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# C1999 - 1999 Physics(3 Solutions Distribution of points...

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Unformatted text preview: 1999 Physics (3 Solutions Distribution of points Mesh. l (15 points) (a) 5 points For cemervatien of momentum 1 point my0 r: (m + [W0 )1) v m me)O m +MO For conservation of mechanical energy l point K +1] 2 + U” or é—MWM m Mmm gh Fer calculating I: in terms of E I point 11 m [(1 - COS 69) For substituting for Mmml and v in the energy equafjon 1 point: 2 5(1‘7’1 + A40 3 (m + .MO W COS For the correct: answer 1 point; m +M yo 7: 0 .lZngwCOSG) m (b) 4 points For Newton’s seeend law (not awarded if net force was set equal to zero) 1 point; If“: = ma For any equatien that indicated that the tension minus the weight is £191; zero l point Y "” Nltotalg : Muﬁala For an expression fer the centripetal foree '1 point a l. 2: or 313. w r 3? U2 T ”"‘ A4ng 3 Mtoml T "‘ For curreetly substituting for Mom; and 1) Ll. point w 2 1 ("2+MO)?(W> =T—(m+M0)g Solving for T T : (i?1+M0)g(3m2COSI9) 1999 Physics C Solutions Mech. 1 (continued) (c) 4 points For all of the following: ‘1) A practical procedure that uses some or all of the apparatus listed and would work 2) Recognition of any assumptions that must be made 3) Indication of the proper mathematical computation using the variables measured Two points were awarded if the description of the procedure was not complete but it would work, or if the mathematical work did not clearly specify the variables used, or any combination of the above. No points were awarded if the procedure would not be feasible in a laboratory situation with the apparatus listed, or if the procedure was merely a repeat of that outlined in part (a) (d) 2 points Fm mma mwbo Ii ., . . a , t do For expressing the acceleration as the time derivative of the speed, a = -—« t) l’ 212 no 0 '2) hr in --—» a: -« -- 1)“ m U m Uggwbt/m or a general expression for the length of the dart in the block as a function of time or for the expression for the total distance L M mun wet/m i? —. Mb (I u— e ) “.2332” L“ 32 Distribution of points 4 points 1 point 1 point 1999 Physics C Solutions Distribution 0f points Mach” 1 (continued) ((11) (continual) Alremate Solution 1 Alternate paints Fur indicating that work equals the: change in kinetic energy I point If? dx 2 1' m2); 2 m £339., n) 4.4 average L A b) q Iii dxzimvg 2 2 0 Zn) 1 » wiLx—mﬁ 2 2 For the correct expression f‘nr the: total distance L I paint mp0 3:: L: Alternate SG/z'ution 2 Alternate poinm Fm At a Ap For the, above: expmssien on f ~51) d: = ~va 0 I paint d ~ 1 . . Using 1) :3— and the fact that 3 goes from zero to L as nme goes dt from zero to inﬁnity \ L j ~22 (IIS :2 mm 0 «('71) = «mu0 Fm the: conect expression for the total distance L I paint ml)0 b Lm 1999 Physics C Solutions Distribution of points Mean. 2 (15 points) (a) 3 points For indicating that the equation for gravitational force is applicable 11 point F x m Gmélm2 r For using the proper expression for the mass of the planet enclosed by the radius 1 point Gm pim3 17‘ m .— 3 P2 For proper cancellation of terms to ShOW the ﬁnal result 1 point 4 v F x - ~3~ mirsz r (b) 4 points F S For drawing a straight line from the origin to a distance R, and not going past R 1 point For having the maximum magnitude occur 21th 1 point For having the curve from R to SR decreasing in magnitude with proper curvature and appearing to reach an asymptote 1 point For recognizing that the force is always negative, Le. the graph is always below the x-axis 1 point 1999 Physics C Solutions Distribution of points Mech. 2 (continued) to) 3 points In this and all subsequent parts, either C or i;— nGpm could be used. For indicating the integral that needs to be calculated to determine the work 1 point W 5 f1? air = j~Cr dr For using the proper limits on the integral (R to zero, not 7') l point 0 W a j ~01» dr R 2 O W 3 ~61- 2 R For the correct answer ll point C 2 W = R 2 Alternate Solution Alternate points For recognition that the work is the area under the curve, which is triangular 1 point For using the corrects limits (zero to R) 1 point For the correct answer 1 point 2 W, :3 CR 2 (d) 2 points For using conservation of energy or workcenergy relationship '1 point AK 2 AU :2 W n l 2 CR2 -—mv = 2 2 For the correct answer 1 point [CR2 1) z m An alternate solution indicating the potential energy as that of a harmonic oscillator also received full credit: (e) 2 points For indicating that the ball will move from the center to the surface of the planet 1 point For indicating that the ball will stop at the surface, return to the center, and continue oscillating in this manner, with no damping i point Describing the motion as simple harmonic oscillation with no damping earned full, credit ...
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C1999 - 1999 Physics(3 Solutions Distribution of points...

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