Prelim_Solutions__1_S05

Prelim_Solutions__1_S05 - P112 Prelim Exam#1 Solutions 1 2...

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Unformatted text preview: P112 Prelim Exam #1 - Solutions 1. 2. 3. 4. 5. (a) C (b) A (2 points each) Spring 2005 F (5 points) (a) (a) (a) (c) [2 points for E, G, or H] (b) A F A (3 points) (c) C (d) B (2 points each) D (2 points) C A (b) (b) (3 points for correct construction + 3 points for proper labeling) A B vx(t) (b) A - v34 t a 4 (t) 2 5 v45 t 6. (a) (c) 4 3 vy(t) (d) B v34 t 7. (a) 0 t 0 t ax(t) ay(t) 0 t 0 t (b) voy = vo sin = (6.0 m/s) sin60 = 5.2 m/s 2 2 voy (5.2 m/s) 2 2 vy = voy - 2gH = 0 H = 2g = 2 = 1.38 m 1.4 m 2(9.8 m/s ) vy = voy - gt = 0 voy 5.2 m/s t = g = 2 = 0.53 s 9.8 m/s (c) (d) vox = vo cos = (6.0 m/s) cos60 = 3.0 m/s vx relative to floor = 3.0 m/s + 2.0 m/s = 5.0 m/s 8. (a) (2 points for Flift + 1 point for w) 2 [ - 2 points for drawing ma or mv /R, - 1 point for each additional incorrect force, 2 - 1 for putting a or v /R right on the diagram, down to a minimum total of 0 ] G (3 points, work NOT needed to be shown) [ Fy = Flift cos - w = may = 0 [1 point for H or K] F lift (b) w or mg (c) w Flift = cos = 1.25 w ] 2 2 v v x (): Fx = max with ax = R Flift sin = m R 2 2 2 2 mv mv v (200 m/s) R = = (mg/cos) sin = g tan = = 5400 m 2 Flift sin (9.8 m/s ) tan37 Block #1: Block #2: Block #3: 9. (a) N 2 on 1 f 2 on 1 w1 or N wedge on 2 T string T string on 2 on 3 f 1 on 2 m1 g N 1 on 2 or w2 m2 g w3 or m3 g [ 1 point for each correct force properly labeled, 0 for a missing force - 1 for each additional incorrect force, - 2 for each instance of "ma", down to a minimum total of 0 ] (b) N2 on 1 & N1 on 2 f2 on 1 & f1 on 2 (2 points for each correct pair) [ - 2 for each additional incorrect pair, - 1 for forces not acting on the blocks, min. total = 0 ] N2 on 1 = m1g cos = N1 on 2 (c) #1: F = N2 on 1 - m1g cos = 0 friction: f = k N1 on 2 = k m1g cos [OR consider #1 + #2] Nwedge on 2 = N1 on 2 + m2g cos = (m2 + m1) g cos (2) #2: F = Nwedge on 2 - N1 on 2 - m2g cos = 0 F|| = T - f - m2g sin = T - k m1g cos - m2g sin = m2a #3: F = m3g - T = m3a Add equations (2) + (3) Solve for a = (3) m3g - k m1g cos - m2g sin = m2a + m3a (m3 - k m1 cos - m2 sin) g m2 + m3 ...
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This note was uploaded on 03/27/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Spring '07 term at Cornell.

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