1993 Physics C Solutions

1993 Physics C Solutions - 1993 Ihysics 6 Solutions...

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Unformatted text preview: 1993 Ihysics 6 Solutions Distribution of Points Mech 1. (a) 3 points Using the expression for the energy stored in a spring: U-%kx2 lpoint Substituting: l E 2 U 2 [400 “J (0.5 m) 1 point U - 50 J 1 point (b) 5 points For recognition of conservation of energy or work-energy theorem 1 point Kinetic energy of block C: K - % mcvcz 1 point work done (or energy dissipated by) friction: Wf - and 1 point K-U—Wf % mcvcz - U — nmcgd 1 point Solving for no: vc _. /;—2 (v - pmcgd) Substituting: 2 ye - /Z—kg- (50 J - (0.4)(4 kg)(10 m/s2)(o.s no) vc - 4.58 m/s 1 point (Full credit also awarded for correct alternate solution computing anet dx, where Fnet - kx — pFn, to find the kinetic energy, and then computing the speed.) 1993 mysics C Solutions Mech 1. (c) (d) (Alternate Solution) UNITS: Distribution of Points (continued) 3 points For any statement of conservation of momentum 1 point mcuc - (mg + mD)uf 1 point Solving for vf: of - mcvc/(mc + up) Substituting: Uf - (a kg)(4.58 m/s)/(a kg + 2 kg) 1 point Uf - 3.05 m/s 3 points I The blobks come to rest when all their kinetic energy has been dissipated, i.e. AKE - Work done by frictional force 1 point %(mc + mD)vf2 - p(mc + mD)gd 1 pOint Solving for d: d - “fa/2M; Substituting: d - (3.05 m/s)2/(2)(0.4)(1o m/sz) d - 1.16 m 1 point (Alternate Points) 2F _ ma (1 point) Zr #(mc + mu»; 2 2 a - 57-- (me + mD) - pg - (0.4)(10 m/s ) - 4 m/s U2 - uoz + Zad (or other appropriate kinematic equations) (1 point) vz-vu2 o—(3 05M)? d-—'28——- 2(_4 m/s) V-1.16m P013” For correct units on all answers 1 point 1993 Physics 0 Solutions Distribution of Points Mach 2. (a) 3 points F kv ° "'3 1 point for F0 correctly drawn and labeled 1 point 1 point for ku correctly drawn and labeled 1 point 1 point for N and mg correctly drawn and labeled 1 point (b) 3 points V Fnet - ma 1 point But Fnet - Fb — kv. therefore: 1 point F0 " ICU - ma Solving for a: a - (F0 — kv)/m 1 point (c) 5 points a - g! 1 ' t dt pain Using the equation from part b: dv (Fb - kv) (1) dc - m Re-arranging and integrating: dv 1 (2) - I E dt 1 point Changing variables by letting u - F0 — kv, du - -k dv: 1 du 1 __E I .5 _ I E d: K 1 point 1n(Fo — kv) — 1nC - - g c, where C is a constant v - i [F0 - Ce-kt/m ] A 1993 Ehysics C Solutions Distribution of Points Mech 2. (continued) (e) 2 points Differentiating the expression for velocity from part (c) to get the expression for acceleration: dv F0 _ -kt/m - dt - k [(0 (-k/m) e F m a 9m 0 I For correct initial value Fb/m 1 point Fbr correct shape of curve 1 point (Alternate Solution) From part (b): a - i (F0 — kv) Substituting for u from part (c): Eh k Pb -kt/m “I‘m—(1‘9 >. F0 a _ __ e-kt/m m 1993 Physics 6 Solutions Distribution of Points Mech 3. (a) 4 points 1 - IF 1 point 27 - O (or rcw - rccw) 1 point Summing torques about the right end of the rod: Fa! — Mg[§] - 0, where Fa is force exerted by axis 1 point Fa-g—a- Fa is directed upward 1 point (Alternate Solution for last two points) (Alternate Points) Sum térques about any other axis and also use 2F - 0. For example, summing torques about left end of rod: Mg[§] - Ft! - 0, where Ft is force exerted by thread Ft _ Hg ' (1 point) 2F - 0, so Ft + F5 - Mg - 0 Fa-Mg-Ft-Mg—I-‘g Frga F8 is directed upward (1 point) (b) 2 points Using Zr - Ia and calculating the torque about the 1 point axis end of the rod: Mg % - M g: a Solving for a: a - g g 1 point 1993 Physics C Solutions Distribution of Points Mach 3 . (continued) (c) 2 points Using the relation between translational and angular acceleration: a - at 1 point Substituting r - g and a from previous part: hlofi £ 2 p I Nu ton» on O 1 point it I (d) 3 points Using Newton's Second Law: 2F — "a 1 point but 2F - Mg - Fr 1 point so Mg - Fr - Ma Substituting 8 - %g and solving for Fr: 1 Fr - 3 M8 1 point (e) 6 points Using conservation of energy the increase in kinetic energy 1 point of rotation Kroc is equal to the decrease in potential energy AU AKrot: ' — Iwz 1 point ' Mg — Sin 6 1 point l Iw2 - Mg g sin 6 2 Solving for w: o) - a? sin 9 Substituting I - mZ/a 3 . w ' is 51“ e 1 point 1993 Physics 6 Solutions Distribution of Points Mech 3. (continued) (e) (continued) (Alternate Solution) (Alternate points Work done by gravitional force Wg equals the increase in (1 point) kinetic energy of rotation Krot Wg - I Mg it cos 9 d9 (1 point) - m sin 9 2 1 2 . AKrot - E Iw (1 pomt) 1 2 l 2 It.) - Mg 2 sin 9 Substituting I - 1422/3 and solving for w: w - 2% sin 9 (1 point) 1993 Physics 0 Solutions Distribution o£ Points E 6: M 1. (a) 2 points If; 1 point for each correct vector 2 points (If both vectors are reversed from correct directions, then partial credit of 1 point awarded) (b) i. 4 points Gauss's Law: § E-dA - Qencl/eo (or 4nernC1) 1 point For 1' > R, using a Gaussian surface that is a cylinder of radius r and length 1: § E-dA - E(21rr£) 1 point Qencl - p(1rR2£) 1 point E(2me) - p(1rR22)/eo 2 2 E - FR [or 27”ch ] 1 point _ Zeor r (b) ii. 2 points For 1' < K, using a similar Gaussian surface as above: E(21rr£) - p(1rr2£)/eo 1 point E - P—r— (or 21rkpr) 1 point 260 1993 thsics c Solutions Distribution of Points E & M 1. (continued) to) 3 points ‘-_°’} 1 point for first correct vector 1 point 2 points for second correct vector 2 points (If vectors are reversed from correct directions or if circular lines of force with counterclockwise arrows shown, then partial credit of 1 point awarded.) (d) 4 points Ampere's Law: § B-dl - p0 Iencl, where Iencl is the current enclosed by the 1 point closed loop of integration (i.e. the current density times the area) For r < R, integrating over a loop of radius I: § Bod! - B(2wr) 1 point Iencl - [;%Z] «r2 - I g: 1 point 8(2wr) - #01 g: "OI r I‘OJr B - —;— fig [ or 2 J 1 point 1993 Physics c Solutions Distribution v of Points E & M 2. (a) i. 2 points Using the expression for the flux of a uniform field: § - BOA 1 point Substituting: ¢ - abBo 1 point (a) 11. 1 point Using the expression relating the emf and flux: d9 5"3 Both the field and area are constant, so E - zero 1 POint (a) iii. 1 point Since there is no emf there is no current in the loop, and thus no magnetic force exerted on the loop. 1 point (b) 2 points 2 points When at - t/2, B - Bo cos «/2 - zero, i.e. the field has been decreasing, and is about to change direction. The induced current will be in a direction to oppose this change, i.e. clockwise. (c) i. 4 points Calculating the flux: Q - abBo cos wt 1 point Calculating the emf: .s - — 99 (negative sign not required) 1 point dt - awao sin wt Using Ohm’s Law: I - E/R 1 point R sin wt 1 point I- 1993 Physics C Solutions E & M 2. (continued) (a) 11. 3 points A w. Fbr a graph showing a period of 2: For a graph consistent with answer to (c)i. Fbr coriect orientation of current at n/Z consistent with answer to (b) (i.e., graph positive at n/2 if clockwise current in (b); graph negative at u/2 if counterclockwise current in (b)). (c) iii. 2 points The maximum value of the current occurs when sin wt - 1. @030 Imax - R Fbr indicating the coefficient of the sin (or cos) term in (c)i. Fbr the correct answer (i.e. the coefficient in (c)i. must be correct) Distribution of Points 1 point 1 point 1 point 1 point 1 point ____________________________________________________________________________________ 1993 Physics C Solutions E & M 3. (a) (b) (d) 2 points The force due to the magnetic field provides the centripetal force that causes the ion to move in the semicircle. F - qv x B, so by the right-hand rule the magnetic field must point into the page (or in the -z direction). For field being perpendicular to the page For direction into the page or in the -z direction 1 point Between the plates, the electric field must exert a force opposite to that of the magnetic field. The magnetic force is to the right, and Felec - qE, so the electric field should point toward the left. Therefore, plate K should have a positive polarity with respect to plate L. 2 points Using the relation between the electric field and potential difference for parallel plates: E - V/d Substituting: E - (1500 V)/(0.012 m) E - 1.25 x 105 V/m 4 points For a particle to pass between the plates undeflected, the forces due to the electric and magnetic fields FE and F3 respectively must be equal in magnitude and in opposite directions. FE-qE EB - qu Therefore, qE - qu Solving for v: u - E/B — (1.25 x 105 V/m)/(0.20 T) u - 6.25 x 105 m/s DiStribution of Pbints 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1993 Physics c Solutions E & M 3. (continued) (e) 3 points The centripetal force Fc is equal to the force qu due to the magnetic'field. 2 am Fc'za— 2 an R qu Solving for m: m - qBR/v Substituting: m - (1.6 x 10'19 C)(0.20 T)(O.SO m)/(6.25 x 105 m/s) 1!: - 2.56 x 10-26 kg (f) 2 points Substituting 2q into the force equation from part (e) and solving for the new radius R': _ .22 2qB Substituting the expression for m from part (e): RI .__ti_ 3% _ R ZqB u 12/2 R’ - 0.25 m UNITS: Additional 1 point awarded if all units are correct Distribution of Points 1 point 1 point 1 point 1 point 1 point 1 point ...
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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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1993 Physics C Solutions - 1993 Ihysics 6 Solutions...

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