# Chapter 37 - Relativity - Problems - Page 1146: 12b

$\gamma = 2$

#### Work Step by Step

From part (a), we knew that $\beta = 0.866$. The formula for lorenz factor is $\gamma = \frac{1}{\sqrt (1-(\frac{v}{c})^2)}$. Substituting the above value of $\beta$ in the formula and solving: $\gamma = \frac{1}{\sqrt (1-\beta^2)}$ $\gamma = \frac{1}{\sqrt (1-0.866^2)}$ $\gamma = 2$

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