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+ | ~~SLIDESHOW~~ | ||

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+ | ======Midterm 1 Practice Exam 2 Solutions====== | ||

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+ | ===== Question 1 ===== | ||

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+ | A. Write Newton's 2nd Law for each body | ||

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+ | For the skier $m_{s}a=m_{s}g\sin\theta-T$ | ||

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+ | For the tree $m_{t}a=T+m_{t}g\sin\theta-\mu_{t}m_{t}g\cos\theta$ | ||

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+ | Add the two equations together | ||

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

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

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

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+ | $(m_{s}+m_{t})g\sin\theta-\mu_{t}m_{t}g\cos\theta=0$ | ||

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

+ | |||

+ | $\tan\theta=\frac{\mu_{t}m_{t}}{(m_{s}+m_{t})}=\frac{0.2\times400}{400+80}=\frac{1}{6}$ | ||

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+ | $\theta=9.46^{o}$ | ||

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

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+ | $T=80\mathrm{kg}\times9.8\mathrm{ms^{-2}}\times\sin(9.46^{0})=129N$ | ||

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+ | ===== Question 2 ===== | ||

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+ | A freedom loving American in the back of a stationary pickup truck fires his rifle in to the air at an angle $\theta$ to the horizontal. Assume that the bullet is not subject to drag forces. At the same time as the gun is fired the driver of the pickup truck begins to accelerate the truck to the right with a constant acceleration $a$. | ||

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+ | A. If after 10s the pickup truck is 250m down the road from where it started, what is the value of the acceleration $a$? | ||

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+ | $\frac{1}{2}at^{2}=250$ | ||

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+ | $a=\frac{250\times 2}{100}=5\mathrm{m\,s^{-2}}$ | ||

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+ | B. How fast is the pickup truck going at this point? | ||

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+ | $5\times 10= 50 \mathrm{m\,s^{-1}}$ | ||

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+ | C. At this time the freedom loving American is hit by the bullet he fired into the air. What is the initial velocity of the bullet? Express your answer using unit vector notation, using $\hat{i}$ for the direction the truck moves in and $\hat{j}$ for vertically upward. | ||

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+ | For the bullet | ||

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+ | $x=v_{x0}t$ | ||

+ | |||

+ | $y=v_{y0}t-\frac{1}{2}gt^{2}$ | ||

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+ | At t=10s the bullet must be at (250,0). | ||

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+ | $250=v_{x0}10$ | ||

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+ | $v_{x0}=25\mathrm{m\,s^{-1}}$ | ||

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+ | $0=v_{y0}-\frac{1}{2}g\times100$ | ||

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+ | $v_{y0}=\frac{\frac{1}{2}\times9.8\times100}{10}$ | ||

+ | |||

+ | $v_{y0}=49\mathrm{m\,s^{-1}}$ | ||

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+ | $\vec{v_{0}}=25\hat{i}+49\hat{j}\,\mathrm{m\,s^{-1}}$ | ||

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+ | D. What is the angle $\theta$ at which the rifle is fired? | ||

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+ | $\tan^{-1}(\frac{49}{25})=62.97^{o}$ | ||

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+ | E. What is the maximum height that the bullet reaches before it turns around? | ||

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+ | $y=49\times5-\frac{1}{2}\times9.8\times5^{2}=122.5\mathrm{m}$ | ||

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+ | F. If the bullet fired has mass 100g find the value of it's total mechanical energy at 3 positions: when it is first fired, when it is at the top of it's trajectory, and just before it hits the freedom loving American. | ||

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+ | Mechanical energy is conserved. It is always equal to the initial kinetic energy, which is | ||

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+ | $\frac{1}{2}0.1\times(49^{2}+25^{2})=151.3\mathrm{J}$ | ||

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+ | ===== Question 3 ===== | ||

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+ | A. $\frac{mv^{2}}{r}=\frac{Gmm}{r^{2}}$ | ||

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+ | $\frac{2\pi r}{v}=2\times60\times60=7200\mathrm{s}$ | ||

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+ | $v=\frac{2\pi r}{7200}$ | ||

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+ | $\frac{(2\pi)^2r^{2}}{7200^{2}r}=\frac{GM}{r^{2}}$ | ||

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+ | $r^{3}=\frac{7200^{2}GM}{(2\pi)^{2}}$ | ||

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+ | $r^{3}=5.24\times10^{20}\mathrm{m^{3}}$ | ||

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+ | $r=8061\mathrm{km}$ | ||

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+ | $h=r-r_{e}=1681\mathrm{km}$ | ||

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+ | B. $v=\frac{2\pi r}{7200}=\frac{2\pi 8.061\times10^{6}}{7200}=7034\mathrm{m/s}$ | ||

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+ | C. $-\frac{GMm}{8.061\times10^6}+\frac{GMm}{6.380\times10^6}=1.3\times10^{10}\mathrm{J}$ | ||

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