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# exam2-S06-ans - torque in –Z direction straight line for...

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Physics I – Exam 2 – Spring 2006 Answer Key Part A – 1: B, 2: C, 3: A, 4: B, 5: B, 6: A, 7: A, 8: C 4 pts each. B-1 20 Points The key to this problem is the conservation of mechanical energy and the Work – Kinetic Energy Theorem. Looking For KE curve is mirror image of PE curve KE goes from 1 J to max 2 J and back to 1 J. force curve is a straight line. force = 0 at x = 5 cm. force = +40 N at x = 0 cm. 10 5 PE (J) x (cm) 10 5 0 10 x (cm) x (cm) KE (J) Net Force (N) 0 5 0 0 1 40 3 2 1 4 -40 2

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B-2 24 Points This problem uses a little bit of 2D projectile motion. Vx = 3. Vy = –9.8 t. x = (3) t. y = 4.9 – 4.9 t 2 . Only Z components are non-zero. τ z = (r x F) z = x (0.5) (–9.8) = –14.7 t l z = (r x p) z = x (0.5) Vy – y (0.5) Vx = –7.35 – 7.35 t 2 Looking For X and Y components of torque = 0. magnitude of torque = 0 at t = 0 sec. magnitude of torque = 14.7 N m at t = 1 sec.
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Unformatted text preview: torque in –Z direction straight line for torque. X and Y components of angular momentum = 0. magnitude of angular momentum = 7.35 kg m 2 /s at t = 0 sec. magnitude of angular momentum = 14.7 kg m 2 /s at t = 1 sec. angular momentum in –Z direction parabola for angular momentum with increasing magnitude of slope. l z (kg m^2/s) t (s) 1.0 τ z ( N m ) t (s) 1.0-7.35-14.7-14.7 Part C Must show work to receive credit. C-1 24 points Cons. of mech. energy, then one-dim elastic collision, then cons. of mech. energy. m 1 g r = ½ m 1 v 2 → v = √(2 g r) v 1 = (1–3)/(1+3) v = -½ v = -½ √(2 g r) → h 1 = ½ m 1 v 1 2 / (m 1 g) = ¼ r = 25 cm v 2 = (2)/(1+3) v = +½ v = +½ √(2 g r) → h 2 = ½ m 2 v 2 2 / (m 2 g) = ¼ r = 25 cm...
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exam2-S06-ans - torque in –Z direction straight line for...

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