A 5.00-kgball 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 89P 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 = 2h 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 L length of the bar before collision 1 L length of the bar after collision 2 Hence we can write this as M Ub = I 2 M UR = ( M bar + 2M R )2 b 12 B Substituting values we get 2 5 kg × 15.3 J × 2m = ( 8 kg×(4.0 m + 2 × 5kg × 2 ) 12 2 2 153 kg m = (50.6 kg m ) 2 2 = 153 kg m /50.6 kg m = 3.023 rad /s