Therefore, we have v I A = U cot θ Horizontal-Momentum Conservation. Because the wedge is constrained to move horizontally, momentum is conserved in the x direction, viz., m A v Ax + m B v Bx = m A v I Ax + m B v I Bx Substituting for the velocity components, we find − mU = − λ mv I B Therefore, v I B = 1 λ U Impact Relation. For a coefficient of restitution equal to e ,wehave v I Bn − v I An = e ( v An − v Bn ) Hence, substituting for the velocity components yields − v I B sin θ − v I A cos θ = e ( − U sin θ − 0) Thus, we conclude that eU = v I B + v I A cot θ To complete the solution, we substitute for v I A and v I B , which yields eU = 1 λ U + U cot 2 θ = ⇒ cot 2 θ = e − 1 λ = λ e − 1 λ Therefore, the wedge angle is θ =tan − 1 5 λ λ e − 1 (b) The kinetic energy before the impact,
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This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.