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 - Problems - Page 194: 46

Answer

$v'_A=3.7\frac{m}{s}$ to the right $v'_B=2.0\frac{m}{s}$

Work Step by Step

In the y direction $mv_A=mv_B'+mv'_{Ay}$ $v_A=v_B'+v'_{Ay}$ In the x direction $mv_B=mv'_{Ax}$ $v_B=v'_{Ax}=3.7\frac{m}{s}$ $\frac{mv_A^2}{2}+\frac{mv_B^2}{2}=\frac{mv_A'^2}{2}+\frac{mv_B'^2}{2}$ $v_A^2+v_B^2=v_A'^2+v_B'^2$ $(v_B'+v'_{Ay})^2+(v'_{Ax})^2=v_A'^2+v_B'^2$ $(v_B')^2+2(v_B')(v'_{Ay})+(v'_{Ay})^2+(v'_{Ax})^2=v_A'^2+v_B'^2$ $2(v_B')(v'_{Ay})=0$ $v'_{Ay}=0$ $v'_A=v'_{Ax}=3.7\frac{m}{s}$
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