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Prelim_Exam__1_S05_Solns

# Prelim_Exam__1_S05_Solns - P112 Prelim Exam#1 Solutions 1 2...

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3 4 5 v t 3 4 4 5 v t v t 3 4 - a ( t) 4 2 P112 Prelim Exam #1 - Solutions Spring 2005 1. (a) C (b) A (2 points each) 2. F (5 points) [2 points for E, G, or H] 3. (a) D (2 points) (b) A (3 points) 4. (a) C (b) A (c) C (d) B (2 points each) 5. (a) A (b) F (c) (3 points for correct construction + 3 points for proper labeling) 6. (a) A (b) A (c) B (d) B 7. (a) v x (t) v y (t) a x (t) a y (t) (b) v oy = v o sin θ = (6.0 m/s) sin60° = 5.2 m/s v y 2 = v oy 2 - 2gH = 0 H = v oy 2 2g = (5.2 m/s) 2 2(9.8 m/s 2 ) = 1.38 m 1.4 m (c) v y = v oy - gt = 0 t = v oy g = 5.2 m/s 9.8 m/s 2 = 0.53 s (d) v ox = v o cos θ = (6.0 m/s) cos60° = 3.0 m/s v x relative to floor = 3.0 m/s + 2.0 m/s = 5.0 m/s 0 t 0 t 0 t 0 t

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w or mg F lift θ 8. (a) (2 points for F lift + 1 point for w) [ - 2 points for drawing ma or mv 2 /R, - 1 point for each additional incorrect force, - 1 for putting a or v 2 /R right on the diagram, down to a minimum total of 0 ] (b) G (3 points, work NOT needed to be shown) [1 point for H or K] [ Σ F y = F lift cos θ - w = ma y = 0 F lift = w cos θ = 1.25 w ] (c) x ( ): Σ F x = ma x with a x = v 2 R F lift sin θ = m v 2 R R = mv 2 F lift sin θ = mv 2 (mg/cos θ ) sin θ = v 2 g tan θ = (200 m/s) 2 (9.8 m/s 2 ) tan37° = 5400 m 9. (a) Block #1: Block #2: Block #3: [ 1 point for each correct force properly labeled, 0 for a missing force
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Prelim_Exam__1_S05_Solns - P112 Prelim Exam#1 Solutions 1 2...

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