9-1 할리데이 11판 솔루션 일반물리학

$$\put \begin{cases} A:m_1\\ B:m_2\\ C:m_3\\ \end{cases}$$ $$\put \begin{cases} 0:\text{Start}\\ 1:\text{After A,B Crash}\\ 2:\text{After B,C Crash}\\ \end{cases}$$ $$\begin{cases} m_A&=m\\ m_B&=2.00m\\ m_C&=2.00m_B=4.00m \end{cases}$$ $$\begin{cases} v_{A0}&=2.00\ut{m/s}\\ v_{B0}&=0\\ v_{C0}&=0\\ \end{cases}$$ $$\Delta \Sigma \vec P=0,$$ $$\text{Elastic Collision}\Harr\vec v_{\text{in}}+ \vec v_{\text{out}}=0,$$ $$\vec v_{Ai}-\vec v_{Bi}+ \vec v_{Af}-\vec v_{Bf}=0,$$ $$\begin{cases} \vec P_{Ai}+\vec P_{Bi}&=\vec P_{Af}+\vec P_{Bf}\\ \vec v_{Ai}+\vec v_{Af}&=\vec v_{Bi}+\vec v_{Bf}\\ \end{cases}$$ $$\begin{cases} \vec P_{Ai}&=\vec P_{Af}+\vec P_{Bf}\\ \vec v_{Ai}+\vec v_{Af}&=\vec v_{Bf}\\ \end{cases}$$ $$\begin{cases} m_A\vec v_{Ai}&=m_A\vec v_{Af}+m_B\vec v_{Bf}\\ \vec v_{Ai}+\vec v_{Af}&=\vec v_{Bf}\\ \end{cases}$$ $$\therefore \begin{cases} \vec v_{Af}&=\cfrac{m_A-m_B}{m_A+m_B}\vec v_{Ai}\\ \vec v_{Bf}&=\cfrac{2m_A}{m_A+m_B}\vec v_{Ai}\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} \vec v_{B1}&=\frac{2m_A}{m_A+m_B}\vec v_{A0}\\ &=\frac{2m}{m+2m}\vec v_{A0}\\ &=\frac{2}{3}\vec v_{A0} \end{aligned}$$ $$\begin{aligned} \vec v_{C2}&=\frac{2m_B}{m_B+m_C}\vec v_{B1}\\ &=\frac{2\cdot 2m}{2m+4m}\cdot\(\frac{2}{3}\vec v_{A0}\)\\ &=\frac{4}{9}\vec v_{A0}\\ &=\frac{8}{9}\ut{m/s}\\ &\approx 0.8888888888888888\ut{m/s}\\ &\approx 0.889\ut{m/s}\\ &\approx 88.9\ut{cm/s}\\ \end{aligned}$$ $$\ab{b-1}$$ $$v_{C2} \lt v_{A0}~~\(\because v_{C2}=\frac{4}{9}\vec v_{A0}\)$$ $$\ab{b-2}$$ $$\begin{aligned} \KE_{C2}&=\frac{1}{2}m_C{v_{C2}}^2\\ &=\frac{1}{2}(4m)\(\frac{4}{9}\vec v_{A0}\)^2\\ &=\frac{64}{81}\(\frac{1}{2}m{v_{A0}}^2\)\\ &=\frac{64}{81}\KE_{A0} \end{aligned}$$ $$\therefore \KE_{C2} \lt \KE_{A0}$$ $$\ab{b-3}$$ $$\begin{aligned} \vec P_{C2}&=m_Cv_{C2}\\ &=(4m)\(\frac{4}{9}\vec v_{A0}\)\\ &=\frac{8}{9}\(m\vec v_{A0}\)\\ &=\frac{8}{9}\vec P_{A0}\\ \end{aligned}$$ $$\therefore \vec P_{C2} \lt \vec P_{A0}$$