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phy141:examprep:m1f10sols [2012/09/20 20:35] (current)
mdawber created
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 +====== Fall 2010 Midterm 1 Solutions ======
  
 +===== Midterm Exam Q1 Solution (Avg score: 14.6/25) =====
 +
 +|Vertical motion of stone<​html>&​nbsp;&​nbsp;&​nbsp;&​nbsp;&​nbsp;</​html>​\\ $y=v_{0}\sin\theta t-\frac{1}{2}gt^2$ | 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 = $r-R_{M}=r-\frac{R_{E}}{2}=9.11\times10^6\mathrm{m}$
phy141/examprep/m1f10sols.txt · Last modified: 2012/09/20 20:35 by mdawber
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