A ball of mass 0.5 kg was dropped from an initial height 3.0 m above a hard floor. During the inelastic bounce, 11.6 J of energy is dissipated. If the time the ball is in contact with the floor is 10.0 ms, with what speed does it rebound.
The ball of mass 0.5 kg was dropped from a height of 3 m. The potential energy of the ball at a height of 3 m is 0.5*3*9.8 = 14.7 J
When it is dropped on the floor, there is an inelastic collision which results in a dissipation of energy equal to 11.6 J. The energy remaining for the ball to rebound is 14.7 - 11.6 = 3.1 J
The kinetic energy in the ball would be converted to potential energy as it rises again after the rebound. Due to the acceleration due to gravity the speed of the ball will reduce as it rises higher. Its speed just when it rebounds can be determined from the kinetic energy left.
(1/2)*m*v^2 = 3.1
=> (1/2)*0.5*v^2 = 3.1
=> v^2 = 3.1*4
=> v = 3.52 m/s
The speed of the ball just when it rebounds is 3.52 m/s