9-41 할리데이 11판 솔루션 일반물리학 put {KE:Kinetic Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\\end{cases} put {KE:Kinetic Energy{m=150[g]vi=5.2[m/s]KEf=12KEi \begin{cases} m&=150\ut{g}\\v_i&=5.2\ut{m/s}\\\KE_f&=\frac{1}{2}\KE_i\end{cases} ⎩⎨⎧mviKEf=150[g]=5.2[m/s]=21KEi(a)\ab{a}(a)KEf=12KEi12mvf2=12⋅12mvi2 \begin{aligned}\KE_f&=\frac{1}{2}\KE_i\\\frac{1}{2}m{v_f}^2&=\frac{1}{2}\cdot\frac{1}{2}m{v_i}^2\\\end{aligned} KEf21mvf2=21KEi=21⋅21mvi2$$ \begin{aligned}v_f&=\sqrt\frac{1}{2}\cdot{v_i}\\&=\frac{v_i}{\sqrt2}\\&=\frac{13\sqrt2}{5}\ut{m/s}\.. 11판/9. 질량중심과 선운동량 2024.04.24
9-40 할리데이 11판 솔루션 일반물리학 put {r:redb:blue\put \begin{cases} r:\text{red}\\b:\text{blue}\end{cases} put {r:redb:blue{vi=4.00[m/s]θ=40.0° \begin{cases} v_i&=4.00\ut{m/s}\\\theta&=40.0\degree\\\end{cases} {viθ=4.00[m/s]=40.0°{vri=vicosθi^+visinθj^vbi=vicosθi^−visinθj^ \begin{cases} v_{ri}&=v_i\cos\theta\i+v_i\sin\theta\j\\v_{bi}&=v_i\cos\theta\i-v_i\sin\theta\j\\\end{cases} {vrivbi=vicosθi^+visinθj^=vicosθi^−visinθj^Δpx=0, \Delta p_x = 0, Δpx=0,{prix=prfxpbix=pbfx \begin{cases} p_{rix}=p_{rfx}\\p_{bix}=p_{bfx}\\\end{cases} {prix=prfxpbix=pbfx$$ \put \begin{cases} v_{\text{in}}&=\vec v_{.. 11판/9. 질량중심과 선운동량 2024.04.24
9-39 할리데이 11판 솔루션 일반물리학 {v1i=10[m/s] \begin{cases} v_{1i}&=10\ut{m/s} \end{cases} {v1i=10[m/s] {θ1=0θ2=+30°θ3=−30° \begin{cases} \theta_1&=0\\ \theta_2&=+30\degree\\ \theta_3&=-30\degree\\ \end{cases} ⎩⎨⎧θ1θ2θ3=0=+30°=−30° ∵△123=Equilateral Triangle\because \triangle 123=\text{Equilateral Triangle}∵△123=Equilateral Triangle ΔΣp⃗=0,\Delta \Sigma \vec p=0,ΔΣp=0, $$ \begin{aligned} \vec p_{1i}&=\vec p_{1f}+\vec p_{2f}+\vec p_{3f}\\ m\vec v_{1i}&=m\vec v_{1f}+m\vec v_{2f}+m\vec v_{3f}\\ \vec v_{1i}&=\vec v_{1f}+\vec v_{2f}+\vec v_{3f.. 11판/9. 질량중심과 선운동량 2024.04.23
9-38 할리데이 11판 솔루션 일반물리학 put {0:Start1:After Turn2:Stop \put \begin{cases} 0:\text{Start}\\ 1:\text{After Turn}\\ 2:\text{Stop} \end{cases} put ⎩⎨⎧0:Start1:After Turn2:Stop {m=1400[kg]v0=5.3[m/s]v⃗0=v0j^Δt0→1=4.6[s]v⃗1=v0i^Δt1→2=350[ms]=0.35[s]v⃗2=0 \begin{cases} m&=1400\ut{kg}\\ v_0&=5.3\ut{m/s}\\ \vec v_0&=v_0\j\\ \Delta t_{0\rarr1} &= 4.6\ut{s}\\ \vec v_1&=v_0\i\\ \Delta t_{1\rarr2} &= 350\ut{ms}=0.35\ut{s}\\ \vec v_2&=0 \end{cases} ⎩⎨⎧mv0v0Δt0→1v1Δt1→2v2=1400[kg]=5.3[m/s]=v0j^=4.6[s]=v0i^=350[ms]=0.35[s]=0 (a)\ab{a}(a) $$ \begin{aligned} \vec J_{0\rarr1} &=\Delta \vec p_{0\rarr1}\\ &=\vec p_1-\vec p_0\\ &=m v_0\i-m v_0.. 11판/9. 질량중심과 선운동량 2024.04.23
9-37 할리데이 11판 솔루션 일반물리학 {mL=1.00[kg]mR=0.500[kg] \begin{cases} m_L&=1.00\ut{kg}\\ m_R&=0.500\ut{kg}\\ \end{cases} {mLmR=1.00[kg]=0.500[kg] ΔΣp⃗=0,\Delta \Sigma \vec p=0,ΔΣp=0, mLvL+mRvR=0m_Lv_L+m_Rv_R=0mLvL+mRvR=0 (a)\ab{a}(a) vL=−1.20[m/s],v_L=-1.20\ut{m/s},vL=−1.20[m/s], vR=−mLmRvL=125[m/s] \begin{aligned} v_R&=-\frac{m_L}{m_R}v_L\\ &=\frac{12}{5}\ut{m/s} \end{aligned} vR=−mRmLvL=512[m/s] S=vt=4825[m]=1.92[m] \begin{aligned} S&=vt\\ &=\frac{48}{25}\ut{m}\\ &=1.92\ut{m} \end{aligned} S=vt=2548[m]=1.92[m] (b)\ab{b}(b) vL←R=−1.20[m/s],v_{L\larr R}=-1.20\ut{m/s},vL←R=−1.20[m/s], $$ \begin{cases} m_Lv_L+m_Rv_R=0\\ .. 11판/9. 질량중심과 선운동량 2024.04.23
9-36 할리데이 11판 솔루션 일반물리학 {mα=4.00[u]mO=16.0[u]θα=−64.0°vOf=1.20×105[m/s]θO=40.0° \begin{cases} m_\alpha&=4.00\ut{u}\\ m_{\text{O}}&=16.0\ut{u}\\ \theta_\alpha&=-64.0\degree\\ v_{\text{O}f}&=1.20\times10^5\ut{m/s}\\ \theta_{\text{O}}&=40.0\degree\\ \end{cases} ⎩⎨⎧mαmOθαvOfθO=4.00[u]=16.0[u]=−64.0°=1.20×105[m/s]=40.0° ΔΣp⃗=0,\Delta \Sigma \vec p=0,ΔΣp=0, {pxi=Σpxf0=Σpyf \begin{cases} p_{xi}&=\Sigma p_{xf}\\ 0&=\Sigma p_{yf}\\ \end{cases} {pxi0=Σpxf=Σpyf $$ \begin{cases} m_\alpha v_{\alpha i}&=m_\alpha v_{\alpha x f}+m_{\text{O}} v_{\text{O} x f}\\ 0&=m_\alpha v_.. 11판/9. 질량중심과 선운동량 2024.04.23
9-35 할리데이 11판 솔루션 일반물리학 {r⃗(t[s])=(3500−160t)i^+2700j^+300k^[m]m=250[kg] \begin{cases} \vec r(t\ut{s})&=(3500-160t)\i+2700\j+300\k\ut{m}\\ m&=250\ut{kg} \end{cases} {r(t[s])m=(3500−160t)i^+2700j^+300k^[m]=250[kg] v⃗= dr⃗ dt=−160i^[m/s] \begin{aligned} \vec v&=\dyt{\vec r}\\ &=-160\i\ut{m/s} \end{aligned} v=dtdr=−160i^[m/s] (a)\ab{a}(a) p⃗=mv⃗=−40000i^[kg⋅m/s] \begin{aligned} \vec p&=m\vec v\\ &=-40000\i\ut{kg\cdot m/s}\\ \end{aligned} p=mv=−40000i^[kg⋅m/s] (b)\ab{b}(b) v⃗=−160i^[m/s], \begin{aligned} \vec v&=-160\i\ut{m/s}, \end{aligned} v=−160i^[m/s], West\text{West}West (c)\ab{c}(c) $$ \begin{aligned} \vec F&=m\vec.. 11판/9. 질량중심과 선운동량 2024.04.21
9-34 할리데이 11판 솔루션 일반물리학 {h=4.0[m]Δt=1.5[ms]=1.5×10−3[s]g=9.80665[m/s2] \begin{cases} h&=4.0\ut{m}\\ \Delta t&=1.5\ut{ms}=1.5\times10^{-3}\ut{s}\\ g&=9.80665\ut{m/s^2} \end{cases} ⎩⎨⎧hΔtg=4.0[m]=1.5[ms]=1.5×10−3[s]=9.80665[m/s2] 2aS=v2−v02,2aS=v^2-{v_0}^2,2aS=v2−v02, 2(−g)(−h)=v2−02v=2gh \begin{aligned} 2(-g)(-h)&=v^2-{0}^2\\ v&=\sqrt{2gh}\\ \end{aligned} 2(−g)(−h)v=v2−02=2gh ∣F∣=∣Δp⃗Δt∣∣s(mg)∣=∣m(v⃗f−v⃗i)Δt∣smg=m2ghΔt \begin{aligned} \abs{F}&=\abs{\frac{\Delta \vec p}{\Delta t}}\\ \abs{s(mg)}&=\abs{\frac{m(\vec v_f-\vec v_i)}{\Delta t}}\\ smg&=\frac{m\sqrt{2gh}}{\Delta t}\\ \end{aligned} ∣F∣∣s(mg)∣smg=ΔtΔp=Δtm(vf−vi)=Δtm2gh $$ \begin.. 11판/9. 질량중심과 선운동량 2024.04.21
9-33 할리데이 11판 솔루션 일반물리학 {mA=1.0[kg]mB=3.0[kg]vA=2.5[m/s]vcom=0 \begin{cases} m_A&=1.0\ut{kg}\\ m_B&=3.0\ut{kg}\\ v_A&=2.5\ut{m/s}\\ v_\com&=0\\ \end{cases} ⎩⎨⎧mAmBvAvcom=1.0[kg]=3.0[kg]=2.5[m/s]=0 rcom=ΣrmM,r_\com=\frac{\Sigma rm}{M},rcom=MΣrm, vcom= drcom dt= d dt(ΣrmM)=1M∑ drm dt=1M∑(m dr dt)=1M∑(mv)=ΣmvM0=mAvA+mBvBM \begin{aligned} v_\com&=\dyt{r_\com}\\ &=\dt\(\frac{\Sigma rm}{M}\)\\ &=\frac{1}{M}\sum\dyt{rm}\\ &=\frac{1}{M}\sum \(m\dyt{r}\)\\ &=\frac{1}{M}\sum \(mv\)\\ &=\frac{\Sigma mv}{M}\\ 0&=\frac{m_Av_A+m_Bv_B}{M}\\ \end{aligned} vcom0=dtdrcom=dtd(MΣrm)=M1∑dtdrm=M1∑(mdtdr)=M1∑(mv)=MΣmv=MmAvA+mBvB $$ \begin{aligned} \abs{v_B}&=\abs{-.. 11판/9. 질량중심과 선운동량 2024.04.21
9-32 할리데이 11판 솔루션 일반물리학 {mA=1200[kg]mB=1800[kg]J⃗=300[N⋅s] \begin{cases} m_A&=1200\ut{kg}\\ m_B&=1800\ut{kg}\\ \vec J &= 300\ut{N\cdot s}\\ \end{cases} ⎩⎨⎧mAmBJ=1200[kg]=1800[kg]=300[N⋅s] {ΔΣp⃗=0J⃗=Δp⃗ \begin{cases} \Delta \Sigma \vec p=0\\ \vec J =\Delta \vec p \end{cases} {ΔΣp=0J=Δp {mBv⃗B−mAv⃗A=J⃗mBv⃗B+mAv⃗A=0 \begin{cases} m_B\vec v_B-m_A\vec v_A&=\vec J\\ m_B\vec v_B+m_A\vec v_A&=0 \end{cases} {mBvB−mAvAmBvB+mAvA=J=0 {v⃗A=−J⃗2mAv⃗B=J⃗2mB \begin{cases} \vec v_A&=-\frac{\vec J}{2m_A}\\ \vec v_B&=\frac{\vec J}{2m_B}\\ \end{cases} {vAvB=−2mAJ=2mBJ $$ \begin{aligned} \Ans &= \ve.. 11판/9. 질량중심과 선운동량 2024.04.21