Unformatted text preview: Problem Set 6: Problem 2.
Problem: A ball impacts the ground at Point A with speed V at an angle as shown. After the impact, the ball reaches Point B, where its velocity vector is exactly horizontal. Its speed, U , at Point B is not given. If the ball's coefficient of restitution is e, determine the distances h and L. Express your answers in terms of e, V , , and gravitational acceleration, g. Solution: Because the surface is immovable, in terms of the impact, only momentum in the tangential direction is conserved. The impact relation holds in its standard form with the velocity of the surface being zero. TangentialVelocity Invariance. The tangential direction is horizontal, i.e., parallel to the x axis, so that vAx = V cos Impact Relation. Letting subscript W denote the wall, we have vW n  vAn = e (vAn  vW n ) Since the surface is stationary, and the normal direction is vertical, i.e., parallel to the z axis, necessarily 0  vAz = e (V sin  0) Therefore, the velocity of the ball just after the impact is v = V cos i + eV sin k Completing the Solution. After the impact, Newton's Second Law tells us that the ball's motion is governed by d2 x x(0) = L, x(0) = V cos m 2 = 0, dt d2 z m 2 = mg, z(0) = 0, z(0) = eV sin dt The solution for x(t and z(t) is x(t) = L  V cos t and 1 z(t) = eV sin t  gt2 2 = vAz = V sin Focusing first on the horizontal distance, L, we use the solution for x(t) to determine the time, tf , at which the ball reaches the upper surface. Hence, since x(tf ) = 0, L  V cos tf = 0 = tf = L V cos ...
View
Full
Document
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
 Shiflett

Click to edit the document details