1051_PartUniversity Physics Solution

# 1051_PartUniversity Physics Solution - 37-14 Chapter 37...

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37-14 Chapter 37 Solve this equation for v : 2 2 22 1/ vF t vc m ⎛⎞ = ⎜⎟ ⎝⎠ and 2 2 (1 / ) Ft vv c m =− 2 1 Ft Ft v mc m += so 2 2 2 (/) 1( / ) Ft m Ft Ft mc m c F t == ++ As 2 2 , 1, Ft Ft t mc Ft →∞ + so . EVALUATE: Note that Ft + is always less than 1, so < always and v approaches c only when . t 37.64. Setting 0 x = in Eq.(37.21), the first equation becomes xu t γ ′=− and the last, upon multiplication by , c becomes . ct ct ′ = Squaring and subtracting gives 2 2 γ () , ct x ut ′′ −= = or 8 4.53 10 m. xc t t = × 37.65. (a) IDENTIFY and SET UP: Use the Lorentz coordinate transformation (Eq.37.21) for 11 (,) xt and : 1 , t x uc ′ = 2 t x ′ = 2 1 / , tu t ′ = 2 2 / t ′ = Same point in S implies 12 . xx = What then is 21 ? ttt ′′′ Δ= − EXECUTE: = implies 112 2 t xu t −=− 2 1 ut t x x and xx x u tt t −Δ From the time transformation equations, 2 1 (/ ) x c Δ= − = Δ−Δ Using the result that x u t Δ = Δ gives 2 1 (( ) / ( ( ) ) ) 1( ) / ( ( ) ) x t c c Δ= Δ Δ 2 ) / ( ( ) ) ) () () / t x t c tx c Δ Δ Δ− Δ 2 2 / () ( / ) , / c x c c Δ = Δ Δ as was to be shown. This equation doesn’t have a physical solution (because of a negative square root) if ) () t Δ> Δ or . t Δ≥Δ (b) IDENTIFY and SET UP: Now require that = (the two events are simultaneous in S ) and use the Lorentz coordinate transformation equations. EXECUTE: = implies / / xc tu 2 u c so 2 x c Δ and 2 u x Δ = Δ From the Lorentz transformation equations, 1 . xxx t Using the result that 2 / uct x =ΔΔ gives 2 1 ) / ) 1( ) / ( ) c t x ct x Δ Δ −Δ Δ 22 2 ) / ) x c t x t Δ Δ Δ t c t t Δ (c) IDENTIFY and SET UP: The result from part (b) is c t Δ= Δ Δ

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Relativity 37-15 Solve for 22 2 2 : ( ) ( ) ( ) tx x c t ΔΔ= Δ−Δ EXECUTE: 2 2 8 8 ( ) ( ) (5.00 m) (2.50 m) 1.44 10 s 2.998 10 m/s xx t c Δ− Δ Δ= = = × × EVALUATE: This provides another illustration of the concept of simultaneity (Section 37.2): events observed to be simultaneous in one frame are not simultaneous in another frame that is moving relative to the first. 37.66. (a) 80.0 m s is non-relativistic, and 2 1 186 J. 2 Km v == (b) 21 5 (1 ) 1 . 3 1 1 0 J . mc γ −= × (c) In Eq. (37.23), c) 88 7 2.20 10 m s, 1.80 10 m s,and so 7.14 10 m s. vu v = × (d) 20.0 m 13.6 m. = (e) 8 8 20.0 m 9.09 10 s. 2.20 10 m s × (f) 8 13.6 m 6.18 10 s, or 6.18 10 s. 2.20 10 m s t tt −− ′′ × = = × × 37.67. IDENTIFY and SET UP: An increase in wavelength corresponds to a decrease in frequency (/ ) , fc λ = so the atoms are moving away from the earth. Receding, so use Eq.(37.26): 0 cu ff = + EXECUTE: Solve for u : 2 0 (/ ) ( ) ff cu cu +=− and 2 0 2 0 1(/ ) uc ⎛⎞ = ⎜⎟ + ⎝⎠ 00 / , / fc f c λλ so / / = 8 0 0 1 ( / ) 1 (656.3/953.4) 0.357 1.07 10 m/s 1 ( / ) 1 (656.3/953.4) c c = = × ++ EVALUATE: The relative speed is large, 36% of c . The cosmological implication of such observations will be discussed in Section 44.6. 37.68. The baseball had better be moving non-relativistically, so the Doppler shift formula (Eq.(37.25)) becomes 0 (1 ( )).
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1051_PartUniversity Physics Solution - 37-14 Chapter 37...

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