260_Physics ProblemsTechnical Physics

260_Physics ProblemsTechnical Physics - 262 P9.29 Linear...

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262 Linear Momentum and Collisions P9.29 pp xf xi = mv mv m OY ms cos . cos . . 37 0 53 0 5 00 °+ °= bg 0799 0602 500 ... vv += (1) mv mv yf yi = °− sin . sin . 37 0 53 0 0 .. = (2) Solving (1) and (2) simultaneously, v O = 399 . and v Y = 301 FIG. P9.29 P9.30 xf xi = : mv mv mv i cos cos . θθ = 90 0 af vvv i cos sin (1) yf yi = : mv mv sin sin . −° = 90 0 0 sin cos = (2) From equation (2), = F H G I K J cos sin θ (3) Substituting into equation (1), v i YY cos sin sin 2 F H G I K J so i Y cos sin sin 22 ej , and i Y = sin . FIG. P9.30 Then, from equation (3), Oi = cos . We did not need to write down an equation expressing conservation of mechanical energy. In the
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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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