A 5.00-kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center. The bar has mass 8.00 kg and is 4.00 m in length. At the other end of the bar sits another 5.00-kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision. How high will the other ball go after the collision?

Solution 88P Step 1 : Data given Mass of the ball M b = 5 kg Distance before collision h = 12.0 m Mass of the bar M bar = 8 kg Length of the bar L = 4.0 m We need to find the distance travelled by the ball at the other end of the bar when a ball is collided with bar at front end We shall find the potential energy of the ball It is given by U = 2gh Substituting values we get U = 2 × 9.8 m/s × 12.0 m U = 235.2 J U = 15.3J Hence the potential energy of the ball is 15.3 J Step 2 : We shall angular velocity of the ball as it collides with the bar It is given by Consider L1= L 2y law of conservation L1 length of the bar before collision L2 length of the bar after collision Hence we can write this as M UR = I b M bar2 2 M Ub = ( 12 + 2M RB) Substituting values we get 8 kg×(4.0 m ) 2 5 kg × 15.3 J × 2m = ( 12 + 2 × 5kg × 2 ) 153 kg m = (50.6 kg m ) 2 = 153 kg m /50.6 kg m 2 = 3.023 rad /s