11판/9. 질량중심과 선운동량 61

9-61 할리데이 11판 솔루션 일반물리학

put {s:shipg:gas \put \begin{cases} s:\text{ship}\\g:\text{gas}\\\end{cases} put y˙= ⁣dy ⁣dt,\put \dot y=\dyt{y},{vi=6.0×103[m/s]vgs=3.0×103[m/s]ms=2.5×104[kg]as=2.0[m/s2]M=ms+mg \begin{cases} v_i&=6.0\times10^3\ut{m/s}\\v_{g\larr s}&=-3.0\times10^3\ut{m/s}\\m_s&=2.5\times10^4\ut{kg}\\a_s&=2.0\ut{m/s^2}\\M&=m_s+m_g\\\end{cases} ΔΣp=0,\Delta \Sigma \vec p=0,{Mvi=mgvg+msvsvgs=vgvsΔvs=vsvi \begin{cases} Mv_i&=m_gv_{g}+m_sv_s\\v_{g\larr s}&=v_g-v_s\\\Delta v_s&=v_s-v_i\end{cases} $$ \begin{aligned}-m_gv_{g\lar..

9-60 할리데이 11판 솔루션 일반물리학

{vi=2.0[m/s]m=4.0[kg]ΔΣKE=16[J] \begin{cases} v_i&=2.0\ut{m/s}\\m&=4.0\ut{kg}\\\Delta \Sigma \KE &=16\ut{J}\\\end{cases} {ΔΣp=0,ΔΣKE=16[J] \begin{cases} \Delta \Sigma \vec p&=0,\\\Delta \Sigma \KE &=16\ut{J}\\\end{cases} {2mvi=mvA+mvB12(2m)vi2+16=12mvA2+12mvB2 \begin{cases} 2mv_i&=mv_A+mv_B\\\frac{1}{2}(2m) {v_i}^2+16&=\frac{1}{2}m {v_A}^2+\frac{1}{2}m {v_B}^2\end{cases} {vA=vi+4mvB=vi4m \begin{cases} v_A&=v_i+\frac{4}{\sqrt m}\\v_B&=v_i-\frac{4}{\sqrt m}\\\end{cases} $$ \begin{ca..

9-59 할리데이 11판 솔루션 일반물리학

{m2=1.0[kg]v2i=0k=490[N/m]m1=2.0[kg]M=m1+m2=3[kg]v1i=4.0[m/s] \begin{cases} m_2&=1.0\ut{kg}\\v_{2i}&=0\\k&=490\ut{N/m}\\m_1&=2.0\ut{kg}\\M&=m_1+m_2=3\ut{kg}\\v_{1i}&=4.0\ut{m/s}\\\end{cases} ΔΣp=0,\Delta \Sigma \vec p=0,m1v1i=Mvf,vf=m1Mv1i \begin{aligned}m_1v_{1i}&=Mv_f,\\v_f&=\frac{m_1}{M}v_{1i}\end{aligned} ΣΔE=0,\Sigma \Delta E=0,12Mvf2=12kx2\frac{1}{2}M{v_f}^2=\frac{1}{2}kx^2$$ \begin{aligned}x&=v_f\sqrt\frac{M}{k}\\&=\frac{m_1v_{1i}}{\sqrt{Mk}}\\&=\frac{4}{7}\sqrt{\frac{2}..

9-58 할리데이 11판 솔루션 일반물리학

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} $$ \begin{case..

9-57 할리데이 11판 솔루션 일반물리학

{vi=vM=mA+mB=mmA=4mBvB=0 \begin{cases} v_i&=v\\M&=m_A+m_B=m\\m_A&=4m_B\\v_B&=0\\\end{cases} {mA=45MmB=15M \begin{cases} m_A&=\frac{4}{5}M\\m_B&=\frac{1}{5}M\\\end{cases} ΔΣp=0,\Delta \Sigma \vec p=0,Mvi=mAvA+mBvB=mAvA \begin{aligned}Mv_i&=m_Av_A+m_Bv_B\\&=m_Av_A\end{aligned} vA=MmAvi=5mAmAvi=5vi \begin{aligned}v_A&=\frac{M}{m_A}v_i\\&=\frac{5m_A}{m_A}v_i\\&=5v_i\\\end{aligned} $$ \begin{aligned}\Delta \Sigma \KE&=\Delta \sum \(\frac{1}{2}mv^2\)\\&=\(\frac..

9-56 할리데이 11판 솔루션 일반물리학

{m=140[g]=0.140[kg]vi=9.30[m/s]vf=vi=9.30[m/s]Δt=3.80[ms]=3.80×103[s] \begin{cases} m&=140\ut{g}=0.140\ut{kg}\\v_i&=9.30\ut{m/s}\\v_f&=-v_i=-9.30\ut{m/s}\\\Delta t&=3.80\ut{ms}=3.80\times10^{-3}\ut{s}\\\end{cases} (a)\ab{a}J=Δp=Δmv=mΔv=m(vfvi)=m(2vi) \begin{aligned}\vec J&=\Delta \vec p\\&=\Delta m\vec v\\&=m\Delta \vec v\\&=m (\vec v_f-\vec v_i)\\&=m (-2\vec v_i)\\\end{aligned} $$ \begin{aligned}J&=2mv_i\\&=\frac{651}{250}\ut{N\cdot s}\\&=2.604\ut{N\cdot s}\\&\approx 2.60\ut{N\cd..

9-55 할리데이 11판 솔루션 일반물리학

{vi=7600[m/s]mA=290.0[kg]mB=150.0[kg]M=mA+mB=440[kg]vAB=910.0[m/s] \begin{cases} v_i&=7600\ut{m/s}\\m_A&=290.0\ut{kg}\\m_B&=150.0\ut{kg}\\M&=m_A+m_B=440\ut{kg}\\v_{A\larr B}&=910.0\ut{m/s}\end{cases} {vAB=vAvBΔΣp=0 \begin{cases} v_{A\larr B}&=v_A-v_B\\\Delta \Sigma \vec p&=0\end{cases} {vAB=vAvBMvi=mAvA+mBvB \begin{cases} v_{A\larr B}&=v_A-v_B\\Mv_i&=m_Av_A+m_Bv_B\end{cases} {vA=vi+mBMvABvB=vimAMvAB \begin{cases} v_A&=v_i+\cfrac{m_B}{M}v_{A\larr B}\\v_B&=v_i-\cfrac{m_A}{M}v_{A\larr B}\\\end{cases} $$\ab..

9-54 할리데이 11판 솔루션 일반물리학

{mA=3.18×104[kg]vBi=0 \begin{cases} m_A&=3.18\times10^4\ut{kg}\\v_{Bi}&=0\\\end{cases} {ΔΣp=0ΔKE=127100 \begin{cases} \Delta \Sigma \vec p&=0\\\Delta \KE&=1-\frac{27}{100}\end{cases} {mAvi=(mA+mB)vf7310012mAvi2=12(mA+mB)vf2 \begin{cases} m_Av_i&=(m_A+m_B)v_f\\\frac{73}{100}\cdot\frac{1}{2}m_A{v_{i}}^2&=\frac{1}{2}(m_A+m_B){v_{f}}^2\end{cases} $$ \begin{aligned}m_B&= \frac{27}{73}m_A \\&=\frac{4293}{3650}\times10^4\ut{kg}\\&\approx 1.1761643835616438\times10^4..

9-53 할리데이 11판 솔루션 일반물리학

{mA=80[kg]mB=320[kg]vi=0vAB=2.5[m/s]M=mA+mB=400[kg] \begin{cases} m_A&=80\ut{kg}\\m_B&=320\ut{kg}\\v_{i}&=0\\v_{A\larr B}&=2.5\ut{m/s}\\M&=m_A+m_B=400\ut{kg}\\\end{cases} ΔΣp=0,\Delta \Sigma \vec p=0,{0=mAvA+mBvBvAB=vAvB, \begin{cases} 0&=m_Av_A+m_Bv_B\\v_{A\larr B}&=v_A-v_B,\\\end{cases} vB=mAMvAB=12[m/s]=0.5[m/s] \begin{aligned}v_B&=-\frac{m_A}{M}v_{A\larr B}\\&=-\frac{1}{2}\ut{m/s}\\&=-0.5\ut{m/s}\\\end{aligned} (a)\ab{a}Dpwn\text{Dpwn}(b)\ab{b}vB=0.5[m/s]v_B=0.5\ut{m/s}(c)\ab{c}$$\..

9-52 할리데이 11판 솔루션 일반물리학

put {S:ShipG:gas \put \begin{cases} S:\text{Ship}\\G:\text{gas}\end{cases} {M=mS+mG=6100[kg]vGS=1200[m/s]g=9.80665[m/s2] \begin{cases} M&=m_S+m_G=6100\ut{kg}\\v_{G\larr S}&=-1200\ut{m/s}\\g&=9.80665\ut{m/s^2}\end{cases} put y˙= ⁣dy ⁣dt,\put \dot y=\dyt{y},ΔΣp=0,\Delta \Sigma \vec p=0,{Mvi=mGvGf+mSvSfvGS=vGvSΔvS=vSvi \begin{cases} Mv_i&=m_Gv_{Gf}+m_Sv_{Sf}\\v_{G\larr S}&=v_G-v_S\\\Delta v_S&=v_S-v_i\end{cases} $$ \begin{aligned}-m_Gv_{G\larr S}&=M\Delta v_S\\\dt(-m_Gv_{G\larr S})&=\dt (M\D..