## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 9 - Impulse and Momentum - Exercises and Problems - Page 240: 16

#### Answer

The velocity of the glider just after the skydiver lets go is still 30 m/s

#### Work Step by Step

Let $m_g$ be the mass of the glider after the skydiver lets go. Let $m_s$ be the mass of the skydiver. We can use conservation of momentum to find the velocity $v_{fx}$ of the glider after the skydiver lets go. Note that after the skydiver lets go of the glider, the skydiver still has a horizontal velocity of 30 m/s. Let's consider the system of the glider and the skydiver; $p_{fx} = p_{0x}$ $m_g~v_{fx}+m_s~(30~m/s) = (m_g+m_s)~(30~m/s)$ $m_g~v_{fx} = (m_g)~(30~m/s)$ $v_{fx} = 30~m/s$ The velocity of the glider just after the skydiver lets go is still 30 m/s.

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