Dynamics_Part36

Dynamics_Part36 -

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Unformatted text preview: Problem Set 5: Problem 2. Problem: A ball of mass m and coefficient of restitution e is dropped from a height H above a fixed incline of angle to the horizontal as shown. The height of the point of impact relative to the ground is h. (a) Determine the ball's velocity, v = vx i + vz k, immediately after bouncing from the incline. Express your answer in terms of e, H, and gravitational acceleration, g. 5 (b) Now, assume the impact is perfectly elastic, = 30o and h = 16 H. Compute the time at which the ball first strikes the ground. Express your answer in terms of g and H. . . . . . . . . . . . . . ... . . .... . .. . . . . . . . . . . . .. .. . .. . .. . .. .. . .. .. . . .. .. .. .. . .. .. . . . .. . .. . .. . . . .. . .. . .. . .. . . . . .. . . . .. . . . .. . .. . . . ... .. . . . . .. . . . . . .. . . . . .. . .. . . . .. ...... .. . .. . . . ..... ......... .. ..... .. .... ........... . .... .... ........... .. . . . .. . .. ................... . .. ... . ............. .... . . . . .............. . . . . . ..................... ... . ...................... ... . . . . . . . . . ........................ . ..... .. ........................... .............................. ...... .. . . . . . . . ................................. . . ... .. ....................................... . ...... ................................... . .. .. . . . . . . .. ........................................... . .. .......................................... . .. . ... .... .. ................................................................................................................................................................................................................................................ . .. . .......... . .. . .. . ............................ .. ............................................................................................................................................................................................................. ......... .............................................................................................................................................................................................................................................................. ............. ......................................................................................................................................................................................................... ... . .......................... ........................................................ ........................................................................................................................................................................................ .... ................................................................................................................................................................................................................................................... . .................. . ........ ....... . .. . . . . . . . . . . . . . . . . . . . . f v H z h x Solution: To solve this problem, we first use energy conservation to determine the velocity of the ball just prior to its impact with the fixed incline. Next, we determine the velocity of the ball just after the impact. This is the initial velocity for the ball's subsequent motion. Finally, using Newton's Second Law for the ball's flight after bouncing from the incline, we compute the time at which the ball first strikes the ground. (a) The line of impact is normal to the inclined surface. Before proceeding, we must establish the unit vectors along and normal to the line of impact. This will facilitate determining the horizontal and vertical velocity components for computation of the ball's trajectory after it bounces from the incline. Referring to the figure below, we find n = i sin + k cos and t = i cos - k sin To transform from nt coordinates to xz coordinates, we can use the fact that i = n sin + t cos and n k = n cos - t sin k ............................................................................................ ........................................................................................ . ...................................................................................................... ................................................................................................ . .......................................................................................................... .. .................................................................................................................... .............................................................................................................. ........................................................................................................................ ................................................................................................................................. ............................................................................................................................. ..................................................................................................................................... . ............................................................................................................................................... ........................................................................................................................................ .................................................................................................................................................... ............................................................................................................................................................ ................................................................................................................................................................ .................................................................................................................................................... ......................................................................................................................................................................... ................................................................................................................................................................................. ......................................................................................................................................................................... . ...................................................................................................................................................................................... ......................................................................................................................................................................... As he ba moves from s n a pos on o he ns an us before energy s conserved e T1 + V1 = T2 + V2 mpac s he nc ne o a mechan ca ...
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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.

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