## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 26 - Problems - Page 1011: 31

#### Answer

The speed of electron B in a frame of reference in which electron A is at rest is $\frac{5}{13}~c$

#### Work Step by Step

Let west be the positive direction. Let $v_{BL}$ be the velocity of electron B relative to the lab. Then $v_{BL} = \frac{4}{5}~c$ Let $v_{AL}$ be the velocity of electron A relative to the lab. Then $v_{AL} = \frac{3}{5}~c$. Then $v_{LA} = -\frac{3}{5}~c$ We can find $v_{BA}$: $v_{BA} = \frac{V_{BL}~+~v_{LA}}{1+\frac{(v_{BL})~(v_{LA})}{c^2}}$ $v_{BA} = \frac{(\frac{4}{5}~c)~+~(-\frac{3}{5}~c)}{1+\frac{(\frac{4}{5}~c)~(-\frac{3}{5}~c)}{c^2}}$ $v_{BA} = \frac{\frac{1}{5}~c}{\frac{13}{25}}$ $v_{BA} = \frac{5}{13}~c$ The velocity of electron B relative to electron A is $\frac{5}{13}~c$ Therefore, the speed of electron B in a frame of reference in which electron A is at rest is $\frac{5}{13}~c$ (to the west)

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