College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 7 - Problems - Page 263: 25


The recoil speed of the submarine is $1.0~cm/s$

Work Step by Step

We can assume the initial momentum of the system is zero. By conservation of momentum, if the torpedo gains a certain momentum $p_t$ in one direction, the submarine will gain the same magnitude of momentum $p_s$ in the opposite direction. We can find the recoil speed $v_r$ of the submarine: $p_s = p_t$ $m_s~v_r = m_t~v_t$ $v_r = \frac{m_t~v_t}{m_s}$ $v_r = \frac{(250~kg)(100.0~m/s)}{2.5\times 10^6~kg}$ $v_r = 0.010~m/s = 1.0~cm/s$ The recoil speed of the submarine is $1.0~cm/s$.
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