exam1_03_solns

exam1_03_solns - (0.36) =1/0.6 =5/3 = 1.67 So t = 1.67...

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PHYSICS 6 HOUR EXAM 1 SPRING 2003 SOLUTIONS I. 1d, 2c, 3b, 4e, 5d, 6c,7c, 8e. II. a. accelerates b. distance it falls = (1/2) g t 2 = (1/2) × (9.8 m/s 2 ) × (5 s) 2 = 4.9 × 25 m = 120 m c. Ball takes 5 sec, whether thrown horizontally or not, since vertical (which accelerates) and horizontal (no acceleration) are independent. b d. Since horizontal motion is at constant 10 m/s, horizontal distance in 5 sec c will be 10 m/s × 5 s = 50 m III. a. Proper time is the given 9x10 -11 sec. In lab, which moves at 0.8c relative to Kaon, time will be longer (dilated) by factor 1/ (1- (0.8) 2 ) = 1/ (1- 0.64) = 1/
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Unformatted text preview: (0.36) =1/0.6 =5/3 = 1.67 So t = 1.67 9x10-11 s = 1.5x10-10 s b. In that time, traveling at 0.8c, the Kaon will cover 0.8 (3.0x10 8 m/s) (1.5x10-10 s) = 3.6x10-2 m = 3.6 cm. c. Velocity addition with u = 0.75c, the pion's velocity in the Kaon rest frame, and V=0.8c, the velocity of the Kaon rest frame relative to the lab. d. E=m c 2 = 8.85x10-28 Kg (3.0x10 8 m/s) 2 = 7.97x10-11 J or 7.97x10-11 J / (1.60x10-19 J/eV) = 4.98x10 8 eV = 498 MeV. 1 u = u +V 1 + u V c 2 = 0.75c +0.8c 1 +0.75 0.8 = 1.55c 1.60 = 0.97c...
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This note was uploaded on 03/29/2008 for the course PHY 006 taught by Professor Goldstein during the Spring '08 term at Tufts.

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