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Solutions to Fall 2012 Final Exam

Question 1 (Ave Score: 20/25)

A.

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

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

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

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

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

$\frac{2\pi r}{v}$ = 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. $v=\sqrt{\frac{GM_{E}}{r}}=\sqrt{\frac{6.67\times10^{-11}\times5.98\times10^{24}}{4.23\times10^7}}=3071\mathrm{m\,s^{-1}}$

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

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

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

D. 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 (Ave Score: 23/25)

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 (Ave Score 24.75/30)

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 (Ave Score: 15.9/20)

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 (Ave Score: 16.7//25)

A. $x_{0}=(m_{2}-m_{1}\sin\theta)\frac{g}{k}$

B. $W_{net}=\frac{1}{2}kA^{2}$

C. Consider the sum of the forces on each object

$m_{2}a=m_{2}g-T_{2}$

$\frac{1}{2}m_{3}a=T_{2}-T_{1}$

$m_{1}a=T_{1}-m_{1}g\sin\theta-kx$

$(m_{1}+m_{2}+\frac{1}{2}m_{3})a=m_{2}g-m_{1}g\sin\theta-kx$

The solution to this equation of motion is an oscillation centered around $x_{0}$ with $\omega^{2}=\frac{k}{m_{1}+m_{2}+\frac{1}{2}m_{3}}$

and the frequency is

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m_{1}+m_{2}+\frac{1}{2}m_{3}}}$

D. $v_{max}=\omega A = \sqrt{\frac{k}{m_{1}+m_{2}+\frac{1}{2}m_{3}}}A$

This occurs when the displacement from $x_{0}=0$

E. $a_{max}= \omega^{2} A = \frac{k}{m_{1}+m_{2}+\frac{1}{2}m_{3}}A$

This occurs when the displacement from $x_{0}=\pm A$

Question 6 (Ave Score: 21.75/25)

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 7 (Ave Score 19.67/25)

A. For both processes

$\Delta E_{int}=\frac{3}{2}nR\Delta T=\frac{3}{2}\times3\times8.314\times100=3741.3\mathrm{J}$

B. For an ideal monatomic gas expanded at constant volume

$Q=\frac{3}{2}nR\Delta T=\frac{3}{2}\times3\times8.314\times100=3741.3\mathrm{J}$

At constant pressure

$Q=\frac{5}{2}nR\Delta T=\frac{5}{2}\times3\times8.314\times100=6235.5\mathrm{J}$

C. At constant volume $W=0J$

At constant pressure $W=nR\Delta T=2494.2\mathrm{J}$

D.

For the constant volume process

$\Delta S=\int \frac{dQ}{T}=\frac{3}{2}nR\int_{T_{1}}^{T_{2}}\frac{dT}{T}=\frac{3}{2}nR\ln\frac{T_{2}}{T_{1}}=\frac{3}{2}\times3\times8.314\ln\frac{393}{293}=10.985\mathrm{J/K}$

For the constant pressure process

$\Delta S=\int \frac{dQ}{T}=\frac{5}{2}nR\int_{T_{1}}^{T_{2}}\frac{dT}{T}=\frac{5}{2}nR\ln\frac{T_{2}}{T_{1}}=\frac{5}{2}\times3\times8.314\ln\frac{493}{393}=14.136\mathrm{J/K}$

E. $PV=nRT$

$P_{initial}=1.013\times10^{5}\mathrm{Pa}$

$V_{initial}=\frac{3\times8.314\times293}{1.013\times10^{5}}=0.072\mathrm{m^{3}}$

$P_{final}=\frac{3\times8.314\times393}{V_{initIal}}=1.386\times10^{5}\mathrm{Pa}$

$V_{final}=\frac{3\times8.314\times493}{P_{final}}=0.090\mathrm{m^{3}}$

Question 8 (Ave Score 19.67/25)

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

phy141/examprep/final12sols.txt · Last modified: 2014/12/03 08:08 by mdawber
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