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15 November, 10:09

What is the percent change in momentum of a proton that accelerates from 0.41c to 0.85c?

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  1. 15 November, 10:15
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    At these speeds, we can't ignore the relativistic masses.

    At 0.41c: m = m₀ / √ (1-.41²) = m₀ / √0.8319 = m₀ / 0.9121 = 1.096 m₀

    At 0.85c: m = m₀ / √ (1-.85²) = m₀ / √0.2775 = m₀ / 0.5268 = 1.898 m₀

    The change in momentum is (1.898m₀ x 0.85c) - (1.096m₀ x 0.41c)

    = (1.613 - 0.449) m₀ c = 1.164 m₀ c.

    As a fraction of the original momentum, that's (1.164 / 0.449)

    = (2.592) = 159% increase.
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