# Fall 2010 Midterm 1 Solutions

## 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 $m_{t}a=T+m_{t}g\sin\theta-\mu_{t}m_{t}g\cos\theta$

$(m_{s}+m_{t})a=(m_{s}+m_{t})g\sin\theta-\mu_{t}m_{t}g\cos\theta$

$a=\frac{(m_{s}+m_{t})g\sin\theta-\mu_{t}m_{t}g\cos\theta}{m_{s}+m_{t}}$

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