Question

A 4.00 kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The has mass 9.00 kg and is 4.00 m in length. At the other end of the bar sits another 4.00 kg ball, unattached to the bar. the dropped ball sticks to the bar after the collision. Assume that the bar is horizontal when the dropped ball hits it. How high will the other ball go after the collision?

Answer 1

Here we will apply momentum conservation during collision, energy conservation before and after collision.

Velocity of first ball before striking, v = sqrt(2gh) = sqrt(2*9.8*15) = 17.146 m/s

By conservation of angular momentum , mvL/2 = I*w

mvL/2 = [ML^2/12+2*m(L/2)^2]w

4*17.146*4/2 = [9*4^2/12+2*4*(4/2)^2]*w

w = 3.117 rad/s

initial velocity of second ball, u = wL/2 = 3.117*4/2 = 6.234 m/s

Height it rises H = u^2/2g = [6.234 m/s]^2/(2*9.8 m/s^2) = 1.983 m answer