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# Solutions to Fall 2015 Practice Final Exam 1

## Question 1

A satellite in a circular geostationary orbit above the Earth's surface (meaning it has the same angular velocity as the Earth) explodes in to three pieces of equal mass. One piece remains in a circular orbit at the same height as the satellite's orbit after the collision, but the velocity of this piece is in the opposite direction to the velocity of the satellite before the explosion. After the explosion the second piece falls from rest in a straight line directly to the Earth. The third piece has an initial velocity in the same direction as the original direction of the satellite, but now has a different magnitude of velocity. The diagram is obviously not remotely to scale!

A. (10 points) At what height above the Earth does the satellite orbit?

$G\frac{mM_{E}}{r^{2}}=\frac{mv^{2}}{r}$

$v_{circular}=\sqrt{\frac{GM_{E}}{r}}$

$M_{E}=5.98\times10^{24}\mathrm{kg}$

$R_{E}=6380\mathrm{km}$

$G=6.67\times10^{-11}\mathrm{Nm^{2}kg^{-2}}$

$\frac{2\pi r}{v}$ = 1 day = 86,400 s

Using the equation for the velocity $v=\sqrt{\frac{GM_{E}}{r}}$

$2\pi\frac{r^{3/2}}{\sqrt{GM_{E}}}$ = 86,400 s

$r=(86,400\frac{\sqrt{GM_{E}}}{2\pi})^{2/3}=4.23\times10^{7}\,\mathrm{m}=42,300\,\mathrm{km}$

(35,900 km above surface)

B. (5 points) What is the magnitude of the velocity of the satellite along it's original circular path?

$v=\sqrt{\frac{GM_{E}}{r}}=\sqrt{\frac{6.67\times10^{-11}\times5.98\times10^{24}}{4.23\times10^7}}=3071\mathrm{m\,s^{-1}}$

C. (5 points) For the piece that falls to the ground, what is the magnitude of the velocity when it strikes the ground? You may neglect any effects due to air resistance.

$\frac{1}{2}mv^{2}=\frac{-GMm}{4.23\times10^{7}}-\frac{-GMm}{6.38\times10^{6}}$

$v^{2}=1.06\times10^{8}$

$v=1.03\times10^{4}\mathrm{m\,s^{-1}}$

D. (5 points) What happens to the third piece? Does it fall to the ground, continue in an orbit around the Earth (not necessarily a circular one), or escape the gravitational pull of the Earth? Justify your answer.

From conservation of momentum the velocity of the third piece must be 4 times the original velocity. The escape velocity is $\sqrt{2}$ times the circular velocity, so the third piece escapes the gravitational pull of the earth.

## Question 2

A. The condition for when static friction and the gravitational force down the plane exactly cancel is given by

$mg\sin\theta=\mu_{s}mg\cos\theta$

$\tan\theta=\mu_{s}=0.4$

$\theta=21.8^{o}$

B. For rolling

$KE=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}\frac{2}{5}mr^{2}\frac{v^{2}}{r^{2}}=\frac{7}{10}mv^{2}$

$\frac{7}{10}mv^{2}=mg1.5\sin\theta$

$v=\sqrt{\frac{10}{7}\times1.5\times9.8\times\sin21.8^{o}}$

$v=2.79\mathrm{m\,s^{-1}}$

C. $\frac{\frac{2}{5}}{\frac{2}{5}+1}=0.29=29\%$

## Question 3

A.

$v_{x}=100-30\cos(10^{o})=70.46\mathrm{ms^{-1}}$

$v_{y}=-30\sin(10^{o})=-5.21\mathrm{ms^{-1}}$

$\vec{v}=70.46\mathrm{ms^{-1}}\hat{i}-5.21\mathrm{ms^{-1}}\hat{j}$

B.

$x=30\cos(10^{o})t=29.54t\,\mathrm{m}$

$y=30\sin(10^{o})t=5.21t\,\mathrm{m}$

C.

$x=100t-l\,\mathrm{m}$

$y=h-\frac{1}{2}gt^{2}\,\mathrm{m}$

D.

$y=52.1\mathrm{m}$

$52.1=h-\frac{1}{2}g10^{2}$

$h=52.1+50\times9.81=542.61\mathrm{m}$

E.

$0\mathrm{m}$

F.

$\frac{1}{2}mv_{0}^{2}+24525=\frac{1}{2}\times50\times100^{2}+240590=250000+240590=490590\mathrm{J}$

## Question 4

A. $\pi \frac{d^{2}}{4}h=0.2$

$d=0.53\mathrm{m}$

B. The internal volume of the drum is $190\mathrm{L}=0.19\mathrm{m^{3}}$, which means that the steel has a volume of $0.01\mathrm{m^{3}}$ and thus a density of $1500 \mathrm{kg/m^{3}}$. The average density of the barrel is then

$\frac{1500\times0.01+900\times0.19}{0.01+0.19}=930\mathrm{kg/m^{3}}$

The equilibrium is found when the bouyant force equals the gravitational force.

$\rho_{water}(h-x)\pi r^{2}g=930h\pi r^{2}g$

$0.9-x=0.9\frac{930}{1000}$

$x=0.063\mathrm{m}=6.3\mathrm{cm}$

C. $P=P_{atm}+\rho gh=1.013\times10^{5}+1000\times9.8\times0.837=1.095\times10^{5}\mathrm{Pa}$

## Question 5

A. $v_{sound}=343\mathrm{m\,s^{-1}}$

$f=\frac{v}{2l}=\frac{343}{2\times0.6}=285.8\mathrm{Hz}$

B. See here.

C. $\Delta f=1\mathrm{Hz}$

$f_{T}=286.8\mathrm{Hz}$

$v_{T}=286.8\times2\times0.6=344.2\mathrm{m\,s^{-1}}$

$T=\frac{344.2-331}{0.6}=22\mathrm{^{o}C}$

D. $f'=\frac{v_{sound}+v_{obs}}{v_{sound}}f$

$286.8=\frac{343+v_{obs}}{343}285.8$

$v_{obs}=1.2\mathrm{m\,s^{-1}}$

E. $f=2\times285.8=571.6\mathrm{Hz}$

## Question 6

A. $e=\frac{W}{Q_{H}}$

$Q_{H}=\frac{1}{0.4}=2.5\mathrm{J}$

B.$Q_{H}=W+Q_{L}$

$Q_{L}=1.5\mathrm{J}$

C. $P_{B}V_{B}=nRT_{H}$

$P_{D}V_{D}=nRT_{L}$

$\frac{P_{B}V_{B}}{P_{D}V_{D}}=\frac{T_{H}}{T_{L}}$

$P_{B}=2P_{atm}\frac{T_{H}}{T_{L}}$

As $e=1-\frac{T_{L}}{T_{H}}=0.4$

$\frac{T_{L}}{T_{H}}=0.6$

$P_{B}=3.33P_{atm}$

D. The expansion from B to C is adiabatic so $P_{B}V_{B}^{\gamma}=P_{C}V_{C}^{\gamma}$

For a diatomic gas $\gamma=\frac{7}{5}$

We also know that $P_{B}V_{B}=nRT_{H}$ and $P_{C}V_{C}=nRT_{L}$

so

$nRT_{H}V_{B}^{\gamma-1}=nRT_{L}V_{C}^{\gamma-1}$

$V_{C}^{\gamma-1}=\frac{T_{H}}{T_{L}}V_{B}^{\gamma-1}$

$V_{C}^{2/5}=\frac{10}{6}$

$V_{C}=3.6\mathrm{L}$

E. $\Delta S=0\mathrm{\frac{J}{K}}$