2015 PhysicsBowl Solutions Ans Ans Ans Ans Ans B 2. C METHOD #1: METHOD #2: 3. A 4.

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1 05 PhyicBowl Solution # An # An # An # An # An B B B 3 D 4 A C D A 3 D 4 C 3 A 3 C 3 A 33 C 43 B 4 B 4 D 4 C 34 A 44 E 5 E 5 E 5 E 35 E 45 B 6 D 6 A 6 A 36 B 46 E 7 A 7 D 7 D 37 A 47 C 8 E 8 C 8 B 38 D 48 B 9 C 9 B 9 E 39 E 49 A 0 E 0 C 30 B 40 C 50 D. B milli repreent 0 3. C METHOD #: Uing contant acceleration kinematic, we have the average velocity computed a v = x = 7.5 = 0.65 m which lead to v = (v t 0 + v f ) 0.65 = (v 0 + 0) v 0 =.5 m. METHOD #: One can graph velocity v time for the box. Knowing that the area under the curve i the change in poition, we have A = bh 7.5 = (.0)v 0 v 0 =.5 m 3. A Vector quantitie point and peed i a calar. 4. B Location of complete contructive interference are known a antinode on the tanding wave. 5. E The gravitational force i alway directed downward!! 6. D Converting unit 0 mile hr min 600 m = 8.9 m hr 60 min 60 mi 7. A Object that move in uniform circular have a velocity direction change toward the center of the circle being traced out. 8. E After the balloon obtain charge from being rubbed, it caue polarization in the wall allowing like charge to be cloer to each other when they come in contact, providing an attractive force to prevent the balloon from falling to the ground. 9. C Since v car > v car at the tart of the interval, the ditance between the car increae. However, noting at that the end of the econd that the peed of the car are v car = v 0 + at = 3 + (.5)() = 4.5 m and v car = (5.5)() = 4.8 m, by the interval end, car i moving fater than car and catching up to it. Hence, the ditance between car i now decreaing. Since the peed of car i only lightly greater than car at the end and tarted off much lower, then car will actually increae it ditance from car. To check, one can compute the poition change a x = v t + a t = (3)() + (.5)() = 3.75 m with x = v t + a t = (9.3)() + (5.5)() =.05 m. 0. E When the witch i cloed, a wire of zero reitance i added to the circuit. We note, though, that the wire i connected to the left ide of the battery. In other word, we have effectively added a horting wire in parallel to the other reitance-le wire on the left ide of the circuit. There i no effect on any of the bulb a a reult becaue the circuit i effectively unchanged.. B When adding/ubtracting, the key i to line up the value and look at the lat column of digit for which each quantity ha a ignificant digit. That i, L + L = 8.0 cm and L 3 + L 4 = cm. Upon ubtracting, we have then = 5.0 cm ince we cannot keep the thouandth place in the reult ince 8.0 only ha value to the hundredth place!. D METHOD #: Knowing that the linear momentum i computed a p = mv, we need to find the peed. Thi i obtained from the kinetic energy a KE = mv = ()v v = m. So, p = mv = ()( ) = 7.0 kg m.

2 METHOD #: By rewriting the kinetic energy lightly, one ha the form KE = p and o p = m KE = ()() = 7.0 kg m 3. C If we rotate our coordinate lightly to take advantage of the incline, we find in the direction perpendicular to the incline that the total normal force act off the incline and only a portion of the gravitational force act that way. That i, n < w. Alo, along the incline urface, there are two downward force (applied and a piece of the gravitational) and one upward force. Since there i no acceleration, thee force um to zero N. Hence, F + w along incline = f meaning that F < f. 4. D When the poition take the form given, we have contant acceleration with x = x 0 + v 0 t + at. By identifying term, we ee that a = 0 or that a = 0 m. 5. E The molecule in the air exert a force upward on the kydiver. The kydiver exert a force downward on the molecule in the air. 6. A Thi i the de Broglie hypothei. 7. D With the poitive velocity, the object i moving to the right arbitrarily. The total change in poition i from the area under the curve. Hence, a long a the velocity i poitive, the object i moving away from the origin. The negative lope from 5 to 7 econd imply indicate that the object move forward but i lowing down (like a vehicle approaching a top ign). 8. C... For an open tube, we can ue that f n = nv nv L = L = (3)(340) = 0.75 m L f n (680) 9. B From PV = nrt, rearrange to compute the volume of the ga a V = nrt = 0.38 m3. The total ma i m = nm = (0) (0.004 kg ) = kg. Finally, ρ = m = kg = 0. mol V 0.38 m C The Nobel Prize wa awarded to Shuji Nakamura, Hirohi Amano, and Iamu Akaaki for the invention of the blue LED.. B METHOD #: By breaking the initial velocity into component, we have v x = 0 co 60 = 0 m and v y = 0 in 60 = 7.3 m. And o, ince the acceleration i only in the y-direction, v x = 0 m and v y = (v 0y + v y ) = ( ) = 8.66 m. So, the total average velocity i computed uing the Pythagorean Theorem giving v = v x + v y = = 3. m. METHOD #: Again, we tart by breaking the velocity into component, but go about finding the location of the highet point in the trajectory. From the vertical component of motion, we have v y = v 0y + at 0 = ( 0)t t =.73. Now, y = y 0 + v 0y t + a yt y = 0 + (7.3)(.73) + ( 0)(.73) = 5.0 m and x = x 0 + v 0x t + a xt x = 0 + (0)(.73) + 0 = 7.3 m. Hence, the magnitude of the diplacement of the object i =.9 m. And finally from the definition of average velocity, we compute v = r.9 m = = 3. m t.73. A Making the free body for each atellite and writing Newton Second Law give F net = ma GmM = ma a = GM r r. Since the radiu of atellite i twice a great, the acceleration i ¼ a large compared to atellite. A for the peed, we write a = v r P and dicover that v r m = GM r v = GM. Hence, atellite will be lower by a factor of. r 3. A From the definition of average acceleration, we compute a = v. Looking only at direction of vector, we have a = (v t f v 0 ) = ( ( )) = ( + ) =. t t 4. C In order for the um of 3 equal-ized vector to be zero, the angle between each vector with any other need to 0. Since the gravitational force i traight downward, the normal force i directed 30 above the horizontal. A picture i ueful! From the t

3 contruction, we ee that the angle θ = 60 making the applied force half-way between the horizontal and the level of the incline. So the applied force would have to be 30 below the incline urface. Thi matche anwer C. 5. E Since the preure and volume decreaed, from PV = nrt, the temperature mut decreae. Thi mean that the internal energy change i negative ince the internal energy depend on the temperature for an ideal ga. Since the volume decreae, the work done on the ga by the urrounding i poitive (force i directed inward on ga and movable piton of container move inward). Finally, from the Firt Law of Thermodynamic, U = Q + W ( ) = Q + (+). Thi mean that the quantity of heat in thi proce will have to be negative Q = ( ) + ( ) = ( ). 6. A The maximum peed during ocillation i v max = ωa = πa which could be found in multiple T way, including equating the maximum KE during ocillation to the maximum potential energy ( ka = mv m v m = k m A = ωa). Now, for one full ocillation, the ma move from full extenion to equilibrium (A), then to full compreion (A) and back again (A) for a total ditance traveled of 4A. Conequently, v avg = d = 4A = 4 t T (Aω) = 4 π (v max ) = v π π max 7. D The area under the force-time curve give the impule on ma. By Newton Third Law, the force on ma ha the ame magnitude but i in the oppoite direction. The area i computed a (0kN)(3m) = 5 N. So, the impule for ma i 5 N. Uing the impule-momentum theorem, p = 5 = m (v f v i ) 5 = v 3.50 f 7.0 v f =.7 m. So, uing the impule on ma, we find p = 5 = m (v f v i ) m = 5 = 5.54 kg B The initial kinetic energy of the ytem i m v = (3.50)(7.0) = J. After the colliion, the kinetic energy i computed a (m + m )v f = ( )(.7) = 33.6 J. The difference in thee quantitie i KE = KE f KE i = = 5.6 J. 9. E By uing the right hand rule with the thumb directed along the current, we find that the wire on the right produce a field directed out of the plane of the page at the electron location. Performing the ame procedure for the left wire, that field alo i directed out of the plane of the page. So, the total field i out of the plane of the page and from F = qv B, the magnetic force i directed to the right (the cro term i to the left, but the electron make the force to the right). Hence, we now need an electric force directed to the left to balance the magnetic force. From F = qe E = F, we ee that if the force i to the left, then the field mut be to the right ince an electron i a negative charge. 30. B The free body diagram of the olid ma in the water ha three force acting gravitational, a pring force from the cale, and a buoyant force from the water. Writing Newton Second Law, we have F net = ma B + T mg = 0. We have from the meaurement in the air (ince the buoyant force on the mall ma from the air will be effectively negligible), that mg =.50 N. Alo, we have that T =.58 N. Solving for the buoyant force yield B = = 0.9 N. The buoyant force i computed a ρ water g V = 0.9 V = 0.9 = 9. (000)(0) 0 5 m 3. The ma of the object i found a m =.50 = 0.50 kg leading to a denity of the ma a ρ = m = 0.50 kg 0 V =.7 03 m 3 3. D Approximately 70% of the Earth urface i water. Approximating the volume of water with the urface area of the earth multiplied by a depth of about 3 mile (5000 meter approximately), we have V water = 7 0 4πr (3 mi) = 8 π( ) (5000) = m 3. The denity of water i 000 kg and o the total ma of water in the ocean i.8 m =.8 0 kg. The molar ma of water i 8 g mol (H =, O = 6) yielding n = q = mol. There are 3

4 hydrogen atom per water molecule, and o we need the number of molecule from Avogadro N = nn A = ( )( ) = giving ( ) = H atom. *** i where thi quetion wa vetted. 3. D For the average peed to be 0 m, a total ditance of 00 m mut be traveled. In the firt econd, a total of x = v t = (6.0 m ) () = 7 m ha been traveled. Thi mean that in the lat 8 econd, a ditance of 8 m need to be travered. Let draw a graph of velocity v time. The haded region have to add to a total ditance of 8 m. In the firt T econd, the object low to ret. Mathematically, a = 6 with area (6)(T) = 3T. For the remaining time, the area T i (8 T)(a(8 T)) = a(8 T). Hence, 3T + a(8 T) = 8. Subtituting the expreion for the acceleration and multiplying by T yield 3T + 3(8 T) = 8 T 3T 88T + 96 = 0. Solving thi expreion yield time of T =.3, 8.. Uing T =.3 give a = 6 = 5.3 m C The wavelength of the ound i v = fλ λ = v = 34 = 6.0 m. Thi mean that the peaker f 57 are eparated by exactly 5 wavelength. For contructive interference, the path difference between the wave from the peaker mut differ than an integer number of wavelength. Hence, there are location at or between the peaker where thi happen 5λ, 4λ, 3λ, 3λ, 4λ, 5λ. A place of detructive interference occur directly between the contructive interference location hence, there are 0 of them. 34. A The image from the firt len i found from = + = + q = 30 cm. f p q 0 5 q Far from t len (more than 40 cm from the firt len) - we have that the image from the firt len i the object for the econd len and that the object i greater than a focal length from the nd len. Thi mean that we will end up with a real image from the nd len and the image will be flipped from the object. Since the t len formed a real image, that image wa flipped from the original object. In other word, for large ditance between the lene, the image i real and pointing upward. Medium range from t len (between 30 and 40 cm from the t len) - the image from the firt len i now inide the focal length of the nd len. Hence, between 30 and 40 cm, the image formed i virtual and pointing downward (there i till the inverion from the t len). Cloe range (le than 30 cm from the firt len) - we have a virtual object! For implicity, let the nd len be placed 0 cm from the firt len making p = 0 cm for the econd len. Thi give = + = + q = 5 cm. Becaue we have a poitive image poition, the image i f p q 0 0 q real. Further, the magnification of the image from the nd len i M = q = ( 5 ) =. Becaue p 0 it i poitive, the image ha the ame orientation a the object. The object for thi len wa the upide-down image from the firt len. Hence, the image i real and pointing downward. 35. E From the angular impule-momentum theorem, the change in angular momentum i the torque multiplied by the time. A the force are equal but at different ditance from the center of each object, τ X < τ Y meaning that L X < L Y ince the force are applied for the ame time. A for the kinetic energy, we note that KE = Iω = L = τ t rft = I ( Mr ) = F t Mr M. All of thee quantitie (force, time, ma) were equal for the two dik thereby making the kinetic energy the ame for the dik! 36. B METHOD #: algebra - One way to handle thi problem i to make little picture and ue ubcript. We write v rain wrt Earth = (0)x + ( )y, v car wrt Earth = (v car co 40)x + ( v car in 40)y = (0.776 v car )x + ( v car )y. Now, to find the velocity of the rain with repect to the car, we compute v rain wrt car + v car wrt Earth = v rain wrt Earth. Hence, we rearrange thi expreion to find v rain wrt car = v rain wrt Earth v car wrt Earth = ( v car )x + ( v car )y. Since thee component of the velocity are known to make a 9 degree angle 4

5 with the vertical, we can write tan 9 0 = ( v car ). Uing a little algebra, we can v car olve thi equation for the unknown v car. Doing thi lead to v car = 5.93 m. METHOD #: pictorial Uing v rain wrt car + v car wrt Earth = v rain wrt Earth, we contruct the picture hown to the right for the velocitie. Hence, from the Law of in 0 = v car in 9 in 9 in 0 Sine, v car = = 5.93 m 37. A The energy tored by an inductor take the form U = LI and be rearranging, the unit of inductance are found from U I J A = Nm A = (kg ) m kg m A = A 38. D To create contructive interference with the 540 nm light, we need to be ure that the interfering ray from the reflection off the alcohol and the ray that pae into the alcohol and i reflected back at the gla urface are in phae. Since the light i traveling from air to alcohol to gla, the index of refraction increae at each interface, meaning that the reflected light i phae-hifted by λ. Thi mean that for light traveling down through the alcohol a ditance t to the gla urface and then traveling an additional ditance t back to the air, we need thi extra path length to be an integer number of wavelength to put the our two wave in phae. That i, t = m λ mλ. Solving for t = = n n m ( 540 ) = 00 m. Since m i an integer, the only poible choice would be t = 400 nm. ()(.35) Likewie, detructive interference i needed for the 43 nm light and that condition it = (p + ) λ n. Solving thi expreion with p = alo yield that t = 400 nm. 39. E By ue a Kirchhoff loop with the battery and bulb 3, we ee that there i no difference in the voltage or current through it whether the witch i open or cloed. By looking at what i left, from Kirchhoff Loop Rule, the potential difference acro bulb i ξ with the witch open. After cloing the witch, the effective reitance of the bulb -4 branch decreae reulting in more current through bulb (and P). Thi increae the potential difference acro bulb thereby decreaing the potential difference acro bulb and the branch from W to X! 40. C From the free body diagram of the ma, there are two force: gravitational and tenion. Writing Newton Second Law for each component of motion, we have F nety = ma y T co θ mg = 0 T co θ = mg and F netx = ma x T in θ = m ( v ) = m ( v ) T r L in θ in θ = mv. To find the L tenion directly, we ll take the y-component expreion and quare it. After thi, we multiply the x- m component by T leading to T in θ = mv T and L T co θ = (mg). Adding thee relation (note that in θ + co θ = ) give T ( mv ) T L (mg) = 0 T.8 T 400 = 0. Uing the Pythagorean Theorem yield T = 7.4 N; 4.6 N. The phyical anwer therefore i T = 7.4 N. 4. A The average peed i ditance divided by time. The truck move at contant rate from the tart to end poition while the car initially move backward and then forward to the ame ending location of the truck. Conequently, the car travel a greater ditance than the truck and ha a higher average peed. 4. C The lope of the poition v, time graph. The car move fater than the truck initially and eventually come to ret, which mean that at ome point, the peed of the car and truck are the ame. From ret, the car accelerate and catche up to the truck by eventually moving fater. Thi mean that the car and truck have the ame peed again before time T. Finally, for time T < t < T, the car low down cloe to ret and the truck catche up. Hence, the car peed i again equal to that of the truck at ome time. Thi make C the correct anwer. 5

6 + p y m m + p y (4) 43. B Since K = p, we can write that K = p x (4) 00 = 800 = p m (4) y p y = 4. Hence, p y = 4 = 4.97 kg m = 5.0 kg m 44. E Jacque Charle i the cientit aociated with the obervation that volume i proportional to temperature at contant preure. 45. B The work done by the peron inerting the dielectric would be equal to the change in the potential energy for the capacitor. Since the battery wa diconnected, the charge on the plate remain contant and W peron = U = Q V. The charge on the plate would be Q = CV = (6.00 μf)(.0 v) = 7.0 μc. The final capacitance can be found by moving the dielectric to any location in the pace between plate and treating the ytem a capacitor in erie. The potential difference through the air-filled portion would be 6 volt ince the field trength i unchanged and we are croing half the original capacitor (which had volt). For the dielectric portion, the dielectric decreae the field trength by a factor of which mean that the potential difference acro the dielectric would be 3 volt for a grand total of 9 volt acro the entire capacitor. Hence, W peron = Q V = (7μC)( 3v) = 08 μj. 46. E We write τ net = Iα and note that the moment of inertia of the rod i I = ML about an axi through the center of ma. To find the moment of inertia about the pivot, we need the parallel axi theorem to obtain I = ML + md where d = L (the ditance between the center of the ma and the 6 pivot) leading to I = ML + m ( L 6 ) = 9 ML. Calculating the torque from the gravitational force at the center of the tick give τ = Fd in θ = Mg ( L ) in 6 90 = MgL. (The minu ign for clockwie). So, τ net = Iα MgL = 6 9 ML α α = 3 g. Uing a L t = rα lead to a t = ( L ) ( 3 g ) a 6 L t = g C From the Firt Law of Thermodynamic, U = Q + W, and ince it i an adiabatic proce Q = 0. We can therefore find the work done by looking at the internal energy change with U = n ( 5 R) T for a diatomic ga. Uing PV = nrt T = PV = (.0 atm)(30 L) = K. For an adiabat, nr K) PV γ = contant and P = nrt V 6 ( mol)(0.08 L atm mol which upon ubtitution give TVγ = Cont. Hence, (730.8)(30) 5 = T f (5) 5 T f = K. Finally, W = U = n ( 5 R) ( ) = 4850 J 48. B Outide the hell of radiu, the total field will be that of a point charge and o will the potential giving V = kq. Between a and, there i no field interior to the conductor (ince the hell are in free pace and there i no mention of any other charge nearby the ytem quickly attain tatic equilibrium) and o the potential at a i given a V a = kq. Noting that there i a field between a and a, we have a potential difference computed a V = V f V i where V f = V a and V i = V a and on the left ide we write V = V(r) V(a) = kq in kq in. And o, we have kq in kq in = 0 kq. Thi a a a a lead to kq in = kq Q in = Q 3 a 49. A From relativity, we know KE = (γ )m 0 c and o, m 0 c = (γ )m 0 c γ = 3. From thi, = 3 = v v 9 c v = 8 9 c v = 8 c = m c 6

7 50. D By travering a loop around the outide, we encloe an entire emf ξ which would be equally ditributed acro each identical reitor ince the outer loop i the only one with current. Hence, the voltage i ξ for any reitor. By travering the top triangular 3 loop, we encloe ξ and have reitor dropping potential ξ, we have ξ = 3 (ξ) + 3 V voltmeter V voltmeter = ξ 6. A a check, the lower triangular loop give ξ = ( ξ 3 ) V voltmeter V voltmeter = ξ 6. The magnitude of the voltage i therefore ξ 6. 7