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hw21 - Overton Mays Homework 21 Due 4:00 am Inst Turner...

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Overton, Mays – Homework 21 – Due: Mar 23 2005, 4:00 am – Inst: Turner 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A triangular wedge of height H = 1 . 404 m, base length L = 2 . 34 m and mass M = 11 . 6 kg is placed on a frictionless table. A small block of mass m = 4 . 71 kg is placed on top of the wedge as shown on the picture below: M m All surfaces are frictionless, so the block slides down the wedge while the wedge slides side- wise on the table. By the time the block slides all the way down to the bottom of the wedge, how far does the wedge slide to the right? Correct answer: 0 . 675745 m. Explanation: 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 = const . Furthermore, we start from rest = center of mass is not moving, and therefore the X coordinate of the center of mass will remain constant while the wedge slides to the right and the block slides down and to the left, X cm = mX block + MX wedge m + M = const . Note that only the X coordinate of the cen- ter of mass is a constant of motion: 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 center of mass; 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 = mL m + M = 0 . 675745 m .
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