9-10 할리데이 11판 솔루션 일반물리학

$$\begin{cases} m_A&=500.0\ut{kg}\\ m_B&=400.0\ut{kg}\\ M&=m_A+m_B\\ \end{cases}$$ $$\begin{cases} v_{i}&=1000\ut{m/s}\\ v&=v_{A\larr Bf}=-100.0\ut{m/s} \end{cases}$$ $$\Delta \Sigma \vec P=0,$$ $$\begin{cases} Mv_i&=m_Av_{Af}+m_Bv_{Bf}\\ v_{A\larr Bf}&=v_{Af}-v_{Bf}\\ \end{cases}$$ $$\begin{cases} v_{Af}&=v_i+\cfrac{m_B}{M}v\\ v_{Bf}&=v_i-\cfrac{m_A}{M}v\\ \end{cases}$$ $$\begin{cases} v_{Af}&=\cfrac{8600}{9}\ut{m/s}\\ v_{Bf}&=\cfrac{9500}{9}\ut{m/s}\\ \end{cases}$$ $$\begin{aligned} \Ans_A&=\frac{\KE_{Af}}{\KE_{Ai}}\\ &=\frac{\frac{1}{2}m_A{v_{Af}}^2}{\frac{1}{2}m_A{v_{Ai}}^2}\\ &=\(\frac{v_{Af}}{v_{i}}\)^2\\ &=\frac{1849}{2025}\\ &\approx 0.9130864197530865\\ &\approx 0.913\\ \end{aligned}$$ $$\begin{aligned} \Ans_B&=\frac{\KE_{Bf}}{\KE_{Bi}}\\ &=\frac{\frac{1}{2}m_B{v_{Bf}}^2}{\frac{1}{2}m_B{v_{Bi}}^2}\\ &=\(\frac{v_{Bf}}{v_{i}}\)^2\\ &=\frac{361}{324}\\ &\approx 1.1141975308641976\\ &\approx 1.11\\ \end{aligned}$$ $$\begin{aligned} \Ans_S&=\frac{\Sigma \KE_{f}}{\Sigma \KE_{i}}\\ &=\frac{\frac{1}{2}m_A{v_{Af}}^2+\frac{1}{2}m_B{v_{Bf}}^2}{\frac{1}{2}M{v_{i}}^2}\\ &=\frac{406}{405}\\ &\approx 1.00 \end{aligned}$$