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Unformatted text preview: Problem Set 5: Problem 3.
Problem: Two identical hockey pucks moving with initial speeds vA = 6 v and vB = v collide as 5 shown. The coefficient of restitution of the pucks is e. You can ignore the friction of the surface on which the pucks move. (a) Determine the velocities of the pucks after impact. (b) Experimentation shows that Puck A is at rest after the impact when = 80.13o , which corresponds to cos = 6/35. Using your result from Part (a), determine the value of e. Solution: (a) The line of impact for the hockey pucks is the x axis. So, there is no need to distinguish between nt and xy coordinates. Tangential-Velocity Invariance. For pucks A and B, we have the following. vAy = vAy and vBy = vBy The initial velocities of the pucks are such that vAy = 0 and vBy = v sin . Therefore, the tangential velocity components after the impact are vAy = 0 and vBy = v sin Normal-Momentum Conservation. Using the fact that the initial velocity components for pucks A and B are vAx = 6 v and vBx = -v cos , momentum conservation along the line of impact tells us that 5 6 mv - mv cos = mvAx + mvBx 5 Dividing through by m yields vAx + vBx = 6 v - v cos 5 Impact Relation. For a coefficient of restitution equal to e, we have vBx - vAx = e (vAx - vBx ) = e 6 v + v cos 5 To complete the solution, we add the equations resulting from normal-momentum conservation and the impact relation, which yields 2vBx = 6 (1 + e)v - (1 - e)v cos 5 = vBx = 1 3 (1 + e)v - (1 - e)v cos 5 2 ...
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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.
- Spring '06