# C1995 - Mechanics'1 Scoring Guide Solutions Note For this...

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Unformatted text preview: Mechanics '1 Scoring Guide Solutions Note: For this salution, vectm quantities to the right are deﬁnﬁd as positiva. (a) 3 points Far a correct exprcasion relating impulse and force a 1:: 1M: 5 0R FA: F0: any indication that the impulse is the area "under: the curve I mag-)(z x 30:“ MM x 10” s) + (2 x 103 NXZZ x 10” s) For the correct answar I a 12 N w s (b) 3 points Far a caveat bxpressian relating impulxe and momentum “z Imfmdv OR A}? 1; F01" a correct axpmssien for the impuise in terms 0f the mass and 5;)ch ofme ball I m 3719(1)“ w vb.) ! Sinca thE: bail is initially at rest, 1)“ == 0 . Substituting and solving for v M: 1)“ == 1 /mb 1)“ = (MN ~ s)/(5 kg) For the correct answer 1) M a 2.4 mfg 8O Distribution of points 1 point 1 point I. point 1 {mint 3; point 1 {mini Physics C m Mechanics Distribution Solutions 0f points (c) i. 2 points For aither an cxpressien for the impulse cm the cube, with an indication that the: value for the cube is equal and opposite that of the ball, or conservation of mumcnmm I m mom“ w v“) and I: «~12N*s OR mtvci 2"“ mbvbl + mtvaf Selving for thﬁ 596% cf the, cube afmr the collisicm: v” m {I + mg!) ,)/mg OR vﬁf : (mava - mbvwﬂmg (711 1 point map-12 N . s + (as kg)(26 m/s)}/(0.5 kg) OR m [(05 WNW) m/s) - (Skg)(2.4 m/s)]/(0.5 kg) For a cormct answer 1)an “Mn/s L 1 paint it I point The spew 0131311135! a‘beve has a positive: value, so the cube is moving :9 the right. 1 point: ((1) 3 points Fm a ammo: axpression for the kinetic: energy dissipated I paint AKchi+KM-K‘ 1?} For 21 came: general exprcssion for kinetic enargy: 1 K 2:: *2” mt)2 ipaim I ’2 ii 2 AK w "5(mcvﬂ + mb‘uM » may“) “W” ghmkgm mi 92 + <5 kg>c234 m M - (0.5 km miSﬂ Fm the: camera: answer 1 pwim AK 2:154} 81 Machanics 1 Scoring Guide Solutions (a) 3 points Beth objects takt: the same amount of time to reach the ﬂoor. For the kinematic aquation fer the vertica} motion 1 2 37% vﬂté‘ Egt Th3 initial velocity i3 zeroﬁ Solving for t : t w Jilly/g a JE(1.2m)/(10m/sﬁ) a 402433 must Fm a cartect axpregsion for hcrizontai motion with no acceleration x m vxt Applying this mutation to the two objects-2: xb :2 (2411113)“).55) z 1.2m xc m (2 m!s)(0.5 3) == Tim Ax m 31,, w xc For the: correct answar Ax :- 02 m 82 Digtribuﬁon of points 1 paint I paint 1 paint WV. Physics C m Mechanics Mechanics 2 ﬂeeting Guide Solutions Distribution of points (a) i. 3 points . . d U The trummum occurs when --- m 0. er For reaming that one must take the derivative of U 1 point For getting the dcrivaxive equal m zero 1 point, Evaluating the darivative: a? U a ( - 1 Mb (if rﬂ l? r2 =0 a ab w...sz 2 b r9 Solving for re: .2. r- .953. b w r; 2 2 ramb For {ha comet answer "1 point r0 m 19 ii. I paint The minimum potential energy is found by evaluating U (rO ). b b UﬂwUUQ matz; + For the correct answer 1 point Uamla 93 Mechanics 2 Scoring Guide Salutions (b) 3 points . . w (W * Uttmg the rclatmntahtp F r- - W, the shape of the graph can be detertmnad d U from the gamma} expteﬁsimn for 72-;- found in part (ah, or qualitatively fmm the slope: of the U (I) graph. Fm a positive fame for F < r0 , smooth and momtonicany decreasing as r increases, that times not cross the: y axis For a zero force at r m I}; , with the curve passing smhothly through the): axitt Fat 3 negative force for r > I}; , smooth and monotonic, that approaches a canstant value (In the abs/cum 0f credit fer the graph, indicating that F a «- 7 mccived 1 point.) 94 Distribution at“ points 1 point 1 point: 1 point Physics C m Mechanics Solutions Distribution of points (0) 4 points For indicating that energy is conserved 1 point E m K +1] a constant For recognizing that the initial kinetic energy is new 1 point rO ‘ 1 2 = “amt? + (/05) , r0 . For evaluating U “f i paint U .332; t. .2... 23?. .. 22. :2. "" ‘1 2b + b " 2 Substituting: .1. “Dz a» 3g.” 2 w E. 2 m “” 2 a "' 2 Solving fer the speed: 1) w J a/ m 1 point For using U€3 m Zn from part (a)ii, and substituting for a v w «I UQ {’2m (ti) 2 points By conservatien of energy, the partieie wiii again come to rest (instantanenusiy have new velocity) at a point r1 where its petential energy equate that at r0 / ’2,» r0 Utiri) : UL?) Au 2 paints (Some students interpreted “at rest” to mean permanently at rest. Fuli credit was awarded for saying that the particle never comes to rest.) (e) 2 points The particle will cseillate, with the end points of the motion being I'm/2 and my 1 point For indicating that the particie will csciiiate 1 point For indicating, imphcitiy 0r explicitly, that energy is not lost 95 Mechanic-s '3 Scoring Guide Solutians Distribution of points (a) 3 points For a 00mm expression for centripetal acceleration 1 point; 2 1) a a: «7 OR at) 2:“ For expressing v a: m in terms 0f given quantities 1 paint in r 22: u w W OR a) a -- T T For the cormct angwer 1 paint, 4ﬁ2ﬂ a; w W T 4.4 (b) 4 points The centripetal force, on star/1. is due to the gravitational force excnad by star 3 . Fm: a currect axpression for the oeniripetal force on star A 1 paint Fa a Mia. OR M‘uf/r. Fm: a comet expreszaion for the gravitational force: (1354“?!b v FG m 2 ! pmm: r F01" using r m r“ + rb 1 point Fm setting F c and F 0 equal 1 point 41: 2Va GM bM a M: T2 In 2 (r. + r13) Sawing far Mb: 3, 2 M x 431: rn(r‘ + r93) b GT2 Solutions Distribution of points (c) 2 points The same calculatioas as above can be performed with the roles of star A and star B switched. For realization of this symmetry 1 point For tho correct answer I point 415% (r. + rt)2 M m l ‘ GT (Alternate solution} Qéltemare points) Writing {ha expression for the center of mass with respect to point P : - M at“ + M,m r" 0 Mi 4» Mb “ For the simpliﬁed version of this aquarion (1 point) Ml rat a Mb r?) r Ma a M b “2-. rl Substituting the expression for M b from part (b) to obtain the ﬁnal answer (1 point) 4:53:13 {rh + “)2 M: g x 2 GT (:1) 3 points I a 2 mi:2 For indicating that I 0‘ mrz 1 point For knowing that in this case the constant of proportionality is 1 for each star. 1 point For the corroct answer 1 point 2 2 I - MJ" + Marl, Mechanics» 3 Scoring Guide Snluﬁons Distribution of points (6) 3 points For indicating that the tom! angular momentum is the sum of the individual stars’ momenta 1 point L m L1I + Lb For a correct general axpmssion for angular momentum 1 paint L w m1) 1' OR 10) L a My.“ + Mbvbrb OR (My: +Mbr:)£o 2m. 2mb 23': L m M‘ T r“ + MbTrb OR (Mari +Mbri)? Fm Wk: 00mm anﬁwer I paint ‘21:: a ?(Miri +Mbr:) ...
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C1995 - Mechanics'1 Scoring Guide Solutions Note For this...

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