chapter26_hw_solution

chapter26_hw_solution - 26.1(a Observers on Earth measure...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 26.1 (a) Observers on Earth measure the time for the astronauts to reach Alpha Centauri as NE 2 4.42 yr. But these observers are moving relative to the astronaut’s internal biological clock and hence experience a dilated version of the proper time interval Np measured on that clock. From NE 2 y Np, we find mp =ArE/y2 Arm/I—(v/c)2 =(4.42 yr)./1—(0.950)2 = (b) The astronauts are moving relative to the span of space separating Earth and Alpha Cen— tauri. Hence, they measure a length—contracted version of the proper distance, LP 2 4.20 ly. The distance measured by the astronauts is L = Lp/y = LIN/1 —(v/c)3 =(4.201y)./1—(o.950)3 = 26.2 26.3 (a) The length of the meter stick measured by the observer moving at speed 0 = 0.9006 relative to the meter stick is L = LP/y = LP 1—(v/c)2 = (1.00 m) 1—(0.900)2 = (b) If the observer moves relative to Earth in the direction opposite the motion of the meter stick relative to Earth, the velocity of the observer relative to the meter stick is greater than that in part (a). The measured length of the meter stick will be less than 0.436 in under these conditions. If the moving clock is observed to “run at a rate one-half the rate of an identical clock at rest,” this means that an observer moving relative to the clock measures the period T of the clocks oscillating mechanism to be double the proper period Tp measured by an observer at rest relative to the clock (i.e., the moving observer sees the clock “tick” half as often). Thus, T = yTp 2 2T!” or y =1/J1—(v/c)3 = 2, giving fill—(v/C)2 2% or (0/6)2 =1— ’U/C=\/§/2 1 3 —=— and 4 4 A - —3 1/1—(v/C)‘ «ll—(0.98)- (b) d=v(Ar)=[0.98(3.0><108 [n/s)](1.3x10—7’ s): (c) d’=v(Arp)=[0.98(3.0>-<103 1n/s)](2.6><10_8 s): 26.6 (a) As measured by observers in the ship (that is, at rest relative to the astronaut), the time required for 750 pulses is At!) = 1.00 min. The time interval required for 750 pulses as measured by the Earth observer is At = 7/ At!) = (1.00 min)/\/1 — (0500):, so the Earth observer measures a pulse rate of 75.0 1— 0.500 3 1.00 min 1.00 min 1—(0.990)2 and the pulse rate observed on Earth is (b) If v = 0.9906, then At = 3/ Ar!) = 75.0 1— 0.990 2 1.00 min That is, the life span of the astronaut (reckoned by the total number of his heartbeats) is much longer as measured by an Earth clock than by a clock aboard the space vehicle. 2 lMeV 26.16 ‘ E = 2= 1.67x10’27ko 3.00><108 L —. = 939MV w R < ax mm ER N/l—(U/Cf 2M: 3.01x103 MeV= 1— (0.950)2 (C) KE =E—ER = 3.0] X 103 M6V—939 MeVz 2.07X 103 MeV2 2.07 GeV (b) E 2 7/27162 2 yER = 26.19 The nonrelativistic expression for kinetic energy is KE 2 mvz, while the relativistic expression is KE 2 E— ER 2 (y —1)ER = (y — l)mc3, where 3/ = l/‘fl — (ti/c)2 . Thus, when the relativistic kinetic energy is twice the predicted nonrelativistic value, we have %—1 17162 :2[lmvgj 01‘ 1:|:1+[3] l—(v/c)" 2 c 2 l—(v/c)2 3 4 2 Squaring both sides of the last result and simplifying gives {3) + (a) — I] = 0 c c c v 4 v 3 Ignoring the trivial solution 19/6 2 0, we must have [—j + [—J — l = 0 c c This is a quadratic equation of the form x2 + x — l = 0 with x = (ti/(3)2. Applying the quadratic 41mg 2 , we ignore the negative solution and find formula gives x = ’2 Since x =(v/c) v 3 —1+J§ x: — = 20.618 wh'*h 'elds vzc 0.618: 0.786c [j 2 u. w J— C 26.22 Let ml be the mass of the fragment moving at 721 = 0.868 c, and m2 be the mass moving at at = 0.987 C. From conservation of mass-energy, _ mic2 r2136: 3 E = —1 + —q = me 1— (0.868)‘ 1— (0.987)‘ giving 2.01m] + 6.22 m2 = 3.34 ><10_27 kg [1] The momenta of the two fragments must add to zero, so the magnitudes must be equal, giving P] = p: or jzlmlvI = yzmzvz. This yields 2 .01 ml (0.868 c) = 6.22 m2 (0.987 c), or m] = 3.52 m2 [2] Substituting Equation [2] into [1] gives (7.07 +6.22)n~i2 = 3.34 x1047 kg, or m: = 2.51X10—28 kg Equation [2] then yields m] : 3.52(2.51><10_38 kg) : 8.84X10-33 kg I R 0.511MeV lGeV 2.525 (a) yzgzm[m}=3mw p E L 3. 3 _ ., _ (b) L=_=m=mxlo—-m= y 3.91x104 ...
View Full Document

{[ snackBarMessage ]}

Page1 / 8

chapter26_hw_solution - 26.1(a Observers on Earth measure...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online