ࡱ> qG x bjbjَ {!Q&]xxxx444HHHHH\H |p555] _ _ _ _ _ _ $NB 45"555 Exx|EEE5|x4] HHxxxx5] EE] 64] p CTHH] Physics I Exam 1 Fall 2006 Answer Key Part A 1: C, 2: D, 3: G, 4: C, 5: A, 6: A 4 pts each. Give credit for 6 if it agrees with 5 even if 5 is wrong. B-1 20 Points The key to this problem is the Impulse-Momentum Theorem and also Newtons Second Law in the form of Equation 11. There are 3 sections to the curve (or two counting the upward and downward part on the right as one), each a section of a parabola. Looking For curves are sections of parabolas. maximum at t = 4 sec. maximum value = 2 kg m/s. value at t = 2 sec is half the maximum value (whatever it was). symmetric curve around t = 4 sec for the interval (2,6) seconds.  EMBED Zoner.Draw.Document  B-2 24 Points This problem is an application of conservation of momentum, equations 44x and 44y. The total momentum in the X direction is +1.2 kg m/s and in the Y direction is 0. Looking For trying to use conservation of momentum in some form, right or wrong. using conservation of momentum separately in X and Y directions. VBx flat where VAx is flat and a sloped line where VAx is a sloped line. VBx final value = 3 m/s. VBy flat where VAy is flat and a sloped line where VAy is a sloped line. VBy final value = 1 m/s. Part C Must show work to receive credit. C-1 32 points Step 1: X and Y components of initial velocity:  EMBED Equation.3   EMBED Equation.3  Step 2: Find time to reach impact point using X equation (no acceleration):  EMBED Equation.3  seconds after the target drops. 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