Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 9 - Linear Momentum and Collisions - Problems and Conceptual Exercises - Page 294: 71

Answer

$2.20m$

Work Step by Step

According to law of conservation of momentum $mv_{\circ}+M(0)=(m+M)(v_f)$ This can be rearranged as: $v_f=(\frac{m}{m+M})(v_{\circ})$ We plug in the known values to obtain: $v_f=\frac{0.0105}{0.0105+1.35}(715)=5.52\frac{m}{s}$ Now the distance can be calculated as $x=(v_f)(\sqrt{\frac{2h}{g}})$ We plug in the known values to obtain: $x=(5.52)(\sqrt{\frac{(2)(0.782)}{9.81}})$ $x=2.20m$
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