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Unformatted text preview: ard on the Web: www.collegeboard.com. AP PHYSICS B 2008 SCORING GUIDELINES
10 points total (a) 4 points For the correct value of t1 , the time car A is accelerating 1 point Distribution of points u f = ui + at1
t1 = u f - ui t1 = 2.0 s ( ) a = ( 5.0 m s - 2.0 m s) (1.5 m s ) For a correct value of x1 , the distance car A travels while accelerating 1 2 x1 = ui t1 + at1 2 1 2 x1 = ( 2.0 m s)(2.0 s) + 1.5 m s2 (2.0 s) 2 x1 = 7.0 m 1 point ( ) Note: The equation u 2 = ui2 + 2ax1 could also be used. f For a correct value of ( x - x1 ) , the distance car A travels at constant velocity ( x - x1 ) = (15.0 m - 7.0 m ) = 8.0 m
t2 = ( x - x1 ) u f = (15.0 m - 7.0 m ) 5.0 m s
t2 = 1.6 s ttot = t1 + t2 = 2.0 s + 1.6 s ttot = 3.6 s 1 point 1 point For correctly calculating t2 , the time car A travels at constant velocity x = x1 + u f t2 (b) (i) 2 points For any clear statement that momentum is conserved m Au Ai = m Au Af + mBuB 1 point m Au Ai - mBuB (250 kg)(5.0 m s) - (200 kg)( 4.8 m s) = 250 kg mA For a correct answer u Af = 1.2 m s u Af =
1 point 1 point (ii) For indicating a direction of car A after the collision that is consistent with the calculation of u Af Note: A correct calculation yields a direction to the right. 1 point 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP PHYSICS B 2008 SCORING GUIDELINES
Question 1 (continued)
Distribution of points (c) 3 points For correctly indicating that the collision is not elastic For a statement that kinetic energy is not conserved For clearly showing that K f < Ki , implying the collision is non-elastic 1 point 1 point 1 point 1 1 2 m u 2 = ( 250 kg )( 5.0 m s) 2 A Ai 2 K i = 3125 J Ki = 1 1 1 1 2 2 m u 2 + m u = ( 250 kg )(1.2 m s) + ( 200 kg )( 4.8 m s) 2 A Af 2 B B 2 2 K f = 2484 J Kf =
Note: Two points were awarded for checking "yes" with a clear, correct explanation that it is a partially elastic collision. 2008 The College Board. All rights re...
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- Spring '11