11판/9. 질량중심과 선운동량

9-16 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 4. 16. 21:46
put {0:Start1:After Crash2:Landing Ground \put\begin{cases} 0:\text{Start}\\ 1:\text{After Crash}\\ 2:\text{Landing Ground} \end{cases} put {KE:Kinetic EnergyGE:Gravitational Potential Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} {mA=3.2[kg]mB=2.0[kg]h=0.30[m]v0=3.0[m/s] \begin{cases} m_A&=3.2\ut{kg}\\ m_B&=2.0\ut{kg}\\ h&=0.30\ut{m}\\ v_{0}&=3.0\ut{m/s} \end{cases} put M=mA+mB=5.2[kg]\put M = m_A+m_B=5.2\ut{kg} ΔΣP=0,\Delta \Sigma \vec P=0, mAvA0=Mv1,m_Av_{A0}=Mv_1, v1=mAMv0 \begin{aligned} v_1&=\frac{m_A}{M}v_0 \end{aligned} ΣΔE=0,\Sigma \Delta E=0, ΔKE+ΔGE=0\Delta \KE+\Delta \GE=0 KE2=KE1ΔGE=12Mv12+Mg(Δy)=12M(mAv0M)2+Mgh=(mAv0)22M+Mgh=39g25+5766524.15991246153846[J]24[J] \begin{aligned} \KE_2&=\KE_1-\Delta \GE\\ &=\frac{1}{2}M{v_1}^2+Mg(-\Delta y)\\ &=\frac{1}{2}M\(\frac{m_Av_0}{M}\)^2+Mgh\\ &=\frac{(m_Av_0)^2}{2M}+Mgh\\ &=\frac{39 g}{25}+\frac{576}{65}\\ &\approx 24.15991246153846\ut{J}\\ &\approx 24\ut{J}\\ \end{aligned}