Prelim_Exam__2_S05_Solns

# Prelim_Exam__2_S05_Solns - o/2 2 = 13v o 2/16 = 13 4 v o =...

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p 1i p 1f J 1 J 2 v 2f Physics 112 - Prelim Exam #2 Solutions Spring 2005 1. (a) -0.2 m, 0.2 m (b) -0.5 m to -0.2 m and 0 to 0.2 m [4 points] (c) -0.1 m, 0.1 m (d) 5 J (e) 4 m/s 2. (a) T 3. (a) B (b) T (c) F (b) E [1 point for A, B, C, D, F] (d) T (e) T (c) M [1 point for H, J, K, L, N] (f) F (g) F (h) F 4. B (i) F (j) T 5. C [2 points for A, B, D, E, F] 6. (a) F thrust = v ex dM dt = Ma = (100 kg)(0.50 m/s 2 ) = 50 N dM dt = Ma v ex = (100 kg)(0.50 m/s) 500 m/s = 50 N 500 m/s = 0.10 kg/s (b) Cons. of Energy: K o + U go = K f + U gf 1 2 mv 2 - GMm r o = 0 - GMm r f v = 2GM 1 r o - 1 r f = 2(6.67 × 10 -11 N-m 2 /kg 2 )(3.0 × 10 15 kg) 1 5000 m - 1 8000 m = 5.5 m/s 7. (a) & (b) [See diagram at far right ] (c) J 1 = p 1 (d) J 2 = p 2 = - p 1 = - J 1 (e) (f) Cons. of Momentum: Σ p x = 2mv o + 0 = 2mv x + mv o /2 v x = 3v o /4 Σ p y = 0 + mv o = 2mv y + 0 v y = v o /2 v = v x 2 + v y 2 = (3v o /4) 2 + (v

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Unformatted text preview: o /2) 2 = 13v o 2 /16 = 13 4 v o = 0.90 v o tan θ = v y v x = v o /2 3v o /4 = 2 3 = 0.667 ⇒ θ = 34° +x +y 2m m m Before After v o v o v /2 o 8. (a) Cons. of Energy: K o + U go = K 1 + U g1 ⇒ 0 + mgH = 1 2 mv 1 2 + 1 2 I ω 1 2 Insert I = 2 5 MR 2 and ω 1 = v 1 R to get: mgH = 1 2 mv 1 2 + 1 5 mv 1 2 = 7 10 mv 1 2 Solve for H = 7v 1 2 10g (b) Ball enters box (quick inelastic collision): Cons. of horizontal Momentum ( → ): Σ p x = mv 1 cos37° = (m + 3m)v 2 ⇒ v 2 = v 1 cos37° 4 = v 1 (4/5) 4 = v 1 5 Spring compresses: Cons. of Energy: K 1 + U s1 = K f + U sf ⇒ 1 2 (m + 3m)v 2 2 = 1 2 kD 2 Solve for D = 2v 2 m k = 2v 1 5 m k...
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## This test prep was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell.

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Prelim_Exam__2_S05_Solns - o/2 2 = 13v o 2/16 = 13 4 v o =...

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