1994 Physics C Solutions

1994 Physics C Solutions - 1994 Physics C Solutions...

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Unformatted text preview: 1994 Physics C Solutions Distribution of Points Mech l. (a) 3 points For the correct expression for the energy stored in a spring: 1 U = 5—ka 1 point For substituting the correct values: 1 U . 3(200 N /m)(O.4 my - 1 point For the correct answer (including units): U =- 16 J 1 point (b) 3 points For applying conservation of energy 1 point Kinetic energy of ball/block combination = energy stored in spring For a correct expression for kinetic energy 1 point 1 2 3‘le = U v1 - .IZU/M = .J' 2(16J)/(2.l kg) For the correct answer. u1 =- 3.9 m/s lpoint (c) 3 points For applying conservation of momentum 1 point For a correct expression for momentum p 1 point mv = le u - le /m = (2.1 kg)(3.9 m/s)/(0.l kg) For the correct answer: v = 81.9m/s lpoint (d) 2 points For indicating that the maximum compression of the spring will be less than 0.4 meters. 1 point For a correct explanation referring to the motion of the blocks or their kinetic energy 1 point Some examples: The 0.8-kg block will begin to move before the spring can compress as much, due to the force that the spring exerts on it . The kinetic energy of the center of mass of the system in this case is non-zero, so some of the initial energy must remain as kinetic energy. Less energy is available as potential energy, so the spring compression is less. Distribution of Points Mech 1. (cont) (e) 4 points Two principles must be applied to calculate the velocity of the 0.8 kg block at the instant the spring is no longer compressed: conservation of momentum and conservation of energy. For indicating conservation of momentum is applicable 1 point le - Mv2 + sza For substituting: (2.1 kg) (3.9 m/s) [or 8.2 kg - m/sz] - (2.1 kg)v2 + (8 kg)v3 1 point For indicating conservation of energy is applicable 1 point 1 z 1 2 1 2 3M1)l - 5M1;z +3M2 Us For substituting: 1 1 1 -2-(2.1 kg) (3.9 m /s)2 [or 16 J] = 3(21 kg)v22 + E(8 kg)v32 1 point The equation 3.9 m/s = 123 — v2 or — (v2 - v3),whichcanbeobtained from the two equations above, was accepted as a substitute equation, but only if one of those equations was also present. ' ——_—-————___—______—_—__—__ Mech 2. (a) 4 points Ktolzl=K nal+K malaria rotational For a correct expression for translational kinetic energy For a correct expression for rotational kinetic energy 1 2 1 ‘2 Km] - Emu + 31w . . v For substituting w = —r- i 2 i E 2 v2 7 2 + 2(5mr){,2)— = 775(25kg)<10m/s)2 K‘om a In the absence of other credit, one point was awarded for indicating that the total kinetic energy is the sum of translational and rotational kinetic energies. (*0 . 1. 4pomts For an indication that conservation of energy is applicable K m, (leaving incline) = K tom (bottom of incline) — U For correct use of U = mgh Tami): =- K ,°Q(bottom of incline) — mgh = 1,7501 — (25 kg)(10m/sz) (3 m) a 1,000J v -1/170(1,000 J)/(25 kg) For correct substitutions v - 7.56 m /s Full credit could be earned for this part if rotation was omitted both here and in (a). If rotation was included in (a) but omitted here, or if translation was included in (a) but omitted here, a maximum of 2 points could be earned. 1 1 . The statementfmu2 a mgh or mgh = é—mvz + 310:)2 was awarded 1 pomt. The statement vf2 = v02 - 2g}; was accepted as equivalent to the energy equation without rotation. Distribution of Points 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point Mech 2. (cont) (b) (cont) ii. 1 point As it leaves the top of the incline the sphere’s velocity will be directed parallel to the incline, i.e. at 25° above the horizontal. For any clear indication that student realized this fact (c) 4 points For a correct kinematic equation for the vertical motion, indicating use of the component of the velocity . r 1 2 2 y, = y0 +voyt + 3th (or vy -= 120, — 2g(y — yo» For recognizing that y f - y0 - —3 m, or otherwise correctly accounting for the difference in height Substituting the appropriate values: )2, a 0,):0 = 3m, on = 7.56m/s(sin25°),g - —10m/s, and solving the quadratic equation: 12 = 1.16 s, - 0.52 s, where the positive solution is the applicable one. (Separate calculations for appropriate intervals could also be done, and added to obtain!2 - 1.16 s.) ' For a correct equation for the horizontal distance, indicating use of the component of the velocity x = vmt - vocos 25°t -= (7.56 m/s) cos 25° (1.157 s) x - 7.93 m (d) 2 points For indicating that the speed of the sphere as it leaves the top of the incline would be less than in (b). For a correct explanation For example: All of the potential energy gained is at the loss of the translational kinetic energy. Distribution of Points 1 point 1 point 1 point 1 point 1 point 1 point 1 point Distribution of Points Mech 3. (a) 3 points Using the general relationship between potential energy and force: U - I F dR f GM,m U - I R, dR For correct limits on the integral 1 point For an expression in which the force is proportional to 1 /R2 1 point For having the remainder of the force expression correct 1 point G M m 1 point was awarded for the expression F = R; in the absence of any other correct work. (b) 2 points The total energy of the satellite at pointA is the sum of the kinetic and potential energies. 1 2 GM em E A = —2—mvo — a For a correct expression for the kinetic energy (with v0) 1 point For a correct expression for the potential energy (with a) 1 point (c) 2 points For a correct expression for angular momentum: L-Iw ORmvr ORpxr ‘ lpoint For correctly substituting v - v0 and r a a 1 point L - mvoa (d) 3 points For any indication of use of conservation of angular momentum: 1 point For correctly substituting velocities and radii 1 point muBb a mvoa For the correct solution: voa ” " b 1 point B Mech 3. (cont) (d) (cont) (Alternate solution) For any indication of use of conservation of energy For correctly substituting velocities and radii GM‘m GM‘m mvo f-mvB - For the correct answer (i.e. solving the equation for v or v2) v3 - [002 + 26M,[ ll” In a circular orbit, the centripetal force is provided by gravity. For any expression of Newton’s second Law, F - ma (e) 3 points For the correct expression for the centripetal acceleration, a - — mvz GMem For the correct answer, with r a a GMe ’U- (0 2 points The work done is equal to the change in the total energy of the satellite. Since the potential energy remains constant, the change in total energy is equal to the change in kinetic energy. For any indication that the work equals the change in kinetic energy W - AK - —m For the correct answer with all substitutions made: W—-—m (.2 - 1.02) . GM, 2 _v0 Distribution of Points (Alternate points) (1 point) (1 point) (1 point) 1 point 1 point 1 point 1 point 1 point Distribution of Points E & M 1. (a) 4 points For a correct expression for the potential: V: 1 J‘dq 1 Q ._._.. or -— 1 point r r 4m: 0 41:80 Since all of the charge islat the same distance, simply substitute the total charge and the appropriate distance. For correct expressions for calculating the total charge: Q = M 1 point 2 == 23R lpoint 1 23R A 41teo R For a correct answer: Va A V = —— OR Zak?» - 280 1 pomt (b) 2 points The electric field from each infinitesimal piece of the ring is canceled by the field from a piece on the opposite side of the ring. ' Therefore, E a 0. 2 points (c) 2 points The total charge on the rod is simply the charge per unit length times the length of the rod. Q - h! For using the correct expression for the length of the rod: Z :- 2RB 1 point For the correct answer: Q " 29m 1 point (d) 2 points Once again, since all of the charge is at the same distance, simply substitute the total charge and the appropriate distance in the expression for the potential. For using the answer to (c) for the total charge 1 point 1 20ml 43:20 R For the correct answer. 6}» . V = OR 21(9)» 1 pomt 2m: 0 E & M 1. (cont) (e) 5 points For a correct expression {or the electric field: 1 d l E =- f ~—qz— or £2 43:30 43tso R Since all the charge is at the same distance, calculating the electric field requires integration over the angle only. Also, the vertical components of the field cancel, so the integration can be limited to the horizontal components. dq - M? d4) For incltiding the cos 4) term in the horizontal component 9 E 1 AR 4) dd) = -—2— cos 43tso _9 R 1 9» . a - — sm (1) 41120 R I 1 A . . = 4“ f (Sine — sm(—0» For a correct expression for the electric field: A sine 2k x sin 6 OR —— 2.11:8 oR R For correctly indicating the direction of the field Distribution of Points 1 point 1 point 1 point 1 point 1 point Distribution of Points E & M 2. (a) v i. lpomt The shuttle end is negative. 1 point Due to the motion of the wire, the free electrons in it have a velocity to the right. Using the right-hand rule, the force on the electrons due to the magnetic field is toward the shuttle. ii. 2 points Electrons in the wire stop moving when the force due to the electric field created by the differencein charge between the ends of the wire counteracts the magnetic force. qE = qu E = v8 The emf V( potential difference between the ends of the wire) is the electric field multiplied by the length 3 of the wire. V = 128! 1 point - (7,600 m/s) (3.3 x 10“ T) (20 x 103 m) V - 5,016 V 1 point (b) 2 points For using Ohm’s Law: I = V/R lpoint = (5,016 V)/(10,000 $2) I =- 0.5016 A 1 point (C) . 1. 2p01nts For a correct expression for the magnetic force on a current: F - I d! x B 1 point The current is perpendicular to the field, so the magnitude of the force is: F - MB a (0.5016 A) (20 x 103 m) (3.3 x 10'5 T) F - 0.331 N 1 point (If a student used 20 m for the length of the tether instead of 20 x 103 m in both (a) and (c), only one answer point was lost.) ii. 1 point By the right—hand rule, the direction of the magnetic force is opposite the shuttle’s velocity. 1 point Distribution of Points E & M 2. (cont) (d) 4 points The change in the shuttle’s energy is equal to the energy dissipated due to the resistance of the wire. AU - Pt ' 1 point P - 12R _ ' 1 point AU=Pm For correct substitution: AU - (0.5016 A)2 (10,000 Q) (7 d) (24 h /d) (60 min/h) (60 5 /min) 1 point AU - 1.52 x 10° J lpoint (Alternate solution) (Alternate points) The change in energy equals the work done on the shuttle. . AE - F -d (1 point) F -d = th (1 point) For correct substitution: AE = (0.33 N)(7,600 m/s)(7 d)(24 hr/d)(60 min/hr)(60 s/ min) (1 point) AE — 1.5 x 109 J I (1 point) (e) 2 points If current was forced to flow the other way, the direction of the magnetic force on the current would be reversed. This force would do work on the shuttle, and the resulting gain of energy would cause an increase in the radius of the orbit (it would “boost”the orbit). ‘ For correctly indicating that the radius would increase 1 point For a correct explanation of the cause ‘ 1 point For correct units in three answers and no incorrect units 1 point —————————-————————____________—_—______—_ E&M3. (a) . 1. 4pomts For a correct expression for Ampere’s Law: §B - d! = uoIm For indicating the correct path of integration, either by drawing a circle on the figure or using 2m in Ampere’s Law The enclosed current is the current per unit area for the inner conductor multiplied by the area inside the distance r. 2 1m - 1172 = 1’7 m2 0 Ir2 3(2117) - Ito—2" a u Ir B .- ° 2 2m ii. 2 points For a < r< b, the enclosed current is the total current in the inner conductor. I m a I B (237R) = no] pt I B .. _°_ 23R (b) lpoint For r >c, the net current enclosed is zero. Therefore, the field is also zero. Distribution of Points 1 point 1 point 1 point 1 point 1 point 1 point 1 point Distn'bution of Points E & M 3. (cont) (c) 4 points 0 a b i c For a linear increase for O < r < a (or answer consistent with (a)i) 1 point For a l/r decrease for a < r < b (or answer consistent with (a)ii) 1 point For any decreasing function for b < r < c 1 point For a zero function for r > c (or answer consistent with (b)) 1 point A maximum of three points could be earned if the graph was discontinuous (‘9 . 1. 2 pomts For a correct expression for the magnetic force on a moving point charge: F-qvaorFaqu lpoint Substituting the expression for B from (a)iz qvp. I F - ° 1 point 2:1:R ii. 1 point Cross Section For correctly indicating the direction of the force 1 point Distribution of Points E & M 3. (cont) (e) lpoint The answers to (d) would not change. 1 point Only the current inside the radius r has any effect on the magnetic field and hence the charge. M— ...
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This note was uploaded on 01/22/2012 for the course PHYS 2425 taught by Professor Park during the Spring '09 term at Cy-Fair College.

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1994 Physics C Solutions - 1994 Physics C Solutions...

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