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Unformatted text preview: 735136.383 eaters Hememm—Rit'Q Sacriﬁceo 119. By analogy with Equation 123, (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.
126. (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 126) the time interval in the laboratory IL = yr ; therefore; 1 ZLP lwvz/c2 c L 134. 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 (Problem134 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 5316 of?  °'c‘3"/<:%2(°35cxoﬁc§)
At =75(At—23Ax]: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 ' 136. (a) Using Equation 128 and Problem 120(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 . . 140.‘ (a) Alpha Centauri is 4 cy away, so the traveler Went L = 1/1—[32 (8c y) in 6 y,' or 8cym=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 .146. (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 1 243. (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 24. 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“? '.25. 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 29)
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 =0.551 —+ u=0.74c (b) E 2 = (pa)2 4 (me2 )2 (Equation 232) :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(y1)=0.511MeV(1/ l—u fr: —1)=0.5110_(2.6s9x10 )
=1.3741x105MeV
1 1 Ek(classi¢a1)= —2mu2 = 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=EIE:[[ll1/l—(2.2x106{3.0x108)zJ «1] =0.5110MeV/c(7.33x1’03)=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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 Summer '10
 ERBILL
 Physics, Work

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