P237HW2sol

# P237HW2sol - Physics 237 Solution to Homework Assignment#2...

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Unformatted text preview: Physics 237 Solution to Homework Assignment #2 2-3. (a) γ u ( ) = 1 - u 2 c 2- 1/2 = 1 - 0.36 ( )- 1/2 = 1.25 (b) p = γ υ ( 29 μ υ = 1.25 μ υχ χ = 1.25 μ χ 2 0.6 ( 29 χ = 0.383 MeV/ c . (c ) E = γ υ ( 29 μ χ 2 = 0.639 MeV. (d) E K = Ε- μ χ 2 = 0.128 MeV. 2.8. When you change the speed from u 1 to u 2 , the work required is just the difference in kinetic energy: W = γ υ 2 ( 29 - γ υ 1 ( 29 ( 29 μ χ 2 . For a proton, mc 2 = 938.27 MeV. Using these equations, we get: (a) For u 1 = 0.15 c & u 2 = 0.16 c , W = 1.51 MeV. (b) For u 1 = 0.85 c & u 2 = 0.86 c , W = 57.6 MeV. (c) For u 1 = 0.95 c & u 2 = 0.96 c , W = 346 MeV. 2.13 (a) We are given that: E = γ υ ( 29 μ χ 2 = 2 μ χ 2 . γ u ( ) = 1 1 - u 2 c 2 = 2 u = 0.866 c . (b) We can also state the given information as: E = π 2 χ 2 + μ 2 χ 4 = 2 μ χ 2 . Hence, p 2 c 2 + μ 2 χ 4 = 4 μ 2 χ 4 π = 3 μ χ 2.19....
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## This note was uploaded on 10/15/2008 for the course PHYS 237 taught by Professor Stephonalexander during the Spring '08 term at Penn State.

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P237HW2sol - Physics 237 Solution to Homework Assignment#2...

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