11판/9. 질량중심과 선운동량

9-37 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 4. 23. 21:04
{mL=1.00[kg]mR=0.500[kg] \begin{cases} m_L&=1.00\ut{kg}\\ m_R&=0.500\ut{kg}\\ \end{cases} ΔΣp=0,\Delta \Sigma \vec p=0, mLvL+mRvR=0m_Lv_L+m_Rv_R=0 (a)\ab{a} vL=1.20[m/s],v_L=-1.20\ut{m/s}, vR=mLmRvL=125[m/s] \begin{aligned} v_R&=-\frac{m_L}{m_R}v_L\\ &=\frac{12}{5}\ut{m/s} \end{aligned} S=vt=4825[m]=1.92[m] \begin{aligned} S&=vt\\ &=\frac{48}{25}\ut{m}\\ &=1.92\ut{m} \end{aligned} (b)\ab{b} vLR=1.20[m/s],v_{L\larr R}=-1.20\ut{m/s}, {mLvL+mRvR=0vLR=vLvR \begin{cases} m_Lv_L+m_Rv_R=0\\ v_{L\larr R}=v_L-v_R \end{cases} vL=mRmLmRvLR=65[m/s] \begin{aligned} v_L&=-\frac{m_R}{m_L-m_R}v_{L\larr R}\\ &=\frac{6}{5}\ut{m/s}\\ \end{aligned} S=vt=2425[m]=0.96[m] \begin{aligned} S&=vt\\ &=\frac{24}{25}\ut{m}\\ &=0.96\ut{m} \end{aligned}