11판/9. 질량중심과 선운동량

9-23 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 4. 18. 22:01
{mA=2mB=2mmB=mLEi=90[J] \begin{cases} m_A&=2m_B=2m\\ m_B&=m\\ \LE_i&=90\ut{J} \end{cases} put {KE:Kinetic EnergyLE:Elastic Potential Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\ \LE : \text{Elastic Potential Energy}\\ \end{cases} {ΔΣP=0ΔΣE=0 \begin{cases} \Delta \Sigma \vec P=0\\ \Delta \Sigma E=0\\ \end{cases} 0=mAvAf+mBvBf=2vAf+vBf \begin{aligned} 0&=m_Av_{Af}+m_Bv_{Bf}\\ &=2v_{Af}+v_{Bf}\\ \end{aligned} {KEA=12mAvAf2KEB=12mBvBf2 \begin{cases} \KE_A&=\frac{1}{2}m_A{v_{Af}}^2\\ \KE_B&=\frac{1}{2}m_B{v_{Bf}}^2\\ \end{cases} {KEA=122mvAf2KEB=12m(2vAf)2 \begin{cases} \KE_A&=\frac{1}{2}\cdot 2m{v_{Af}}^2\\ \KE_B&=\frac{1}{2}\cdot m{(-2v_{Af})}^2\\ \end{cases} {KEA=122mvAf2KEB=124mvAf2 \begin{cases} \KE_A&=\frac{1}{2}\cdot 2m{v_{Af}}^2\\ \KE_B&=\frac{1}{2}\cdot 4m{v_{Af}}^2\\ \end{cases} KEB=2KEA\KE_B=2\KE_A 90[J]=KEA+KEB=KEA+2KEA=3KEA \begin{aligned} 90\ut{J}&=\KE_A+\KE_B\\ &=\KE_A+2\KE_A\\ &=3\KE_A\\ \end{aligned} (a,b)\ab{a,b} {KEA=30[J]KEB=60[J] \begin{cases} \KE_A&=30\ut{J}\\ \KE_B&=60\ut{J}\\ \end{cases}