Hw8sol[1] - Chapter 8 209 Divide equation(2 by(1 tan 0 = = 0.633 2.43 From which Then either(I or(2 gives V =(b K =(9o.o kg(5.oo m/s)2(95.o kg(3.oo

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Unformatted text preview: Chapter 8 209 Divide equation (2) by (1) tan 0 = = 0.633 2.43 From which Then, either (I) or (2} gives V = (b) K,- =%(9o.o kg)(5.oo m/s)2 +%(95.o kg)(3.oo m/s)2 = 1.55x 103 I K, =%(185 kg)(2.88 111/5): = 7.67x 102 J Thus, the kinetic energy lost is 783 Jinto internal energy. p1]: = pxi: moo cos 19+ mvy cos(90.0°—-0) = mo,- 61 o ----®---- _ Before 00 cos 0+ vY sm9= v,- (1) ©/io pyf = pyi: moo sinB—mvy sin(90.0°—6’) =0 (/4; _ . _ - — _ q _ . _ _ _ -_ ql...__—_-_..__. v0 sin6=vycost9 (2) ‘m" _ \(_ -9 «From equation (2), u R cost? I g; = 3 Y z’o “([3ng ( ) a FIG. P825 Substituting into equation (1), . - v 0052 9 +0 sinfl- :2- Y sinfi Y ' so oficos2 9+ sin2 6)=v,-sin6, and . Then, from equation (3), 00 = v,- cosfl . We dld not need to wnte down an equation expressing conservation of mechanical energy. In the problem situation, the requirement of perpendicular final velocities is equivalent to the condition of elasticity. ' 210 P826 P827 P828 Momentum and Coliisfons We use conservation of momentum for the system of two vehicles for both northward and eastward components. For the eastward direction: M(13.0 m/s) = 21WJr cos 550° For the northward direction: 5.00 m/s + 0 = (4.33 III/S) cos 30.0°+ 02!: = 1.25 m/s 0 =(4.33 m/s)sin30.0°+v2fy ‘0ny G -2.16 m/s 62f = 2.50 m/s at -6o.o° “Hz-f: (a) §i=l3f 3° pn=paf and 3% = PM‘ m7),- = mv cos 9+ mvcos ¢ (1) O” 0=mvsin0+mosin¢ (2) From (2.), sin0=~sin¢ so 6: -¢. Furthermore, energy conservation for the syetem “6' P828 of two protons requires émvi2 =§mv2 +—mv2 212 Momentum and Collisions (b) E=1 2 1 2 mlv1+ 2 =21[(5.oox10-27)(cs_00><10“)2 +( 13:439x10‘” 1 Section 8.5 P832 _ 2mg,- _ (2.00 ~37??— the middle square, and A3 .‘lCM 1ch = 2 The Center of Mass (2-50 kg)(0)+(4.oo ng—osoo In) W %M(5.00 cm)+g-M(15.D cm)-+ @MXZSD cm) W 1 2 may; 8.40x 10*”)(400 x105)2 +(3.60x 10-”)(125 x10 : N kg)(3.00 m) +(3.00 kg)(2..50 m) + = 13.3 cm 220 Momentum and Collisions (d) person: my —me,. =60.0 kg(1.33—4.00)§ m/s= cart: 120 kg(1.33 m/s)-— 0 = 1 1 (e) x f —x,. = 3(vi + 3,): = E[(4,00+ 1.33) m/s]0.680 s = 1 1 ' (f) xf—xi =§(v,-+vf)t=§(0+1.33 m/s}0.6805=m ‘PBAB Using conservation of momentum from just before to just after the impact of the bullet with the block: 222 Momentum and Collisions P851 (a) The initial momentum of the system is zero, which remains constant throughout the motion. Therefore, when m1 leaves the wedge, we must have mszedge +mlvblock =0 or (3.00 1(3):;wedge +(0.500 kg)(+4.00 111/5) = 0 (b) Using conservation of energy for the block~wedge- Earth system as the block slides down the smooth (frictionless) wedge, we have FIG. P351 [Kbiock +usysteml. +[KwedgeL =[Kblock +usystem]f +[Kwedgelf I or [0+ mlgh]+0 = [5415(400)2 + 0]+%m2 (—0.667)2 which gives . P852 T =d—m dt (Avmel)+%rt£(Avah)z T=(3.00 kg/s)(600 III/s)+(80.0 kg/s)(600 m/s—ZZB m/s) T=1800N+3.02x104 N= 3.20x104 N The delivered power is the force (or thrust) multiplied by the velocity, 3' 5P=Tv=(3.20x104 N)(2.23 m/s): *P8.53 It is interesting to compare this problem with Problem P650. The force exerted by the spting on ea block is in magnitude IE4 = 1:2: = (3.85 N/m)(0.08 m) = 0.308 N. (a) With no friction, the elastic energy in the Spring becomes kinetic energy of the blocks, w ' have momenta of equal magnitude in opposite directions. The blocks move with constant speed after they leave the spring. (K+LI)I.=(K+U)f 1' 2 1 ~2—kx =Em1vff+§mzv§f l 1 ' 1 3(305 N/m)(0.08 m)z =E(0.25 1(3):]; +—2—(0.50 kg)v§, "1161; + "1262'- : m1?!” + "1262! 0 = (0.25 kg)vlf + (0.50 kg)vzfi U” = 2102f 1 2 1 1 0.012 3 1:5(025 ngszf) +§(0.50 kg)v§, =§(1.5 1:90;, 1/2 0.123 I _. c 92! =[0I75 kg] =0.128 m/s v2, =0.128 m/s: v1, =2(0.128 m/s)=0.256 m/s 0” =0.256 m/s(—i) continued on next page Chapter 8 225 —1 "2y .1 1/50,- 3 = 9: -— = (b) a tan (Uh) tan [ 3 Each object swings dOWn according to ng =%mv12 MgR =§Mu12 01 = 1{2gR Swinging up: gun + "on; = (M + m)gR(1 — 005.35“) :12 = J 2gR(1 — cos 35°) I 1[233M] — cos 35°) (M + m). = (M — _m)1/2.gR 0.425M +0.425m = M— m 1.42511: = 0.575M P11“ = pn': mshell val-tell cos 45‘0°+mca1mon Dread] = 0 (200K125) cos 45.0°+(5 000)!)recoil = 0 Use conservation of energy for the system of the cannon, right after the cannon is fired to the instant when the the carriage, and the spring from cannon comes to rest. ‘ 1 1 1r<f+u‘ng +113). =Ki+um+um¢ owing” 2 m‘ kl: 2.00x10‘ m: Page 260 Rotational Motion SOLUTIONS TO PROBLEMS Section 10.1 Angular Position, Speed, and Acceleration P10.1 m 61H, = dB 011% =EL=O =10.0-r-4.00£1t=u = 10.0 rad/0 _ dd), _ 2 a,=0 — dt i=0 — 400 rad/s (b) 61Hum s = 5.00 + 30.0 + 18.0 = d0 df=3£o s = dCL=3Im 5 = + 4.001‘Lz100 s = t t d6?) 2 _ = ‘I = 4.00 re. s “11-3.00 s d i=3”) 5 tics—The ngld Object Under Constant Angul Acceleration P10.2 of = 251x104 rev/min = 2.63 x 103 rad/s I I w -a:- .6 1 3 _ . (a) a=—f—t—_'=-z—§i-g—2—r§_/i_9= 8.21x102 ta 52 . S . 1 (b) 0f=oJ,.r+§.spr"=0+—;_}-(8.21x102 rad/52)(3.2.s)2= 421x103 rad 100 rev 1min erad 10:: P103 w-=———__ ____ ._.__ I =0 ‘ 1.00min(60.0sII.00rev) 3 rad/s mi w 0-103 (a) f: f I __ az+w- 10 1 0» 9r=ar=[—’-z——']t=(-z’5mws )= a),- = 3 600 rev/min =3.77>< 102 rad/s "PIOA 49:50.0 rev=3.14><102 rad and a1, =0 50% = a),-2 +2a9 Chapter 10 261 a; = 5.00 rev/s = 10.01: rad/s. We wiil break the motion into two stages: (1) a period during which the tub speeds up and (2) a period during which it slows down. 0 + 10.07: rad/s . P105 While speeding up, 91 = a = 2 (8.00 s) = 40.0:rrad While slowing down, 92 = ‘ait = WOZD s) = 60.0lrrad so, em = 01 + 62 = 1001'rrad = 9r —9,- =a),-t+%arz and to}: =a1,-+at aretwo equationsintwo unknowns mi and a: — ~m- a —9 -(m ~at)t+latz~wt lmz wi'wf - f i"- f 2 - f ‘2 2.1: rad 1 rev 61.5rad ' 232 d=294 d- 4.50 2a: a: -= ra ra ( s) 45,052 m At? lrev anad _5 =--= = = 7.27x10 (3) w M May 864005 A0 107° anad 4 _ At=—-—=-——— = 2.57x10 3 or428mm (b) . a, 7.27x10'5 rad/Si 350°] M 37.0 rev( J=9s.o rad/s (3.00 s)—%a(3.00 s)2 $0.3 Relations Between Rotational and Translational Quantities 2 Estimate the tire's radius at 0.250 m and miles driven as 10 000 per year. 9_§_1.00x104 mi[1609m =6.44 107 r 0.250111 J x rad/yr lmi lrev 644x rad/yr[2”rad] 10 x 0 rev/yr or 0 rev/yr _Zflrad[1200rev v a at = (126 rad/s)(3.00 x 10'2 m) = =w2r=<126>2(8-00x10'2)=1260 m/sz soar = s 2 r6: m = (126 rad/s)(8.00 x10‘2 m)(2.oo s) = (d) (6) Chapter 10 263 _ v_1.30 In/s_ .-—;———-—0_023m — 1.30 m/s 1 “i=m= ‘ 2.2.4 rad/s—56.5 rad/s —34J rad/s _3 2 =w-+at:a= =————= —7.63x10 5 “'1' ' (74(60)+33)s 44735 rad/ —1 —1 565 224 4473 —1 105 d Bf—Bi—E(a)f+a)i)t—E(( . + .)rad/s)( s)— x =0: =(1.30 m/s)(4 473 s): 5.81 x103 m = 400 kg, r1 =|y1|'= 3.00 m; = 2.00 kg, 72 =ly2] = 2.00 m; =3-00 kg; r3 =Iy3| = 400 m; _;_:-2.00 rad/s about the x-axis Ix=m1 I, = 40003.00)2 + 200(200)2 + 3.000400)2 = page 2_1 2_ 2wa -§(9z.0)(2.00) _ The force of static friction must act forward and then Inore and more inward on the tires, to produce both tangential and centripetal acceleration. Its tangential component is m(1.70 III/32). Its radially 2 . . mv . . ‘ mward component 15 —. 'I'l'us takes the maxunum value 1‘ mwfir = mr(aJ,-2 +2m56) = mr(0 + 20%] = mind: mmt = mrr(1.70 m/sz). With skidding impending we have 2F}, = may, +n — mg = 0, n = mg f5 zflsn =flsmg=1/m2(1.70 m/sz)2 +m27rz(1.70 m/sz)2 1.70 2 fls=———;‘/s - Rotational Kinetic Energy 1'12 4' marzz + M3732 264 Rotational Motion 1 1 (b) 121 = r150: 3.00(2.00) = KI = Em! of = Emmxaoof = 720 1 U2 =T2w= K2 251-71123: = 03 =r3w=4.00(2.00)= K3 =-21—m31)32 1<=I<I +1<2 +K3 =72.0+16.0+96.0==211wa P10.15 From conservation of energy for the object-mmtable-cylinder- Earth system, 1 v 2 1 2 +3"!!! —mgh U2 2 I;— = 2mgh — my P10.16 The moment of merha of a flun rod aboutan 3x15 through one end IS I -§ML2. kmehc energy is 3mm as K R =zllhw§ +2154»: 2 2 “nth bl = :11th = 60.0 kg(32.70 n1) =146 kg m2 2 2 and 1,. ELIng = 100 kg: 50m) =675 kg'm2 erad 1 h In ddm ————. =1 1 4 a on 0,, 12h [36003] 45x 0 rad/s anad 1 h _3 = —-—-— = 1 0 whfle 0),” 1h {36005) 75x1 rad/s Therefore, K R = 5-(146)(1.45 x 10“)2 +-— N '3‘ {31‘ 1—: E‘n X H Q d; M II ...
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This note was uploaded on 01/24/2011 for the course PHYSICS 6A taught by Professor Koskeshian during the Fall '10 term at UCLA.

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Hw8sol[1] - Chapter 8 209 Divide equation(2 by(1 tan 0 = = 0.633 2.43 From which Then either(I or(2 gives V =(b K =(9o.o kg(5.oo m/s)2(95.o kg(3.oo

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