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**Unformatted text preview: **Solution for Long-answer Homework 12 Interactions problems Solution to Long-answer Homework Problem 12.1(A non-isolated collision) Problem: A hockey puck of mass . 25kg travels at 10 m s to the right and collides with a lump of Silly Putty (inertia . 1kg ) at rest. The combined lump of puck and putty moves at 3 . 5 m s to the right. (a)What is the energy dissipated in the collision? (b)Is this collision isolated? Why or why not? (c)What would be the total convertible energy in this collision, assuming it was isolated? (Do you want to change your answer to (b)?) Solution to part (a) Even if this is an isolated collision, it is an inelastic collision with obvious deformation of the Silly Putty (due to its squishy-ness), so the change in kinetic energy will not be zero, and this value is the energy dissipated. KE f- KE i = Δ E diss Find the kinetic energies: KE i = 1 2 m 1 v 2 01 + 1 2 m 2 v 2 02 = 1 2 (0 . 25kg) parenleftBig 10 m s parenrightBig 2 + 0 = 12 . 5J KE f = 1 2 ( m 1 + m 2 ) v 2 f = 1 2 (0 . 35kg) parenleftBig 3 . 5 m s parenrightBig 2 = 2 . 1J Now compute E diss =- Δ K : E diss = 12 . 5J- 2 . 1J = 10 . 4J Grading Key: Part (a) 5 Points 2 point(s) : E diss = Δ KE 2 point(s) : Used correct velocities (1pt for initial, 1 point for final). 1 point(s) : answer with unit Solution to part (b) In an inelastic collision all the energy that can be dissipated in an isolated system is dissipated: K ′ = 1 2 m 1 × m 2 ( m 1 + m 2 ) v 2 rel,i This is only 3 . 6J . So, this situation is not physically possible, unless a huge amount of energy was dissipated in unsticking the silly putty from the floor. This certainly was not an isolated collision. We also could have told this since momentum is not conserved. Grading Key: 3 Points 1 point(s) : not isolated 2 point(s) : either show momentum is not conserved or too much energy is dissipated. 1 Solution to part (c) K ′ = 1 2 m 1 × m 2 ( m 1 + m 2 ) v 2 rel,i This is only 3 . 6J . Grading Key: 4 Points 1 point(s) : correct formula 1 point(s) : correctly calculated μ 1 point(s) : correctly calculated v rel,i 1 point(s) : correctly substitutes it all in (can get full credit if calculate v f for totally inelastic collision and use that to calculate correct change in K. Total Points for Problem: 12 Points Solution to Long-answer Homework Problem 12.2(General Potential Energy to find kinetic energy) Problem: A particle is in a region where the potential energy has the form U = 10 x 2 Jm 2 . The particle initially is at rest at x = 1 . 0m ....

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