Activity ID: First Letter of Last Name: The University of British Columbia Midterm 2 - November 9, 2016 Physics 157 Elements of Physics I TIME: 1.0 hour CANDIDATE’S NAME: Last Name First Name or Initials Please place your UBC Student card on your desk. SIGNATURE: STUDENT NUMBER: THIS EXAMINATION CONSISTS OF 3 QUESTIONS AND A TOTAL OF 6 PAGES. CHECK TO ENSURE THAT THIS PAPER IS COMPLETE. INSTRUCTOR’S NAME: (circle one) 101–Witt 102–Folk 103–Kotlicki TO GET FULL MARKS YOU MUST SHOW YOUR WORK (a) Each candidate should be prepared to produce, upon request, his/her UBC card. (b) Read and observe the following additional rules: No candidate shall be permitted to enter the examination room after the expiration of one-half hour, or leave during the first half-hour of the examination. Candidates are not permitted to ask questions of the invigilators, except in cases of supposed errors or ambiguities in examination questions. CAUTION: Candidates guilty of any of the following, or similar, dishonest practices shall be immediately dismissed from the examination and shall be liable to disciplinary action. (a) Making use of any books, memoranda, calculators, cell phones, audio or visual players or other memory aid devices and electronics, other than authorized by the examiners. (b) Speaking or communicating with other candidates. Any calculator is allowed. Each Question is worth the points indicated below: 1 2 10 3 10 10 Total: 30 1 Name: Activity ID: 1. The photograph below shows a person jumping on a pneumatic pogo stick. In case you have not seen one before, a person on a pogo stick bounces up and down as the stick absorbs the energy and releases it again like a spring. Many pogo sticks do use springs, but in this problem the “bounce” comes from compression of a cylinder of gas as the person hits the minimum height of their bounce. The cross section of a pneumatic pogo stick is shown on the right: the cylinder of gas, with a piston that is connected by a shaft to the foot of the pogo that hits the ground. When a pogo stick is up in the air, the cylinder is expanded to its maximum volume of Vi = 1.3 l, the pressure in the cylinder is pi =200 kPa, and the temperature Ti is 20 C. At the top of a jump, the center of gravity of the person and pogo stick together (mass 60 kg) are h1 = 2.1 m above the ground. At the lowest point, the air in the cylinder of the pogo stick is maximally compressed and the center of gravity is h2 =1.2m above the ground. a) How many moles of gas are there in the cylinder? b) What is the temperature inside the cylinder at maximal compression? c) What is the volume of air in the cylinder at maximal compression? cylinder piston shaft platform to stand on foot of pogo that lands on the ground (compressed position) extended position The photograph above on the left shows a person jumping on a pneumatic pogo stick. The cross section of the stick is shown above on the right. When the person is in the air the pressure in the cylinder is pi =200 kPa and the temperature Ti is a room temperature. At the top of her jump this person, weighing m = 60 kg with the stick, has her center of gravity h1= 2.1 m above the ground. When she lands, and the air in the cylinder of the pogo stick is fully compressed, her center of gravity is h2=1.2m above ground. The volume of the uncompressed gas in the cylinder is Vi = 1.3 l. 1. 2. 3. 4. How many molls of gas there are in the cylinder? What is the final temperature inside the cylinder after she lands? What is the final volume of air in the cylinder after she lands? What is the final pressure inside the cylinder after she lands? 1. n = pi Vi/TiR = 0.11 2. mg(h1-h2) = work done on the gas in adiabatic process (fast). ΔU = ncvΔT = Q - W = -W W = mg(h1-h2) ΔT = mg(h1-h2)/ncv = 254K Ti= 293K Tf = Ti +ΔT = 547 K (γ−1) (γ−1) 3. TiVi = Tf Vf Vf = (TiVi (γ−1)/Tf )1/(γ−1)= 0.27 l 4. pf = nRTf / Vf = 1850 kPa 1p 2p 2p 2p 2p 1p 2 Name: Activity ID: 1. (continued) 3 Name: Activity ID: 2. A thermodynamic system undergoes a cyclic process as shown in the figure below. The cycle consists of two closed loops which are right triangles. The cycle is the path c ! a ! b ! c then c ! d ! e ! c. Given Ta = 400K, Tb = Tc = Te = 200K, and Td = 100K. Sa = Sb = 5 J/K and Sc = 4 J/K Sd = Se = 2 J/K. a) Find the work for the whole process, including both loops. b) Find the heat input to the system, QH , for the whole process including both loops. c) What is the efficiency of system? T a e c b d S 4 Name: Activity ID: 3. As part of your job at the patent office, you are asked to evaluate the six designs shown in the figures below for innovative new heat engines. a) Which of the designs violate(s) the first law of thermodynamics? b) Which of the remaining designs violate(s) the second law of thermodynamics by implying an impossible efficiency? (Show your work to obtain full credit. Carefully note the hot and cold temperatures, and various di↵erent powers, for each process shown.) 2 1 1000W 500K 1000W 250W 250K 4 1000W 1000W 200K 300K 6 5 600K 500K 500W 500W 200K 600K 1000W 600K 1000W 400W 600W 600W 1000W 650W 450W 550W 3 600K 600W 300K 0W 300K 5 Useful Constants R = 8.31451J/mol K, 1 atm = 1.013 ⇥ 105 Pa, 1 atm litre = 101.3 J One can use 1 atm ⇡ 1 ⇥ 105 Pa and 1 atm litre ⇡ 100 J. StefanBoltzmann constant = 5.6704 ⇥ 10 8 W m 2 K 4 . 1 K 1 , The density of water is 1 gram/cm3 =1000 kg/m3 . air = 1.4, CV air = 20.8Jmol Mechanics Linear Motion: x = x0 + 12 (vx0 + vx )t, x = x0 + vx0 t + 12 ax t2 , vx = vx0 + ax t, vx 2 = vx0 2 + 2ax (x x0 ) 2 Circular Motion: ac = vr d Forces: F~ = m~a, F~ = dt p~, Friction: F = µN , Spring: F = kx, Damping: F~ = b~v Bouyant: FB = ⇢V g R ~r ~ v. W = ~rif F~ · d~r, W = F~ · ~r, K = 12 mv 2 , U gravity = mgy, U spring = 12 kx2 , P = dW dt , P = F · ~ Thermodynamics L Thermal Expansion: L = ↵L T Stress and Strain: F A =Y L 3 1 Ideal Gas Law: P V = nRT , Kav = 2 kT , 2 kT for each degree of freedom. Thermal Conductivity: I = Qt = kA Tx Black Body Radiation: P = e AT 4 , max T = 2.8977685 ⇥ 10 3 m · K Internal Energy: U = nCV T First Law: dQ = dU + dW for an ideal gas dW = P dV . Work for isothermal process W = nRT ln(Vf /Vi ). For adiabatic expansion T V 1 = constant, if the number of moles is constant P V constant and = Cp /Cv . Work for adiabatic process W = Z V2 P dV = C V1 Z V2 V1 dV C = (V2 1 V 1 V1 1 = C where C is a ) Q = mc T , Q = mL. CP = CV + R, CV = f2 R where f =degrees of freedom. f = 3 for monatomic and f = 5 for diatomic. dS = dQ/T e = W/QH COPCooling = |QC |/|W | COPHeating = |QH |/|W | eCarnot = 1 TC /TH Integrals Z xn+1 + constant n+1 xn dx = n 6= Trig sin ✓1 + sin ✓2 = 2 cos( 1 ✓1 Z ✓2 2 x 1 ) sin( dx = ln x + constant ✓1 + ✓2 ) 2 Area and Volume Area of a sphere A = 4⇡r2 . Area of a cylinder A = 2⇡rl. Area of a circle A = ⇡r2 . Volume of a cylinder V = l⇡r2 . Volume of a sphere V = 43 ⇡r3 . Oscillations ! = 2⇡f , T = f1 , x = Acos(!t + ), ! 2 = Damped Oscillations: x = A0 e bt m bt 2m k m. cos(!t + ), where ! = q !0 2 b 2 ( 2m ) , Q = 2⇡ Energy for damped E = E0 e Waves q v = Tµ , k = 2⇡ , v = f , P = 12 µ! 2 A2 v, p0 = ⇢!vs0 q v (1± D ) RT Pav v= = 10dB log10 ( II0 ) Doppler E↵ect f 0 = f0 (1⌥ vvs ) M , I = 4⇡r 2 , v y = A cos(kx ⌥ !t + ) Interference k x + = 2⇡n or ⇡(2n + 1) n = 0, ±1, ±2, ±3, ±4, . . . mv Standing Waves fm = mv 2L m = 1, 2, 3, . . . fm = 4L m = 1, 3, 5, . . . Beats E E. f = f2 f1