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Unformatted text preview: This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A triangular wedge 5 m high, 13 m base length, and with a 14 kg mass is placed on a frictionless table. A small block with a 8 kg mass (and negligible size) is placed on top of the wedge as shown in the figure below. 14 kg 13 m 5m 8 k g Δ X wedge 14 kg 13 m 5m 8 k g All surfaces are frictionless, so the block slides down the wedge while the wedge slides sidewise on the table. By the time the block slides all the way down to the bottom of the wedge, how far Δ X wedge does the wedge slide to the right? Correct answer: 4 . 72727 m. Explanation: Let : M = 14 kg , m = 8 kg , L = 13 m , and H = 5 m . Consider the wedge and the block as a two body system. The external forces acting on this system — the weight of the wedge, the weight of the block and the normal force from the table — are all vertical, hence the net hor izontal momentum of the system is conserved, P wedge x + P block x = constant . Furthermore, we start from rest = ⇒ center ofmass is not moving, and therefore the X coordinate of the centerofmass will remain constant while the wedge slides to the right and the block slides down and to the left, X cm = mX block + M X wedge m + M = constant . Note: Only the X coordinate of the center ofmass is a constant of motion; i.e. , the Y cm accelerates downward because the P y compo nent of the net momentum is not conserved. Constant X cm means Δ X cm = 0 and there fore m Δ X block + M Δ X wedge = 0 . Note that this formula does not depend on where the wedge has its own centerofmass; as long as the wedge is rigid, its overall dis placement Δ X wedge is all we need to know. Finally, consider the geometry of the prob lem: By the time the block slides all the way down, its displacement relative to the wedge is equal to the wedge length L , or rather L because the block moves to the left of the wedge. In terms of displacements relative to the inertial frame of the table, this means Δ X block Δ X wedge = L. Consequently, 0 = m Δ X block + M Δ X wedge = m ( L + Δ X wedge ) + M Δ X wedge and therefore Δ X wedge = m m + M L = (8 kg) (8 kg) + (14 kg) (13 m) = 4 . 72727 m . 002 10.0 points Pat builds a track for his model car out of wood. The track is 3 cm wide (along the z coordinate), 1 m high (along the y coordi nate) and 3 . 1 m long (along the x coordinate starting from x = 0)....
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This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
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