A proton, which has a mass of $1.67 \times 10 ^ { - 27 }$, is moving at $0.80 c$ relative to the laboratory. What does a physicist in that lab measure for its momentum? $\left( c = 3.0 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } \right)$
A) $6.7 \times 10 ^ { - 19 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
B) $5.8 \times 10 ^ { - 19 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
C) $4.0 \times 10 ^ { - 19 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
D) $5.0 \times 10 ^ { - 19 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
E) $4.4 \times 10 ^ { - 19 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$

Question

Quizplus · Accepted Answer

The momentum of a relativistic particle is given by $p = \gamma mv$, where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$, $m$ is the rest mass of the particle, and $v$ is its velocity. For a proton moving at $0.80c$, $\gamma = \frac{1}{\sqrt{1 - (0.80)^2}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} = \frac{5}{3}$. Thus, $p = \frac{5}{3} \times 1.67 \times 10^{-27} \times 0.80 \times 3.0 \times 10^8 = 6.7 \times 10^{-19} \, \text{kg} \cdot \text{m/s}$.

A Proton, Which Has a Mass Of 1.67×10−271.67 \times 10 ^ { - 27 }1.67×10−27

A Proton, Which Has a Mass Of $1.67 \times 10 ^ { - 27 }$