# hw10 - 25.7 a The iength is Ezﬁz—iO—mzﬂr—ZOﬁOm...

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Unformatted text preview: 25.7 a} The iength is Ezﬁz—iO—mzﬂr—ZOﬁOm. 7 w ' 7,} 2 1—-(0.se)2 “(El b) Since you measure the iength to be 0.60 m and it is traveling past you at a speed of 0.80c, the time it takes to pass you is ' 0.60 In : 0.60 m x 2-5 X 104, S. 0.804: 080(300 x 105 m/s) Zoom! R? 25.13 at) Your roommate calculates your speed by dividing the distance traveled by the time as measured by his rulers and clocks at rest in his reference frame. So, v _ 100.0 rn 0 5.00 x 10-7 s = 2.00 X108 m/s. 956 CHAPTER 25. THE SPECIAL THEORY OF- RELATJVITY I b) and c) In your reference frame you are at rest, and the corridor is moving past you at a speed 1'; = 2.00x 108 m/s Consequently, relative to you, the length of the corridor is ' V g_£_0_100.0m 3; 7 “r ' i So, to'ﬁnd 2 we ﬁrst need to ﬁnd 7: Ar : ——-——2 : z 1 _ v_ #1 _ (2.00 x 198 m/s)2 C? t (3.00 x 108 m/s )2 Therefore, relative to you, the length of the moving corridor is __ Q 4 100.0 m ’y H 1.34 According to you, the corridor is 74.6 m long and is traveling past you at a speed of 2.00 x 108 m/s. Therefore, you say the time it takes to pass is this distance divided by the speed: If, __ 74.6 m m 2.00 x 108 m/s 3 274.6300. = 3.73 ><-10_7 s = 373 ns. 25.17 a) Assume that a-Centauri and the Earth are at rest relative to one another You see yourself at rest; and the Earth leaving you at a speed ’0‘. Therefore: relative to o-Centauri approaching you at a speed 1), you, the distance from Earth to a—C‘entauri is contracted; and is f 30 _ 4.2 LY a 7 “r ' In your reference frame, it takes you 3.0 y to travel this distance. Therefore, in your reference frame, 4.2 LY 4.2 LY LY distance : (speed)(time) :> If: o(3.0 y) z? 7 : o(3.0 y) :> 71; z 3 0 y = 1.4 . 959 25.18. . . . . LY Now one light year per year 13 gust the speed c of light. So, 1.4 m 1.41:. Thus Y 71.1 : 1.4c. We want to solve this equation for 1). Since 7 also depends upon 1;, we rewrite “f in terms of v and turn the proverbial mathematics}. crank: l 71; : 1.4a :4; 2 1) 2"- 1.4a => 2 2 (1 4)2c2 1.... 0—. 1 —- HE" C2 0 v2 112 => 25 2 — v2 __ {1.4)2 c2 m l + (1.4)2 :s 3 = 0.82 c So 'u :: 0.826 = U.82(3.00'x 105 m/S ) : 2.5 x 108 m/s. b) The distance in your friends’ reference frame is 4.2 LY and they see you traveling at a constant speed of 0.81c. Hence, in their reference frame, the time required for the journey is 4.2 LY LY = 5. —~»- : 5. . 0.82c 1 ( c > 1 y 25.18 a) Let Event 1 be the event that occurs ﬁrst in S, and let 3:1 and 7:1 be its position and time. Let Event 2 be the one that occurs 1.00 5 later. Assume that the x-axis in S is oriented so that the m-coordinate of Event 2 is greater than the :c-coordinate of Event 1. Then in 5' the ttvo events are described by (Event 1) a: = \$1 and t = t1, and (Event 2) a: = \$2 = 2:1 —E 2.40 X 108 m and t: t2 = t1 + 1.08 5. Remark. Had we assumed that the :c—coordinate of Event 2 was less than the cc—coordinate of Event 1, then the position equation for Event 2 wonid have been \$=x2:\$i—2.4{)x108m, VVe’ll see below that, in the standard geometry, this is impossible. b) Transform the two events to the 3’ reference frame with the Lorentz transformation, Equations 25.33 and 25.36 on page 11645 of the text. 33’ i 3’1 : nan — vii) (Event 1) U 15" St; 37(t1—E'é‘1‘1). m’ : 3’2 2 7(322 — 11752) = 7(351-E— 2.40 x 108 In —— v(t1+ 1.00 s)] (Event 2) n. t’ :t’g 27(t2—Cigxg) :7(t1+1.0[}s —612(a:1+2.40 X 108m). CHAPTER 25. THE SPECIAL THEORY OF RELATIVITY 960 We are told that in Sf, both events are at the same position x’. Hence, m; z 3; 2a 7931 _ Us) : 7m + 2.40 x 108 m a o(t1 + 1.00 5)) :2)» 0m 22.40 x108 m “111.008 => 0:2.40x 108 m/s. then the quantity 2.40 X 108 In in the above ' Remark. Had we assumed in part a) that 3:1 < 33;, v : ~2.4Ox108 rn/s. computation would have been replaced by m2.40>< 108 In, and we would have found that Since speed can’t be negative, our initiai assumption that \$2 > \$1 has to be correct. c) In S", the time between the two events is I r . o ’U n r; w t3 : 7(t’1—I— 1.00 s — 55% + 2.40 x :08 111)) — 7(31 — C—Exl) : '7 (1.00 s — E(2.4.0 x 10?: an) 8 1 ' (1.00 s — wean x 105 111)) = 0.600 s. X 108 m/S)2 X m/S)‘ 27 (3.00 x 108 Isl/s)? 25.22 3.) Choose the origin where the Ferrari is when t : O s'. The Ferrari is evidently not accelerating, so you describe its posrtion by 3(3) a 30+ mt : 0 m + (0.20%): = (0.200c)t. Similarly, you describe the position of the Yugo by \$0?) : (0.5806}{tw10.03). The Yugo catches up to the Ferrari when they have the same position: (0.2000)t = (0.500c)[t — 10.0 s) r: 2: 10.73. b) Determine the position of either car when t : \$6.7 s. For example, the position of the Ferrari is 30:) = (0.2003) : 0.200(300 x 103 m/s){is.7 s) = 1.00 x109 m g one billion meters away! 25.26 Here’s the picture a». B «Mali .1 e We’ve chosen 6' at rest with respect to space ship B, and S” at rest with respect to the Earth. ' Let u be the speed of the Earth in the frame 3, and um the x—component of velocity of space ship A in frame 5’. We are given that um z. 0.90s and we want to ﬁnd '0. We also know that in the frame 5" , the speed of space ship A is n —— since both space ships are approaching the Earth with the same speed. Therefore, it; 2 1). Substitute 0.90c for um and v for u; in the relative velocity addition equation (Equation 25.45 on page 1172 of the text) and simplify. ' ’ ' i; ’U 1) 2:0 2vc2 - — um : It’s—\$17 : 090C = iw = 2 = 2 2 ::> [190((22 +212) 2 2’UC I 15111 1+ _ ' 'U C +1) 3. '1' 2 C? 1 "75' c c~ 2} 0.90112 H 261; + 0.90::2 = 0 1112/52; 25,27. 7 Apply the quadratic formula ‘0' x —-{—20) i «4—22)? m 4(0.90)(0.90c2) : 2 i1/4 — 4(090)? _ “2(030) ‘_ "We = (1.11 i 0.48); - __ {Choose the root ‘ '0 2 (1,11 _ 0453):: : 0.63c, 25.28 Choose one of the protons, and let S be a frame at rest relative to it. Let S’ be the frame of the lab. We’ll use the relative veiocity addition equation (Equation 25.45 on page 1i72 of the text) with "u = 0.990 0000, the speed of the lab in the chosen proton‘s frame, u; : 0.990 000.2, the speed of the other proton in the lab’s frame, and ﬁnd use the speed of the other proton in the chosen proton’s frame. u __ ug-iv __ 0990000c+—0990000c __ 201900000) C__09999496 I *'1 I age *ﬂl 1 (09900000)Ul§90000c) "14+(0.990000)2 ‘ ‘ ‘ T T m______._ 02 02 25.29 Here is a picture of the situation, viewed from S. a) Use the relativistic velocity component addition equation: umr-i—v ﬂ (0m/s)+e Ex: —— ’U 2 0990:. 1+ 1+ (0 Irish For the other velocity component, use 1:53,: = uyiiuxl ﬂ m/S) “ % 7(1+ Cg) A,(1+ c2 ) / 958 - CHAPTER 25. THE SPECIAL THEORY OF RELATIVITY Now we need the numerical value of 7. ’y —- 71 W : 7.1 v2 1 — {0.990}? 1 * —2 0 Hence f u 2 ﬂ = 0.9950 : O‘MC y "f 7.1 b) The angle gz') that the velocity 13 makes with the m—axis in S is found from uy 0.14c -— — z 2 =§> rad 3 8.0g. tw¢—u\$ 0wm 0 _ ¢ , ...
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hw10 - 25.7 a The iength is Ezﬁz—iO—mzﬂr—ZOﬁOm...

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