# Prelim_Exam__2_Answers_SP04 - = 1 2 (2m)v 2 + 1 2 I 2 + 0...

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P112 Prelim Exam #2 Solutions 1. (a) B (b) C 2. A 3. B 4. C 5. (a) None (b) 3.5 cm (c) 2.0 to 3.5 cm (d) 4.5 cm (e) 4.0 J 7. (a) +x +y 1 2 1 6. (a) U(t) t 0 K(t) t 0 U(t) + K(t) t 0 (b) p 1o p 1f p = J 1 1 (c) Σ p x = (150 kg)(4.0 m/s) + 0 = 0 + (250 kg)v x v x = (150 kg)(4.0 m/s) 250 kg = 2.4 m.s Σ p y = 0 + (250 kg)(2.5 m/s) = (150 kg)((2.0 m/s) + (250 kg)v y v y = (250 kg)(2.5 m/s) - (150 kg)(2.0 m/s) 250 m/s = 1.3 m/s

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8. (a) Conservation of Energy: K o + U go = K 1 + U g1 (Let y = 0 at the bananas.) 0 + MgH = 0 + 1 2 Mv 1 2 v 1 = 2gH (b) Grabbing bananas (inelastic collision), use Conservation of Momentum: Σ p x = Mv 1 = (M + m)v 2 v 2 = M M + m v 1 = M M + m 2gH Swing up, use Conservation of Energy: K 2 + U g2 = K f + U gf 1 2 (M + m)v 2 2 + 0 = 0 + (M + m)g H 2 v 2 = gH = M M + m 2gH Solve for m: M + m = 2 M m = ( 2 - 1) M = 0.41 M 9. (a) I = 1 2 mR 2 + 1 2 mR 2 = mR 2 (b) Conservation of Energy: K o + U go = K f + U gf (Let y = 0 at final position.) 0 + (2m)gH
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Unformatted text preview: = 1 2 (2m)v 2 + 1 2 I 2 + 0 Rolling: v = r = m 2 r 2 + 1 2 (mR 2 ) 2 = r 2 + R 2 2 m 2 = 2gH r 2 + R 2 2 10. Energy Conservation: K f + U sf = K o + U so + W fric (Let x = 0 for relaxed spring.) 0 + 1 2 kx 2 = 0 + 1 2 kd 2- f k (d + x) = 1 2 kd 2- k mg(d + x) (using f k = k N = k mg) Rearrange: 1 2 kx 2 + ( k mg) x + k mgd - 1 2 kd 2 = 1 2 (100 N/m) x 2 + (0.40)(50 N) x + (0.40)(50 N)(0.50 m) - 1 2 (100 N/m)(0.50 m) = (50 N/m) x 2 + (20 N) x - 2.5 N-m = Solve for x = - 20 (20) 2- 4(50)(-2.5) 2(50) = - 0.50 m OR + 0.10 m 1st root gives original position, so choose 2nd root: x = + 0.10 m Distance from wall = L o + x = 1.00 m + 0.10 m = 1.10 m...
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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 University (Engineering School).

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Prelim_Exam__2_Answers_SP04 - = 1 2 (2m)v 2 + 1 2 I 2 + 0...

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