You drop a ball from a height of 2.8 m, and it bounces back to a height of 1.0 m. (Ignore air resistance.)
(a) What fraction of its initial energy is lost during the bounce?
b) What is the ball's speed just as it leaves the ground after the bounce?
(c) Where did the energy go?
What is the ratio of the potential energy between h=1.0 and h=2.8 m?
v = sqrt(2*H*g)
To answer these questions, we need to consider the concepts of potential energy, kinetic energy, and the conservation of energy. Let's break it down step by step:
(a) To determine what fraction of the ball's initial energy is lost during the bounce, we can compare the potential energy before and after the bounce. The potential energy of an object in this case is given by the formula: PE = m * g * h, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Before the bounce, the ball's potential energy is PE_initial = m * g * h_initial = m * 9.8 * 2.8.
After the bounce, the ball's potential energy is PE_final = m * g * h_final = m * 9.8 * 1.0.
The fraction of energy lost can be calculated by dividing the potential energy lost during the bounce by the initial potential energy:
Fraction of energy lost = (PE_initial - PE_final) / PE_initial.
(b) To find the ball's speed just as it leaves the ground after the bounce, we can use the principle of conservation of energy. Since there is no air resistance, the total mechanical energy is conserved. This means that the sum of potential energy and kinetic energy before the bounce is equal to the sum of potential energy and kinetic energy after the bounce.
Initially, the ball has gravitational potential energy, but no kinetic energy. As it falls, its potential energy decreases while its kinetic energy increases. At the bottom of the bounce, when the ball reaches its maximum speed, all of the potential energy is converted into kinetic energy. When the ball reaches its maximum height again, its kinetic energy is zero, and it has converted all of its energy back into potential energy.
To find the speed just as it leaves the ground after the bounce, we can equate the potential energy before the bounce (m * g * h_initial) to the kinetic energy just as it leaves the ground (0.5 * m * v^2). Solving for v (velocity), we have:
v = sqrt(2 * g * (h_initial - h_final)).
(c) The energy lost during the bounce goes into various forms. In this case, it's primarily converted into heat and sound. When the ball hits the ground, some of the energy is dissipated as heat due to inelastic collisions between the ball and the ground. Additionally, the impact of the bouncing ball also produces sound waves, which carry away some energy.
Note: It is important to remember that these answers may vary depending on the assumptions and limitations given for the problem.