= 16.7 m. Using constant acceleration kinematics then yields a = v v E The expression for the resistance of a resistor is given as R = ρl 4 )

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1 016 PhyicBowl Solution # An # An # An # An # An 1 C 11 C 1 B 31 E 41 D A 1 B E 3 D 4 B 3 D 13 A 3 C 33 B 43 C 4 D 14 E 4 B 34 C 44 E 5 B 15 B 5 A 35 A 45 D 6 D 16 C 6 C 36 B 46 A 7 E 17 A 7 D 37 E 47 C 8 A 18 A 8 D 38 D 48 B 9 B 19 D 9 E 39 A 49 A 10 C 0 E 30 A 40 D 50 E 1. C 1 econd < 1 minute < 1 day < 1 week < 1 year. A he velocity of the object i found from the lope of the line tangent to the point on the poition v. time graph. o have the greatet peed, we are looking for the larget magnitude of lope to the line tangent. he teepet curve on the poition v. time graph hown occur at time t = D Firt, we convert unit to obtain 60 km 1000 m 1 hr 1 min = 16.7 m. Uing contant hr 1 km 60 min 60 acceleration kinematic then yield a = v v 0 = = 3.7 m t D he implet way to look at the addition of quantitie i to line up the decimal place. hat i, we are computing = he 4 i bolded a it i the only digit known in the hundredth place ince three of the quantitie top at the tenth. Hence, the final value i only kept to the tenth place making = m the correct anwer. 5. B By performing the free body diagram on the ma m, we have the tenion in the tring and the gravitational force. Since everything i at ret and not accelerating, we find from Newton Second Law that F net = ma + ( mg) = 0 = 60 N. Note that for the ma on the ground, there i an additional normal force acting! 6. D Uing contant acceleration kinematic, we have y = y 0 + v 0 t + 1 at 0 = H + ( 0)() + 1 ( 10)() H = = 60 m. 7. E he expreion for the reitance of a reitor i given a R = ρl. So, computing in turn, we find R 1 = ρl ; R πr = ρ( L ) = ρl ; R π( r ) πr 3 = ρ( L 4 ) = ρl π( r ) πr. Hence, R 1 = R 3 < R. 8. A Copernicu publihed the heliocentric theory in 151 while Galileo did a lot of work related to thi theory about a century later. Iaac Newton tarted producing work in the 1660 while Maxwell contribution to phyic had to wait until the B Since the conventional current in the circuit i directed to the right through the reitor, the electron are flowing in the oppoite direction (to the left). he electric field in the interior of the reitor would be in the direction of conventional current and hence point to the right. 10. C By Newton Second Law, in order to accelerate an object, there mut be an imbalance in the force. o accelerate an object upward, a force mut exit on the object that exceed the gravitational force acting on it. Hence, the required force i F > W. 11. C From the information, we tart by finding the acceleration a a = F net = 1.0 = 4.0 m m 3.0. Uing contant acceleration kinematic, we then have v = v 0 + at v = 0 + (4)(5) = 0 m. 1. B Kinetic energy i computed a KE = 1 mv = 1 (3.0)(0) = 600 J. A 1

2 13. A he ideal mechanical advantage for an inclined plane i found a the length of the hypotenue divided by the height. Hence, in thi ituation, we find IMA = h y = = Nm kg 14. E From the equation heet, we find the unit for thee quantitie. Doing thi give Nm C kg. In other word, thi ha unit of charge divided by ma. C = C kg = 15. B Since all EM wave travel at the peed of light, we have v = fλ λ = v = f = 3.0 m 16. C From the firt quarter rotation, we determine the angular acceleration a θ = ω 0 t + 1 αt π = α(1) α = π rad. And o, for one full revolution, we have ω = ω 0 + α θ ω = 0 + (π)(π) = (π). Hence, ω = (π) rad. 17. A hi i a baic tatement of Kepler Firt Law. 18. A By definition, average velocity i diplacement divided by time. Hence, thee quantitie mut be directed the ame way. 19. D Uing the ideal ga equation in the form PV = Nk B, we compute (1.0 atm)(v) = ( )( )(7 + 73) = 4968 Nm. So, we need to convert the preure a 1.0 atm = Pa. Hence, V = 4968 = m3. 0. E From the original location, we can balance the torque about the center of the plank to find τ net = 0 (M 1 )g(5) (M )g(4) = 0 M = 5 M 4 1 where the ditance of the ma M to the rotation axi i 4.0 m. So, by moving the mae, we have the following torque condition: τ net = (M 1 )g(3) (M )g(). Replacing the ma M = 5 4 M 1 into the new expreion give τ net = (M 1 )g(3) ( 5 M 4 1) g() = 1 Mg > 0. A we aumed counterclockwie torque about the center are poitive, the net torque being poitive mean that the right ide will rie and the left ide will fall. Since there i a torque imbalance, thi motion occur with a non-zero angular acceleration! 1. B he minimum peed after launch for a projectile i when it reache it apex (highet point) when the y-component of the projectile velocity i 0 m. From contant acceleration kinematic, we find v y = v 0y + a y t 0 = v 0y + ( 10)(1.94) v 0y = 19.4 m. At the highet point, the total velocity of the object i the x-component of the velocity (which i contant). Making a right triangle uing the initial velocity component give the angle above the horizontal at launch to be tan θ = v y θ = v x tan 1 ( 19.4 ) = E In order to balance the reaction, we write 19K e A 0 Z X + 0 ν e. hi mean that we have = A + 0 A = 40. Further, 19 + ( 1) = Z + 0 Z = 18. By having Z = 18, thi mean that the element i different from Potaium a that ha 19 proton. In other word, from the choice provided, we ee that A 40 Z X = 18Ar. 3. C Buoyant force i computed a the weight of the fluid diplaced. he balloon have the ame ize and therefore diplace the ame amount of air. In other word, the buoyant force are the ame! he difference between the balloon would be een on releae when the helium balloon rie and the xenon balloon fall. 4. B For the proton to move with contant velocity, the net force on it mut be zero. Since the proton move at a right angle to the magnetic field, there i a magnetic force on the proton. By the right-hand rule, with the velocity up the plane of the page and the field into the place, the magnetic force i directed to the left. In order to cancel thi force, the electric force acting on the proton mut be directed to the right. By F = qe, the field and force are in the ame direction for a poitive charge. he field i directed to the right.

3 5. A he force on the charged particle i computed a F = qe. he electric field between the plate of a capacitor can be found from E = V 4.0 V = = 480 V = 480 N. So, we find the charge between the d 0.05 m m C plate a q = F = 1.00 = C =.08 mc. E C By adding the momenta of the mae together, we have the total momentum of the center of ma of the ytem. hat i, m 1 v 1 + m v = (m 1 + m )v cm (10)(8) + (5)( 7) = (10 + 5)v cm 45 = 15 v cm v cm = 3.00 m. 7. D For the initial ituation, the third lowet frequency tanding wave produced by the tube would be the fifth harmonic. ube cloed at one end only have odd harmonic. hi mean that we have f = 5v. Since there are only odd harmonic, the next harmonic for the tube would be the eventh which 4L ha frequency f new = 7v = 7 4L 5 (5v) = 7 f. 4L 5 8. D It wa recently announced that gravitational wave have been detected! 9. E he len hown would be converging becaue of it hape. With the incident ray coming from the right ide, the light will refract through the focu on the oppoite ide of the len. hi refracted ray i hown in the figure. 30. A In order to melt a material, the property deired i the latent heat of fuion. 31. E By removing bulb #, the current for thi branch become zero Amp meaning that the effective reitance of the circuit ha rien. By increaing the effective reitance of the circuit mean that the total current in the circuit will decreae. Since bulb #1 alway receive the total current in the circuit, it now ha a maller potential difference from Ohm Law V = RI a the reitance of bulb #1 i unchanged. By Kirchhoff Loop Rule, if the battery tay the ame (which it did) and there i le potential difference acro bulb #1, then there i more potential difference in the ret of the circuit (from Q to X)! 3. D Since the ma i in imple harmonic motion, the angle from the vertical upon firt releae would be fairly mall (< 15 ). When the ma wing through it lowet location, the acceleration of the ma would be traight up to the pivot of the pendulum. hi mean that > W to have more upward force compared to downward force. o check on the ize of the net force, let u aume that it i equal to the gravitational force and determine the angle at which the ma i releaed. In other word, F = W = mv = mg mv = mgl where L i the length of the pendulum. From mechanical energy L conervation during the fall of the ma to the lowet point from the highet, we write KE + PE = 0 ( 1 mv 0) + mg( L(1 co θ)) = 0 where θ i the maximum angle that the tring make with the vertical. Uing our ubtitution, we have 1 mgl mg(l(1 co θ)) = 0 co θ = 1 θ = 60. In other word, ince we are nowhere near that initial angle, F < W and F < W <. 33. B he electric potential i computed a V = kq for a point charge. A a reult, at the point P, we r have V p = kq + kq + k( Q) = a a a (k) (Q Q) = ( ) a (kq ) > 0 A for the field, we need to a vector add. By the ymmetry, the field from the poitive charge are directed at a 45 upward from the left. In other word, it i in the direction oppoite to that from the negative charge. he total field from the poitive charge come to ( kq a ) + ( kq a ) = kq k(q) a. he field from the negative charge i kq ( a) = a. Since the field from the negative charge ha a maller magnitude that the field from the poitive charge, the reultant field point up and to the left. 34. C Object 1 mut be catching up to object from behind becaue 1) the intantaneou momentum of neither object ever reached 0 kg m and ) object 1 loe momentum while object gain momentum. 3

4 Becaue object i in front of object 1 with m 1 v 1 = m v after colliion then v v 1. hi mean that m m 1. In the limiting cae of m = m 1, then anwer (A) and (D) would be true and (E) would be fale. Anwer B i incorrect from Newton hird Law. 35. A Iridecence ha to do with the angle at which one i looking at an object and that the color of the object appear to change with it uch a what happen with a peacock! 36. B MEHOD #1: One approach to the problem i to graph the velocity of each racer a a function of time. hi i hown in the figure. he robot lead by it maximum amount when the peed of the object are the ame. hi i indicated at time t wherea time t 1 indicate when the robot reache it maximum peed. From the figure, we can find the change in poition of each object with the area under the curve. hi mean that the ditance that the robot lead the car i equal to the beige-colored triangle in the graph. All that i needed i to find the time of interet. hi i done a follow: v robot = v 0 + a robot t 1 t 1 = v robot = 1.76 and a robot 5.60 v robot = v 0 + a car t t = v robot =.68. a car 5.60 he area of the triangle i found a A = 1 bh = 1 (t t 1 )v robot = 1 ( )(15) = 6.9 m. MEHOD #: Alternatively, we can approach the problem from an algebraic point of view. In order to determine when the robot lead by the greatet ditance, we need to find where the object are when they have the ame peed! In other word, it take the car v robot = v 0 + a car t t = v robot a car 5.60 =.68. By looking at contant acceleration kinematic for the car, we have it poition a x c = x 0c + v 0 t + 1 a cart x c = (5.60)(.68) = 0.1 m. he motion for the robot occur in two tage a it accelerate at 8.50 m for ome time and then travel at a contant rate thereafter. he robot reache it maximum peed after v robot = v 0 + a robot t robot t robot = v robot = So, a robot 5.60 we compute the poition of the robot after.68 a x 0 to x 1.76 to.68. For the firt part, we write x 0 to 1.76 = v 0 t + 1 a robott = (8.50)(1.76) = 13. m and x 1.76 to.68 = v 0 t + 1 a robott = (15)( ) + 0 = 13.8 m. Hence, the robot i at a poition of = 7.0 m a full 6.9 m ahead of the car. 37. E From the figure, we ee that the tring length from the ocillator to the pulley went from being one full wavelength on the left to being two full wavelength on the right. hi mean that the wavelength i now ½ a large with the ma ubmerged. Since the ocillator i not changing the frequency of vibration, thi mean that the wave peed for the tring i now ½ a large from v = fλ. he wavepeed on a tring i determined a v = /μ which mean to cut the peed in half reulted from a tenion now ¼ a large. 38. D From the free body diagram of the ma hanging on the tring in the figure on the left, we find F net = Ma Mg = 0 = 80 N. Since we determined in the previou quetion that the tenion i now 0 Nwith the ma ubmerged. hi mean that there i a buoyant force of 60 N acting on the ma from the water. So, B = ρ w gv di 60 = (1000)(10)V di V di = m 3 where V di i the volume of diplaced water (which i the ame a the volume of the ma). Since the ma i a cube, L 3 = m 3 L = 0.18 m. 4

5 39. A MEHOD #1: By definition, a = v. Call the final peed of the object to be V and o we can write v i = 17 x + 0 y indicating it i moving to the left at the tart and then v f = V co 55 x V in 55 y. Alo, we have a = 0 x 9.8 y. And o, by looking at the x-component, we ee a x = v fx v ix V = 17 co 55 = 9.6 m t 0 = ( V co 55 ) ( 17) MEHOD #: One can contruct a vector picture repreenting the velocitie a hown V = v i + v. Noting that the reultant change in velocity i traight downward (direction of acceleration), we ee from the contruction that V = 17 = 9.6 m. co D From the previou anwer, we now conider the y-component of the acceleration to find the time. hat i, we have a y = v fy v iy = = D MEHOD #1: Let find thing uing kinematic. In the vertical direction, we write v y = v 0y 9.8 = ( V in 55 ) (0) 17 co 55 in tan 55 = a y y v y = (15 in 30 ) + ( 10)( 10) v y = 16.0 m. he x-component of the velocity i a contant v x = 15 co 30 = 13.0 m. Conequently, upon landing, the angle made by the object with the horizontal i computed a tan θ = v y θ = tan 1 ( 16 ) = v x 13 MEHOD #: Uing mechanical energy conervation, we can find the peed of the object when it reache the ground a KE + PE = 0 ( 1 mv f 1 mv i ) + mg y = 0 v f = v 0 g y Hence, v f = (15) (10)( 10) v f = 0.6 m. Since the x-component of the velocity i v x = 15 co 30 = 13.0 m, we can find the angle below the horizontal a co θ = v x v θ = co 1 ( B Uing the len equation, we have 1 = f p q 1 = q q = 16 cm. he magnification i found a M = q p = 16 1 = 6. A the magnification i negative, thi make the image real and inverted. Since M > 1, the image i alo larger than the object. 43. C A the object i in tatic equilibrium, the um of force and the um of torque mut be zero. he force from the cable i upward and rightward. Becaue the gravitational force i downward, the pivot mut be providing a force to the left in order to cancel the rightward force from the cable. A for the vertical component, by chooing an axi of rotation through the direction of the gravitational force at the vertical location of the pivot, the only non-zero torque are from the cable and the vertical component of the pivot force. A the torque from the cable would be oriented counterclockwie from thi axi of rotation, the vertical force at the pivot need to provide a clockwie torque. hi i poible only if that vertical component i upward. 44. E A the object move in circular path, it i becaue of the magnetic force which ha a magnitude of qvb. aking a ratio of the force, we find F He = q Hev He B 1 = () F p q p v p B (v He ) (1) ( v He ) = 1. v p v p 45. D he kinetic energy of the electron emitted range from 0 J up to 1 mv = 1 ( )( ) = J = 0.67 ev. he energy of the incoming photon i determined to be E = hc = ( )( ) = 4.97 ev. he work function i determined by taking λ the difference between the maximum kinetic energy and the energy of the incoming photon, meaning φ = = 4.30 ev. + 5

6 46. A he ecape peed from a planet i found uing mechanical energy conervation. hat i, ΔKE + ΔPE = 0 (0 1 mv ec ) + (0 ( GmM )) = 0 v ec = GM. he ma of a planet would be R R computed a M = ρv = ρ ( 4 3 πr3 ). With thi ubtitution, we have v ec = Gρ(4 3 πr3 ) = R 8 ρgπ. 3 hi mean that v X v y = R X R Y = C Lenz Law i required to anwer thi quetion. he magnetic field aociated with the long wire i into the page to the right of the wire and out of the page to the left of the wire. A the loop on the right i moved away, there are weaker field line penetrating into the loop. A a reult, there i an induced current oriented to try to replace the miing field line. hi mean that the current in the right loop i oriented clockwie to put field line into the plane in the interior of the loop. Likewie, on the left ide of the long wire, there are again weaker field line coming out of the plane of the loop. Here, the induction would be for a current oriented counterclockwie to have additional field line out of the plane through the interior of the loop. 48. B he heat aociated with the adiabatic proce i zero Joule (a that i what it mean to be reveribly adiabatic). For iobaric procee, the heat i Q p = nc p and for iovolumic procee, the heat i Q v = nc v. Alo, we need the ideal ga equation to determine what happen to the temperature. For (B), the temperature double reulting in Q B = n ( 5 R) ( ) = 5 nr, for (C), we have Q C = n ( 3 R) ( ) = 3 nr, and for (E) we compute Q E = n ( 5 R) ( ) = 5 nr. 4 Latly, becaue there i no internal energy change for an iothermal proce, Q = W for thi proce. For the iobar in (B) there i a doubling in the internal energy with more negative work done than for the iotherm meaning that Q B > Q E from U = Q + W. 49. A Becaue the pacehip pilot i in one location, he ha a ingle reting clock. hi mean that the time he record to cro the platform will be a proper time. From the pilot perpective, the platform i length contracted to be of length L = L p γ = 1 ( v c ) L p = 1 ( ) (500) = 400 m. So, her clock will read a time of t = L v = =.μ 50. E We will ue mechanical energy conervation to approach thi problem, ΔKE + ΔPE = 0 ( 1 Iω 0) + (Mg y) = 0. o find the moment of inertia about the pivot, we need the parallel axi theorem to obtain I = 1 1 ML + md where d = L (the ditance between the center of the ma and the 6 pivot) leading to I = 1 1 ML + m ( L 6 ) = 1 9 ML. Alo, the center of ma will have y = L when the 6 tick i horizontal. Upon ubtitution, we obtain ( 1 Iω 0) + (Mg y) = ML ω Mg L = Lω g = 0 ω = 3g. Finally, the peed of the center of ma will be computed a L v = ωr with r = L 6 leading to v = 3g L L 6 = 3gL 36 = 1 1 gl R 6

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

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