Bouncing Ball Dissipative Interaction

[B3NR4Y]

Homework Statement

A 0.62-kg basketball dropped on a hardwood floor rises back up to 64% of its original height. If the basketball is dropped from a height of 1.6m , how much energy is dissipated in the first bounce? How much energy is dissipated in the fourth bounce?

Homework Equations

U(x) = mgh

The Attempt at a Solution

I answered the first part correctly, and found that the change in Energy, or dissipated energy, after the first bounce is 3.5 J. I am having trouble answering the second part however, I was able to convince myself that U(x) = 0.62-kg*9.8*(1.6*0.64^n) where n=the number of bounces by thinking about it for a while. So I found the initial potential energy to be 9.7216 J and when I calculated the final potential energy after four bounces I got ~1.63 J. Subtracting final from initial, I got an answer of ~8.1 J as the change in energy, but that is wrong. I don't know how, can anyone steer me in the right direction (preferably in the next 20 minutes (I promise I didn't procrastinate ))

 

[olivermsun]

You got the right change in energy, but that was for the total change from initial to after the 4th bounce.

What about the change in the 4th bounce (difference of 3rd to 4th)?

 

[B3NR4Y]

You got the right change in energy, but that was for the total change from initial to after the 4th bounce.

What about the change in the 4th bounce (difference of 3rd to 4th)?

Hm, so the energy after the 3rd bounce is U(x) = ~2.54 and then after the forth bounce it would be ~1.63 so I subtract the two and get ~0.92 J

 

[olivermsun]

Looks right.

 

[HalfLostAndSearching]

?I would approach this problem by first considering the concept of dissipative interaction. This refers to the loss of energy due to non-conservative forces, such as friction or air resistance, during a physical process. In this case, the basketball experiences dissipative interaction as it bounces on the hardwood floor.

To calculate the amount of energy dissipated in the first bounce, we can use the equation U(x) = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the ball. Since the ball rises back up to 64% of its original height, we can calculate the height after the first bounce to be 1.6m * 0.64 = 1.024m. Therefore, the change in potential energy is U(1.024m) - U(1.6m) = 0.62kg * 9.8m/s^2 * (1.024m - 1.6m) = 3.5J.

To calculate the energy dissipated in the fourth bounce, we can use the same equation, but with the height after four bounces being (1.6m * 0.64)^4 = 0.262m. Therefore, the change in potential energy after four bounces is U(0.262m) - U(1.6m) = 0.62kg * 9.8m/s^2 * (0.262m - 1.6m) = -7.1J.

It is important to note that in the fourth bounce, the ball does not rise back up to its original height, but rather falls to a lower height. This results in a negative change in potential energy, indicating that energy is being dissipated rather than stored. Therefore, the total energy dissipated in the first four bounces is 3.5J + (-7.1J) = -3.6J.

In conclusion, the amount of energy dissipated in the first bounce is 3.5J, and the amount of energy dissipated in the fourth bounce is -7.1J. The total energy dissipated in the first four bounces is -3.6J. This demonstrates how dissipative interaction can lead to a decrease in the overall energy of a system.