AP PHYSICS C: ELECTRICITY AND MAGNETISM 2007 SCORING GUIDELINES Gnral Nots About 2007 AP Physics Scoring Guidlins 1. Th solutions contain th most common mthod of solving th fr-rspons qustions and th allocation of points for this solution. Som also contain a common altrnat solution. Othr mthods of solution also rciv appropriat crdit for corrct work. 2. Gnrally, doubl pnalty for rrors is avoidd. For xampl, if an incorrct answr to part (a) is corrctly substitutd into an othrwis corrct solution to part (b), full crdit will usually b awardd. On xcption to this may b cass whn th numrical answr to a latr part should b asily rcognizd as wrong,.g., a spd fastr than th spd of light in vacuum. 3. Implicit statmnts of concpts normally rciv crdit. For xampl, if us of th quation xprssing a particular concpt is worth on point, and a studnt s solution contains th application of that quation to th problm but th studnt dos not writ th basic quation, th point is still awardd. Howvr, whn studnts ar askd to driv and xprssion it is normally xpctd that thy will bgin by writing on or mor fundamntal quations, such as thos givn on th AP Physics xam quation sht. For a dscription of th us of such trms as driv and calculat on th xams, and what is xpctd for ach, s Th Fr-Rspons Sctions Studnt Prsntation in th AP Physics Cours Dscription. 2 4. Th scoring guidlins typically show numrical rsults using th valu g = 9.8 m s, but us of 2 10 m s is of cours also accptabl. Solutions usually show numrical answrs using both valus whn thy ar significantly diffrnt. 5. Strict ruls rgarding significant digits ar usually not applid to numrical answrs. Howvr, in som cass answrs containing too many digits may b pnalizd. In gnral, two to four significant digits ar accptabl. Numrical answrs that diffr from th publishd answr du to diffrncs in rounding throughout th qustion typically rciv full crdit. Excptions to ths guidlins usually occur whn rounding maks a diffrnc in obtaining a rasonabl answr. For xampl, suppos a solution rquirs subtracting two numbrs that should hav fiv significant figurs and that diffr starting with th fourth digit (.g., 20.295 and 20.278). Rounding to thr digits will los th accuracy rquird to dtrmin th diffrnc in th numbrs, and som crdit may b lost. 2007 Th Collg Board. All rights rsrvd. Visit apcntral.collgboard.com (for AP profssionals) and www.collgboard.com/apstudnts (for AP studnts and parnts).
AP PHYSICS C: ELECTRICITY AND MAGNETISM 2007 SCORING GUIDELINES Qustion 1 15 points total Distribution of points (a) 3 points Writing th gnral loop rul for th circuit = IR + V C V C = 0 at tim = 0, so = I0R For corrct substitution of a valu for I 0 and a valu for R into Ohm s law or th loop rul with th rcognition that V C = 0 at tim = 0 For corrctly rading th magnitud of I 0 from th graph and using it in a valid quation 2.2 I 2.3 0 For corrctly using th currnt in ma 2.2 ma I 2.3 ma 0 ( )( W) = 2.25 10 A 550 = 1.24 V (b) 3 points For a corrct loop rul quation = IR + V C For corrctly rading I( t = 4 s) from th graph 0.3 ma I( t = 4 s) 0.4 ma For corrct substitution of from part (a) into a corrct quation VC ( )( = 1.24 V - 0.35 10 A 550 W), using th middl of th rang of accptabl valus for I V = 1.05 V (or valu consistnt with th valu of I chosn within th accptabl rang) C Not: Us of a valu for currnt that was not corrctly xprssd in ma was accptabl if th ma point was not awardd in part (a) (ithr for using an incorrctly xprssd valu or no valu at all). Altrnat solution For corrctly using an xprssion for VC VC t t ( 1 ) - = - or quivalnt For corrct substitution for th tim and th tim constant t t 4.0 s 4.0 s = = 6 = = 1.82 t RC - 550 W 4000 10 F 2.2 s ( )( ) For corrct substitution of from part (a) into a corrct quation 1.82 V = 1.24 V 1 - - C V C = ( )( ) 1.04 V Altrnat points 2007 Th Collg Board. All rights rsrvd. 3
AP PHYSICS C: ELECTRICITY AND MAGNETISM 2007 SCORING GUIDELINES Qustion 1 (continud) Distribution of points (c) 2 points For using VC from part (b) in a corrct quation ( 1.05 V) Q = CV = C For corrct substitution of C ( 4000 10-6 F )( 1.05 V ) Q = -6 C Q = 4.20 10 C or 4200 10 C (or valu consistnt with V from part (b)) Altrnat solution For a corrct substitution of or I0-1 t t - 1.24 V 1 t t ( ) ( )( ) Q = C - = C - 4 -t t OR Q = Ú ( 2.25 10 A)( ) 0.82 ( 4000 10-6 F )( 1.24 V )( 1-1 ) into a corrct quation For a corrct substitution of C or t into a corrct quation Q = - 4 3 OR ( 2.2s Q = ) Ú ( 2.25 10 - A) ( -t ) 0 Q = 4.16 10 C or 4160 m C Not: If th answr to (c) was corrct using on of th xponntial quations, it could b substitutd into V = Q C in part (b) for full crdit. dt dt Altrnat points (d) 2 points For starting th graph at th origin For a sktch that is approximatly xponntial and approaching an asymptot 2007 Th Collg Board. All rights rsrvd. 4
AP PHYSICS C: ELECTRICITY AND MAGNETISM 2007 SCORING GUIDELINES () 2 points Qustion 1 (continud) Distribution of points 2 2 or R P = IVR or I R V R, whr VR = - VC For th substitution of on corrct quantity into a corrct quation for th instantanous powr For th substitution of th scond corrct quantity into a corrct quation for th instantanous powr 3 3 ( 0.35 10 - A)( 0.19 V ) or ( 0.35 10 - A) 2 ( 550 ) or ( 0.19 V) 2 ( 550 ) P = W W -5 P = 6.7 10 W (or valu consistnt with arlir valus of I and V ) Not: Us of a valu for currnt that was not corrctly xprssd in ma was accptabl if th ma part was not awardd in part (a) (ithr by using an incorrctly xprssd valu or no valu at all), or if th valu usd in part () was consistnt with th valu usd in part (b). C (f) 3 points For marking th Gratr than choic For any indication that C incrass C = k0a d or Cnw = 3C0 For xplicitly and corrctly addrssing tim dpndnc -t kt -0.61 Q 3 ( t 4 s) kc( 1 ) 3( 1 k= = - - ) = = = 1.63, t 3 3 RC, Q ( ) 1.82 1 t 4 s -t t - k= = or quivalnt k= = C 1-1- ( ) ( ) Nots: If Gratr than was not chckd, no points wr awardd for part (f) rgardlss of th justification. If Gratr than was chckd, a corrct valu of Q both justification points. k ( t ) = 3 = 4 s = 6.7 10 C arnd 2007 Th Collg Board. All rights rsrvd. 5
AP PHYSICS C: ELECTRICITY AND MAGNETISM 2007 SCORING COMMENTARY Qustion 1 Ovrviw This qustion was dsignd to assss studnts ability to intrprt and utiliz information prsntd graphically, as wll as thir undrstanding of th charging of a simpl RC circuit, including th dissipation of nrgy in th rsistor and th consquncs of changing th capacitanc. Sampl: E1A Scor: 14 This clarly writtn rspons lost only in th justification in part (f) for failing to considr th tim dpndnc of th charg on th capacitor as it is chargd. Sampl: E1B Scor: 10 Part (a) rcivd full crdit. In part (b) th voltag across th rsistor is computd, so only for th corrct valu of I at 4 sconds was arnd. In part (c) only was awardd for th corrct substitution of th tim constant into th quation, but th substitution for th maximum valu of Q is incorrct, so th othr point was not awardd. Part (d) rcivd full crdit. Part () arnd only for using th corrct valu of I at 4 sconds in th corrct quation; th us of th voltag from part (b) is incorrct. Part (f) lost for not addrssing tim dpndnc. Sampl: E1C Scor: 4 Part (a) rcivd 2 points but lost th final point for not using milliamps as th unit for currnt, substituting for th currnt as if th unit wr amprs. Part (b) was awardd only th point for using a valu for th currnt at 4 sconds that is within th accptabl rang. Sinc th point for not xprssing th currnt in milliamps was lost in part (a), thr was no furthr pnalty for th sam rror in part (b). No crdit was givn for part (c), and in part (d) th graph must start at th origin to rciv th scond point, so no crdit was givn for part (d) ithr. In part () was givn for a corrct substitution for th currnt into a valid quation, but th voltag is incorrct. In part (f) th wrong box is chckd, so no points wr awardd. 2007 Th Collg Board. All rights rsrvd.