7aMidterm2Rev_sunday_sol

# 7aMidterm2Rev_sunday_sol - Midterm 2 Review Solutions Ruza...

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Midterm 2 Review Solutions Ruza Markov Problem 1. Before the block is released all the energy is stored in the spring. After the launch, the spring returns to its equilibrium length and all the po- tential energy of the spring is transfered into the kinetic energy of the block which moves with speed v : kx 2 2 = mv 2 2 v = r k m x. (1) a) If the ramp is attached to the table it won’t move and we can choose only the block to be our system. The normal force on the block due to the ramp and the gravity are the external forces on the system and the net external force will have both horizontal and vertical components. Therefore momentum of the block won’t be conserved neither in the x nor in y direction. But, the energy of the block will be conserved be- cause there is no friction or any other dissipative forces. When the block is at the maximal vertical height h a it is not moving any more — all its kinetic energy became gravitational potential energy: mv 2 2 = mgh a h a = kx 2 2 mg . (2) b) If the ramp is free to slide across the table a part of the blocks kinetic energy will be spent on making the ramp move. If we want to use conservation of energy we must treat the ramp and the block together as our system. Energy of this system is conserved. Although there will exist both horizontal and ver- tical components of the force between the block and the ramp we don’t have to worry about those when thinking about momentum conservation as those forces are not external. The only external forces are the the weights of the block and the ramp and the normal force due to the table. All these external forces are vertical so that the momentum of the system in the x direction is conserved. When the block is at the maximal vertical height h b its velocity does not have a vertical component and it is not moving with respect to the ramp any more. The block and the ramp are moving together with speed u so the conservation of momentum in the x direction is: mv = ( m + M ) u, (3) while the conservation of energy reads: mv 2 2 = ( m + M ) u 2 2 + mgh b . (4) Solving this system of equations gives: h b = M m + M kx 2 2 mg < h a . (5) Problem 2. a) The curved boundaries of the object are given by y ( x ) = C | x | 1 / 3 and y ( L ) = H give C = H L - 1 / 3 so that y ( x ) = H ( | x | /L ) 1 / 3 .

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