Physics 43 HW 2 Chapter 39 Problems given from 7 th Edition

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1 Physis 3 HW Chater 39 Problems gien from 7 th Edition Problems:, 7,, 9, 1, 0,,, 9, 33, 35, 3, 0, 5,. How fast must a meter stik be moing if its length is measured to shrink to m? P39. L = L L Taking L = where L = 1.00 m gies L = L L 1 = = = 0.66 L 7. An atomi lok moes at km/h for 1.00 h as measured by an idential lok on the Earth. How many nanoseonds slow will the moing lok be omared with the Earth lok, at the end of the 1.00 h interal? P39.7 Δ t = γδ t = Δt so Δ t = Δt 1 t Δ and If then Δ t Δ t = Δt 1.00 = km h = = 77. m s s = 9.6 and Δt Δ t = s = 1.5 s = 1.5 ns 7. A muon formed high in the Earth s atmoshere traels at seed = for a distane of.60 km before it deays into an eletron, a neutrino, and an antineutrino ( μ + + ) e. (a) How long does the muon lie, as measured in its referene frame? How far does the muon trael, as measured in its frame? P39. For =, γ = (a) The muon s lifetime as measured in the Earth s rest frame is.60 km Δ t = and the lifetime measured in the muon s rest frame is 3 Δ t 1.60 Δ t = = = γ ( 3.00 s).1 μs 3 L.60 L = L = = = 69 m γ 7.09

2 9. A saeraft with a roer length of 300 m takes μs to ass an Earth obserer. Determine the seed of the saeraft as measured by the Earth obserer. P39.9 The saeshi is measured by the Earth obserer to be length-ontrated to oerhead by: L = L or L = L Also, the ontrated length is related to the time required to ass L = t or L = t = ( t) Equating these two exressions gies L L = ( t) or + = L t L Using the gien alues: L = 300 m and 7 t = 7.50 s this beomes giing = = The idential twins Seedo and Goslo join a migration from the Earth to Planet X. It is 0.0 ly away in a referene frame in whih both lanets are at rest. The twins, of the same age, deart at the same time on different saeraft. Seedo s raft traels steadily at 0.950, and Goslo s at Calulate the age differene between the twins after Goslo s saeraft lands on Planet X. Whih twin is the older? P39.1 In the Earth frame, Seedo s tri lasts for a time Δx 0.0 ly Δ t = = = 1.05 yr ly yr Seedo s age adanes only by the roer time interal Δt Δ t = = 1.05 yr 0.95 = 6.57 yr during his tri γ Similarly for Goslo, Δx 0.0 ly Δ t = = 0.75 = 17.6 yr ly yr While Seedo has landed on Planet X and is waiting for his brother, he ages by 0.0 ly 0.0 ly = 5.61 yr ly yr ly yr Then Goslo ends u older by 17.6 yr ( 6.57 yr yr ) = 5.5 yr.

3 0. A moing rod is obsered to hae a length of.00 m and to be oriented at an angle of 30.0 with reset to the diretion of motion, as shown in Figure P39.3. The rod has a seed of (a) What is the roer length of the rod? What is the orientation angle in the roer frame? 1 1 P39.0 γ = = = We are also gien: L 1 =.00 m, and θ = 30.0 (both measured in a referene frame moing relatie to the rod). L = L osθ =.00 m 0.67 = 1.73 m x Thus, 1x 1 1 and L = L sin θ = (.00 m )( 0.500) = 1.00 m 1y 1 1 L is a roer length, related to by x L1x L1x = L γ Therefore, L =.0L = 17.3 m x 1x and L = L1 = 1.00 m (Lengths erendiular to the motion are unhanged). (a) L ( L ) ( L ) = + gies L = 17. m x y y y L 1 y θ = tan gies θ = 3.30 L x A red light flashes at osition xr = 3.00 m and time tr = s, and a blue light flashes at xb B = 5.00 m and tbb = s, all measured in the S referene frame. Referene frame S has its origin at the same oint as S at t = t = 0; frame S moes uniformly to the right. Both flashes are obsered to our at the same lae in S. (a) Find the relatie seed between S and S. Find the loation of the two flashes in frame S. () At what time does the red flash our in the S frame? P39. (a) From the Lorentz transformation, the searations between the blue-light and red-light eents are desribed by Δ x = γ ( Δx Δ t) 0 = γ.00 m (.00 s).00 m = =.50 s.00 s 1 γ = = s 3.00 s Again from the Lorentz transformation, x = γ ( x t) : x = m (.50 s)( 1.00 s) x =.97 m () t = γ t x : (.50 s) ( 3.00 s) t = s ( 3.00 m ) t = 1.33 s

4 . A Klingon saeraft moes away from the Earth at a seed of 0.00 (Fig. P39.6). The starshi Enterrise ursues at a seed of relatie to the Earth. Obserers on the Earth see the Enterrise oertaking the Klingon raft at a relatie seed of 0.0. With what seed is the Enterrise oertaking the Klingon raft as seen by the rew of the Enterrise? P39. u x = Enterrise eloity = Klingon eloity From Equation ux u x = = = u x An unstable artile at rest breaks into two fragments of unequal mass. The mass of the first fragment is.50 kg, and that of the other is kg. If the lighter fragment has a seed of 0.93 after the breaku, what is the seed of the heaier fragment? P39.9 Relatiisti momentum of the system of fragments must be onsered. For total momentum to be zero after as it was before, we must hae, with subsrit referring to the heaier fragment, and subsrit 1 to the lighter, = 1 or or.50 kg γmu = γ1mu 1 1 = ( 1.67 kg) ( u ) Proeeding to sole, we find u (.960 kg) = u = and u = ( 0.93 ) 7 u u 1 = 33. Find the momentum of a roton in MeV/ units assuming its total energy is twie its rest energy. P39.33 E= γ m = m or γ = Thus, u 1 3 = = or γ 3 u = The momentum is then 3 m = γ mu = m = M ev 3 M ev = 3 = 1.63

5 35. A roton moes at Calulate its (a) rest energy, total energy, and () kineti energy. P39.35 (a) E m R 7 = = 1.67 kg.99 s = 1.50 J= 93 M ev 1.50 J 1 E= γ m = =.1 J = 3.00 M ev ( 0.950) 3 () 3 K = E m =.1 J 1.50 J= 3.31 J=.07 M ev 3. In a tyial olor teleision iture tube, the eletrons are aelerated through a otential differene of V. (a) What seed do the eletrons hae when they strike the sreen? What is their kineti energy in joules? P39.3 (a) ( Δ ) = = ( γ ) 1 e q V K m 1 q ΔV ev Thus, γ = = 1+ = 1+ = 1.09 u m e ev so 1 (u/) = and u= K ( γ ) me q( V) = 1 = Δ = 1.60 C.50 J C =.00 J Consider eletrons aelerated to an energy of 0.0 GeV in the 3.00 km long Stanford Linear Aelerator. (a) What is the γ fator for the eletrons? What is their seed? () How long does the aelerator aear to them? P39.0 (a) E= γ m = 0.0 GeV with m = M ev for eletrons. 0.0 ev Thus, γ = = ev γ = = 3.91 from whih u= u () 3 u L 3.00 L = L = = = 7.67 = γ m 5. The ower outut of the Sun is W. How muh mass is onerted to energy in the Sun eah seond? P39.5 de dm dm dt dt dt 6 P = = = = 3.5 W Thus, dm dt J s 9 = =. kg s ( 3.00 s)

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