# prac exam 1 - General Physics ll - Exam l Solutions -...

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Unformatted text preview: General Physics ll - Exam l Solutions - Spring 2002 t'xleJ/Ml/My Web Filesheactheeclt/ng-examI-soln-sO2.htm lg, ’ GENERAL PHYSICS II - EXAM I SOLUTIONS - SPRING 2002 PROBLEM SCORE l. x 25 2. z 25 3. ____/ 25 TOTAL SCORE / mo 4. x 25 ,1, _ ‘59,)" PROBLEMS ‘74,)» l. A mass m = no g is attached to a spring with a spring constant I: = 3.00 N/m. The mass oscillates / along a horizontal frictionless surface. At time t - 0 the spring is stretched LOO cm from its equilibrium position and the mass has a velocity of+ 5.00 cm/s. a) What is the period of oscillation of the mass? b) What is the maximum speed of the mass? c) Calculate the amplitude and the phase constant for the oscillation of the mass. d) If everything is kept the same. except that one were to double the initial displacement of the mass. would the following quantities increase. decrease or remain the same: (i) frequency of oscillation. (ii) phase constant. (iii) total energy of the syStem. and (iv) maximum acceleration. _ _ _,__ “I”-.. /v a "j W ._._ A l : wt : 5: D ra‘J/g " i " .0. l1 mtg Tﬂti ; 5",” ': -J- 2 G 5- 5.0}; \ 1 2. 17-) Er WM: 2 {we 4—;Lx, -g.‘ _ if 3 p __ I, )7 A“: ; fr”,+ 1%";- = Y”? * “3x0 3 7 5:95.- + 5/9053 Hz \J/ [4‘ = 101:»); /’ I‘ I'll ,’ 1",r (I . , ‘ r . .1 - ~ \ ms ; -. ~, . 4 *_ > - ( (3/ C A ’2’,” "’44 vi. '1, 5:”: a g -, A." 1.,,__« #1 { [Acuity-my y, é - , H l ' 'r‘J/ "' i I 7 ,_ .. ,1 .l' I 1 l,,‘ 7;, [1‘7‘ 7 “y »" ' Ir,7,"\ , , ,‘ Tum ‘ 7."' ‘7'. ~ I I‘ I I "Qt xZ/ (Hr, ,p "1/?" . «J .... c. ' _rvr cv’ ~ ) .7, \/ 2! ‘ \$:r.¢'€ E .»,. ire“, lily“: / ., .zzy-ruaci‘? X‘, €KCY?:>;: | ‘ trrrgl'r, - z. . - I , ~- \ \_// Id, “ﬂax - L) XM H‘CrFLI).A} /, ;ACr(‘:.U.:-,L c I ( >1. Mir-Hr nCif-‘JZd‘GX : Z(, n [.1 l of 5 4/29/2002 1:36 PM General Physics ll - Exam l Solution: - Spring 2002 ﬁleJI/lﬂ/My Web Files/teach/teach/gp2-examl-solnvso2.htm 3 ,xyﬁ'z. Whrcc charges ol‘ equal magnitude. q. are ﬁxed at the corners of an equilateral triangle with sides of length a. The charge at the upper corner of the triangle is positive and the other two charges ue negative. (:1) Calculate the electric force. F. (in unit vector notation) on I charge of +3q at point P (the origin of the xy coordinate system) shown in the diagram. Calculate F in terms ofq. a. a nudge. lb) Where could one place a fourth charge of -4q. such that the force on the charge at P would be zero? .2 — 7 ‘7 , ,- , (“'4' i5, = r. a re J t 47 3 5 p: 2 7:4}? )0 Hill" ‘3‘; ' 170/94“, k’Ve 5“““ CJ—a—yc. 4-13 {-7.1 55"": d:\$'f,t€""¢' -5? _ I .— ,,. A. . f3 1",; 4rvvs ;,:' r -? l '1" 2~ «J I i A ‘r an: " I" ’ Ti. 79:" ,1 4' ,. I 5' k3" ~ ’ "l3 (3 g viz l Nu Law-#4; m l '11! {a r; \ .: l , - t\,\ Ir. _ /’,t.. ‘i (“i ’ ’" awn/«>04 V “w L A / r: r - 4”; (j \ 1,1,0» _/ '7 " + -: .— (3’) v-ngj ‘7"; VIJ‘Lf & 51.1 5 " {V6}! r-K / F2; 1 “a l ‘ ﬂ 7 7 r 5d“ at» 3.5 rig-55H”. and. win “The? ILL 4’10 CA‘ﬂa /' ' "‘5 ? WM} 7, Fl¢c¢ T ALoVb 7 1.17;? +y-awt5_ \J/ﬂ 4, t,’ 19a tum-.1 : I!th 432L339, : “Le " - MM; . : mm in, ‘ {a a 7 ﬂat; = ;: J ,— \ ,_A.1»~ .1 TI 4,. I if a" 4 rt, (, (.7 to I ‘~ { \UA \ L 1 z / _ 4 " V J i V’ I) h} ‘ \ {l1 1 {Ev 9’ t 9 _ ‘ - v k \ “an —1-; gum" [7 X-O I. - V3 \ 1 F \l \ ' A \ \ ( ' w; twill, L. JKST a" 0‘? ("MM l7 ‘ - 2 of 5 4/29/2002 l:36 PM General Physics ll - Exam l Solutions - Spring 2002 ﬁleJI/Hl/My Web Flies/teacMcuch/ng-examl-soln-sOZhun 6 , “S' 9 3. A thin glass rod is hent into a semicircle of radius r. A charge +q is uniformly distributed along the upper half 2/ and a charge of —q is unifomtly distributed along the lower half. as shown in the ﬁgure. a) What are the charge densities for the upper and lower halves of the semicircle? b) What is the magnitude and direction of the electric ﬁeld at P. the center of the semicircle? 4 4 ﬁ/ . -, o- ‘x {" . I In.i+- \ :’ 1 r, (J) .1 1m! /+ L‘, ‘27-) WT i Mi in” 19"” )"k‘; A " 17 = 5 k/ 1' (gr) pr X f: f (,1. :/.. “I” a Jfflr l‘ajsr In}; J!M;Clr(}( 'T r" . IV : r r‘ '3- .54- / gi't'v‘r'l; ""11: I. a «.4 "a; 4 ’0’“; j l " ' . d7 1 i 1 x I J 7 .3 J, r 7103‘: x -[p/m w} i- f 52 (A / 91:} r57.“ 170‘ yv/V .fpt Jv-v’ 5 011,6. i- A 7 l ._ 0 _ J T: c 'r:"L-,l:'£ ai-‘JS any») ,y , 71-.vrmar 5 f0). .4 29"” ,P "I, f. , - aL ‘ — J \/r K v _ ‘ FAQ, 7'51] Willi-7 y” "“ (J A», 2‘; 71—; -(J 1.19. Mt! 4M j‘m" 04 E a 1- f, ‘ Jr}? ,' - 1;? 5c: '2. - , » 1/; M? ~ r g , i: .21 l?’ A V *7 .-. tin "p t," ' , l » J/f I :- QVQ’ “M a?) uni (h r a J //. f , 0 ",4 w A)?" r ' I 9"" K 7,; r v” ’ it / — ‘ rr57 I” ‘1 15, : L )Jp/ ' 7- )JE ,1 (-1» ‘ 1’. ( , 1 a ' I I \ pm . ’ U, .7 ,r J — 2i (’9 i f 9 — J | g ' ail . = A A5 _ l 9/] ) f — VA?" \ f r T: .4 ' v) f I“ 1 \ I ’5' V r 9 1 w ‘ 7.. a 05 .. \ V / ‘. C 10 \ A’J-ﬁ f z __:LM’-a:o'- ’- 1 g WI I, «Y1 "1'59 7 o (H , 117:, I .1, x?- V . i 5 M! r t'jf) ' r 1. ,. r « e , .71 i ,: , i - l as ".77; ‘w’ r J, 77"”, /- ~r ‘ '3 ( ) e I 1 -5- — 34.; \\ , r’r .( .__,/ ' A, a: i f '_,_;.,-r r ’ " — 1. i ll C ‘\ \‘\. 3 of5 4/29/2002 1:36 PM General Physics ll - Exam I Solutions - Spring 2002 ﬁkJ/leMy Web Files/moh/teach/ng-exunl-soln-sOthm O c Q 4. a. A small portion near the center of I very large nonconducting slab of thickness d w ' and positive uniform volume charge ’4, 6370\‘35 density +p,, is shown. A cylindrical L!” Gaussian surface of radius a enclous part / of the slab. Let EL and 5‘ represent the magnitude of the electric ﬁeld at the left and right ends of the cylinder. which are at different distances from the slab. / i. Find the charge enclosed by the fa Gaussian surface. ‘ _ r~ __ 9'} //’/_\r ﬂirt ' ?"'«1/'N~2’)M: Li/rLOJLWOIJ’J 5 ; I ~ 1 a“ “3. Side vtew t\/ m T7,,» _ ii. In the side view drawing. sketch electric ﬁeld lines outside the slab. , a l ls the magnitude of EL greater than, less than. wequmo 5.? E r..\ ’; ~——’ 1 (z; a: t J i . in. Find the ﬂux through the entire G ' ' i . a - .. dimensions. ’ ‘ .usstan surface tn terms of E. And the given ‘ r ‘ t I 7.‘ - 'ﬂ 'I‘1 ,r-r- ‘Iijr' " M" ' EL [‘L Jr LIL/'9 7 Edi/'7’" "4 f7" '0' I E I (— t ,2 I z : _ /" (:3: 3 2W9" ER ‘ iv. Use Gauss' law to ﬁnd the electric ﬁeld E. in terms of the given dimensions and the volume charge density p. a a , e —- .1 ./ 5‘ a 755 (1' to; ._,a ’ 4" -0a1 ’ h I A ‘ D [‘i' : if 1- 0 b. A positive point charge +Q, is now placed to the left of the charged slab and the Gaussian surface as shown. (Assume the charge distribution within the slab is ,» - ‘ ‘ unaffected by the point charge +Q,) ' d L‘E ~’ A '7 s y .7 i " > i. Will the absolute value of the Hunt u h the right ' l “9; end cap of the Gaussian surfacefﬁu a: decrease. I l l _ or remain the same? Explain. " " , r- ‘\ , , _ _. ' 5 " .)J7}l r‘ ,, é], ‘5 Jet» l‘.".-ie“ ,Ji : '- '0"?! 040 - ' ' .' ., ._ 4 r. - New I _ .,. _ I w sf" ,»- end ca I , [,1 Le: ;~),’-,i{. £12.“; ‘, D lair/t 4.. A‘ﬁlﬁ, 9.1.7'4140', -~‘ 4 l.— p w.) l r l p 1‘ ' U I 0 ‘ ' H e 4,. «ma/1 1;. 1".17/9 191) u —... /' Tar TIL rth. 2-4 {if} s.d . it. Will the absolute value of the ﬂux through th _n" V I a new / I . 9M” surface increase. dureau. charisma E) (’ 1/; f ~"5arneZ'Explaini "\ . .~_ . ‘. 3'2; ."uit due 70 2.0. w i” (it"ww In"; I” “4‘ 1 . ‘1 "‘ ' L V” . ~I e- L- a. >1 "" “Y’a's‘f' I L”,th 1/2 _f"’«U-j_'-, £747, W9 On :07'!‘ .IYI.‘! " lrijryega Jr”; a ~, .t an," 7,: .' urw’d r‘l'-."--1'ri'm' 5W face. M “'“l' I r61!” f Lg” .( f 7""1' i'Jx reivv ,: 7.1-. 5.: r-<¢ Fiht'e’ “Why ""0‘ A?“ aim/’2 “rah, UL, ,. .;: ,:1 / : .. Fry“ (35¢ ycz' Lav 7‘} ."T “'1‘ “my” 17/ k (/9. in 3in Jr: 5 1‘9v5‘ 470)" any Cit/r13. .‘v 3'44»- - ’ : 7 PM 4°f5 4/29/2002 I 3 ...
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## This note was uploaded on 10/12/2009 for the course PHYS physics 2 taught by Professor Kogan during the Spring '08 term at University of Cincinnati.

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prac exam 1 - General Physics ll - Exam l Solutions -...

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