Describe (both quanititavely and qualitiatively) the energy changes of a ball of mass 0.5kg, dropped from a height of 10m and left to bounce. Make use of the law of conservation of energy.

At the drop height, the ball possesses 0J of Kinetic Energy (KE), and only possesses Gravitational Potential Energy(GPE). From GPE=mgh...GPE=O.5x9.91x10=49.05J. As it drops, its GPE decreases and its KE increases. At the instant the ball touches the ground, it will possess 49.05J of KE and 0J of GPE. From KE=0.5mv^2, the KE of the ball at the instant it touches the ground can be found as 14m/s.

As it travels back up, the GPE begins to increase, and KE decreases, but this time the ball does not gain the initial height is once had. This is due to the law of conservation of energy. On impact with the ground, energy is converted to heat and sound which reduces the total energy the ball possesses, thus on each bounce its energy reduces, and it can never regain its original height.

Answered by David M. Physics tutor

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