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qz6sol_3705s11

# qz6sol_3705s11 - of the plateau When the hollow ball hits...

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TA: Tomoyuki Nakayama Monday, March 14, 2011 PHY 2048: Physic 1, Discussion Section 3705 Quiz 6 (Homework Set #8) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ In the figure below right, a small, hollow, uniform ball is shot from point P so that it rolls smoothly along a horizontal path, up along a ramp, and onto a plateau. Then it leaves the plateau horizontally to land on a game board, at a horizontal distance d = 20.0 cm from the right edge of the plateau. The vertical heights are h 1 = 5.00 cm and h 2 = 3.00 cm. The rotational inertia of a hollow ball is I = (2/3)mr 2 . a) What is the speed of the ball at the edge
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Unformatted text preview: of the plateau? When the hollow ball hits the game board, the vertical displacement of the ball is –h 2 . The flight time of the hollow ball is -h = -(1/2)gt 2 ⇒ t = √ (2h 2 /g) = 0.0782 s During this time interval, the horizontal displacement is d. The velocity of the ball at the edge is d = v t v = d/t = 2.56 m/s b) What is the speed of the ball at point P? While the ball rolls on the surface, the mechanical energy of the ball is conserved. The ball rolls without slipping. Hence the velocity of the center of mass of the ball and its angular velocity are related as v = r ω . The speed ot the ball at point P is E i = E f (1/2)mv 2 + (1/2)I ω 2 = (1/2)mv 2 + (1/2)I ω 2 + mgh 1 (1/2)mv 2 + (1/2)(2/3)mr 2 (v/r) 2 = (1/2)mv 2 + (1/2)(2/3)mr 2 (v/r) 2 + mgh 1 v = √ (v 2 + 6gh 1 /5) = 2.67 m/s...
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