11판/9. 질량중심과 선운동량

9-58 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 4. 30. 20:17
put {0:Start1:After Burst2:Stop \put \begin{cases} 0:\text{Start}\\ 1:\text{After Burst}\\ 2:\text{Stop} \end{cases} {mL=3.0[kg]μL=0.40μR=0.50dL=0.15[m]dR=0.25[m]g=9.80665[m/s2] \begin{cases} m_L&=3.0\ut{kg}\\ \mu_L&=0.40\\ \mu_R&=0.50\\ d_L&=0.15\ut{m}\\ d_R&=0.25\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} ΔΣp=0,\Delta \Sigma \vec p=0, 0=mLvL1+mRvR10=m_Lv_{L1}+m_Rv_{R1} W=ΔKE,W=\Delta \KE, fd=12mv2,f d=\frac{1}{2}mv^2, v12=2μgd \begin{aligned} {v_1}^2&=2\mu gd\\ \end{aligned} {vL12=2μLgdLvR12=2μRgdR0=mLvL1+mRvR1 \begin{cases} {v_{L1}}^2&=2\mu_L gd_L\\ {v_{R1}}^2&=2\mu_R gd_R\\ 0&=m_Lv_{L1}+m_Rv_{R1} \end{cases} mR=mLdLdRμLμR=653[kg]2.0784609690826525[kg]2.1[kg] \begin{aligned} m_R&=m_L\sqrt{\frac{d_L}{d_R}\cdot\frac{\mu_L}{\mu_R}}\\ &=\frac{6}{5}\sqrt3\ut{kg}\\ &\approx 2.0784609690826525\ut{kg}\\ &\approx 2.1\ut{kg}\\ \end{aligned}