11판/9. 질량중심과 선운동량

9-6 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 4. 13. 12:38
{mA=2140[kg]mB=242[kg]vAi=0vBi=5.3[m/s] \begin{cases} m_A&=2140\ut{kg}\\ m_B&=242\ut{kg}\\ v_{Ai}&=0\\ v_{Bi}&=5.3\ut{m/s}\\ \end{cases} ΔΣP=0,\Delta \Sigma \vec P=0, Pi=PAf+PBf,\vec P_i=\vec P_{Af}+\vec P_{Bf}, {mBvi=mAvAf+mBvBfvBAf=vBfvAf \begin{cases} m_Bv_i&=m_Av_{Af}+m_Bv_{Bf}\\ v_{B\larr Af}&=v_{Bf}-v_{Af} \end{cases} vAf=mBmA+mB(vivBAf)v_{Af}=\frac{m_B}{m_A+m_B}(v_i-v_{B\larr Af}) (a)\ab{a} vBAf=0v_{B\larr Af}=0 vAf=mBmA+mBvi=641311910[m/s]0.5384550797649035[m/s]0.54[m/s]54[cm/s] \begin{aligned} v_{Af}&=\frac{m_B}{m_A+m_B}v_i\\ &=\frac{6413}{11910}\ut{m/s}\\ &\approx 0.5384550797649035\ut{m/s}\\ &\approx 0.54\ut{m/s}\\ &\approx 54\ut{cm/s}\\ \end{aligned} (b)\ab{b} vBAf=5.3[m/s]=viv_{B\larr Af}=5.3\ut{m/s}=v_i vAf=0 \begin{aligned} v_{Af}&=0 \end{aligned} (c)\ab{c} vBAf=5.3[m/s]=viv_{B\larr Af}=-5.3\ut{m/s}=-v_i vAf=2mBmA+mBvi=64135955[m/s]1.076910159529807[m/s]1.1[m/s] \begin{aligned} v_{Af}&=\frac{2m_B}{m_A+m_B}v_i\\ &=\frac{6413}{5955}\ut{m/s}\\ &\approx 1.076910159529807\ut{m/s}\\ &\approx 1.1\ut{m/s}\\ \end{aligned}