273_Physics ProblemsTechnical Physics

273_Physics ProblemsTechnical Physics - Chapter 9 *P9.59...

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Chapter 9 275 *P9.59 (a) Conservation of momentum: 05 2 3 1 15 1 2 3 1 3 8 4 4 0 2 2 . ±±± . . . . ± . ±± . ± . . kg m s kg m s kg m s kg kg m s kg m s kg ij k ijk v v k k −+ + −+− =− + + = + −⋅ + +⋅ = ej e j e j f f The original kinetic energy is 1 2 2 3 1 1 2 1 5 123 1 4 0 222 .. . kg m s kg m s J 22 ++ + = The final kinetic energy is 1 2 1 3 8 0 185 kg m s J += different from the original energy so the collision is inelastic . (b) We follow the same steps as in part (a): = + + = + + + 4 025 075 2 0 5 1 5 4 0 125 0 375 1 0 250 0 750 2 00 2 2 . ± . ± . . . ± . . ± . . . ± . ± . ± k i j k v v k i j k k e j e j kg m s kg m s kg kg m s kg m s kg ms f f We see vv 21 ff = , so the collision is perfectly inelastic . (c) Conservation of momentum: = + + + = + −− =− − 4 1 3 4 267 0333 2 2 . ± . . . . ± . . ± . ± . ± . ± k i j k v v k k k e j e j af kg m s kg m s kg kg m s kg m s kg a a a f f Conservation of energy: 14 0 1 2 1 3 1 2 2 5 0 25 5 33 1 33 0 083 3 2 . . . . . . J kg m s kg m s J J =+ + + + + + + aa a 0 0 333 1 33 6 167 1 33 1 33 4 0 333 6 167 0667 274 674 2 2
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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