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Physics I – Exam 1 – Spring 2006 Answer Key Part A – 1: C, 2: C, 3: B, 4: A, 5: B, 6: B, 7: D, 8: B 4 pts each. B-1 20 Points The key to this problem is the Impulse-Momentum Theorem and also Newton’s Second Law in the form of Equation 11. There are 4 sections to the curve, each a section of a parabola. Looking For curves are sections of parabolas. maximum at t = 4 sec. maximum value = 4 kg m/s. value at t = 2 sec is half the maximum value (whatever it was). symmetric curve around t = 4 sec. Fnet (N) 2 1 0 -1 t (sec) 2 4 6 8 -2 p (kg m/s) 4 3 2 1 0 t (sec) 2 4 6 8 B-2 24 Points Vx and Vy (m/s) This problem is an application of equations 1-5 for twodimensional motion. Looking For Vx is constant = +9.8 m/s Vy is straight line. Vy starts at 0 and ends at –19.6 m/s. X starts at 0 and ends at 19.6 m. X must be a straight line. Y is a parabola opening down. Y = 14.7 at t = 1 sec. 9.8 Vx 0 t (s) -9.8 1.0 Vy 2.0 -19.6 X (m) 19.6 9.8 t (s) 0 Part C Must show work to receive credit. C-1 24 points This is a straightforward conservation of momentum problem in two dimensions. P 10 m i p1 p 2 p1 9.24 cos( 22.5) m i 9.24 sin( 22.5) m j p1 8.54 m i 3.54 m j p 2 P p1 (10 8.54) m i (0 3.54) m j v 2 p 2 m 1.46 i (3.54) j 1.0 2.0 1.0 2.0 Y (m) 19.6 14.7 9.8 4.9 t (s) 0