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phy131studiof15:lectures:finalp1sol [2015/11/30 14:47] mdawber created |
phy131studiof15:lectures:finalp1sol [2015/12/02 09:36] (current) mdawber [Question 6] |
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===== Question 1 ===== | ===== Question 1 ===== | ||

- | A. | + | {{phy141f12finalq1fig.png}} |

+ | | ||

+ | 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? | ||

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

Line 27: | Line 31: | ||

(35,900 km above surface) | (35,900 km above surface) | ||

- | B. $v=\sqrt{\frac{GM_{E}}{r}}=\sqrt{\frac{6.67\times10^{-11}\times5.98\times10^{24}}{4.23\times10^7}}=3071\mathrm{m\,s^{-1}}$ | + | B. (5 points) What is the magnitude of the velocity of the satellite along it's original circular path? |

- | C.$\frac{1}{2}mv^{2}=\frac{-GMm}{4.23\times10^{7}}-\frac{-GMm}{6.38\times10^{6}}$ | + | $v=\sqrt{\frac{GM_{E}}{r}}=\sqrt{\frac{6.67\times10^{-11}\times5.98\times10^{24}}{4.23\times10^7}}=3071\mathrm{m\,s^{-1}}$ |

+ | | ||

+ | 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. | ||

+ | | ||

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

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

Line 35: | Line 43: | ||

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

- | 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. | + | 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 ===== | ===== Question 2 ===== | ||

- | A. The condition for when static friction and the gravitational force down the plane exactly cancel is given by | + | {{phy141f12finalq2fig.png}} |

+ | | ||

+ | 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 | ||

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

Line 47: | Line 64: | ||

$\theta=21.8^{o}$ | $\theta=21.8^{o}$ | ||

- | B. For rolling | + | 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 | ||

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

Line 57: | Line 77: | ||

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

- | C. $\frac{\frac{2}{5}}{\frac{2}{5}+1}=0.29=29\%$ | + | C. (10 points) What percentage of the kinetic energy of the ball is rotational while it is rolling down the incline? |

+ | | ||

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

===== Question 3 ===== | ===== Question 3 ===== | ||

- | A. | + | {{m1fig2_2011.png}} |

+ | | ||

+ | 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. | ||

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

Line 70: | Line 96: | ||

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

- | B. | + | 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. |

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

Line 76: | Line 102: | ||

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

- | C. | + | 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. |

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

Line 82: | Line 108: | ||

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

- | D. | + | 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? |

$y=52.1\mathrm{m}$ | $y=52.1\mathrm{m}$ | ||

Line 90: | Line 116: | ||

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

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

$0\mathrm{m}$ | $0\mathrm{m}$ | ||

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

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

Line 102: | Line 128: | ||

===== Question 4 ===== | ===== Question 4 ===== | ||

- | A. $\pi \frac{d^{2}}{4}h=0.2$ | + | {{phy141f12finalq4fig.png}} |

+ | | ||

+ | A 90cm tall cylindrical steel oil drum weighs 15kg and has a external volume of 0.2m$^{3}$. When full the drum contains 190L of crude oil. The density of crude oil is 900kg/m$^{3}$ and of water is 1000kg/m$^{3}$. | ||

+ | | ||

+ | | ||

+ | A. (5 points) What is the external diameter of the oil drum? | ||

+ | | ||

+ | $\pi \frac{d^{2}}{4}h=0.2$ | ||

$d=0.53\mathrm{m}$ | $d=0.53\mathrm{m}$ | ||

- | 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 | + | B. (10 points) If the oil drum has fallen overboard and is floating upright in the sea, what length of the oil drum $x$ sticks up above the water? As a simplifying approximation you may assume that the all the steel is on the sides of the drum, and the top and bottom do not contribute to the mass of the drum. |

+ | | ||

+ | 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 | ||

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

Line 118: | Line 153: | ||

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

- | C. $P=P_{atm}+\rho gh=1.013\times10^{5}+1000\times9.8\times0.837=1.095\times10^{5}\mathrm{Pa}$ | + | C. (5 points) What is the pressure at the bottom of the drum? You may assume standard atmospheric pressure for the air above the drum. |

+ | | ||

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

===== Question 5 ===== | ===== Question 5 ===== | ||

- | A. $v_{sound}=343\mathrm{m\,s^{-1}}$ | + | {{phy141f12finalq6fig.png}} |

+ | | ||

+ | The speed of sound in air is $v=331+0.6T\, \mathrm{ms^{-1}}$ where $T$ is the temperature in $^{\circ}$C. Two organ pipes of length 0.6m which are open at both ends are used to produce sounds of slightly different frequencies by heating one of the tubes above the room temperature of 20$^{\circ}$C. | ||

+ | | ||

+ | A. (5 points) What is the lowest frequency sound that can be produced in the room temperature pipe? | ||

+ | | ||

+ | $v_{sound}=343\mathrm{m\,s^{-1}}$ | ||

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

- | B. See [[phy141:lectures:29&#open_and_closed_pipes|here]]. | + | B. (5 points) Draw on the figure the form of the standing wave amplitude for both displacement and pressure that produces this sound. |

- | C. $\Delta f=1\mathrm{Hz}$ | + | See [[phy131studiof15:lectures:chapter18&#open_and_closed_pipes|here]]. |

+ | | ||

+ | C. (5 points) To produce beats with a beat frequency of 1Hz with the sound produced in (a) using the first harmonic of the heated pipe what should the temperature of the heated pipe be? | ||

+ | | ||

+ | $\Delta f=1\mathrm{Hz}$ | ||

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

Line 137: | Line 184: | ||

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

- | D. $f'=\frac{v_{sound}+v_{obs}}{v_{sound}}f$ | + | D. (5 points) At what speed should a person running toward the tubes be moving so that the sound from the room temperature tube has an apparent frequency equal to the actual frequency of sound produced by the heated tube. |

+ | | ||

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

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

Line 143: | Line 192: | ||

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

- | E. $f=2\times285.8=571.6\mathrm{Hz}$ | + | E. (5 points) What is the frequency of the second harmonic that the room temperature pipe produces? |

+ | | ||

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

===== Question 6 ===== | ===== Question 6 ===== | ||

- | A. $e=\frac{W}{Q_{H}}$ | + | {{phy141f12finalq8fig.png}} |

+ | | ||

+ | The diagram shows the P-V diagram for a 40% efficient ideal Carnot engine. Assume the gas used in this Carnot engine is an ideal diatomic gas. | ||

+ | | ||

+ | A. (5 points) For every Joule of work obtained from the engine, how much heat needs to be added to engine? | ||

+ | | ||

+ | $e=\frac{W}{Q_{H}}$ | ||

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

- | B.$Q_{H}=W+Q_{L}$ | + | B. (b) (5 points) For every Joule of work obtained from the engine how much heat is lost to the environment? |

+ | | ||

+ | $Q_{H}=W+Q_{L}$ | ||

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

- | C. $P_{B}V_{B}=nRT_{H}$ | + | C. (5 points) At points B and D the gas in the system has the same volume, but different temperatures. If the gas at point D is at twice atmospheric pressure, what is the pressure of the gas at point B? |

+ | | ||

+ | $P_{B}V_{B}=nRT_{H}$ | ||

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

Line 170: | Line 231: | ||

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

- | D. The expansion from B to C is adiabatic so $P_{B}V_{B}^{\gamma}=P_{C}V_{C}^{\gamma}$ | + | D. (5 points) If the volume of the gas at point B is 1L what is the volume of the gas at point C? |

+ | | ||

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

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

Line 186: | Line 249: | ||

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

- | E. $\Delta S=0\mathrm{\frac{J}{K}}$ | + | E. (5 points) How much does the net entropy of the engine and the environment change for every Joule of work done by this Carnot engine? |

+ | | ||

+ | | ||

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