homework2 - 735136.383 eaters Hememm—Rit'Q Sacrifice-o...

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Unformatted text preview: 735136.383 eaters Hememm—Rit'Q Sacrifice-o 1-19. By analogy with Equation 1-23, (a) “L: ux+v2= 0.9c+0.9c 2=fl==09945c 1+vuf/c I+(0.9c)(0.9c)/c '1.81 (b) u 2 1H,, 3 (1.8c/181)+0.9c _1..8+(09)(1.81) _3_4296_ 0.99970 1+vu; lc 1+(1. 8c/1 .81)(o. 9c)/c 1.81+(1. 8)(0 9) 3. 430 .. ._ .. L L 21. 1-26. (a) In the spaceshlp the lengthL = the proper length LP; therefore, t5 = ——”—+-1 = p . C C C (b) In the laboratory frame the length is contracted to L = L1: / y and the round trip time is (0) Yes. The time ts measured in the spaceship is the proper time interval 1. From time dilation (Equation 1-26) the time interval in the laboratory IL = yr ; therefore; 1 ZLP lw-vz/c2 c L 1-34. Let S be the rest frame of Earth, .5" be Heide’s rest frame, and S" be Han’s rest time. 1 = 1.1 198 y . cu—I-_—'.‘_‘_ s 1fl—(OASC/c)2 9/5. = ——-—1—_ a 3.206 ,/1—(0.95c/c)2 When they meet, each will have traveled a distance d from Earth. Heide: d: 0.45c 1‘” .d . _ _1) "' - 035: (’9 . Hans: d = 0.95c 1M Q'J'SCfi “his, =1 ““4”: and tHans : [Heia‘e _1 in years' M/ Therefore, (0.95c — 0.4SC)IHME = 0.95c and 1M = 1.90 y; rm = 0.90 y ' (a) In her reference frame S ' , Heide has aged A1.” when she and Hans meet. [I (Problem1-34 continued) 0.45c 2 C At'zys.[At—-1;—Ax] =1.1198[1.90— (0.450x1.90)] C =1.198(1.90y)[1-—(0.45)2:l =1.697y In his reference frame S " , Hans has aged At" when he and Heide meet. . v 5-316 of? - °'c‘3"/<:%2(°35cxofic§) At =75-(At—23-Ax]:3 Wy)Ll—(0.95) =0.290y agmmfl) I- (0.13) a 0.3%,) ' - ' The difference in their ages will be 1.697 — 0.290 = 1 gm}, 3 1.4 y _ I 2: L, 7, W ' (b) Heide will be the older. H 2 ' 1-36. (a) Using Equation 1-28 and Problem 1-20(b); At'=At/y =Az(1—v2 l2c2)=At—Atv2 Izc2 whereAt=3.15x'107s/y -- - — v = 2mg. IT =(27:)(637xIO°m)/(108min)(6OS/min) v=6.177x103m/s =2.06x10'5c Time lost by satellite clock = sz #2 = (3.15 ><107 s)(2.06><10“5)2 12 = 0.00668: = 6.68m 0’) ls=At(v2 /2c2) At = 2/(v2 lcz) = 2!(2.06><‘10'5)2 = 4.71x109s =150y . . 1-40.‘ (a) Alpha Centauri is 4 c-y away, so the traveler Went L = 1/1—[32 (8c- y) in 6 y,' or 8c-ym=v(6y) m xv.(6/8c)=(3/4)(v/c) #1—[32 =(3/4)2 163 (3/4)2,62 +32 =1 I [32 =1/(1 +0.5625) v=0.8c At=yAto =y(6y) and y=1l l—fiz =1.667 (b) , At = 1 .667 (6 y) = 10y or 4y older than the other traveler. =3 Alpha Centauri .1-46. (a) L=Lply =Lp'x/1—u2/cz =_100m‘/I—(0.85)2 =52.7m u +u _ 0L85c+0.85c _ 1.706 1+uu/c2 1+(o_gs)2 1.72 (c) L’=Lp/y'=Lp\/1—u’2/cz =1()01v111,f1—(1.70/1.72)2 =16.1m (b) u'= =0.987c (d) As viewed from Earth, the ships pass in the time required for one ship to move its _ own contracted length. At = £1 2 JE— — 2.1 x 10% (e) 2'2' d (ymu) = m(udy + ydu) 1 l 1 2-43. (3) y=———--——-—-—=——————-——-:——=I.25 ’lwuz [02 f1—(0.6c)2/62 0.8 t ‘50-“ M (b) p = ymu = Ame): (u/c)/c =1.25(0.511MeV)(0.6)/c = 0.383 MeV/c (c) E = ymcz :1.25(0.511MeV)=0.639 MeV (d) 15* =(y—1)mc2 =0.25(0.511,MeV):0.128 MeV 2-4. The quantity required is the kinetic energy. Ek 4-: (y “11)mc2 = [(1 -—u2 [.62 Tm -—1]mc2 2 —-ln’2 2 2 _(a) Ek =[(1—(0.5)) -—l]mc =0.155mc (b) E,e =[(1_(0.9)2)_"2 4]ch =1.29mc2 (c) 13,,=‘[(1—-(0.99)2)_"2 4] me2 = 6.095“?- '.2-5. AEzAmc2 Am=AElc2= 10.17 2 =1.1x10‘16kg (3.03x103m/s) _ Because work is done on the system, the mass increases by this amount. (a) v=(Earth-— Moon distance)/time=3.8x108m/1:5s=0.84c (b) E1: = gvmc2 ——,mc2 2 me2 (1/ —1) (Equation 2—9) mc2 (proton) = 938.3MeV y =1/1/1—(034)2 = 1.87 Eh =938.3MeV(I.87——1)=813MeV . _ 2 (c) m(u)=—~—T-——=W=l.730x103MeV/cz =1.73oc;eV/c2 (d) Classically, E,‘ = gm}?1 = %(93’:8.31MeV/c2 )(0.84c)2 = 331MeV %mor=Wxioo=s9% 9 2—8. 7 Ek 2 me2 (7—1) (Equation 2-9) Ek (u2)—Ek (ul)=W2, =-—mc2 (y(u2)—1)—mc2(y(ul)—i) 01', W51 Smcz [7(u2)_7(u1)] I (a) W21 = 938.3MeV[(1—0.162)_m —(1—0.152)””2] =1.51MeV _ (b) W21 = 938.3MeV[(1— 0.862 )'”2 —(1—0.852)"”2] =57.6MeV ‘ (c) WZI = 938.3MeV[(1—m0.962)_m —(1L~0.952)‘”2] = 3.35x10’MeV = 3.35GeV 212 E = ymc2 (Equation 2—10) ' y = El me2 = 1400MeV /938MeV = 1.4925 (a) y=1/x/1—u2/c2=1.4925 ~+ 1~uZ/¢.~2=1/(1.4925)2 u2 /c2 =1—1/(1.4925)2 =-0.5-51 —+- u=0.74c (b) E 2 = (pa)2 4- (me2 )2 (Equation 2-32) :1 2- 21:11 ' '2 2 =1040MV/ g 113 (me) c[( 4001145211) (938MeV)] e c 2—14. u=2.2:><105m/sandy=1/\/1—uZ/c2 (a) I 2 2 -s Ek=0.511MeV(y-1)=0.511MeV(1/ l—u fr: —1)=0.5110_(2.6s9x10 ) =1.3741x10-5MeV 1 1 Ek(classi¢a1)= —2-mu2 = Emcz(1:2lcz):(0.51IOMeV/2)(2.2><106 lcz)2 =1.374x10"5MeV .9 % difference = —1—X10—x 100 = 0.0073% 1.3741x10's . _._m w mm 4132 —(mc:2)2 =%1f(ymc2)2 -—(mc2)2 A . , ' 2 1:2 =mcq/f—l=EI-E:[[ll1/l—(2.2x106{3.0x108)zJ «1] =0.5110MeV/c(7.33x1’0-3)=3.74745x10‘3'MeV’/c ' (b) 1 p:.... , C mc2 ' p(classical) = mu = [E] = (0.51 1()1|/I¢2V/c)(2.2>~<106 /3.0x105) . C CZ =3.74733x10‘3MeV/c 1.2x10—7 ' % dlfference = m ' .. . X x100 = 0.003% ...
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