Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 7 - Linear Momentum - General Problems - Page 195: 67

Answer

The two astronauts are 20 meters apart.

Work Step by Step

Let $v_1$ be the speed of the lighter astronaut, and let $d_1$ be the distance traveled by this astronaut: $d_1 = v_1t$ $v_1 = d_1/t$ Let $v_2$ be the speed of the heavier astronaut. We can use conservation of momentum to find an expression for $v_2$: $m_2~v_2 = m_1~v_1$ $v_2 = \frac{m_1~v_1}{m_2} = \frac{(55~kg)~v_1}{85~kg}$ We can use $v_2$ to find $d_2$: $d_2 = v_2~t = (\frac{(55~kg)~v_1}{85~kg})(t) = \frac{55~kg}{85~kg}(d_1) = \frac{(55~kg)(12~m)}{85~kg}$ $d_2 = 7.8~m$ The total distance between the astronauts is $d_1 + d_2$ which is 12 m + 7.8 m = 20 meters. The two astronauts are 20 meters apart.
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