# Impulse... steel ball falling on steel slab

by offbeatjumi
Tags: ball, falling, impulse, slab, steel
 HW Helper PF Gold P: 3,443 Impulse... steel ball falling on steel slab The complete kinematic equation is $$v^{2}_{f}=v^{2}_{i}+2a(x_{f}-x_{i})$$ Here vf is the speed when the object is at position xf and vi is the speed when the object is at xi. We assume that position is measured from the slab. For step 1, xi=2.00 m, xf=0, vi=0, a = -9.8 m/s2 and we are looking for vf, the speed of the ball just before it hits the slab on its way down. This gives $$v^{2}_{f}=0^{2}+2(-9.8\;m/s^2)(0 - 2.00\;m)$$ If you do the calculation, you get -.2504 kg*m/s for the momentum as you have already found. The problem with your solution is the identification of the variables after the bounce. After the bounce the speed of the ball at xi = 0 is the unknown speed vi. What is the speed vf when the position is xf = 1.60 m?