Why Do Bouncing Balls Lose Height While Upholding Energy Conservation?

[butz3]

if you drop any kind of ball onto a hard surface, it will never bounce as high as the height it was dropped from. show algebraically that michanical energy has been lost between the maximum height on hte one bounce and the maximum height on the next bounce, and explain why this does not disprove the energy conservation law. any help would be appreacted thanks.

 

[arildno]

Suppose you drop the ball from initial height h_{0} , and in the subsequent bounces, the maximally reached heights satisfy the inequality:
h_{0}\geq{h}_{1}\geq{h}_{2}\geq{h}_{3}...
1)Now, what is the kinetic energy of the ball at these heights?
2)Using your answer on that, what must then the total mechanical energy at each stage be?
3)How can mechanical energy be lost from a system?

 

[wais]

The energy conservation law states that energy cannot be created or destroyed, but can only be transferred or transformed from one form to another. In the case of a bouncing ball, the energy is initially in the form of potential energy due to its position at a certain height. As the ball falls, this potential energy is converted into kinetic energy, which is the energy of motion. When the ball hits the ground, some of this kinetic energy is transferred to the ground, causing it to deform and dissipate as heat and sound energy. The remaining energy is then transformed back into potential energy as the ball bounces back up.

To show algebraically that mechanical energy is lost between bounces, we can use the equation for potential energy (PE) and kinetic energy (KE):

PE = mgh (where m is the mass of the ball, g is the acceleration due to gravity, and h is the height)

KE = 1/2 mv^2 (where v is the velocity)

At the maximum height on the first bounce, all of the initial potential energy is converted into kinetic energy, so we can equate the two equations:

mgh = 1/2 mv^2

Solving for v, we get:

v = √(2gh)

On the next bounce, the ball will not reach the same maximum height as it did on the first bounce. This means that the velocity at the maximum height on the second bounce will be less than the velocity on the first bounce. Using the same equation for KE, we can calculate the velocity at the maximum height on the second bounce:

KE = 1/2 mv^2

At this point, we can see that the mass and velocity are the same, but the height is different. This means that there is less kinetic energy, and therefore less mechanical energy, at the maximum height on the second bounce compared to the first bounce. This loss of energy can be attributed to the energy dissipated as heat and sound during the first bounce.

However, this does not disprove the energy conservation law. The total amount of energy in the system (ball and ground) remains constant. The energy may be transformed into different forms, but the total amount remains the same. In this case, the energy lost as heat and sound is still accounted for in the total energy of the system. Therefore, the energy conservation law is still applicable and valid in this scenario.