9-52 할리데이 11판 솔루션 일반물리학

$$\put \begin{cases} S:\text{Ship}\\ G:\text{gas} \end{cases}$$ $$\begin{cases} M&=m_S+m_G=6100\ut{kg}\\ v_{G\larr S}&=-1200\ut{m/s}\\ g&=9.80665\ut{m/s^2} \end{cases}$$ $$\put \dot y=\dyt{y},$$ $$\Delta \Sigma \vec p=0,$$ $$\begin{cases} Mv_i&=m_Gv_{Gf}+m_Sv_{Sf}\\ v_{G\larr S}&=v_G-v_S\\ \Delta v_S&=v_S-v_i \end{cases}$$ $$\begin{aligned} -m_Gv_{G\larr S}&=M\Delta v_S\\ \dt(-m_Gv_{G\larr S})&=\dt (M\Delta v_S)\\ -\dyt{m_G}v_{G\larr S}&=M\dyt {v_S}\\ -\dot{m_G}v_{G\larr S}&=Ma_S=F_{\text{Thrust}}\\ \therefore \dot{m_G}&=\frac{Ma_S}{-v_{G\larr S}}\\ &=\frac{61}{12}a_S \end{aligned}$$ $$\ab{a}$$ $$F_{\text{Thrust}}=Mg,$$ $$\begin{aligned} \therefore \dot{m_G}&=\frac{61}{12}a_S\\ &=\frac{61}{12}g\\ &\approx 49.85047083333333\ut{kg/s}\\ &\approx 5.0\times10\ut{kg/s}\\ \end{aligned}$$ $$\ab{b}$$ $$F_{\text{Thrust}}=M(g+a_b),$$ $$\begin{aligned} \therefore \dot{m_G}&=\frac{61}{12}(g+a_b)\\ &\approx 156.60047083333333\ut{kg/s}\\ &\approx 1.6\times10^2\ut{kg/s}\\ \end{aligned}$$