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Final 2011 Exam Solutions

Q1 (Ave Score: 21.6/25)

a) $\frac{mv^{2}}{r}=\frac{Gmm}{r^{2}}$

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

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

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

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

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

$r=8061\mathrm{km}$

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

b) $v=\frac{2\pi r}{7200}=\frac{2\pi 8.061\times10^{6}}{7200}=7034\mathrm{m/s}$

c) $-\frac{GMm}{8.061\times10^6}+\frac{GMm}{6.380\times10^6}=1.3\times10^{10}\mathrm{J}$

Q2 (Ave Score: 21.5/25)

a) $a=0.5\mathrm{ms^{-2}}$

$Fr-Tr=\frac{1}{2}m_{pulley}r^{2}\alpha=\frac{1}{2}m_{pulley}r^{2}\frac{a}{r}$

$F-T=\frac{1}{2}m_{pulley}a$

$m_{gift}a=T-m_{gift}g\to T=m_{gift}(a+g)$

$F=m_{gift}(a+g)+\frac{1}{2}m_{pulley}a$

$F=4(10.3)+\frac{1}{2}\times2\times0.5=41.7\mathrm{N}$

b) $\alpha=\frac{a}{r}=\frac{0.5}{0.2}=25\mathrm{s^{-2}}$

c) $F_{N}=m_{elf}g-F=20\times9.8-41.7=154.3\mathrm{N}$

d) $F=m_{gift}g=4\times9.8=39.2\mathrm{N}$

e) $\omega=\frac{v}{r}=\frac{0.5}{0.2}=2.5\mathrm{s^{-1}}$

Q3 (Ave Score:21.9/25)

a) $\theta=\tan^{-1}\frac{30}{60}=26.57^{o}$

b)

Vertical forces, up is positive $T\sin\theta+Fh_{y}=m_{sign}g+m_{rod}g$

Horizontal forces, right is positive $-T\cos\theta+Fh_{x}=0$

Torques $0.6T\sin\theta=0.8m_{sign}g+0.4m_{rod}g$

$T=\frac{0.8\times3\times9.8+0.4\times1\times9.8}{0.6\times\sin26.57^{o}}=102.24\mathrm{N}$

c) $Fh_{y}=4.981-102.24\sin26.57^{o}=-6.53\mathrm{N}$

So force points down.

d) $Fh_{x}=102.24\cos26.57^{o}=91.4\mathrm{N}$

Force points to the right

Q4 (Ave Score: 20.3/25)

Weight of displaced fluid

$(\frac{1}{2}1000+\frac{1}{2}1.2)\frac{4}{3}\pi0.1^{3}$

Weight of ball including dense air

$1800\frac{4}{3}\pi(0.1^{3}-0.09^{3})+\rho_{air}\frac{4}{3}\pi0.09^{3}$

At equilibrium these two are equal, so

$\rho_{air}=\frac{(\frac{1}{2}1000+\frac{1}{2}1.2)\frac{4}{3}\pi0.1^{3}-1800\frac{4}{3}\pi(0.1^{3}-0.09^{3})}{\frac{4}{3}\pi0.09^{3}}$

$\rho_{air}=17.55\mathrm{kg\,m^{-3}}$

b) $m_{air}=17.55\times\frac{4}{3}\pi0.09^{3}=.054\mathrm{kg}$

c) $n=\frac{54}{29}=1.85 \mathrm{moles}$

d) $PV=nRT$

$P=\frac{1.85\times8.314\times293}{\frac{4\pi}{3}0.09^{3}}=1471\mathrm{kPa}=14.6\mathrm{atm}$

Q5 (Ave Score: 21.9/25)

a) $kx_{0}=mg$

$x_{0}=\frac{0.5\times9.8}{10}=0.49\mathrm{m}$

b) $\Delta PE=\int_{x_{0}}^0 F_{Net}\,dx=\int_{x_{0}}^0(kx-mg)\,dx=-\frac{1}{2}kx_{0}^{2}+mgx_{0}=-\frac{1}{2}\times10\times0.49^{2}+0.5\times9.8\times0.49=1.2\mathrm{J}$

If we took the spring in the other direction then

$\Delta PE=\int_{x_{0}}^{2x_{0}} F_{Net}\,dx=\int_{x_{0}}^{2x_{0}}(kx-mg)\,dx=\frac{3}{2}kx_{0}^{2}-mgx_{0}=\frac{3}{2}\times10\times0.49^{2}-0.5\times9.8\times0.49=1.2\mathrm{J}$

So you can see that there is the same amount of energy stored in either direction, which is important for simple harmonic motion to occur.

c) $f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}=\frac{1}{2\pi}\sqrt{\frac{10}{0.5}}=0.71\mathrm{Hz}$

d) $v_{max}=\omega A= 2\pi\times0.71\times0.49=2.19\mathrm{m/s}$ occurs at displacement $x=0\mathrm{m}$ from $x_{0}$

e) $a_{max}=\omega^{2} A=(2\pi\times0.71)^{2}\times0.49=9.8\mathrm{m/s^{2}}$ occurs at displacements $x=-0.49\mathrm{m},x=0.49\mathrm{m}$ from $x_{0}$

Q6 (Ave Score: 22.1/25)

a) Diagram should have pressure nodes wherever there are displacement antinodes and vice versa.

b) $v=f\lambda$

$\lambda=0.4\mathrm{m}$

$f=\frac{330}{0.4}=825\mathrm{Hz}$

c) To get node at each ear the wavelength must be halved, hence frequency is doubled, $f=1650\mathrm{Hz}$

d) Another half wavelength, 0.2m in front of his ears/nose.

Q7 (Ave Score: 19.2/25)

a) For an adiabatic process

$P_{i}V_{i}^{\gamma}=P_{f}V_{f}^{\gamma}$ where in this case $\gamma=\frac{7}{5}$

$P_{f}=P_{i}(\frac{V_{i}}{V_{f}})^{\gamma}=101.3\mathrm{kPa}\times{5}^{7/5}=964.2\mathrm{kPa}$

b) As the gas is ideal

$P_{i}V_{i}=nRT_{i}$

$P_{f}V_{f}=nRT_{f}$

$\frac{T_{f}}{T_{i}}=\frac{P_{f}V_{f}}{P_{i}V_{i}}$

$T_{f}=\frac{964.2}{101.3}\times\frac{1}{5}\times275=523.5\mathrm{K}$

c) For an isothermal process

$W=nRT\ln{\frac{V_{f}}{V_{i}}}$

$PV=nRT$

$W=964.2\times10^{3}\times 2\ln(5)=3103\mathrm{kJ}$

d) For an isothermal process

$P_{i}V_{i}=P_{f}V_{f}$

$P_{f}=\frac{V_{f}}{V_{i}}P_{i}=\frac{964.2}{5}=192.8\mathrm{kPa}$

e) Red is first process, green is second

Q8 (Ave Score: 18.5/25)

a) $COP=\frac{Q_{L}}{W}=\frac{T_{L}}{T_{H}-T_{L}}=274/24=11.42$

$Q_{L}=11.42W=11.42\mathrm{kJ}$

b) 1st Law gives $Q_{H}=Q_{L}+W=12.42\mathrm{kJ}$

c)$\Delta S= -\frac{Q_{L}}{T_{L}}=\frac{11420}{274}=-41.67\mathrm{J/K}$

d)$\Delta S= \frac{Q_{H}}{T_{H}}=\frac{12420}{298}=41.67\mathrm{J/K}$

e) $\Delta S= 0\mathrm{J/K}$

phy141/examprep/final11sols.txt · Last modified: 2014/12/05 13:05 by mdawber
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