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Midterm Exam 1 Information

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Midterm Exam Q1 Solution (Avg score: 14.6/25)

Vertical motion of stone     
$y=v_{0}\sin\theta t-\frac{1}{2}gt^2$
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)

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 = $r-R_{M}=r-\frac{R_{E}}{2}=9.11\times10^6\mathrm{m}$

Video of Review Session

Unfortunately the battery in the camera ran out, so there is about 10-15 minutes missing from the end of the video. But I don't think anything too important is missing. If you are having trouble watching the video within the page you can download the video and play it in Quicktime.

Worked solutions to Practice Exams

Midterm 1 Practice Exam 1 Solutions

Note: For some reason I wrote $\frac{v}{d}=t$ in the solution to Q1, when of course it should be $\frac{d}{v}=t$. As we only make use of the equality of $\frac{v}{d}$ or $\frac{d}{v}$ for x and y the method still works, but it doesn't make my error any less silly, sorry about that!

Midterm 1 Practice Exam 2 Solutions

phy141/examprep/m1.txt · Last modified: 2010/10/06 22:25 by mdawber
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