9-55 할리데이 11판 솔루션 일반물리학

$$\begin{cases} v_i&=7600\ut{m/s}\\ m_A&=290.0\ut{kg}\\ m_B&=150.0\ut{kg}\\ M&=m_A+m_B=440\ut{kg}\\ v_{A\larr B}&=910.0\ut{m/s} \end{cases}$$ $$\begin{cases} v_{A\larr B}&=v_A-v_B\\ \Delta \Sigma \vec p&=0 \end{cases}$$ $$\begin{cases} v_{A\larr B}&=v_A-v_B\\ Mv_i&=m_Av_A+m_Bv_B \end{cases}$$ $$\begin{cases} v_A&=v_i+\cfrac{m_B}{M}v_{A\larr B}\\ v_B&=v_i-\cfrac{m_A}{M}v_{A\larr B}\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} v_A&=v_i+\cfrac{m_B}{M}v_{A\larr B}\\ &=\frac{174025}{22}\ut{m/s}\\ &\approx 7910.227272727273\ut{m/s}\\ &\approx 7910\ut{m/s}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} v_A&=v_i-\cfrac{m_A}{M}v_{A\larr B}\\ &=\frac{154005}{22}\ut{m/s}\\ &\approx 7000.227272727273\ut{m/s}\\ &\approx 7000\ut{m/s}\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} \KE_{Ai}&=\frac{1}{2}m_A{v_i}^2\\ &=8.3752\times10^9\ut{J}\\ &\approx 8.375\times10^9\ut{J}\\ &\approx 8375\ut{MJ}\\ \end{aligned}$$ $$\begin{aligned} \KE_{Bi}&=\frac{1}{2}m_B{v_i}^2\\ &=4.332\times10^9\ut{J}\\ &= 4332\ut{MJ}\\ \end{aligned}$$ $$\begin{aligned} \KE_{i}&=\frac{1}{2}M{v_i}^2\\ &=1.27072\times10^{10}\ut{J}\\ &\approx 1.271\times10^{10}\ut{J}\\ &\approx 12.71\ut{GJ}\\ \end{aligned}$$ $$\ab{d}$$ $$\begin{aligned} \KE_{Af}&=\frac{1}{2}m_A{v_A}^2\\ &=\frac{4391281590625}{484}\ut{J}\\ &\approx 9.07289584839876\times10^9\ut{J}\\ &\approx 9.073\times10^9\ut{J}\\ &\approx 9073\ut{MJ}\\ \end{aligned}$$ $$\begin{aligned} \KE_{Bf}&=\frac{1}{2}m_B{v_B}^2\\ &=\frac{3439043303625}{484}\ut{J}\\ &\approx 7.105461371126033\times10^9\ut{J}\\ &\approx 7.105\times10^9\ut{J}\\ &\approx 7105\ut{MJ}\\ \end{aligned}$$ $$\begin{aligned} \KE_{f}&=\frac{1}{2}m_A{v_A}^2+\frac{1}{2}m_B{v_B}^2\\ &=\frac{3915162447125}{242}\ut{J}\\ &\approx 1.6178357219524794\times10^{10}\ut{J}\\ &\approx 1.618\times10^{10}\ut{J}\\ &\approx 16.18\ut{GJ}\\ \end{aligned}$$ $$\ab{e}$$ $$\text{by Spring Energy}$$