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 $t_{i}$

$0=v_{0}\sin\theta t_{i}-\frac{1}{2}gt_{i}^{2}$


$100-10t_{i}=v_{0}\cos\theta\, t_{i}$






$v_{0}=\frac{-10}{\sqrt{2}}\pm \sqrt{100g+50}$

$v_{0}=25.0 \mathrm{ms^{-1}}$ 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. $v_{BWx}^2+v_{BWy}^2=25$


This can either be solved by inspection (ie, if you see that its just a 3,4,5 triangle) or




$v_{Bwx}=3$ (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



B. If $v$ is constant $a=0$





C. From our first equation, when $a=0$, $m_{s}g\sin\theta-T=0$


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 $M$


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!)

$\frac{4\pi^2 r^2}{T^2}=\frac{GM}{r}$



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

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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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