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# MECHANICS (WITH LAB) PHYS 101

Rice University

GPA 3.68

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This 43 page Class Notes was uploaded by Ms. Janae Huels on Monday October 19, 2015. The Class Notes belongs to PHYS 101 at Rice University taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/225018/phys-101-rice-university in Physics 2 at Rice University.

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Date Created: 10/19/15

Physics 102 epledged Problem 3 Trme alluwed z hmsztz single smug w r 4 mm Yuu may use yuur uwn textbuuk yuur miss and anunrpmgrzmmed calmlatur Yuu may alsu ebnsuu Shaw huwyuu zmved at yuur answer the current answer by xtselfmzy nut be suf mem Further msh ucnuns a Wntelegbly nu ma sde um 5quot x 1 1quot whne urhghdy hntedpzper b w Mm m u Ow m m wl Van wl un v TM Mr G o fulluwedbyyuur FIRST AME a Beluwyuur name pnrrnbe phrase edged Prublem 3quot fulluwed bythe due date 3 Alsu mrbeaze Standing and mrl time 0 h perumt vulume grven by 7 A where A15 2 cunstant and the ream arsrenee 39nm the center 1 rs smuunded by a cuncenmc cundumlng shell ufmner rams b and uuta ad us cluud 2 Use Gauss39 Lawtu damnethe elem eld at all pumts m wane Inmcatethe Gaussian surfaces used b Shuwun a carefullylabeled gur the ram dependence quhe elecmc eld e What are the ebarge dansmes er and be un the mar and Enter faces quhe eundumurv z Cunslder a spherical cluud uf ebarge uf umfurm charge dansxty e and radms a cuntammg a whencal earmy ufradxus l2Z as shuwn m the gure Calculate the elecmc eld P e large cluud Solution Pledged Problems 3 l a This problem involves applying Gauss Law Because the charge distributions have radial symmetry the charge densities for both the cloud and conducting shell depend only on r the radial distance from the center then one will use Gaussian surfaces having the same symmetry namely concentric spheres of varying r Applying Gauss Law first to a sphere inside the cloud with radius r lt a one gets gt gt 1 r 1 r A 411A r 411Ar E dA M f Pr411r 2dr f 2411r 2dr dr 60 60 o 60 o F39 o o 60 Again symmetry indicates that on the Gaussian surface the electric eld E will be constant and parallel to dill and so can be removed from the surface integration The surface area is simply 41tr2 so that for r lt a E A A r r lt a no For a S r lt b the charge enclosed will be the total charge in the inner cloud or Q9 41TAa and so Gauss Law indicates that d E4m2M 60 60 and therefore if a r28 1 aSrltb 0 Within the conducting shell for b S r S C sufficient charge will accumulate on the inner radius equal in magnitude to the total charge in the cloud but of opposite sign so that the eld inside the conductor will be zero E 0 b s r s 6 Outside the conducting shell for C lt r EdjE41r2 M w 60 60 and therefore q 4nAa E lt 4nsor2 r C r c 6 produce atota1 charge equal to m magmtude but opposrte m srgh to the charge m the cloud Thxs charge densxty wru be umforrh over the rhher face As deterrmhedm part a the total 2 2 rh Therefore the charge dehsrty on the mherface wru be 7 mm c All c v my 7 Sxmxlarly 6a on the nun t rq on the shell dmded by the total surface area of that outerface 41ch2 Therefore the charge dehsrty on the outer face wru be r AntAa c 47th m Thrs problem rhvolves the Law of rposmon ofelecm elds The key to P charge density shown m the gure can be trea ed k P uhrforrh charge The rst havrhg a uhrforrh charge densxty ofp an raddus a rs centered at a e on se g aumform e ge 10 us a2 r centered at a2 When the charged dehsrhes m the eavrcy are added they produce ahe charge density ofU rh Lhatvolume The elecm eld associated thh either charge 39 Law or can be Qm Wtsnrz where Qheus the m1 charge m the cloud andns measured from the center ofthe center ofthe cloud Z am T Wu aw d 7 Annisz 7 47an X 7 lsnxzx be Z 7 am if 7 WNW3 E 7 ap E Z Wean new 7 39Iij 2490 7 39Iij The total freldwru be the sum ofthese elds 7 a e e e p EtptEEZ3X n Physics 102 e Pledged Problem 7 Tlme alluwed z hrmrsztz single smug Nb h7A shdw huw yuu zmved at yuur answer the current answerby ltselfmzy hdl he suf clent Further lnsh ucnuns a Wnte legbly dh dhe slde uf a 5quotx 11quotwhlte urllghdy tinted paper w Vule m w e mm mwl ml dd uh v by yuur FIRST NAME B uwyuur heme phhl the phrase edged Prublem 739 fulluwed by the due date 3 Alsu mdleaze grandma and Bud time 0 1 Two pameles othe same charge q 5 and m2 6 x 1039 kg are a elerated from rest through a potentlal dlfference AV looov They are then lnjected slmultaneously lth areglon o unlform magnetlc fleld B UlTasshoWn 10 x10 c and dlfferentmasses ml x 10quot5kg x x x 13mm the paper Assume that at t n bath pameles are at the uhgh ufthe cuurdlnate sySLem x u and y u and Lhatthevelunty ufeachls dlrected alanth usmveyrax ls a Desmhelhe subsequenturhlts quhetvm pzmcles glvmg sueh dumddes as relilus angular veluclty mdpmud ufrevulunun DEW a dlzgrzm carefully labeled h Alla what dme mtel39val wlll bath pzmcles rst andquot than mammum separanur e Alla what dme mtel39val wlll bath pzmcles Smultaneuusly retum m the uhgh x n y n he the rstume t gt m 2 Consldex There ls no charge on the lower face The am ofthls problem ls to ealeulate the total magnetle moment assoelateolvnth the rotatang else t t a elemental nng ofradws r and thckness oh b What ls the magnetae moment producedby ths element7 e What ls the total magnetae moment of the alsev Solution 7 Pledged Problems 7 1 L circniar motion 39 nnaii R 7 it n nnaii u vR and the period of revolution T 2mq3 27rm The only vanabies missing are the particles velocities L39 L L 39 39 39 39 acceleration from rest through a potential 1 difference oflOOOV will provide each particle with a kinetic energy RE Emuz a v 21000Vqm So particiemi with amass ofS gtlt lO39iSk g will have V 2ooox1o gtlt 10quot S x 1 0 15 63 x 103ms R mduJqs 5 x 1015kgxc3 gtlt 103m510 gtlt 10105mm 315m cc1 uiRJ 63 x 103m391315 m 2 x 103 711115quot 71 2111 2112 x 103 314 x 1035 Particlemz with amass of6 gtlt lO39iSk g will have 2 a ZOOOXID X 1039 6 X 10 15 58 X 103m5 R2 mZvZqB a x 1015kg58 gtlt 103ms1o gtlt 10100mm 348m nil uZRZ 583 x 103m3913AS m 167 x 103 711115quot T2 211ch 211167 x 103 376 x 1035 B into the paper b From looking at the particles paths it is clear the particles will attain their maximum separation when M is at the origin and M2 is halfway around its path at the farthest distance from the origin In determining the time interval when this will first occur it should be noted that the times of revolution T1 and T2 are proportional to the particle s masses T1 56 T2 they do not depend upon the particles velocities So the condition that must be satisfied is that M must have traveled n times around its path while W12 will have traveled only nl2 times around its path 1 5 1 5 1 nT1 ltn T2 gtnlt T2 ltn T2 gt nn gtn 3 2 6 2 6 2 This will occur at time nT1 3314 x 10 33 942 x 10 3s c An approach similar to that described in b can be used to determine when the particles will simultaneously return to the origin for the first time This will occur when the particle with the shorter time of revolution m1 travels one additional revolution compared to the number of revolutions traveled by m2 5 5 m1T1mT2 gtm1ltgT2mT2 gtgm1m gtm5 This will occur at time m 1T1 6314 X 10 33 188 X 10 35 H 1mA V c 2 density quota r whteh the charge ows eumeht of 7 d1 my aomxmj 17 dJmenslonal face ofwxdth 11 across which the charge ows b The magnetic moment assoemted with this nng is simply d m a7wd7mzj aw rnadv Where A 15 the area enclosedby the elemental nng em 0 mtegmte over 139 from Oak 7 7 amtR amkz 1 totifduiiawm d7 4 4 Where Q WERsz the total charge on the 645 101 Fall 2007 Pledged PI39C 1 Problem 1 12 pts a 2 pts ii 1 ptl Correct expression for A Each component of vector A ptl iii 1 ptl Correct expression for A Each component of vector 1 ptl b 2 pts ii 1 ptl Correct expression for iii 1 ptl Correct expression for Dr c 4 pts ii 1 ptl Magnitude of iii 1 ptl Direction of Cl iiil 1 ptl Magnitude of Dl ivl 1 ptl Direction of Dr d 2 pts e 2 pts ii 1 ptl for each orthogonal vector 2 Problem 2 18 pts a 8 pts ii 3 pts graph for z of the train vs time The graph needs to be parabolic and concave upwardl iii 3 pts graph for z of the person vs time The graph needs to be linear with positive iiil 2 pts for the initial separation between 1mm and zperson equal to d b 5 pts ii 4 pts making the connection between vperson vuainl iii 1 pt correct answer for t 10 s c 5 pts ii 4 pts making the connection between prerson z rainl The nal positions of the train and the person must be the same for the two to meet iii 1 pt correct answer for d 50 ft 101 Fall 2007 P1 and 1D Kinelnat Time allowed 2 hours in one sitting 39 Monday September 10 at 5PM in the box marked Phy5101102 in the Physics Lounge You may use your own textbook your notes and a noneprogrammed calculator For the purposes of this problem set you may also use the online solutions to the corresponding suggested problems You should consult no other help Please follow the standard format I Write legibly on one side of 8712 white or lightly tinted paper I Staple all sheets including this one together in the upper left corner I lVIa al 139 I On the outside staple side up on successive lines 7 PRINT your last name in CAPITAL letters 7 PRINT your rst name 7 Print the phrase Pledged Problems 17 and the due date 7 Print the times at which you started and nished the problems 7 Write and sign the Pledge with the understanding you may consult the resources described above 1 Consider two displacements vectors A and E that lie in the z 7 y planer Vector A has a magnitude of 10 cm and is directed 30 degrees clockwise from the zaxis Vector B has a magnitude of 8 cm and is directed 45 degrees counterclockwise from the zaxis a Write each vector in component form using the unit vectors Q and lal Begin by drawing out the vectors because this will help with the decomposition into I and y components 031 45 30 From the gure above h A75 cos 30 10cm lt Ag my sin 30 10cm gt V10 1 2 118r66cm 75cm Us 75 cos 45 8cm mm By sin 45 8cm lt 5r66cm b Determine the vector sum 6 A E and the vector difference 13 1 1b We now sum the components of the vectors A and E A2Bx3AyByJ e 3 14r32cm 066 cm BxiAx3ByAyf e 13 730001112 1066 cm A 1 2 3 4 5 6 7 8 9 10 c Find the magnitude and direction of C and U1 1c1 We use the pythagorean theorem to nd the magnitude of the vectors7 and the de nition of the tangent function to nd the direction of the vectors 1511Cx20y2 161 14132 cm2 0166 cm2 161 14134cm tan 190 010461 E 190 tan 1010461 W 216 counterclockwise from the zaxis 11311Dx2 Dy 1131 x73cm2 101660111 e 1131 1111crn D tan19D Di 012814 11 19D tan 1012814 W 151720 counterclockwise from the yaxis 11 12 13 14 15 16 17 18 19 20 d Find the vector F3 that satis es the equation 2 A 3 1 F3 0 1d Perform the required arithmetic to the components of A and 1 E 7 3E 7 2 A 21 E3BE72AE 3By72Ay3 22 E 7 1698 717i32cm216i98 103 23 E 7 70 34cm34r26i98 cm 24 e Find two vectors lying in the rig plane whose magnitudes are the same as A but that are perpendicular to A lei Since A forms a 30 angle with the zaxis clockwise a vector that is perpendicular to A will form a 60 angle with the zaxis counterclockwise These new vectors need to have a magnitude of 10 cmi F 10cm cos60 10cm sin60 25 F 5i0cm 8i66cm 26 This ful lls one of the two vectors needed but what about the other vector By multiplying the vector F by a negative sign we change the direction but not the magnitude therefore we have a new vector 37E77100m cos6027100msin603 27 6 7 750011127 866 cm 28 2 A person is running on at7 horizontal ground at a maximum velocity of 10 fts to catch a train When the person is a distance d from the nearest entry to the train7 the train starts from rest with a constant acceleration of 1 fts2 away from the person If the person just catches the train a plot a graph of position vs time for the train and person on a common graphi x ft t s b how long will it take the person to catch the train IF the person just catches the train7 the train must have the same speed as the person at that instant in time From the zversus time graph7 the slope of the train at the time the two meet is the same as the slope of the line governing the person s position versus time vperson Utrain 29 v0 at 30 31 c what is the distance d For the two objects to meet7 we must equate the nal 1 positions ram 1mm 32 1 2 votd at 33 34 Time allowed D nent 11111 101 Fall 2007 P1 2 hours in one sittlng Monday November 5 at 5PM in the box marked PhysiOl102 in the Physics Lounge You may use your own textbook your notes and a noneprogrammed calculator For the purposes of this problem set you may also use the online solutions to the corresponding suggested problems You should consult no other help Please follow the standard format I Write legibly on one side of 8712 white or lightly tinted paper I Staple all sheets including this one together in the upper left corner I IVIa al 139 I On the outside staple side up on successive lines 7 PRINT your last name in CAPITAL letters 7 PRINT your rst name 7 Print the phrase Pledged Problems 87 and the due date 7 Print the times at which you started and nished the problems 7 Write and sign the Pledge with the understanding you may consult the resources described above 1 An object at rest on a at horizontal surface explodes into two fragments one seven times as massive as the otheri The heavier fragment slides 82 m before stopping How far does the the lighter fragment slide Assume that both fragments have the same coefficient of kinetic frictioni SOLUTIONS Since there is an external force friction acting along the zdirection horizontal direction the zcomponent of momentum is NOT conserved for the system of two agments but at the instant the two pieces begin to y apart the momentum zcomponent is conserved p10 pr 1 07va7mvm 2 m 7 lt3 Since friction decelerates the masses to zero speed we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block I WNET7m 7 My d1 ii 7 m 1134 4 v34 2 Mk 9 d1 5 1 2 WNETm 7Mkmg d2 ii mvm 6 vi 2 Mk 9 d2 7 ed 7 1 d s 1 7 E 2 d2 49 d1 4018 m 9 2 A reworks rocket is launched vertically upward with an initial speed of 40 ms At the peak of its trajectory it explodes into two equalmass fragmentsi One fragment reaches the ground 28 s after the explosion When does the second fragment reach the ground SOLUTION Since the rework rocket is launched vertically upward we need to calculate the maximum height relative to the ground the rocket achievesi 1 9f y0v0y At EgAtQ 10 1 2 or WorkEnergy 7 mg yf 7E mvo 11 v2 yf 2i 81155721 height before explosioni 12 g The momentum of the system is NOT conserved since gravity is acting on the systemi Recalling the behavior of a projectile the y component of velocity and hence momentum will determine the ight times Assume that v1y points upward and that v2y is directed toward the earth The time it takes a massive body to fall to the earth from an initial height of 8155 m is 411 s verify this number therefore the 287s corresponds to the time it takes a projectile thrown down toward the earth to reach the earth pf pgy Immediately before and after the explosion 13 vay 7 mugy 0 14 v1y v2y 15 yf yov2y At 7 gAtQ 16 v2y 14 33ms 17 1114 14 33ms away from the earth 18 The time it takes a projectile to return back to its original launch point assuming the projectile was launched vertically with speed 1 is t 2i 292 s Once the projectile reaches its original launch point it now has a vertical speed equal to v pointing down toward the earth We know from the statement of the problem that it takes 287s for this projectile to hit the earth The total time it takes the second mass to reach the ground is T 21923 21873 517981 3 A cookie sheet moves along a conveyor belt towards an oven7 With mounds of unbaked cookie dough dropping vertically onto it at the rate of one 12gram mound every two seconds What average force must the conveyor belt exert on the cookie sheet to keep it moving at a constant speed of 50 cms SOLUTION To nd the average force the conveyor belt exerts on the cookie sheet to keep it moving at a constant speed7 rst consider the average force exerted on the cookie dough Newton7s third law ensures that the magnitude of the force exerted on the cookie dough by the cookie sheet is equal to the magnitude of the force exerted on the cookie sheet by the cookie dough The conveyor belt Will have to balance this force in order to keep the sheet moving a constant speed A101 At Favgx Apxdough Apxsheet Apxdough mdough O Sms 7 0 mdwgh 0 006kgs gtlt Ati Fm 0 006 05 N 3 0 X 103 N 19gt 20gt 21gt 22gt 23 Time allowed D You may use your own textbook your notes and a non nentum 81 RL 2 hours in one sittlng Monday November 12 at 5PM in the box marked Phys101102 in the Physics Lounge eprogrammed calculator For the purposes of this problem set you may also use the online solutions to the corresponding suggested problems You should consult no other help Please follow the standard format I Write legibly on one side of 8712 White or lightly tinted paper I Staple all sheets including this one together in the upper left corner IVIa all f I On the outside staple side up on successive lines 7 PRINT your last name in CAPITAL letters 7 PRINT your rst name 7 Print the phrase Pledged Problems 97 and the due date 7 Print the times at Which you started and nished the problems 7 Write and sign the Pledge With the understanding you may consult the resources described above H A spaceship in deep space far from any other massive bodies initially at rest explodes breaking into three pieces TWO pieces having equal mass y off perpendicular to one another With the same speed of 30 ms The third piece has three times the mass of each other piece SOLUTION a What is the direction and magnitude of its velocity immediately after the explosion I Will assume that a piece of mass m ies off horizontally Which implies that the other mass m ies off verticallyi I Will choose one piece to y off in the positive z direction and the other mass m piece to y off in the negative y direction To nd the direction and magnitude of the third piece consider momentum conservation Am 0 1 APy 0 2 3 mvo 7 3721mm 0 voe 1v0 IOms To the left 4 3 5 mvo 7 3mv3y 0 gt Ugy 1vo IOms Up 6 3 v3 mm 7 vs IOms 72 1 8 b How much energy was released in the explosion Compare the kinetic energy of the system immediately before and after the explosion AKE KEf KE0 9 AKEKEf400m 10 Where 777 represents the total mass of the spaceship 2 A stream of glass beads each with a mass of 05 g comes out of a horizontal tube at a rate of 100 per second The beads fall a vertical distance of 05 m to a balance pan and bounce back to their original height as shown in the gure below How much mass must be placed in the other pan of the balance to keep the pointer at zero SOLUTION To determine the mass that must be placed in the other pan of the balance we need to determine the average force exerted on the pan by the falling stream of glass beads We can nd the average force exerted on one glass bead by the pan and use Newton s third law to nd the force we need A13 A10 F we y Apy pyaf py70 210 13 10y0 mB Wyn 14 By using energy conservation on a glass bead the speed of a glass bead immediately before it collides with the pan can be determined AKE AU 01 15 gtvy0 WT 3113ms 16 gtApy 3113 gtlt10 3 kgms 17 At 7 i 18 7 100 S Ap vae T 3113 X 10 1N 19 Since this is the average force exerted on the glass beads by the pan7 Newton7s 3rd law ensures that the glass beads exert a force of equal magnitude on the pan down This means that we need the weight on the other pan to be 3113 gtlt 10 1N1 W3113gtlt10 1Nmg 20 a 2132 X10 2kg329 21 Time allowed 2 hours lu oue slttlug Due Monday notoher 1 at 5m lu the box marked Phy5101102 lu the Physlcs Lounge You may use your own extbook your notes nd a noneprogrammed calculator Fox the pulp see of thls pzoblem set you may also use the onllne solutlons to the conespondmg suggested pzoblems You should consult no other help Please follow the standard fozmat erte leglhly on one slde of gel2quot whlte ox llghtly tlnted paper Staple all sheets lncludlng thls one together ln the uppez left corner 1 c On the oulslde staple slde up on successlye llnes c e PRINT your last name ln CAPITAL letters 7 PRINT your rst name 7 Punt the phase Pledged Pxoblems 4 and the due date 7 Punt the tlmes at whlch you started and nlshed the pxoblems e erte and slgn the Pledge Wlth the understandlng you may consult the xesouxces descnhed above 1 A block of mass m ls placed on a frlctlonless lncllne mahlng an angle a wlth respect to the horlzontal A horlzontal force ls applled to the block as shown ln the gure below F e B a Draw a ReerBodyrDlagram for the block of mass m Flrst ldentlfy the number of forces actlng on the block of mass m There are time external forces actlng on the block of mass m l have chosen a coordlnate system parallel and perpendlcular to the lncllned plane b What force F ls requlred to move the block up the lncllne at asteady speed 39u The key to the statement ls steady speed 39u whlch lmplles NO acceleratlon along the lncllne FFcossemgslns0 1 FN7Fsln97mgcoss0 2 y Solving Eq 1 for F yields an 2 A Rice student is standing still and whirling her keys of mass m around her head in a horizontal circle of radius R and height h above the ground The string of negligible mass attached to her keys is tight no slack exists in the string and makes an angle of 0 below the horizontal as shown in the gure below The string attached to her keys suddenly breaks sending the keys ying off initially moving tangent to their original circular path and strike the ground after traveling a horizontal distance d a While the keys are attached to the string draw a F ree Body Diagram for the keys Ht m b Find an expression for the tension in the string while the keys are in circular motion Express your answer in terms of some or all of the following m g d and 0 The important concept to realize for this problem is that a net force exists on the keys The x component horizontal of tension is the force supplying the centripetal force keeping the keys in circular motion This can be summarized with Newton s 2 law as follows 2 my F 7T 0 if 4 Ex cos B ZFTsin07mg0 5 y Solving Eq 5 for T yields 7 W T T sin0 6 c Find an expression for the centripetal acceleration of the keys while in circular motion Express your answer in terms of only 11 g h and R Since we are asked to nd the centripetal acceleration of the keys one needs to recall how centripetal acceleration is related to the speed of the circulating object and the radius of the circle traced out by the rotating object For this problem7 1 was not given7 but we know that the keys ew off horizontally when the string broker This problem now becomes a twodimensional projectile problem with the projectile thrown horizontally a distance h above the ground If 10 vogct 8 d Ut igm 9 1 2 yfy0v0yt 9t 10 1 2 0 h igt ight 11 Solving for t igm in Eq 11 and substituting into Eq 9 yields the following expression for v v d 21h 12 ng ac m 13 d Calculate the centripetal acceleration given the following values h 1 75 n17 R 05 m and d 5 mi Using Eqi 137 the following value is obtained ac 14299 140i1ms2 14 Physics 101 Fall 2007 Pledged Problems 6 7 Energy Time allowed 2 hours in one sitting Due Monday October 22 at 5PM in the box marked Phys101102 in the Physics Lounge You may use your own textbook your notes and a non programmed calculator For the purposes of this problem set you may also use the online solutions to the corresponding suggested problems You should consult no other help Please follow the standard format 0 Write legibly on one side of 8 12 white or lightly tinted paper 0 Staple all sheets including this one together in the upper left corner 0 Make one vertical fold o On the outside staple side up on successive lines 7 PRINT your last name in CAPITAL letters 7 PRINT your rst name 7 Print the phrase Pledged Problems 67 and the due date 7 Print the times at which you started and nished the problems 7 Write and sign the Pledge with the understanding you may consult the resources described above 1 A roller coaster car on a frictionless track shown in the gure below starts at rest at height h The track is straight until point A Between points A and D the track consists of circleshaped segments of radius R What is the maximum height hm from which the car can start so as not to y off the track when going over the hill at point C Give your answer in terms of the radius R Solution If the roller coaster car is moving too fast at point C the car will y off the track N gt 0 Applying Newt0n7s 2nd law to the car at point C will give information pertaining to the normal force N 2 m U N 1 mg R N gt 0 The car just leaves the track 2 2 m U f max 3 mg R Umaac g R Since we were asked to nd the maximum height from which the car should be released from v 0 we need to consider the work done by gravity on the car We neglect the normal force7s contribution because the normal force is always perpendicular to the cars displacement hence the normal force does no work on the car WNET A K E WNET ngmy F d ngvity mg hmw 7 R NOTE The work done by gravity is positive AKE mghma 7R lmvELax m9hma1 R 2 3 hmw ER 5 6 7 8 9 10 2 A freight company uses a compressed spring to launch a 2 kg package up a 1 m high frictionless ramp into a truck7 as illustrated in the gure below The spring constant k is 500 Nm and the spring is initially compressed 30 cm If the package is released from rest7 what is its speed immediately upon entering the truck7 ie7 at the top of the ramp Solution First7 nd how many forces are doing work on the box There are two forces doing work on the box spring forces vary with position7 and gravitational forces WNET AKE 11 WNET Wspring Wgrauity FTS Fag 39 0 1 Wsprmg k x dx 5 k Ax2 The work by the spring is positive 13 ALE ngm y mg 1m NOTE The work done by gravity is negative 14 1 2 AKE mg1m kAx 15 1 2 1 2 Emu mg1m kAx 16 With m 2 kg7 Ax 03 m7 and k 500 Nm the speed of the of the box at the top of the ramp is gtv 17ms 18 Physics 102 7 Pledged Problem 4 Tlme allowed 2 1111an at a single sitting DUE AFMMUN D w h H mm p l Youmayuseyour at th uru t A L L Show Further m strueu ohs a whte leglbly oh hue slde of 8 5quot x 11quot whrte or llghdy tmted pape b Staple all sheets together meludmg thrs ohe m the upperleft eomer and make one vemcal fold e on the outsrde staple slde up Prmt your name m eaprtal letters your LAST NAME rst followed by y ur FIRST NAME d Eelow your name prmt the phrase Pledged Problem 4quot followed by the due date 3 Also mdleate start time and end time 1 4 pts A eharge Mqls loeated at the ongm and a eharge eq ls oh the x axls at xa a whte ah expresslon for the potehtral oh the x axls for x gt a b Fmd a pomt m ths reglon where V e 0 e Use the result of a to fmd the eleetrre eldFoh the x axls for x gt a and d Fmd a pomt where E 0 p p eharge dehsrty 9 that hes m theXZplahe ahdhas threlmess b see gure 4E3 a Use Gauss39 Law to ealeulate the eleetrre eld veetor as a functlon of the y eoordmater e eehter ofthe sheet b Caleulate the resultmg potehtral as a functlon ofy Set the potehual at y e 0 to be zero Sketeh thrs potehual Solution 7 Pledged Problems 4 0 a sum othe potentrals assoerateolwrth eaeh eharge element sourerng the potentaal e 1 vp So for the eharge olstnbutaon m the problem vp for x gt Ills we r 1 Art 1 y pm 47t nx Antsnx 41 1 a W sn 7 araj b ma ls zero when 7 a 1 w en 139 43 a Plugglng th5 value forx baekrnto the equataon con rms ths ls the nght answer 0 Taklng the denvanve of ma wrth respeet to x ylelds e l 7 q d l 1 dx Antst arts Antsndxx arts q 4 1 Mrsn x1 artsy whenLrj u 4xrajzrx 3xzr Sax4az 0 aemxaaem u The loeataon where a ls therefore x 2a Agam plugglng th5 value u furx gt a for rrnto the equatron for a ennlarmsthtsrsthenght answer Gauss39 Law As we39ve dlscussedln elass andm the book aplanarunlform eharge magmde aglven dlstance from the surface In all reglons the duectlon owalll be away from the xraxls so for y gt 0 the duectlon ls 3 and for ylt0 wlll be 7 Thls symmetry ls true lnslde and outsrole the planar sheet so we wlll ehoose a y strueture whose sldes are parallel to the electxl elds and whose top and bottom are equrdlstantfrom theyraxls sheet l 1 g So using Gauss Law 475 95 E dl QM for bZ lt y lt bZ 0 5 E 471 damp 47th asides EA EA 0 ZEA 0 Where A is the area of the top and bottom of the surface and Q9 p2yA Solving for E gives W E b2lty ltb2 For lyl gt b2 the expression for 472 will be substantially the same as for 271 472 47ml 27th distdeg EA EA 0 ZEA but now Q9 pbA so that pb E E ylt b2 andygtb2 o b Obtaining potentials from electric elds involves the inverse process to that used in Problem 1 where we took the partial derivative of the potential in a certain direction to nd the electric eld in that direction Now we will integrate over the electric eld along a given path in this case in the y direction to nd the potential difference between the endpoints ofthe integration AV Violml VimHal dj Because we have designated the potential at y 0 to be zero this will be the initial point of integration The ending point will be some distance in the y direction from y 0 Because we are integrating in the y direction and have determined that the electric eld is pointing only in the y direction the term being integrated will become E d3 Eydy So for y lt b2 AV Vfinal Vfinal Vy 0 g dy 0 0 2 2 p y W b V 0 lt y 0lt2 gt 250 for lyl 2 For y gt bZ the term for the electric eld changes at y bZ so the integration will have 2 components one for 0 gt b2 the other from b 2 gt y bZpyd y pb d 20 bZ 220 y y AV Vy 0 f0Eydy f0 quotb22 I719 b pby p192 2 20 y 2 2 58 20 Symmetry requires the potential fory lt r 17210 be the same as y gt 172 So a sketch 0ny Vy Wm looklxke Physics 102 7 Pledged Problem 6 Tlrne allowed 2 hnllrs at a single sitting DU39E AFM ann M h I mm 1 1 You ur v V A L l show how you amved at your answer the eorreet answer by ltself rnay notbe sumelent Further lnstrueta ons a wrlte leglbly on nne slde of8 quotx11quotwhlt b Staple all sheets to 5 e or hghtly trn gethe mclud ng th5 one ln the upper e on the outslole staple sld nntyo teol paper lelt eorner anol make one vertaeal fold urnarne ln eapltal letters your LAST NAME rst a Determine the value ofaz 2 A In the circuit shown to the right the current through the ammeter A doubles when switch 5 is closed Determine the emf 82 B In the circuit shown to the right the until equilibrium is achieved Determine the total charge not the cunent that ows through the ammeter S 511 711 Solution 7 Pledged Problems 6 la When both switches are open no current flows through the right loop and the circuit simplifies to a single loop with three resistors in series as shown The currentI AAV through the ammeter can be determined by determining 100 Q the equivalent resistance R R1R2R3 3009 21009 2009 1009 1009 S 4009 and applying the equla onYZ IR to get 81 12V I 0030A R 400 b A er the switches are closed the circuit appears as shown below the 1009 resistor in the left hand loop is shorted by the closing of the switch and so no longer is included in the loop The value of 22 can then be determined by applying Kirchhofi s Junction and Loop rules First the direction of 39 labeled consistent with the ow of current current will ow from the positive or high end ofa resistor to the negative or low end Then 3 independent equations can be generated and the results from part a applied 13 I from part a 11 12 I 12 20011 10013 0 2 30012 10013 0 With the substitution 1 0030A the second equation produces a value of 11 0045A Those results plugged into the rst equation yields a value ofIz 0015A Finally plugging those resulm into the third equation yields 82 30012 10013 300 0015 1000030 2 15V 2a When the switch is open no current ows along e center line and so that part of the circuit can be removed leaving the circuit shown below 81 The currentIthrough the ammeter can be 12 V 6 Q determined by determining the equivalent resistance 18 Q R R1 R2 189 69 249 and applying the equation V IR to get V 12 Imt 05A A er the switch is closed the circuit appears as below Then 3 independent equations can be generated and the results from the initial part applied 11 2 X1m10A 21 0 12313 sz 18110 2 313 6120 After substituting the value for 12 into the third equation 2 7 313 7 6V which can then be plugged into the second equation to solve for h 12 313 7 2 7 18 1 18 7 o 1 7 6 I 033A with this result the iirst equation indicates that 13 7 067A and therefore according to the third equation 2 7 30367 7 61 7 o 2 7 av ne potential L capacitor At that point no more charge will ow through the ammeter but will continue to ow through the two resistors in accordance with the equation V 7 Km R2 The charge that will L a A L 39 L 39 39 in 11 until in 7c 39 39 39 capacitor in be negative and with the lower capacitor will be positive The potential V across eac resistor L r 439 r 39 L 39 fmT 39 Using this approach one nds that I 7 Wm R2 IZVIOQ 309 7 03A The voltage V 39 39 39 the voltage V2 across the bottom resistor and capacitor is V2 7 03A300 7 9V e amount of charge that will have owed from the top capacitor will be 7Q 7 7 cm 5x10396F3V 7 15x10395c and the amount that will have owedto the bottom capacitor will be Q2 7 C39sz 7 7x10396F9V 7 53x10395c The total charge that will have tlowed therefore will be m 7 Q QZ 715x1075c 63X10395C SXIO SC tot Physics 102 7 Pledged Problem 5 Tlme allowed 2 hnllrs at a single sitting DU39E AFM ann w h 7 mm Dh l l trr n t Show Further mstxuch ons a wrrte leglbly on one slde of8 5quot n 11quot whrte or hghtly trnted paper b Staple all sheets together rneludrng th5 one m the upperlelt eorner and make one verheal fold e on the outsrde staple slde up Pnnty eaprtal letters your LAST NAME rst followed by your FIRST N AME d Belowyourname prrnt the phrase Pledged Problem 5quot followed by the due date 3 Also rndreate start time and end time 0 our name 1 l 5 pts Two eapaertors 12hr and our rnrhally uneharged an open swrteh and abattery 10 V Th 0 V uh l A The swrteh ls elosed andthe eapaertors beeome eharged Determrne the potentral at P B Wrth the swrteh shll elosed a slab ofdreleetrre 12hr onstant 2 0 ls rnserted between the plates ofthe eapaertor lllng the spaee between the plates D t rmrne the potenhal atP 1 T c Wrth the slab strll m plaee the swrteh ls opened m HF lab 15 then removed Determlne the potenual z Aputmual dx ermce um szmmally applied armss AB on 4512 V and the suurce uf shuwn beluw A Calculate the charge un each capamm The switch isnnw dnsa l u p VA 3W B c 12V 4 mm s A D Solution Pledged Problems 5 1 A Note 7 there are a number of ways to solve this problem The following is one way to approach and solve it After the switch is closed an amount of charge iQ will accumulate on each of the two capacitors on the upper plate on the lower plate Q can be determined by calculating the value for Cm the single capacitor equivalent to the two original capacitors in series and then applying the equation Q Cqum where Vm 10V The voltage across each capacitor can then be determined by applying the equation Q CV 139 1 2 So 1 1 1 1 1 3 1 ceq c1 c2 12x106 6x106 12x106 4x106 or Ceq 4uF Therefore Q Cqum 4 X 10396 CV10V 4011C and V2 QCz 40uC6uF 667V B The same approach is used to determine the potential at P after the dielectric has been inserted recognizing that C2 will increase proportionally with the dielectric constant K ofthe inserted material or C239 KCZ 2 61F 121F Therefore 1 1 1 1 1 2 1 ng C1 C 12X10 6 12X10 6 12X10 6 6X106 or C39eq 6uF Therefore Q39 C39qum 6 X 10396 CV10V 6011C and 392 Q39C39z 40uC6tzF 50V One could have also noted that because the 2 capacitors now have the same 12uF capacitance the voltage drop across each will be the same or 12 of the total 10V C Opening the switch has the effect of eliminating the ow of charge so that Q on the lower capacitor before and after removal of the dielectric remains the same In part B Q was determined to be 6011C The voltage across the capacitor after the dielectric is removed will be related to the charge Q and original capacitance by the equation 392 Q39C2 60uC6MF 10V A The charge on the 6nF capacitor which we will call Q1 is related to the 12 V potential drop across it by the equation Q CV so that Q1 6nF X 12V 72pC Because the remaining capacitors are in series the charge on each of them which we will call Q2 will be the same and can be determined by calculating the equivalent capacitance of a single capacitor 111111141 Cal C1 c2 C38X10 6 4x106 8X10 68X10 62X10 6 or Ceq ZMF Therefore Q2 Cqu 2 X 10396 CV12V 24pC B After the switch is closed Points A and C will be at the same potential Therefore there will be no potential drop across the 2 capacitors on the right side of the system the 4nF capacitor between C and D and the 8nF capacitor between D and A Point D will be at the same potential as Point A and therefore the potential difference VD VA 0 The potential difference VB 7 VA can be determined by recognizing that because the system is isolated the total charge on the tops and bottoms of the two remaining capacitors Q1 Q 72nC 24nC 96nC will remain constant although it will redistribute itself between the two capacitors because of the closing of the switch To determine the new potential difference one first calculates the equivalent capacitance for the two capacitors and then uses that together with the known total charge of 96MC to derive the new voltage difference across the two capacitors Because the capacitors are parallel Ceq 6uF 8uF l4uF and Vnew QCeq 96pC14pC 69V C The amount of charge which owed through the switch S after it was closed will have 2 components the charge that flowed as a result of discharging the two capacitors on the right side of the circuit and the charge that owed as part of the redistribution of the 96uC on the two capacitors on the left side of the circuit As determined in part A the amount of charge on the two capacitors on the right equaled 24uC That charge would have owed from C to A to discharge those two capacitors The new charge distribution for the capacitors on the left can be determined by using the voltage VBA found in part B and the equation Q CV Thus the charge on the 8uF capacitor will be Calm SHF x 69V 55pC and the charge on the 6 HF capacitor will be CWF 6pF x 69V 41 C A quick check confirms that the total charge remains 96 uC 55uC 41 uC The initial charge on the 8uF capacitor was 24uC as determined in part A so the amount of charge that owed through the switch associated with the left two capacitors will be 55uC 24uC 3luC owing from A to C The net charge owing from A to C will therefore be 31pC 2411C 7pC 101 Fall 2007 P1 Time allowed 2 hours in one sitti D n5 Monday September 24 at 5PM in the box marked Phy5101102 in the Physics Loun e 2D Kinelnat 5 You may use your own textbook your notes and a noneprogrammed calculatorv For the purposes of this problem set you may also use the online solutions to the corresponding suggested problems You should consult no other help Please follow the standard format I Write legibly on one side of 8712 White or lightly tinted paper I Staple all sheets including this one together in the upper left corner I lVIake one ve I On the outside staple side up on successive lines PRINT your last name in CAPITAL letters PRINT your rst name Print the phrase Pledged Problems 37 and the due date Print the times at Which you started and nished the problems Write and sign the Pledge With the understanding you may consult the resources described above H An archer sh launches a droplet of water from the surface of a small lake at an angle of 60 above the horizontal He is ainnng at a spider sitting on a leaf 50 cm to the right of the sh and on a branch 25 cm above the water surface The sh is trying to knock the spider into the water so that the sh can eat the spider a What must the speed of the water droplet be for the sh to be successful To determine the speed of the water droplet7 we need to rely on the z and y coordinate of the trajectory formed by the water droplet If 7 IO v0 cos 60 t 1 05m v0 t t7 l W iyo v0 sin60 t 7 Eth 1 02572170107 t7 g Using Eq 3 in Eq 5 yields the following equation for v0 v0 282 ms t2 1 2 3 4 5 6 7 8 b When it hits the spider With the speed given in a7 is the droplet rising or falling To determine Whether the water droplet is rising or falling7 one needs to compare the time it takes to hit the spider With the time it takes to reach the maximum height If the time it takes to hit the spider is shorter than the time it takes for the droplet to reach its maximum height7 then the droplet is rising First7 determine the time it takes the droplet to reach its maximum height vyj 0i vyj 340 ayt 9 0 v0 111600 ythighest 10 x v i thighesz 0 2 11 g thigh 02493 12 By comparing the time thigh With t 0354s one sees7 that the drop is falling 2i Catapults date from thousands of years ago and were used historically to launch everything from stones to horses During a battle in what is now Bavaria inventive artillerymen from the united German clans launched giant spaetzle from their catapults toward a Roman forti cation whose walls were 85 m high The catapults launched the spaetzle projectiles from a height of 40 m above the ground and a distance of 38 m from the walls at an angle of 60 above the horizontal If the projectiles were to hit the top of the wall ignore any effects due to air resistance a what launch speed was necessary We start this problem just as we started the previous problem treat the z and y components of the trajectory as separate entities connected together through time If 7 IO v0 cos60 t 13 38m ltv0 t 14 76 t i 15 e W o 1 2 yf7y0v0 s1n60t7 igt 16 7 Q 2 1 sismi 4m 7 38 7 i 76 7 17 1 0 2888 v3 7 g 18gt ags 7 4 5 09gt b How long was the spaetzle in the air An expression connecting time and initial launch speed has already been established via Eq 15 7 76 v0 21 c At what speed did the projectiles hit the wall Since this question is asking about speed one needs to nd the components of the nal velocity of the spaetzlei t 20 yaw v v0 cos60O 10i75ms 22 vyj vyj 7 gt igm v0 sin60 7 g 3533 l6iOlms 23 2 4 m f 715 19i32m3 24 lv 1932 ms 25

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