# Differences

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 — phy141:examprep:m1f10sols [2012/09/20 20:35] (current)mdawber created 2012/09/20 20:35 mdawber created 2012/09/20 20:35 mdawber created Line 1: Line 1: + ====== Fall 2010 Midterm 1 Solutions ====== + ===== Midterm Exam Q1 Solution (Avg score: 14.6/25) ===== + + |Vertical motion of stone<​html>&​nbsp;&​nbsp;&​nbsp;&​nbsp;&​nbsp;​\\ $y=v_{0}\sin\theta t-\frac{1}{2}gt^2$ | Horizontal motion of stone\\ $x=v_{0}\cos\theta\,​ t$ | + + At time of impact $t_{i}$ + + $0=v_{0}\sin\theta t_{i}-\frac{1}{2}gt_{i}^{2}$ + + $t_{i}=\frac{2v_{0}\sin\theta}{g}$ + + $100-10t_{i}=v_{0}\cos\theta\,​ t_{i}$ + + $(v_{0}\cos\theta+10)\frac{2v_{0}\sin\theta}{g}=100$ + + $\frac{2v_{0}^{2}\sin\theta\cos\theta}{g}+\frac{20v_{0}\sin\theta}{g}=100$ + + $v_{0}^{2}+\frac{20}{\sqrt{2}}v_{0}=100g$ + + $v_{0}^{2}+\frac{20}{\sqrt{2}}v_{0}+50=100g+50$ + + $(v_{0}+\frac{10}{\sqrt{2}})^2=100g+50$ + + $v_{0}=\frac{-10}{\sqrt{2}}\pm \sqrt{100g+50}$ + + $v_{0}=25.0 \mathrm{ms^{-1}}$ or $v_{0}=-39.2\mathrm{ms^{-1}}$ but we discard the negative solution. + + ===== Midterm Exam Q2 Solution (Avg Score 19.1/30) ===== + + {{phy141m1q2sol.png}} + + A. $\tan\theta=\frac{400}{400}$ → $\theta=45^{o}$ + + B. $v_{BWx}^2+v_{BWy}^2=25$ + + $v_{BWx}+1=v_{BWy}$ + + This can either be solved by inspection (ie, if you see that its just a 3,4,5 triangle) or + + $2v_{BWx}^2+2v_{BWx}+1=25$ + + $v_{BWx}^2+v_{BWx}-12=0$ + + $(v_{BWx}-3)(V_{BWx}+4)=0$ + + $v_{Bwx}=3$ (negative solution doesn'​t work) and it follows that $v_{Bwy}=4$ + + so $\vec{v}_{BW}=(3\hat{i}+4\hat{j})\mathrm{ms^{-1}}$ + + C. $\tan\phi=\frac{4}{3}=53.13^{o}$ + + D. Total work done is proportional to change in velocity, which is zero, so no work is done. + + ===== Midterm Exam Q3 Solution (Avg Score 17.3/20) ===== + + A. Write Newton'​s 2nd Law for each body + + For the skier $m_{s}a=m_{s}g\sin\theta-T$ + + For the tree $m_{t}a=T+m_{t}g\sin\theta-\mu_{t}m_{t}g\cos\theta$ + + Add the two equations together + + $(m_{s}+m_{t})a=(m_{s}+m_{t})g\sin\theta-\mu_{t}m_{t}g\cos\theta$ + + $a=\frac{(m_{s}+m_{t})g\sin\theta-\mu_{t}m_{t}g\cos\theta}{m_{s}+m_{t}}$ + + B. If $v$ is constant $a=0$ + + $(m_{s}+m_{t})g\sin\theta-\mu_{t}m_{t}g\cos\theta=0$ + + $(m_{s}+m_{t})g\sin\theta=\mu_{t}m_{t}g\cos\theta$ + + $\tan\theta=\frac{\mu_{t}m_{t}}{(m_{s}+m_{t})}=\frac{0.2\times400}{400+80}=\frac{1}{6}$ + + $\theta=9.46^{o}$ + + C. From our first equation, when $a=0$, $m_{s}g\sin\theta-T=0$ + + $T=80\mathrm{kg}\times9.8\mathrm{ms^{-2}}\times\sin(9.46^{0})=129N$ + + ===== Midterm Q4 solution (Avg Score 16.4/25) ===== + + For an orbit of a satellite of mass of $m$ at distance $r$ around a body of mass $M$ + + $ma=\frac{GMm}{r^2}$ + + where $a=\frac{mv^{2}}{r}$ + + $\frac{v^2}{r}=\frac{GM}{r^2}$ → $v^2=\frac{GM}{r}$ + + We need to find out what $v$ should be. + + The period of the motion is $\frac{24\times60\times60}{2}\mathrm{s}=43,​200\mathrm{s}$ + + $v=\frac{2\pi r}{T}$ (Watch out, here r is the distance from the center of the planet NOT the radius of the planet!) + + $\frac{4\pi^2 r^2}{T^2}=\frac{GM}{r}$ + + $r^{3}=\frac{GMT^2}{4\pi^2}=\frac{G\times0.1M_{E}\times(43,​200)^2}{4\pi^2}$ + + $r=1.23\times10^{7}$ + + height above surface = $r-R_{M}=r-\frac{R_{E}}{2}=9.11\times10^6\mathrm{m}$