11판/9. 질량중심과 선운동량

9-49 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 4. 25. 20:29
put {b:boatc:car \put \begin{cases} b:\text{boat}\\ c:\text{car}\\ \end{cases} {mc=1500[kg]Lc=3.0[m]mb=4000[kg]Lb=14[m]M=mb+mc \begin{cases} m_c&=1500\ut{kg}\\ L_c&=3.0\ut{m}\\ m_b&=4000\ut{kg}\\ L_b&=14\ut{m}\\ M&=m_b+m_c\\ \end{cases} Δxcb=ΔxcΔxb=11[m] \begin{aligned} \Delta x_{c\larr b}&=\Delta x_c-\Delta x_b\\ &=-11\ut{m} \end{aligned} ΔΣp=0,\Delta \Sigma \vec p=0, 0=Δpcom=MΔvcom \begin{aligned} 0&=\Delta \vec p_\com\\ &=M\Delta \vec v_\com\\ \end{aligned} xcomf=xcomiΣxfmM=ΣximMΣxfm=Σximmbxbf+mcxcf=mbxbi+mcxci \begin{aligned} x_{\com f}&=x_{\com i}\\ \frac{\Sigma \vec x_fm}{M}&=\frac{\Sigma \vec x_im}{M}\\ \Sigma \vec x_fm&=\Sigma \vec x_im\\ m_bx_{bf}+m_cx_{cf}&=m_bx_{bi}+m_cx_{ci}\\ \end{aligned} mbΔxb+mcΔxc=0 m_b\Delta x_{b}+m_c\Delta x_{c}=0 Δxb=mcMΔxcb=3[m]=3.0[m] \begin{aligned} \Delta x_b&=-\frac{m_c}{M}\Delta x_{c\larr b}\\ &=3\ut{m}\\ &=3.0\ut{m} \end{aligned}