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HW_1solutions

# HW_1solutions - 13%Aic’s QQiSB""...

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Unformatted text preview: 13%Aic’s. QQiSB "" Homemeﬁk‘ﬁl SQCLC‘TLeNﬁ- 1-2. 'l-4. _ 2L _ 2(2.74x104m) _—————8——~——— =l.83x10'4s c 3.00x10 m/S , . 2L 2 (’0) From Equation 1-6 the correctlon 6: = —-—x 1:2— _ .. ‘ c c at=(1.33><1()*4s)'(10“)2 =1.83x10—‘2s ( ) F ' ta] measurements 55—6- — w; c mm exPenmen c # 299, 796km Is No, the relativistic correction of order 10'8 is three orders of magnitude smaller than =l.3x10_5 the experimentai uncertainty. (a) This is an exact analog of Example 1-1 with'L = 12.5 m, c = 130 mph, and v = 20 mph. Calling the plane ﬂying perpendicular to the wind plane #1 and the one ﬂying parallel to the wind plane #2, plane #1 will win by At where va (12.5mi)(20mi/h)2 At 2—";— 2m...”— x = 0.00231: = 8.2s c (130mm)- (b) Pilot #1 must use a heading 9 = sin‘l (20/130) = 8.8° relative to his course on both legs. Pilot #2 Inust use a heading of 0° relative to the course on both legs. 2L 2- .__._ . .. 'D.‘ MI 3 Ac: (qui'éition11'10) where a = 590 nm, L = 11 m, and AN = 0-01 ﬁ‘inge ANA 2 v2 = 2L6 =(0_01ﬁinge)(590x10—9m)(3.00>< 1081n/s)i.é(1 1m) v=4.91x103m/Sw5kmls 1-12 , vx' , ’ ' 7 7 77* k _ __ _ 1: l’(‘1 4-0—29] 2 = y[t2 4p?) (from Equation 1-19) , - 7- W, __ _ m ﬂ ”WV Vi ' vxl’ I Vx' I ' 7 _ _ .,-.-.-.,, (a) t2 —t, z: {:2 +—(-:—Zl—irl “720) _ y(t2 — I _ ’ v * (b) The quantities x: and x; in Equation 1—19 are each equal to x; , but x, and x2 in ’ ' ' Equation 1-18 are different and unknown. _ " _ ' 2 ”2 ___W 1 13‘ (a) y = 1/ (1 -—v2 /c2 )u = 1/[1 —(0.85.;-)2 [CI] = 1.898 W’ "WWWWW x' = y(x~vt) =1.898[75m —(0.85c)(2.0x10‘53)] = —-9.537x1o3m " " W ._ ...................... W 777777777 y! = y 2 18m 7,7,. WWW, WW .., ,.-,-,--...,-..+.VW. z' = z = 4.0m - .. .. - _ WWW 1' =_ 7(t — vx/ c2) =1.898[2.0x10‘5s ~(0.85c)(75m)/c2] = 3.756 x 10-53 f , W 7 7 ______ _ Cb) x ; y(x' + vt') =1.89s[—9.537 x103m +(0.85c)(3.756x10“55)] = 75.8-m difference is due to rounding of y, x', and t'. y = y' = 18m _ _——___ ___._ "I z : 2' = 4"011I ._ H _. "”— ""————'—""““""“'““ W" z = y (t' + tax" / c2) = 1.898 [3.756 x1043 + (0.85c) (—9.537 X 103 m)", 6‘2] = 2-0 X104, S " ' EEM__..__-M-J,_ _,___ 1—15. (a) Let frame S be the rest ﬂame of Bath and frame .5" be the spaceship moving at speed _ ﬁg v to the right relative to Earth; The other spaceship moving to the left relative to Earth iiiﬁ at speed It is the “particle”. Then v = 0.9:: and u, = —0.9c. iﬂﬁ~--——-~—~-—-—-- j. =_ if? (Equation 1-22) . ' Wemee I — uxv / c ‘"'""’"“"""""WW’”‘ . : —-0.9c-O.9c 2 = —1.8c = _0'9945c ,.,..,,-W, ______________;..an--..a 1—(‘—O.9c)(0.9c)/c 1.81 7 - 7 7 W W— ______ _ (b) Calculating as above with v = 3.0x 104ml 5 = —uJr _ ___ _ _ 4 _ 4 4 E w inf -n _ u; =W =;.6_'9X_10___§”_‘:/_S_ = —6.0xlp4m/s -. . W 1 (—3.0x10 m/s)(3.0x10 m/s) 1+10 — “___ W “ W (3.0x108m / sf, _ “WWW—— 1—21. At = yAt’ (Equation 1-26) At—At’ yAr'——At' l v2 1 v2 , = =7—1z——=7"lz‘“‘4‘ At Al" 2 c 20 , - , IIZ v2 =2c2 mg? v=c(2>< mtg—I'M] =c(2x{’).0})”2 =0.I4c t I ub-V‘ ' a)Notethat =1/\/1—v2/c2 ml and 1c-y=c- 3.15x1073 . 7 a From Equation 1-27: At =1!2 -—z‘I = 0 , since the novas are simultaneous in system S . (Earth). Therefore, in S’ (the aircraﬁ) Ar'=‘é—t{=-fz—(x{~xf) _(AJ(:Q)X=1)L:~ I 106 1;. ME = Ah ~"‘°/ca(°°5.“"°0 rmmjxwi’ —2.5x103)(c)(3.15x1073) S C (k =1.46><105s = 40.5}; (3-8.) , .(b) Since Ali-is positive, :;,..> I,’ ; therefofe, the nova in Lyra is detected on the aircraﬁ before the nova in Orion. \/1~—v2 lc2 ‘3 -8 ,,,,,,, , . __._________.. ____ _ _ __.__.._. ____.W,. = 2.6x10 5' =2.6x10 s =S.96x10‘3s . [1—(0.9c)2/c1]1-’2 m 1-24. (a) At=mf= (Bquationl—26) ﬂ , _ **** (b) s =‘vAt.= (0.9)(3.0x108mIs)'(6.0x1o-*s) =16.1m ——~ * * (c) s = vAt = (0.9)(3.0x108m/S)(2.6x10—85) = 7.0m , (\$1) (As)2=(cAt)2_-(Ax)2 (Equation 1731) J 7 [c(6.0x"10‘3)]2—(16.1-m)’=324—259'=65 —> As=7.8m ' ——"*" 1—25. From Equation 1-28, L =Lp [7 = LPW where L = 85m and Lp =100m ,, , - -- T‘Tmf'TM,".54#""""f"" , W = L/Lp = 85/100 _ ——- Squaring " 1*1’2/62 =(85/100)2 v2 = [1-(85/100)2]cz = 0.2775c2 and v = 0.527c =1.58x-108 m ls‘ _.___m__r ,,,,,, 1‘28* In 5’ ; Ax’ -_— 1 .0m cos 30.0 = 0‘ 866m ‘, , W ._.... ._____.______-__.__-,_m,,,,,,, Ay' = 1 .0m sin 30° = 0.500m where (9' = 30° (39“?5 WW7 __ , Rim {9&5 M13, M...,__._._____.____.__..___.__~.,_,,,,.,W, W¢____~ ,,,,,, In S : Ax = Axk/I #— 132 = 0.866m,/1—(0.8)2 : 0.520m ,,,,,,,7W,.,_______ Ay: Ay' = 0.500m , where 6 ==tan‘I 0500 =43.9° i , W,ﬁ,,,,,,___ __ " ' mw—u— #"ﬁmﬁrw'krﬁﬁiﬁﬁ ' V .520 . ._:- . ..-.. -,__,..__.. _u,u,ww._____ __ W_______.______ , L = "(my +(Ay)2 = “0520”,): +(0_500)2 = 0.721,” _rw ,,,,,,,,,,,, 1'30' A'=£=—€—=1/1_VICAO (Equation 1-36) , f' f 1+)3 1+v/c _ 0 —'ﬂ r 2 , 2 i =l——v/c -> 51— (l+v!c)=1——v/c 1‘, 1+v/c ,1" _ - 2 2 W 2 ' ' 1— 17/1 Solving forv/c, Z i] +1 =1_[i) :1=__(___.L).§ 6 lo 3., c 1+(1’llo) r 1— 590 [650 2 '21., = 650nm. For yellow ,1' = 590nm. 3 = __(_’1"i___i‘_’"_)_ = 0.097 c 1+(590m/650nm)2 ‘ Similarly, for green 1’ = 525nm —> 3 a 0.210 c 0.333 c ‘ , and for Blue A’ = 460111}: -—> I 1’. = .141 ”ti _{Ltﬁzltﬁ- fan/31:, —> 2,— lfﬂzo 1§ﬂ(656.3nm) . ' —3 For ﬁ :103 : ,1 = (656.3nm) \$13; = 657.0nm —2 For ,6 =10'2 : A = (656.3nm) _ 1+3: '=' 662.9nm .’ ‘ “L ,8 =10“l : ,1 =(6563nm) ‘ 1+1” = 725.6nm 1—10'I ...
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HW_1solutions - 13%Aic’s QQiSB""...

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