Long-Answer HW 12 - Interactions

# Long-Answer HW 12 - Interactions - Solution for Long-answer...

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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 inertia . 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. If we look at all kinds of energy that we know about, we know: E = 0 = K + U + E diss + E source Now, since it is not mentioned that it is not a level surface, and no springs or anything are used, and things stay squished after the collision, there is no potential energy, so U = 0 . Nothing here can eat or burn fuel, so E source = 0 . This just leaves kinetic energy and dissipated energy. The change in dissipated energy is always positive (you dont end up with dissipated energy going into other forms, that is sort of the point, for dissipated energy), so you only get dissipated if kinetic energy decreases, without getting stored in any form of potential. 0 = KE + E diss = KE f- KE i + E diss Rearranging:- ( KE f- KE i ) = ( KE i- KE f ) = 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 substitute in to our equation E diss = K i- K f : E diss = 12 . 5J- 2 . 1J = 10 . 4J Grading Key: Part (a) 5 Points 2 point(s) : | E diss | = | KE | 2 point(s) : Used correct velocities (1pt for initial, 1 point for final). 1 point(s) : answer with unit Solution to part (b) 1 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 , the rest of the initial kinetic energy would be required to conserve momentum! 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....
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## Long-Answer HW 12 - Interactions - Solution for Long-answer...

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