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hw2_solutions

# hw2_solutions - PHY 251 Fall 2009 homework problem set 2...

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PHY 251 Fall 2009: homework problem set 2, due PHY 251 drop box in room A-129 by noon on Friday, Sep. 18. 1. Dick and Jane are each carrying an exact meter stick (according to them in their frame of reference). Dick goes past you at mach ten or 3000 m/s, while Jane goes past you at half the speed of light. How much shorter than 1 meter does Jane’s meter stick appear to you? How much shorter than 1 meter does Dick’s meter stick look to you? Answer: In both cases, you see a length = 0 relative to what they see, or a length differ- ence of 0 = 0 (1 1 ) . In the case of Jane, we have β = 0 . 5 and γ = 1 / radicalBig 1 (1 / 2) 2 = 2 / 3 giving 0 = 0 (1 1 γ ) = (1 m )(1 3 2 ) = 0 . 134 meters . In the case of Dick, we have β = ( v/c ) = (3 × 10 3 / 3 × 10 8 ) = 10 5 so we have to use a low- β expansion for the Lorentz factor γ : 0 = 0 (1 1 γ ) = 0 parenleftBig 1 (1 β 2 ) 1 / 2 parenrightBig 0 parenleftbigg 1 (1 1 2 β 2 ) parenrightbigg = 0 1 2 β 2 = (1 m ) 1 2 (10 5 ) 2 = 5 × 10 10 m . Given that atoms have a size of around 2 × 10 10 m, this is going to be hard to detect! 2. A smug city slicker bets a farmer that he can’t get his 10 m long ladder into a 8 m long shed. The farmer, who reads Einstein each day after milking his cows, takes him up on the bet. He tells the city slicker to stand to the side of the shed and look in the windows at each end, and the farmer then runs fast as he can through the shed while carrying the ladder. How fast does the farmer have to run to win the bet? While on the run, how long does the shed appear to the farmer, and does the farmer ever think his ladder is entirely inside the shed? (3 answers required). Answer: For the city slicker in the stationary frame S 1 to observe the ladder as being com- pletely inside the shed while the farmer runs by in the farmer’s frame S 2 , we must have a Lorentz contraction calculated from L 1 = (1 ) L 2 . Therefore γ = L 2 L 1 = 10 m 8 m = 5 4 . Now γ 2 = 1 / (1 β 2 ) so β = radicalBig 1 1 2 = radicalBig 1 4 2 / 5 2 = radicalBig 9 / 25 = 3 / 5 or v = 0 . 60 c . Now, to the farmer the shed appears to be moving toward him at v = 0 . 60 c , so its length is Lorentz contracted according to γ = 5 / 4 , giving an apparent shed length of (8 m ) / (5 / 4) = 6 . 4 m. As a result, the farmer never sees the ladder as being completely

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hw2_solutions - PHY 251 Fall 2009 homework problem set 2...

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