Question

A rubber ball is dropped vertically on the ground from a height of 1.0 m.

After colliding with the ground, the ball bounces to a height of 60 cm. The next bounce it only reaches a height of 36 cm.
A. What form is the energy in when the ball is in contact with the ground?
B. It lists the energy transfers that take place from when the ball is dropped to the maximum of the first bounce?
C. Calculate the speed of the ball just before and just after hitting the ground during the first bounce.
D. Predict what the height of the third bounce will be. What assumptions have you considered?

Answer 1

The potential energy of the ball when it has fallen 3 m cannot be calculated accurately without knowing the mass of the ball. When the ball is dropped, it starts with potential energy which is converted into kinetic energy as it falls. The speed of the ball just before hitting the ground during the first bounce can be calculated using the equation v = √(2gh), where v is the speed, g is the acceleration due to gravity, and h is the height. The height of the third bounce can be predicted by assuming that the ball loses 10% of its height with each bounce.

Step-by-step explanation:

a. The potential energy of the ball can be calculated using the formula PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height. However, the mass of the ball is not given in the question, so we cannot calculate the potential energy precisely.

b. When the ball is dropped, it starts with potential energy. As it falls, this potential energy is converted into kinetic energy. When the ball reaches its maximum height after bouncing, all of its kinetic energy has been converted back into potential energy. The energy transfers can be summarized as follows:

• Initial potential energy -> Kinetic energy
• Kinetic energy -> Potential energy
• Potential energy -> Kinetic energy
• Kinetic energy -> Potential energy

c. The speed of the ball just before hitting the ground can be calculated using the equation v = √(2gh), where v is the speed, g is the acceleration due to gravity, and h is the height. Since the ball is dropped from a height of 1.0 m, the speed just before hitting the ground can be calculated as v = √(2 * 9.8 * 1) = √(19.6) ≈ 4.43 m/s. Just after the ball hits the ground, it rebounds, so its speed is reversed. Therefore, the speed just after hitting the ground during the first bounce is -4.43 m/s.

d. The height of the third bounce can be predicted using the concept of energy conservation. Since the ball loses 10% of its height with each bounce, we can assume that the potential energy after each bounce is 90% of the previous one. Therefore, the height of the third bounce can be calculated as 0.9 * 0.6 m = 0.54 m.