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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.

An inclined plane has a coefficient of static friction $\mu_{s}=0.4$ and kinetic friction $\mu_{k}=0.3$.

A. (5 points) What is the maximum inclination angle $\theta$ for which a stationary block on the plane will remain at rest?

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. (10 points) If a ball ($I=\frac{2}{5}Mr^{2}$) is placed on the incline when it has the inclination angle you found in part (a) and rolls through a distance of 1.5m, what is the translational velocity of the ball when it reaches the bottom of the incline?

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. (10 points) What percentage of the kinetic energy of the ball is rotational while it is rolling down the incline?

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

A plane is flying horizontally with a constant speed of 100m/s at a height $h$ above the ground, and drops a 50kg bomb with the intention of hitting a car that has just begun driving up a 10$^{\circ}$ incline which starts a distance $l$ in front of the plane. The speed of the car is a constant 30m/s. For the following questions use the coordinate axes defined in the figure, where the origin is taken to be the initial position of the car.

A. (5 points) What is the initial velocity of the bomb relative to the car? Write your answer in unit vector notation.

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

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

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

B. (5 points) Write equations for both components ($x$ and $y$) of the car's displacement as a function of time, taking t=0s to be the time the bomb is released.

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

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

C. (5 points) Write equations for both components ($x$ and $y$) of the bomb's displacement as a function of time, taking t=0s to be the time the bomb is released.

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

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

D. (5 points) If the bomb hits the car at time t=10s what was the height of the plane above the ground $h$ when it dropped the bomb?

$y=52.1\mathrm{m}$

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

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

E. (5 points) What is the horizontal displacement of the plane relative to the car when the bomb hits the car at t=10s.

$0\mathrm{m}$

F. (5 points) How much kinetic energy does the bomb have when it hits the car?

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

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

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

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