Question

# A certain particle of mass $$m$$ has momentum of magnitude $$m c .$$ What are $$(\mathrm{a}) \beta,(\mathrm{b}) \gamma,$$ and $$(\mathrm{c})$$ the ratio $$K / E_{0} ?$$

Solution

## (a) We set Eq. 37-41 equal to $$\mathrm{mc}$$, as required by the problem, and solve for the speed. Thus,$$\dfrac{m v}{\sqrt{1-v^{2} / c^{2}}}=m c\\$$leads to $$\beta=1 / \sqrt{2}=0.707\\$$(b) Substituting $$\beta=1 / \sqrt{2}$$ into the definition of $$\gamma$$, we obtain$$\gamma=\dfrac{1}{\sqrt{1-v^{2} / c^{2}}}=\dfrac{1}{\sqrt{1-(1 / 2)}}=\sqrt{2} \approx 1.41\\$$(c) The kinetic energy is$$K=(\gamma-1) m c^{2}=(\sqrt{2}-1) m c^{2}=0.414 m c^{2}=0.414 E_{0}\\$$which implies $$K / E_{0}=0.414$$Physics

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