Prelim_Exam__2_F04_Solns

# Prelim_Exam__2_F04_Solns - go = K U g ⇒ 1 2 mv o 2 0 = 1...

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Physics 112 — Exam #2 Solutions 1. (a) - 0.1 m, 0.1 m (b) 0 (c) 0.1 m (d) 1 J (e) 3 J 2. (a) F (b) F (c) T (d) F (e) T 3. B 4. (a) - - + 0 + 0 (b) F (c) D 5. Cons. of Energy: K o + U go = K f + U gf 1 2 mv o 2 - GMm = 1 2 mv f 2 - GMm R E Solve for v f = v o 2 + 2GM R E = (20 × 10 3 m/s) 2 + 2(6.67 × 10 -11 N-m 2 /kg 2 )(6.0 × 10 24 kg) 6.4 × 10 6 m = 2.3 × 10 4 m/s = 23 km/s 6. (a) At top of loop, the normal contact force = 0 for minimum speed: Σ F y = ma y mg = m v 2 D/2 v 2 = gD/2 Cons. of Energy: K o + U go = K + U g 0 + mgH = 1 2 mv 2 + 1 2 I ω 2 + mgD = 1 2 mv 2 + 1 2 2 5 mR 2 v R 2 + mgD = 7 10 mv 2 + mgD = 7 10 m(gD/2) + mgD = 27 20 mgD H = 27 20 D (b) E 7. (a) (c) Cons. of Momentum: x: 0 + 2mv o sin30° = m v o 2 + 2mv x y: 2mv o cos30° - mv o = 2mv y v x = v o 4 & v y = 3 - 1 2 v o = 0.366v o v = v x 2 + v y 2 = 17 - 8 3 4 v o = 0.443 v o (b) (d) m w = mg m 2m m Before After +x +y v o v o v /2 o 30° mv /2 o mv o p = J m m J = - J 2m m

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8. (a) At the instant the block is at its highest height h on the wedge, the block and wedge have the same speed v. First, find v: Cons. of Momentum: mv o + 0 = (m + 3m)v v = v o 4 Then use Cons. of Energy to find h: K o + U
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Unformatted text preview: go = K + U g ⇒ 1 2 mv o 2 + 0 = 1 2 (m + 3m)v 2 + mgh = 1 2 (4m) v o 4 2 + mgh = 1 8 mv o 2 + mgh Solve for h = 3v o 2 8g (b) Cons. of Momentum: mv o + 0 = mv b + 3mv w ⇒ v o + 0 = v b + 3v w (1) Elastic 1-D Collision ⇒ v o- 0 = v w- v b (2) (1) + (2) ⇒ 2v o = 4v w ⇒ v w = v o 2 = v o 2 → (2) ⇒ v b = v w- v o = v o 2- v o = - v o 2 = v o 2 ← OR use the result equations for 1-D Elastic Collisions: v wf = m w- m b m w + m b v wo + 2m b m w + m b v bo = 2m 4m (0) + 2m 4m v o = v o 2 v bf = 2m w m b + m w v wo + m b- m w m b + m w v bo = 2m 4m (0) + - 2m 4m = - v o 2 Same results as above!...
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Prelim_Exam__2_F04_Solns - go = K U g ⇒ 1 2 mv o 2 0 = 1...

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