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: 47

Answer

a) $30^o$ b) $v'_T=-v'_N=\frac{v}{\sqrt{3}}$ c) $\frac{2}{3}$ of total kinetic energy

Work Step by Step

a) In the horizontal direction $m_Nv_N=m_Tv'_{Tx}$ $v_N=2v'_{Tx}$ In the vertical direction $0=m_Nv'_N+m_Tv'_{Ty}$ $v'_N=-2v'_{Ty}$ $\frac{m_Nv_N^2}{2}+\frac{m_Tv_T^2}{2}=\frac{m_Nv_N^{'2}}{2}+\frac{m_Tv_T^{'2}}{2}$ $v_N^2+2v_T^2=v_N^{'2}+2v_T^{'2}$ $(2v'_{Tx})^2+2v_T^2=(-2v'_{Ty})^{2}+2v_T^{'2}$ $4v_{Tx}^{'2}=4v_{Ty}^{'2}+2v_{Tx}^{'2}+2v_{Ty}^{'2}$ $2v_{Tx}^{'2}=6v_{Ty}^{'2}$ $\sqrt{2}v'_{Tx}=\sqrt{6}v'_{Ty}$ $\frac{v'_{Ty}}{v'_{Tx}}=\frac{1}{\sqrt{3}}$ $\arctan\Big(\frac{1}{\sqrt{3}}\Big)=30.0^o$ b) $v'_{Tx}=\frac{v}{2}$ $v'_T=\frac{\frac{v}{2}}{\cos(30^o)}=\frac{v}{\sqrt{3}}$ $v'_{Ty}=\frac{v}{\sqrt{3}}\sin(30^o)=\frac{v}{2\sqrt{3}}$ $v'_N=-2\frac{v}{2\sqrt{3}}=-\frac{v}{\sqrt{3}}$ c) $v^2=\frac{v^2}{3}+2\frac{v^2}{3}$ $\frac{2}{3}$ of total kinetic energy
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.