=m‘mh +312“ ’2‘ (+/) “waft: afifififi—Zfi] Solving, (vA‘ ), = 0.550 fps l (v,_ )2 = -l.95 TA =1.95ftls +- BallA: (+\) \ "Ii (VA,)i =”'A (V4,): 23 3 ‘_ 2 ‘m‘4’(3)‘-3—23("~)2 (Vb): =—2.40 ftls Thus. l (VA )2 = J(0.550)1 +(2.40)z = 2.46 ft/s Ans (v.); = ,/(1.95)1+(2.40)2 =3.09 ftls Ans ”15(V I)! =nla (VI, )2 ——»<;)==g-2<va.>z (m) =-2.40ftls .Vl 15-89. Ball A strikbs ball B with an initial velocity of (VA), as shown. If both balls have the same mass and the collision is perfectlyielastic, determine the angle 0 after collision. Ball B is originally at rest. Neglect the size of each ball. ‘ ‘ Velocity before impact : l (UM). =(v4).c05¢ (UAy)1=(vA)lSin¢ (Uni): =0 (03,). =0 Velocity after impwt : (DA: )2 =(DA )2 C050] ('04,): =(U4h5in91 (vi, )2 = (v: )1 «>st (115,), =-(v,)2 sine; Conservation of " y " momemim: "'5 (”By )I = ”'5 (”By )1 0=m[—(un);sin0;] i9z =0° ”no a Conservation of " x " monientum: "IA (04. )i “"5 (”31)! ="IA (Duh +"ln(vax)2 3 i5 MivAll €05¢+0=m(u,4)1cos01 +m(v.)1coso° A A (v4).c05¢=(vA)zcosei+(vn)z (l) Coefficientof Restitution (s direction) 2 ’d e- ‘(Unxh‘muh . l- (Ua)2¢050°‘(ll4 )2 C059: _ M _ “a (uni): “(Dull ' (UA)1COS¢-‘0 (v4). cus¢=—(u4)zcos0. Hull): (2) Subtracting Eq.(l) from Eq (2) yields: ZivA house. =0 Since 201,4); :0 C0591=0 6| =90° io=91+92=90°+0°=90“ Ans ...
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