9-21 할리데이 11판 솔루션 일반물리학

$$ \put \begin{cases} m_A=m_1=m\\ m_B=m_2=2m\\ M=m_A+m_B=3m \end{cases} $$ $$ \begin{cases} h&=2.50\ut{m}\\ m_B&=2.00m_A=2m\\ \mu&=0.600\\ \end{cases} $$ $$ \put \begin{cases} 0:\text{Start}\\ 1:\text{A Landing Before A B Crash}\\ 2:\text{After A B Crash}\\ 3:\text{Stop} \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \TE : \text{Thermal Energy}\\ \end{cases} $$ $$\Sigma \Delta E=0,$$ $$ \begin{aligned} \KE_{A1}&=\GE_{A0}\\ \frac{1}{2}m{v_1}^2&=mgh_0\\ {v_1}^2&=2gh_0\taag1\\ \end{aligned} $$ $$ \put \begin{cases} v_{\text{in}}&=v_{A1}-v_{B1}=v_1\\ v_{\text{out}}&=v_{A2}-v_{B2}\\ \end{cases} $$ $$\Delta \Sigma \vec P=0,$$ $$ \begin{aligned} m_Av_1&=m_Av_{A2}+m_Bv_{B2}\\ v_{B2}&=\frac{m_A}{M}(v_{\text{in}}-v_{\text{out}})\taag2 \end{aligned} $$ $$\Sigma \Delta E=0,$$ $$ \begin{aligned} \TE&=\KE\\ fd&=\frac{1}{2}m{v_{B2}}^2\\ \mu m g d &=\frac{1}{2}m{v_{B2}}^2\\ \end{aligned} $$ $$ \begin{aligned} d &=\frac{{v_{B2}}^2}{2\mu g}\\ &=\(\frac{m_A}{M}\)^2\cdot\frac{(v_{\text{in}}-v_{\text{out}})^2}{2\mu g}\\ &=\frac{5}{54 g}(v_{\text{in}}-v_{\text{out}})^2\\ \end{aligned} $$ $$\ab{a}$$ $$ \text{Elastic Collision}\Harr\vec v_{\text{in}}+ \vec v_{\text{out}}=0,$$ $$ \begin{aligned} d&=\frac{5}{54 g}(v_{\text{in}}-v_{\text{out}})^2\\ &=\frac{5}{54 g}(v_{\text{in}}+v_{\text{in}})^2\\ &=\frac{5}{54 g}(2v_1)^2\\ &=\frac{20}{27}h\\ &=\frac{50}{27}\ut{m}\\ &\approx 1.8518518518518519\ut{m}\\ &\approx 1.85\ut{m}\\ \end{aligned} $$ $$\ab{b}$$ $$ \text{Perfectly Inelastic Collision}\Harr\vec v_{\text{out}}=0,$$ $$ \begin{aligned} d&=\frac{5}{54 g}(v_{\text{in}}-v_{\text{out}})^2\\ &=\frac{5}{54 g}{v_1}^2\\ &=\frac{5}{27}h\\ &=\frac{25}{54}\ut{m}\\ &\approx 0.46296296296296297\ut{m}\\ &\approx 0.463\ut{m}\\ &\approx 46.3\ut{cm}\\ \end{aligned} $$