# Fall 2010 Final Exam Solutions

## Q1 solution (Avg Score 18.4/25)

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

## Q2 Solution (Ave: 17/25)

a) $v=at$

$1=a\times0.5$ $a=2\mathrm{ms^{-2}}$

$Fr-Tr=\frac{1}{2}Mr^2\alpha$ → $F-T=\frac{1}{2}Ma$

$T-mg=ma$ → $T=mg+ma$

$F=\frac{1}{2}Ma+mg+ma$

$F=0.5\times 0.2\times 2+1\times(9.8+2)=12\mathrm{N}$

b)$\alpha=\frac{a}{r}=\frac{2}{0.1}=20\mathrm{rad\,s^{-2}}$

c) At constant speed $a=0$ so $F=mg=1\times9.8=9.8\mathrm{N}$

d) $\omega=\frac{v}{r}=\frac{1}{0.1}=10\mathrm{rad\,s^{-1}}=10\times\frac{60}{2\pi}=95.5\mathrm{rpm}$

e) $L=I\omega=\frac{1}{2}{M}r^{2}\omega=0.5\times0.2\times0.1^{2}\times10=0.01 \mathrm{kg\,m^{2}s^{-1}}$ out of page by right hand rule

## Q3 solution (Ave: 19.95/25)

a) $\frac{1}{2}kx^2=mgh$

$x=\sqrt{\frac{2mgh}{k}}$

$h=1\mathrm{m}\tan10^{o}=0.176\mathrm{m}$

$x=\sqrt{\frac{2\times.08\times9.81\times0.176}{15}}=0.136\mathrm{m}$

b) $KE_{Rot}=\frac{1}{2}I\omega^{2}=\frac{1}{2}\frac{2}{5}MR^{2}v^{2}R^{2}=1/5Mv^{2}$

$KE_{Tot}=\frac{1}{2}Mv^{2}+KE_{Rot}=\frac{7}{10}Mv^{2}$

$\frac{KE_{Rot}}{KE_{Tot}}=\frac{2/10}{7/10}=\frac{2}{7}=28.6\%$

## Q4 solution (Ave: 17.4/20)

The condition for the balloon to rise is that the bouyant force must equal the weight

$m_{b}g+\rho_{He}Vg=\rho_{Air}Vg$

$m_{b}=(\rho_{Air}-\rho_{He})\frac{4\pi}{3}r^{3}=(\rho_{Air}-\rho_{He})\frac{\pi}{6}d^{3}$

$d^{3}=\frac{6m_{b}}{(\rho_{Air}-\rho_{He})\pi}$

$d=(\frac{6\times0.003}{(1.2-0.18)\pi})^{1/3}=0.178\mathrm{m}$

## Q5 solution (Ave: 20.76/25)

a) $mv=(m+M)v'$

$v'=\frac{m}{m+M}v=\frac{0.1}{10}100=1\mathrm{m/s}$

b) $T=2\pi\sqrt{\frac{l}{g}}=2\pi\sqrt{\frac{1.5}{9.81}}=2.46\mathrm{s}$

There are many ways to solve c) and d). These are the ones I think are easiest.

c) $a_{max}=v_{max}\omega=1\times\frac{2\pi}{2.46}=2.56m\,s^{-2}$

d) $s_{max}=\frac{v_{max}}{\omega}=1\times\frac{2.46}{2\pi}=0.391\mathrm{m}$

This is the arc length that the block moves through

$s=r\theta$

$\theta=\frac{0.391}{1.5}=0.261\mathrm{rads}=14.94^{o}$

Alternatively, you can, and most of you did, do this by conservation of energy to get h and then a bit of trig.

## Q6 Solution (Ave: 15.4/25)

a) The speed of sound is given by $v=\sqrt{\frac{\beta}{\rho}}$ where $\beta$ is the bulk modulus and $\rho$ is the mass density.

If we treat the gas as ideal we can find the molar density using the ideal gas law $PV=nRT$ → $\frac{n}{V}=\frac{P}{RT}=\frac{1.1\times10^{5}}{8.314\times293}=45.16\mathrm{mol\,m^{-3}}$.

To get the mass density $\rho$ we times the molar density by the molar mass $\rho=45.15*44.1\times10^{-3}=1.99\mathrm{kg\,m^{-3}}$

The speed of sound in the tube is then $v=\sqrt{\frac{1.1\times10{5}}{1.99}}=235\mathrm{m/s}$

b) The ends of the tube experience maximum pressure variation. The displacement variation at each end must be zero as the ends are sealed, and the pressure and displacement are out of phase with each other.

c)There are 3 flame maxima visible, which means there are 3 pressure nodes and 4 pressure antinodes. The 2 m length of tube thus contains 3/2 wavelengths, so $\lambda=2/3l=4/3\mathrm{m}$. $f=\frac{v}{\lambda}=\frac{235}{4/3}=176\mathrm{Hz}$.

d)There are 6 flame maxima visible, which means there are 6 pressure nodes and 7 pressure antinodes. The 2 m length of tube thus contains 3 wavelengths, so $\lambda=1/3=2/3\mathrm{m}=\frac{235}{2/3}=352\mathrm{Hz}$.

Some you misinterpreted my comment on the holes stopping before the end of the tube, this was intended simply to alert you to the fact that the flame height at the end of the tube should be a minimum even though in the bottom picture it looks like a maximum on the end. I did not deduct points if you used 1.9m for the length of the tube.

## Q7 Solution (Ave:17/25)

a)

The isothermal process does more work (greater area under PV curve).

b) In an isothermal expansion T is constant, so from $PV=nRT$, $P_{0}V_{0}=P_{1}V_{1}$, $P_{1}=\frac{V_{0}}{V_{1}}P_{0}=\frac{P_{0}}{2}$.

c) In an adiabatic process $PV^{\gamma}$ is constant, where for a diatomic gas $\gamma=\frac{7}{5}$. $P_{0}V_{0}^{7/5}=P_{1}V_{1}^{7/5}$→$P_{1}=\frac{V_{0}^{7/5}}{V_{1}^{7/5}}P_{0}=\frac{P_{0}}{2^{7/5}}$

d) $P_{0}V_{0}=nRT_{0}$

$P_{1}V_{1}=nRT_{1}$→$\frac{P_{0}}{2^{7/5}}2V_{0}=nRT_{1}$

Divide second equation by first to get

$\frac{T_{1}}{T_{0}}=\frac{2}{2^{7/5}}$→$T_{1}=\frac{T_{0}}{2^{2/5}}$

e) In an adiabatic process $Q=0$, and as $\Delta S=\frac{Q}{T}$, $\Delta S = 0 \mathrm{\frac{J}{K}}$

## Q8 Solution (Ave:14.9/30)

a) $Q_{L}=m_{w}l_{f}=1\times3.34\times10^{5}=3.34\times10^{5}\mathrm{J}$

For a Carnot cycle

$\frac{T_{L}}{T_{H}}=\frac{Q_{L}}{Q_{H}}$ → $Q_{H}=\frac{T_{H}}{T_{L}}Q_{L}=\frac{293}{273}\times3.34\times10^{5}=3.58\times10^{5}J$

b) $W=Q_{H}-Q_{L}=3.58\times10^{5}-3.34\times10^{5}=2.4\times10^{4}J$

c) Entropy change of water $\frac{-3.34\times10^{5}}{273}=-1223\mathrm{J/K}$

Entropy change of room $\frac{3.58\times10^{5}}{293}=1223\mathrm{J/K}$

Total entropy change = 0 J/K (This is true for all Carnot Cycles)